Last time we saw that a series representation is like "unzipping" a function and looking at its "DNA."
A series representation reveals much about how a function behaves and is related to other functions.
Today, we utilize the power of series representations to do things we would otherwise not be able to do.
Recall: Taylor Series
If $f(x)$ has derivatives of all orders at $x = a,$ then the Taylor series expansion for the function $f(x)$ at $a$ is $$\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$$ The Taylor series for $f(x)$ at $a=0$ is known as the Maclaurin series for $f(x).$
Some Common Series Representations
$\displaystyle \frac{1}{1-x} =\sum_{n=1}^{\infty}x^{n}$ on $(-1,1)$
$\displaystyle \sin x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ on $(-\infty,\infty)$
$\displaystyle \cos x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$ on $(-\infty,\infty)$
$\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$ on $(-\infty,\infty)$
$\displaystyle \ln x =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(x-1)^{n}$ on $(0,2]$
$\displaystyle \sinh x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$ on $(-\infty,\infty)$
$\displaystyle \cosh x=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$ on $(-\infty,\infty)$
$\displaystyle \tan^{-1} x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$ on $[-1,1]$
Example: Finding the Sum of a Series
Use your knowledge of series to find the exact sum $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n 2^{n-1}}{n!}$
The series resembles the Maclaurin series for $e^x.$
We'll begin by letting $S$ represent the sum of the series. Then, $$ \begin{array}{llll} &S=\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n 2^{n-1}}{n!}&\mbox{}\\ \implies &2S=\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n 2^{n}}{n!}&\mbox{}\\ \implies &2S=\displaystyle \sum_{n=0}^{\infty}\frac{(-2)^{n}}{n!}&\mbox{}\\ \implies &2S=\displaystyle e^{-2}&\mbox{}\\ \implies &S=\displaystyle \frac{e^{-2}}{2}&\mbox{}\\ \end{array} $$
We'll begin by letting $S$ represent the sum of the series. Then, $$ \begin{array}{llll} &S=\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n 2^{n-1}}{n!}&\mbox{}\\ \implies &2S=\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n 2^{n}}{n!}&\mbox{}\\ \implies &2S=\displaystyle \sum_{n=0}^{\infty}\frac{(-2)^{n}}{n!}&\mbox{}\\ \implies &2S=\displaystyle e^{-2}&\mbox{}\\ \implies &S=\displaystyle \frac{e^{-2}}{2}&\mbox{}\\ \end{array} $$
Example: Finding a Tricky Limit
Evaluate the limit $\displaystyle \lim_{x \rightarrow 0} \frac{1-\cos(5x)}{x^2}.$
Since $\displaystyle \cos x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!},$ we see that
$$
\begin{array}{lll}
\displaystyle \lim_{x \rightarrow 0} \frac{1-\cos(5x)}{x^2}
&=\displaystyle \frac{\displaystyle 1-\sum_{n=0}^{\infty}\frac{(-1)^n(5x)^{2n}}{(2n)!}}{x^2}&\mbox{}\\
&=\displaystyle \frac{\displaystyle 1-\sum_{n=0}^{\infty}\frac{(-1)^n 5^{2n}x^{2n}}{(2n)!}}{x^2}&\mbox{}\\
&=\displaystyle \frac{\displaystyle 1-\left(1-\frac{5^2 x^2}{2!}+\frac{5^4 x^4}{4!}-\frac{5^6 x^6}{6!}+\cdots\right)}{x^2}&\mbox{}\\
&=\displaystyle \frac{\displaystyle \frac{5^2 x^2}{2!}+\frac{5^4 x^4}{4!}-\frac{5^6 x^6}{6!}+\cdots}{x^2}&\mbox{}\\
&=\displaystyle \frac{5^2}{2!}+\frac{5^4}{4!}x^2-\frac{5^6}{6!}x^6+\cdots&\mbox{}\\
&=\displaystyle \frac{25}{2}+\frac{625}{24}x^2-\frac{3125}{24}x^6+\cdots&\mbox{}\\
\end{array}
$$
Thus,
$$
\lim_{x \rightarrow 0} \frac{1-\cos(5x)}{x^2}= \lim_{x \rightarrow 0}\left(\frac{25}{2}+\frac{625}{24}x^2-\frac{3125}{24}x^6+\cdots\right)=\frac{25}{2}
$$
Huge Application: Finding Antiderivatives with No Closed Form
Recall from last term that functions like $\displaystyle f(x)=e^{-x^2}$ have no antiderivative which can be stated in terms of elementary functions.
