- I updated the list of topics that we covered in the class and that can appear on the final exam.
- Today we deduced the formula for the surface area of the surface of revolution produced by the graph of $y = f(x)$, $x \in (a,b)$, as it rotates about the $x$-axis. Here we assume that $f(x) \geq 0$ for all $x \in (a,b)$.
-
A very nice vase can be produced by rotating the graph of $y = 2+\sin x$, $x \in (0,2 \pi)$, about the $x$-axis. Below I illustrate how the approximation process works for the surface area of this vase.
Place the cursor over the image to start the animation.
-
The animation below illustrates how revolution creates a vase.
Place the cursor over the image to start the animation.
- The surface area of the surface of revolution created by the graph of $y = f(x)$, $x \in (a,b)$, as it rotates about the $x$-axis is given by \[ \int_a^b 2\pi f(x) \sqrt{1+\bigl(f'(x)\bigr)^2} dx. \] Thus the surface area of the vase above is \[ \int_0^{2 \pi} 2\pi \bigl(2+\sin x \bigr) \sqrt{1+\bigl(\cos x\bigr)^2} dx = 4 \pi \int_0^{2 \pi} \sqrt{1+\bigl(\cos x\bigr)^2} dx. \] The last integral Mathematica evaluates as 16 2^(1/2) Pi EllipticE[1/2] which is approximately equal to $96.012$.
- An important application of definite integrals is the calculation of the length of a curve given by its parametric equations $x = u(t), y= v(t)$, $t\in (a,b)$. We deduced that the length of such curve is given by \[ \int_a^b \sqrt{\bigl(u'(t)\bigr)^2 + \bigl(v'(t)\bigr)^2} dt. \]
-
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 \[ x = u(t) = (1-\cos t)\cos t, \quad y = v(t) = (1-\cos t) \sin t \quad \text{where} \quad 0 \leq t \leq 2\pi. \]
-
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.
-
It is almost irresistible at this point not to think of the surface of revolution created by the cardioid. Consider the cardioid deduced in class
\[
x = u_1(t) = 2 \cos(t) -\cos(2t), \quad y = v_1(t) = 2 \sin(t) -\sin(2t).
\]
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
\[
2 \pi \int_0^\pi v_1(t) \sqrt{\bigl(u_1'(t)\bigr)^2 + \bigl(v_1'(t)\bigr)^2 } dt = 8 \pi \int_0^\pi \bigl(2 \sin(t) -\sin(2t)\bigr) \sin(t/2) dt.
\]
Here we used the fact that for $t \in (0,\pi)$ we have
\[
\sqrt{\bigl(u_1'(t)\bigr)^2 + \bigl(v_1'(t)\bigr)^2 } = 4 \sin(t/2),
\]
what was proved in class. 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.
- Nice interesting exercises with applications of definite integrals are at the end of Section 6.1 in Notes on Calculus. In particular I liked 10, 12, 13, 15, 16.
-
In Problem 4 on Assignment 6 you are asked to determine the volume enclosed by two and three specially positioned cylinders. A related question is to determine the volume of the intersection of two and three specially positioned cylinders. I am posting images of the solids formed by the intersection of two and three cylinders. It is remarkable how easy it is to produce these images in Mathematica; one uses the command RegionPlot3D[]. Specifically, for the intersection of two cylinders I used
RegionPlot3D[
and for for the intersection of three cylinders I used
And[y^2 + z^2 < 1, z^2 + x^2 < 1], {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, PlotPoints -> {200, 200, 200}, Mesh -> None, PlotStyle -> {Opacity[1]}, Ticks -> {Range[-2, 2, 1], Range[-2, 2, 1], Range[-2, 2, 1]}, ImageSize -> 500
]RegionPlot3D[
And[x^2+y^2 < 1, y^2 + z^2 < 1, z^2 + x^2 < 1], {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, PlotPoints -> {200, 200, 200}, Mesh -> None, PlotStyle -> {Opacity[1]}, Ticks -> {Range[-2, 2, 1], Range[-2, 2, 1], Range[-2, 2, 1]}, ImageSize -> 500
] -
The following six examples illustrate solids in Problem 2 on Assignment 6. They inspired by the subsection "Volumes of Regions of Known Cross-Section" of Section 8.2. Here the base is always a unit disk and the cross-sections are
regular polygons: equilateral triangle, square, regular pentagon, regular hexagon, regular heptagon and regular octagon. There are twelve animations below.
Place the cursor over the image to start the animation.
A cross-section is an equilateral triangle.
Place the cursor over the image to start the animation.
A cross-section is a square.
Place the cursor over the image to start the animation.
A cross-section is a regular pentagon.
Place the cursor over the image to start the animation.
A cross-section is a regular hexagon.
Place the cursor over the image to start the animation.
A cross-section is a regular heptagon.
Place the cursor over the image to start the animation.
A cross-section is a regular octagon.
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- I mentioned in class that my guiding slogan for an integral calculus course could be "How long is a smile?" As one of the applications of definite integrals we will learn how to calculate the length of a given graph of a function. The slogan is inspired by the fact that a graph of a parabola can be used as a smile in a smiley face, as is shown in the animations below.
Place the cursor over the image to start the animation.
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- Here is a list of topics that will be covered on the exam on Friday.
- Today in class I created this Mathematica notebook with calculations of approximations of a definite integral based on Left, Right, Middle Riemann sums and on trapezoidal and Simpson's rule.
- Here we are dealing with definite integrals. Each definite integral is a real number. I consider it important to always distinguish between the exact value of certain number and its approximation. In the textbook Calculus: Single Variable, 5th Edition, Deborah Hughes-Hallett, Andrew M. Gleason, W at all, the authors sometimes do not make this distinction. For example on page 36 they write: $\cos\bigl(\pi/5\bigr) = 0.809$. I consider this unacceptable. Interestingly, they never write $\pi = 3.14159$; they respect $\pi$, but \[ \cos\bigl(\pi/5\bigr) = \frac{1+\sqrt{5}}{4} = \frac{\phi}{2} \] did not earn authors' respect. Here $\phi = \bigl(1+\sqrt{5}\bigr)/2$ is the golden ratio. An appropriate way to write decimal approximation is \[ \pi \approx 3.14159265359 \quad \text{or} \quad \pi = 3.14159265359\ldots \] Here is a proof of the fact that $\cos\bigl(\pi/5\bigr)$ is exactly one half of the golden ratio.
- We covered much of the material listed here. We will spend the rest of the week doing problems and reviewing these topics. Reading from the Notes on Calculus: Sections 5.1, 5.2 and 5.5. Reading and exercises from the Notes on Calculus: Section 5.3, do exercises 1(b), 8, 10. Section 5.4 up to 5.4.4, do exercises 1, 3, 4, and in particular 6 which has a typo: $c$ needs to be replaced by $f'(c)$ in 5.4.3. Section 5.6, do the exercises 1, 2, 3, 4, 5. Section 5.7, do exercises 1, 2, 3. Section 5.7, do exercises 1, 2, 3, 4, 5, 6 (answer this in two different ways), 9, 10, 11.
- In this item we list several definitions and properties of defined objects.
- 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, \[ \|\mathcal P \| = \max_{1\leq j \leq n} \bigl(x_j - x_{j-1}\bigr). \]
- A partition $\bigl(x_0,x_1,\ldots,x_n\bigr)$ of $[a,b]$ together with a sequence $\bigl(t_1,\ldots,t_n\bigr)$ such that \[ t_j \in \bigl[ x_{j-1}, x_j \bigr] \quad \text{for all} \quad j \in \{1,\ldots,n\} \] is called a tagged partition of the interval $[a,b]$. This tagged partition is denoted by $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$.
- Let $f:[a,b] \to \mathbb R$ be a function defined on $[a,b]$ and let $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ be a tagged partition of $[a,b]$. The sum \[ \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \] is called the Riemann sum of $f$ relative to the tagged partition $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ of $[a,b]$.
