text:
# L’Hôpital’s Rule

If two functions $f(x)$ and $g(x)$ both approach zero as $x\to a$, the function $\frac{f(x)}{g(x)}$ is said to assume the indeterminate form $\frac{0}{0}$ at $x=a$. Regardless of the term, $\frac{f(x)}{g(x)}$ may approach a limit as $x$ approaches $a$. The following theorem named after the French mathematician G. F. A. de L’Hôpital (1661-1704) is useful for the process of determining this limit if it exists.

Theorem (L’Hôpital’s Rule). If the functions $f(x)$ and $g(x)$ are continuous in an interval containing $a$ and if $f'(x)$ and $g'(x)$ exist such that $g'(x)\ne 0$ in this interval (except possibly at $x=a$), then when $\lim_{x\to a}f(x)=\lim_{x\to 0}g(x)=0$ (or equivalently $f(a)=g(a)=0$), we have $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ provided the limit on the right exists.

Example. Evaluate $\lim_{x\to 0}\frac{\tan x}{x}$.

Solution. The limit is an indeterminate form $\frac{0}{0}$. By applying L’Hôpital’s Rule, we obtain $$\lim_{x\to 0}\frac{\tan x}{x}=\lim_{x\to 0}\frac{\sec^2x}{1}=1$$

The Indeterminate form $\frac{\infty}{\infty}$

If $f(x)\to\infty$ and  $g(x)\to \infty$ as $x\to a$ (or $x\to\pm\infty$), the function $\frac{f(x)}{g(x)}$ is said to assume the indeterminate form $\frac{\infty}{\infty}$ at $x=a$ (or at $x=\pm\infty$). The limit of $\frac{\infty}{\infty}$ as $x\to a$ (or $x\to\pm\infty$) may still be found by L’Hôpital’s Rule if it exists.

Example. $$\lim_{x\to\infty}\frac{x^2}{e^x}=\lim_{x\to\infty}\frac{2x}{e^2}=\lim_{x\to\infty}\frac{2}{e^x}=0$$

Example. Evaluate $\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}$.

Solution. Applying L’Hôpital’s Rule, we have $$\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}=\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x^2}$$ While L’Hôpital’s rule can still be applied, we would accomplish nothing by doing so. This example shows that L’Hôpital’s rule may not necessarily leads to a desirable result. For the above example, the limit can be found by $$\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}=\lim_{x\to 0+}\frac{\frac{1}{x}}{e^{\frac{1}{x}}}=\lim_{z\to\infty}\frac{z}{e^z}=\lim_{z\to\infty}\frac{1}{e^z}=0$$

The indeterminate form $0\cdot\infty$

If $f(x)\to 0$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), the function $f(x)g(x)$ is said to assume the indeterminate form $0\cdot\infty$ at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)g(x)$ as $x\to a$ (or $x\to\pm\infty$) exists, it may be found by writing $f(x)g(x)$ as $$\frac{f(x)}{\frac{1}{g(x)}}\ \mbox{or}\ \frac{g(x)}{\frac{1}{f(x)}}$$ and applying L’Hôpital’s rule.

Example.

1. $\lim_{x\to\infty}xe^{-x}=\lim_{x\to\infty}\frac{x}{e^x}=\lim_{x\to\infty}\frac{1}{e^x}=0$.
2. $\lim_{x\to 0}\sin 3x\cot 2x=\lim_{x\to 0}\frac{\sin 3x}{\tan 2x}=\lim_{x\to 0}\frac{3\cos 3x}{2\sec^22x}=\frac{3}{2}$.

The Indeterminate Form $\infty-\infty$

If $f(x)\to\infty$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), the difference $f(x)-g(x)$ is said to assume the indeterminate form $\infty-\infty$  at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)-g(x)$ as $x\to a$ (or as $x\to\pm\infty$) exists, it may be found by transforming the difference into a fraction by algebraic means and applying L’Hôpital’s rule.

Example. $\lim_{x\to 0}(\csc x-\cot x)=\lim_{x\to 0}\frac{1-\cos x}{\sin x}=\lim_{x\to 0}\frac{\sin x}{\cos x}=0$

Example. Evaluate $\lim_{x\to\infty}(x-\ln x)$.

