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## Power Series

A power series in $$x$$ about $$x_0$$ is a series of the form $\sum_{n=0}^{\infty}a_n(x-x_0)^n=a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots,$ where $$a_0,a_1,\ldots$$ are real numbers. If $$x_0=0$$, it becomes $\sum_{n=0}^{\infty}a_n x^n=a_0+a_1 x+a_2 x^2+\cdots$ Example.

1. $$\displaystyle 1+\frac{x-1}{1!}+\frac{(x-1)^2}{2!}+\frac{(x-1)^3}{3!}+\cdots \;\;\;\; (=e^{x-1})$$

2. $$\displaystyle 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \;\;\;\;\;\;\;\; (=e^x)$$

3. $$\displaystyle 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots \;\;\;\;\;\;\;\; (=\cos x)$$

4. $$\displaystyle \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots \;\;\;\;\;\;\;\; (=\sin x)$$

Convergence: Power series $$\sum_{n=0}^{\infty}a_n(x-x_0)^n$$ converges to $$L$$ at $$x=c$$ if $\lim_{n\to \infty}\sum_{k=0}^{n}a_k(c-x_0)^k=L,$ and we write $$\sum_{n=0}^{\infty}a_n(c-x_0)^n=L$$. It diverges if it does not converge.

Example.

1. $$\sum_{n=0}^{\infty}a_n(x-x_0)^n$$ converges to $$a_0$$ at $$x=x_0$$.

2. $$\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ converges to $$e^x$$ at each $$x$$.

Radius of Convergence: For a power series $$\sum_{n=0}^{\infty}a_n(x-x_0)^n$$, there is number $$R\geq 0$$, possibly $$R=\infty$$, such that it converges for each $$x$$ in $$(x_0-R,x_0+R)$$, i.e., $$|x-x_0| < R$$, and diverges for each $$x$$ in $$(-\infty,x_0-R)\cup(x_0+R,\infty)$$, i.e., $$|x-x_0| > R$$. This $$R$$ is called the radius of convergence of the power series.

Note that the power series may converge or diverge at $$x_0-R$$ and $$x_0+R$$. So the interval of convergence would be one of $$(x_0-R,x_0+R),\; (x_0-R,x_0+R],\; [x_0-R,x_0+R)$$, and $$[x_0-R,x_0+R]$$. Also note that $$R=\infty$$ if and only if the power series converges for all $$x$$.

Ratio Test: For a power series $$\sum_{n=0}^{\infty}a_n(x-x_0)^n$$, suppose $L=\lim_{n\to \infty}\;\left\vert\frac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}\right\vert=|x-x_0|\lim_{n\to \infty}\;\left\vert\frac{a_{n+1}}{a_n}\right\vert\;.$ The power series converges if $$L < 1$$, diverges if $$L > 1$$ and may converge or diverge if $$L=1$$. Also $$R=1/\displaystyle\lim_{n\to \infty}\;\left\vert\frac{a_{n+1}}{a_n}\right\vert$$ is the radius of convergence.

Example. Find the radius and interval of convergence of $\sum_{n=0}^{\infty}\frac{(-3)^n}{n2^n}(x-4)^n.$ Solution. Here $$a_n=(-3)^n/n2^n$$. So $L=|x-x_0|\lim_{n\to \infty}\;\left\vert\frac{a_{n+1}}{a_n}\right\vert =|x-4|\lim_{n\to \infty}\;\left\vert\frac{(-3)^{n+1}/(n+1)2^{n+1}}{(-3)^n/n2^n}\right\vert =|x-4|\lim_{n\to \infty}\frac{3n}{2(n+1)}=\frac{3}{2}|x-4|.$ So the power series converges when $$L=\frac{3}{2}|x-4| < 1$$, i.e., $$|x-4| < \frac{2}{3}$$, i.e., for all $$x$$ in $$(4-\frac{2}{3},4+\frac{2}{3})=(\frac{10}{3},\frac{14}{3})$$. It diverges when $$x < \frac{10}{3}$$ or $$x > \frac{14}{3}$$. So the radius of convergence is $$\frac{2}{3}$$. It can be verified that the power series diverges when $$x=\frac{10}{3}$$ and $$x=\frac{14}{3}$$. Thus the interval of convergence is $$\left(\frac{10}{3},\frac{14}{3}\right)$$.

Taylor Series: Suppose $$f$$ is infinitely differentiable function at $$x_0$$, i.e., $$f^{(n)}(x_0)$$ exists for $$n=1,2,\ldots$$. A Taylor series of $$f$$ about $$x_0$$ is a series of the form $\sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n=f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots$ $$f$$ is called analytic at $$x=x_0$$ if the Taylor series of $$f$$ about $$x_0$$ has a positive radius of convergence, i.e., $$R > 0$$. When is the Taylor series of $$f$$ equal to $$f(x)$$? $f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n=f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots,$ for all $$x$$ in $$(x_0-R,x_0+R)$$ whenever $\lim_{n\to \infty}\sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!} (x-x_0)^k=f(x),$ for all $$x$$ in $$(x_0-R,x_0+R)$$.

Example.

1. $$e^x=\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}= 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$ and $$R=\infty$$.

2. $$\cos x=\displaystyle\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}=\displaystyle 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$ and $$R=\infty$$.

3. $$\sin x=\displaystyle\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{(2n+1)!}=\displaystyle \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$ and $$R=\infty$$.

4. $$\ln(1+x)=\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$ and $$R=1$$.

5. $$\displaystyle\frac{1}{1-x}=\displaystyle\sum_{n=0}^{\infty} x^n= 1+x+x^2+x^3+\cdots$$ and $$R=1$$.

Calculus on power series:
On the interval $$(x_0-R,x_0+R)$$, power series $$\sum_{n=0}^{\infty}a_n(x-x_0)^n$$ can be differentiated and integrated term by term.

Example. \begin{align*} \displaystyle\frac{d}{dx}\cos x=\frac{d}{dx}\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!} &=\frac{d}{dx}\left( 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)\\ &=-\frac{x}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots\\ &=-\sin x. \end{align*} Index shift:
Suppose $$y=\displaystyle\sum_{n=0}^{\infty}a_nx^n=a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots$$. Its derivative is $y'=a_1 +2a_2 x+3a_3 x^2+4a_4x^3+\cdots=\sum_{n=1}^{\infty}na_nx^{n-1}.$ Note that the generic term of the series for $$y'$$ involves $$x^{n-1}$$ instead of $$x^n$$. Our goal is to write the generic term involving $$x^n$$. To do that substitute $$m=n-1$$. Then $$n=m+1$$ and $$n=1$$ corresponds $$m=0$$. So we get $y'=\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{m=0}^{\infty}(m+1)a_{m+1}x^{m}.$ Finally we can replace the dummy index $$m$$ by $$n$$ and get $y'=\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n}.$ Similarly verify the index shift for $$y''$$: $y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}.$

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