- Let $f:[a,b] \to \mathbb R$ be a bounded function defined on $[a,b]$ and let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ be a partition of of $[a,b]$. For every $j \in \{1,\ldots,n\}$ denote by $m_j$ the greatest number such that $m_j \leq f(x)$ for every $x \in \bigl[x_{j-1},x_j \bigr]$ and denote by $M_j$ the least number such that $f(x) \leq M_j$ for every $x \in \bigl[x_{j-1},x_j \bigr]$. Then the sums \[ L(f,\mathcal P) = \sum_{j=1}^n m_j \bigl(x_j - x_{j-1}\bigr) \qquad \text{and} \qquad U(f,\mathcal P) = \sum_{j=1}^n M_j \bigl(x_j - x_{j-1}\bigr) \] are, respectively, called lower Riemann sum of $f$ relative to the partition $\mathcal P$ and upper Riemann sum of $f$ relative to the partition $\mathcal P$.
- Let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ and let $\bigl(x_0,x_1,\ldots,x_n; t_1,\ldots,t_n\bigr)$ be a accompanying tagged partition of $[a,b]$. Then \[ \tag{LRU} L(f,\mathcal P) \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \leq U(f,\mathcal P). \]
- Let $f:[a,b] \to \mathbb R$ be a function defined on $[a,b]$ and let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ be a partition of of $[a,b]$. The sum \[ {\rm Mi}(f,\mathcal P) = \sum_{j=1}^n f\left(\tfrac{x_{j-1}+x_j}{2}\right) \bigl(x_j - x_{j-1}\bigr) \] is called the middle Riemann sum of $f$ relative to the partition $\mathcal P$; the sum \[ {\rm Le}(f,\mathcal P) = \sum_{j=1}^n f(x_{j-1}) \bigl(x_j - x_{j-1}\bigr) \] is called the left Riemann sum of $f$ relative to the partition $\mathcal P$; ; the sum \[ {\rm Ri}(f,\mathcal P) = \sum_{j=1}^n f(x_{j}) \bigl(x_j - x_{j-1}\bigr) \] is called the right Riemann sum of $f$ relative to the partition $\mathcal P$
- Let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ be a partition of $[a,b]$. A particular case of inequality (LRU) is \[ \tag{LMU} L(f,\mathcal P) \leq {\rm Mi}(f,\mathcal P) \leq U(f,\mathcal P). \]
- 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 \[ \max_{1\leq j \leq n} \bigl(x_j - x_{j-1}\bigr) \lt \delta(\epsilon) \qquad \text{implies} \qquad \left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - S \right| \lt \epsilon. \] If $f: [a,b] \to \mathbb R$ is Riemann integrable on $[a,b]$, then the number $S$ is called the definite integral of $f$ on $[a,b]$; the Riemann integral of $f$ on $[a,b]$ is denoted by \[ \int_a^b \!\! f(x) dx. \]
- Exercise. Let $C \in \mathbb R$. Prove that the constant function $f(x) = C$, $x \in \mathbb R$, 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 hint $f(x) = x$. Let $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ of $[a,b]$ be an arbitrary partition of $[a,b]$. Then we have \[ U(f,\mathcal P) + L(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j + x_{j-1} \bigr) \bigl(x_j - x_{j-1}\bigr) = \sum_{j=1}^n \bigl((x_j)^2 - (x_{j-1})^2 \bigr) = b^2 - a^2. \] Also \[ U(f,\mathcal P) - L(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j - x_{j-1} \bigr) \bigl(x_j - x_{j-1}\bigr) \leq \| \mathcal P \| \sum_{j=1}^n \bigl( x_j - x_{j-1} \bigr) = (b - a) \|\mathcal P\|. \] Adding the equality \[ U(f,\mathcal P) + L(f,\mathcal P) = b^2 - a^2. \] and the inequality \[ U(f,\mathcal P) - L(f,\mathcal P) \leq (b - a) \|\mathcal P\|. \] we get \[ 2 U(f,\mathcal P) \leq b^2 - a^2 + (b - a) \|\mathcal P\| \] and, subtracting them we get \[ 2 L(f,\mathcal P) \geq b^2 - a^2 - (b - a) \|\mathcal P\|. \] Together with inequality (LRU) we thus have \[ \frac{b^2 - a^2}{2} - \frac{b - a}{2} \|\mathcal P\| \leq L(f,\mathcal P) \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \leq U(f,\mathcal P) \leq \frac{b^2 - a^2}{2} + \frac{b - a}{2} \|\mathcal P\|. \] This hence proves \[ \left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - \frac{b^2 - a^2}{2} \right| \leq \frac{b - a}{2} \|\mathcal P\|. \] Thus, if we choose a partition $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ such that $\| \mathcal P \| \lt 2 \epsilon /(b-a)$, then we must have \[ \left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - \frac{b^2 - a^2}{2} \right| \lt \epsilon. \]
Remark. In the above proof we used the fact that for the function $f(x) = x$ we have $L(f,\mathcal P) = {\rm Le}(f,\mathcal P)$ and $U(f,\mathcal P) = {\rm Ri}(f,\mathcal P)$. Also the identity \[ U(f,\mathcal P) + L(f,\mathcal P) = {\rm Ri}(f,\mathcal P) + {\rm Le}(f,\mathcal P) = b^2 - a^2 \] shows that the average of the left and right Riemann sums gives the exact value of the integral. This is the case for every linear function $f(x) = m x + k$. -
Exercise. Prove that the square function $f(x) = x^2$, $x \in \mathbb R$, is integrable on any interval $[a,b]$ with $0 \leq a \lt b$ and
\[
\int_a^b \! x^2 \, dx = \frac{b^3 - a^3}{3}.
\]
Proof. In this hint $f(x) = x^2$ and $0 \leq a \lt b$. Based on the identity
\[
\left(u^2 + v^2 + 4 \left( \frac{u+v}{2} \right)^2 \right) (v - u) = 2\bigl( v^3 - u^3 \bigr)
\]
which holds for arbitrary reals $u$ and $v$, one can prove that for an arbitrary partition $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ of $[a,b]$ we have
\[ \tag{1}
L(f,\mathcal P) + U(f,\mathcal P) + 4 {\rm Mi}(f,\mathcal P) = 2 \bigl( b^3 - a^3 \bigr).
\]
Since $0 \leq a$ we have $x_j^2 - x_{j-1}^2 \gt 0$ for all $j \in \{1,\ldots,n\}$. Therefore,
\[
U(f,\mathcal P) - L(f,\mathcal P) = \sum_{j=1}^n \bigl( x_j^2 - x_{j-1}^2 \bigr) \bigl(x_j - x_{j-1}\bigr) \leq \| \mathcal P \| \sum_{j=1}^n \bigl( x_j^2 - x_{j-1}^2 \bigr) = (b^2 - a^2) \|\mathcal P\|.
\]
Thus,
\[ \tag{2}
U(f,\mathcal P) - L(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|.
\]
Using inequality (LMU), (2) implies
\[ \tag{3}
U(f,\mathcal P) - {\rm Mi}(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|
\]
and
\[ \tag{4}
{\rm Mi}(f,\mathcal P) - L(f,\mathcal P) \leq (b^2 - a^2) \|\mathcal P\|.
\]
Adding inequalities (1), (2) and (3) multiplied by $4$ and dividing the resulting inequality by 6 we get
\[
U(f,\mathcal P) \leq \frac{b^3 - a^3}{3} + \frac{5}{6}(b^2 - a^2) \|\mathcal P\|
\]
Adding inequalites (1), (2) multiplied by $-1$ and (3) multiplied by $-4$ and dividing the resulting inequality by 6 we get
\[
L(f,\mathcal P) \geq \frac{b^3 - a^3}{3} - \frac{5}{6}(b^2 - a^2) \|\mathcal P\|.