But, with the mighty power of power series, we can now easily find an antiderivative.
The series for $e^x$ is
$$
\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}
$$
Replacing $x$ with $-x^2$ we may write
$$
\begin{array}{ll}
\displaystyle e^{-x^2}&= \displaystyle \sum_{n=0}^{\infty}\frac{\left(-x^2\right)^{n}}{n!}
&= \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}\\
\end{array}
$$
We may now integrate term by term.
$$
\displaystyle \int e^{-x^2} \,dx = C+\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)n!}
$$
Huge Example
Approximate the integral $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{0}^{1} e^{-x^2/2} \, dx$ to within $0.00001$
The series for $e^{-x^2/2}$ is
$$
\begin{array}{ll}
\displaystyle e^{-x^2/2}&= \displaystyle \sum_{n=0}^{\infty}\frac{\left(\frac{-x^2}{2}\right)^{n}}{n!}
&= \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{2^n \cdot n!}\\
\end{array}
$$
We may now compute the antiderivative using term-by-term integration.
$$
\int e^{-x^2/2}\, dx =C+\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)2^n \cdot n!}
$$
Then,
$$
\int_{0}^{1} e^{-x^2/2}\, dx =\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)2^n \cdot n!}
$$
We then have that
$$
\frac{1}{\sqrt{2\pi}}\int_{0}^{1} e^{-x^2/2}\, dx =\sum_{n=0}^{\infty}\frac{(-1)^n}{\sqrt{2\pi}(2n+1)2^n \cdot n!}
$$
Since this is an alternating series, we may use the alternating series test and bound the error of the $(N-1)\mbox{st}$ partial sum by
$$b_{N}=\frac{1}{\sqrt{2\pi}(2N+1)2^{N}\cdot N!}$$
We now simply find a value of $N$ for which $\displaystyle \frac{1}{\sqrt{2\pi}(2N+1)2^{N} \cdot N!}\leq 0.00001.$
With some trial and error, we see that $N=5$ fits the bill.
Thus, our approximation is
$$
\begin{array}{ll}
\displaystyle \frac{1}{\sqrt{2\pi}}\int_{0}^{1} e^{-x^2/2} \, dx
&\displaystyle \approx \sum_{n=0}^{4}\frac{(-1)^n}{\sqrt{2\pi}(2n+1)2^n \cdot n!}\\
&= \frac{(-1)^0}{\sqrt{2\pi}(2\cdot 0+1)2^0 \cdot 0!}+\frac{(-1)^1}{\sqrt{2\pi}(2\cdot 1+1)2^1 \cdot 1!}+\frac{(-1)^2}{\sqrt{2\pi}(2\cdot 2+1)2^2 \cdot 2!}+\frac{(-1)^3}{\sqrt{2\pi}(2\cdot 3+1)2^3 \cdot 3!}+\frac{(-1)^4}{\sqrt{2\pi}(2\cdot 4+1)2^4 \cdot 4!}\\
&=0.3413535638...
\end{array}
$$
We know that the approximation is within $0.00001$ of the true value. That is,
$$
\left|\frac{1}{\sqrt{2\pi}}\int_{0}^{1} e^{-x^2/2} \, dx-0.3413535638...\right|\leq 0.00001
$$
or,
$$
0.3413435638... \leq \frac{1}{\sqrt{2\pi}}\int_{0}^{1} e^{-x^2/2} \, dx \leq 0.3413635638...
$$
With some trial and error, we see that $N=5$ fits the bill.
