- 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 $f$ is a differentiable function and $f(x) \geq 0$ 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. The animation below illustrates how revolution creates a vase.
Place the cursor over the image to start the animation.
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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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To get an approximation for the surface area we need to calculate the lateral surface area of a part of a cone between two parallel planes as in the picture below.
Thus the lateral surface area of this part of a cone equals
Lateral surface area of a part of a cone.
Unwrap the area, cut in pieces and rearrange. Hover the cursor to animate.
Easy area to calculate.
(1/2) (blue length + red length) * black length
- Let $y = f(x)$, $x \in [a,b],$ be a differentiable positive function with a continuous derivative. Let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ be a partition of the interval $[a,b]$. An approximation for the surface area formed as the graph of $f$ rotates about the $x$-axis is given by \[ \sum_{k=1}^n 2 \pi \frac{f(x_{k-1}) + f(x_k)}{2} \sqrt{(x_k-x_{k-1})^2 + \bigl(f(x_k) -f(x_{k-1})\bigr)^2 } \] In the last formula we have $n$ lateral surface areas of narrow parts of cones. In each of these cones the red length and the blue lengths are \[ 2 \pi f(x_{k-1}) \quad \text{and} \quad 2 \pi f(x_{k}), \] respectively, while the black length is \[ \sqrt{(x_k-x_{k-1})^2 + \bigl(f(x_k) -f(x_{k-1})\bigr)^2} = \sqrt{1 + \bigl( f'(t_k)^2 \bigr)} (x_k-x_{k-1}), \] for some $t_k \in (x_{k-1},x_k).$ Thus the above approximation can be written as \[ \sum_{k=1}^n 2 \pi \frac{f(x_{k-1}) + f(x_k)}{2} \sqrt{1 + \bigl( f'(t_k)^2 \bigr)} (x_k-x_{k-1}). \] The last sum is not a Riemann sum of an integrable function. However, when the mesh of our partition is very small we can use the following approximation \[ \frac{f(x_{k-1}) + f(x_k)}{2} \approx f(t_k). \] With this approximation the last sum becomes \[ \sum_{k=1}^n 2 \pi f(t_k) \sqrt{1 + \bigl( f'(t_k)^2 \bigr)} (x_k-x_{k-1}). \] which is a Riemann sum of the integral \[ \int_a^b 2\pi f(x) \sqrt{1+\bigl(f'(x)\bigr)^2} dx. \]
- Thus, 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 \[ \int_a^b 2\pi f(x) \sqrt{1+\bigl(f'(x)\bigr)^2} dx. \]
- Applying this formula to the surface area of the vase above is \[ \int_0^{2 \pi} 2\pi \bigl(2+\sin x \bigr) \sqrt{1+\bigl(\cos x\bigr)^2} dx = 16 \pi \int_0^{\pi/2} \sqrt{1+\bigl(\cos x\bigr)^2} dx = 16 \pi E(-1). \] Here $E(-1)$ is the value of the complete elliptic integral of the second kind at $-1.$ We have $E(-1) \approx 1.9101.$ So, the surface area 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.
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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
]
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- 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 \[ \int_0^h \frac{x^2}{h^2} {\operatorname{area}}({\rm Base}) dx = \frac{1}{h^2} {\operatorname{area}}({\rm Base}) \int_0^h x^2 dx = \frac{1}{3} {\operatorname{area}}({\rm Base}) \, h. \]
- I wrote a webpage about with all the necessary information about families of ellipses and Elliptic Integrals which should be helpful for the second part of Problem 5 on Assignment 5,
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This is the
Mathematica notebook that I created in class yesterday and today. We calculated three (families of) involutes in this file:
- involutes of the curve $\displaystyle y = \frac{1}{2} \left( 1 +3 \sqrt[3]{\frac{x^2}{2}} \right)$ (Since this is the evolute of the parabola $y=x^2$, we know that one of its involutes will be exactly this parabola.);
- involutes of the parabola $y=x^2$;
- involutes of the unit circle $(\cos t, \sin t),$ $t \geq 0.$
- This is the Mathematica notebook in which I explored a reconstruction of a curve from its evolute (which is a definition of an involute). This notebook is not particularly well organized but it has some interesting stuff in it. There is some repetitiveness. I sometimes copy and paste a section and then try to improve it. The most important content in this notebook is a proof that the derivative of the length of the evolute equals the derivative of the radius of the osculating circle. I will post this proof on this page later.
- I have written up a proof of two remarkable properties of evolutes here. I hope that this proof is simpler than one that I presented in class.
- Today we started with applications of definite integrals. 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 arc length of the curve \[ C = \bigl\{ \bigl(u(t), v(t) \bigr) \in {\mathbb R}\!\times\!{\mathbb R} \, | \, \alpha \leq t \leq \beta \bigr\} \] is given by \[ \int_\alpha^\beta \sqrt{\bigl(u'(t)\bigr)^2 + \bigl(v'(t)\bigr)^2} dt. \]
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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.
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The following six examples illustrate solids in Problem 2 on Assignment 5. They are 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.
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In this example a cross-section is an equilateral triangle.
Place the cursor over the image to start the animation.
A cross-section is an equilateral triangle.
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In this example a cross-section is a square.
Place the cursor over the image to start the animation.
A cross-section is a square.
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In this example a cross-section is a regular pentagon.
Place the cursor over the image to start the animation.
A cross-section is a regular pentagon.
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In this example a cross-section is a regular hexagon.
Place the cursor over the image to start the animation.
A cross-section is a regular hexagon.
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In this example a cross-section is a regular heptagon.
Place the cursor over the image to start the animation.
A cross-section is a regular heptagon.
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In this example a cross-section is a regular octagon.
Place the cursor over the image to start the animation.
A cross-section is a regular octagon.
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In this example a cross-section is an equilateral triangle.
- I mentioned in class that a guiding slogan for an integral calculus course could be "How long is a smile?" 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.
- 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.
Place the cursor over the image to start the animation.
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Fundamental Theorem of Calculus. Let $a,b \in \mathbb R$ be such that $a\lt b.$ Let $F: [a,b]\to \mathbb R$ and $f: [a,b]\to \mathbb R$ be functions with the following four properties:
- $f$ is Riemann integrable on $[a,b]$.
- $F$ is continuous on $[a,b]$
- $F$ is differentiable on $(a,b)$.
- $F'(x) = f(x)$ for all $x \in (a,b)$.
- We recall the definitions related to the Riemann Integral from yesterday and we introduce some new ones.
- 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 $f$ corresponding to the tagged partition $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ of $[a,b].$
-
There are five important Riemann sums. 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 \[ \operatorname{Le}(f,\mathcal P) = \sum_{j=1}^n f(x_{j-1}) \bigl(x_j - x_{j-1}\bigr) \] is called the left Riemann sum of $f$ corresponding to the partition $\mathcal P$;
- The sum \[ \operatorname{Ri}(f,\mathcal P) = \sum_{j=1}^n f(x_{j}) \bigl(x_j - x_{j-1}\bigr) \] is called the right Riemann sum of $f$ corresponding to the partition $\mathcal P$;
- The sum \[ \operatorname{Mi}(f,\mathcal P) = \sum_{j=1}^n f\left(\tfrac{x_{j-1}+x_j}{2}\right) \bigl(x_j - x_{j-1}\bigr) \] is called the middle Riemann sum of $f$ corresponding to the partition $\mathcal P$;
- For each $j \in \{1,\ldots,n\}$ denote by $m_j$ the greatest number such that $m_j \leq f(x)$ for every $x \in \bigl[x_{j-1},x_j \bigr]$. The sum \[ \operatorname{Lo}(f,\mathcal P) = \sum_{j=1}^n m_j \bigl(x_j - x_{j-1}\bigr) \] is called the lower Riemann sum of $f$ corresponding to the partition $\mathcal P$.
- For each $j \in \{1,\ldots,n\}$ denote by $M_j$ the least number such that $f(x) \leq M_j$ for every $x \in \bigl[x_{j-1},x_j \bigr].$ The sum \[ \operatorname{Up}(f,\mathcal P) = \sum_{j=1}^n M_j \bigl(x_j - x_{j-1}\bigr) \] is called the upper Riemann sum of $f$ corresponding to the partition $\mathcal P$.
-
The following inequalities relate the above introduced Riemann sums.