\]
Together with inequality (LRU) we thus have
\[
\frac{b^3 - a^3}{3} - \frac{5}{6}(b^2 - a^2) \|\mathcal P\| \leq L(f,\mathcal P) \leq \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) \leq U(f,\mathcal P) \leq \frac{b^3 - a^3}{3} + \frac{5}{6}(b^2 - a^2) \|\mathcal P\|,
\]
for arbitrary choice of $t_j \in [x_{j-1},x_j]$ for $j \in \{1,\ldots,n\}$. This hence proves
\[
\left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - \frac{b^3 - a^3}{3} \right| \leq \frac{5}{6}(b^2 - a^2) \|\mathcal P\|.
\]
Thus, if we choose a partition $\mathcal P = \bigl(x_0,x_1,\ldots,x_n\bigr)$ such that $\| \mathcal P \| \lt \tfrac{6}{5} \epsilon /(b^2-a^2)$, then we must have
\[
\left| \sum_{j=1}^n f(t_j) \bigl(x_j - x_{j-1}\bigr) - \frac{b^3 - a^3}{3} \right| \lt \epsilon.
\]
Remark. In the above proof we used the fact that for the function $f(x) = x^2$ and for $a \geq 0$ we have $L(f,\mathcal P) = {\rm Le}(f,\mathcal P)$ and $U(f,\mathcal P) = {\rm Ri}(f,\mathcal P)$. Also the identity \[ 2 \bigl( b^3 - a^3 \bigr) = L(f,\mathcal P) + U(f,\mathcal P) + 4 {\rm Mi}(f,\mathcal P) = {\rm Le}(f,\mathcal P) + {\rm Ri}(f,\mathcal P) + 4 {\rm Mi}(f,\mathcal P) \] shows that the exact value of the integral of $f$ over $[a,b]$ is a weighted average of the left, the right and the middle Riemann sum: \[ \int_a^b \! f(x) dx = \frac{1}{6} {\rm Le}(f,\mathcal P) + \frac{1}{6} {\rm Ri}(f,\mathcal P) + \frac{2}{3} {\rm Mi}(f,\mathcal P) \] -
Theorem. If $: f[a,b]\to \mathbb R$ is Riemann integrable on $[a,b]$ and $\alpha \in \mathbb R$, then the function $(\alpha \, f)$ defined as $(\alpha f)(x) = \alpha f(x)$ is also integrable on $[a,b]$ and
\[
\int_a^b \! \alpha f(x) \, dx = \alpha \int_a^b \! f(x) \, dx.
\]
-
Theorem. If $f: [a,b]\to \mathbb R$ and $g: [a,b]\to \mathbb R$ are Riemann integrable on $[a,b]$, then the function $f+g$ defined as $(f + g)(x) =f(x) + g(x)$ is also integrable on $[a,b]$ and
\[
\int_a^b \! \bigl( f(x) + g(x) \bigr) \, dx = \int_a^b \! f(x) \, dx + \int_a^b \! g(x) \, dx.
\]
-
Theorem. Let $f: [a,b]\to \mathbb R$ be a bounded function. The function $f$ is Riemann integrable on $[a,b]$ if and only if there exists a sequence of partitions $\mathcal P_n$ of $[a,b]$ such that
\[
\lim_{n\to \infty} \bigl( U(f,\mathcal P_n) - L(f,\mathcal P_n) \bigr) =0.
\]
In this case
\[
\int_a^b \! f(x) \, dx = \lim_{n\to \infty} U(f,\mathcal P_n) = \lim_{n\to \infty} L(f,\mathcal P_n).
\]
Remark. This theorem shows that our definition of an integrable function is equivalent to the definition in the Notes on Calculus.
Remark. It is interesting to point out that this theorem can be used to give a different proof that the function $f(x) = x^2$ is integrable. We did this in class for the interval $[0,1]$. This proof is based on the fact that $1^2 + 2^2 + \cdots + n^2 = n(n+1)(2n+1)/6$. -
Theorem. If $f: [a,b]\to \mathbb R$ is a monotinic function, then $f$ is Riemann integrable on $[a,b]$.
-
Theorem. If $f: [a,b]\to \mathbb R$ is a continuous function, then $f$ is Riemann integrable on $[a,b]$.
-
Fundamental Theorem of Calculus. Let $F: [a,b]\to \mathbb R$ and $f: [a,b]\to \mathbb R$ be functions with the following three properties:
- $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$.
- $f$ is Riemann integrable on $[a,b]$.
- $F'(x) = f(x)$ for all $x \in (a,b)$.
- Today we talked about parametric equations of curves, Section 3.7 in the Notes on Calculus and Section 4.8 in the Calculus textbook (do 1-4, 6, 7, 10, 18, 20, 21, 24, 31, 34, 39, 40, 42, 46).
- Today in class I created this Mathematica notebook with a manipulation of a cycloid.
- I used this Mathematica notebook to create all the figures for this website. For the figures with many osculating circles I used this Mathematica notebook.
- I wrote a webpage introducing Higher_order_approximations of differentiable functions. In particular I introduce the concept of an osculating circle for a graph of function. This is done in complete analogy with how we introduced the concept of the tangent line.
- We discussed many optimization problems from the Notes on Calculus Section 4.4 and the Calculus textbook Section 4.4.
- I pointed out that Problem 1 on Assignment 5 is not stated well. Its statement would be improved if we add the definition of a crease: By definition, a crease is a paper fold which has one end-point on the bottom edge of the paper. Also, this problem would be clearer if we add the adjective "vertical" between the adjective "opposite" and the noun "edge".
- This is an updated Mathematica file which I emailed to you yesterday. It should be helpful for Problem 3 (d) on Assignment 4. The name of the file is Finding_inverses.nb.
-
Below is an animation related to Problem 1 on Assignment 4. 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.
- I am hoping that you can easily calculate the important transition points (marked by red disks) in the above animation. This calculation does not involve the discussion that I presented today in class.
- When choosing colors for geometric objects I always try to follow some logic. My guide is that Mathematica uses Red-Green-Blue color function in which the value for each of the components is a number between 0 and 1, including 0 and 1. More about this you can find on my linear algebra page. In the above animation I had three normal lines and three tangents. I decided to use the colors complementary to Red, Green, Blue, that is Cyan, Magenta, Yellow for the tangents and the colors which are half-way between these colors and Black, that is Teal, Purple, Olive for the normals.
- I updated the Mathematica file which I posted yesterday.
-
Today we did implicit differentiation. The relevant section in the Calculus textbook is 3.7 (do 30, 31, 32, 34, 35). Problem 34 asks for a common tangent line to two circles. Inspired by this one can ask for a common tangent lines to different kind of functions. For example
- Find the common tangent lines to the graphs of $y = 2 x^2$ and $y = -x^2$.
-
In class we looked at the famous curve Folium of Descartes. The folium of Decartes is the following subset of $\mathbb R^2$:
\[
F = \bigl\{ (x,y) \in \mathbb R^2 \, | \, x^3 + y^3 = 6 x y \bigr\}.
\]
Clearly $(3,3) \in F$. It is also clear that $F$ is symmetric with respect to the line $y=x$. That is, we have $(a,b) \in F$ if and only if $(b,a) \in F$. This suggests a natural question:
- Find the involution function $y = f(x)$ such that $3 = f(3)$ and whose graph is a subset of $F$.
Other interesting questions related to $F$ are- Find all 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 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 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 Decartes $x^3+y^3 = 6xy$
- The linked Mathematica file contains some examples of code that might be useful for Problem 3 on Assignment 4. The name of the file is Circles_in_graphs.nb. As before download this file to a computer which has Mathematica 8 on it and open the file from Mathematica.
- We finished Chapter 3, Computing Derivatives in Notes on Calculus. Do 1, 3, 4, 5, 6, 9, 11, 12 in Section 3.5. The relevant sections in the textbook are 3.1 (do 57, 58, 68, 69), 3.2 (do 44, 45, 46, 47) , 3.3 (do 31-39, 55, 56, 57, 58, 59, 60), 3.4 (do 51-58,61, 62, 79, 85), 3.5 (do 42, 50, 51), 3.6 (do 43, 44, 45, 46, 47-55)
-
Below is the animation for Problem 2 on Assignment 4
Place the cursor over the image to start the animation.