Huge Application (Continued)
The mean height of women in their 20's in the United States is $\mu=64.3$ inches with a standard deviation of $\sigma=2.7$ inches. What is the probability that a randomly chosen woman in her 20's is between $64.3$ and $67$ inches tall? That is, what is the probability that her height will be within $0$ to $1$ standard deviations above the mean?
$$
\begin{array}{ll}
\displaystyle P(64.3 \lt \mbox{Height} \lt 67)
&=\displaystyle P\left(\frac{64.3-64.3}{2.7} \lt z \lt \frac{67-64.3}{2.7}\right)\\
&=\displaystyle P(0 \lt z \lt 1)\\
&=\displaystyle \frac{1}{\sqrt{2\pi}}\int_{0}^{1} e^{-z^2/2} \, dz \\
&\approx 0.3413435638\\
\end{array}
$$
Another Huge Application: The Period of a Pendulum
The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length $L$ that makes a maximum angle $\theta_{max}$ with the vertical, its period $T$ is given by $ \displaystyle T=4\sqrt{\frac{L}{g}}\int_{0}^{\pi/2} \frac{d\theta}{\sqrt{1-k^2\sin^2 \theta}} $ where $g$ is the acceleration due to gravity and $\displaystyle k = \sin\left(\frac{\theta_{max}}{2}\right).$
Use the series $\displaystyle \frac{1}{\sqrt{1+x}}=1+\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n}x^n$ to estimate the period of the pendulum using one term and two terms in the above series.
From the series $\displaystyle \frac{1}{\sqrt{1+x}}=1+\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n}x^n$
we can replace $x$ with $-k^2 \sin^2 \theta$ and write
$$
\begin{array}{ll}
\displaystyle \frac{1}{\sqrt{1-k^2\sin^2 \theta}}
&=\displaystyle 1+\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n}(-k^2\sin^2 \theta)^n\\
&=\displaystyle 1+\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n}(-1)^n k^{2n} \sin^{2n} \theta\\
&=\displaystyle 1+\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{n!2^n}k^{2n}\sin^{2n} \theta\\
\end{array}
$$
Then,
$$
\begin{array}{lll}
\displaystyle T
&=\displaystyle 4\sqrt{\frac{L}{g}}\int_{0}^{\pi/2} \frac{d\theta}{\sqrt{1-k^2\sin^2 \theta}}&\mbox{}\\
&=\displaystyle 4\sqrt{\frac{L}{g}}\int_{0}^{\pi/2} \left(1+\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{n!2^n}k^{2n}\sin^{2n} \theta \right) \,d\theta&\mbox{}\\
&=\displaystyle 4\sqrt{\frac{L}{g}}\left( \int_{0}^{\pi/2} 1 \,d\theta + \sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{n!2^n}k^{2n} \int_{0}^{\pi/2} \sin^{2n} \theta \,d\theta\right)&\mbox{}\\
&=\displaystyle 4\sqrt{\frac{L}{g}}\left( \frac{\pi}{2} + \frac{1}{2} k^{2}\int_{0}^{\pi/2} \sin^{2} \theta \,d\theta+\frac{3}{8} k^{4}\int_{0}^{\pi/2} \sin^{4} \theta \,d\theta+\frac{5}{16} k^{6}\int_{0}^{\pi/2} \sin^{6} \theta \,d\theta+\cdots \right)&\mbox{}\\
&=\displaystyle 4\sqrt{\frac{L}{g}}\left( \frac{\pi}{2} + \frac{1}{2} k^{2}\frac{\pi}{4}+\frac{3}{8} k^{4}\frac{3\pi}{16}+\frac{5}{16} k^{6}\frac{5\pi}{32}+\cdots \right)&\mbox{}\\
&=\displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}+\frac{9}{64} k^{4}+\frac{25}{256} k^{6}+\cdots \right)&\mbox{}\\
%&=\displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}+\frac{1^2\cdot 3^2}{2^2\cdot 4^2} k^{4}+\frac{1^2 \cdot 3^2 \cdot 5^2}{2^2 \cdot 4^2 \cdot 6^2} k^{6}+\cdots \right)&\mbox{}\\
\end{array}
$$
The one-term approximation is
$$
T \approx 2\pi\sqrt{\frac{L}{g}}
$$
and works well for small values of $\theta_{max}$ of about $0.1$ radians $(6^{\circ})$ or less.