- For an arbitrary Rieman sum of a tagged partition ${\mathcal P} = (x_0,x_1,\ldots,x_n;t_1,\ldots,t_n)$ we have \[ \tag{LRU} \operatorname{Lo}(f,\mathcal P) \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \leq \operatorname{Up}(f,\mathcal P). \]
- A particular case of inequality (LRU) is \[ \tag{LMU} \operatorname{Lo}(f,\mathcal P) \leq \operatorname{Mi}(f,\mathcal P) \leq \operatorname{Up}(f,\mathcal P). \]
- 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 $C \in \mathbb R$ and let $a, b \in\mathbb R$ be such that $a \lt b.$. Prove that the constant function $f(x) = C$, $x \in \mathbb R$, is integrable on any interval $[a,b]$ and \[ \int_a^b C \, dx = C(b-a). \]
-
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 \[ \operatorname{Up}(f,\mathcal P) + \operatorname{Lo}(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j + x_{j-1} \bigr) \bigl(x_j - x_{j-1}\bigr) = \sum_{j=1}^n \bigl((x_j)^2 - (x_{j-1})^2 \bigr) = b^2 - a^2. \] Also \[ \operatorname{Up}(f,\mathcal P) - \operatorname{Lo}(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j - x_{j-1} \bigr) \bigl(x_j - x_{j-1}\bigr) \leq \| \mathcal P \| \sum_{j=1}^n \bigl( x_j - x_{j-1} \bigr) = (b - a) \|\mathcal P\|. \] Adding the equality \[ \operatorname{Up}(f,\mathcal P) + \operatorname{Lo}(f,\mathcal P) = b^2 - a^2. \] and the inequality \[ \operatorname{Up}(f,\mathcal P) - \operatorname{Lo}(f,\mathcal P) \leq (b - a) \|\mathcal P\|. \] we get \[ 2 \operatorname{Up}(f,\mathcal P) \leq b^2 - a^2 + (b - a) \|\mathcal P\| \] and, subtracting them we get \[ 2 \operatorname{Lo}(f,\mathcal P) \geq b^2 - a^2 - (b - a) \|\mathcal P\|. \] Together with inequality (LRU) we thus have \[ \frac{b^2 - a^2}{2} - \frac{b - a}{2} \|\mathcal P\| \leq \operatorname{Lo}(f,\mathcal P) \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \leq \operatorname{Up}(f,\mathcal P) \leq \frac{b^2 - a^2}{2} + \frac{b - a}{2} \|\mathcal P\|. \] 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 $\operatorname{Lo}(f,\mathcal P) = {\operatorname{Le}}(f,\mathcal P)$ and $\operatorname{Up}(f,\mathcal P) = {\operatorname{Ri}}(f,\mathcal P).$ Also the identity \[ \operatorname{Up}(f,\mathcal P) + \operatorname{Lo}(f,\mathcal P) = {\operatorname{Ri}}(f,\mathcal P) + {\operatorname{Le}}(f,\mathcal P) = b^2 - a^2 \] 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 $f(x) = m x + k$. - I did not did the proof below in class. In spirit, the proof below is similar to the proof above, just a little bit more involved algebraically.
-
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 real numbers $u$ 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
\[ \tag{1}
\operatorname{Lo}(f,\mathcal P) + \operatorname{Up}(f,\mathcal P) + 4 {\operatorname{Mi}}(f,\mathcal P) = 2 \bigl( b^3 - a^3 \bigr).
\]
Since $0 \leq a$ we have $x_j^2 - x_{j-1}^2 \gt 0$ for all $j \in \{1,\ldots,n\}$. Therefore,
\[
\operatorname{Up}(f,\mathcal P) - \operatorname{Lo}(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j^2 - x_{j-1}^2 \bigr) \bigl(x_j - x_{j-1}\bigr) \leq \| \mathcal P \| \sum_{j=1}^n \bigl( x_j^2 - x_{j-1}^2 \bigr) = (b^2 - a^2) \|\mathcal P\|.
\]
Thus,
\[ \tag{2}
\operatorname{Up}(f,\mathcal P) - \operatorname{Lo}(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|.
\]
Using inequality (LMU), (2) implies
\[ \tag{3}
\operatorname{Up}(f,\mathcal P) - {\operatorname{Mi}}(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|
\]
and
\[ \tag{4}
{\operatorname{Mi}}(f,\mathcal P) - \operatorname{Lo}(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|.
\]
Adding inequalities (1), (2) and (3) multiplied by $4$ and dividing the resulting inequality by 6 we get
\[
\operatorname{Up}(f,\mathcal P) \leq \frac{b^3 - a^3}{3} + \frac{5}{6}(b^2 - a^2) \|\mathcal P\|
\]
Adding inequalites (1), (2) multiplied by $-1$ and (3) multiplied by $-4$ and dividing the resulting inequality by 6 we get
\[
\operatorname{Lo}(f,\mathcal P) \geq \frac{b^3 - a^3}{3} - \frac{5}{6}(b^2 - a^2) \|\mathcal P\|.
\]
Together with inequality (LRU) we thus have
\begin{align*}
\frac{b^3 - a^3}{3} - \frac{5}{6}(b^2 - a^2) \|\mathcal P\|
& \leq \operatorname{Lo}(f,\mathcal P) \\
& \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \\
& \leq \operatorname{Up}(f,\mathcal P) \\
& \leq \frac{b^3 - a^3}{3} + \frac{5}{6}(b^2 - a^2) \|\mathcal P\|,
\end{align*}
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 $\operatorname{Lo}(f,\mathcal P) = {\operatorname{Le}}(f,\mathcal P)$ and $\operatorname{Up}(f,\mathcal P) = {\operatorname{Ri}}(f,\mathcal P)$. Also the identity \[ 2 \bigl( b^3 - a^3 \bigr) = \operatorname{Lo}(f,\mathcal P) + \operatorname{Up}(f,\mathcal P) + 4 {\operatorname{Mi}}(f,\mathcal P) = {\operatorname{Le}}(f,\mathcal P) + {\operatorname{Ri}}(f,\mathcal P) + 4 {\operatorname{Mi}}(f,\mathcal P) \] 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: \[ \int_a^b \! f(x) dx = \frac{1}{6} \operatorname{Le}(f,\mathcal P) + \frac{1}{6} {\operatorname{Ri}}(f,\mathcal P) + \frac{2}{3} {\operatorname{Mi}}(f,\mathcal P). \]
- 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 $a, b \in \mathbb R$ be such that $a \lt b$ 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 $[a,b]$.
- 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 $f$ corresponding to the tagged partition $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ of $[a,b]$.
- 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 \[ \int_a^b \!\! f(x) dx. \]
- As specified in the information sheet the next exam is on Tuesday, November 20. Ryan wrote a very good summary of what might appear on this exam.
-
Here are some of the notebooks that I created in class during last few weeks.
- Mathematica file with an example of Taylor polynomial approximations.
- Mathematica file with an example of an evolute.
- Mathematica file with the envelope of normals to cardioid.
- Mathematica file with several examples of envelopes of families of normals. I did not show this notebook in class.
- Mathematica file with several examples of circles in graphs. I did not show this notebook in class.
- We derived a formula for the derivative of an inverse function. Under the appropriate conditions on $f$ and $x$ the following formula holds: \[ \frac{d}{dx} f^{-1}(x) = \frac{1}{f'\bigl(f^{-1}(x)\bigr)} \]
-
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 $(3,3) \in F$ and $(0,0) \in F.$ 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 $y = f(x)$ such that $3 = f(3)$ and whose graph is a subset of $F$.
Another interesting question about $F$ is:- Find the continuous involution $y = g(x)$ such that $0 = g(0)$ 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 $x^3+y^3 = 6xy$
-
To determine the asymptote of the Folium of Descartes we can consider all the lines $y=b-x$ and we can determine 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 of intersection and for some $b$ there are two points of intersection of the line and the Folium. It turns 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.
-
There is another way of determining the asymptote of the Folium of Descartes. What distinguishes the asymptote is the fact that it is very close to the the graph of $y = g(x).$ As we know these points satisfy
\[
x^3 + y^3 - 6xy = 0.
\]
Let $m$ and $b$ be arbitrary real numbers. Consider the line $y = mx + b$ and determine $m$ and $b$ such that the quantity
\[
x^3 + (mx+b)^3 - 6x(mx+b)
\]
is for all $x$ as small as possible. We have
\[
x^3 + (mx+b)^3 - 6x(mx+b) = \left(m^3+1\right) x^3 + 3 m \left( b m - 2\right) x^2 + 3 b \left(b m - 2 \right) x + b^3.
\]
To make the last expression as small as possible for all $x$ we have to choose
\[
m = -1 \qquad \text{and} \qquad b = -2.
\]
We deduce that along the line $y = -x - 2$ the quantity $x^3 + y^3 - 6xy$ evaluates to $-8.$ This is almost enough to believe that this line is the asymptote. However, I will give some calculations that further substantiate this claim.