Solution. While $x-\ln x$ assumes the indeterminate form $\infty-\infty$ at $x=\infty$, there is no algebraic means to transform it to a fraction. An indeterminate of the form $\infty-\infty$ such as the one in consideration may be evaluated by finding the limit of its exponential. Let $y=x-\ln x$. Then $$e^y=e^{x-\ln x}=\frac{e^x}{e^{\ln x}}=\frac{e^x}{x}$$ Hence, $$\lim_{x\to\infty}e^y=\lim_{x\to\infty}\frac{e^x}{x}=\lim_{x\to\infty}\frac{e^x}{1}=\infty$$ Since $y\to\infty$ when $e^y\to\infty$, $$\lim_{x\to\infty}(x-\ln x)=\infty$$

The Indeterminate Forms $0^0$, $\infty^0$, $1^\infty$

If $f(x)\to 0$ and $g(x)\to 0$, or $f(x)\to\infty$ and $g(x)\to 0$, or $f(x)\to 1$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), $f(x)^{g(x)}$ is said to assume the indeterminate form $0^0$, $\infty^0$, or $1^\infty$, respectively at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)^{g(x)}$ exists as $x\to a$ (or $x\to\pm\infty$), it may be found by denoting $f(x)^{g(x)}$ by $y$ and investigating the limit approached by the logarithm $$\ln y=g(x)\ln f(x)$$ If $\lim_{x\to a}\ln y=k$, then $\lim_{x\to a}y=e^k$.

Example. Evaluate $\lim_{x\to 0}x^x$.

Solution. $x^x$ assumes the indeterminate form $0^0$ at $x=0$. Let $y=x^x$. Then $\ln y=x\ln x$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to 0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to 0}(-x)=0$$ Hence $\lim_{x\to 0}y=\lim_{x\to 0}x^x=e^0=1$.

Example. Evaluate $\lim_{x\to 0}(1-\sin x)^{\frac{1}{x}}$.

Solution. The function assumes the indeterminate form $1^\infty$ at $x=0$. Let $y=(1-\sin x)^{\frac{1}{x}}$. Then $\ln y=\frac{\ln(1-\sin x)}{x}$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln(1-\sin x)}{x}=\lim_{x\to 0}\frac{\frac{-\cos x}{1-\sin x}}{1}=-1$$ Hence, $\lim_{x\to 0}(1-\sin x)^{\frac{1}{x}}=e^{-1}=\frac{1}{e}$.

Example. Use L’Hôpital’s rule to show the limit $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

Proof. Let $y=(1+x)^{\frac{1}{x}}$. Then $y$ assumes the indeterminate form $1^\infty$ at $x=0$. $\ln y=\frac{\ln(1+x)}{x}$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln(1+x)}{x}=\lim_{x\to 0}\frac{1}{1+x}=1$$ Hence, $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

# Taylor series

Suppose that $f$ is differentiable infinitely many times on an open interval containing $a$. Then $$\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}(a)}{3!}(x-a)^3+\cdots$$ is called the Taylor series of $f$ centered at $a$. A Taylor series centered at 0 is called a Maclaurin series.

Example. Find the Maclaurin series of $f(x)=e^x$ and its radius of convergence $R$.

Solution. The Maclaurin series is $\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{x^n}{n!}$. By the ratio test $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{|x|}{n+1}=0<1$$ for all $x$ so the series converges for all $x$ i.e. the radius of convergence is $R=\infty$.

The following theorem tells when a function $f(x)$ can be represented by its Taylor series. Recall that $f(x)=T_n(x)+R_n(x)$ where $T_n(x)$ is the $n$-th degree Taylor polynomial of $f$ at $a$ and $R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ where $a<\xi<x$ or $x<\xi<a$.

Theorem. $f=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$ for $|x-a|<R$ if and only if $\lim_{n\to\infty}R_n(x)=0$ for $|x-a|<R$.

