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I got several questions about the wording in Problem 4 on Assignment 5. I agree the wording is not clear. In the textbook in Project~3 on page 488 the authors ask "What is the volume of the solid that the two cylinders enclose?" They point out to Figure 8.119. Since Figure 8.119 shows the union of two cylinders I assumed that what they mean by "enclose" is in fact what we in mathematics call union. One of the important legacies of development of mathematics over millennia is introduction of very precise language. It is disappointing that the text-book that we are using does not promote mathematically precise language.
Problem 4. In this problem by cylinder we mean a solid cylinder. We consider three cylinders $A$, $B$ and $C$ defined by \begin{align*} A & = \bigl\{ (x,y,z) \in {\mathbb R}\!\times\!{\mathbb R}\!\times\!{\mathbb R} \, | \, y^2 + z^2 \leq 1 \ \text{and} \ x \in [-1,1] \bigr\}, \\ B & = \bigl\{ (x,y,z) \in {\mathbb R}\!\times\!{\mathbb R}\!\times\!{\mathbb R} \, | \, z^2 + x^2 \leq 1 \ \text{and} \ y \in [-1,1] \bigr\}, \\ C & = \bigl\{ (x,y,z) \in {\mathbb R}\!\times\!{\mathbb R}\!\times\!{\mathbb R} \, | \, x^2 + y^2 \leq 1 \ \text{and} \ z \in [-1,1] \bigr\}. \end{align*} (a) Calculate the exact values of the following two volumes determined by the cylinders $A$ and $B$: \[ A \cup B \qquad \text{and} \qquad A\cap B. \] (b) Calculate the exact values of the following two volumes determined by the cylinders $A$, $B$ and $C$: \[ A \cup B \cup C \qquad \text{and} \qquad A \cap B \cap C. \]
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It would have also been interesting to ask the for the following volumes:
\[
\bigl(A \setminus (B \cup C)\bigr) \cup \bigl(B \setminus (C \cup A)\bigr) \cup \bigl(C \setminus (A \cup B)\bigr)
\]
and
\[
\bigl((A \cap B) \setminus C \bigr) \cup \bigl((B \cap C) \setminus A \bigr) \cup \bigl((C \cap A) \setminus B \bigr).
\]
It is not so easy to visualize all these sets. Below I show the sets $(A \cap B) \setminus C$ and $B \setminus (C \cup A)$.
The set $(A\cap B)\setminus C$
The set $B \setminus (C \cup A)$
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Since it is difficult to visualize these sets in three dimensions I will illustrate these concepts with three ellipses. Here we consider ellipse to be an area. Specifically we consider
\begin{align*}
A & = \bigl\{ (x,y) \in {\mathbb R}\!\times\!{\mathbb R} \, | \, x^2 + 9 y^2 \leq 9 \bigr\}, \\
B & = \bigl\{ (x,y) \in {\mathbb R}\!\times\!{\mathbb R} \, | \, 7 x^2 - 4\sqrt{3} x y + 3 y^2 \leq 9 \bigr\}, \\
C & = \bigl\{ (x,y) \in {\mathbb R}\!\times\!{\mathbb R} \, | \, 7 x^2 + 4\sqrt{3} x y + 3 y^2 \leq 9 \bigr\}.
\end{align*}
The boundaries of these three ellipses intersect at 12 points. These 12 points form two regular hexagons, the smaller one of radius $3/\sqrt{7}$ and the larger one of radius $\sqrt{3}$. The points on the smaller hexagon are in counterclockwise direction:
\[
\left( \frac{3}{\sqrt{7}}, 0 \right), \quad \left( \frac{3}{2 \sqrt{7}}, \frac{3\sqrt{3}}{2 \sqrt{7}} \right), \quad \left( - \frac{3}{2 \sqrt{7}}, \frac{3\sqrt{3}}{2 \sqrt{7}} \right), \quad \left(- \frac{3}{\sqrt{7}}, 0 \right), \quad \left( - \frac{3}{2 \sqrt{7}}, - \frac{3\sqrt{3}}{2 \sqrt{7}} \right), \quad \left( \frac{3}{2 \sqrt{7}}, - \frac{3\sqrt{3}}{2 \sqrt{7}} \right).
\]
The points on the larger hexagon are in counterclockwise direction:
\[
\left( \frac{3}{2}, \frac{\sqrt{3}}{2} \right), \quad \left(0, \sqrt{3} \right), \quad \left( -\frac{3}{2}, \frac{\sqrt{3}}{2} \right), \quad \left( - \frac{3}{2}, - \frac{\sqrt{3}}{2} \right), \quad \left(0, -\sqrt{3} \right), \quad \left( \frac{3}{2}, -\frac{\sqrt{3}}{2} \right).
\]
The set $A\cup B \cup C$ The area is $6\pi \approx 18.8496$
The set $A\cap B \cap C$ The area is $6 \arcsin\bigl(10/(7\sqrt{7})\bigr) \approx 3.42226$
$\bigl((A \cap B) \setminus C \bigr) \cup \bigl((B \cap C) \setminus A \bigr) \cup \bigl((C \cap A) \setminus B \bigr)$ The area is $18 \arcsin\bigl(1/7\bigr) \approx 2.58026$
$\bigl(A \setminus (B \cup C)\bigr) \cup \bigl(B \setminus (C \cup A)\bigr) \cup \bigl(C \setminus (A \cup B)\bigr)$ The area is $6 \pi -18 \arcsin\bigl(\sqrt{3} /(2\sqrt{7}) \bigr) \approx 12.847$
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Few remarks about the final exam
- On the final exam I expect you to know how to use Trapezoidal rule, Midpoint rule and Simpson's rule to approximate integrals. To do this you will need a scientific calculator to perform calculations with decimal numbers.
- On the final exam there will be a problem in which you will have to find the derivative of a function of the form $f^g$ ($f$ to the power $g$) where $f$ is a positive function and $g$ is a nonnegative function. This is called a "generalized power rule" As an example you can study the function $x^x$ with x>0. It is interesting to look for a minimum of this function.
- I expect you to know parametric formulas for cycloid and the cardioid that we derived in class.
- Here is a list of topics that we covered in the class and that can appear on the final exam.
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Today we talked about calculating volumes using cylindrical shells. Here is an animation of this process using the torus. We consider the torus obtained by rotating a circle of radius $r$ whose center is at the distance $R \gt r$ from the $z$-axis about the $z$-axis.
Place the cursor over the image to start the animation.
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The above animation inspires a way of unwrapping the torus by unwrapping each of the cylinders presented in the above animation. This leads to the picture below.
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You can use a bagel to visually confirm the fact that the volume of a torus is equal to the volume of the cylinder whose base is the cross section circle of the torus and whose height is the average of the circumference of the smallest and the largest circle on the torus. This is done by cutting a bagel in a special way and rearranging its pieces.
This construction also shows that the surface area of the torus is the same as the surface area of the cylinder pictured above, that is $2 R\pi \times 2 r \pi.$ -
It is also worth showing a torus created by a rotating cardioid
Place the cursor over the image to start the animation.
- Today we deduced the formula for the surface area of the surface of revolution produced by the graph of $y = f(x)$, $x \in (a,b)$, as it rotates about the $x$-axis. Here we assume that $f(x) \geq 0$ for all $x \in (a,b)$.
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A very nice vase can be produced by rotating the graph of $y = 2+\sin x$, $x \in (0,2 \pi)$, about the $x$-axis. Below I illustrate how the approximation process works for the surface area of this vase.
Place the cursor over the image to start the animation.
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The animation below illustrates how revolution creates a vase.
Place the cursor over the image to start the animation.
- The surface area of the surface of revolution created by the graph of $y = f(x)$, $x \in (a,b)$, as it rotates about the $x$-axis is given by \[ \int_a^b 2\pi f(x) \sqrt{1+\bigl(f'(x)\bigr)^2} dx. \] Thus the surface area of the vase above is \[ \int_0^{2 \pi} 2\pi \bigl(2+\sin x \bigr) \sqrt{1+\bigl(\cos x\bigr)^2} dx = 4 \pi \int_0^{2 \pi} \sqrt{1+\bigl(\cos x\bigr)^2} dx. \] The last integral Mathematica evaluates as 16 2^(1/2) Pi EllipticE[1/2] which is approximately equal to $96.012$.
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Next we calculate the surface area of the surface of revolution created by the cardioid. Consider the cardioid deduced in class
\[
x = u(t) = 2 \cos(t) -\cos(2t), \quad y = v(t) = 2 \sin(t) -\sin(2t).
\]
Here the cardioid rotates about $x$-axis. Using the same reasoning as today in class one can deduce that the surface area of the surface of revolution created by the cardioid is given by the integral
\[
2 \pi \int_0^\pi v(t) \sqrt{\bigl(u'(t)\bigr)^2 + \bigl(v'(t)\bigr)^2 } dt = 8 \pi \int_0^\pi \bigl(2 \sin(t) -\sin(2t)\bigr) \sin(t/2) dt.
\]
Here we used the fact that for $t \in (0,\pi)$ we have
\[
\sqrt{\bigl(u'(t)\bigr)^2 + \bigl(v'(t)\bigr)^2 } = 4 \sin(t/2).
\]
Now we need to calculate the integrals
\[
\int_0^\pi \sin(t) \, \sin(t/2) dt = \frac{4}{3} \quad \text{and} \quad \int_0^\pi \sin(2t) \, \sin(t/2) dt = - \frac{8}{15}.
\]
Finally, put everything together to obtain that the surface area of the surface of revolution created by the cardioid is
\[
8 \pi \left(2 \, \frac{4}{3} + \frac{8}{15} \right) = \frac{128}{5} \pi.
\]
Place the cursor over the image to start the animation.
- Today I posted a summary of what we did yesterday.
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What is the area enclosed by one arc of the cycloid and the $x$-axis?
- Recall that the parametric equations of one arc of the cycloid are \[ x = u(t) = t - \sin(t), \qquad y = v(t) = 1 - \cos(t), \qquad t \in [0,2 \pi]. \]
- It is clear that the graph of the cycloid \[ C = \bigl\{ \bigl(t-\sin(t), 1- \cos(t) \bigr) \in {\mathbb R}\!\times{\mathbb R} \, | \, t \in [0,2\pi] \bigr\} \] is a function.
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The problem is that we do not have a convenient formula for this function. However, the function $u:[0,2 \pi] \to [0,2 \pi]$ defined by $u(t) = t - \sin(t)$ is a bijection. We do not have a formula for $u^{-1}$ but it is certainly a function $u^{-1}:[0,2 \pi] \to [0,2 \pi].$ On the graph below $u$ is shown in navy blue and $u^{-1}$ is shown in maroon.
$u(t) = t-\sin(t)$ in navy blue and its inverse $u^{-1}$ in maroon
- Thus, the formula for the function $C$ (the graph of the cycloid) is \[ y = f(x) = v\bigl(u^{-1}(x)\bigr), \quad x \in [0,2\pi]. \]
- Now we can use the substitution $t = u^{-1}(x)$ to calculate the area enclosed by the cycloid and the $x$-axes. Then $x = u(t)$ and $dx = u'(t)dt.$ Here are the details \begin{align*} \int_0^{2\pi} f(x) dx & = \int_0^{2\pi} v\bigl(u^{-1}(x)\bigr) dx \\ & = \int_0^{2\pi} v(t)\, u'(t) dt \\ & = \int_0^{2\pi} (1-\cos t)\, (1- \cos t) dt \\ & = \int_0^{2\pi} ( 1 - 2 \cos t +(\cos t)^2 ) dt \\ & = 3 \pi. \end{align*}
- Notice that in the previous item we deduced a general formula for calculating the area enclosed by a curve $C$ given by its parametric equations \[ C = \bigl\{ \bigl( u(t), v(t) \bigr) \in {\mathbb R}\!\times{\mathbb R} \, | \, t \in [\alpha,\beta] \bigr\}. \] The only condition is that $C$ is a function. This condition is equivalent to the condition that $u: [\alpha,\beta] \to [a,b]$ is a bijection. Here $a = \min\{u(\alpha),u(\beta)\}$ and $b = \max\{u(\alpha),u(\beta)\}.$ The formula is \[ \left| \int_\alpha^\beta v(t) \, u'(t) dt \right|. \] In class I explained how this formula can be deduced using approximations of the area by rectangles, the Mean Value Theorem and the Riemann sums of the above integral.
- One can apply the above method to calculate the area enclosed by the cardioid. It is sufficient to consider the part of the cardioid above the $x$-axis. However, the the part of the cardioid above the $x$-axis is not a function. Therefore, we need to split it in two parts which are functions and calculate the related ares.
- We calculated the volume of the body obtained by rotation of the cycloid about the $x$-axis.
- We calculated the volume of the body obtained by rotation of the cycloid about the $y$-axis. This is more like an upside down baking dish.
- An important application of definite integrals is the calculation of the length of a curve given by its parametric equations $x = u(t), y= v(t)$, $t\in (a,b)$. We deduced that the length of such curve is given by \[ \int_a^b \sqrt{\bigl(u'(t)\bigr)^2 + \bigl(v'(t)\bigr)^2} dt. \]
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Below I list few problems in which arc length of a curve can be calculated explicitly.
- Problem 1. Calculate the arc length of the graph of the function $y = x^{3/2}$ between the points $(0,0)$ and $(1,1)$.
- Problem 2. Calculate the arc length of the graph of the function $y= (1/4) x^2-(1/2) \ln x$, between the points $(1,1/4)$ and $\bigl(e,(e^2-2)/4\bigr)$.
- Problem 3. Calculate the arc length of the cycloid given by the parametric equations \[ x = u(t) = t- \sin t , \quad y = v(t) = 1-\cos t \quad \text{where} \quad 0 \leq t \leq 2 \pi. \]
- Problem 4. Calculate the arc length of the astroid curve given by the parametric equations \[ x = u(t) = (\cos t)^3, \quad y = v(t) = (\sin t)^3 \quad \text{where} \quad 0 \leq t \leq 2 \pi. \]
- Problem 5. Calculate the arc length of the spiral given by the parametric equations \[ x = u(t) = (\exp t)(\cos t), \quad y = v(t) = (\exp t)(\sin t) \quad \text{where} \quad -\pi \leq t \leq \pi. \]
- Problem 6. Calculate the arc length of the cardioid given by the parametric equations \[ x = u(t) = (1-\cos t)\cos t, \quad y = v(t) = (1-\cos t) \sin t \quad \text{where} \quad 0 \leq t \leq 2\pi. \]
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You will probably notice that what I call cardioid in Problem 6 differs from the cardioid that we deduced in class. The parametric equations that we deduced for a cardioid in class were
\[
x = u_1(t) = 2 \cos(t) -\cos(2t), \quad y = v_1(t) = 2 \sin(t) -\sin(2t).
\]
The cardioid that we deduced is the cardioid in Problem 6 scaled by two and translated to the point $(1,0)$. This is justified by the following calculations:
\begin{align*}
u_1(t) & = 2 \cos(t) -\cos(2t) \\
& = 2 \cos(t) -(\cos t)^2 +(\sin t)^2 - (\cos t)^2 - (\sin t)^2 + 1 \\
& = 2 \cos(t) - 2(\cos t)^2 + 1 \\
& = 2(1-\cos t) \cos t + 1 \\
& = 2 u(t) + 1, \\
v_1(t) & = 2 \sin(t) -\sin(2t) \\
& = 2 \sin(t) - 2 (\sin t) (\cos t)\\
& = 2(1-\cos t) \sin t \\
& = 2 v(t).
\end{align*}
Hence
\[
\bigl(u_1(t),v_1(t)\bigr) = 2\bigl(u(t),v(t)\bigr) + (1,0).
\]
Or, in pictures,
The cardioid from Problem 6
The cardioid from Problem 6 scaled by 2
The cardioid from Problem 6 scaled by 2 and translated.
- One important application of the definite integral is to prove that the volume of an arbitrary cone is exactly one third of the volume of the cylinder which is circumscribed around that cone. In other words, the volume of an arbitrary cone is \[ \frac{1}{3} \times {\operatorname{area}}({\rm Base}) \times {\rm height}. \]
- The easiest setting in which one can prove this formula is the case of a right pyramid with a square base and the hight which equals to the side of the base. Three such pyramids make a cube. Here is a blueprint for a model of such a pyramid. You can make three of them and put them together in a cube. That is a quite convincing proof.
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Here we consider a pyramid with a rectangular base $ABCD$ and the apex $V$ as in the picture below. The height $h$ of this pyramid is the length of the line segment $\overline{VS}$ where $S$ is the closest point to $V$ in the plane of the base $ABCD.$ The line $VS$ is orthogonal to the plane $ABCD.$ Let $x$ be real number between $0$ and $h.$ Let $S_x$ be a point on the line segment $\overline{VS}$ such that the length of the line segment $\overline{VS_x}$ equals to $x.$ Next we will calculate the area of the intersection of this pyramid and the plane through $S_x$ and parallel to the base $ABCD.$ This intersection is the rectangle denoted by $A_xB_xC_xD_x$ in the picture below.
The right triangles $\triangle VSA$ and $\triangle VS_xA_x$ are similar. Therefore \[ \frac{\overline{VA_x}}{\overline{VA}} = \frac{x}{h}. \] The triangles $\triangle VAB$ and $\triangle VA_xB_x$ are similar. Therefore \[ \frac{\overline{VA_x}}{\overline{VA}} = \frac{\overline{A_xB_x}}{\overline{AB}}. \] The last two equalities yield \[ \frac{\overline{A_xB_x}}{\overline{AB}} = \frac{x}{h}. \] Similarly \[ \frac{\overline{A_xC_x}}{\overline{AC}} = \frac{x}{h}. \] Thus we have \[ {\operatorname{area}}(A_xB_xC_xD_x) = \overline{A_xB_x} \times \overline{A_xC_x} = \left(\frac{x}{h}\right)^2 \overline{AB} \times \overline{AC} = \frac{x^2}{h^2} {\operatorname{area}}(ABCD) \]
Place the cursor over the image to start the animation.