Let now $x = x_0$ be a fixed with large $|x_0|$. To be specific assume that $x_0 \gt 0.$ Consider the following function of $y$ \[ h(y) = y^3 - 6 x_0 y + x_0^3. \] Then \[ h'(y) =3 y^2 - 6 x_0. \] At the value $y = - x_0 -2$ we have \[ h(- x_0 -2) = -8 \qquad \text{and} \qquad h'(-x_0-2) =3 x_0^2 + 6 x_0 + 12. \] The linear approximation of the function $h(y)$ at the point $(-x_0-2,-8)$ is \[ \ell(y) = \bigl(3 x_0^2 + 6 x_0 + 12\bigr) \bigl(y - (-x_0-2\bigr) - 8. \] The $y$-intercept of the line $\ell(y)$ is \[ y_0 = -x_0 - 2 + \frac{8}{3x_0^2 + 6x_0 +12}. \] For this value of $y$ the function $h$ is close to $0$: \[ h(y_0) = - \left( \frac{4}{3} \right)^{\!\!3} \frac{64 + 72 x_0 + 36 x_0^2 + 9 x_0^3}{\bigl(4 + 2 x_0 + x_0^2\bigr)^3} = - \left( \frac{4}{3} \right)^{\!\!3} \frac{\frac{64}{x_0^3} + \frac{72}{x_0^2} + \frac{36}{x_0} + 9}{\bigl(\frac{4}{x_0} + 2 + x_0 \bigr)^3} \lt 0 \] Thus, for large $x_0$ the value of $h(y_0)$ is almost $0.$ Therefore the point $(x_0, y_0)$ is very close to the point $\bigl(x_0,g(x_0)\bigr)$ which is on the Folium of Decartes. Since the point $(x_0, -x_0-2)$ is very close to the point $(x_0, y_0)$ we conclude that the point $(x_0, -x_0-2)$ is very close to $\bigl(x_0,g(x_0)\bigr)$ which is on the Folium of Decartes.
With a little bit of effort we could give the upper bound for the distance between these two points.
The following choice is somewhat random. Consider the line through the point $(-x_0-2,-8)$ with a smaller slope than $\ell$: \[ \ell_1(y) = \bigl(3 x_0^2 + 6 x_0 + 4 \bigr) \bigl(y - (-x_0-2\bigr) - 8. \] The $y$-intercept of this line is \[ y_1 = -x_0 - 2 + \frac{8}{3x_0^2 + 6x_0 +4}. \] For this value of $y$ the function $h$ is close to $0$, but positive: \[ h(y_1) = \frac{ 192 x_0^2 \bigl( 8 + 9 x_0 + 3 x_0^2 \bigr) }{\bigl(4 + 6 x_0 + 3 x_0^2\bigr)^3} = 192 x_0 \frac{\frac{8}{x_0^2} + \frac{9}{x_0} + 3}{\bigl(\frac{4}{x_0} + 6 + 3 x_0 \bigr)^3} \gt 0 \] Thus, for large $x_0$ the value of $h(y_1)$ is almost $0$ but positive. Since $g(x_0)$ is a negative root of $h(y)$ and since both $y_0$ and $y_1$ are negative numbers and $h(y_0) \lt 0$ and $h(y_1) \gt 0$ and $h$ is a continuous function we conclude from the Intermediate Value Theorem that \[ -x_0 - 2 \lt y_0 \lt g(x_0) \lt y_1 = -x_0 - 2 + \frac{8}{3x_0^2 + 6x_0 +4}. \] Therefore \[ g(x_0) - \bigl(-x_0 - 2 \bigr) \lt \frac{8}{3x_0^2 + 6x_0 +4}. \] Since the last quantity aproaches $0$ as $x_0$ approaches $+\infty$, this proves that the line $y = -x - 2$ is the asymptote to the graph of the involution $y = g(x).$
- I added a paragraph to our webpage on osculating circles in which I derive the parametric equation for the evolute of a curve given by its parameric equation.
- Today we studied some famous examples of parametric curves.
- When working with parameterized curves it is very useful to think of these equations as vector functions. So, we first reviewed the concept of a vector in the three dimensional space. I think that the best application of vector is the rgb code for colors.
- All colors are identified in Mathematica by a vector $\displaystyle \begin{bmatrix}x \\ y \\ z\end{bmatrix}$ with $0 \leq x \leq 1$, $0 \leq y \leq 1$, $0 \leq z \leq 1$. In other words, Mathematica identifies colors with the points in the unit cube in the $xyz$-space. In this setting the unit cube is called The Color Cube.
- Below is an image of the color cube with 27 colors emphasized.
- Some of the colors emphasized below have common names. For others I tried to find appropriate names.
- Here I adopt following mathematical definitions of dark and light adjectives for colors: For a specific COLOR we define the dark COLOR to be the color which is half-way between COLOR and BLACK, that is the vector corresponding to the COLOR scaled by $1/2$. For a specific COLOR we define the light COLOR to be the color which is half-way between COLOR and WHITE, that is the sum of the vector $(1/2,1/2,1/2)$ and the vector corresponding to the COLOR scaled by $1/2$.
-
In this terminology maroon is just dark red, navy is dark blue, teal is dark cyan, purple is dark magenta, olive is dark yellow, gray is light black, or gray is dark white, salmon is light red, ultra pink is light magenta.
$\displaystyle \begin{bmatrix}0 \\ 0 \\0\end{bmatrix}$ Black $\displaystyle \begin{bmatrix}1/2 \\ 1/2 \\ 1/2\end{bmatrix}$ Gray $\displaystyle \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ White $\displaystyle \begin{bmatrix}1 \\ 0 \\0\end{bmatrix}$ Red $\displaystyle \begin{bmatrix}1/2 \\ 0 \\0\end{bmatrix}$ Maroon $\displaystyle \begin{bmatrix}1/2 \\ 1/2 \\0\end{bmatrix}$ Olive $\displaystyle \begin{bmatrix}1 \\ 1/2 \\0\end{bmatrix}$ Orange $\displaystyle \begin{bmatrix}1 \\ 1 \\0\end{bmatrix}$ Yellow $\displaystyle \begin{bmatrix}0 \\ 1 \\0\end{bmatrix}$ Green $\displaystyle \begin{bmatrix}0 \\ 1/2 \\0\end{bmatrix}$ Dark Green $\displaystyle \begin{bmatrix}1/2 \\ 1 \\0\end{bmatrix}$ Chartruse $\displaystyle \begin{bmatrix}0 \\ 1/2 \\ 1/2\end{bmatrix}$ Teal $\displaystyle \begin{bmatrix}0 \\ 1 \\1/2\end{bmatrix}$ Spring Green $\displaystyle \begin{bmatrix}0 \\ 0 \\1\end{bmatrix}$ Blue $\displaystyle \begin{bmatrix}0 \\ 0 \\ 1/2\end{bmatrix}$ Navy $\displaystyle \begin{bmatrix}1/2 \\ 0 \\ 1/2\end{bmatrix}$ Purple $\displaystyle \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ Magenta $\displaystyle \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}$ Cyan $\displaystyle \begin{bmatrix}1/2 \\ 0 \\ 1\end{bmatrix}$ Dark Violet $\displaystyle \begin{bmatrix}0 \\ 1/2 \\1\end{bmatrix}$ Sky Blue $\displaystyle \begin{bmatrix}1/2 \\ 1/2 \\ 1\end{bmatrix}$ Light Blue $\displaystyle \begin{bmatrix} 1 \\ 1/2 \\1\end{bmatrix}$ Ultra Pink $\displaystyle \begin{bmatrix} 1/2 \\ 1 \\1 \end{bmatrix}$ Light Cyan $\displaystyle \begin{bmatrix}1 \\ 0 \\ 1/2 \end{bmatrix}$ Magenta Red $\displaystyle \begin{bmatrix} 1 \\ 1/2 \\ 1/2 \end{bmatrix}$ Salmon $\displaystyle \begin{bmatrix}1 \\ 1 \\ 1/2 \end{bmatrix}$ Light Yellow $\displaystyle \begin{bmatrix}1/2 \\ 1 \\ 1/2 \end{bmatrix}$ Light Green Place the cursor over the image to start the animation.
-
Thinking of colors as vectors helps us to understand a transition between two colors. Below I show three ways of interpreting a transition between
Teal and Yellow.
Place the cursor over the image to start the animation.
Since Teal and Yellow are the heads of particular vectors in the Color Cube, to construct a transition I connected the heads with a line segment. Points on this line segment are the heads of special linear combinations of the vectors representing Teal and Yellow. As an exercise write the linear combinations which are used in the above transition.
Place the cursor over the image to start the animation.
