 # Power series for derivatives a. Differentiate the Taylor series centered at 0 for the following functions. b. Identify the function represented by the differentiated series. c. Give the interval of convergence of the power series for the derivative. f(x)=ln(1+x) Jaya Legge 2021-02-18 Answered
Power series for derivatives
a. Differentiate the Taylor series centered at 0 for the following functions.
b. Identify the function represented by the differentiated series.
c. Give the interval of convergence of the power series for the derivative.
$f\left(x\right)=\mathrm{ln}\left(1+x\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Clara Reese

Taylor series of function f(x) at a is defined as:
$f\left(x\right)=f\left(a\right)+\frac{{f}^{\prime }\left(a\right)}{1!}\left(x-a\right)+\frac{{f}^{″}\left(a\right)}{2!}\left(x-a{\right)}^{2}+\frac{{f}^{‴}\left(a\right)}{3!}\left(x-a{\right)}^{3}+...$
Consider the given:
$f\left(x\right)=\mathrm{ln}\left(1+x\right)$
$f\left(x\right)\approx 0+\frac{\frac{d}{dx}\left(\mathrm{ln}\left(1+x\right)\right)\left(0\right)}{1!}x+\frac{\frac{{d}^{2}}{d{x}^{2}}\left(\mathrm{ln}\left(1+x\right)\right)\left(0\right)}{2!}{x}^{2}+\frac{\frac{{d}^{3}}{d{x}^{3}}\left(\mathrm{ln}\left(1+x\right)\right)\left(0\right)}{3!}{x}^{3}+...$
$f\left(x\right)\approx 0+\frac{1}{1!}x+\frac{-1}{2!}{x}^{2}+\frac{2}{3!}{x}^{3}+\frac{-6}{4!}{x}^{4}+\frac{24}{5!}{x}^{5}+...$
$f\left(x\right)\approx x-\frac{1}{2}{x}^{2}+\frac{1}{3}{x}^{3}-\frac{1}{4}{x}^{4}+\frac{1}{5}{x}^{5}+...$
b) The series sum representation-
$\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n+1}\frac{{x}^{n}}{n}$
Apply ratio test and find interval of convergence.
If $|\frac{{a}_{n+1}}{{a}_{n}}|\le q$ eventually for some $\sum _{n=1}^{\mathrm{\infty }}|{a}_{n}|$ converges
If $|\frac{{a}_{n+1}}{{a}_{n}}|>1$ eventually then $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ diverges.
Consider the given series:
$f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n+1}\frac{{x}^{n}}{n}$
${a}_{n}=\left(-1{\right)}^{n+1}\frac{{x}^{n}}{n}$
${a}_{k+1}=\left(-1{\right)}^{n+2}\frac{{x}^{n+1}}{n+1}$
$\frac{{a}_{n+1}}{{a}_{n}}=\frac{\left(-1{\right)}^{n+2}\frac{{x}^{n+1}}{n+1}}{\left(-1{\right)}^{n+2}\frac{{x}^{n}}{n}}$
$=\left(-1\right)\frac{nx}{\left(n+1\right)}$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\left(-1\right)\frac{x}{\left(1+\frac{1}{n}\right)}|$
$=|-x|$
From ratio test, if L > 1 series diverges.
L< 1 series converges.
||x||<1 series converges.
Find the interval of convergence. $-1
Interval of convergence $=\left\{x:x=-1
Interval of convergence =(-1,1]

###### Not exactly what you’re looking for? Jeffrey Jordon Jeffrey Jordon Jeffrey Jordon