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Consider a cone with an arbitrary base and the apex $V$ as in the picture below. The volume of this cone is
\[
\frac{1}{3} \times {\operatorname{area}}({\rm Base}) \times {\rm height}.
\]
Place the cursor over the image to start the animation.
Another justification of this formula is obtained by calculation the area of the cross section of the cone and the light blue plane which is parallel to the plane of the base and which goes through the point $S_x.$ Denote this cross section by ${\rm Base}_x.$ Each square $R_j,$ $j \in \{1,\ldots,n\},$ has a corresponding square in the light blue plane. Denote this square by $R_{j,x}$. In the previous item we proved that \[ {\operatorname{area}}({R_{j,x}}) = {\operatorname{area}}({R_j})\, \frac{x^2}{h^2}. \] We have \[ {\operatorname{area}}({{\rm Base}_{x}}) \approx \sum_{j=1}^n {\operatorname{area}}({R_{j,x}}) = \sum_{j=1}^n {\operatorname{area}}({R_j})\, \frac{x^2}{h^2} \approx \frac{x^2}{h^2} {\operatorname{area}}({\rm Base}). \] Therefore \[ {\operatorname{area}}({{\rm Base}_{x}}) = \frac{x^2}{h^2} {\operatorname{area}}({\rm Base}). \] Now a standard argument with Riemann sums leads to the fact that the volume of the cone is given by the integral \[ \int_0^h \frac{x^2}{h^2} {\operatorname{area}}({\rm Base}) dx = \frac{1}{h^2} {\operatorname{area}}({\rm Base}) \int_0^h x^2 dx = \frac{1}{3} {\operatorname{area}}({\rm Base}) \, h. \]
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In Problem 4 on Assignment 5 you are asked to determine the volume enclosed by two and three specially positioned cylinders. A related question is to determine the volume of the intersection of two and three specially positioned cylinders. I am posting images of the solids formed by the intersection of two and three cylinders. It is remarkable how easy it is to produce these images in Mathematica; one uses the command RegionPlot3D[]. Specifically, for the intersection of two cylinders I used
RegionPlot3D[
and for for the intersection of three cylinders I used
And[y^2 + z^2 < 1, z^2 + x^2 < 1], {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, PlotPoints -> {200, 200, 200}, Mesh -> None, PlotStyle -> {Opacity[1]}, Ticks -> {Range[-2, 2, 1], Range[-2, 2, 1], Range[-2, 2, 1]}, ImageSize -> 500
]RegionPlot3D[
And[x^2+y^2 < 1, y^2 + z^2 < 1, z^2 + x^2 < 1], {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, PlotPoints -> {200, 200, 200}, Mesh -> None, PlotStyle -> {Opacity[1]}, Ticks -> {Range[-2, 2, 1], Range[-2, 2, 1], Range[-2, 2, 1]}, ImageSize -> 500
] -
The following six examples illustrate solids in Problem 2 on Assignment 5. They inspired by the subsection "Volumes of Regions of Known Cross-Section" of Section 8.2. Here the base is always a unit disk and the cross-sections are
regular polygons: equilateral triangle, square, regular pentagon, regular hexagon, regular heptagon and regular octagon. There are twelve animations below.
Place the cursor over the image to start the animation.
A cross-section is an equilateral triangle.
Place the cursor over the image to start the animation.
A cross-section is a square.
Place the cursor over the image to start the animation.
A cross-section is a regular pentagon.
Place the cursor over the image to start the animation.
A cross-section is a regular hexagon.
Place the cursor over the image to start the animation.
A cross-section is a regular heptagon.
Place the cursor over the image to start the animation.
A cross-section is a regular octagon.
- I mentioned in class that my guiding slogan for an integral calculus course could be "How long is a smile?" As one of the applications of definite integrals we will learn how to calculate the length of a given graph of a function. The slogan is inspired by the fact that a graph of a parabola can be used as a smile in a smiley face, as is shown in the animations below.
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Place the cursor over the image to start the animation.
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- Relevant reading related to the definition of Riemann integral in the Calculus textbook is Chapter 5. Recommended exercises and problems are 5.2: 15, 29, 30, 31, 34, 45; 5.3: 33-34, 37; 5.4: 11, 21, 29, 30, 33, 37, 39-42, 43, 44, 45, 47, 48, 53, 54-56. Review of Chapter 5: 49, 66.
- Here is a list of topics that will be covered on the exam on Monday.
- Using the definition of Riemann integral one can prove that some familiar kinds of functions are Riemann integrable.
-
Theorem. If $f: [a,b]\to \mathbb R$ is a monotonic function, then $f$ is Riemann integrable on $[a,b]$.
-
Theorem. If $f: [a,b]\to \mathbb R$ is a continuous function, then $f$ is Riemann integrable on $[a,b]$.
- The following abstract theorem gives a characterization of Riemann integrability just in terms of the lower and upper Riemann sums.
-
Theorem. Let $f: [a,b]\to \mathbb R$ be a bounded function. The function $f$ is Riemann integrable on $[a,b]$ if and only if there exists a sequence of partitions $\mathcal P_n$ of $[a,b]$ such that
\[
\lim_{n\to \infty} \bigl( U(f,\mathcal P_n) - L(f,\mathcal P_n) \bigr) = 0.
\]
In this case
\[
\int_a^b \! f(x) \, dx = \lim_{n\to \infty} U(f,\mathcal P_n) = \lim_{n\to \infty} L(f,\mathcal P_n).
\]
Remark. It is interesting to point out that this theorem can be used to give a different proof that the function $f(x) = x^2$ is integrable. This proof is based on the fact that \[ 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}. \] For example, to prove that $\int_0^1 x^2 dx = 1/3$ consider the following sequence of partitions of $[0,1].$ For $n \in \mathbb N$ set \[ x_0 = 0 \lt x_1 = \frac{1}{n} \lt x_2 = \frac{2}{n} \lt \cdots \lt x_{n-1} = \frac{n-1}{n} \lt x_n = \frac{n}{n} = 1. \] Denote this partition by $\mathcal P_n.$ Then \[ L(x^2,\mathcal P_n) = \sum_{j=1}^n (x_{j-1})^2 (x_j - x_{j-1}) = \sum_{j=1}^n \left(\frac{j-1}{n}\right)^2 \frac{1}{n} = \frac{(n-1)n(2n-1)}{6n^3} \] and \[ U(x^2,\mathcal P_n) = \sum_{j=1}^n (x_{j})^2 (x_j - x_{j-1}) = \sum_{j=1}^n \left(\frac{j}{n}\right)^2 \frac{1}{n} = \frac{n(n+1)(2n+1)}{6n^3}. \] Therefore \[ U(x^2,\mathcal P_n) - L(x^2,\mathcal P_n) = \frac{1}{n}. \] Hence \[ \lim_{n\to\infty} \bigl(U(x^2,\mathcal P_n) - L(x^2,\mathcal P_n)\bigr) = 0 \] and \[ \lim_{n\to\infty} U(x^2,\mathcal P_n) = \lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3} = \lim_{n\to\infty} \frac{1}{3} \left( 1 + \frac{1}{n} \right) \left( 1 + \frac{1}{2n} \right) = \frac{1}{3}. \] - The following theorems provide algebra of definite integrals.
-
Theorem. If $: f[a,b]\to \mathbb R$ is Riemann integrable on $[a,b]$ and $\alpha \in \mathbb R$, then the function $(\alpha \, f)$ defined as $(\alpha f)(x) = \alpha f(x)$ is also integrable on $[a,b]$ and
\[
\int_a^b \! \alpha f(x) \, dx = \alpha \int_a^b \! f(x) \, dx.
\]
-
Theorem. If $f: [a,b]\to \mathbb R$ and $g: [a,b]\to \mathbb R$ are Riemann integrable on $[a,b]$, then the function $f+g$ defined as $(f + g)(x) =f(x) + g(x)$ is also integrable on $[a,b]$ and
\[
\int_a^b \! \bigl( f(x) + g(x) \bigr) \, dx = \int_a^b \! f(x) \, dx + \int_a^b \! g(x) \, dx.
\]
-
Fundamental Theorem of Calculus. Let $F: [a,b]\to \mathbb R$ and $f: [a,b]\to \mathbb R$ be functions with the following three properties:
- $f$ is Riemann integrable on $[a,b]$.
- $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$.
- $F'(x) = f(x)$ for all $x \in (a,b)$.
- The second part of the Fundamental Theorem of Calculus involves the concept of Maya's carpet. For a continuous function $f:[a,b] \to \mathbb R$ Maya's carpet of $f$ is the following function: \[ F(x) = \int_a^x f(t)dt. \]
- Before illustrating Maya's carpet I illustrate the values of a function f :
- Let f be a continuous function represented in the figure below by its graph, the black curve. Fix a number a and let x be an arbitrary number such that x > a. Then the cyan area in the animated figure below changes with x. In other words, the cyan area is a function of x. In an old calculus book I saw that the cyan area was called Maya's carpet. I like this name since you can think of the animation below as a cyan carpet being unrolled in the xy-plane.
- Fundamental Theorem of Calculus. Let $f: [a,b]\to \mathbb R$ be a continuous function. For every $x \in [a,b]$ set \[ F(x) = \int_a^x f(t)dt. \] Then $F$ is differentiable on $(a,b)$ and for all $x \in (a,b)$ we have $F'(x) = f(x).$
- The proof of the second part of the Fundamental Theorem of Calculus uses the following Mean Value Theorem for Integrals.
- Mean Value Theorem for Integrals. Let $f: [a,b]\to \mathbb R$ be a continuous function. Then there exists $c \in [a,b]$ such that \[ \int_a^b f(x) dx = f(c) (b-a). \] Proof. Since $f$ is continuous on $[a,b]$ the Extreme Values Theorem applies. Therefore there exist $x_0, x_1 \in [a,b]$ such that for all $x \in [a,b]$ we have \[ f(x_0) \leq f(x) \leq f(x_1). \] The last inequalities imply that \[ f(x_0) \bigl(b-a\bigr) \leq \int_a^b f(x) dx \leq f(x_1) \bigl(b-a\bigr). \] Now dividing by $b-a \gt 0$ we get \[ f(x_0) \leq \frac{1}{b-a} \int_a^b f(x) dx \leq f(x_1) . \] Thus the real number \[ \frac{1}{b-a} \int_a^b f(x) dx \] is between two values of a continuous function. Now the Intermediate Value Theorem implies that there exists $c \in [a,b]$ such that \[ f(c) = \frac{1}{b-a} \int_a^b f(x) dx. \]
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
- Definition. A function $f: [a,b] \to \mathbb R$ is said to be Riemann integrable on $[a,b]$ if there exists a number $S$ such that for every $\epsilon \gt 0$ there exists $\delta(\epsilon) \gt 0$ with the following property: for every tagged partition $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ of $[a,b]$ we have that \[ \max \bigl\{ x_j - x_{j-1} \, |\, j \in \{1,\ldots,n\} \bigr\} \lt \delta(\epsilon) \qquad \text{implies} \qquad \left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - S \right| \lt \epsilon. \] If $f: [a,b] \to \mathbb R$ is Riemann integrable on $[a,b],$ then the number $S$ is called the definite integral of $f$ on $[a,b]$; the Riemann integral of $f$ on $[a,b]$ is denoted by \[ \int_a^b \!\! f(x) dx. \]
- Exercise. Let $C \in \mathbb R$. Prove that the constant function $f(x) = 1$, $x \in \mathbb R$, is integrable on any interval $[a,b]$ and \[ \int_a^b 1 \, dx = b-a. \]
-
Exercise. Prove that the linear function $f(x) = x$, $x \in \mathbb R$, is integrable on any interval $[a,b]$ and
\[
\int_a^b \! x \, dx = \frac{b^2 - a^2}{2}.
\]
Proof. In this proof $f(x) = x.$ Let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ of $[a,b]$ be an arbitrary partition of $[a,b].$ Then we have \[ U(f,\mathcal P) + L(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j + x_{j-1} \bigr) \bigl(x_j - x_{j-1}\bigr) = \sum_{j=1}^n \bigl((x_j)^2 - (x_{j-1})^2 \bigr) = b^2 - a^2. \] Also \[ U(f,\mathcal P) - L(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j - x_{j-1} \bigr) \bigl(x_j - x_{j-1}\bigr) \leq \| \mathcal P \| \sum_{j=1}^n \bigl( x_j - x_{j-1} \bigr) = (b - a) \|\mathcal P\|. \] Adding the equality \[ U(f,\mathcal P) + L(f,\mathcal P) = b^2 - a^2. \] and the inequality \[ U(f,\mathcal P) - L(f,\mathcal P) \leq (b - a) \|\mathcal P\|. \] we get \[ 2 U(f,\mathcal P) \leq b^2 - a^2 + (b - a) \|\mathcal P\| \] and, subtracting them we get \[ 2 L(f,\mathcal P) \geq b^2 - a^2 - (b - a) \|\mathcal P\|. \] Together with inequality (LRU) we thus have \[ \frac{b^2 - a^2}{2} - \frac{b - a}{2} \|\mathcal P\| \leq L(f,\mathcal P) \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \leq U(f,\mathcal P) \leq \frac{b^2 - a^2}{2} + \frac{b - a}{2} \|\mathcal P\|. \] This hence proves \[ \left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - \frac{b^2 - a^2}{2} \right| \leq \frac{b - a}{2} \|\mathcal P\|. \] Thus, if we choose a partition $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ such that $\| \mathcal P \| \lt 2 \epsilon /(b-a)$, then we must have \[ \left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - \frac{b^2 - a^2}{2} \right| \lt \epsilon. \]
Remark. In the above proof we used the fact that for the function $f(x) = x$ we have $L(f,\mathcal P) = {\operatorname{Le}}(f,\mathcal P)$ and $U(f,\mathcal P) = {\operatorname{Ri}}(f,\mathcal P)$. Also the identity \[ U(f,\mathcal P) + L(f,\mathcal P) = {\operatorname{Ri}}(f,\mathcal P) + {\operatorname{Le}}(f,\mathcal P) = b^2 - a^2 \] shows that the average of the left and right Riemann sums gives the exact value of the integral. This is the case for every linear function $f(x) = m x + k$. -
Exercise. Prove that the square function $f(x) = x^2$, $x \in \mathbb R$, is integrable on any interval $[a,b]$ with $0 \leq a \lt b$ and
\[
\int_a^b \! x^2 \, dx = \frac{b^3 - a^3}{3}.
\]
Proof. In this proof $f(x) = x^2$ and $0 \leq a \lt b$. Based on the identity
\[
\left(u^2 + v^2 + 4 \left( \frac{u+v}{2} \right)^2 \right) (v - u) = 2\bigl( v^3 - u^3 \bigr)
\]
which holds for arbitrary reals $u$ and $v$, one can prove that for an arbitrary partition $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ of $[a,b]$ we have
\[ \tag{1}
L(f,\mathcal P) + U(f,\mathcal P) + 4 {\operatorname{Mi}}(f,\mathcal P) = 2 \bigl( b^3 - a^3 \bigr).
\]
Since $0 \leq a$ we have $x_j^2 - x_{j-1}^2 \gt 0$ for all $j \in \{1,\ldots,n\}$. Therefore,
\[
U(f,\mathcal P) - L(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j^2 - x_{j-1}^2 \bigr) \bigl(x_j - x_{j-1}\bigr) \leq \| \mathcal P \| \sum_{j=1}^n \bigl( x_j^2 - x_{j-1}^2 \bigr) = (b^2 - a^2) \|\mathcal P\|.
\]
Thus,
\[ \tag{2}
U(f,\mathcal P) - L(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|.
\]
Using inequality (LMU), (2) implies
\[ \tag{3}
U(f,\mathcal P) - {\operatorname{Mi}}(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|
\]
and
\[ \tag{4}
{\operatorname{Mi}}(f,\mathcal P) - L(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|.
\]
Adding inequalities (1), (2) and (3) multiplied by $4$ and dividing the resulting inequality by 6 we get
\[
U(f,\mathcal P) \leq \frac{b^3 - a^3}{3} + \frac{5}{6}(b^2 - a^2) \|\mathcal P\|
\]
Adding inequalites (1), (2) multiplied by $-1$ and (3) multiplied by $-4$ and dividing the resulting inequality by 6 we get
\[
L(f,\mathcal P) \geq \frac{b^3 - a^3}{3} - \frac{5}{6}(b^2 - a^2) \|\mathcal P\|.
\]
Together with inequality (LRU) we thus have
\[
\frac{b^3 - a^3}{3} - \frac{5}{6}(b^2 - a^2) \|\mathcal P\| \leq L(f,\mathcal P) \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \leq U(f,\mathcal P) \leq \frac{b^3 - a^3}{3} + \frac{5}{6}(b^2 - a^2) \|\mathcal P\|,
\]
for arbitrary choice of $t_j \in [x_{j-1},x_j]$ for $j \in \{1,\ldots,n\}$. This hence proves
\[
\left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - \frac{b^3 - a^3}{3} \right| \leq \frac{5}{6}(b^2 - a^2) \|\mathcal P\|.
\]
Thus, if we choose a partition $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ such that $\| \mathcal P \| \lt \tfrac{6}{5} \epsilon /(b^2-a^2),$ then we must have
\[
\left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - \frac{b^3 - a^3}{3} \right| \lt \epsilon.
