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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