This approximation works well for small-amplitude applications such as mechanical pendulum clocks.
For larger values of $\theta_{max},$ the two-term approximation is $$ \displaystyle T \approx \displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}\right) $$ If we need more accuracy, we can simply add more terms.
We note that $\displaystyle 2\pi\sqrt{\frac{L}{g}}$ and $\displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}\right)$ are lower bounds for $T.$
We shall now obtain an upper bound for $T.$ First, we note that $$ 1 + \frac{1}{4} k^{2}+\frac{9}{64} k^{4}+\frac{25}{256} k^{6}+\cdots = 1 + \frac{1}{4} k^{2}+\frac{1^2\cdot 3^2}{2^2\cdot 4^2} k^{4}+\frac{1^2 \cdot 3^2 \cdot 5^2}{2^2 \cdot 4^2 \cdot 6^2} k^{6}+\cdots $$ From the above, it's clear that each coefficient in the above series is less than $\frac{1}{4}.$ Then $$ \begin{array}{ll} \displaystyle 1 + \frac{1}{4} k^{2}+\frac{9}{64} k^{4}+\frac{25}{256} k^{6}+\cdots &\\ \lt \displaystyle 1 + \frac{1}{4} k^{2}+\frac{1}{4} k^{4}+\frac{1}{4} k^{6}+\cdots &\mbox{}\\ = \displaystyle 1 + \frac{1}{4} k^{2}\left(1+k^{2}+k^{4}+\cdots\right) &\mbox{}\\ = \displaystyle 1 + \frac{1}{4} k^{2}\frac{1}{1-k^2} &\mbox{}\\ = \displaystyle \frac{4-4k^2}{4-4k^2} +\frac{k^2}{4-4k^2} &\mbox{}\\ = \displaystyle \frac{4-3k^2}{4-4k^2} &\mbox{}\\ \end{array} $$ Thus, multiplying the above by $\displaystyle 2\pi\sqrt{\frac{L}{g}},$ we have $$ \displaystyle T = \displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}+\frac{9}{64} k^{4}+\frac{25}{256} k^{6}+\cdots \right) \leq 2\pi \frac{4-3k^2}{4-4k^2} \sqrt{\frac{L}{g}} $$ Putting it all together, we have $$ \displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}\right)\leq T \leq 2\pi\frac{4-3k^2}{4-4k^2}\sqrt{\frac{L}{g}} $$
For larger values of $\theta_{max},$ the two-term approximation is $$ \displaystyle T \approx \displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}\right) $$ If we need more accuracy, we can simply add more terms.
We note that $\displaystyle 2\pi\sqrt{\frac{L}{g}}$ and $\displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}\right)$ are lower bounds for $T.$
We shall now obtain an upper bound for $T.$ First, we note that $$ 1 + \frac{1}{4} k^{2}+\frac{9}{64} k^{4}+\frac{25}{256} k^{6}+\cdots = 1 + \frac{1}{4} k^{2}+\frac{1^2\cdot 3^2}{2^2\cdot 4^2} k^{4}+\frac{1^2 \cdot 3^2 \cdot 5^2}{2^2 \cdot 4^2 \cdot 6^2} k^{6}+\cdots $$ From the above, it's clear that each coefficient in the above series is less than $\frac{1}{4}.