\]
Remark. In the above proof we used the fact that for the function $f(x) = x^2$ and for $a \geq 0$ we have $L(f,\mathcal P) = {\operatorname{Le}}(f,\mathcal P)$ and $U(f,\mathcal P) = {\operatorname{Ri}}(f,\mathcal P)$. Also the identity \[ 2 \bigl( b^3 - a^3 \bigr) = L(f,\mathcal P) + U(f,\mathcal P) + 4 {\operatorname{Mi}}(f,\mathcal P) = {\operatorname{Le}}(f,\mathcal P) + {\operatorname{Ri}}(f,\mathcal P) + 4 {\operatorname{Mi}}(f,\mathcal P) \] shows that the exact value of the integral of $f$ over $[a,b]$ is a weighted average of the left, the right and the middle Riemann sum: \[ \int_a^b \! f(x) dx = \frac{1}{6} {\rm Le}(f,\mathcal P) + \frac{1}{6} {\operatorname{Ri}}(f,\mathcal P) + \frac{2}{3} {\operatorname{Mi}}(f,\mathcal P) \] -
Fundamental Theorem of Calculus. Let $F: [a,b]\to \mathbb R$ and $f: [a,b]\to \mathbb R$ be functions with the following three properties:
- $f$ is Riemann integrable on $[a,b]$.
- $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$.
- $F'(x) = f(x)$ for all $x \in (a,b)$.
- Today we list several definitions. A motivation for these definitions is in this pdf file (right click on this file and download it to your computer and open it in Acrobat) in which I tried to present a calculation of a common simple area in as simple terms as possible.
- Let $a, b \in \mathbb R$ and let $n$ be a positive integer. A finite sequence of numbers $\bigl(x_0,x_1,\ldots,x_n\bigr)$ such that \[ a = x_0 \lt x_1 \lt \cdots \lt x_{n-1} \lt x_n = b \] is called a partition of the interval of $[a,b]$.
- Let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ be a partition of of $[a,b]$. The mash or norm of the partition $\mathcal P$, denoted by $\|\mathcal P \|$, is defined to be the length of the longest subinterval, that is, \[ \|\mathcal P \| = \max \bigl\{ x_j - x_{j-1} \, |\, j \in \{1,\ldots,n \} \bigr\} . \]
- A partition $\bigl(x_0,x_1,\ldots,x_n\bigr)$ of $[a,b]$ together with a sequence $\bigl(t_1,\ldots,t_n\bigr)$ such that \[ t_j \in \bigl[ x_{j-1}, x_j \bigr] \quad \text{for all} \quad j \in \{1,\ldots,n\} \] is called a tagged partition of the interval $[a,b]$. This tagged partition is denoted by $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$.
- Let $f:[a,b] \to \mathbb R$ be a function defined on $[a,b]$ and let $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ be a tagged partition of $[a,b]$. The sum \[ \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \] is called the Riemann sum of $f$ relative to the tagged partition $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ of $[a,b]$.
- Let $f:[a,b] \to \mathbb R$ be a bounded function defined on $[a,b]$ and let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ be a partition of $[a,b]$. For each $j \in \{1,\ldots,n\}$ denote by $m_j$ the greatest number such that $m_j \leq f(x)$ for every $x \in \bigl[x_{j-1},x_j \bigr]$. For each $j \in \{1,\ldots,n\}$ denote by $M_j$ the least number such that $f(x) \leq M_j$ for every $x \in \bigl[x_{j-1},x_j \bigr].$ Then the sums \[ L(f,\mathcal P) = \sum_{j=1}^n m_j \bigl(x_j - x_{j-1}\bigr) \qquad \text{and} \qquad U(f,\mathcal P) = \sum_{j=1}^n M_j \bigl(x_j - x_{j-1}\bigr) \] are, respectively, called lower Riemann sum of $f$ relative to the partition $\mathcal P$ and upper Riemann sum of $f$ relative to the partition $\mathcal P$.
- Let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ and let $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ be a accompanying tagged partition of $[a,b]$. Then \[ \tag{LRU} L(f,\mathcal P) \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \leq U(f,\mathcal P). \]
- Let $f:[a,b] \to \mathbb R$ be a function defined on $[a,b]$ and let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ be a partition of $[a,b]$. The sum \[ {\rm Mi}(f,\mathcal P) = \sum_{j=1}^n f\left(\tfrac{x_{j-1}+x_j}{2}\right) \bigl(x_j - x_{j-1}\bigr) \] is called the middle Riemann sum of $f$ relative to the partition $\mathcal P$; the sum \[ {\rm Le}(f,\mathcal P) = \sum_{j=1}^n f(x_{j-1}) \bigl(x_j - x_{j-1}\bigr) \] is called the left Riemann sum of $f$ relative to the partition $\mathcal P$; the sum \[ {\rm Ri}(f,\mathcal P) = \sum_{j=1}^n f(x_{j}) \bigl(x_j - x_{j-1}\bigr) \] is called the right Riemann sum of $f$ relative to the partition $\mathcal P$
- Let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ be a partition of $[a,b]$. A particular case of inequality (LRU) is \[ \tag{LMU} L(f,\mathcal P) \leq {\rm Mi}(f,\mathcal P) \leq U(f,\mathcal P). \]
- On Wednesday and Thursday we did some famous examples of parametric curves.
- The most famous example of a parametric curve is the unit circle: \[ \bigl\{ (\cos t, \sin t) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in [0,2\pi) \bigr\} \]
- Given two distinct points $(x_0,y_0), (x_1,y_1) \in {\mathbb R}\!\times\!{\mathbb R}$ parametric equations of the line through these two points is \[ \bigl\{ \bigl( x_0 + t(x_1 - x_0) , y_0 + t ( y_1 - y_0) \bigr) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in \mathbb R \bigr\} \] Alternatively, given a point $(x_0,y_0) \in {\mathbb R}\!\times\!{\mathbb R}$ and $a, b \in \mathbb R$ such that $a^2+ b^2 \gt 0,$ then \[ \bigl\{ \bigl( x_0 + a\, t , y_0 + b\, t \bigr) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in \mathbb R \bigr\} \] represents a line through the point $(x_0,y_0) \in {\mathbb R}\!\times\!{\mathbb R}.$
-
We derived parametric equations of cycloid
\[
\bigl\{ (t - \sin t, 1 - \cos t) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in \mathbb R \bigr\}
\]
Place the cursor over the image to start the animation.
-
We derived parametric equations of cardioid
\[
\bigl\{ \bigl(2 \cos t - \cos(2t), 2 \sin t - \sin(2t) \bigr) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in [0, 2\pi ) \bigr\}
\]
Place the cursor over the image to start the animation.
-
We also derived parametric equations of nephroid
\[
\left\{ \left( \frac{1}{2} \left(3 \cos t - \cos(3t) \right) , \frac{1}{2} \left( 3 \sin t - \sin(3t) \right) \right) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in [0, 2\pi ) \right\}
\]
Here a circle of the radius $1/2$ rolls around the unit circle.
Place the cursor over the image to start the animation.
-
Now it is easy to get parametric equations of any curve obtained by rotating a smaller (or larger) circle around the unit circle. Here we do it with a circle of radius $1/3$
\[
\left\{ \left( \frac{1}{3} \left(4 \cos t - \cos(4t) \right) , \frac{1}{3} \left( 4 \sin t - \sin( 4 t) \right) \right) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in [0, 2\pi ) \right\}
\]
Place the cursor over the image to start the animation.
-
Finally, it does not take much more to get parametric equations of a curve obtained by rotating a smaller circle inside the unit circle. Here we do it with a circle of radius $1/3$
\[
\left\{ \left( \frac{1}{3} \left(2 \cos t + \cos(2t) \right) , \frac{1}{3} \left( 2 \sin t - \sin( 2 t) \right) \right) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in [0, 2\pi ) \right\}
\]
Place the cursor over the image to start the animation.
-
Just for fun we show a circle of the radius $1/5$ rotating inside the unit circle:
\[
\left\{ \left( \frac{1}{5} \left(4 \cos t + \cos(4t) \right) , \frac{1}{5} \left( 4 \sin t - \sin( 4 t) \right) \right) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in [0, 2\pi ) \right\}
\]
Place the cursor over the image to start the animation.
- Let us now define the concept of the tangent line to a curve given by its parametric equations.
- Let $\alpha, \beta \in \mathbb R$ be such that $\alpha \lt \beta$ and let $t_0 \in (\alpha, \beta).$ Let $f:(\alpha,\beta) \to \mathbb R$ and $g:(\alpha,\beta) \to \mathbb R$ be two functions which are both differentiable at $t_0$ and such that \[ \bigl(f'(t_0)\bigr)^2 + \bigl(g'(t_0)\bigr)^2 \gt 0. \] Then the line \[ \Bigl\{ \bigl( f(t_0) + f'(t_0) t, g(t_0) + g'(t_0) t \bigr) \,|\, t \in \mathbb R \Bigr\} \] is said to be the tangent line to the curve given by the parametric equations \[ \Bigl\{ \bigl( f(t), g(t) \bigr) \,|\, t \in (\alpha,\beta) \Bigr\} \] at the point $\bigl(f(t_0),g(t_0)\bigr).$
-
Here is the cardioid
\[
\bigl\{ \bigl(2 \cos t - \cos(2t), 2 \sin t - \sin(2t) \bigr) \in {\mathbb R}\!\times\!{\mathbb R} \,| \, t \in [0, 2\pi ) \bigr\}
\]
with its 255 tangent lines.
-
Just for comparison, here is the unit circle with its 256 tangent lines.
-
It is interesting to look at the family of all normals to the cycloid and the cardioid:
-
Today we did implicit differentiation. The relevant section in the Calculus textbook is 3.7 (do 30, 31, 32, 34, 35, 36, 38, 39). Problem 38 asks for a common tangent line to two circles. Inspired by this one can ask for a common tangent lines to different kind of functions. For example
- Find the common tangent lines to the graphs of $y = x^2+1$ and $y = -2x^2$.
-
We looked at the famous curve Folium of Descartes. The folium of Descartes is the following subset of $\mathbb R^2$:
\[
F = \bigl\{ (x,y) \in \mathbb R\!\times\!\mathbb R \, | \, x^3 + y^3 = 6 x y \bigr\}.
\]
Clearly $(3,3) \in F$. It is also clear that $F$ is symmetric with respect to the line $y=x$. That is, we have $(a,b) \in F$ if and only if $(b,a) \in F$. This suggests a natural question:
- Find the continuous involution function $y = f(x)$ such that $3 = f(3)$ and whose graph is a subset of $F$.
Other interesting questions related to $F$ are- Find all continuous functions $y = f(x)$ with the following two properties: the graph of $f$ is a subset of $F$ and $f$ is not a restriction of a function $g$ whose graph is also a subset of $F$. (I found three such functions.)
- Find all continuous bijections $y = f(x)$ with the following two properties: the graph of $f$ is a subset of $F$ and $f$ is not a restriction of a bijection $g$ whose graph is also a subset of $F$. (I found four such bijections.)
- Find all continuous involutions $y = f(x)$ with the following two properties: the graph of $f$ is a subset of $F$ and $f$ is not a restriction of an involution $g$ whose graph is also a subset of $F$. (I found two such involutions.)
Folium of Descartes $x^3+y^3 = 6xy$
-
We also determined the asymptote of the Folium of Descartes. We did it by considering all the lines $y=b-x$ and determining the trichotomy for the numbers $b$: for some $b$ the line does not cross the Folium of Descartes, for some $b$ there is only one point intersection and for some points $b$ there are two points of intersection of the line and the Folium. It turned out that the following trichotomy holds:
- If $b \in (-2,0)\cup (0,6)$, then the line $y=b-x$ intersects the Folium of Descartes at exactly two distinct points.
- If $b \in \{0,6\}$, then the line $y=b-x$ intersects the Folium of Descartes at only one point.
- If $b \in (-\infty,-2]\cup (6,+\infty)$, then the line $y=b-x$ does not intersect the Folium of Descartes.
- On Thursday and Friday we did my favorite optimization problems. I will post them later.
- Today we did several examples of osculating circles. Let $I \subseteq \mathbb R$ be an open interval, $a \in I$ and let $f: I \to \mathbb R$ be a function which is twice differentiable on $I$. Recall the definition of the osculating circle: If $f''(a) \neq 0$, then the center of the osculation circle to the graph of $f$ at the point $(a,f(a))$ is \[ \left( a - f'(a) \frac{1 + \left(f'(a) \right)^2 }{ f''(a)}, \ f(a) + \frac{1 + \left(f'(a) \right)^2 }{ f''(a)} \right) \] and its radius is \[ \frac{\left( 1 + \left(f'(a) \right)^2 \right)^{3/2}}{\bigl| f''(a) \bigr|}. \] The reciprocal of the radius is called the curvature of the graph of $f$ at the point $(a,f(a)).$
-
We calculated the osculating circle to the graph of the exponential function $\exp(x)$ with the smallest radius: It is the osculating circle at the point $\bigl(-(\ln 2)/2, 1/\sqrt{2} \bigr)$ which is centered at $\bigl(-(3+\ln 2)/2, 2\sqrt{2} \bigr)$ and with the radius $3\sqrt{3}/2.$
the smallest osculating circle in teal
- Finding the osculating circle with a smallest radius is an example of an optimization problem. Next we will do several optimization problems.
- Problem. A cone without base can be constructed from a circular sector cut out from a unit disk. Find the angle of the circular sector which yields the cone with the maximum volume. (Using this disk you can verify your solution with an experiment.)
- Section 4.3 of the textbook deals with optimization problems. Do the following problems: 16, 17, 28, 29, 31, 36, 44, 46, 47, 48.
- I wrote a webpage introducing Higher_order_approximations of differentiable functions. In particular I introduce the concept of an osculating circle for a graph of function. This is done in complete analogy with how we introduced the concept of the tangent line.
- Today we did three problems with one unifying theme: trichotomy.
-
The first trichotomy occurs in counting distinct real roots of the quadratic equation
\[ \tag{qe}
x^2 + \alpha \, x + \beta = 0,
\]
where $\alpha, \beta \in \mathbb R.$ The number of distinct real roots of equation (qe) depends on the pair $(\alpha,\beta) \in \mathbb R\!\times\!\mathbb R.$ There are three possibilities:
- with $(\alpha,\beta)$ in equation (qe) the equation has no real solutions (we will color these points yellow),
- with $(\alpha,\beta)$ in equation (qe) the equation has a unique real solution (we will color these points navy blue),
- with $(\alpha,\beta)$ in equation (qe) the equation has exactly two real solutions (we will color these points light teal),
yellow points correspond to no real solutions of (qe) navy blue points correspond to a unique real solution of (qe) teal points correspond to two distinct real solutions of (qe)
-
The next trichotomy occurs in counting distinct real roots of the cubic equation
\[ \tag{ce}
x^3 + \alpha \, x + \beta = 0,
\]
where $\alpha, \beta \in \mathbb R.$ The number of distinct real roots of equation (ce) depends on the pair $(\alpha,\beta) \in \mathbb R\!\times\!\mathbb R.$ There are three possibilities:
- with $\alpha$ and $\beta$ in equation (ce) the equation has a unique real solution (we will color such points $(\alpha,\beta)$ in yellow),
- with $\alpha$ and $\beta$ in equation (ce) the equation has exactly two real solutions (we will color such points $(\alpha,\beta)$ in navy blue),
- with $\alpha$ and $\beta$ in equation (ce) the equation has exactly three real solutions (we will color such points $(\alpha,\beta)$ in light teal),
First notice that with $\alpha = 0$ equation (ce) has a unique real solution $-\sqrt[3]{\beta}.$ Hence all the points on the $\beta$-axis are colored yellow.Set $f(x) = x^3 + \alpha \, x + \beta.$ Then $f'(x) = 3\, x^2 + \alpha.$ If $\alpha \gt 0,$ then $f'(x) \gt 0$. By the First Monotonicity Test $f$ is an increasing function. It follows from the continuity of $f$ and the Intermediate Value Theorem that there exists $c \in \mathbb R$ such that $f(c) = 0.$ Such $c$ is unique since $f$ is increasing. Hence, all the points in $(\alpha,\beta)$ with $\alpha \gt 0$ are colored yellow.Let $\alpha \lt 0$. If $\beta = 0$, then (ce) has three real solutions $x=0$, $x = -\sqrt{-\alpha}$ and $x = \sqrt{-\alpha}$. Intuitively it is clear that for small values of $\beta$ equation (ce) will have three real solutions. Now we need to identify those values of $\beta$ for which equation (ce) will have three real solutions. It turns out that the local minimum and the local maximum of $f$ play a role here. The local maximum of $f$ is at the point \[ \left( -\frac{|\alpha|^{1/2}}{\sqrt{3}}, \ \beta + 2 \frac{|\alpha|^{3/2}}{3\sqrt{3}} \right) \] and the local minimum is at the point \[ \left( \frac{|\alpha|^{1/2}}{\sqrt{3}}, \ \beta - 2 \frac{|\alpha|^{3/2}}{3\sqrt{3}} \right). \] For fixed $\alpha \lt 0$, cubic equation (ce) will have three distinct real roots if and only if the local maximum value is positive and the local minimum value is negative. That is if and only if \[ \beta + 2 \frac{|\alpha|^{3/2}}{3\sqrt{3}} \gt 0 \qquad \text{and} \qquad \beta - 2 \frac{|\alpha|^{3/2}}{3\sqrt{3}} \lt 0. \] That is, if and only if \[ - 2 \frac{|\alpha|^{3/2}}{3\sqrt{3}} \lt \beta \lt 2 \frac{|\alpha|^{3/2}}{3\sqrt{3}}. \] Thus all the points $(\alpha, \beta)$ where $\alpha \lt 0$ and $\beta$ satisfies the above two inequalities will be colored teal.The points \[ \left( \alpha, - 2 \frac{|\alpha|^{3/2}}{3\sqrt{3}} \right) \quad \text{and} \quad \left( \alpha, 2 \frac{|\alpha|^{3/2}}{3\sqrt{3}} \right) \] where $\alpha \lt 0$ are colored navy blue.Thus the following coloring of the $\alpha\beta$-plane is introduced by the trichotomy of number of real solutions of (ce):
yellow points correspond to a unique real solution of (ce) navy blue points correspond to two distinct real solutions of (ce) teal points correspond to three distinct real solutions of (ce)
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Let $a \gt 1$ and $\beta \gt 0$ and consider the following equation
\[ \tag{ep}
a^x = x^\beta \qquad \text{with} \qquad x \gt 0.