-
To clarify the question (d) in Problem 3 on Assignment 4 I post the relevant figures with inscribed circles spaced evenly along the diagonal.
The functions $x^2$ and $\sqrt{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs.
The functions $x^3$ and $\sqrt[3]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs.
The functions $x^4$ and $\sqrt[4]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs.
The functions $x^5$ and $\sqrt[5]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs.
-
Today we calculated some specific derivatives. To prove that the derivative of $\sin x$ is $\cos x$ we need the following limit
\[
\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}.
\]
Similar to what we did on September 30, this will be proved by using a geometric argument deduced from the unit circle. In the figure below consider $x \in (0,\pi/2)$ and the lengths of the line segments $\overline{AB}$, $\overline{AC}$ and the length of the circular arc from $A$ to $C$. We see that these three lengths are listed in increasing order and they calculate to be:
\[ \tag{#}
\sin x \lt \sqrt{2 - 2 \cos x} \lt x.
\]
Here the length of $\overline{AC}$ is calculated by the Pythagorean theorem
\[
\sqrt{(\sin x)^2 +(1- \cos x)^2} = \sqrt{(\sin x)^2 +1- 2 \cos x +(\cos x)^2} = \sqrt{2 - 2 \cos x}
\]
Squaring the inequalities in (#) we get \[ (\sin x)^2 \lt 2 - 2 \cos x \lt x^2. \] Now, dividing the last inequalities by the positive number $2 x^2$ we get \[ \frac{1}{2} \left(\frac{\sin x}{x}\right)^2 \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}. \] On September 30 we proved \[ 1 - x \lt \frac{\sin x}{x}. \] Therefore \[ \frac{1}{2} \left(1-x\right)^2 \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}. \] Since \[ \frac{1}{2} - x \lt \frac{1}{2} \left(1-x\right)^2, \] we conclude \[ \frac{1}{2} - x \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}. \] From the last displayed relationship we deduce that \[ \frac{1}{2} - \frac{1 - \cos x}{x^2} \lt x. \] For $x \in (-\pi/2,0)$ we have \[ \frac{1}{2} - \frac{1 - \cos x}{x^2} \lt |x|. \] Thus, for all $x \in (-\pi/2,\pi/2)$ we have \[ \left| \frac{1}{2} - \frac{1 - \cos x}{x^2} \right| \lt |x|. \] This shows that \[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}. \]
Figure for $\displaystyle \lim_{x\to 0} \frac{1- \cos x}{x^2} = \frac{1}{2}$
- To prove that \[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0. \] we notice that for $x \in (0,\pi/2)$ we proved \[ \frac{1}{2} \left(1-x\right)^2 \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}. \] Hence for $x \in (0,\pi/2)$ we have \[ 0 \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2} \] and consequently \[ 0 \lt \frac{1 - \cos x}{x} \lt \frac{1}{2} x. \] Thus, for all $x \in (-\pi/2,\pi/2)$ we have \[ 0 \lt \left| \frac{1 - \cos x}{x} \right| \lt \frac{1}{2} |x|. \] This proves that \[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0. \]
- Below are four figures that will appear on the next assignment. The problem consists in finding equations of all circles that touch the given function and its inverse. Then you will be asked to find the maximal disk that touches both graphs.
The functions $x^2$ and $\sqrt{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs.
The functions $x^3$ and $\sqrt[3]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs.
The functions $x^4$ and $\sqrt[4]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs.
The functions $x^5$ and $\sqrt[5]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs.
-
Today we did 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 $(a,b)$.
-
The proof of the Mean Value Theorem uses 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 $(a,b)$,
- $f(a) = f(b)$.
-
The proof of Rolle's Theorem uses Fermat's Interior Extremum Theorem.
- In class we proved Rolle's Theorem and the Mean Value Theorem.
-
Important consequences of the Mean Value Theorem are the following statements.
Constant Function Theorem. Let $a, b \in \mathbb R$, $a \lt b$. Let $f : (a,b) \to \mathbb R$ be a function. If $f$ is differentiable on $(a,b)$ and $f'(x) = 0$ for all $x \in (a,b)$, then there exists $K\in \mathbb R$ such that $f(x) = K$ for all $x \in (a,b)$.
- If $f'(x) \gt 0$ for all $x \in (a,b)$, then $f$ is increasing on $(a,b)$.
- If $f'(x) \lt 0$ for all $x \in (a,b)$, then $f$ is decreasing on $(a,b)$.
- If $f'(x) \geq 0$ for all $x \in (a,b)$, then $f$ is nondecreasing on $(a,b)$.
- If $f'(x) \leq 0$ for all $x \in (a,b)$, then $f$ is nonincreasing on $(a,b)$.
Racetrack Principle. Let $a, b \in \mathbb R$ and $a \lt b$. Let $f : [a,b] \to \mathbb R$ and $g : [a,b] \to \mathbb R$ be functions which satisfy the following three conditions:- $f$ and $g$ are continuous on the closed interval $[a,b]$,
- $f$ and $g$ are differentiable on the open interval $(a,b)$,
- $f'(x) \leq g'(x)$ for all $x \in (a,b)$.
- If $f(a) = g(a)$, then $f(x) \leq g(x)$ for all $x \in (a,b)$.
- If $f(b) = g(b)$, then $f(x) \geq g(x)$ for all $x \in (a,b)$.
- The relevant section in the Calculus textbook is Section 3.10. Do the exercises 1-9 and problems 12-16, 19-23, 26, 27, 28. The relevant section in the Notes on Calculus is Section 2.7. Do the exercises 3-6.
- Here is a list of topics that will be covered on the exam on Wednesday.
- We started discussing derivatives on Friday. We gave the formal definition of a derivative of a function at a point and the definition of differentiability of a function on an interval. We proved that a function which is differentiable at a point $a$ is continuous at the same point. See the proof of Theorem 2.1 on page 106 in the Calculus textbook. We also considered standard examples of derivatives velocity and density. This is covered in Sections 2.1, 2.2 and 2.3 in Notes on Calculus. Do the exercises: in 2.2: 2, 3, 4, 5, 7, 8; in 2.3: 1, 2. Relevant Sections in the Calculus textbook are 2.1, 2.2, 2.3, 2.4 and 2.6. Do the exercises: in 2.1: 20, 21, 25, 26; in 2.2: 13, 15, 16, 17, 33, 34, 36; in 2.3: 40, 41, 42; in 2.6: 16
- Today we talked about local linearity property of differentiable functions. See Notes on Calculus Section 2.4. Do exercises 1, 3.
- Let $f$ be a differentiable function at a point $a$. Note the definition of the tangent line to the graph of $f$ at the point $\bigl(a,f(a)\bigr)$ is the line given in point-slope form by \[ y = f(a) +f'(a)(x-a). \] It follows from the definition of the derivative of $f$ at $a$ that the tangent line to the graph of $f$ at the point $\bigl(a,f(a)\bigr)$ is the unique line with the property that \[ f(a+h) -\bigl(f(a) + f'(a) h \bigr) = o(h) \quad \text{as} \quad h \to 0. \] In other words, the tangent line to the graph of $f$ at the point $\bigl(a,f(a)\bigr)$ is the unique line with the property that \[ \lim_{h\to 0}\frac{f(a+h) -\bigl(f(a) + f'(a) h \bigr)}{h} = 0. \] Or, expressed in terms of the variable $x$, \[ \lim_{x\to a}\frac{f(x) -\bigl(f(a) + f'(a) (x-a) \bigr)}{x-a} = 0. \] See, pages 109 and 110 in Notes on Calculus.