$ Then $$ \begin{array}{ll} \displaystyle 1 + \frac{1}{4} k^{2}+\frac{9}{64} k^{4}+\frac{25}{256} k^{6}+\cdots &\\ \lt \displaystyle 1 + \frac{1}{4} k^{2}+\frac{1}{4} k^{4}+\frac{1}{4} k^{6}+\cdots &\mbox{}\\ = \displaystyle 1 + \frac{1}{4} k^{2}\left(1+k^{2}+k^{4}+\cdots\right) &\mbox{}\\ = \displaystyle 1 + \frac{1}{4} k^{2}\frac{1}{1-k^2} &\mbox{}\\ = \displaystyle \frac{4-4k^2}{4-4k^2} +\frac{k^2}{4-4k^2} &\mbox{}\\ = \displaystyle \frac{4-3k^2}{4-4k^2} &\mbox{}\\ \end{array} $$ Thus, multiplying the above by $\displaystyle 2\pi\sqrt{\frac{L}{g}},$ we have $$ \displaystyle T = \displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}+\frac{9}{64} k^{4}+\frac{25}{256} k^{6}+\cdots \right) \leq 2\pi \frac{4-3k^2}{4-4k^2} \sqrt{\frac{L}{g}} $$ Putting it all together, we have $$ \displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}\right)\leq T \leq 2\pi\frac{4-3k^2}{4-4k^2}\sqrt{\frac{L}{g}} $$
Bounding the Period $T$ of a Pendulum
$L=$$\mbox{ m},$ $g=$$\mbox{ m/s}^2,$ $\theta_{max}=$$\mbox{ rad}$
| $\displaystyle 2\pi\sqrt{\frac{L}{g}}$ | $\leq$ | $\displaystyle 2\pi\sqrt{\frac{L}{g}}\left( 1 + \frac{1}{4} k^{2}\right)$ | $\leq$ | $T$ | $\leq$ | $\displaystyle 2\pi\frac{4-3k^2}{4-4k^2}\sqrt{\frac{L}{g}}$ |
| $\leq$ | $\leq$ | $T$ | $\leq$ |
Scenic Vista: The Numbers $e,$ $\pi,$ and $i$
From the series expansion $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!},$ we see that $\displaystyle e=\sum_{n=0}^{\infty}\frac{1}{n!}$ and $\displaystyle e^{-1}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}.$
We also saw in a previous section that $\displaystyle \tan^{-1} x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$ on $[-1,1]$ from which it follows that $\displaystyle \frac{\pi}{4}=\tan^{-1} 1=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}.$
Substitute $i\theta$ in for $x$ in the series expansion $e^x$ and see what happens!
For any $x$ we have $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}.$
Thus, it should be that
$$
\begin{array}{lll}
\displaystyle e^{i\theta}
&=\displaystyle \sum_{n=0}^{\infty}\frac{(i\theta)^{n}}{n!}&\mbox{}\\
&=\displaystyle \sum_{n=0}^{\infty}i^{n}\frac{\theta^{n}}{n!}&\mbox{}\\
&=\displaystyle i^{0}\frac{\theta^{0}}{0!}+i^{1}\frac{\theta^{1}}{1!}+i^{2}\frac{\theta^{2}}{2!}+i^{3}\frac{\theta^{3}}{3!}+i^{4}\frac{\theta^{4}}{4!}+i^{5}\frac{\theta^{5}}{5!}+i^{6}\frac{\theta^{6}}{6!}+i^{7}\frac{\theta^{7}}{7!}+\cdots&\mbox{}\\
&=\displaystyle 1+i\theta-\frac{\theta^{2}}{2!}-i\frac{\theta^{3}}{3!}+\frac{\theta^{4}}{4!}+i\frac{\theta^{5}}{5!}-\frac{\theta^{6}}{6!}-i\frac{\theta^{7}}{7!}+\cdots&\mbox{}\\
&=\displaystyle 1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}-\frac{\theta^{6}}{6!}+\cdots +i\theta-i\frac{\theta^{3}}{3!}+i\frac{\theta^{5}}{5!}-i\frac{\theta^{7}}{7!}+\cdots&\mbox{}\\
&=\displaystyle 1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}-\frac{\theta^{6}}{6!}+\cdots +i\left(\theta-\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!}-\frac{\theta^{7}}{7!}+\cdots\right)&\mbox{}\\
&=\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n\theta^{2n}}{(2n)!}+i\sum_{n=0}^{\infty}\frac{(-1)^n \theta^{2n+1}}{(2n+1)!}&\mbox{}\\
\end{array}
$$
Do these series look familiar?
$${\Huge e^{i\theta}=\cos \theta +i \sin \theta}$$
$${\Huge e^{i\pi}+1=0}$$
$${\Huge e^{i\theta}=\cos \theta +i \sin \theta}$$
$${\Huge e^{i\pi}+1=0}$$