\]
The number of distinct real roots of equation (ep) depends on the pair $(a,\beta) \in (1,+\infty)\!\times\!(0,+\infty).$ There are three possibilities:
- with $a$ and $\beta$ in equation (ep) the equation no real solutions (we will color such points $(a,\beta)$ in yellow),
- with $a$ and $\beta$ in equation (ep) the equation has a unique real solution (we will color such points $(a,\beta)$ in navy blue),
- with $a$ and $\beta$ in equation (ep) the equation has exactly two real solutions (we will color such points $(a,\beta)$ in light teal),
Next we will prove the above trichotomy and find the exact values of the solutions when they exist. Assume that $a \gt 1$, $\beta \gt 0$ and $x \gt 0.$ Equation (ep) is equivalent to the equation \[ x\, \ln a = \beta \ln x. \] Next substitute $x = e^z$ where $z \in \mathbb R$ \[ e^z \, \ln a = \beta z. \] The preceding equation is equivalent to \[ - \frac{\ln a}{\beta} = - z e^{-z}. \] On the right-hand side of the preceding equation we have an expression of the form $XE^X$ which we encountered in the definition of the Lambert $W$ function. Before proceding further review the definition of the Lambert $W$ functions given on October 25. Since $-(\ln a)/\beta \lt 0$ the preceding equation does not have a real solution if and only if $-(\ln a)/\beta \lt -1/e.$ That is if and only if $e \, (\ln a) \gt \beta.$ The preceding equation has a unique real solution if and only if $-(\ln a)/\beta = -1/e,$ that is if and only if $\beta = e \, (\ln a).$ In this case the unique solution for $z$ is $z=1.$ Finally, the preceding displayed equation has two distinct real solutions if and only if $e \, (\ln a) \lt \beta.$ In this case two distinct solutions are given by \[ z = - W_0\bigl(-(\ln a)/\beta\bigr), \qquad z = - W_{-1}\bigl(-(\ln a)/\beta\bigr). \]For the original equation (ep) we have the same conclusion. However, we will use the following equivalences \begin{align*} \beta \lt e \, (\ln a) \quad & \Leftrightarrow \quad e^\beta \lt a^e \\ \beta = e \, (\ln a) \quad & \Leftrightarrow \quad e^\beta = a^e \\ \beta \gt e \, (\ln a) \quad & \Leftrightarrow \quad e^\beta \gt a^e \end{align*} to formulate the trichotomy for equation (ep) slightly differently. For $(a,\beta) \in (1,+\infty)\!\times\!(0,+\infty)$ we have the following trichotomy:- If $e^\beta \lt a^e$, then equation (ep) has no real solutions. In this case \[ x^\beta \lt a^x \qquad \text{for all} \qquad x \geq 0. \] The points $(a,\beta)$ with $e^\beta \lt a^e$ are yellow on the plot below.
- If $e^\beta = a^e$, then equation (ep) has a unique real solution. That unique real solution is $x = e.$ In this case \[ x^\beta \lt a^x \qquad \text{for all} \qquad x \in [0,e)\cup (e,+\infty). \] The points $(a,\beta)$ with $e^\beta = a^e$ are navy blue on the plot below.
- If $e^\beta \gt a^e$, then equation (ep) has exactly two real solutions \[ x_1 = e^{- W_0\bigl(-(\ln a)/\beta\bigr)}, \qquad x_2 = e^{- W_{-1}\bigl(-(\ln a)/\beta\bigr)}. \] Clearly $x_1 \lt e \lt x_2$ and \[ x^\beta \gt a^x \qquad \text{for all} \qquad x \in (x_1,x_2) \] and \[ x^\beta \lt a^x \qquad \text{for all} \qquad x \in [0,x_1)\cup (x_2,+\infty). \] The points $(a,\beta)$ with $e^\beta \gt a^e$ are teal on the plot below.
In the above plot I placed the line $\beta = a$ in slightly darker teal. This line tells us about the number of positive solutions of the equation $a^x = x^a.$ Since $x = a$ is clearly a solution this equation has at least one real solution. This solution is unique if $a = e$. Otherwise the equation $a^x = x^a$ has two distinct solutions. If $a \lt e$ the second solution is greater than $e$, and if $e \lt a$ the second solution is smaller than $e.$
yellow points correspond to no real solutions of (ep) navy blue points correspond to a unique real solution of (ep) teal points correspond to two distinct real solutions of (ep)
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Below is an animation related to Problem 3 on Assignment 3. The animation shows a point, marked by a navy blue circle, moves along the parabola $y=x^2$ and looks for all normals to the parabola that go through that point. The first scene represents the point at which there are two normals, one teal and one purple. Then the point proceeds towards the region where there are three normals, one teal, one purple and one olive. After a while the point will return and enter the region where there is only the teal normal. Each normal is accompanied by the tangent line at the point of normalcy: the teal normal with the cyan tangent line, the purple normal with the magenta tangent line and the olive normal with the yellow tangent line.
Place the cursor over the image to start the animation.
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Recall two important bijections from yesterday. The bijections are plotted below in navy blue and their inverses are plotted in maroon. The maroon functions are two branches of the Lambert W function. The formulas for maroon functions are given below graphs. Remember that two Lambert $W$ functions are
\[
W_0 : [-1/e, + \infty) \to [-1,+\infty)
\]
and
\[
W_{-1} : (-1/e, 0) \to (-\infty, -1).
\]
$ X = [-1,+\infty), \ Y = [-1/e,+\infty), \ W_0 = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
The Lambert $W_0$ function
$X = (-\infty,-1), \ Y = (-1/e,0), \ W_{-1} = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\}$
The Lambert $W_{-1}$ function
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Below are several examples how Lambert W functions are used to solve equations.
- Solve $2^x = 5 x$ for $x \in \mathbb R$. This is how this equation is being solved. Rewrite the equation using the exponential function $e^x$: \[ e^{x \ln 2} = 5 x. \] Next bring the equation to the form $Y = X e^X$ where $Y$ will be a specific number. Then we can use the Lambert functions to solve this equation. Here we will bring the exponential part together with $x$. \[ \frac{1}{5} = x\,e^{-x \ln 2} . \] Now we recognize that our $X$ must be $-x \ln 2$, so we have to multiply the equation by $-\ln 2$: \[ -\frac{\ln 2}{5} = - (x \ln 2) \,e^{-(x \ln 2)}. \] Since $-1/e \lt -(\ln 2)/5 \lt 0$ the above equation has two solutions. The solutions are \[ - (x_1 \ln 2) = W_{0}\bigl(-(\ln 2)/5\bigr) \quad \text{and} \quad - (x_2 \ln 2) = W_{-1}\bigl(-(\ln 2)/5\bigr). \] Hence, \[ x_1 = - \frac{1}{\ln 2} W_{0}\bigl(-(\ln 2)/5\bigr) \approx 0.235456 \quad \text{and} \quad x_2 = - \frac{1}{\ln 2} W_{-1}\bigl(-(\ln 2)/5\bigr) \approx 4.488001. \]
-
Solve $x^x = y$ for $x \gt 0$ and appropriate $y \gt 0$. This is a little trickier. Apply $\ln$ to both sides to get
\[
x \ln x = \ln y.
\]
Rewrite the equation using the exponential function $X \, e^X$:
\[
(\ln x) \, e^{\ln x} = \ln y.
\]
It just happens that this equation is exactly in the form $X\,e^X = Y$. So this equation has a solution if and only if $\ln y \geq -1/e$. This is the case if and only if $y \geq e^{-1/e}$. The equation has the unique solution if $y = e^{-1/e}$ and that solution is $x = 1/e$. If $e^{-1/e}\lt y \lt 1$, the equation has two solutions given by
\[
\ln (x_1) = W_{-1}\bigl( \ln y \bigr) \quad \text{and} \quad \ln (x_2) = W_{0}\bigl( \ln y \bigr)
\]
Hence
\[
x_1 = \exp\Bigl( W_{-1}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{-1}\bigl( \ln y \bigr)} \quad \text{and} \quad x_2 = \exp\Bigl( W_{0}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{0}\bigl( \ln y \bigr)}.
\]
The last identity might look obscure at first. It follows from the fact that $X = W(Y)$ if and only if $X\,e^X = Y$, with $X$ and $Y$ in the appropriate domains. Therefore, $W(Y)\,e^{W(Y)} = Y$, with $Y$ in the appropriate domain. Hence,
\[
e^{W(Y)} = \frac{Y}{W(Y)}.
\]
Here $W$ stands for either $W_0$ or $W_{-1}$. In each case we have to adjust the domains for $X$ and $Y$ appropriately.
If $y \geq 1$ the equation has a unique solution which is given by \[ x = \exp\Bigl( W_{0}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{0}\bigl( \ln y \bigr)}. \] -
Prove that the function $f:\mathbb R \to \mathbb R$ defined by $f(x) = x \exp(x^2)$ is a bijection. Find a formula for its inverse.
The claim that $f$ is a bijection follows from Monotonicity tests and the fact that $f$ is continuous and $f$ is not bounded. Let $y \gt 0$. Here we will demonstrate how to solve \[ x \exp(x^2) = y. \] Since $y \gt 0$ a solution $x$ must be positive. Therefore the given equation is equivalent to \[ x^2 \exp(2\,x^2) = y^2. \] The last equation is equivalent to \[ 2 x^2 \exp(2\,x^2) = 2 y^2. \] Since $2 y^2 \gt 0$ the last equation is equivalent to \[ 2\, x^2 = W_0(2\, y^2). \] Hence the unique solution of the original equation is \[ x = \sqrt{ \frac{W_0(2\, y^2)}{2} }. \] Since the given function is odd, its inverse will also be odd. Therefore the inverse function of $f$ is \[ f^{-1}(x) = (\operatorname{sgn} x) \sqrt{ \frac{W_0(2\, x^2)}{2} }. \]
$f(x)$ in navy blue and its inverse $f^{-1}(x)$ in maroon
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We also discussed the following problem: Let $a \gt 1$ and $\beta \gt 0$ and consider the following equation
\[ \tag{ep}
a^x = x^\beta \qquad \text{with} \qquad x \gt 0.
\]
Prove the following trichotomy: Either equation (ep) has no solutions, or (ep) has a unique solution, or (ep) has exactly two solution. In the cases when solutions exist find the exact values.
I will post the solution of this problem tomorrow.
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We first prove Fermat's Interior Extremum Theorem.
Fermat's Interior Extremum Theorem. Let $a, b, c \in \mathbb R$ and $a \lt c \lt b.$ Let $f : (a,b) \to \mathbb R$ be a function which is differentiable at $c.$ If $f(c)$ is a maximum or a minimum value of $f$ on $(a,b)$, then $f'(c) = 0.$To toggle the proof of the Fermat's Interior Extremum Theorem clickAssume that $f(c)$ is a maximum value of $f$ on $(a,b)$, that is assume that \[ \forall\, x\in (a,b) \qquad f(x) \leq f(c). \] Further assume that $f$ is differentiable at $c$, that is assume that \[ \lim_{x\to c} \frac{f(x) - f(c)}{x-c} = f'(c). \] The function $\frac{f(x) - f(c)}{x-c}$ is defined on $(a,c)\cup(c,b)$. Therefore we can take $\delta_0 = \min\{c-a,b-c\}$ in the definition of limit. By the definition of limit there exist $\delta_0 \gt 0$ such that the function $\frac{f(x) - f(c)}{x-c}$ is defined for all $x \in (c-\delta_0,c)\cup (c,c+\delta_0)$ and for every $\epsilon \gt 0$ there exists $\delta(\epsilon) \in \mathbb R$ such that $0 \lt \delta(\epsilon) \leq \delta_0$ and \[ 0 \lt |x-c| \lt \delta(\epsilon) \qquad \Rightarrow \qquad \left| \frac{f(x) - f(c)}{x-c} - f'(c) \right| \lt \epsilon. \] From the preceding implication we deduce two inequalities. First, assuming that $0 \lt x-c \lt \delta(\epsilon)$ we deduce that \[ \frac{f(x) - f(c)}{x-c} - f'(c) \lt \epsilon, \] and therefore, since $\frac{f(x) - f(c)}{x-c} \geq 0$, we have $-f'(c) \lt \epsilon.$ (this is the first inequality) Second, assuming that $0 \lt c - x \lt \delta(\epsilon)$ we deduce that \[ f'(c) - \frac{f(x) - f(c)}{x-c} \lt \epsilon, \] and therefore, since $\frac{f(x) - f(c)}{x-c} \leq 0$, we have $f'(c) \lt \epsilon.$ (this is the second inequality)Thus, we have proved that both inequalities are true: $f'(c) \lt \epsilon$ and $-f'(c) \lt \epsilon$ and both of these inequalities are true for all $\epsilon \gt 0.$ This implies that $f'(c) = 0.$Notice that the contrapositive of Fermat's Interior Extremum Theorem isContrapositive of FIET. Let $a, b, c \in \mathbb R$ and $a \lt c \lt b$. Let $f : (a,b) \to \mathbb R$ be a function which is differentiable at $c$. If $f'(c) \neq 0$, then $f(c)$ is neither a maximum nor a minimum value of $f$ on $(a,b).$
-
A consequence of the Fermat's Interior Extremum Theorem is Rolle's Theorem.
Rolle's Theorem. Let $a, b \in \mathbb R$ and $a \lt b$. Let $f : [a,b] \to \mathbb R$ be a function satisfying the following three properties
- $f$ is continuous on the closed interval $[a,b]$,
- $f$ is differentiable on the open interval $(a,b)$,
- $f(a) = f(b)$.
To toggle the proof of the Rolle's Theorem clickTo be completed later .... -
Finally, a consequence of Rolle's Theorem is Mean Value Theorem.
Mean Value Theorem. Let $a, b \in \mathbb R$ and $a \lt b$. Let $f : [a,b] \to \mathbb R$ be a function satisfying the following two properties
- $f$ is continuous on the closed interval $[a,b]$,
- $f$ is differentiable on the open interval $(a,b)$.
To toggle the proof of the Mean Value Theorem clickTo be completed later .... - In class we proved all three: Fermat's Interior Extremum Theorem, Rolle's Theorem and the Mean Value Theorem.
-
Important consequences of the Mean Value Theorem are the following statements.
Constant Function Theorem. Let $a, b \in \mathbb R$, $a \lt b$. Let $f : (a,b) \to \mathbb R$ be a function. If $f$ is differentiable on $(a,b)$ and $f'(x) = 0$ for all $x \in (a,b)$, then there exists $K\in \mathbb R$ such that $f(x) = K$ for all $x \in (a,b)$.
- First. If $f'(x) \gt 0$ for all $x \in (a,b)$, then $f$ is increasing on $(a,b)$.
- Second. If $f'(x) \lt 0$ for all $x \in (a,b)$, then $f$ is decreasing on $(a,b)$.
- Third. If $f'(x) \geq 0$ for all $x \in (a,b)$, then $f$ is nondecreasing on $(a,b)$.
- Fourth. If $f'(x) \leq 0$ for all $x \in (a,b)$, then $f$ is nonincreasing on $(a,b)$.
Racetrack Principle. Let $a, b \in \mathbb R$ and $a \lt b$. Let $f : (a,b) \to \mathbb R$ and $g : (a,b) \to \mathbb R$ be functions which satisfy the following three conditions:- $f$ and $g$ are continuous on the closed interval $[a,b],$
- $f$ and $g$ are differentiable on the open interval $(a,b),$
- $f'(x) \leq g'(x)$ for all $x \in (a,b).$
- If $f(a) = g(a)$, then $f(x) \leq g(x)$ for all $x \in (a,b)$.
- If $f(b) = g(b)$, then $f(x) \geq g(x)$ for all $x \in (a,b)$.
- The Monotonicity Tests are powerful tools to identify bijections. For example, consider the function $f(x) = x \exp(x)$ defined for all $x \in \mathbb R$. It follows from the First Monotonicity Test and the Mean Value Theorem that the function \[ g: [-1.+\infty) \to [-1/e,+\infty), \qquad g(x) = x \exp(x) \] is an increasing bijection. The inverse of this function is called the Lambert $W_0$ function. It follows from the Second Monotonicity Test and the Mean Value Theorem that the function \[ h: (-\infty, -1] \to [-1/e,0), \qquad h(x) = x \exp(x) \] is a decreasing bijection. The inverse of this function is called the Lambert $W_{-1}$ function.
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Here are the plots of two branches of the Lambert W function. The Lambert $W$ functions are plotted in maroon and their inverses are plotted in
plotted below in navy blue. The formulas for maroon functions are given below graphs.
$ X = [-1,+\infty), \ Y = [-1/e,+\infty), \ W_0 = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
The Lambert $W_0$ function
$X = (-\infty,-1), \ Y = (-1/e,0), \ W_{-1} = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\}$
The Lambert $W_{-1}$ function
- Mathematica's notation for $W_0(x)$ function is ProductLog[x], or, equivalently ProductLog[0,x]. Mathematica's notation for $W_{-1}(x)$ function is ProductLog[-1,x].
- The Wikipidia's page on Lambert W function has a lot of information about this function. However, this page deals with the complex $W$ function so it might be difficult for you to extract useful information.
-
Below is the animation for Problem 2 on Assignment 3
Place the cursor over the image to start the animation.
-
One application of the problem that I did in class today is the following animation. Here the unit circle is bouncing in the parabola $y=x^2.$
Place the cursor over the image to start the animation.
-
It is a little bit more difficult to find a unit circle as in the figure below.
the navy blue function is $e^x$, the purple circle is a unit circle
Solution.Denote by $A$ the red point with the coordinates $\bigl(x_0, \exp(x_0) \bigr)$. The dark green line is the tangent line to navy blue graph of $y=e^x$ at the red point $\bigl(x_0, \exp(x_0) \bigr)$. Its equation is \[ y = (x-x_0) \exp(x_0) + \exp(x_0). \] The teal line is the line which is orthogonal to the dark green line at the red point $\bigl(x_0, \exp(x_0) \bigr)$. Its equation is \[ y = -(x-x_0) \exp(-x_0) + \exp(x_0). \] Note- Since the slope of the teal line is $-\exp(-x_0)$ and since the length of the line segment $\overline{AB}$ is $\exp(x_0)$, we conclude that the length of the line segment $\overline{BC}$ is $\exp(2x_0)$.