In the above animation I used the colors from the line segment connecting Teal and Yellow to color the rectangles in the middle of the square.
Above is the unit circle colored using colors from the line segment connecting Teal and Yellow.
- 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.
-
Similarly one can derive 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.
- In class I created this Mathematica file in which I plot the cardioid. The file is called 20181108_Cardioid.nb. Exactly one year ago I created this Mathematica file in which I demonstrate graphs of few other parametric curves. The file is called 20171108_Parametric_Equations.nb. Both files are created in Mathematica version 8. Notice that in file names I always use underscore instead of a blank space.
- The goal today is to develop a method for finding the envelope for a family of lines. Here we write an equation for a line as an implicit equation \[ a x + b y + c = 0. \] Here $a$ and $b$ are real numbers which are not both zero, that is such that $(a,b) \neq (0,0).$
-
We first establish a little piece of theory about implicit equations of lines. Let the green line be given by the equation $ax+by+c=0$ and let the navy blue line be given by $a_1 x+b_1 y+c=0$. Assume that these two lines intersect at exactly one point $(x_0,y_0).$ Then the orange line
\[
(a_1-a) x+(b_1 - b) y+(c_1 - c)= 0
\]
also goes through the point $(x_0,y_0).$ This is verified by simple verification:
Since
\[
ax_0+by_0+c=0 \quad \text{and} \quad a_1x_0+b_1y_0+c_1=0
\]
we clearly have
\[
(a_1-a) x_0+(b_1 - b) y_0+(c_1 - c)= 0.
\]
- Now assume that we are given a family of lines \[ a(s) x + b(s) y + c(s) = 0. \] Here $a(s)$, $b(s)$ and $c(s)$ are differentiable functions defined on a common domain $D.$ In the picture below this family of lines is represented by light gray lines. The picture below indicates that all the lines in the given family are tangent to some curve. This curve is called the envelope of the given family. Our goal is to determine the equation for the envelope.
- We use the following strategy to determine the equation of the envelope. We chose $s_0 \in D$ and focus on the line \[ a(s_0) x + b(s_0) y + c(s_0) = 0. \] This is the green line in the picture below. The goal is to find the point on the green line which is also on the envelope. That is the green point in the picture below.
- To find the green point we explore the lines in the family which are nearby the green line. That is we choose small $\Delta s$ and consider the line \[ a(s_0+\Delta s) x + b(s_0+\Delta s) y + c(s_0+\Delta s) = 0. \] This is the navy blue line in the picture below. The navy blue line and the green line intersect at the navy blue point. Intuitively it is clear that as $\Delta s \to 0$ the navy blue point approaches the green point.
-
Now comes the brilliant idea: Replace the navy blue line with the orange line
\[
\bigl( a(s_0+\Delta s) - a(s_0) \bigr) x + \bigl( b(s_0+\Delta s) - b(s_0) \bigr) y + \bigl( c(s_0+\Delta s) - c(s_0) \bigr) = 0.
\]
The orange line also goes through the navy blue point. But, more importantly, the orange line can be written as follows:
\[
\frac{ a(s_0+\Delta s) - a(s_0) }{\Delta s} x + \frac{ b(s_0+\Delta s) - b(s_0) }{\Delta s} y + \frac{ c(s_0+\Delta s) - c(s_0) }{\Delta s} = 0.
\]
Taking the limit as $\Delta s \to 0$ in the last equation we get
\[
a'(s_0) x + b'(s_0) y + c'(s_0) = 0.
\]
This is the equation of the purple line at the end of the animation below. This also means that the orange lines have a limiting line as $\Delta s \to 0.$ Further this means that the limit of blue points in the animation below is the intersection of the purple line and the green line.
Place the cursor over the image to start the animation.
- Thus, to determine the green point on the green line we intersect the green and the purple line, that is the lines \[ a(s_0) x + b(s_0) y + c(s_0) = 0. \] and \[ a'(s_0) x + b'(s_0) y + c'(s_0) = 0. \] By doing this for every $s$ we get all the points on the envelope.
-
Example. Consider the family of lines given by
\[
s^3 x + s^2 y + s = 0,
\]
where $s \neq 0$ is a real number. Determine the envelope of this family of lines. For a given $s \neq 0$ the point on the envelope is on the intersection of the lines
\[
s^3 x + s^2 y + s = 0 \quad \text{and} \quad 3s^2 x + 2 s y + 1 = 0.
\]
Dividing the first equation by $s \neq 0$ and subtracting so obtained equation from the second one we get
\[
2s^2 x + s y = 0.
\]
Solving for $s$ yields $s = - y/(2x).$ Substituting this in $s^2 x + s y + 1 = 0$ we get that the equation for the envelope is $y^2 = 4x.$
-
Example. In this example we start with the parabola $y = x^2$ and determine the family of normals to the parabola. The normal at the parabola at the point $(s,s^2)$ is given by
\[
y = -\frac{1}{2s} (x-s) + s^2.
\]
Rewrite this equation in the implicit form
\[
x + 2s y -s -2 s^3 = 0.
\]
What is the envelope of this family of lines? The points on the envelope are at the intersections of the lines \[
x + 2s y -s -2 s^3 = 0 \quad \text{and} \quad 2 y - 1 - 6 s^2 = 0.
\]
Solving the second equation for $y$ we get
\[
y = \frac{1}{2} + 3 s^2.
\]
Substituting this expression in the first equation, solving for $s$ and substituting in the last equation we get the equation of the envelope
\[
y = \frac{1}{2} + \frac{3\sqrt[3]{4}}{4} x^{2/3}.
\]
Place the cursor over the image to start the animation.
- This is universal. We will prove that for every twice differentiable function $f$ the envelope of the family of the normal lines to the graph of $f$ is identical to the curve formed by the centers of the osculating circles to the graph $f$. This curve is called the evolute of the graph of $f.$
-
We recall two definitions.
- The evolute of a graph is the set of all the centers of the osculating circles of that graph.
- On our webpage about osculating circles we showed the evolutes of the functions: $y= \sin x$, $y = x^2$, $y = x^3-x$.
- An envelope of a family of lines in the plane is a curve that is tangent to each member of the family at some point.
- Theorem. For a twice differentiable function $f$, the envelope of its family of normals is the evolute of $f$.
- Proof. Let $f:D\to \mathbb R$ be a twice differentiable function. The normal line to the graph of $f$ at the point $\bigl( s, f(s) \bigr)$ is given by \[ y = -\frac{1}{f'(s)} (x-s) + f(s). \] In the implicit form this equation can be written as \[ x + f'(s) y - s - f(s) f'(s) = 0. \] With $s \in D$ the last equation represents the family of all the normals to the graph of $f$. To find the envelope of this family we find the equations of the purple lines corresponding to this family: \[ f''(s) y - 1 - \bigl( f'(s) \bigr)^2 - f(s)f''(s) = 0. \] Solving the last equation for $y$ we get \[ y = f(s) + \frac{1 + \bigl( f'(s) \bigr)^2 }{f''(s)}. \] Substituting this $y$ in the implicit equation for the family of normals and solving for $x$ we get \[ x = s - f'(s) \frac{1 + \bigl( f'(s) \bigr)^2 }{f''(s)}. \] Thus the parametric equations for the envelope of the family of normals are \[ y = f(s) + \frac{1 + \bigl( f'(s) \bigr)^2 }{f''(s)}, \qquad x = s - f'(s) \frac{1 + \bigl( f'(s) \bigr)^2 }{f''(s)}. \] As we calculated on our webpage about osculating circles these are exactly the coordinates of the center of the osculationg circle which touches the graph of $f$ at the point $(s,f(s)).$ This completes the proof.
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Examples. Next few pictures show envelopes which are in fact evolutes.
- Today we did an example 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 $(a,f(a)).$
- The evolute of a graph is the set of all the centers of the osculating circles of that graph.
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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 $\bigl(\ln(1/\sqrt{2}), 1/\sqrt{2} \bigr)$ 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
- A more general version of the above problem would be to find the smallest osculating circle of the exponential function $b^x$ with $b \gt 1.$ Then, we could find the exponential function whose smallest osculating circle is the unit circle.
- 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 book deals with optimization problems. Do the following problems: 16, 17, 28, 29, 31, 36, 41, 42, 44, 46, 47, 48, 51, 53.
- Here is my webpage about Osculating circles.
- Today we talked about higher_order_approximations of differentiable functions.
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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
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Few more examples of many tangents are below:
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And few examples of families of normals are below:
- We will come back to these families of lines soon.
