Power Series, Taylor Series

Power Series: infinite polynomials

A power series about $x=0$ is a series of the form $$\sum_{n=0}^\infty c_nx^n=c_0+c_1x+c_2x^2+\cdots+c_nx^n+\cdots.$$ A power series about $x=a$ is a series of the form $$\sum_{n=0}^\infty c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_n(x-a)^n+\cdots,$$ where the center $a$ and the coefficients $c_0,c_1,c_2,\cdots,c_n,\cdots$ are constants.

Example

Let $c_n=1$ for all $n$, we have $$\sum_{n=0}^\infty x^n$$

We consider $x$ as a constant, it is the geometric series. The series converges when $x\in (-1,1)$, and it converges to ${1\over 1-x}$.

Convergence of a Power Series

• It diverges when $|x|>1$ because $\lim\limits_{n\rightarrow\infty} {x^n\over n}\neq 0$.

• It converges when $|x|<1$.

• It diverges when $x=-1$

• It converges when $x=1$

• It diverges when $|x|>1$ because $\lim {x^{2n-1}\over 2n-1}\neq 0$.

• It converges when $|x|<1$.

• It converges when $x=-1$

• It converges when $x=1$

• It converges for all $x$

Convergence Theorem for Power Series

If the power series $\sum_{n=0}^\infty a_n x^n$ converges at $x=c$ with $|c|> 0$, then it converges absolutely for all $x\in (-|c|,|c|)$, If the series diverges at $x=d$, then it diverges for all $x$ with $|x|>|d|$.

If it converges at $x=c$, then we have that $|a_nc^n|\rightarrow 0$. That is, there exist $N$ such that $a_n<|c|^{-n}$ for all $n>N$.

• Root test, for any $x\in (-|c|,|c|)$, we have $\sqrt[n]{|a_nx^n|}=\sqrt[n]{|a_n|}|x|<1$ for $n>N$, therefore, it convergs for $x\in(-|c|,|c|)$.

If it diverges at $x=d$,then we can prove by contradiction, that it diverges for $|x|>d$.

The convergence of the series $\sum c_n(x - a)^n$ is described by one of the following three cases:

• There is a positive number $R$ such that the series diverges for $x$ with $|x - a|> R$ but converges absolutely for $x$ with $|x - a|< R$. The series may or may not converge at either of the endpoints $x = a - R$ and $x = a + R$.

• The series converges absolutely for every $x$ ($R = \infty$).

• The series converges at $x = a$ and diverges elsewhere ($R = 0$).

$R$ is called the radius of convergence of the power series, and the interval of radius $R$ centered at $x=a$ is called the the interval of convergence. The interval may be open, closed, or half-open.

Steps to test the convergence of a power series

• Use Ratio Test or Root Test to find the radius $R$.

• If $R$ is finite and positive. Test both endpoints using Comparison Test, Interal Test, or Alternating Series Test

Operations on Power Series

If $A(x)=\sum_{n=0}^{\infty}a_nx^n$ and $B(x)=\sum_{n=0}^\infty b_nx^n$ converge absolutely for $|x|<R$, and $$c_n=a_0b_n+a_1b_{n-1}+a_2b_{n-2}+\cdots+a_{n-1}b_1+a_nb_0=\sum_{k=0}^na_kb_{n-k}$$ then $\sum_{n=0}^\infty c_nx^n$ converges absolutely to $A(x)B(x)$ for $|x|<R$: $$\sum_{n=0}^\infty a_nx^n \sum_{n=0}^\infty b_nx^n=\sum_{n=0}^\infty c_n x^n$$

Substitute a function in a convergent power series

If $\sum_{n=0}^\infty a_nx^n$ converges absolutely for $|x|<R$, then $\sum_{n=0}^\infty a_n(f(x))^n$ converges absolutely for any continuous function $f$ on $|f(x)|<R$.

The Term-by-Term Differentiation Theorem

If $\sum c_n(x-a)^n$ has radius of convergence $R>0$, it defines a function $$f(x)=\sum_{n=0}^\infty c_n(x-a)^n,\mbox{ on the interval } a-R<x<a+R$$ The function $f$ has derivatives of all orders inside the interval, and we obtain the derivatives by differentiating the original series terms by term: $$\begin{align*}f'(x)=&\sum_{n=1}^\infty nc_n(x-a)^{n-1}\\ f''(x)=&\sum_{n=2}^\infty n(n-1)c_n(x-a)^{n-2}\end{align*}$$

The Term-by-Term Integration Theorem

If $$f(x)=\sum_{n=0}^\infty c_n(x-a)^n$$ converges for $|x-a|<R$. Then $$\sum_{n=0}^\infty c_n{(x-a)^{n+1}\over n+1}$$ converges for $|x-a|<R$ and $$\int f(x)dx=\sum_{n=0}^\infty c_n{(x-a)^{n+1}\over n+1}+C$$ for $|x-a|<R$.

Example

Taking derivative, we have $$f'(x):=\sum_{n=0}^\infty {(-1)^nx^{2n}} = {1\over 1+x^2}.$$

So we have $f(x)=\arctan(x)$.

