Power series and Taylor series

So far we dealt with number series. This week we add $x$ to the terms: a series like $\sum c_nx^n$ is a separate number series for each value of $x$. It converges for some $x$ and diverges for others. The set of $x$ where it converges is always an interval, and finding this interval is the most dependable question on the exam. In the second half we do the opposite: we write a given function ($e^x$, $\sin x$, $\ln(1+x)$) as an infinite polynomial. This is also the basis of how calculators and programming languages compute Math.sin.

Golden rules:

1. For the radius, apply the ratio test (sometimes the root test), and keep $\lvert x-a\rvert$ outside.
2. The endpoints are always checked separately. The ratio test gives $L=1$ there and does not decide.
3. Derive from a known series: substitute, multiply, differentiate, integrate. Much faster than computing from scratch with derivatives.
4. The Taylor coefficient is $c_n = \dfrac{f^{(n)}(a)}{n!}$. You also use it backwards: from the coefficient to the derivative.

Power series and the radius of convergence

Definition. A power series centered at $a$:

$\sum_{n=0}^{\infty}c_n(x-a)^n = c_0+c_1(x-a)+c_2(x-a)^2+\dots$

Theorem. For every power series one of these three cases holds:

1. It converges only at $x=a$ ($R=0$).
2. It converges for every $x$ ($R=\infty$).
3. There is an $R>0$: for $\lvert x-a\rvert<R$ it converges absolutely, for $\lvert x-a\rvert>R$ it diverges. For $\lvert x-a\rvert = R$ (the endpoints) the theorem says nothing.

$R$ is called the radius of convergence, and the interval of $x$ where it converges is called the interval of convergence.

Template for finding the interval

1) Ratio test: L(x) = lim |a_(n+1)/a_n|   (a_n is the WHOLE term, x included)
2) Solve L(x) < 1  →  |x - a| < R
3) Plug x = a - R and x = a + R into the series; test the resulting NUMBER series
4) Write the interval: ends with [ or ( as appropriate

Example 1. $\displaystyle\sum_{n=1}^\infty\frac{(x-3)^n}{n\,2^n}$.

$\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{\lvert x-3\rvert^{n+1}}{(n+1)2^{n+1}}\cdot\frac{n\,2^n}{\lvert x-3\rvert^n} = \frac{\lvert x-3\rvert}{2}\cdot\frac{n}{n+1}\to\frac{\lvert x-3\rvert}{2}.$

$\lvert x-3\rvert<2$, that is, $1<x<5$, $R = 2$.

• $x=5$: $\sum\dfrac{2^n}{n2^n} = \sum\dfrac1n$, diverges.
• $x=1$: $\sum\dfrac{(-2)^n}{n2^n} = \sum\dfrac{(-1)^n}{n}$, converges by the alternating test.

Interval: $[1,5)$.

Example 2. $\sum\dfrac{x^n}{n!}$: the ratio is $\dfrac{\lvert x\rvert}{n+1}\to0$, for every $x$. $R=\infty$.

Example 3. $\sum n!\,x^n$: the ratio is $(n+1)\lvert x\rvert\to\infty$ ($x\neq0$). $R=0$.

Example 4. $\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{4^n}$. There are only even powers; apply the ratio test to the whole term: $\dfrac{x^2}{4}<1$, that is, $\lvert x\rvert<2$. At the endpoints $x^2 = 4$, the terms are $(-1)^n$, diverges. The interval is $(-2,2)$. In fact this is a geometric series with ratio $-x^2/4$, with sum $\dfrac{1}{1+x^2/4}$.

PITFALL · $x$ with a coefficient. If you see $(2x+1)^n$ or $(3x-6)^n$, the center and radius cannot be read off directly. Solve the inequality $\lvert2x+1\rvert<1$ for $x$: $-1<x<0$. The center is $-\frac12$, $R = \frac12$. Writing "$R=1$" is a common mistake.

Writing functions as power series

The main source is the geometric series:

$\frac{1}{1-x} = \sum_{n=0}^\infty x^n,\qquad\lvert x\rvert<1.$

Substitution. Replace $x$ with $-x^2$: $\dfrac{1}{1+x^2} = \sum(-1)^nx^{2n}$, $\lvert x\rvert<1$.