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Let us do some theory. 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 $f'(c) = 0.$
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Notice that the contrapositive of Fermat's Interior Extremum Theorem is
Contrapositive 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 $(a,b).$
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A consequence of 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 Rolle's Theorem clickAssume that $f:[a,b]\to \mathbb R$ satisfies three conditions of Rolle's Theorem. Then, since $f$ is continuous on $[a,b]$, the Extreme Value Theorem yields that there exist $c_1, d_1 \in [a,b]$ such that \[ \forall \, x \in [a,b] \qquad f(c_1) \leq f(x) \leq f(d_1). \] Since $[a,b] = \{a,b\} \cup (a,b)$ and $\{ c_1, d_1 \} \subseteq [a,b],$ we have the following dichotomy:- Case 1. $\{ c_1, d_1 \} \subseteq \{a,b\}$
- Case 2. $\{ c_1, d_1 \} \cap (a,b) \neq \emptyset$
Now assume that Case 2 holds. Let $c \in \{ c_1, d_1 \} \cap (a,b).$ Then $c \in (a,b)$ and $c$ is an extreme value of $f$ on $[a,b].$ By Fermat's Interior Extremum Theorem we have $f'(c) = 0.$ This completes the proof of Rolle's Theorem. -
A consequence of Rolle's Theorem is the 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 clickAssume that $f:[a,b]\to \mathbb R$ satisfies two conditions of the Mean Value Theorem. Define a new function $g:[a,b] \to \mathbb R$ by \[ \forall \, x \in [a,b] \qquad g(x) = f(x) - \frac{f(b) - f(a)}{b-a} (x-a). \] It is not difficult to establish that the function $g$ satisfies all the conditions of Rolle's Theorem. Consequently, there exists $c \in (a,b)$ such that $g'(c) = 0.$ Since \[ \forall \, x \in (a,b) \qquad g'(x) = f'(x) - \frac{f(b) - f(a)}{b-a}, \] we deduce that \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \] This completes the proof of the Mean Value Theorem. - Important consequences of the Mean Value Theorem are the following theorems: Constant Function Theorem, Monotonicity Tests and Racetrack Principles.
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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 $x \in (a,b)$.To toggle the proof of the Constant Function Theorem clickTo be completed later ....
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Monotonicity Tests. Let $a, b \in \mathbb R$ and $a \lt b$. Let $f : (a,b) \to \mathbb R$ be a differentiable function on $(a,b)$.
- 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 $(a,b)$.
To toggle the proof of the Monotonicity Tests clickTo be completed later .... -
Our textbook proves the Racetrack Principle.
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 $x \in (a,b)$.
- Thank you Sean for asking such a beautiful question today in class.
- In response to Sean's question today in class I created this Mathematica file. The file is called Solutions_of_General_Cubic.nb. The file is created in Mathematica version 8. Notice that in file names I always use underscore instead of just a blank space.
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This was Sean's question: Let $b, c, d \in \mathbb R.$ Consider the general cubic equation
\[
x^3 + b x^2 + c x + d = 0.
\]
In the space of all triples $(b,c,d)$ identify those triples for which the above cubic equation has exactly two real soulutions. Such triples form a surface in three-space. In the above Mathematica notebook I obtained a prametric equation for this surface. Below is the resulting picture:
Place the cursor over the image to start the animation.
- I created a short web page about two Lambert $W$ functions. These function you need for Problem 1 and Problem 2 on Assignment 3.
- The following item will help you with Problem 4 on Assignment 3. Before starting that item I want to point out that Mathematics is full of interesting trichotomies. The most famous is the trichotomy for real numbers: each real number is either negative, zero, or positive.
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In this item we study the trichotomy which arises 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 $(\alpha,\beta) \in \mathbb R\!\times\!\mathbb R.$ It turns out that each pair $(\alpha,\beta)$ falls into one of the following three categories:
- 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 orange),
- with $\alpha$ and $\beta$ in equation (ce) the equation has exactly three real solutions (we will color such points $(\alpha,\beta)$ in light blue).
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 blue.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 orange.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) orange points correspond to two distinct real solutions of (ce) blue points correspond to three distinct real solutions of (ce)
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Below is an animation related to Problem 4 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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This is the additional problem that I want you to do for the exam redo: Find four circles of radius 3 that are tangent to both, the graph of the parabola $y=x^2$ and the $x$-axis; as in the picture below.
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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 $c \in D$ and let $f: D \to \mathbb R$ be a function. We say that the function $f$ is differentiable at $c$ if the following limit exists: \[ \lim_{x\to c} \frac{f(x) - f(c)}{x - c}. \] If $f$ is differentiable at $c$ we define the derivative of $f$ at $c,$ denoted by $f'(c),$ to be \[ f'(c) = \lim_{x\to c} \frac{f(x) - f(c)}{x - c}. \] 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 $c \in \mathbb R$ and for every $c \in \mathbb R$ we have $f'(c) = 2c.$To toggle the proof clickLet $c \in \mathbb R$ be arbitrary. We need to prove that \[ \lim_{x\to c} \frac{x^2 - c^2}{x - c} = 2c. \] First, for an arbitrary $x \neq c$ we simplify the expression \[ \left| \frac{x^2 - c^2}{x - c} - 2c \right| = \left| {x + c} - 2c \right| = \left| x - c \right|. \] Let $\epsilon \gt 0$ be arbitrary. The preceding equality yields that the following implication is true: \[ 0 \lt |x-c| \lt \epsilon \qquad \Rightarrow \qquad \left| \frac{x^2 - c^2}{x - c} - 2c \right| \lt \epsilon. \] Thus we can take $\delta(\epsilon) = \epsilon$. This proves that \[ 2c = \lim_{x\to c} \frac{x^2 - c^2}{x - c}. \] Thus we proved that $f'(c) = 2c.$
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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 $c \in \mathbb R$ and we have $f'(c) = n\, c^{n-1}.$
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Let $f(x) = 1/x$ with $x \gt 0.$ Then $f$ is differentiable at every $c \gt 0$ and we have $f'(c) = - 1/(c^2).$To toggle the proof clickLet $c \gt 0$ be arbitrary. We need to prove that \[ \lim_{x\to c} \frac{\frac{1}{x} - \frac{1}{c}}{x - c} = -\frac{1}{c^2}. \] 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 $\delta_0 = c/2$ and let $x \in (c/2,c) \cup (c,3c/2).$ Next, we simplify the expression \begin{align*} \left| \frac{\frac{1}{x} - \frac{1}{c}}{x - c} - \left( - \frac{1}{c^2} \right) \right| & = \left| \frac{ \frac{c - x}{xa} }{x - c} - \left( - \frac{1}{c^2} \right) \right| \\ & = \left| -\frac{1}{xa} + \frac{1}{c^2} \right| \\ & = \frac{1}{c} \, \left| -\frac{1}{x} + \frac{1}{c} \right| \\ & = \frac{1}{c} \, \left| \frac{x - c}{xa} \right| \\ & = \frac{1}{c} \, \frac{|x - c|}{x \, c } \\ & \leq \frac{1}{c} \, \frac{|x - c|}{ (c/2) \,c } \\ & = \frac{2}{c^3} \, |x - c| \end{align*} Hence we proved that for an arbitrary $c \gt 0$ and for all $x \in (c/2,c) \cup (c,3c/2)$ we have \[ \left| \frac{\frac{1}{x} - \frac{1}{c}}{x - c} - \left( - \frac{1}{c^2} \right) \right| \leq \frac{2}{c^3} |x - c|. \] To complete the proof, let $\epsilon \gt 0$ be arbitrary and set \[ \delta(\epsilon) = \min\left\{ \frac{c}{2}, \frac{c^3}{2} \epsilon \right\}. \] Assume that \[ 0 \lt |x-c| \lt \min\left\{ \frac{c}{2}, \frac{c^3}{2} \epsilon \right\}. \] Then \[ x \in (c/2,c) \cup (c,3c/2) \qquad \text{and} \qquad \frac{2}{c^3} |x-c| \lt \epsilon . \] Since $x \in (c/2,c) \cup (c,3c/2)$, as we proved above we have \[ \left| \frac{\frac{1}{x} - \frac{1}{c}}{x - c} - \left( - \frac{1}{c^2} \right) \right| \leq \frac{2}{c^3} |x - c|. \] As we assume $\frac{2}{c^3} |x-c| \lt \epsilon$, we deduce that \[ \left| \frac{\frac{1}{x} - \frac{1}{c}}{x - c} - \left( - \frac{1}{c^2} \right) \right| \lt \epsilon. \] Thus we have proved \[ 0 \lt |x-c| \lt \min\left\{ \frac{c}{2}, \frac{c^3}{2} \epsilon \right\} \qquad \Rightarrow \qquad \left| \frac{\frac{1}{x} - \frac{1}{c}}{x - c} - \left( - \frac{1}{c^2} \right) \right| \lt \epsilon. \] That is, $f'(c) = -\frac{1}{c^2}.$