- Since the slope of the teal line is $-\exp(-x_0)$ and since the length of the line segment $\overline{DE}$ is $1$, we conclude that the length of the line segment $\overline{EC}$ is $\exp(x_0)$.
- Since the coordinates of the point $B$ are $(x_0,0)$, the previous two items yield that the coordinates of the point $C$ are $\bigl(x_0+\exp(2x_0), 0\bigr)$ and the coordinates of the point $E$ are $\bigl(x_0+\exp(2x_0)-\exp(x_0), 0\bigr)$.
- Consequently, the coordinates of the purple center $D$ are \[ \bigl(x_0+\exp(2x_0)-\exp(x_0),1\bigr). \]
- Since the distance between the purple center $D$ and the red point $A$ is $1$ we must have \[ \bigl(\exp(2x_0)-\exp(x_0)\bigr)^2 +\bigl(\exp(x_0) - 1\bigr)^2 = 1. \] Solving this equation for $x_0$ will solve the problem.
- Simplifying the last equation yields \[ \exp(4 x_0)-2 \exp(3 x_0) + \exp(x_0) + \exp(2 x_0) - 2 \exp(x_0) = 0. \] Dividing by $\exp(x_0)$ we get \[ e^{3x_0} - 2 e^{2x_0} + 2 e^{x_0} - 2 = 0. \] Substituting $z = \exp(x_0)$ yields \[ z^3 - 2 z^2 + 2 z -2 = 0. \] This cubic equation has only one real root which is between $1$ and $2$. Mathematica calculates that root to be \[ \frac{1}{3} \left(2-\frac{2}{\sqrt[3]{17+3 \sqrt{33}}}+\sqrt[3]{17+3 \sqrt{33}}\right). \] Hence for the value of $x_0$ we have \[ x_0 = \ln \left(\frac{1}{3} \left(2-\frac{2}{\sqrt[3]{17+3 \sqrt{33}}}+\sqrt[3]{17+3 \sqrt{33}}\right) \right) \approx 0.434175014665. \] With this $x_0$ we can calculate the approximate coordinates of the purple center $D$ to be \[ \bigl( 1.27346177 , 1 \bigr) \]
-
Here we will prove that a function $f$ which is differentiable at a point $a$ satisfies the Lipschitz condition for continuity at $a$.
Theorem. Let $D \subset \mathbb R$, let $f : D \to \mathbb R$ be a function and let $a \in D$. If $f$ is differentiable at $a$, then there exist $d \gt 0$ and $K \gt 0$ such that $(a-d,a+d) \subseteq D$ and for every $x \in (a-d,a+d)$ we have \[ \bigl| f(x) - f(a) \bigr| \leq K\, |x-a|. \]
Before proving this theorem we prove the following implication involving the absolute value function. For every $u, v, c \in \mathbb R$ we have
\[
|u-v| \lt c \qquad \Rightarrow \qquad |u| \lt |v| + c.
\]
To toggle the proof of this fact click
Let $u, v, c \in \mathbb R$ be arbitrary and assume the
\[
|u-v| \lt c.
\]
Based on "Mt Vernon-B'ham-Blaine" principle, we have
\[
v -c \lt u \lt v+ c.
\]
Since $-|v| \leq v$ and $v \leq |v|$ the last displayed inequality implies
\[
-(|v|+c) = -|v| -c \lt u \lt |v|+ c.
\]
Again, based on "Mt Vernon-B'ham-Blaine" principle, where B'ham is $0$, we have
\[
|u| \lt |v|+c.
\]
Let $D \subseteq \mathbb R$, let $f : D \to \mathbb R$ be a function and let $a \in D$. Assume that $f$ is differentiable at $a$. That is, assume that
\[
\lim_{x\to a} \frac{f(x) - f(a)}{x - a} = f'(a).
\]
By the definition of limit this means that there exists $\delta_0 \gt 0$ such that $(a-\delta_0, a) \cup (a, a+\delta_0) \subseteq D$ and for every $\epsilon \gt 0$ there exists $\delta(\epsilon) \in \mathbb R$ such that
\[
0 \lt \delta(\epsilon) \leq \delta_0 \qquad \text{and} \qquad 0 \lt |x-a| \lt \delta(\epsilon) \quad \Rightarrow \quad \left| \frac{f(x) - f(a)}{x - a} - f'(a) \right| \lt \epsilon.
\]
The preceding statement is true for all $\epsilon \gt 0$. Since $1 \gt 0$, it is true for $\epsilon = 1.$ Thus, there exists $\delta(1) \in \mathbb R$ such that
\[
0 \lt \delta(1) \leq \delta_0 \qquad \text{and} \qquad 0 \lt |x-a| \lt \delta(1) \quad \Rightarrow \quad \left| \frac{f(x) - f(a)}{x - a} - f'(a) \right| \lt 1.
\]
Set $d = \delta(1)$. Then clearly $d \gt 0$. Since $d \leq \delta_0$ we have $(a-d,a)\cup(a,a+d) \subseteq D$. By the assumption $a \in D$. Therefore, $(a-d,a+d) \subseteq D$.
Since we set $d = \delta(1)$ we have
\[
0 \lt |x-a| \lt d \quad \Rightarrow \quad \left| \frac{f(x) - f(a)}{x - a} - f'(a) \right| \lt 1.
\]
The implication involving the absolute value function proved just before this theorem yields
\[
\left| \frac{f(x) - f(a)}{x - a} - f'(a) \right| \lt 1 \quad \Rightarrow \quad \left| \frac{f(x) - f(a)}{x - a} \right| \lt | f'(a) |+1.
\]
Further, multiplying by $|x-a|$ we get
\[
\left| \frac{f(x) - f(a)}{x - a} \right| \lt | f'(a) |+1 \quad \Rightarrow \quad
\left| {f(x) - f(a)} \right| \lt \bigl( | f'(a) |+1 \bigr) \, |x-a|
\]
Putting the preceding three implications together we get
\[
0 \lt |x-a| \lt d \quad \Rightarrow \quad \left| {f(x) - f(a)} \right| \lt \bigl( | f'(a) |+1 \bigr) \, |x-a|.
\]
Since the last implication is also true for $x=a$ we have
\[
|x-a| \lt d \quad \Rightarrow \quad \bigl| {f(x) - f(a)} \bigr| \lt \bigl( | f'(a) |+1 \bigr) \, |x-a|.
\]
Since $|x-a| \lt d$ is equivalent to $x \in (a-d,a+d)$, we have proved that for every $x \in (a-d,a+d)$ we have
\[
\bigl| {f(x) - f(a)} \bigr| \lt \bigl( | f'(a) |+1 \bigr) \, |x-a|.
\]
Thus, we can take $K = | f'(a) |+1$.
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It is interesting to plot a function with a thousand of its tangent lines. Below is the sine function.
Place the cursor over the image to start the animation.
the sine function in navy blue and its many tangents in gray
- Here is a list of topics that will be covered on the exam on Wednesday.
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Today we did intuitive and formal approach to derivatives. Here is a formal definition.
Definition. Let $D$ be a subset of $\mathbb R$, let $a \in D$ and let $f: D \to \mathbb R$ be a function. We say that the function $f$ is differentiable at $a$ if the following limit exists: \[ \lim_{x\to a} \frac{f(x) - f(a)}{x - a}. \] If $f$ is differentiable at $a$ we define the derivative of $f$ at $a,$ denoted by $f'(a),$ to be \[ f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}. \] A function is said to be differentiable if it is differentiable at each point of its domain.
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Next we went to calculate derivatives of some famous functions.
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Let $f(x) = x^2.$ Then, $f$ is differentiable at every $a \in \mathbb R$ and for every $a \in \mathbb R$ we have $f'(a) = 2a.$To toggle the proof clickLet $a \in \mathbb R$ be arbitrary. We need to prove that \[ \lim_{x\to a} \frac{x^2 - a^2}{x - a} = 2a. \] First, for an arbitrary $x \neq a$ we simplify the expression \[ \left| \frac{x^2 - a^2}{x - a} - 2a \right| = \left| {x + a} - 2a \right| = \left| x - a \right|. \] Let $\epsilon \gt 0$ be arbitrary. The preceding equality yields that the following implication is true: \[ 0 \lt |x-a| \lt \epsilon \qquad \Rightarrow \qquad \left| \frac{x^2 - a^2}{x - a} - 2a \right| \lt \epsilon. \] Thus we can take $\delta(\epsilon) = \epsilon$. This proves that \[ 2a = \lim_{x\to a} \frac{x^2 - a^2}{x - a}. \] Thus we proved that $f'(a) = 2a.$
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More generally we have:
Let $n \in \mathbb N$ and let $f(x) = x^n$ with $x \in \mathbb R.$ Then $f$ is differentiable at every $a \in \mathbb R$ and we have $f'(a) = n\, a^{n-1}.$
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Let $f(x) = 1/x$ with $x \gt 0.$ Then $f$ is differentiable at every $a \gt 0$ and we have $f'(a) = - 1/(a^2).$To toggle the proof clickLet $a \gt 0$ be arbitrary. We need to prove that \[ \lim_{x\to a} \frac{\frac{1}{x} - \frac{1}{a}}{x - a} = -\frac{1}{a^2}. \] As before, when we deal with limits it is helpful to specify the domain for the values of $x,$ that is the role of $\delta_0 \gt 0.$ Set $\delta_0 = a/2$ and let $x \in (a/2,a) \cup (a,3a/2).$ Next, we simplify the expression \begin{align*} \left| \frac{\frac{1}{x} - \frac{1}{a}}{x - a} - \left( - \frac{1}{a^2} \right) \right| & = \left| \frac{ \frac{a - x}{xa} }{x - a} - \left( - \frac{1}{a^2} \right) \right| \\ & = \left| -\frac{1}{xa} + \frac{1}{a^2} \right| \\ & = \frac{1}{a} \, \left| -\frac{1}{x} + \frac{1}{a} \right| \\ & = \frac{1}{a} \, \left| \frac{x - a}{xa} \right| \\ & = \frac{1}{a} \, \frac{|x - a|}{x \, a } \\ & \leq \frac{1}{a} \, \frac{|x - a|}{ (a/2) \,a } \\ & = \frac{2}{a^3} \, |x - a| \end{align*} Hence we proved that for an arbitrary $a \gt 0$ and for all $x \in (a/2,a) \cup (a,3a/2)$ we have \[ \left| \frac{\frac{1}{x} - \frac{1}{a}}{x - a} - \left( - \frac{1}{a^2} \right) \right| \leq \frac{2}{a^3} |x - a|. \] This is what we used to call Fact. Let $\epsilon \gt 0$ be arbitrary. The preceding inequality yields that the following implication is true: \[ 0 \lt |x-a| \lt \min\left\{ \frac{a}{2}, \frac{a^3}{2} \epsilon \right\} \qquad \Rightarrow \qquad \left| \frac{\frac{1}{x} - \frac{1}{a}}{x - a} - \left( - \frac{1}{a^2} \right) \right| \lt \epsilon. \] Thus we can take $\delta(\epsilon) = \min\left\{ \frac{a}{2}, \frac{a^3}{2} \epsilon \right\}.$ This proves that \[ -\frac{1}{a^2} = \lim_{x\to a} \frac{\frac{1}{x} - \frac{1}{a}}{x - a}. \] Thus we proved that $f'(a) = -\frac{1}{a^2}.$
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More generally we have:
Let $n \in \mathbb N$ and let $f(x) = x^{-n}$ with $x \in \mathbb R\setminus\{0\}.$ Then $f$ is differentiable at every $a \in \mathbb R\setminus\{0\}$ and we have $f'(a) = -n\, a^{-n-1}.$
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Let $f(x) = \sqrt{x}$ with $x \gt 0.$ Then, $f$ is differentiable at every $a \gt 0$ and we have $f'(a) = \frac{1}{2\sqrt{a}}.$To toggle the proof clickLet $a \gt 0$ be arbitrary. We need to prove that \[ \lim_{x\to a} \frac{\sqrt{x} - \sqrt{a}}{x - a} = \frac{1}{2\,\sqrt{a}}. \] Next, with $x \gt 0$ such that $x \neq a$ we simplify the expression \begin{align*} \left| \frac{\sqrt{x} - \sqrt{a}}{x - a} - \frac{1}{2\,\sqrt{a}} \right| & = \left| \frac{1}{\sqrt{x} + \sqrt{a} } - \frac{1}{2\,\sqrt{a}} \right| \\ & = \left| \frac{2\,\sqrt{a} - \sqrt{x} - \sqrt{a} }{2\,\sqrt{a} \bigl( \sqrt{x} + \sqrt{a} \bigr) } \right| \\ & = \left| \frac{ \sqrt{a} - \sqrt{x} }{2\,\sqrt{a} \bigl( \sqrt{x} + \sqrt{a} \bigr) } \right| \\ & = \left| \frac{ a - x }{2\,\sqrt{a} \bigl( \sqrt{x} + \sqrt{a} \bigr)^2 } \right| \\ & = \frac{ |x - a| }{2\,\sqrt{a} \bigl( \sqrt{x} + \sqrt{a} \bigr)^2 } \\ & \leq \frac{ 1 }{2\, a \sqrt{a} } \, | x - a | \end{align*} Hence we proved that for an arbitrary $a \gt 0$ and for all $x \gt 0$ such that $x \neq a$ we have \[ \left| \frac{\sqrt{x} - \sqrt{a}}{x - a} - \frac{1}{2\,\sqrt{a}} \right| \leq \frac{ 1 }{2\, a \sqrt{a} } \, | x - a |. \] This is what we used to call Fact. Let $\epsilon \gt 0$ be arbitrary. The preceding inequality yields that the following implication is true: \[ 0 \lt |x-a| \lt 2\, a \sqrt{a} \, \epsilon \qquad \Rightarrow \qquad \left| \frac{\frac{1}{x} - \frac{1}{a}}{x - a} - \left( - \frac{1}{a^2} \right) \right| \lt \epsilon. \] Thus we can take $\delta(\epsilon) = 2\, a \sqrt{a} \, \epsilon.$ This proves that \[ \frac{1}{2\,\sqrt{a}} = \lim_{x\to a} \frac{\sqrt{x} - \sqrt{a}}{x - a}. \] Thus we proved that $f'(a) = \frac{1}{2\,\sqrt{a}}$
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More generally we have:
Let $\beta \in (0,+\infty)$ and let $f(x) = x^{\beta}$ with $x \gt 0.$ Then $f$ is differentiable at every $a \gt 0$ and we have $f'(a) = \beta \, a^{-\beta-1}.$
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Let $f(x) = \sin(x)$ with $x \in \mathbb R.$ Then, $f$ is differentiable at every $a \in \mathbb R$ and we have $f'(a) = \cos(a).$
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Let $f(x) = \cos(x)$ with $x \in \mathbb R.$ Then, $f$ is differentiable at every $a \in \mathbb R$ and we have $f'(a) = -\sin(a).$
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Let $f(x) = \exp(x)$ with $x \in \mathbb R.$ Then, $f$ is differentiable at every $a \in \mathbb R$ and we have $f'(a) = \exp(a).$
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Let $f(x) = \ln(x)$ with $x \in (0,+\infty).$ Then, $f$ is differentiable at every $a \in (0,+\infty)$ and we have $f'(a) = 1/a.$
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Let $f(x) = \arccos(x)$ with $x \in [-1,1].$ Then, $f$ is differentiable at every $a \in (-1,1)$ and we have $f'(a) = -\frac{1}{\sqrt{1-x^2}}.$
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Let $f(x) = \arcsin(x)$ with $x \in [-1,1].$ Then, $f$ is differentiable at every $a \in (-1,1)$ and we have $f'(a) = \frac{1}{\sqrt{1-x^2}}.$
-
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Why are functions familiar from precalculus continuous? A simple answer to this question is: The Lipschitz condition for continuity applies to most of them. I will just list the relevant inequalities without proving them.
- For every $a\in \mathbb R$ and for every $x\in \mathbb R$ we have $|\sin(x) - \sin(a)| \leq |x-a|.$
- For every $a\in \mathbb R$ and for every $x\in \mathbb R$ we have $|\cos(x) - \cos(a)| \leq |x-a|.$
- For every $a\gt 0$ and for every $\displaystyle x\in \left(\frac{a}{2},\frac{3a}{2}\right)$ we have $\displaystyle |\ln(x) - \ln(a)| \leq \frac{2}{a}\, |x-a|.$
- For every $a\in \mathbb R$ and for every $\displaystyle x \in \left(a - \frac{1}{2},a+ \frac{1}{2}\right)$ we have $\displaystyle \left| e^x - e^a \right| \leq 2e^{a}\, |x-a|.$
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Another powerful tool for proving continuity of elementary functions (this is a synonym for functions learned in precalculus) is the theorem establishing the algebra of continuous functions and the theorem proving that a composition of continuous functions is continuous.
Algebra of continuous functions. Let $f$ and $g$ be real functions and let $a$ be a real number. Assume that $f$ and $g$ are continuous at the point $a$. The following statements hold.
- If $h = f + g$, then $h$ is continuous at $a$.
- If $h = f g$, then $h$ is continuous at $a$.
- If $\displaystyle h = \dfrac{f}{g}$ and $g(a) \neq 0$, then $h$ is continuous at $a$.
Continuity of composition. Let $f$ and $g$ be real functions and let $a$ be a real number. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f\circ g$ is continuous at $a.$
- Intuitive approach to the continuity of functions is presented in Section 1.7 of the textbook. Recommended exercises are 1, 2, 4, 7, 8, 9, 11-14, 15 and problems 20, 22, 23, 28, 29, 35.
