§8.4 Taylor series

Introduction

Taylor series and the Ratio Test

Taylor series of several important functions

Summary

The series $T(x) = 3-x+\frac{1}{3}x^2-\frac{1}{9}x^3+\cdots$ is geometric with

For the series of $f(x) = \ln(2)-\ln(3-x)$ centered at $1$, consecutive terms have ratios $r_n(x) = \frac{n}{n+1}\cdot\frac{1}{2}(x-1)$. The limit $r(x)$ equals

For the logarithm series above with limiting ratio $r(x) = \frac{1}{2}(x-1)$, solving the convergence condition gives

For the Taylor series of $e^x$, the term ratio is $r_n(x) = \frac{x}{n+1}$, so the limiting ratio is $r(x) = 0$. The Ratio Test then guarantees

For the sine series $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$, consecutive term ratios satisfy $r_n(x) = -\frac{x^2}{(2n+2)(2n+3)}$, whose limit is $0$. The Ratio Test therefore yields

Plots show the degree $10$ Taylor polynomial of a mystery function matching it to pixel accuracy on the interval $(1,3)$, centered at $1$. A natural conjecture is

The series $\sum_{k=0}^{\infty}\frac{x^k}{k!}$ is the Taylor series of

For the series $\sum_{k=0}^{\infty}\frac{x^k}{k!}$, the ratio of consecutive terms gives

Substituting $x^2$ into the sine series gives $\sin(x^2) = x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\cdots$, which converges

For $f(x) = \frac{1}{1+x}$ centered at $a = 1$, writing $f(x) = \frac{1}{2}\cdot\frac{1}{1+\frac{x-1}{2}}$ gives the series

Using the degree $10$ polynomial of $e^x$ at $a=0$ to estimate $e^2$, the Lagrange bound with $M = 9$ gives error at most $\frac{9\cdot2^{11}}{11!}$. Report this bound, rounded to six decimals.

To estimate $\cos(1)$ within $0.00000001$ using the Taylor polynomial at $0$, the Lagrange bound with $M = 1$ requires degree $n = 11$. Report the required degree.

A spreadsheet table listing derivatives at zero shows every value $f^{(k)}(0) = 1$ for the function behind the series $\sum\frac{x^k}{k!}$.

For the series behind these tables, the degree $10$ polynomial is visually indistinguishable from the function across $-4 \lt x \lt 4$, while degree $20$ covers roughly $-8 \lt x \lt 8$.

For $f(x) = \cos(x)$ centered at $a = \frac{\pi}{2}$, the degree $4$ Taylor polynomial contains no $(x-\frac{\pi}{2})^2$ term.