-
Trying to help you with Problem 2 on Assignment 3 I got myself in a little bit of a problem when I decided to find a unit circle as in the figure below.
the navy blue function is $e^x$, the purple circle is a unit circle
Solution.Denote by $A$ the red point with the coordinates $\bigl(x_0, \exp(x_0) \bigr)$. The dark green line is the tangent line to navy blue graph of $y=e^x$ at the red point $\bigl(x_0, \exp(x_0) \bigr)$. Its equation is \[ y = (x-x_0) \exp(x_0) + \exp(x_0). \] The teal line is the line which is orthogonal to the dark green line at the red point $\bigl(x_0, \exp(x_0) \bigr)$. Its equation is \[ y = -(x-x_0) \exp(-x_0) + \exp(x_0). \] Note- Since the slope of the teal line is $-\exp(-x_0)$ and since the length of the line segment $\overline{AB}$ is $\exp(x_0)$, we conclude that the length of the line segment $\overline{BC}$ is $\exp(2x_0)$.
- Since the slope of the teal line is $-\exp(-x_0)$ and since the length of the line segment $\overline{DE}$ is $1$, we conclude that the length of the line segment $\overline{EC}$ is $\exp(x_0)$.
- Since the coordinates of the point $B$ are $(x_0,0)$, the previous two items yield that the coordinates of the point $C$ are $\bigl(x_0+\exp(2x_0), 0\bigr)$ and the coordinates of the point $E$ are $\bigl(x_0+\exp(2x_0)-\exp(x_0), 0\bigr)$.
- Consequently, the coordinates of the purple center $D$ are \[ \bigl(x_0+\exp(2x_0)-\exp(x_0),1\bigr). \]
- Since the distance between the purple center $D$ and the red point $A$ is $1$ we must have \[ \bigl(\exp(2x_0)-\exp(x_0)\bigr)^2 +\bigl(\exp(x_0) - 1\bigr)^2 = 1. \] Solving this equation for $x_0$ will solve the problem.
- Simplifying the last equation yields \[ \exp(4 x_0)-2 \exp(3 x_0) + \exp(x_0) + \exp(2 x_0) - 2 \exp(x_0) = 0. \] Dividing by $\exp(x_0)$ we get \[ e^{3x_0} - 2 e^{2x_0} + 2 e^{x_0} - 2 = 0. \] Substituting $z = \exp(x_0)$ yields \[ z^3 - 2 z^2 + 2 z -2 = 0. \] This cubic equation has only one real root which is between $1$ and $2$. Mathematica calculates that root to be \[ \frac{1}{3} \left(2-\frac{2}{\sqrt[3]{17+3 \sqrt{33}}}+\sqrt[3]{17+3 \sqrt{33}}\right). \] Hence for the value of $x_0$ we have \[ x_0 = \ln \left(\frac{1}{3} \left(2-\frac{2}{\sqrt[3]{17+3 \sqrt{33}}}+\sqrt[3]{17+3 \sqrt{33}}\right) \right) \approx 0.434175014665. \] With this $x_0$ we can calculate the approximate coordinates of the purple center $D$ to be \[ \bigl( 1.27346177 , 1 \bigr) \]
-
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
- Today in class I created Mathematica file in which I explain how to solve equations in Mathematica. The name of the file is 20151008_Finding_roots.nb. As before download this file to a computer which has Mathematica 8 on it and open the file from Mathematica.
-
I cannot get away from the beauty and utility of the Lambert $W$ function. I have to put three more examples of applications of the Lambert $W$ functions.
-
The reasoning presented in this problem will be used in two following problems.
Let $a \in (-1,0)$. Find real $x$ distinct from $a$ such that \[ x e^x = ae^a. \] Think about this for a minute. Please answer this yourself without looking in the spoiler marked by "more". $x = W_{-1}(a e^a)$
Let $a \in (-\infty,-1)$. Find real $x$ distinct from $a$ such that \[ x e^x = a e^a. \] Think about this for a minute. Please answer this yourself without looking in the spoiler marked by "more". $x = W_{0}(a e^a)$ -
Consider the function $f: \mathbb R \to \mathbb R$ defined piecewise by $f(x) = xe^x / \bigl( e^x - 1 \bigr)$, $x \in \mathbb R\!\setminus\!\{0\}$ and $f(0) = 1$. Using derivatives we can prove that this function is increasing. Using limits we can prove that this function is continuous. In fact, it is a bijection $f: \mathbb R \to (0,+\infty)$.
Find the formula for its inverse.
Let $y \in (0,+\infty)$ and $y \neq 1$. We will solve the equation \[ \frac{xe^x}{e^x - 1} = y \] for $x \in \mathbb R\setminus \{0\}$. Solving proceeds by eliminating the fraction: \[ x e^x = y e^x - y. \] Now it is natural to group terms with $e^x$ since our goal is to get an expression of the form $Xe^X$: \[ (x - y) e^x = - y. \] However, to get $Xe^X$ on the left hand side we have to multiply by $e^{-y}$. So, do that: \[ (x - y) e^{x-y} = - y e^{-y}. \] Have in mind that $y$ is a given positive number different from $1$. Thus we have exactly the situation described in the previous problem. We have to distinguish two cases. If $y \in (0,1)$, then the solution for $x-y$ is given by \[ x- y = W_{-1}\bigl( - y e^{-y} \bigr). \] If $y \in (1,+\infty)$, then the solution for $x-y$ is given by \[ x- y = W_{0}\bigl( - y e^{-y} \bigr). \] Hence, the solution is the piece-wise defined function \[ x = \begin{cases} y + W_{-1}\bigl( - y e^{-y} \bigr) & \text{if} \quad y \in (0,1), \\ 0 & \text{if} \quad y = 1, \\ y + W_{0}\bigl( - y e^{-y} \bigr) & \text{if} \quad y \in (1,+\infty). \end{cases} \] -
Consider the function $g: \mathbb R \to \mathbb R$ defined piecewise by $f(x) = \bigl( e^x - 1 \bigr)/x$, $x \in \mathbb R\!\setminus\!\{0\}$ and $f(0) = 1$. Using derivatives we can prove that this function is increasing. Using limits we can prove that this function is continuous. In fact, it is a bijection $f: \mathbb R \to (0,+\infty)$.
Find the formula for its inverse.
The method is very similar to the preceding problem. We use the same tricks. Let $y \in (0,+\infty)$ and $y \neq 1$. We will solve the equation \[ \frac{e^x - 1}{x} = y \] for $x \in \mathbb R\setminus \{0\}$. First eliminate the fraction: \[ e^x -1 = x y. \] Now seeking an expression of the form $Xe^X$ is more obscure. Divide by $y e^x$: \[ \frac{1}{y} - \frac{1}{y} e^{-x} = x e^{-x}. \] Now grouping terms with $e^{-x}$ seems natural: \[ - \Bigl(\frac{1}{y} + x \Bigr) e^{-x} = - \frac{1}{y}. \] However, to get $Xe^X$ on the left hand side we have to multiply by $e^{-1/y}$. So, we do that: \[ - \Bigl(x + \frac{1}{y} \Bigr) e^{-(x+1/y)} = - \frac{1}{y} e^{-1/y}. \] Have in mind that $y$ is a given positive number different from $1$. Again, we have exactly the situation described in the first problem of today. We have to distinguish two cases. If $y \in (0,1)$, then the solution for $-(x+1/y)$ is given by \[ - \Bigl(\frac{1}{y} + x \Bigr) = W_{0}\Bigl( - \frac{1}{y} e^{-1/y} \Bigr). \] If $y \in (1,+\infty)$, then the solution for $x-y$ is given by \[ - \Bigl(\frac{1}{y} + x \Bigr) = W_{-1}\Bigl( - \frac{1}{y} e^{-1/y} \Bigr). \] Hence, the solution is the piece-wise defined function \[ x = \begin{cases} - \frac{1}{y} - W_{0}\Bigl( - \frac{1}{y} e^{-1/y} \Bigr) & \text{if} \quad y \in (0,1), \\ 0 & \text{if} \quad y = 1, \\ - \frac{1}{y} - W_{-1}\Bigl( - \frac{1}{y} e^{-1/y} \Bigr) & \text{if} \quad y \in (1,+\infty). \end{cases} \]
$ X =\mathbb R, \ Y = (0,+\infty), \ f = \bigl\{ \bigl(x, xe^x/(e^x-1) \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
The function $f$ from the example above
$X = \mathbb R, \ Y = (0,+\infty), \ g = \bigl\{ \bigl(x,(e^x - 1)/x \bigr) \ \bigl| \bigr. \ x \in X \bigr\}$
The function $g$ from the example above
-
The reasoning presented in this problem will be used in two following problems.