Series Representation

Assume that $f$ is the sum of a power series about $x=a$: $$f(x)= \sum_{n=0}^\infty a_n(x-a)^n=a_0+a_1(x-a)+a_2(x-a)^2+\cdots+a_n(x-a)^n+\cdots$$

• $f(a)=a_0$

• $f'(a)=1!\cdot a_1$

• $f''(a)=2!\cdot a_2$

• $f'''(a)=3! \cdot a_3$ $$f(x)= f(a)+f'(a)(x-a)+{f''(a)\over 2!}(x-a)^2+\cdots+{f^{(n)}(a)\over n!}(x-a)^n+\cdots$$

Taylor and Maclaurin Series

Let $f$ be a function with derivatives of all orders throughout some interval containing $a$ as an interior point. Then its Taylor series at $x=a$ is $$f(a)+f'(a)(x-a)+{f''(a)\over 2!}(x-a)^2+\cdots+{f^{(n)}\over n!}(x-a)^n+\cdots=\sum_{k=0}^\infty {f^{(k)}(a)\over k!}(x-a)^k.$$ The Maclaurin series of $f$ is the Taylor series at $x=0$, or $$f(0)+f'(0)x+{f''(0)\over 2!}x^2+\cdots+{f^{(n)}(0)\over n!}x^n+\cdots=\sum_{k=0}^\infty {f^{(k)}(0)\over k!}x^k.$$

Find the Taylor series generated by $f(x)={1\over x}$ at $x=2$. When does it converge to $1/x$?

We can find that $$f^{(n)}(x)={(-1)^{n}n!\over x^{n+1}}$$

It converges when $x\in (0,4)$.

Taylor Polynomials

Let $f$ be a function with derivatives of order up to $N$ in some interval containing $a$ as an interior point. Then for any integer $n$ from 0 through $N$, the Taylor polynomial of order $n$ generated by $f$ at $x=a$ is $$P_n(x)=f(a)+f'(a)(x-a)+{f''(a)\over 2!}(x-a)^2+\cdots+{f^{(n)}(a)\over n!}(x-a)^n.$$

Two questions.

When does the Taylor series converge to its generating function?

How accurately does a Taylor polynomial approximatie the function on a given interval?

Taylor's Theorem

If $f$ and its first $n$ derivatives $f'$, $f'', \cdots, f^{(n)}$ are continuous on an open interval $I$ containing $a$, and $f^{(n)}$ is differentiable on the same interval, then for given $x\in I$, there exist a number $c$ between $a$ and $x$ such that $$\begin{align*}f(x) =&{\color{blue}f(a)+f'(a)(x-a)+{f''(a)\over 2!}(x-a)^2+\cdots+{f^{(n)}(a)\over n!}(x-a)^n}\\ &+{\color{red}{f^{(n+1)}(c)\over (n+1)!}(x-a)^{n+1}}.\\ =&{\color{blue}f(a)+f'(a)(x-a)+{f''(a)\over 2!}(x-a)^2+\cdots+{f^{(n)}(a)\over n!}(x-a)^n}\\ &+{\color{red}R_n(x)}.\\ \end{align*}$$ The function $R_n(x)$ is the remainder of order $n$ or the error term for using $P_n(x)$ approximate $f$ over the interval $I$.

If $R_n(x)\rightarrow 0$ as $n\rightarrow \infty$ for all $x\in I$, we say that the Taylor series $P_n$ generated by $f$ at $x=a$ converges to $f$ on $I$, and we write $$f(x)=\sum_{k=0}^\infty {f^{(k)}(a)\over k!}(x-a)^k$$

The Remainder Estimation Theorem

If there is a positive constant $M$ such that $|f^{(n+1)}(t)|\leq M$ for all $t$ between $x$ and $a$, inclusive, then the remainder term $R_n(x)$ in Taylor's Theorem satisfies the inequality $$|R_n(x)|\leq M{|x-a|^{n+1}\over (n+1)!}$$ If this inequality (the same $M$) holds for every $n$ and the order conditions of Taylor's theorem are satisfied by $f$, then the series converges to $f(x)$.

Example

For what $x$ can we replace $\cos x$ by $1-(x^2/2!)$ with an error of magnitude no greater than $3\times 10^{-4}$.

The Binomial Series for Powers and Roots

For a given constant $m$ (can be non-integer), computer the Taylor series generated by $f(x)=(1+x)^m$ at $x=0$. $$\begin{align*} 1+mx+{m(m-1)\over 2!}x^2+&{m(m-1)(m-2)\over 3!}x^3 + \cdots \\ +& {m(m-1)(m-2)\cdots(m-k+1)\over k!}x^k+\cdots \end{align*}$$ This is called binomial series, which converges absolutely for $|x|<1$.

• If $m$ is a positive integer or 0, finite terms

• If $m$ is not a positive integer or 0, infinitely many terms

The Binomial Series

For $|x|<1$, $$(1+x)^m = 1+\sum_{k=1}^\infty \begin{pmatrix} m\\k \end{pmatrix}x^k,$$ where we define

• $\sqrt{1+x}$

We can compute its Taylor series at $x=0$.

• $\sqrt{1-x^2}$

We can compute its Taylor series at $x=0$.

• $\sqrt{1-{1\over x}}$

We can not compute its Taylor series at $x=0$ because it is not defined at $x=0$.

Example

Evaluating Nonelementary Integrals $\int \sin x^2 dx$ and estimate $\int_0^1\sin x^2 dx$

Use Taylor series to find a limit

• $$\lim\limits_{x\rightarrow 1}{\ln x\over x-1}$$

It is ${0\over 0}$, we can just use Taylor series. $$\lim\limits_{x\rightarrow 1}{\ln x\over x-1}=\lim\limits_{x\rightarrow 1}{x-1\over x-1}=1$$

• $$\lim\limits_{x\rightarrow0}\left({1\over \sin x}-{1\over x}\right)$$

We need to change it to ${0\over 0}$ first; it is equivalent to $$\lim\limits_{x\rightarrow0}{x-\sin x\over x\sin x } = \lim\limits_{x\rightarrow0}{{1\over 6}x^3\over x x }=0$$

Frequently Used Taylor Series