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Example 2. Consider the function $f(x) = \sqrt{x}$ defined for all $x \in [0,+\infty).$ Prove that this function is continuous.Proof. I will first prove that $f$ is continuous at the point $a = 0$. I will just do the proof of the implication in the definition; the motivation was explained in class. Let $\epsilon \gt 0$ be arbitrary. Set $\delta(\epsilon) = \epsilon^2$. Then clearly $\delta(\epsilon) \gt 0.$ Now, for every $x \in [0,+\infty)$ we need to prove the implication
$|x - 0| \lt \epsilon^2$ implies $|\sqrt{x} - 0 | \lt \epsilon$.
Now I will prove that the function $f$ is continuous at each point $a$ such that $a \gt 0$. For that we need the following fact.Fact B. For all $a \gt 0$ and all $x \geq 0$ we have $\displaystyle \biggl|\sqrt{x} -\sqrt{a} \biggr| \leq \frac{1}{\sqrt{a}} |x-a|$.
To toggle the proof of Fact B clickLet $a \gt 0$ and $x \geq 0$ be arbitrary. Then \begin{align*} \displaystyle \biggl|\sqrt{x} -\sqrt{a} \biggr| & = \biggl| (\sqrt{x} -\sqrt{a})\, 1 \biggr| \\ & = \left| (\sqrt{x} -\sqrt{a})\, \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}} \right| \\ & = \left| \frac{x - a}{\sqrt{x} + \sqrt{a}} \right| \\ & = \frac{1}{\sqrt{x} + \sqrt{a}} \left| x - a \right| \\ & \leq \frac{1}{\sqrt{a}} \left| x - a \right| \\ \end{align*} The inequality above is a consequence of the "pizza-party" principle and the fact that $x \geq 0.$Let $a\gt 0$ be arbitrary. To prove that $f(x) = \sqrt{x}$ is continuous at $a$ for every $\epsilon \gt 0$ we need to find $\delta(\epsilon) \gt 0$ such that for every $x \in [0,+\infty)$ we can prove the implication$|x - a| \lt \delta(\epsilon)$ implies $|\sqrt{x} - \sqrt{a} | \lt \epsilon$.
$|x - a| \lt \sqrt{a} \, \epsilon$ implies $|\sqrt{x} - \sqrt{a} | \lt \epsilon$.
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While proving limits and continuity we have always worked with some related Facts. What we have done can be formulated in two theorems, one for continuity and one for limits.
Lipschitz condition for continuity. Let $D \subset \mathbb R$ and let $f : D \to \mathbb R$ be a function. Let $a \in D$. If there exist $d \gt 0$ and $K \gt 0$ such that $(a-d,a+d) \subseteq D$ and for every $x \in (a-d,a+d)$ we have \[ \bigl| f(x) - f(a) \bigr| \leq K\, |x-a|, \] then $f$ is continuous at $a$.
To toggle the proof of Lipschitz condition for continuity click
Let $D \subset \mathbb R$ and let $f : D \to \mathbb R$ be a function. Let $a \in D$. Assume that there exist $d \gt 0$ and $K\gt 0$ such that $(a-d,a+d) \subseteq D$ and for every $x \in (a-d,a+d)$ we have
\[
\bigl| f(x) - f(a) \bigr| \leq K\, |x-a|.
\]
Next I will prove that $f$ is continuous at $a.$ Let $\epsilon \gt 0$ be arbitrary. Set
\[
\delta(\epsilon) = \min\left\{ d, \frac{\epsilon}{K} \right\}.
\]
Now, I need to prove the implication
\[
|x-a| \lt \min\left\{ d, \frac{\epsilon}{K} \right\} \qquad \Rightarrow \qquad \bigl| f(x) - f(a) \bigr| \lt \epsilon.
\]
Here is a proof. Assume that $|x-a| \lt \min\left\{ d, \frac{\epsilon}{K} \right\}$. Then $|x-a| \lt d$ and $|x-a| \lt {\epsilon}/{K}.$ Consequently, $x \in (a-d,a+d)$ and $K\,|x-a| \lt {\epsilon}.$ Since $x \in (a-d,a+d)$, by the assumption we have $\bigl| f(x) - f(a) \bigr| \leq K\, |x-a|.$ Since both $\bigl| f(x) - f(a) \bigr| \leq K\, |x-a|$ and $K\,|x-a| \lt {\epsilon}$ are true, by transitivity of order, we have $\bigl| f(x) - f(a) \bigr| \lt \epsilon.$ This proves the continuity of $f$ at $a.$
Lipschitz condition for the existence of limit.
Let $D \subset \mathbb R$ and let $f : D \to \mathbb R$ be a function. Let $a, L \in \mathbb R$. If there exist $d \gt 0$ and $K \gt 0$ such that $(a-d,a)\cup (a,a+d) \subseteq D$ and for every $x \in (a-d,a)\cup (a,a+d)$ we have
\[
\bigl| f(x) - L \bigr| \leq K\, |x-a|,
\]
then $\displaystyle \lim_{x\to a} f(x) = L$.
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Two most important theorems about continuous functions are the Intermediate Value Theorem and the Extreme Values Theorem.
Intermediate Value Theorem. Let $a, b \in \mathbb R$ and $a \lt b$. If $f : [a,b] \to \mathbb R$ is a continuous function on the closed interval $[a,b]$, then
for every real number $y$ between $f(a)$ and $f(b)$ there exists $x \in [a,b]$ such that $y = f(x)$.
Extreme Values Theorem. Let $a, b \in \mathbb R$ and $a \lt b$. If $f : [a,b] \to \mathbb R$ is a continuous function on the closed interval $[a,b]$, thenthere exist $c, d \in [a,b]$ such that $f(c) \leq f(x) \leq f(d)$ for all $x \in [a,b]$.
- As we discussed in class, to get a complete justification for the answer for (b) in Problem 4 on Assignment 1 one needed to use the Intermediate value theorem. Interestingly, we did not need the Extreme values theorem to rigorously answer (c) in the same problem.
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Which kind of Fact is useful if we are given a function $f$ and we want to prove that it is continuous at a point $a$? It is useful to prove an inequality of the form
If $|x-a| \lt d$, then $|f(x) - f(a) | \lt K |x-a|$.
-
Example 1. Consider the function $f(x) = 1/x$ defined for all $x \in (0,+\infty).$ Prove that this function is continuous.
Fact A. Let $a\in\mathbb R$ be such that $a \gt 0$. If $|x-a| \lt a/2$, then $\displaystyle \biggl|\frac{1}{x} -\frac{1}{a} \biggr| \lt \frac{2}{a^2} |x-a|$.
To toggle the proof of Fact A clickAssume that $|x-a| \lt a/2$. That is assume that $a/2 \lt x \lt(3a)/2$. Then \[ \displaystyle \biggl|\frac{1}{x} -\frac{1}{a} \biggr| = \frac{|x-a|}{x\, a} \leq \frac{1}{(a/2)\,a} |x-a| = \frac{2}{a^2} |x-a|. \] The inequality above is a consequence of the "pizza-party" principle and the fact that $x \gt a/2$.To prove that $f(x) = 1/x$ defined for all $x \in (0,+\infty)$ is continuous let $a$ be an arbitrary positive real number. Let $\epsilon \gt 0$ also be arbitrary. To find $\delta(\epsilon) \gt 0$ from the definition of continuity we solve \[ \frac{2}{a^2} |x-a| \lt \epsilon \] for $|x-a|$. We get \[ |x-a| \lt \frac{a^2}{2} \epsilon. \] To assure that Fact A is valid we need \[ |x-a| \lt \frac{a}{2}. \] Now we set \[ \delta(\epsilon) = \min\left\{ \frac{a}{2}, \frac{a^2}{2} \epsilon \right\}. \] With this $\delta(\epsilon)$ we can prove$ \displaystyle |x-a| \lt \min\left\{ \frac{a}{2}, \frac{a^2}{2} \epsilon \right\}$ implies $\displaystyle \biggl|\frac{1}{x} -\frac{1}{a} \biggr| \lt \epsilon$.
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Below is a formal $\epsilon$-$\delta$-definition of continuity of a function. This is one of the most important definitions in mathematics.
Definition. Let $D$ be a subset of $\mathbb R$, let $a \in D$ and let $f: D \to \mathbb R$ be a function. We say that the function $f$ is continuous at $a$ if the following condition is satisfied:
-
For every $\epsilon \gt 0$ there exists $\delta(\epsilon) \gt 0$ such that for all $x \in D$ we have
$|x-a| \lt \delta(\epsilon)$ implies $|f(x) - f(a) | \lt \epsilon$.
-
For every $\epsilon \gt 0$ there exists $\delta(\epsilon) \gt 0$ such that for all $x \in D$ we have
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Today I got an excellent office hour question: What makes a Fact useful when we work with limits? Here is an answer. When we work with
\[
\displaystyle \lim_{x\to a} f(x) = L
\]
it is useful to prove an inequality of the form
If $0 \lt |x-a| \lt d$, then $|f(x) - L | \lt K |x-a|$.
-
The remark in the previous item applies to working with continuity. When we are trying to prove that a function is continuous at a point $a$ it is useful to prove an inequality of the form
If $0 \lt |x-a| \lt d$, then $|f(x) - f(a) | \lt K |x-a|$.
- Based on our work with limits you should be able to prove that the functions $f(x) = x^2,$ $f(x) = 1/x,$ $f(x) = \sqrt{x},$ are continuous on their natural domains.
- In class I created this Mathematica file related to the function that is on Assignment 2. The file is called 20161007.nb. The file is created in Mathematica version 8.
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Example 2. Prove that $\displaystyle \lim_{x\to 0} \cos x = 1$.How does this fit into the definition of limit? Here $D = \mathbb R$, $f(x) = \cos(x)$, $a=0$ and $L = 1$.To prove the given limit, in the definition of limit we can take $\delta_0 = \pi/3,$ since clearly $\cos(x)$ is defined for all $x \in (-\pi/3,0)\cup(0,\pi/3).$Our next objective is to discover $\delta(\epsilon).$ Let $\epsilon \gt 0$ be arbitrary. We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $|\cos(x) - 1 | \lt \epsilon$.
To discover $\delta(\epsilon)$ we need the following fact.Fact B. If $|x| \lt \pi/3$, then $|\cos(x) - 1| \leq |x|$.
To toggle the proof of Fact B clickFirst assume that $0 \lt x \lt \pi/3$. Then we can relay on the figure below. This figure shows a part of the unit circle. The number $x$ is represented by the teal circular arc length between points $A$ and $C$. Then, by the definition of the cosine function the blue length of the line segment $OB$ equals $\cos x$. The triangle $ABC$ is a right triangle whose one leg is the line segment $BC$ and whose hypotenuse is $AC$. Therefore the length $BC$, which equals $1-\cos x$, is less than the length of the line segment $AC$. Further the length of the line segment $AC$ is less than arc length of the teal circular arc between $A$ and $C$. Since the arc length of the teal circular arc equals $x$, this proves $1 - \cos x \lt x$. Similarly, if $-\pi/3 \lt x \lt 0$, we have $1 - \cos x \lt -x$. Thus, in either case $1 - \cos x \lt |x|$. Since $\cos(0) =1$ and $|\cos x - 1| = 1-\cos x$ this proves Fact B.
Figure for $\displaystyle \lim_{x\to 0} \cos(x) = 1$
Our goal is to solve have $| \cos(x) - 1 | \lt \epsilon$ for $|x|.$ This is somewhat complicated. However, Fact B yields that the solutions of $|x| \lt \epsilon$ will also be solutions of $| \cos(x) - 1 | \lt \epsilon.$ But $|x| \lt \epsilon$ is already solved for $|x|.$ Therefore Fact B yields that $|x| \lt \epsilon$ implies $| \cos(x) - 1 | \lt \epsilon.$ But, to use Fact B we need $|x| \lt \pi/3.$ Therefore we set \[ \delta(\epsilon) = \min\{\epsilon,\pi/3\}. \] This $\delta(\epsilon)$ has the following properties:$0 \lt \delta(\epsilon) \leq \pi/3$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $|\cos(x) - 1 | \lt \epsilon$.
Why? Since both $\epsilon$ and $\pi/3$ are positive we have $\delta(\epsilon) \gt 0$. Since $\min\{\epsilon,\pi/3\} \leq \pi/3$ we have $\delta(\epsilon) \leq \pi/3$. Thus $0 \lt \delta(\epsilon) \leq \pi/3$.To prove the implication \[ 0 \lt |x| \lt \delta(\epsilon) \quad \Rightarrow \quad |\cos(x) - 1 | \lt \epsilon \] assume that $0 \lt |x| \lt \min\{\epsilon,\pi/3\}$. Then $|x| \lt \pi/3$. By Fact B this implies $|\cos(x) - 1| \leq |x|.$ The assumption $|x| \lt \min\{\epsilon,\pi/3\}$ also implies $|x| \lt \epsilon.$ Since by Fact B $|\cos(x) - 1| \leq |x|$, we thus have $|\cos(x) - 1| \lt \epsilon.$ -
Example 3. Prove that $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1$.How does this fit into the definition of limit? Here $D = {\mathbb R}\setminus\{0\}$, $\displaystyle f(x) = \frac{\sin x}{x}$, $a=0$ and $L = 1$.To prove the given limit, in the definition of limit we can take $\delta_0 = \pi/3,$ since clearly $\displaystyle \frac{\sin x}{x}$ is defined for all $x \in (-\pi/3,0)\cup(0,\pi/3).$Our next objective is to discover $\delta(\epsilon).$ Let $\epsilon \gt 0$ be arbitrary. We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \left| \frac{\sin x}{x} - 1 \right| \lt \epsilon$.
To discover $\delta(\epsilon)$ we need the following fact.Fact C. If $0 \lt |x| \lt \pi/3$, then $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt |x|$.
To toggle the proof of Fact C clickThe proof of Fact C is slightly more complicated than the proof of Fact B. We first proveFact C-1. If $0 \lt x \lt \pi/3$, then $\displaystyle \cos(x) \lt \frac{\sin x}{x} \lt 1$.
To prove Fact C-1 assume that $0 \lt x \lt \pi/3$ and rely on the figure below. This figure shows a part of the unit circle. The number $x$ is represented by the teal circular arc length between points $A$ and $C$. By the definition of the cosine function the length of the navy blue line segment $OB$ equals $\cos x$ and by the definition of the sine function the green length of the line segment $AB$ equals $\sin x$. Now compare the areas of the triangle $OCA$, the yellow area of the circular sector identified by points $OCA$ and the area of the triangle $OCD$. Clearly these three areas are listed in the increasing order. By calculating each of the three areas we conclude \[ \frac{1}{2} \sin x \lt \frac{1}{2} x \lt \frac{1}{2} \tan x. \] Dividing the preceding inequalities by $\dfrac{1}{2} \sin x \gt 0$ and taking reciprocals yields \[ \cos x \lt \frac{\sin x}{x} \lt 1. \] This proves Fact C-1.To prove Fact C assume again that $0 \lt x \lt \pi/3$. By Fact B we have that $1- \cos(x) \lt |x|,$ that is $1- |x| \lt \cos(x).$ Together with Fact C-1 this yields \[ 1-|x| \lt \frac{\sin x}{x} \lt 1. \] Since all the functions in the preceding inequality are even, the inequality also holds for negative numbers $x \in (-\pi/3,0)$. Now subtracting $1$ in each term of $1-|x| \lt (\sin x)/x \lt 1$ yields $-|x| \lt (\sin x)/x -1 \lt 0,$ and multiplying by $-1$ finally gives $0 \lt 1- (\sin x)/x \lt |x|$. This proves Fact C.
Figure for $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1$
Our goal is to solve have $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt \epsilon$ for $|x|.$ For this we can use Fact C and repeat the reasoning from Example 2 to conclude that in this case we can also set \[ \delta(\epsilon) = \min\{\epsilon,\pi/3\}. \] This $\delta(\epsilon)$ has the following properties:$0 \lt \delta(\epsilon) \leq \pi/3$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt \epsilon$.
Why? The reasoning is identical to the reasoning presented in Example 2. -
Example 4. Prove that $\displaystyle \lim_{x\to 0} \frac{1-\cos(x)}{x^2} = \frac{1}{2}$.How does this fit into the definition of limit? Here $D = {\mathbb R}\setminus\{0\}$, $\displaystyle f(x) = \frac{1-\cos(x)}{x^2}$, $a=0$ and $L = 1/2.$To prove the given limit, in the definition of limit we can take $\delta_0 = 1,$ since clearly $\displaystyle \frac{1-\cos(x)}{x^2}$ is defined for all $x \in (-1,0)\cup(0,1).$Our next objective is to discover $\delta(\epsilon).$ Let $\epsilon \gt 0$ be arbitrary. We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \left| \frac{1-\cos(x)}{x^2} - \frac{1}{2} \right| \lt \epsilon$.
To discover $\delta(\epsilon)$ we need the following fact.Fact D. If $0 \lt |x| \lt 1$, then $\displaystyle \biggl|\frac{1-\cos(x)}{x^2} - \frac{1}{2} \biggr| \lt |x|$.