- Please update your notes on today's lecture. On the next assignment I will assign what we discussed today. I will remind you what we discussed today below. Below I change the notation. In class I used $a \gt 1$ for the base of an exponential function and $\alpha \gt 0$ for the power of a power function. However, now I see that the letters $a$ and $\alpha$ are too similar to each other. Therefore, below I will use $b \gt 1$ to be the base of an exponential function and $a \gt 0$ for the power of a power function. Please adjust your notes accordingly. I will make the appropriate changes in the Mathematica file that I created.
-
Here is a draft of the problem that we discussed today. In this problem you can freely use that the power function $x \mapsto x^a$, $x \geq 0$, is continuous and that the exponential function $x \mapsto b^x$, $x \in \mathbb R$, is also continuous. You can also freely use two Lambert $W$ functions
\[
W_0 : [-1/e, + \infty) \to [-1,+\infty)
\]
and
\[
W_{-1} : (-1/e, 0) \to (-\infty, -1)
\]
which I will review below.
-
Consider real numbers $a \gt 0$ and $b \gt 1$ and the following equation
\[ \tag{*}
x^a = b^x \quad \text{where} \quad x \geq 0.
\]
- Prove that there exists a bijection $\beta: (0,+\infty) \to (1,+\infty)$ such that for every $a \in (0,+\infty)$ the equation \[ \tag{**} x^a = \bigl( \beta(a) \bigr)^x \quad \text{where} \quad x \geq 0, \] has a unique solution. Prove that in this case \[ x^a \leq \bigl( \beta(a) \bigr)^x \quad \text{for all} \quad x \geq 0. \]
- There is something remarkable hiding in the previous question. I hope that you will discover that remarkable fact. In a. we proved that for each $a \gt 0$ equation (**) has a unique solution. For each $a \gt 0$ find that unique solution. Denote that solution by $s(a)$. Do you notice anything remarkable about the function $s(a)$? State it clearly and explain the significance of it in simple terms.
- Prove that for every $a \in (0,+\infty)$ and every real $b$ such that $b \gt \beta(a)$ the equation (*) does not have a solution. Prove that in this case \[ x^a \lt b^x \quad \text{for all} \quad x \geq 0. \]
- Prove that for every $a \in (0,+\infty)$ and every real $b$ such that $b \lt \beta(a)$ the equation (*) has exactly two solutions. Denote these solutions by $x_1$ and $x_2$ with $0 \lt x_1 \lt x_2$. Find explicit formulas for $x_1$ and $x_2$ in terms of the Lambert $W$ functions $W_0$ and $W_{-1}$. Prove \[ x_1 \lt e \lt x_2. \] Prove that \[ x^a \lt b^x \quad \text{for all} \quad x \gt x_2. \]
-
Consider real numbers $a \gt 0$ and $b \gt 1$ and the following equation
\[ \tag{*}
x^a = b^x \quad \text{where} \quad x \geq 0.
\]
- Here are two more important bijections plotted below in navy blue. Their inverses are plotted in maroon. The maroon functions are two branches of the Lambert W function. The formulas for maroon functions are given below graphs.
$ X = [-1,+\infty), \ Y = [-1/e,+\infty), \ W_0 = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
The Lambert $W_0$ function
$X = (-\infty,-1), \ Y = (-1/e,0), \ W_{-1} = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\}$
The Lambert $W_{-1}$ function
- Mathematica's notation for $W_0(x)$ function is ProductLog[x], or, equivalently ProductLog[0,x]. Mathematica's notation for $W_{-1}(x)$ function is ProductLog[-1,x].
- The Wikipidia's page on Lambert W function has a lot of information about this function. However, this page deals with the complex $W$ function so it might be difficult for you to extract useful information.
-
Below are two examples how Lambert W functions are used to solve equations.
- Solve $2^x = 5 x$ for $x \in \mathbb R$. This is how this equation is being solved. Rewrite the equation using the exponential function $e^x$: \[ e^{x \ln 2} = 5 x. \] Next bring the equation to the form $Y = X e^X$ where $Y$ will be a specific number. Then we can use the Lambert functions to solve this equation. Here we will bring the exponential part together with $x$. \[ \frac{1}{5} = x\,e^{-x \ln 2} . \] Now we recognize that our $X$ must be $-x \ln 2$, so we have to multiply the equation by $-\ln 2$: \[ -\frac{\ln 2}{5} = - (x \ln 2) \,e^{-(x \ln 2)}. \] Since $-1/e \lt -(\ln 2)/5 \lt 0$ the above equation has two solutions. The solutions are \[ - (x_1 \ln 2) = W_{0}\bigl(-(\ln 2)/5\bigr) \quad \text{and} \quad - (x_2 \ln 2) = W_{-1}\bigl(-(\ln 2)/5\bigr). \] Hence, \[ x_1 = - \frac{1}{\ln 2} W_{0}\bigl(-(\ln 2)/5\bigr) \approx 0.235456 \quad \text{and} \quad x_2 = - \frac{1}{\ln 2} W_{-1}\bigl(-(\ln 2)/5\bigr) \approx 4.488001. \]
-
Solve $x^x = y$ for $x \gt 0$ and appropriate $y \gt 0$. This is a little trickier. Apply $\ln$ to both sides to get
\[
x \ln x = \ln y.
\]
Rewrite the equation using the exponential function $X \, e^X$:
\[
(\ln x) \, e^{\ln x} = \ln y.
\]
It just happens that this equation is exactly in the form $X\,e^X = Y$. So this equation has a solution if and only if $\ln y \geq -1/e$. This is the case if and only if $y \geq e^{-1/e}$. The equation has the unique solution if $y = e^{-1/e}$ and that solution is $x = 1/e$. If $e^{-1/e}\lt y \lt 1$, the equation has two solutions given by
\[
\ln (x_1) = W_{-1}\bigl( \ln y \bigr) \quad \text{and} \quad \ln (x_2) = W_{0}\bigl( \ln y \bigr)
\]
Hence
\[
x_1 = \exp\Bigl( W_{-1}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{-1}\bigl( \ln y \bigr)} \quad \text{and} \quad x_2 = \exp\Bigl( W_{0}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{0}\bigl( \ln y \bigr)}.
\]
The last identity might look obscure at first. It follows from the fact that $X = W(Y)$ if and only if $X\,e^X = Y$, with $X$ and $Y$ in the appropriate domains. Therefore, $W(Y)\,e^{W(Y)} = Y$, with $Y$ in the appropriate domain. Hence,
\[
e^{W(Y)} = \frac{Y}{W(Y)}.
\]
Here $W$ stands for either $W_0$ or $W_{-1}$. In each case we have to adjust the domains for $X$ and $Y$ appropriately.
If $y \geq 1$ the equation has a unique solution which is given by \[ x = \exp\Bigl( W_{0}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{0}\bigl( \ln y \bigr)}. \]
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I used Mathematica to explore equation (*) in class.
- Today in class I created
this Mathematica file with simple explorations of equations (*) and (**). The file is called 20151007.nb. Right-click on the 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 computers in BH 215. Look in
Start -> All Programs -> Math Applications -> Mathematica. Open Mathematica first, then open 20151007.nb from Mathematica. There are some explanations in the file. - More information on how to use Mathematica version 8 you can find on my Mathematica 8 page.