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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 $c \in \mathbb R\setminus\{0\}$ and we have $f'(c) = -n\, c^{-n-1}.$
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Let $f(x) = \sqrt{x}$ with $x \gt 0.$ Then, $f$ is differentiable at every $c \gt 0$ and we have $f'(c) = \frac{1}{2\sqrt{c}}.$To toggle the proof clickLet $c \gt 0$ be arbitrary. We need to prove that \[ \lim_{x\to c} \frac{\sqrt{x} - \sqrt{c}}{x - c} = \frac{1}{2\,\sqrt{c}}. \] Next, with $x \gt 0$ such that $x \neq c$ we simplify the expression \begin{align*} \left| \frac{\sqrt{x} - \sqrt{c}}{x - c} - \frac{1}{2\,\sqrt{c}} \right| & = \left| \frac{1}{\sqrt{x} + \sqrt{c} } - \frac{1}{2\,\sqrt{c}} \right| \\ & = \left| \frac{2\,\sqrt{c} - \sqrt{x} - \sqrt{c} }{2\,\sqrt{c} \bigl( \sqrt{x} + \sqrt{c} \bigr) } \right| \\ & = \left| \frac{ \sqrt{c} - \sqrt{x} }{2\,\sqrt{c} \bigl( \sqrt{x} + \sqrt{c} \bigr) } \right| \\ & = \left| \frac{ c - x }{2\,\sqrt{c} \bigl( \sqrt{x} + \sqrt{c} \bigr)^2 } \right| \\ & = \frac{ |x - c| }{2\,\sqrt{c} \bigl( \sqrt{x} + \sqrt{c} \bigr)^2 } \\ & \leq \frac{ 1 }{2\, c \sqrt{c} } \, | x - c | \end{align*} Hence we proved that for an arbitrary $c \gt 0$ and for all $x \gt 0$ such that $x \neq c$ we have \[ \left| \frac{\sqrt{x} - \sqrt{c}}{x - c} - \frac{1}{2\,\sqrt{c}} \right| \leq \frac{ 1 }{2\, c \sqrt{c} } \, | x - c |. \] This is what we used to call Fact. Let $\epsilon \gt 0$ be arbitrary. The preceding inequality yields that the following implication is true: \[ 0 \lt |x-c| \lt 2\, c \sqrt{c} \, \epsilon \qquad \Rightarrow \qquad \left| \frac{\frac{1}{x} - \frac{1}{c}}{x - c} - \left( - \frac{1}{c^2} \right) \right| \lt \epsilon. \] Thus we can take $\delta(\epsilon) = 2\, c \sqrt{c} \, \epsilon.$ This proves that \[ \frac{1}{2\,\sqrt{c}} = \lim_{x\to c} \frac{\sqrt{x} - \sqrt{c}}{x - c}. \] Thus we proved that $f'(c) = \frac{1}{2\,\sqrt{c}}$
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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 $c \gt 0$ and we have $f'(c) = \beta \, c^{-\beta-1}.$
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Let $f(x) = \sin(x)$ with $x \in \mathbb R.$ Then, $f$ is differentiable at every $c \in \mathbb R$ and we have $f'(c) = \cos(c).$
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Let $f(x) = \cos(x)$ with $x \in \mathbb R.$ Then, $f$ is differentiable at every $c \in \mathbb R$ and we have $f'(c) = -\sin(c).$
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Let $f(x) = \exp(x)$ with $x \in \mathbb R.$ Then, $f$ is differentiable at every $c \in \mathbb R$ and we have $f'(c) = \exp(c).$
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Let $f(x) = \ln(x)$ with $x \in (0,+\infty).$ Then, $f$ is differentiable at every $c \in (0,+\infty)$ and we have $f'(c) = 1/c.$
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Let $f(x) = \arccos(x)$ with $x \in [-1,1].$ Then, $f$ is differentiable at every $c \in (-1,1)$ and we have $f'(c) = -\frac{1}{\sqrt{1-x^2}}.$
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Let $f(x) = \arcsin(x)$ with $x \in [-1,1].$ Then, $f$ is differentiable at every $c \in (-1,1)$ and we have $f'(c) = \frac{1}{\sqrt{1-x^2}}.$
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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.
- In class I created this Mathematica file in which I created graphs related to the preceding item. The file is called 20181022_Circles_in_parabola.nb. The file is created in Mathematica version 8. Notice that in file names I always use underscore instead of just a blank space.
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Problem 3 on Assignment 3 is in fact an exercise about derivatives.
Place the cursor over the image to start the animation.
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We stated earlier the Lipschitz condition for the existence of limit. This condition can be restated as a condition for continuity of a function at a point.
Lipschitz condition for continuity. Let $D \subset \mathbb R,$ let $c \in D$ and let $f : D \to \mathbb R$ be a function. If there exist $\delta_0 \gt 0$ and $K \gt 0$ such that $(c-\delta_0,c+\delta_0) \subseteq D$ and for every $x \in (c-\delta_0,c+\delta_0)$ we have \[ \bigl| f(x) - f(c) \bigr| \leq K\, |x-c|, \] then $f$ is continuous at $c$.
To toggle the proof of Lipschitz condition for continuity click
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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 $c \in \mathbb R$ and for every $x\in \mathbb R$ we have $|\sin(x) - \sin(c)| \leq |x-c|.$
- For every $c \in \mathbb R$ and for every $x\in \mathbb R$ we have $|\cos(x) - \cos(c)| \leq |x-c|.$
- For every $c \gt 0$ and for every $\displaystyle x\in \left(\frac{c}{2},\frac{3c}{2}\right)$ we have $\displaystyle |\ln(x) - \ln(c)| \leq \frac{2}{c}\, |x-c|.$
- For every $c\in \mathbb R$ and for every $\displaystyle x \in \left(a - \frac{1}{2},c+ \frac{1}{2}\right)$ we have $\displaystyle \left| e^x - e^c \right| \leq 2e^{c}\, |x-c|.$
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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 $c$ be a real number. Assume that $f$ and $g$ are continuous at the point $c.$ The following statements hold.
- If $h = f + g$, then $h$ is continuous at $c.$
- If $h = f g$, then $h$ is continuous at $c.$
- If $\displaystyle h = \dfrac{f}{g}$ and $g(c) \neq 0$, then $h$ is continuous at $c.$
Continuity of composition. Let $f$ and $g$ be real functions and let $c$ be a real number. If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then $f\circ g$ is continuous at $c.$ -
Today I handed out Assignment 3. It deals mostly with derivatives that we will do next week. Below is the animation for Problem 3 on Assignment 3
Place the cursor over the image to start the animation.
Let $D \subset \mathbb R,$ let $c\in D$ and let $f : D \to \mathbb R$ be a function. Assume that there exist $\delta_0 \gt 0$ and $K\gt 0$ such that $(c-\delta_0,c+\delta_0) \subseteq D$ and for every $x \in (c-\delta_0,c+\delta_0)$ we have
\[
\bigl| f(x) - f(c) \bigr| \leq K\, |x-c|.
\]
Next I will prove that $f$ is continuous at $c.$ Let $\epsilon \gt 0$ be arbitrary. Set
\[
\delta(\epsilon) = \min\left\{ \delta_0 , \frac{\epsilon}{K} \right\}.
\]
Now, I need to prove the implication
\[
|x-c| \lt \min\left\{ \delta_0 , \frac{\epsilon}{K} \right\} \qquad \Rightarrow \qquad \bigl| f(x) - f(c) \bigr| \lt \epsilon.
\]
Here is a proof. Assume that $|x-c| \lt \min\left\{ \delta_0, \frac{\epsilon}{K} \right\}$. Then $|x-c| \lt \delta_0$ and $|x-c| \lt {\epsilon}/{K}.$ Consequently, $x \in (c-\delta_0,c+\delta_0)$ and $K\,|x-c| \lt {\epsilon}.$ Since $x \in (c-\delta_0,c+\delta_0)$, by the assumption we have $\bigl| f(x) - f(c) \bigr| \leq K\, |x-c|.$ Since both $\bigl| f(x) - f(c) \bigr| \leq K\, |x-c|$ and $K\,|x-c| \lt {\epsilon}$ are true, by transitivity of order, we have $\bigl| f(x) - f(c) \bigr| \lt \epsilon.$ This proves the continuity of $f$ at $c.$
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Below is a more general $\epsilon$-$\delta$-definition of continuity at a point. For a function defined on a closed interval $[a,b]$ this definition applies to the endpoints $a$ and $b$.
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
- Two most important theorems about continuous functions are the Intermediate Value Theorem and the Extreme Values Theorem.