To toggle the proof of Fact D clickThe proof of Fact D is slightly more complicated than the proof of Fact C. As in the proofs of Fact B and Fact C we will use a geometric argument deduced from the unit circle. In the figure below we consider $x \in (0,1)$. Further we consider the lengths of the line segments $\overline{AB}$, $\overline{AC}$ and the length of the circular arc from $A$ to $C$. We see that these three lengths are listed in increasing order and they calculate to be: \[ \tag{#} \sin x \lt \sqrt{2 - 2 \cos x} \lt x. \] Here, the length of $\overline{AC}$ is calculated by the Pythagorean theorem \[ \sqrt{(\sin x)^2 +(1- \cos x)^2} = \sqrt{(\sin x)^2 +1- 2 \cos x +(\cos x)^2} = \sqrt{2 - 2 \cos x} \]
Squaring the inequalities in (#) we get \[ (\sin x)^2 \lt 2 - 2 \cos x \lt x^2. \] Now, dividing the preceding inequalities by the positive number $2 x^2$ we get \[ \tag{*} \frac{1}{2} \left(\frac{\sin x}{x}\right)^2 \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}. \] In the proof of Fact C we proved that \[ 1 - |x| \lt \frac{\sin x}{x}. \] Since $x \in (0,1)$ both quantities in the preceding inequality are positive. Therefore we can square both sides of the inequality and divide by $2$ to get \[ \frac{1}{2} \left(1-|x|\right)^2 \lt \frac{1}{2} \left(\frac{\sin x}{x}\right)^2. \] Together with inequalities (*), the preceding inequality yields \[ \frac{1}{2} \left(1-|x|\right)^2 \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}. \] Since \[ \frac{1}{2} - |x| \lt \frac{1}{2} \left(1-|x|\right)^2, \] we conclude \[ \frac{1}{2} - |x| \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}. \] In our reasoning so far we assumed that $x \in (0,1).$ But, since all the functions in the preceding expression are even, the inequalities also hold for $x \in (-1,0)$ as well. That is, for all $x \in (-1,0) \cup (0,1)$ we have \[ \frac{1}{2} - |x| \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}. \] Using the last two inequalities, for all $x \in (-1,0) \cup (0,1)$ we deduce \begin{align*} \left| \frac{1-\cos(x)}{x^2} - \frac{1}{2} \right| & = \frac{1}{2} - \frac{1 - \cos x}{x^2} \\ & \lt \frac{1}{2} - \left( \frac{1}{2} - |x| \right) \\ & = |x|. \end{align*} This proves Fact D.
Figure for $\displaystyle \lim_{x\to 0} \frac{1- \cos x}{x^2} = \frac{1}{2}$
Our goal is to solve have $\displaystyle \biggl| \frac{1- \cos(x)}{x^2} -\frac{1}{2} \biggr| \lt \epsilon$ for $|x|.$ For this we can use Fact D and repeat the reasoning from Example 2 to conclude that in this case we can also set \[ \delta(\epsilon) = \min\{\epsilon,1\}. \] This $\delta(\epsilon)$ has the following properties:$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{1- \cos(x)}{x^2} - \frac{1}{2} \biggr| \lt \epsilon$.
Why? The reasoning is identical to the reasoning presented in Example 2. -
Example 5. Prove that $\displaystyle \lim_{x\to 0} \frac{1-\cos(x)}{x} = 0$.How does this fit into the definition of limit? Here $D = {\mathbb R}\setminus\{0\}$, $\displaystyle f(x) = \frac{1-\cos(x)}{x}$, $a=0$ and $L = 0.$To prove the given limit, in the definition of limit we can take $\delta_0 = 1,$ since clearly $\displaystyle \frac{1-\cos(x)}{x}$ is defined for all $x \in (-1,0)\cup(0,1).$Our next objective is to discover $\delta(\epsilon).$ Let $\epsilon \gt 0$ be arbitrary. We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \left| \frac{1-\cos(x)}{x} - 0 \right| \lt \epsilon$.
To discover $\delta(\epsilon)$ we need the following fact.Fact E. If $0 \lt |x| \lt 1$, then $\displaystyle \biggl|\frac{1-\cos(x)}{x} \biggr| \lt \frac{1}{2} |x|$.
To toggle the proof of Fact E clickThe proof of Fact E uses an inequality that was established in the proof of Fact D. Let $0 \lt |x| \lt 1$, that is let $x \in (-1,0)\cup(0,1)$. In the proof of Fact D we proved that \[ \frac{1-\cos(x)}{x^2} \lt \frac{1}{2}. \] Clearly \[ 0 \lt \frac{1-\cos(x)}{x^2}, \] we have \[ 0 \lt \frac{1-\cos(x)}{x^2} \lt \frac{1}{2}. \] Multiplying the preceding expression by $|x| \gt 0$ we get \[ 0 \lt \frac{1-\cos(x)}{|x|} \lt \frac{1}{2}|x|. \] Since \[ \biggl|\frac{1-\cos(x)}{x} \biggr| = \frac{1-\cos(x)}{|x|}, \] this proves Fact E.Our goal is to solve have $\displaystyle \biggl| \frac{1- \cos(x)}{x} -0 \biggr| \lt \epsilon$ for $|x|.$ For this we can use Fact E and discover that we need to set \[ \delta(\epsilon) = \min\{2\epsilon,1\}. \] This $\delta(\epsilon)$ has the following properties:$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{1- \cos(x)}{x} - 0 \biggr| \lt \epsilon$.
Why? The reasoning is quite similar to the reasoning presented in Example 2. You just need to use Fact E instead of Fact B.
-
Here is my definition of limit at $a$.
- I. There exists $\delta_0 \gt 0$ such that $f(x)$ is defined for every $x \in (a-\delta_0,a)\cup(a,a+\delta_0)$.
- II.
For every $\epsilon \gt 0$ there exists $\delta(\epsilon)$ such that $0 \lt \delta(\epsilon) \leq \delta_0$ and
$0 \lt |x-a| \lt \delta(\epsilon)$ implies $|f(x) - L | \lt \epsilon$.
- In the Calculus textbook read Sections 1.7 and 1.8. Do the problems: 1.7 19, 20, 21, 22, 23, 24, 30, 1.8 1, 7, 8, 10, 14, 20, 46, 47, 57.
- I will illustrate the definition of limit with four examples, one today and three tomorrow. Below is the first example which involves only algebra. Tomorrow I will do three examples which involve geometry, or rather, trigonometry.
-
Example 1. Prove that $\displaystyle \lim_{x\to 2} x^2 = 4$.How does this fit into the definition of limit? Here $D = \mathbb R$, $f(x) = x^2$, $a=2$ and $L = 4$.To prove the given limit, in the definition of limit we can take $\delta_0 = 1,$ since clearly $f(x)$ is defined for all $x \in (1,2)\cup(2,3).$Our next objective is to discover $\delta(\epsilon).$ Let $\epsilon \gt 0$ be arbitrary. We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x-2| \lt \delta(\epsilon)$ implies $|x^2 - 4 | \lt \epsilon$.
To discover $\delta(\epsilon)$ we need the following fact.Fact A. If $|x-2| \lt 1$, then $|x^2 - 4| \leq 5 |x - 2|$.
To toggle the proof of Fact A clickAssume that $|x-2| \lt 1.$ Then $1 \lt x \lt 3.$ Consequently, $3 \lt x+2 \lt 5.$ Since $x+2 \gt 0,$ we have $|x+2| = x+2$ and hence $|x+2| \lt 5.$ Next recall the difference of two squares formula and rules for the absolute value function to simplify \[ | x^2 - 4 | = |(x+2)(x-2)| = |x+2|\, |x-2|. \] Since $|x+2| \lt 5$ the last identity yields $|x^2 - 4| \leq 5 |x - 2|.$ This proves Fact A.Our goal is to solve have $| x^2 - 4 | \lt \epsilon$ for $|x-2|.$ This is somewhat complicated. However, Fact A yields that the solutions of $5 |x-2| \lt \epsilon$ will also be solutions of $| x^2 - 4 | \lt \epsilon.$ Solving $5 |x-2| \lt \epsilon$ for $|x-2|$ is easy. The solution is $|x-2| \lt \epsilon/5.$ Therefore we can choose \[ \delta(\epsilon) = \min\{\epsilon/5,1\} = \begin{cases} 1 & \text{if} \ \ \epsilon \geq 5 \\ \epsilon/5 & \text{if} \ \ \epsilon \lt 5 \end{cases}. \] This $\delta(\epsilon)$ has the following properties:$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x-2| \lt \delta(\epsilon)$ implies $|x^2 - 4 | \lt \epsilon$.
Why? Since both $\epsilon/5$ and $1$ are positive we have $\delta(\epsilon) \gt 0$. Since $\min\{\epsilon/5,1\} \leq 1$ we have $\delta(\epsilon) \leq 1$. Thus $0 \lt \delta(\epsilon) \leq 1$.To prove the implication \[ 0 \lt |x-2| \lt \delta(\epsilon) \quad \Rightarrow \quad |x^2 - 4 | \lt \epsilon \] assume that $0 \lt |x-2| \lt \min\{\epsilon/5,1\}$. Then $|x-2| \lt 1$. By Fact A this implies $|x^2 - 4| \leq 5 |x - 2|.$ The assumption $|x-2| \lt \min\{\epsilon/5,1\}$ also implies $|x-2| \lt \epsilon/5$ and therefore $5 |x-2| \lt \epsilon.$ Since we already proved $|x^2 - 4| \leq 5 |x - 2|$, we thus have $|x^2 - 4 | \lt \epsilon.$
-
First I state Problem 0 for Assignment 2 that I just hended out in class. In this problem I ask you to explore the inverse functions related to the hyperbolic sine and the hyperbolic cosine. Notice that in the post on Sunday, October 2, 2016 I solved the analogous problem for the hyperbolic tangent.
Problem 0. In this problem we study the inverse functions related to hyperbolic sine $\sinh$ and the hyperbolic cosine $\cosh$.
- I. Prove that the function $\sinh: \mathbb R \to \mathbb R$ is a bijection and for an arbitrary $x\in\mathbb R$ find an expression for $\sinh^{-1}(x)$ in terms of known functions.
- II. Define the function ${\operatorname{arccosh}}$, a natural bijection related to $\cosh: \mathbb R\to \mathbb R$. Provide all necessary proofs.
-
Now I will use the definition of the limit at infinity to prove
\[
\lim_{x\to+\infty} \frac{x}{x+\sin(x)} = 1.
\]
Clearly the function $x/\bigl(x+\sin(x)\bigr)$ is defined for all $x \geq 2$. So, we can take $X_0 = 2$ in the definition of limit.
Let $\epsilon \gt 0$. We need to solve \[ \left| \frac{x}{x+\sin(x)} - 1 \right| \lt \epsilon \] to find $X(\epsilon)$. First simplify (here we assume that $x \geq 2$) \[ \left| \frac{x}{x+\sin(x)} - 1 \right| = \left| \frac{x-x-\sin(x)}{x+\sin(x)} \right| = \frac{|-\sin(x)|}{x+\sin(x)} \] Still, solving \[ \frac{|-\sin(x)|}{x+\sin(x)}\lt \epsilon \] is not possible. Therefore we replace $\frac{|-\sin(x)|}{x+\sin(x)}$ with something slightly bigger and solvable: \[ \frac{|-\sin(x)|}{x+\sin(x)} \leq \frac{1}{x-1} \lt \epsilon \] Now, solving \[ \frac{1}{x-1} \lt \epsilon \] for $x \geq 2$ will lead to $X(\epsilon)$: Multiply be $x-1 \gt 0$ and divide by $\epsilon \gt 0$ to get \[ \frac{1}{\epsilon} \lt x - 1. \] Thus, the solution is \[ x \gt \frac{1}{\epsilon} + 1. \] Since we need $X(\epsilon) \geq 2$, we will define \[ X(\epsilon) = \max\left\{\frac{1}{\epsilon} + 1,2 \right\}. \]
-
Here is my definition of limit at infinity.
- I. There exists $X_0 \in \mathbb R$ such that $f(x)$ is defined for every $x \geq X_0$.
- II.
For every $\epsilon \gt 0$ there exists $X(\epsilon) \in \mathbb R $ such that $X(\epsilon) \geq X_0$ and
$x \gt X(\epsilon)$ implies $|f(x) - L | \lt \epsilon$.
- I mentioned in class that I have seen an engineering paper in which the author states $\tanh(8) = 1.$ But, we proved that for all real numbers $x$ we have $-1 \lt \tanh(x) \lt 1.$ In some sense the engineer was correct if an error of $10^{-6}$ was tolerable. This is since Mathematica gives the approximation for $\tanh(8)$ to $7$ decimal places to be $0.9999998$. Therefore \[ 0.99999974 \lt \tanh(8) \lt 0.99999985 \] and consequently \[ 1- \tanh(8) \lt 1- 0.99999985 = 1.5\times 10^{-7} \lt 10^{-6}. \] Since the hyperbolic tangent is an increasing function (see below) we also have that \[ 1- \tanh(x) \lt 10^{-6} \] for all $x \gt 8$.
- Now we could ask whether a more accurate engineer could also write $\tanh(x) = 1$ for some $x$? This bring us close to the definition of limit. Roughly speaking the definition of limit tells that no matter how accurate an engineer (this is measured by $\epsilon \gt 0$) there exist real numbers $x$ (this is determined by $X(\epsilon)$ in the definition) for which that engineer could write $\tanh(x) = 1.$
- Here I prove that $\tanh$ is an increasing function. Let $x_1, x_2$ be real numbers such that $x_1 \lt x_2$. To prove that $\tanh$ is increasing I will simplify the expression $\tanh(x_2) - \tanh(x_1):$ \begin{align*} \tanh(x_2) - \tanh(x_1) & = \frac{e^{x_2} - e^{-x_2}}{e^{x_2} + e^{-x_2}} - \frac{e^{x_1} - e^{-x_1}}{e^{x_1} + e^{-x_1}} \\ & = \frac{e^{2x_2} - 1}{e^{2x_2} + 1} - \frac{e^{2x_1} - 1}{e^{2 x_1} + 1} \\ & = \frac{\bigl(e^{2x_2} - 1\bigr)\bigl(e^{2 x_1} + 1\bigr) - \bigl(e^{2x_1} - 1\bigr)\bigl(e^{2 x_2} + 1\bigr)}{\bigl(e^{2x_2} + 1\bigr)\bigl(e^{2 x_1} + 1\bigr)} \\ & = 2 \frac{e^{2x_2} - e^{2x_1}}{\bigl(e^{2x_2} + 1\bigr)\bigl(e^{2 x_1} + 1\bigr)}. \end{align*} Since the exponential function is increasing and $2x_2 \gt 2x_1$, the last expression is positive. Hence, $\tanh$ is an increasing function.
-
Since $\tanh$ is an increasing function it is a bijection. Let us calculate its inverse $\tanh^{-1}$. This function is sometimes denoted arctanh. Let $y \in (-1,1)$ and let us solve $\tanh(x) = y$. That is solve
\[
\frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} = y.
\]
Multiplying both sides by $e^{x} + e^{-x} \gt 0$ we get
\[
e^{x} - e^{-x} = y \bigl( e^{x} + e^{-x} \bigr).
\]
Multiplying by $e^{-x} \gt 0$ we get
\[
e^{2x} - 1 = y \bigl( e^{2x} + 1 \bigr)
\]
and simplifying yields
\[
e^{2x} \bigl(1-y\bigr) = y + 1.
\]
Separating $x$ and $y$ in the last equality yields
\[
e^{2x} = \frac{1+y}{1-y}.
\]
Since $y \in (-1,1)$ we have $0 \lt 1-y \lt 2$, $0 \lt 1+y \lt 2$ and $\bigl(1+y\bigr)/\bigl(1-y\bigr) \gt 0$.
Hence
\[
x = \frac{1}{2}\, \ln \frac{1+y}{1-y}.
\]
This proves
\[
\tanh^{-1}(x) = \frac{1}{2}\, \ln \frac{1+x}{1-x} \quad \text{for all} \quad x \in (-1,1).
\]
$\tanh$ in navy blue and its inverse $\tanh^{-1}$ in maroon
-
Now I will use the definition of limit to prove that
\[
\lim_{x\to+\infty} \tanh(x) = 1.
\]
Let $\epsilon \gt 0$ be arbitrary. If $\epsilon \geq 2$, then for all $x \in \mathbb R$ we have
\[
|\tanh(x) - 1| = 1 - \tanh(x) \lt 2 \leq \epsilon.
\]
Therefore we can take $X(\epsilon) = 1$. Let $\epsilon \in (0,2)$. Then $1-\epsilon \in (-1,1)$. We need to solve
\[
|\tanh(x) - 1| = 1 - \tanh(x) \lt \epsilon.
\]
Simplifying the last inequality is equivalent to
\[
1-\epsilon \lt \tanh(x).
\]
Since $\tanh^{-1}$ is also an increasing function, the last inequality is equivalent to
\[
\tanh^{-1}\bigl(1-\epsilon\bigr) \lt x.
\]
Since
\[
\tanh^{-1}\bigl(1-\epsilon\bigr) = \frac{1}{2}\, \ln \frac{2 - \epsilon}{\epsilon},
\]
we have proved that
\[
x \gt \frac{1}{2}\, \ln \frac{2 - \epsilon}{\epsilon} \qquad \text{implies} \qquad |\tanh(x) - 1| \lt \epsilon.
\]
This shows that for a given $\epsilon \gt 0$ we can take $X(\epsilon) = 1$ if $\epsilon \geq 2$ and $X(\epsilon) = \frac{1}{2}\, \ln \bigl(\bigl(2 - \epsilon\bigr)/\epsilon\bigr)$ if $\epsilon \lt 2$.
Let us test this conclusion on $\epsilon = 10^{-6}$. Then, approximated to 10 decimal places \[ X(10^{-6}) = \frac{1}{2}\, \ln \frac{2 - 10^{-6}}{10^{-6}} \approx 7.2543286193. \] Hence \[ x \gt X(10^{-6}) \approx 7.2543286193 \qquad \text{implies} \qquad 1 - \tanh(x) \lt 10^{-6}. \] Since $8 \gt X(10^{-6})$ the last implication justifies sloppy engineer's (one who tolerates $10^{-6}$ error) writing $\tanh(8) = 1$.
- Below is a formal definition of a composition of two functions.
Place the cursor over the image to start the animation.