- Today in class I created
this Mathematica file with simple explorations of equations (*) and (**). The file is called 20151007.nb. Right-click on the 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 computers in BH 215. Look in
- Today we discussed Section 1.9 in Notes on Calculus. We almost rigorously proved that $x = o\bigl(2^x\bigr)$ as $x \to +\infty$. Read this section and do all the exercises.
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Two most important theorems about continuous functions are the Intermediate Value Theorem and the Extreme Values Theorem.
Intermediate Value Theorem. Let $a, b \in \mathbb R$ and $a \lt b$. If $f : [a,b] \to \mathbb R$ is a continuous function on the closed interval $[a,b]$, then
for every real number $y$ between $f(a)$ and $f(b)$ there exists $x \in [a,b]$ such that $y = f(x)$.
Extreme Values Theorem. Let $a, b \in \mathbb R$ and $a \lt b$. If $f : [a,b] \to \mathbb R$ is a continuous function on the closed interval $[a,b]$, thenthere exist $x_0, x_1 \in [a,b]$ such that $f(x_0) \leq f(x) \leq f(x_1)$ for all $x \in [a,b]$.
- As we discussed in class, to get a complete justification for the answer for (b) in Problem 4 on Assignment 1 one needed to use the Intermediate values theorem. Interestingly, we did not need the Extreme values theorem to rigorously answer (c) in the same problem.
- Today we discussed Problem 4 on Assignment 1. The conclusion of the discussion was that proving the following implication would be very helpful: Let $a, b \in \mathbb R$. Prove \[ -1 \lt a \lt 1 \ \ \text{and} \ \ -1 \leq b \leq 1 \ \ \text{imply} \ \ 0 \lt 1 + ab \lt 2 \ \ \text{and} \ \ -1 \leq \frac{a+b}{1+ab} \leq 1. \] Please present your proof neatly.
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Although we will not work with the formal $\epsilon$-$\delta$-definition of continuity of a function, it is good for you to see this definition. In more advanced courses you will gain appreciations for its clarity.
Definition. Let $X$ be a subset of $\mathbb R$, let $a \in X$ and let $f: X \to \mathbb R$ be a function. We say that the function $f$ is continuous at $a$ if the following condition is satisfied:
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For every $\epsilon \gt 0$ there exists $\delta(\epsilon) \gt 0$ such that for all $x \in X$ we have
$|x-a| \lt \delta(\epsilon)$ implies $|f(x) - f(a) | \lt \epsilon$.
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For every $\epsilon \gt 0$ there exists $\delta(\epsilon) \gt 0$ such that for all $x \in X$ we have
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I will illustrate the definition of limit with three examples.
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Prove that $\displaystyle \lim_{x\to 2} x^2 = 4$. Here $X = \mathbb R$, $f(x) = x^2$, $a=2$ and $L = 4$. We can take $\delta_0 = 1$, since clearly $(1,2)\cup(2,3) \subseteq \mathbb R$. Let $\epsilon \gt 0$ be arbitrary. Our goal is to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x-2| \lt \delta(\epsilon)$ implies $|x^2 - 4 | \lt \epsilon$.
- If $|x-2| \lt 1$, then $|x+2| \lt 5$. To justify this fact assume that $|x-2| \lt 1$. Then $1 \lt x \lt 3$. Then $3 \lt x + 2 \lt 5$. Then $x + 2 = |x+2| \lt 5$.
- If $|x-2| \lt 1$, then $|x^2 - 4| \lt 5 |x - 2|$. First recall that using the difference of two squares formula we have $| x^2 - 4 | = |(x+2)(x-2)| = |x+2|\, |x-2|$. To justify this fact assume that $|x-2| \lt 1$. By fact F1 we have that $|x+2| \lt 5$. Multiplying the last inequality by a nonnegative number $|x-2|$ we get $|x+2|\, |x-2| \leq 5 \, |x-2|$. Thus $|x^2 - 4| \lt 5 |x - 2|$.
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x-2| \lt \delta(\epsilon)$ implies $|x^2 - 4 | \lt \epsilon$.
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Prove that $\displaystyle \lim_{x\to 0} \cos x = 1$. Here $X = \mathbb R$, $f(x) = \cos x$, $a=0$ and $L = 1$. We can take $\delta_0 = \pi/2$, since clearly $(-\pi/2,0)\cup(0,\pi/2) \subseteq \mathbb R$. Let $\epsilon \gt 0$ be arbitrary. Our goal is to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq \pi/2$ and $0 \lt |x-0| = |x| \lt \delta(\epsilon)$ implies $|\cos x -1 | \lt \epsilon$.
- If $0 \lt |x| \lt \pi/2$, then $|\cos x - 1| \lt |x|$. First assume that $0 \lt x \lt \pi/2$. Then we can rely on the figure below on the left. This figure shows a part of the unit circle. The number $x$ is represented by the circular arc length between points $A$ and $C$. Then, by the definition of the cosine function the length of the line segment $OB$ equals $\cos x$. The triangle $ABC$ is a right triangle whose one leg is the line segment $BC$ and whose hypotenuse is $AC$. Therefore the length $BC$, which equals $1-\cos x$, is less than the length of the line segment $AC$. Further the length of the line segment $AC$ is less than arc length of the teal circular arc between $A$ and $C$. Since the arc length of the teal circular arc equals to $x$, this proves $1 - \cos x \lt x$. Similarly, if $-\pi/2 \lt x \lt 0$, we have $1 - \cos x \lt -x$. Thus, in either case $1 - \cos x \lt |x|$. Since $|\cos x - 1| = 1-\cos x$, F3 is proved.
$0 \lt \delta(\epsilon) \leq \pi/2$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $|\cos x - 1 | \lt \epsilon$.
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Prove that $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1$. Here $X = \mathbb R\setminus\{0\}$, $f(x) = (\sin x)/x$, $a=0$ and $L = 1$. We can take $\delta_0 = \pi/2$, since clearly $(-\pi/2,0)\cup(0,\pi/2) \subseteq \mathbb R\setminus\{0\}$. Let $\epsilon \gt 0$ be arbitrary. Our goal is to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq \pi/2$ and $0 \lt |x-0| = |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt \epsilon$.
- If $0 \lt x \lt \pi/2$, then $\displaystyle \cos x \lt \frac{\sin x}{x} \lt 1$. Assume that $0 \lt x \lt \pi/2$. To prove this inequality we rely on the figure above on the right. This figure shows a part of the unit circle. The number $x$ is represented by the circular arc length between points $A$ and $C$. Then, by the definition of the cosine function the length of the line segment $OB$ equals $\cos x$ and by the definition of the sine function the length of the line segment $AB$ equals $\sin x$. Compare the areas of the triangle $OCA$, the yellow area of the circular sector identified by points $OCA$ and the area of the triangle $OCD$. Clearly these three areas are listed in the increasing order. By calculating each of the three areas we conclude \[ \frac{1}{2} \sin x \lt \frac{1}{2} x \lt \frac{1}{2} \tan x. \] Dividing the above inequalities by $\dfrac{1}{2} \sin x \gt 0$ and taking reciprocals yields \[ \cos x \lt \frac{\sin x}{x} \lt 1. \] This proves F4.
- If $0 \lt |x| \lt \pi/2$, then $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt |x|$. Here is a proof. Assume first that $0 \lt x \lt \pi/2$. In F3 we proved that $1- |x| \lt \cos x$. Together with F4 this yields $1-|x| \lt (\sin x)/x \lt 1$. Since all the functions in the last inequality are even, the inequality also holds for negative numbers $x \in (-\pi/2,0)$. Now subtracting $1$ in each term of $1-|x| \lt (\sin x)/x \lt 1$ we get $-|x| \lt (\sin x)/x -1 \lt 0$. Now multiplying by $-1$ yields $0 \lt 1- (\sin x)/x \lt |x|$. This proves F5.
$0 \lt \delta(\epsilon) \leq \pi/2$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt \epsilon$.