-
In this item I will present a proof of the Intermediate Value Theorem which uses algebra, basic set theory and, most importantly, the definition of continuity and Completeness Axiom as given here.
for every real number $v$ between $f(a)$ and $f(b)$ there exists $u \in [a,b]$ such that $v = f(u)$.
Let $v \in \bigl[f(a),f(b)\bigr]$ be arbitrary. If $v = f(a)$ we can take $u=a$ and if $v = f(b)$ we can take $u=b.$ So, the core of the proof is the case when $f(a) \lt v \lt f(b).$
Let $v \in \bigl(f(a),f(b)\bigr)$ be arbitrary. Set \[ A = \bigl\{ x \in [a,b] \, \bigl|\bigr. \, f(x) \lt v \bigr\}. \] Before we continue we need to establish a fact about the set $A:$
To prove the existence of $\beta$ we use the fact that $f$ is continuous (from the left) at $b.$ Set \[ \epsilon_2 = \frac{1}{2} \bigl(f(b) - v \bigr). \] Since $v \lt f(b),$ we have $\epsilon_2 \gt 0.$ Since $f$ is continuous at $b,$ there exists $\delta_2 = \delta(\epsilon_2) \gt 0$ such that $\delta_2 \lt b-a$ and \[ 0 \leq b - x \lt \delta_2 \quad \Rightarrow \quad \bigl| f(x) - f(b) \bigr| \lt \frac{1}{2} \bigl(f(b) - v\bigr). \] The last displayed implication can be restated as \[ x \in \bigl(b-\delta_2,b \bigr] \quad \Rightarrow \quad f(b) - \frac{1}{2} \bigl(f(b) - v\bigr) \lt f(x). \] Since \[ f(b) - \frac{1}{2} \bigl(f(b) - v\bigr) = \frac{1}{2} \bigl(v+f(b)\bigr) \gt v, \] we have \[ x \in \bigl(b-\delta_2,b \bigr] \quad \Rightarrow \quad v \lt f(x). \] The contrapositive of the last implication for $x \in [a,b]$ is \[ f(x) \leq v \quad \Rightarrow \quad x \in [a, b-\delta_2\bigr]. \] This means that with $\beta = b-\delta_2$ we have $\beta \in (a,b)$ and $A \subseteq [a,\beta].$
Since $[a,\alpha) \subseteq A,$ we have that $A$ is not an empty set.
Set \[ B = \bigl\{ y \in [a,b] \, \bigl|\bigr. \, \forall\, x \in A \ \, x \leq y \bigr\}. \] Since $A \subseteq [a,\beta]$ we have that $[\beta,b] \subseteq B.$ Thus $B$ is a nonempty subset of $\mathbb R.$ By definition of $B$ we have \[ \forall \, x\in A \quad \forall \, y \in B \qquad x \leq y. \] The sets $A$ and $B$ satisfy all the hypotheses of the Completeness Axiom. Hence, by the Completeness Axiom there exists $c \in \mathbb R$ such that \[ \forall \, x\in A \quad \forall \, y \in B \qquad x \leq c \leq y. \]
Let us observe three facts about this super special number $c$ that we discovered using the Completeness Axiom:
Finally we will use the continuity of the function $f$ at $c$ to prove that $v=f(c)$. Let $\epsilon \gt 0$ be arbitrary. By the definition of continuity at $c$ there exists $\delta(\epsilon) \gt 0$ such that \[ x \in [a,b] \ \text{and} \ | x-c | \lt \delta(\epsilon) \quad \Rightarrow \quad \bigl| f(x) - f(c) \bigr| \lt \epsilon. \] As we mentioned many times in class, the above statement is equivalent to \[ x \in [a,b] \ \text{and} \ c-\delta(\epsilon) \lt x \lt c+\delta(\epsilon) \quad \Rightarrow \quad f(c) - \epsilon \lt f(x) \lt f(c) + \epsilon. \] By Fact 1 we have $c \lt b$. Choose $s \in [a,b]$ such that $s \lt b$ and $c \lt s \lt c+\delta(\epsilon).$ Then, by Fact 2 we have $v \leq f(s)$ and by the last displayed statement we have $f(s) \lt f(c) + \epsilon.$ Hence $v \lt f(c) + \epsilon.$
By Fact 1 we have $a \lt c$. Choose $s \in [a,b]$ such that $a \lt s$, $c-\delta(\epsilon) \lt s \lt c.$ Then, by Fact 3 there exists $x \in [a,b]$ such that $s \lt x \leq c$ and $f(x) \lt v.$ Since $c-\delta(\epsilon) \lt s \lt x \leq c$, the last displayed statement implies $f(c) - \epsilon \lt f(x)$. Thus, $f(c) - \epsilon \lt v.$
In conclusion, for an arbitrary $\epsilon \gt 0$ we have proved that \[ f(c) - \epsilon \lt v \lt f(c) + \epsilon. \] In other words we have proved that \[ \forall \, \epsilon \gt 0 \quad \bigl| f(c) - v \bigr| \lt \epsilon. \] The last statement implies that $v = f(c).$ Thus, we can take $u = c$. QED -
My next goal is to give a proof of the Extreme Values Theorem which is in spirit similar to the above proof.
there exist $c, d \in [a,b]$ such that $f(c) \leq f(x) \leq f(d)$ for all $x \in [a,b]$.
- 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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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 $c \in D$ and let $f: D \to \mathbb R$ be a function. We say that the function $f$ is continuous at $c$ if the following two condition is satisfied:
- I. There exists $\delta_0 \gt 0$ such that $(c-\delta_0, c+\delta_0) \subset D.$
- II.
For every $\epsilon \gt 0$ there exists $\delta(\epsilon)$ such that $0 \lt \delta(\epsilon) \leq \delta_0$ and
$|x-c| \lt \delta(\epsilon)$ implies $|f(x) - f(c) | \lt \epsilon$.
-
Example 1. Consider the function $f(x) = 1/x$ defined for all $x \in (0,+\infty).$ Prove that this function is continuous.Proof. To prove that $f(x) = 1/x$ defined for all $x \in (0,+\infty)$ is continuous let $c$ be an arbitrary positive real number. To satisfy I. in the definition we set $\delta_0 =c/2.$
Fact A. If $|x-c| \lt c/2$, then $\displaystyle \biggl|\frac{1}{x} -\frac{1}{c} \biggr| \leq \frac{2}{c^2} |x-c|$.
To toggle the proof of Fact A clickAssume that $|x-c| \lt c/2$. That is assume that $c/2 \lt x \lt(3c)/2$. Then \[ \displaystyle \biggl|\frac{1}{x} -\frac{1}{c} \biggr| = \frac{|x-c|}{x\, c} \leq \frac{1}{(c/2)\,c} |x-c| = \frac{2}{c^2} |x-c|. \] The inequality above is a consequence of the "pizza-party" principle and the fact that $x \gt c/2$.Next we prove II. Let $\epsilon \gt 0$ be arbitrary. To find $\delta(\epsilon) \gt 0$ we solve \[ \frac{2}{c^2} |x-c| \lt \epsilon \] for $|x-c|$. We get \[ |x-c| \lt \frac{c^2}{2} \epsilon. \] To assure that Fact A is valid we need \[ |x-c| \lt \frac{c}{2}. \] Now we set \[ \delta(\epsilon) = \min\left\{ \frac{c}{2}, \frac{c^2}{2} \epsilon \right\}. \] With this $\delta(\epsilon)$ we can prove$ \displaystyle |x-c| \lt \min\left\{ \frac{c}{2}, \frac{c^2}{2} \epsilon \right\}$ implies $\displaystyle \biggl|\frac{1}{x} -\frac{1}{c} \biggr| \lt \epsilon$.
-
Example 2. Consider the function $f(x) = x^2$ defined for all $x \in \mathbb R.$ Prove that this function is continuous.Proof. To prove that $f(x) = x^2$ defined on $\mathbb R$ is continuous let $c$ be an arbitrary real number. To satisfy I. in the definition we set $\delta_0 =1.$
Fact B. If $|x-c| \lt 1$, then $\displaystyle \bigl|x^2 -c^2 \bigr| \leq \bigl(1+2 |c| \bigr) |x-c|$.
