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Chapter 1: Problem 15

Find the Taylor series centered at \(x=a\) and its corresponding radius of convergence for the given function. In most cases, you need not employ the direct method of computation of the Taylor coefficients. a. \(f(x)=\sinh x, a=0\) b. \(f(x)=\sqrt{1+x}, a=0\). c. \(f(x)=\ln \frac{1+x}{1-x}, a=0\) d. \(f(x)=x e^{x}, a=1\). e. \(f(x)=\frac{1}{\sqrt{x}}, a=1\) f. \(f(x)=x^{4}+x-2, a=2\) g. \(f(x)=\frac{x-1}{2+x}, a=1\)

Short Answer

Expert verified
a. Taylor series is \(sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+...\) and the radius of convergence is \(+\infty\). b. Taylor series is \(1+\frac{x}{2}-\frac{x^2}{8}+...\) and the radius of convergence is 1. c. Taylor series is \(2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}\) and the radius of convergence is 1. d. Taylor series is \(2- (x-1)+ \frac{(x-1)^2}{2!}-...\) and the radius of convergence is \(+\infty\). e. Taylor series is \(1- \frac{1}{2}(x-1) + \frac{3}{8}(x-1)^2 -...\) and the radius of convergence is 1. f. Taylor series is \(x^{4}+x-2\) and the radius of convergence is \(+\infty\). g. Taylor series is \(1-3 +\frac{3x}{2} - \frac{3x^2}{4}+...\) and the radius of convergence is 2.

Step by step solution

01

Problem a. \(f(x) = sinh(x), a=0\)

The sinh function expands directly into a Taylor series. \(sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+...\). It represents the function for all real numbers, so the radius of convergence is \(+\infty\).
02

Problem b. \(f(x) = \sqrt{1+x}, a=0\).

We can use the binomial series formula: \((1+x)^k = 1 + kx + \frac{k(k-1)x^2}{2!}+...\) If we set k=1/2, we have the Taylor series for the function. Hence, \(f(x)=\sqrt{1+x} = 1+\frac{x}{2}-\frac{x^2}{8}+...\). Radius of convergence is 1.
03

Problem c. \(f(x)=\ln \frac{1+x}{1-x}, a=0\).

Simplify the function as the difference of natural logarithms. \[f(x)=\ln(1+x)-\ln(1-x)=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}\]. With this representation, the radius of convergence is 1.
04

Problem d. \(f(x)=xe^{x}, a=1\).

Here we can use the known Taylor Series for \(e^x\) which is \(e^x = \sum_{n=0}^{+\infty} \frac{x^n}{n!}\). For \(f(x)=xe^{x}\), shift x to (x-1) considering a=1 and apply the above Taylor series. Hence, \(f(x)= (x-1)e^{x-1} = 2- (x-1)+ \frac{(x-1)^2}{2!}-...\). The radius of convergence is \(+\infty\).
05

Problem e. \(f(x)=\frac{1}{\sqrt{x}}, a=1\).

Here also apply the binomial series expansion with k=-1/2. Hence, \(\frac{1}{\sqrt{x}}=(1-(x-1))^{-1/2} = 1- \frac{1}{2}(x-1) + \frac{3}{8}(x-1)^2 -...\). Radius of convergence is 1.
06

Problem f. \(f(x)=x^{4}+x-2, a=2\).

This is a polynomial and hence its Taylor series is itself. So, \(f(x)=x^{4}+x-2\). The radius of convergence is \(+\infty\).
07

Problem g. \(f(x)=\frac{x-1}{2+x}, a=1\).

Here, simplify the function using partial fractions. This will result in a geometric series. So, \(f(x)=\frac{x-1}{2+x}=1-\frac{3}{2+x}=1-3(1-\frac{x}{2})^{-1} = 1-3 +\frac{3x}{2} - \frac{3x^2}{4}+...\). The radius of convergence is 2.

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Most popular questions from this chapter

Solve the following equations for \(x\) : a. \(\cosh (x+\ln 3)=3\) b. \(2 \tanh ^{-1} \frac{x-2}{x-1}=\ln 2\). c. \(\sinh ^{2} x-7 \cosh x+13=0\).

Find the sum for each of the series: a. \(5+\frac{25}{7}+\frac{125}{49}+\frac{625}{343}+\cdots\) b. \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 3}{4^{n}}\) c. \(\sum_{n=2}^{\infty} \frac{2}{5^{n}}\). d. \(\sum_{n=-1}^{\infty}(-1)^{n+1}\left(\frac{e}{\pi}\right)^{n}\). e. \(\sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}+\frac{1}{3^{n}}\right)\). f. \(\sum_{n=1}^{\infty} \frac{3}{n(n+3)}\) g. What is \(0.569 ?\)

Write the following in terms of logarithms: a. \(\cosh ^{-1} \frac{4}{3}\). b. \(\tanh ^{-1} \frac{1}{2}\). c. \(\sinh ^{-1} 2\).

Determine the exact values of a. \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\). b. \(\tan \left(\sin ^{-1} \frac{x}{7}\right)\). c. \(\sin ^{-1}\left(\sin \frac{3 \pi}{2}\right)\).

In the event that a series converges uniformly, one can consider the derivative of the series to arrive at the summation of other infinite series. a. Differentiate the series representation for \(f(x)=\frac{1}{1-x}\) to sum the series \(\sum_{n=1}^{\infty} n x^{n},|x|<1\) b. Use the result from part a to sum the series \(\sum_{n=1}^{\infty} \frac{n}{5^{n}}\) c. Sum the series \(\sum_{n=2}^{\infty} n(n-1) x^{n},|x|<1\) d. Use the result from part \(c\) to sum the series \(\sum_{n=2}^{\infty} \frac{n^{2}-n}{5^{n}}\) e. Use the results from this problem to sum the series \(\sum_{n=4}^{\infty} \frac{n^{2}}{5^{n}}\)
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