Since $\sum_{n=0}^\infty\frac{x^n}{n!}$ converges for all $x$, $$\label{eq:explim}\lim_{n\to\infty}\frac{x^n}{n!}=0$$ Now we show that $$\label{eq:expx}e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ for all $x$. $R_n(x)=\frac{e^\xi}{(n+1)!}x^{n+1}$ where $0<\xi<x$ or $x<\xi<0$. If $0<\xi<x$ then $0<R_n(x)<\frac{e^x}{(n+1)!}x^{n+1}\to 0$ as $n\to\infty$ by \eqref{eq:explim}. If $x<\xi<0$ then $0\leq |R_n(x)|<\frac{|x|^{n+1}}{(n+1)!}\to 0$ as $n\to\infty$ again by \eqref{eq:explim}. This completes the proof of \eqref{eq:expx}. For $x=1$ we have a definition of the Euler number $e$ in terms of a series as $$\label{eq:eulernum}e=\sum_{n=0}^\infty\frac{1}{n!}=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots$$

Example. Find the Taylor series of $f(x)=e^x$ at $a=2$.

Solution. By the same manner, one can show that $$e^x=\sum_{n=0}^\infty\frac{e^2}{n!}(x-2)^n$$ for all $x$.

Example. Find the Maclaurin series of $\sin x$ and show that it represents $\sin x$ for all $x$.

Solution. The Maclaurin series is $\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}$ and by the ratio test one can show that it converges for all $x$. Since $|f^{(n+1)}(\xi)|\leq 1$, $0\leq |R_n(x)|\leq\frac{|x|^{n+1}}{(n+1)!}\to 0$ as $n\to\infty$. Hence, $$\label{eq:sinx}\sin x=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}$$ for all $x$.

Similarly one can also show that $$\label{eq:cosx}\cos x=\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}$$ for all $x$.

Euler’s Formula

If $x$ is replaced by $ix$ in \eqref{eq:expx}, we obtain \begin{align*}e^{ix}&=\sum_{n=0}^\infty\frac{(ix)^n}{n!}\\&=\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}+i\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}\\&=\cos x+i\sin x\end{align*}$$\label{eq:eulerformula}e^{ix}=\cos x+i\sin x$$ is called the Euler’s Formula and it represents the point on the unit circle centered at the origin corresponding to the angle $x$. The formula in \eqref{eq:eulerformula} comes in handy in so many places of mathematics from trigonometry, calculus, differential equations to abstract algebra, topology, geometry to name a few. It is also a useful tool in physics on many occasions. For example, one can derive the sine sum and the cosine sum formulas using \eqref{eq:eulerformula}: $e^{i(\theta_1+\theta_2)}=\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)$.  On the other hand, \begin{align*}e^{i(\theta_1+\theta_2)}&=e^{i\theta_1}e^{i\theta_2}\\&=(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)\\&=(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\cos\theta_1\sin\theta_2+\sin\theta_1\cos\theta_2)\end{align*} Hence we obtain the sine and the cosine sum formulas \begin{align*}\sin(\theta_1+\theta_2)&=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2\\\cos(\theta_1+\theta_2)&=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\end{align*} When $\theta=\pi$, we obtain the so-called the Euler’s identity $$\label{eq:euleridentity}e^{\pi i}+1=0$$ which contains the most fundamental constants of mathematics $\pi$, $e$, $i$, 0, and 1.

The Binomial Series

Let us find the Maclaurin series of $f(x)=(1+x)^k$ where $k$ is any real number. First, we find $$f^{(n)}(x)=k(k-1)\cdots(k-n+1)(1+x)^{k-n}$$ and so $$f^{(n)}(0)=k(k-1)\cdots(k-n+1)$$ Hence the Maclaurin series is given by $$\sum_{n=0}^\infty\frac{k(k-1)\cdots(k-n+1)}{n!}x^n$$ The coefficients $\frac{k(k-1)\cdots(k-n+1)}{n!}$, $n=0,1,2,\cdots$ are denoted by ${}_n\mathrm{C}_k$ or $\begin{pmatrix}k\\n\end{pmatrix}$ and are called binomial coefficients. $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{k-n}{n+1}\right||x|=\lim_{n\to\infty} |x|$$ So by the ratio test if $|x|<1$ then the series converges and if $|x|>1$ then it diverges. Furthermore it can be shown (we omit the proof) that $$\label{eq:binomialseries}(1+x)^k=\sum_{n=0}^\infty\frac{k(k-1)\cdots(k-n+1)}{n!}x^n$$ for $|x|<1$. The series in \eqref{eq:binomialseries} is called the binomial series.