To toggle the proof of Fact B clickIn this proof we use the triangle inequality: For arbitrary real numbers $u, v \in \mathbb R$ we have \[ | u+v | \leq | u | + | v |. \] Assume that $|x-c| \lt 1.$ Then \[ | x+c | = | x-c + 2 c | \leq | x-c | + 2 | c | \leq 1 + 2 | c |. \] Using the difference of squares identity and properties of the absolute value we have \[ \displaystyle \bigl| x^2 - c^2 \bigr| = | (x - c) (x + c) | = | x - c | \, | x + c | . \] Since \[ | x+c | \leq 1 + 2 | c |, \] the last equality yields \[ \displaystyle \bigl| x^2 - c^2 \bigr| \leq \bigl( 1 + 2 | c | \bigr) | x - c | . \]Next we prove II. Let $\epsilon \gt 0$ be arbitrary. To find $\delta(\epsilon) \gt 0$ we solve \[ \bigl( 1 + 2 | c | \bigr) | x - c | \lt \epsilon \] for $|x-c|$. We get \[ |x-c| \lt \frac{ \epsilon }{1 + 2 | c |}. \] To assure that Fact B is valid we need \[ |x-c| \lt 1. \] Therefore we set \[ \delta(\epsilon) = \min\left\{ 1, \frac{\epsilon}{1 + 2 | c |} \right\}. \] With this $\delta(\epsilon)$ we can prove$ \displaystyle |x-c| \lt \min\left\{ 1, \frac{ \epsilon }{1 + 2 \bigl| c \bigr|} \right\}$ implies $\displaystyle \bigl| x^2 - c^2 \bigr| \lt \epsilon$.
-
Example 3. Consider the function $f(x) = \sqrt{x}$ defined for all $x \in (0,+\infty).$ Prove that this function is continuous.Proof. To prove that $f(x) = \sqrt{x}$ defined for all $x \in (0,+\infty)$ is continuous, let $c$ be an arbitrary positive real number. To satisfy I. in the definition we set $\delta_0 =c/2.$
Fact C. If $x \in (0,+\infty)$, then $\displaystyle \bigl| \sqrt{x} - \sqrt{c} \bigr| \leq \frac{1}{ \sqrt{c}} |x-c|$.
To toggle the proof of Fact C clickAssume that $x \in (0,+\infty)$. Then \[ \displaystyle \bigl| \sqrt{x} - \sqrt{c} \bigr| = \left| \bigl(\sqrt{x} - \sqrt{c}\bigr) \frac{\sqrt{x} + \sqrt{c}}{\sqrt{x} + \sqrt{c}} \right| = \frac{ |x - c | }{\sqrt{x} + \sqrt{c}} \leq \frac{1}{\sqrt{c}} |x-c|. \] The last inequality is a consequence of the "pizza-party" principle.Next we prove II. Let $\epsilon \gt 0$ be arbitrary. To find $\delta(\epsilon) \gt 0$ we solve \[ \frac{1}{\sqrt{c}} |x-c| \lt \epsilon \] for $|x-c|$. We get \[ |x-c| \lt \sqrt{c} \epsilon. \] Now we set \[ \delta(\epsilon) = \min\left\{ \frac{c}{2}, \sqrt{c} \epsilon \right\}. \] With this $\delta(\epsilon)$ we can prove$ \displaystyle |x-c| \lt \min\left\{ \frac{c}{2}, \sqrt{c} \epsilon \right\}$ implies $\displaystyle \bigl| \sqrt{x} - \sqrt{c} \bigr| \lt \epsilon$.
- In class I created this Mathematica file in which I show how to use Mathematica to confirm that the inequalities that we studied here are valid. The file is called Examples_continuity.nb. The file is created in Mathematica version 8.
- Today I commented about yesterday's post and we did two examples of limits from my Limits webpage.
- We did the definition of $\displaystyle \lim_{x\to c}f(x) = L$ and several examples today. Remember that I wrote a webpage dedicated to Limits.
-
Today I got inspired to introduce
Lipschitz condition. Let $c \in \mathbb R$ and $D \subset \mathbb R$. Let $f : D \to \mathbb R$ be a function. We say that $f$ satisfies Lipschitz condition at $c$ if there exist $\delta_0 \gt 0$, $K \gt 0$ and $L \in \mathbb R$ such that $(c-\delta_0,c)\cup (c,c+\delta_0) \subseteq D$ and for every $x \in (c-\delta_0,c)\cup (c,c+\delta_0)$ we have \[ \bigl| f(x) - L \bigr| \leq K\, |x-c|. \]It was not my intention to start with abstract theorems this early. However, I hope that you will understand the meaning of the following theorem:Theorem. Let $c \in \mathbb R$ and $D \subset \mathbb R$. Let $f : D \to \mathbb R$ be a function. Assume that $f$ satisfies Lipschitz condition at $c$ with constants $\delta_0 \gt 0$, $K \gt 0$ and $L \in \mathbb R$. Then \[ \lim_{x\to c}f(x) = L. \]Can you prove this theorem?
- I shared the following hint with several of you during my office hours. For Problem 3 on Assignment 2 it is very useful to prove the following implication: Let $a, b \in \mathbb R$. The following implication holds \[ a \in [-1,1] \ \ \text{and} \ \ b \in (-1,1) \quad \text{imply} \quad 1 + ab \in(0,2) \ \ \text{and} \ \ \frac{a+b}{1+ab} \in [-1,1]. \]
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For Problem 4 on Assignment 2 you need to provide approximate solutions to some equations which do not have explicit exact solutions. I demonstrated how to do this in Mathematica version 8. In class I created
this Mathematica file. The file is called Find_root.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. You can use your U:\ directory. That is your private directory at Western. Make sure that you save the file with exactly the same name. After saving the file you can open it with Mathematica 8. You will find Mathematica 8 on all computers in BH 215 and in the Math Center. Look for Mathematica in the Start menu in the directory named Wolfram Mathematica. Open Mathematica first, then open Find_Root.nb from Mathematica.
- More information on how to use Mathematica version 8 you can find on my Mathematica 8 page.
- We finished discussing the limit at $+\infty$ today; see Limits page.
-
Here are some general remarks about your papers that I finished grading today:
- Presentation is important. Please make your papers professionally presented.
- Make your response to a problem well organized and clear. If a problem contains a specific question, it is appropriate to start with a specific answer to that question. Then proceed with a justification of your answer.
- Presentation of your solution and the reasoning process with which you arrived to that solution are completely different tasks. After you have solved a problem, think creatively of a best way of presenting that solution.
- Make clear which facts you use in your argument. The facts which you use and which are not well-known need to be justified in the paper.
- I wrote a webpage dedicated to Limits. Before reading this page I present a narrative which might help with the understanding of the formal definition.
- 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 $x \geq 8$.
- 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 (in the definition this is measured by $\epsilon \gt 0$), there exist real numbers $x$ (this is determined by the number $X(\epsilon)$ in the definition) for which that engineer could write $\tanh(x) = 1.$
- We finished discussing the Function page.
- Here is the link to my Math 224 web page with a proof that the graph of $y = \sqrt{x^2-1}$ with $x \geq 1$ when rotated counterclockwise by $\pi/4$ coincides with the graph of $y=1/(2x)$ with $0\lt x \leq \sqrt{2}/2.$ The proof is posted on Tuesday, October 2, 2018.
- Before continuing to read my Function page, try to formulate your own conditions under which a flip of a function $f$ will also be a function.
- Today we talked about the rigourous definition of a function. I wrote my Function page. It has all the topics that we will study. I am working on improving the presentation.
- We also talked about the basics of mathematical logic. The concepts of implication, its contrapositive and its converse. I think that I offered a strong evidence that an implication and its contrapositive are equivalent. All these consents were illustrated with my favorite example of an implication: If it rains, then the Red Square is wet.
- It is a good idea to familiarize yourself with the software package Mathematica. I demonstrated few calculations in Mathematica version 8. In class I created this Mathematica file. The file is called Basic_calculations.nb. Please open this file in Mathematica and start learning how to use this amazing tool. Please notice that in file names I always use the underscore symbol instead of just a blank space.
- 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 Basic_calculations.nb from Mathematica.
- More information on how to use Mathematica version 8 you can find on my Mathematica 8 page.
- In this class we will use the set notation. Therefore we reviewed some set terminology and notation. The Wikipedia entry Set (mathematics) is a good source. In particular review the following items:
-
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.
-
We introduce one more set operation: Cartesian product of two sets.
- The information sheet
- Some useful Wikipedia links (You are familiar with most of these concepts. However, it is very useful to read Wikipedia entries and enjoy the breadth and depth of their scope.):
- 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.
-
Surprisingly, the question: What is the definition of a real number? is not easy to answer.
- 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.
- Notice that the only specific numbers that appear in the axioms of the set of real numbers are the number $1$ and the number $0$.
- I handed out the first assignment today. The assignment is due in class on Friday, October 5. The class on Friday, October 5 will be shorter than usual. The class will end at 11:40am.
- When writing up your assignment have in mind the following: 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" might lead to interesting articles on this topic.