Figure for $\displaystyle \lim_{x\to 0} \cos x = 1$
Figure for $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1$
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Prove that $\displaystyle \lim_{x\to 2} x^2 = 4$. Here $X = \mathbb R$, $f(x) = x^2$, $a=2$ and $L = 4$. We can take $\delta_0 = 1$, since clearly $(1,2)\cup(2,3) \subseteq \mathbb R$. Let $\epsilon \gt 0$ be arbitrary. Our goal is to discover $\delta(\epsilon)$ such that
- Below is a formal definition of a composition of two functions.
Place the cursor over the image to start the animation.
The composition at a generic point $x$
The composition at another generic point $x$
Finding a minimum of the composition
Finding a maximum of the composition
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Here is my somewhat detailed definition of limit.
- I. There exists $\delta_0 \gt 0$ such that $(a-\delta_0,a)\cup(a,a+\delta_0) \subset X$.
- II.
For every $\epsilon \gt 0$ there exists $\delta(\epsilon)$ such that $0 \lt \delta(\epsilon) \leq \delta_0$ and
$0 \lt |x-a| \lt \delta(\epsilon)$ implies $|f(x) - L | \lt \epsilon$.
- In the Calculus textbook read Sections 1.7 and 1.8. Do the problems: 1.7 19, 20, 21, 22, 23, 24, 30, 1.8 1, 7, 8, 10, 14, 20, 46, 47, 57. Read Section 1.6 in Notes on Calculus and do exercises 4, 6, 7, 11, 13, 14, 16, 17.
- Today I handed out Assignment 1. It is due on October 5. To receive the full credit for a problem your solution needs to be well written. Guidelines for Good Mathematical Writing written by Professor Francis Su from Harvey Mudd College can help you improve your mathematical writing. Prof. Su's article was published in MAA FOCUS, Vol. 35. No. 4, 2015, pages 20-22. Googleing "Good Mathematical Writing" will lead to more interesting articles on this topic.
- Today we defined four piecewise defined functions:
The sign function
The unit step function
The floor function
The ceiling function
- We also discussed hyperbolic functions sinh, cosh and tanh.
- I moved all stuff related to functions here. I will keep updating today's post.
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We discussed the formal definition of a function. Here is the definition from Wikipedia's function page:
- Fun 1. For every $x \in X$ there exists $y \in Y$ such that the pair $(x,y)$ belongs to the function $f$.
- Fun 2. If pairs $(x,y_1)$ and $(x,y_2)$ belong to the function $f$, then $y_1= y_2$.
- A function $f$ from $X$ to $Y$ is often denoted by $f: X \to Y$. If $f$ is a function then instead of $(x,y) \in f$ we write $y = f(x)$. The set $X$ is called the domain of $f$ and the set $Y$ is called the codomain of $f$. The set \[ \bigl\{ y \in Y \ \bigl| \bigr. \ y = f(x) \ \text{for some} \ x \in X \bigr\} \] is called the range of $f$.
- If $X$ and $Y$ are nonempty sets of $\mathbb R$ we can rephrase the definition of a function as follows: A subset $f$ of the Cartesian product $\mathbb R\!\times\!\mathbb R$ is a function from $X$ to $Y$ if every vertical line that goes through a point in $X\!\times\!\{0\}$ crosses $f$ at exactly one point. This formulation is sometimes called the vertical line test.
- Unfortunately, in precalculus and calculus classes functions are often defined by formulas and the sets $X$ and $Y$ are not specified explicitly. If this is the case, we consider that the domain of such a function is the maximal set of values for which the formula is defined. This domain is called the natural domain of a function.
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We distinguish two important kinds of functions: injections and surjections.
Definition. A function $f$ from $X$ to $Y$ is an injection if $f(x_1) \neq f(x_2)$ whenever $x_1, x_2 \in X$ and $x_1 \neq x_2$.Definition. A function $f$ from $X$ to $Y$ is a surjection if for every $y \in Y$ there exists $x \in X$ such that $y = f(x)$. In other words, a function $f$ from $X$ to $Y$ is a surjection if the range of $f$ equals $Y$.Definition. A function $f$ from $X$ to $Y$ is a bijection if it is both an injection and a surjection.
- A synonym for "injection" is "one-to-one function". A synonym for "surjection" is "onto function". I strongly encourage you to ignore these synonyms.
- If $X$ and $Y$ are nonempty subsets of $\mathbb R$ we can rephrase the definitions of an injection and the definition of a surjection as follows: A function $f:X \to Y$ is an injection (a surjection) if every horizontal line that goes through a point in $\{0\}\!\times\!Y$ crosses $f$ at at most (at least) one point. A function $f:X \to Y$ is a bijection if every horizontal line that goes through a point in $\{0\}\!\times\!Y$ crosses $f$ at at exactly one point. These formulations are sometimes called the horizontal line tests.
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The importance of bijections comes from the following theorem:
- Precalculus is full of important bijections. Below I list several bijections and you can figure out why they are important: \begin{alignat*}{2} X & = [0,+\infty), & \quad Y & = [0,+\infty), & \qquad f_1 & = \bigl\{ (x,x^2) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = \mathbb R, & \quad Y & = \mathbb R, & \qquad f_2 & = \bigl\{ (x,x^3) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = \mathbb R, & \quad Y & = (0,+\infty), & \qquad f_3 & = \bigl\{ (x,e^x) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = (-\pi/2,\pi/2), & \quad Y & = \mathbb R, & \qquad f_4 & = \bigl\{ \bigl( x,\tan(x) \bigr) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = [-\pi/2,\pi/2], & \quad Y & = [-1,1], & \qquad f_5 & = \bigl\{ \bigl(x,\sin(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} \\ X & = [0,\pi], & \quad Y & = [-1,1], & \qquad f_6 & = \bigl\{ \bigl(x,\cos(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} \\ \end{alignat*}
- These six bijections are plotted below in navy blue. Their inverses are plotted in maroon.
$ X = [0,+\infty), \ Y = [0,+\infty), \ f_1 = \bigl\{ (x,x^2) \ \bigl| \bigr. \ x \in X \bigr\} $
$X = \mathbb R, \ Y = \mathbb R, \ f_2 = \bigl\{ (x,x^3) \ \bigl| \bigr. \ x \in X \bigr\}$
$X= \mathbb R, \ Y = (0,+\infty), \ f_3 = \bigl\{ (x,e^x) \ \bigl| \bigr. \ x \in X \bigr\}$
$X = (-\pi/2,\pi/2), \ Y = \mathbb R, \ f_4 = \bigl\{ \bigl( x,\tan(x) \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
$ X = [-\pi/2,\pi/2], \ Y = [-1,1], \ f_5 = \bigl\{ \bigl(x,\sin(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
$ X = [0,\pi], \ Y = [-1,1], \ f_6 = \bigl\{ \bigl(x,\cos(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} $
- Read Sections 1.2 and 1.8 in Notes on Calculus. Do the following exercises: 1.2 1, 2, 3, 4, 7, 8, 9, 10, 11, 15, 16, 17 and 1.8 1, 2, 3, 4, 5, 6, 7, 8
- The information sheet
- Some useful Wikipedia links:
- 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.
- In this class we will use the set notation. 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 rational numbers \[ \mathbb Q = \Bigl\{x \in \mathbb R \, \bigl| \bigr. \, x = p/q \ \text{for some} \ p, q \in \mathbb Z \ \text{with} \ q \neq 0 \Bigr\}. \] Further we list several kinds of intervals of real numbers. For example, for $a,b \in \mathbb R$ with $a \lt b$, we have 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 \[ (a,b] = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \lt x \ \text{and} \ x \leq b \bigr\}, \] and the closed interval \[ [a,b] = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, a \leq x \ \text{and} \ x \leq b \bigr\}. \] For $a \in \mathbb R$ we have 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 \[ (-\infty,a] = \bigl\{x \in \mathbb R \, \bigl| \bigr. \, x \leq a \bigr\}. \]