Example. FInd the Maclaurin series of $f(x)=\frac{1}{\sqrt{4-x}}$.

Solution. \begin{align*}\frac{1}{\sqrt{4-x}}&=\frac{1}{2}\frac{1}{\sqrt{1-\frac{x}{4}}}\\&=\frac{1}{2}\left(1-\frac{x}{4}\right)^{-\frac{1}{2}}\\&=\frac{1}{2}\sum_{n=0}^\infty\begin{pmatrix}-\frac{1}{2}\\n\end{pmatrix}\left(-\frac{x}{4}\right)^n\\&=\frac{1}{2}\left\{1+\left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{x}{4}\right)^2+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(-\frac{x}{4}\right)^3+\cdots\\+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)}{n!}\left(-\frac{x}{4}\right)^n+\cdots\right\}\\&=\frac{1}{2}\left\{1+\frac{x}{8}+\frac{1\cdot 3}{2!8^2}x^2+\frac{1\cdot\cdot 5}{3!8^3}x^3+\cdots+\frac{1\cdot 3\cdot 5\cdots(2n-1)}{n!8^n}x^n+\cdots\right\}\end{align*} This series converges if $\left|-\frac{x}{4}\right|<1$ i.e. $|x|<4$. The radius of convergence is 4.

Working with Taylor Series

Example. Use Taylor series to evaluate $\lim_{x\to 0}\frac{x^2+2\cos x-2}{3x^4}$.

Solution. \begin{align*}\lim_{x\to 0}\frac{x^2+2\cos x-2}{3x^4}&=\lim_{x\to 0}\frac{x^2+2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)-2}{3x^4}\\&=\lim_{x\to 0}\frac{2\left(\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)}{3x^4}\\&=\frac{2}{4!3}=\frac{1}{36}\end{align*}

Remark. The above limit can be also calculated using the L’Hôpital’s rule as it is an indeterminate form of type $\frac{0}{0}$.

Example. Approximate $\int_0^1 e^{-x^2}dx$ with an error no greater than $5\times 10^{-4}$.

Solution. \begin{align*}\int_0^1 e^{-x^2}dx&=\int_0^1\left(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots+\frac{(-1)^nx^{2n}}{n!}+\cdots\right)dx\\&=\left[\left(x-\frac{x^3}{3}+\frac{x^5}{2!5}-\frac{x^7}{3!7}+\cdots+\frac{(-1)^nx^{2n+1}}{n!(2n+1)}+\cdots\right)\right]_0^1\\&=1-\frac{1}{3}+\frac{1}{2!5}-\frac{1}{3!7}+\cdots+\frac{(-1)^n}{n!(2n+1)}+\cdots\end{align*} Recall that for an alternating series $\sum_{n=0}^\infty(-1)^n a_n$, $|R_n|\leq a_{n+1}$. For $n=4$, $|R_4|\leq a_5=\frac{1}{5!11}=7.6\times 10^{-3}>5\times 10^{-4}$. For $n=5$, $|R_5|\leq a_6=\frac{1}{6!13}=1.07\times 10^{-4}<5\times 10^{-4}$. Thus the error is less than $5\times 10^{-4}$ if $n\geq 5$. The approximation with $n=5$ is given by \begin{align*}\int_0^1 e^{-x^2}dx&\approx 1-\frac{1}{3}+\frac{1}{2!5}-\frac{1}{3!7}+\frac{1}{5!11}\\&\approx 0.747\end{align*}

# Applications of Definite Integrals: Surfaces of Revolution

Suppose that a continuously differentiable curve $y=f(x)\geq 0$ on $[a,b]$ is revolved about the $x$-axis as shown in Figure 1.

Figure 1

The area $A$ of the resulting surface of revolution can be obtained by adding (i.e. integrating) the areas of infinitesimally thin circular ribbons of radius $f(x)$ with thickness $ds$ at each $a\leq x\leq b$. $$\label{eq:surfrev}A=\int_{x=a}^{x=b}2\pi f(x)ds=\int_a^b 2\pi f(x)\sqrt{1+[f'(x)]^2}dx$$

Example. Find the area of the surface generated by revolving $y=2\sqrt{x}$, $1\leq x\leq 2$ about the $x$-axis.