Changing the center. Write $\dfrac1x$ around $a=2$: $\dfrac1x = \dfrac{1}{2+(x-2)} = \dfrac12\cdot\dfrac{1}{1+\frac{x-2}{2}} = \sum_{n=0}^\infty\dfrac{(-1)^n(x-2)^n}{2^{n+1}}$, $\lvert x-2\rvert<2$.

Term-by-term differentiation and integration. Inside the interval of convergence, the derivative and integral of a power series can be taken term by term and the radius does not change (the endpoints may).

• Integrate $\dfrac{1}{1+x}$: $\ln(1+x) = \sum_{n=1}^\infty\dfrac{(-1)^{n+1}x^n}{n}$, $-1<x\le1$.
• Integrate $\dfrac{1}{1+x^2}$: $\arctan x = \sum_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{2n+1}$, $-1\le x\le1$.
• The derivative of $\dfrac{1}{1-x}$: $\dfrac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1}$.

After integrating, do not forget the constant: determine it by plugging in $x=0$. Since $\ln(1+0) = 0$, the constant is $0$.

Taylor and Maclaurin series

Definition. If $f$ is differentiable to every order near $a$, the Taylor series of $f$ centered at $a$ is:

$\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n = f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots$

If $a=0$ it is called the Maclaurin series. The sum of the first $n+1$ terms is called the Taylor polynomial of degree $n$, $P_n(x)$.

Intuition. $P_n$ is the unique polynomial of degree $n$ that shares with $f$ at the point $a$ the same value, the same slope, the same curvature (everything up to the $n$th derivative). As the degree increases, it hugs $f$ better near $a$.

Maclaurin series to memorize

$\sin$ is an odd function, and its series has only odd powers. $\cos$ is even, with only even powers. This is a checking tool.

Example 5 (Taylor from the definition). $f(x) = \ln x$, $a=1$. $f(1) = 0$, $f' = x^{-1}$, $f'' = -x^{-2}$, $f''' = 2x^{-3}$, and in general $f^{(n)}(x) = (-1)^{n+1}(n-1)!\,x^{-n}$. At $a=1$, $f^{(n)}(1) = (-1)^{n+1}(n-1)!$, and the coefficient is $\dfrac{(-1)^{n+1}(n-1)!}{n!} = \dfrac{(-1)^{n+1}}{n}$:

$\ln x = (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\dots$

Example 6 (from a known series). $e^{-x^2}$: in the series of $e^u$, put $u=-x^2$: $1-x^2+\dfrac{x^4}{2!}-\dfrac{x^6}{3!}+\dots$ One line instead of taking six derivatives.

Example 7 (approximation with a Taylor polynomial). The polynomial $P_4$ of $\cos x$ is $1-\dfrac{x^2}{2}+\dfrac{x^4}{24}$. For $x = 0.5$ it gives $1-0.125+0.0026042 = 0.8776042$. The true value is $\cos0.5 = 0.8775826$. Four correct digits with only three terms. As you move away from the center (like $x=3$) the same polynomial gets worse; more terms are needed. Estimating how large the error is is the topic of next week.

Example 8 (product). The first three nonzero terms of the Maclaurin series of $e^x\sin x$: multiply the two series like polynomials and add up the terms up to $x^3$.

$\left(1+x+\frac{x^2}{2}+\dots\right)\left(x-\frac{x^3}{6}+\dots\right) = x+x^2+\left(\frac12-\frac16\right)x^3+\dots = x+x^2+\frac{x^3}{3}+\dots$

When multiplying, add each degree separately: the $x^3$ term comes from $1\cdot(-\frac{x^3}{6})$ and $\frac{x^2}{2}\cdot x$.

Radius with the root test

If the whole term is an $n$th power, the root test is cleaner: for $\displaystyle\sum\frac{(x-1)^n}{n^n}$, $\sqrt[n]{\lvert a_n\rvert} = \dfrac{\lvert x-1\rvert}{n}\to0$, for every $x$. $R = \infty$. For $\displaystyle\sum\left(\frac{n+1}{2n}\right)^nx^n$ the root is $\dfrac{n+1}{2n}\lvert x\rvert\to\dfrac{\lvert x\rvert}{2}$, $R = 2$.

TIP · from the coefficient to the derivative. $f^{(k)}(a) = k!\times[\text{coefficient of }(x-a)^k]$. In questions like “find $f^{(20)}(0)$”, build the series from a known series, read off the coefficient of $x^{20}$, and multiply by $20!$.

Common mistakes