The composition at a generic point $x$
The composition at another generic point $x$
Finding a minimum of the composition
Finding a maximum of the composition
- Today we discussed Problem 4 on Assignment 1. The conclusion of the discussion was that proving the following implication would be very helpful: Let $a, b \in \mathbb R$. Prove the following implication: \[ -1 \lt a \lt 1 \ \quad \text{and} \ \quad -1 \leq b \leq 1 \ \qquad \text{imply} \ \qquad 0 \lt 1 + ab \lt 2 \ \quad \text{and} \ \quad -1 \leq \frac{a+b}{1+ab} \leq 1. \] I presented a proof of this inequality in class. Please understand my proof, try to improve it and present your proof neatly as a part of your work on the assignment.
-
We also discussed hyperbolic functions. I proved that for all $x \in \mathbb R$ we have $-1 \lt \tanh(x) \lt 1$. I showed that $\tanh(x)$ is very, very close to $1$ for a large positive real number $x$. I did the explorations in Mathematica version 8. In class I created
this Mathematica file. The file is called 20160928.nb. Please open this file in Mathematica and start learning how to use this amazing tool.
- First download the file to your computer. Right-click on this underlined link. In the pop-up menu that appears your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 8. You will find Mathematica 8 on all computers in BH 215. Open Mathematica first, then open 20160928.nb from Mathematica.
- More information on how to use Mathematica version 8 you can find on my Mathematica 8 page.
-
Today we discussed six famous bijections that I posted yesterday. In fact, I should have started the discussion with linear functions.
-
Below I plot several linear bijections and their inverses. As before, the functions are in navy blue and their inverses are plotted in maroon. For each example you can guess what the values of $m$ and $b$ are used.
- Since you inspired me to talk about sine with yarn today, I wanted to make a record of that concept. I would like to make an animation in Mathematica inspired by this. Stay tuned.
Sine with yarn
-
The most important hyperbolic functions are Sinh, Cosh and Tanh. They are defined as follows:
$\color{blue}{\cosh(x)} = \color{green}{\dfrac{1}{2} e^x} + \color{red}{\dfrac{1}{2} e^{-x}}, \ x \in \mathbb R $
$ \color{blue}{\sinh(x)} = \color{green}{\dfrac{1}{2} e^x} \color{red}{- \dfrac{1}{2} e^{-x}}, \ x \in \mathbb R $
$ \tanh(x) = \dfrac{\sinh(x)}{\cosh(x)} = \dfrac{e^x - e^{-x}}{e^x + e^{-x}}, \ x \in \mathbb R $
-
There are several reasons why hyperbolic functions are important. One reason is that they appear naturally in important applications. Since we can not go into these applications in this class, I should offer a simpler reasons. Here are some.
- From your mathematical experience in high school you are aware that the exponential function $f(x) = \exp(x)$ is important. One could argue that a drawback to the exponential function is that it is neither even, not odd function. So, one could ask: Does there exist an even function, call it $e(x)$, and an odd function, call it $o(x)$, such that \[ \exp(x) = e(x) + o(x) \quad \text{for all} \quad x \in \mathbb R. \] The answer to this question is yes and we can calculate these two functions. Since the equality that we want holds for all $x \in \mathbb R$ we can replace $x$ by $-x$ and still the equality will hold: \[ \exp(-x) = e(-x) + o(-x) \quad \text{for all} \quad x \in \mathbb R. \] But, not remember that $e$ is an even function, so $e(-x) = e(x)$ and $o$ is an odd function, so $o(-x) = -o(x)$. Hence, the last equality can be rewritten as \[ \exp(-x) = e(x) - o(x) \quad \text{for all} \quad x \in \mathbb R. \] Now adding the first equality and the last equality we get \[ \exp(x) + \exp(-x) = 2 e(x) \quad \text{for all} \quad x \in \mathbb R. \] Hence $e(x) = \bigl(\exp(x) + \exp(-x)\bigr)/2 = \cosh(x)$. Similarly, subtracting the equality $\exp(-x) = e(-x) + o(-x)$ from the equality $\exp(x) = e(x) + o(x)$ we get \[ \exp(x) - \exp(-x) = 2 o(x) \quad \text{for all} \quad x \in \mathbb R. \] Hence $o(x) = \bigl(\exp(x) - \exp(-x)\bigr)/2 = \sinh(x)$. Thus, hyperbolic cosine and hyperbolic sine form a unique pair of an even and an odd function whose sum is the most important function: the exponential function.
-
Why the names hyperbolic cosine and hyperbolic sine? Calculate the expression
\[
\bigl(\cosh(x)\bigr)^2 - \bigl(\sinh(x)\bigr)^2.
\]
You will see that
\[
\bigl(\cosh(x)\bigr)^2 - \bigl(\sinh(x)\bigr)^2 = 1 \quad \text{for all} \quad x\in \mathbb R.
\]
Now, I am not sure whether you learned in high school that the set
\[
\bigl\{ (x,y) \in \mathbb R\!\times \!\mathbb R \, \bigl| \bigr. \, x^2 - y^2 = 1\bigr\}
\]
is called hyperbola. Here is its graph
Thus, by setting \[ \bigl\{ (\cosh(t), \sinh(t)) \in \mathbb R\!\times \!\mathbb R \, \bigl| \bigr. \ t \in \mathbb R\bigr\} \] we get the right branch of the above hyperbola and by setting \[ \bigl\{ (-\cosh(t), \sinh(t)) \in \mathbb R\!\times \!\mathbb R \, \bigl| \bigr. \ t \in \mathbb R\bigr\} \] we get the left branch of the above hyperbola. This is analogous to the fact that the set \[ \bigl\{ (\cos(t), \sin(t)) \in \mathbb R\!\times \!\mathbb R \, \bigl| \bigr. \ t \in [0, 2 \pi) \bigr\} \] equals to the unit circle, since \[ \bigl(\cos(x)\bigr)^2 + \bigl(\sin(x)\bigr)^2 = 1. \]
Hyperbola $\bigl\{ (x,y) \in \mathbb R\!\times \!\mathbb R \, \bigl| \bigr. \, x^2 - y^2 = 1\bigr\}$
-
Consider a pair of functions $f : \mathbb R \to \mathbb R$ and $g : \mathbb R \to \mathbb R$ and ask the following question:
Change the question slightly.
- It is important to point out that there are identities for hyperbolic functions which are quite similar to identities that you learned for trigonometric functions. For example, for all $x,y \in \mathbb R$ we have \begin{align*} \sinh(x + y) &= \sinh (x) \cosh (y) + \cosh (x) \sinh (y) \\ \cosh(x + y) &= \cosh (x) \cosh (y) + \sinh (x) \sinh (y) \\ \tanh(x + y) &= \frac{\tanh(x) +\tanh(y)}{1+ \tanh(x) \tanh(y) } \\ \end{align*} Many more properties of the hyperbolic functions you can find at the Wikipedia page Hyperbolic functions.
- I found an old book The Application of Hyperbolic Functions to Electrical Engineering Problems by 'Arthur Edwin Kennell. It is now a free ebook.
- I am pleased that several of you have decided to look into learning LaTeX. The best way to learn LaTeX is to look at some sample documents. Here is a simple LaTeX sample document in which I prove an interesting inequality. Here is the LaTeX file of the first assignment without pictures.
- I handed out Assignment 1 on Friday. It is due on October 3. To receive the full credit for a problem your solution needs to be well written. Guidelines for Good Mathematical Writing written by Professor Francis Su from Harvey Mudd College can help you improve your mathematical writing. Prof. Su's article was published in MAA FOCUS, Vol. 35. No. 4, 2015, pages 20-22. Googleing "Good Mathematical Writing" will lead to more interesting articles on this topic.
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Today we further reviewed the concepts: of a function, of an injection, of a surjection; see the definitions given on Friday, September 23. We informally proved the following theorem:
- Precalculus is full of important bijections. Below I list several bijections and you can figure out why they are important: \begin{alignat*}{2} X & = [0,+\infty), & \quad Y & = [0,+\infty), & \qquad f_1 & = \bigl\{ (x,x^2) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = \mathbb R, & \quad Y & = \mathbb R, & \qquad f_2 & = \bigl\{ (x,x^3) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = \mathbb R, & \quad Y & = (0,+\infty), & \qquad f_3 & = \bigl\{ (x,e^x) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = (-\pi/2,\pi/2), & \quad Y & = \mathbb R, & \qquad f_4 & = \bigl\{ \bigl( x,\tan(x) \bigr) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = [-\pi/2,\pi/2], & \quad Y & = [-1,1], & \qquad f_5 & = \bigl\{ \bigl(x,\sin(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = [0,\pi], & \quad Y & = [-1,1], & \qquad f_6 & = \bigl\{ \bigl(x,\cos(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} \\ \end{alignat*}
- These six bijections are plotted below in navy blue. Their inverses are plotted in maroon.
$ X = [0,+\infty), \ Y = [0,+\infty), \ f_1 = \bigl\{ (x,x^2) \ \bigl| \bigr. \ x \in X \bigr\} $
$X = \mathbb R, \ Y = \mathbb R, \ f_2 = \bigl\{ (x,x^3) \ \bigl| \bigr. \ x \in X \bigr\}$
$X= \mathbb R, \ Y = (0,+\infty), \ f_3 = \bigl\{ (x,e^x) \ \bigl| \bigr. \ x \in X \bigr\}$
$X = (-\pi/2,\pi/2), \ Y = \mathbb R, \ f_4 = \bigl\{ \bigl( x,\tan(x) \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
$ X = [-\pi/2,\pi/2], \ Y = [-1,1], \ f_5 = \bigl\{ \bigl(x,\sin(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
$ X = [0,\pi], \ Y = [-1,1], \ f_6 = \bigl\{ \bigl(x,\cos(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
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Here is a problem related to the concepts of a function and a bijection.
- Is the subset $F$ a function from ${\mathbb R}$ to ${\mathbb R}$? Give a detailed answer to this question. Does $F$ satisfy Fun 1? Does $F$ satisfy Fun 2?
- Find the largest subsets $X\subseteq {\mathbb R}$ and $Y\subseteq {\mathbb R}$ such that the subset \[ f = \bigl\{(x,y) \in F \, | \, x\in X, y \in Y\bigr\} \] is a function from $X$ to $Y$. Find all such sets $X$ and $Y$.
- Find the largest subsets $A\subseteq {\mathbb R}$ and $B\subseteq {\mathbb R}$ such that the subset \[ g = \bigl\{(x,y) \in F \, | \, x\in A, y \in B \bigr\} \] is a bijection from $A$ to $B$. Find all such sets $A$ and $B$.
The Unit Circle $\bigl\{ (x,y) \in {\mathbb R}\times {\mathbb R} \, | \, x^2+y^2=1 \bigr\}$
- Today we mentioned the floor, an important piecewise defined functions. Here are three more important piecewise defined functions.
The sign function
The unit step function
The floor function
The ceiling function
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We introduce one more set operation: Cartesian product of two sets.
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We discussed the formal definition of a function. Here is the definition from Wikipedia's function page:
- Fun 1. For every $x \in X$ there exists $y \in Y$ such that the pair $(x,y)$ belongs to $f$.
- Fun 2. If pairs $(x,y_1)$ and $(x,y_2)$ belong to $f$, then $y_1= y_2$.
- A function $f$ from $X$ to $Y$ is often denoted by $f: X \to Y$. If $f$ is a function then instead of $(x,y) \in f$ we write $y = f(x)$. The set $X$ is called the domain of $f$ and the set $Y$ is called the codomain of $f$. The set \[ \bigl\{ y \in Y \ \bigl| \bigr. \ y = f(x) \ \text{for some} \ x \in X \bigr\} \] is called the range of $f$.
- If $X$ and $Y$ are nonempty subsets of $\mathbb R$ we can rephrase the definition of a function as follows: A subset $f$ of the Cartesian product $\mathbb R\!\times\!\mathbb R$ is a function from $X$ to $Y$ if every vertical line that goes through a point in $X\!\times\!\{0\}$ crosses $f$ at exactly one point. This formulation is sometimes called the vertical line test.
- In precalculus and calculus classes functions are often defined by formulas and the sets $X$ and $Y$ are not specified explicitly. If this is the case, we consider that the domain of such a function is the largest subset of $\mathbb R$ on which the formula is defined. This domain is called the natural domain of a function. For example, the natural domain of the reciprocal function $r(x) = 1/x$ is the set ${\mathbb R}\setminus\{0\}$.
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We distinguish three important kinds of functions: injections, surjections and bijections.
Definition. A function $f$ from $X$ to $Y$ is an injection if $f(x_1) \neq f(x_2)$ whenever $x_1, x_2 \in X$ and $x_1 \neq x_2$.Definition. A function $f$ from $X$ to $Y$ is a surjection if for every $y \in Y$ there exists $x \in X$ such that $y = f(x)$. In other words, a function $f$ from $X$ to $Y$ is a surjection if the range of $f$ equals $Y$.Definition. A function $f$ from $X$ to $Y$ is a bijection if it is both an injection and a surjection.
- A synonym for "injection" is "one-to-one function". A synonym for "surjection" is "onto function". I strongly encourage you to ignore these synonyms.
- If $X$ and $Y$ are nonempty subsets of $\mathbb R$ we can rephrase the definition of an injection as follows: A function $f : X \to Y$ is an injection if every horizontal line that goes through a point in $\{0\}\!\times\!Y$ crosses $f$ at at most one point.
- If $X$ and $Y$ are nonempty subsets of $\mathbb R$ we can rephrase the definition of a surjection as follows: A function $f:X \to Y$ is a surjection if every horizontal line that goes through a point in $\{0\}\!\times\!Y$ crosses $f$ at at least one point.
- If $X$ and $Y$ are nonempty subsets of $\mathbb R$ we can rephrase the definition a bijection as follows: A function $f:X \to Y$ is a bijection if every horizontal line that goes through a point in $\{0\}\!\times\!Y$ crosses $f$ at exactly one point.
- The formulations in the preceding three bullets are sometimes called the horizontal line tests.
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Here are some important subsets of $\mathbb R$:
- The set of integers \[ \mathbb Z = \bigl\{\ldots, -3,-2,-1,0,1,2,3,\ldots \bigr\} \]
- The set of positive integers \[ \mathbb N = \bigl\{1,2,3,\ldots \bigr\} = \bigl\{ {x \in \mathbb R} \, \bigl| \bigr. \, {x \in \mathbb Z} \ \text{and} \ x \gt 0 \bigr\} \] Sometimes the term "natural number" is used as a synonym for positive integer. However, this is not universal. Sometimes the set of natural numbers includes the number $0$; see the Wikipedia entry for Natural numbers.
- the set of rational numbers \[ {\mathbb Q} = \Bigl\{x \in \mathbb R \, \bigl| \bigr. \, x = \frac{p}{q} \ \ \text{for some} \ \ p, q \in \mathbb Z \ \ \text{with} \ \ q \neq 0 \Bigr\}. \] The set of rational numbers could be equivalently defined by \[ {\mathbb Q} = \Bigl\{x \in \mathbb R \, \bigl| \bigr. \, x = \frac{p}{q} \ \ \text{for some} \ \ {p \in \mathbb Z} \ \ \text{and} \ \ q \in \mathbb N \Bigr\}. \]
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Further we list several kinds of intervals of real numbers. For $a,b \in \mathbb R$ with $a \lt b$, we have the following four kinds of bounded intervals:
- the open interval \[ (a,b) = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \lt x \ \text{and} \ x \lt b \bigr\}, \]
- the left-closed, right-open interval \[ [a,b) = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \leq x \ \text{and} \ x \lt b \bigr\}, \]
- the left-open, right-closed interval \[ (a,b] = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \lt x \ \text{and} \ x \leq b \bigr\}, \]
- the closed interval \[ [a,b] = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \leq x \ \text{and} \ x \leq b \bigr\}. \]
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For $a \in \mathbb R$ we have the following four kinds of unbounded intervals
- open, left-bounded, right-unbounded interval \[ (a,+\infty) = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \lt x \bigr\}, \]
- closed, left-bounded, right-unbounded interval \[ [a,+\infty) = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \leq x \bigr\}, \]
- open, left-unbounded, right-bounded interval \[ (-\infty,a) = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, x \lt a \bigr\}, \]
- closed, left-unbounded, right-bounded interval \[ (-\infty,a] = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, x \leq a \bigr\}. \]
- In the above definitions we used so called the set-builder notation. For example, the formula \[ \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \lt x \bigr\}, \] means that in this set we include all real numbers $x$ which are larger than a given real number $a$. The curly brackets announce that the formula specifies a set.
- Since in this class we use the set notation we reviewed some set terminology and notation. The Wikipedia entry Set (mathematics) is a good source. In particular review the following items:
- The information sheet
- Some points from today's class:
- The development of calculus rests on the properties of real numbers. Therefore it is useful to review the properties of real numbers. We will use the standard notation $\mathbb R$ for the set of real numbers.
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I got a very good question today: What is the definition of a real number?
- The above Wikipedia page provides several answers to this question.
- Traditionally this question is not discussed in calculus classes. It is assumed that students have sufficient intuitive understanding of the concept of a real number from their prior mathematical experience.
- In my Math 226 and Math 312 I define the set of real numbers $\mathbb R$ by listing its axioms. At the following link you can find sixteen axioms of $\mathbb R$. There are five axioms related to the addition of real numbers, five axioms related to the multiplication of real numbers, the distributive law which provides a connection between multiplication and addition, four axioms about the order among real numbers and the most subtle axiom - the Completeness Axiom.
- The power of axiomatic approach is that all statements about real numbers that are studied in math courses can be proved using these sixteen axioms.
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The only specific numbers that appear in the axioms of the set of real numbers are the number $1$ and the number $0$. In class I made an argument that the number $\pi$ and the number $e$ are the next two most important real numbers. You should be familiar with the geometric definition of $\pi$. In class I gave the "business algebra" definition of $e$:
The number $e$ is the value of the account in which 1 dollar has been invested for 1 year in a savings account paying 100% annual interest compounded continuously.
These four real numbers together with the imaginary unit $i$ form the set of numbers which I call "the Hall of Fame of Numbers": \[ \bigl\{ 1, 0, \pi, e, i \bigr\}. \]