Solution. \begin{align*}A&=2\pi\int_1^2 2\sqrt{x}\sqrt{1+\left(\frac{1}{\sqrt{x}}\right)^2}dx\\&=4\pi\int_1^2 \sqrt{1+x}dx\\&=4\pi\left(\frac{2}{3}\right)[(1+x)^{\frac{3}{2}}]_1^2\\&=\frac{8\pi}{3}(3\sqrt{3}-2\sqrt{2})\end{align*}

Revolution about the $y$-axis

If $x=g(y)\leq 0$ is continuously differentiable on $[c,d]$, the area of the surface generated by revolving the curve $x=g(y)$ about the $y$-axis is $$\label{eq:surfrev2}A=\int_c^d 2\pi g(y)\sqrt{1+[g'(y)]^2}dy$$

Example. The line segment $x=1-y$, $0\leq y\leq 1$ is revolved about the $y$-axis to generate a cone. Find its lateral surface area.

Solution. \begin{align*}A&=\int_0^1 2\pi(1-y)\sqrt{1+(-1)^2}dy\\&=2\sqrt{2}\pi\int_0^1(1-y)dy\\&=\sqrt{2}\pi\end{align*}

Surface are of revolution for parametrized curves

If a smooth curve $r(t)=(x(t),y(t))$, $\alpha\leq t\leq\beta$ is traversed exactly once as $t$ increases from $\alpha$ to $\beta$, then the surface area of revolution is given by:

1. Revolution about the $x$-axis ($y(t)\geq 0$) $$\label{eq:surfrev3}A=\int_\alpha^\beta 2\pi y(t)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$
2. Revolution about the $y$-axis ($x(t)\geq 0$) $$\label{eq:surfrev4}A=\int_\alpha^\beta 2\pi x(t)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$

Example. The circle $x^2+(y-1)^2=1$ of radius 1 centered at $(0,1)$ is parametrized as $$x=\cos t,\ y=\sin t+1,\ 0\leq t\leq 2\pi$$ Find the are of the surface swept out by revolving the circle about the $x$-axis.

Solution. Figure 2 shows the surface of revolution

Figure 2

and Figure 3 the same surface but for $-1\leq x\leq 0$

Figure 3

It is a doughnut shaped surface which is called a torus in mathematics. Using \eqref{eq:surfrev3} the area is given by \begin{align*} A&=2\pi\int_0^{2\pi}(1+\sin t)\sqrt{(-\sin t)^2+(\cos t)^2}dt\\&=2\pi\int_0^{2\pi}(1+\sin t)dt\\&=2\pi[t-\cos t]_0^{2\pi}\\&=4\pi^2\end{align*}

# Applications of Definite Integrals: Arc Length

Suppose that $y=f(x)$ is continuously differentiable on $[a,b]$.

Figure 1

Figure 1 shows the curve $y=f(x)$ is approximated by a PL (Piecewise Linear) curve (in red). Suppose that the line segments of the PL curve are infinitesimal and denote by $dL$ a line element. Then it is given by \begin{align*}dL&=\sqrt{(dx)^2+(dy)^2}\\&=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\\&=\sqrt{1+[f'(x)]^2}dx\end{align*} Hence the length of the curve $y=f(x)$, $a\leq x\leq b$ is $$\label{eq:arclength}L=\int_{x=a}^{x=b}dL=\int_a^b\sqrt{1+[f'(x)]^2}dx$$

Often a curve is represented by a parametric equation $r(t)=(x(t),y(t))$, $\alpha\leq t\leq\beta$ such that $x(t)$ and $y(t)$ are continuously differentiable. Such a curve is called smooth. A typical example is a motion in the plane (or in space). For a smooth curve $r(t)=(x(t),y(t))$, $\alpha\leq t\leq\beta$, $dL$ is \begin{align*}dL&=\sqrt{(dx)^2+(dy)^2}\\&=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\end{align*} and hence $$\label{eq:arclength2}\int_\alpha^\beta\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$ If $r(t)$ is the motion of a particle, $\left|\frac{dr(t)}{dt}\right|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$ is the speed of the particle and so \eqref{eq:arclength2} represents the distance traveled by the particle over the time interval $\alpha\leq t\leq\beta$.

Example (The circumference of a circle). Find the length of the circle of radius $r$.

Solution.  The equation of the upper semicircle of radius r centered at the origin is $y=\sqrt{r^2-x^2}$, $-r\leq x\leq r$. $\frac{dy}{dx}=-\frac{x}{\sqrt{r^2-x^2}}$ and so the circumference is \begin{align*}2\int_{-r}^r\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx&=2\int_{-r}^r\frac{r}{\sqrt{r^2-x^2}}dx\\&=4\int_0^r\frac{r}{\sqrt{r^2-x^2}}dx\\&=4\int_0^r\frac{1}{\sqrt{1-\left(\frac{x}{r}\right)^2}}dx\\&=4r\int_0^1\frac{1}{\sqrt{1-u^2}}du\\&=4r[\sin^{-1}(u)]_0^1\\&=4r\left(\frac{\pi}{2}\right)=2\pi r\end{align*}

The circle $x^2+y^2=r^2$ of radius $r$ centered at the origin can be described by the parametric equations $$x=r\cos t,\ y=r\sin t,\ 0\leq t\leq 2\pi$$ $\frac{dx}{dt}=-r\sin t$ and $\frac{dy}{dt}=r\cos t$ so $\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=r$. Therefore the circumference is $\int_0^{2\pi}rdt=2\pi r$. we find that using parametric equations of the circle comes with a much simpler calculation.

Example. Find the length of the asteroid $$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$$

Solution. The name asteroid is due to its graph as seen in Figure 2.

Figure 2. Asteroid

The asteroid can be parametrized as $$x=\cos^3 t,\ y=\sin^3 t,\ 0\leq t\leq 2\pi$$ $\frac{dx}{dt}=-3\cos^2 t\sin t$ and $\frac{dy}{dt}=3\sin^2 t\cos t$ so $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=9\cos^2t\sin^2t$. For $0\leq t\leq\frac{\pi}{2}$, $\sin t\geq 0$ and $\cos t\geq 0$. Hence \begin{align*}L&=4\int_0^{\frac{\pi}{2}}\sqrt{9\cos^2t\sin^2t}dt\\&=12\int_0^{\frac{\pi}{2}}\cos t\sin tdt\\&=6\int_0^{\frac{\pi}{2}}\sin 2tdt\\&=3[-\cos 2t]_0^{\frac{\pi}{2}}=6\end{align*} The integral $\int_0^{\frac{\pi}{2}}\cos t\sin tdt$ also can be done using a substitution. Let $u=\sin t$. Then $du=\cos tdt$ so \begin{align*}\int_0^{\frac{\pi}{2}}\cos t\sin tdt&=\int_0^1udu\\&=\frac{1}{2}\end{align*}

Example. Find the length of the curve $$y=\frac{4\sqrt{2}}{3}x^{\frac{3}{2}}-1,\ 0\leq x\leq 1$$

Solution. $\frac{dy}{dx}=2\sqrt{2x}$ so $\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+8x}$ and \begin{align*}L&=\int_0^1\sqrt{1+8x}dx\\&=\frac{2}{3}\cdot\frac{1}{8}[(1+8x)^{\frac{3}{2}}]_0^1\\&=\frac{13}{6}\end{align*}

Example. Find the length of the curve $y=\left(\frac{x}{2}\right)^{\frac{2}{3}}$, $0\leq x\leq 2$.

Solution. $\frac{dy}{dx}=\frac{1}{3}\left(\frac{2}{x}\right)^{\frac{1}{3}}$. Note that this derivative does not exist at $x=0$. Rewrite the function as $x=2y^{\frac{3}{2}}$, $0\leq y\leq 1$. $\frac{dx}{dy}=3\sqrt{y}$ and \begin{align*}L&=\int_{y=0}^{y=1}dL\\&=\int_{y=0}^{y=1}\sqrt{(dx)^2+(dy)^2}\\&=\int_0^1\sqrt{\left(\frac{dx}{dy}\right)^2+1}dy\\&=\int_0^1\sqrt{9y+1}dy\\&=\frac{1}{9}\cdot\frac{2}{3}[(9y+1)^{\frac{3}{2}}]_0^1\\&=\frac{2}{27}(10\sqrt{10}-1)\end{align*}

# Properties of Power Series

A power series is a series of the form $$\label{eq:powerseries}\sum_{n=0}^\infty c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\cdots$$ The power series in \eqref{eq:powerseries} is particularly called a power series in $x-a$, a power series centered at $a$, or a power series about $a$. The number $a$ is called the center of the power series. The set of the values of $x$ for which the series converges is called the interval of convergence and the radius of the interval of convergence is called the radius of convergence. Since a power series in general does not converge everywhere, it is important to find the interval of convergence or the radius of convergence (if you find one, you consequently find the other). In the study of the convergence of power series, we mostly use the ratio test but the root test may come in handy sometimes.

Example. Find the interval and the radius of convergence for each of the following series.

1. $\sum_{n=0}^\infty\frac{x^n}{n!}$
2. $\sum_{n=0}^\infty n!x^n$
3. $\sum_{n=1}^\infty\frac{(x-3)^n}{n}$

Solution.

1.  \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{\frac{|x|^{n+1}}{(n+1)!}}{\frac{|x|^n}{n!}}\\&=\lim_{n\to\infty}\frac{|x|}{n+1}\\&=0<1\end{align*} for all $x$. This means that the interval of convergence is $(-\infty,\infty)$ and the radius of convergence is $\infty$.
2. \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{(n+1)!|x|^{n+1}}{n!|x|^n}\\&=\lim_{n\to\infty}(n+1)|x|\\&=\infty\end{align*} unless $x=0$. The interval of convergence is $\{0\}$ and the radius of convergence is 0.
3. \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{\frac{|x-3|^{n+1}}{n+1}}{\frac{|x-3|^n}{n}}\\&=\lim_{n\to\infty}\frac{n}{n+1}|x-3|\\&=|x-3|\end{align*} In order for the series to converge, we require that $|x-3|<1$ or $-1<x-3<1$ i.e. $2<x<4$. For the last step, we must check if the series converges at the end points $x=2$ and $x=4$. If $x=2$, the series becomes $\sum_{n=1}^\infty\frac{(-1)^n}{n}$. This is an alternating harmonic series and we learned here that it converges. If $x=4$, the series becomes the harmonic series $\sum_{n=1}^\infty\frac{1}{n}$ which diverges. Therefore, the interval of convergence is $[2,4)$. Since the center of the power series is 3, the radius is 1.

Representing Functions as Power Series

As we will study later, functions that are infinitely many times differentiable (such functions are simply called smooth) in a open interval containing $a$ can be represented a power series centered at $a$ called a Taylor series. But even without knowing Taylor series expansion of a function, using the geometric series $$\label{eq:geomseries}\frac{1}{1-x}=1+x+x^2+x^3+\cdots=\sum_{n=0}^\infty x^n\ \mbox{if}\ |x|<1$$ we can represent many functions as power series.

Example. Express $\frac{1}{1+x^2}$ as a power series and find the interval of convergence.

Solution. \begin{align*}\frac{1}{1+x^2}&=\frac{1}{1-(-x^2)}\\&=\sum_{n=0}^\infty(-x^2)^n\\&=\sum_{n=0}^\infty(-1)^nx^{2n}\end{align*} provided $|x|^2<1$. Hence, the interval of convergence is $|x|<1$ or $(-1,1)$.

Example. Find a power series representation of $\frac{x^3}{x+2}$.

Solution. First we find a power series representation of $\frac{1}{x+2}$. \begin{align*}\frac{1}{x+2}&=\frac{1}{2}\frac{1}{1+\frac{x}{2}}\\&=\frac{1}{2}\frac{1}{1-\left(-\frac{x}{2}\right)}\\&=\frac{1}{2}\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^n}\\&=\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^{n+1}}\end{align*} provided $\left|\frac{x}{2}\right|<1$ or $|x|<2$. Therefore, $\frac{x^3}{x+2}=\sum_{n=0}^\infty(-1)^n\frac{x^{n+3}}{2^{n+1}}$ for $-2<x<2$.

Note: If one replaces $n$ by $n-3$, then the series can be rewritten equivalently as $\sum_{n=3}^\infty(-1)^{n-1}\frac{x^n}{2^{n-2}}$.

Differentiation and Integration of Power Series

It turns out that one can differentiate and integrate a power series term by term as he would’ve done with polynomials.

Theorem. If the power series $\sum_{n=0}^\infty c_n(x-a)^n$ has radius of convergence $R>0$ then the function $f$ defined by $$f(x)=\sum_{n=0}^\infty c_n(x-a)^n$$ is differentiable on the interval of convergence $(a-R,a+R)$ and $$f'(x)=\sum_{n=1}^\infty nc_n(x-a)^{n-1}$$ and $$\int f(x)dx=C+\sum_{n=0}^\infty\frac{c_n}{n+1}(x-a)^{n+1}$$ where $C$ is a constant. The two power series $f'(x)$ and $\int f(x)dx$ both have the same radius of convergence $R$.

This theorem along with \eqref{eq:geomseries} allows us to find the power series representations of a broader class of functions as we will see in the following examples.

Example. Express $\frac{1}{(1-x)^2}$ as a power series. What is the radius of convergence?

Solution. \begin{align*}\frac{1}{(1-x)^2}&=\frac{d}{dx}\frac{1}{1-x}\\&=\frac{d}{dx}\sum_{n=0}^\infty x^n\ (|x|<1)\\&=\sum_{n=1}^\infty nx^{n-1}\\&=\sum_{n=0}^\infty(n+1)x^n\end{align*} The radius of convergence is 1.

Example. Find a power series representation for $\ln (1-x)$ and its radius of convergence.

Solution. \begin{align*}\ln(1-x)&=-\int\frac{dx}{1-x}\\&=-\int\sum_{n=0}^\infty x^ndx\ (|x|<1)\\&=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}+C\\&=-\sum_{n=1}^\infty\frac{x^n}{n}+C\end{align*} where $C$ is a constant. When $x=0$, $\ln(1-x)=0$ so we find $C=0$. Therefore, $$\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$$

Example. Find a power series representation for $f(x)=\tan^{-1}x$.

Solution. \begin{align*}\tan^{-1} x&=\int\frac{1}{1+x^2}dx\\&=\int\sum_{n=0}^\infty (-1)^nx^{2n}\ (|x|<1)\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}+C\end{align*} where $C$ is a constant. Since $\tan^{-1}0=0$, we find $C=0$. Hence, $$\tan^{-1}x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$

Example. Find a power series representation of $\ln\left(\frac{1+x}{1-x}\right)$.

Solution.  First $\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x)$. \begin{align*}\ln(1+x)&=\int\frac{dx}{1+x}\\&=\int\sum_{n=0}^\infty (-1)^nx^ndx\ (|x|<1)\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}+C\end{align*} Using $\ln(1+x)=0$ for $x=0$, we find $C=0$ and so $\ln(1+x)=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}$. Recall that we found a power series representation $\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$ in an example above. Hence, \begin{align*}\ln\left(\frac{1+x}{1-x}\right)&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}\\&=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n}+\sum_{n=1}^\infty\frac{x^n}{n}\\&=\sum_{n=1}^\infty[(-1)^{n-1}+1]\frac{x^n}{n}\end{align*} Note $$(-1)^{n-1}+1=\left\{\begin{array}{ccc} 2 & \mbox{if} & n\ \mbox{is odd}\\0 & \mbox{if} & n\ \mbox{is even}\end{array}\right.$$ Therefore, $$\ln\left(\frac{1+x}{1-x}\right)=2\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1}=2\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$$

Remark. Since we already know that $\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$ and $\ln(1+x)=\ln(1-(-x))$, we can find a power series representation of $\ln(1+x)$ by replacing $x$ by $-x$ in the power series representation of $\ln(1-x)$: \begin{align*}\ln(1+x)&=-\sum_{n=0}^\infty\frac{(-x)^{n+1}}{n+1}\\&=-\sum_{n=0}^\infty\frac{(-1)^{n+1}x^{n+1}}{n+1}\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}\end{align*}