Taylor series and interval of convergence a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. f(x)=log_3(x+1),a=0

Taylor series and interval of convergence a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. f(x)=log_3(x+1),a=0

Question
Series
asked 2021-03-11
Taylor series and interval of convergence
a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.
b. Write the power series using summation notation.
c. Determine the interval of convergence of the series.
\(f(x)=\log_3(x+1),a=0\)

Answers (1)

2021-03-12

A) The Taylor series for a function f(x) continuously differentiable at \(x=a\) is given by,
\(f(x)=f(a)+\frac{(x-a)}{1!}f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)...\)
Here
\(f(x)=\log_3(x+1)\)
\(a=0,\)
Substitute the values,
\(f(x)=f(a)+\frac{(x-a)}{1!}f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)...\)
\(\log_3(x+1)=\log_3(x+1)+\frac{x}{1!}(\frac{1}{(x+1)\log3})+\frac{x^2}{2\times1}(\frac{-1}{(x+1)^2\log3})+\frac{x^3}{3\times2\times1}[-(\frac{-2}{(x+1)^3\log3})]...\)
\(=\frac{\log(x+1)}{\log(3)}+\frac{x}{(x+1)\log3}-\frac{x^2}{2(x+1)^2\log3}+\frac{x^3}{3(x+1)^3\log3}...\)
The first four terms of the Taylor's expansion of \(f(x)=\log_3(x+1)\) are respectively,
\(\frac{\log(x+1)}{\log_3},\frac{x}{(x+1)\log3},\frac{x^2}{2(x+1)^2\log3},\frac{x^3}{3(x+1)^3\log3}\)
B) The power series for the above Taylor's expansion is given by,
\(\log_3(x+1)=\sum_{n=0}^\infty\left\{\frac{(x-a)^n}{n!}\cdot f^n(a)\right\}\)
Re-arrange the series,
\(\log_3(x+1)=\frac{x}{(x+1)\log(3)}+\frac{x}{(x+1)\log(3)}-\frac{1}{\log(3)}[\frac{x^2}{2(x+1)^2}-\frac{x^3}{3(x+1)^3}+\frac{x^4}{4(x+1)^4}-\frac{x^5}{5(x+1)^5}+\frac{x^6}{6(x+1)^6}...]\)
\(=\frac{x}{(x+1)\log(3)}+\frac{x}{(x+1)\log(3)}-\frac{1}{\log(3)}\sum_{n=2}^\infty[\frac{x^n}{n(x+1)^n}\cdot(-1)^n]\)
The series will be converging if the term \(\sum_{n=2}^\infty[\frac{x^n}{n(x+1)^n}\cdot(-1)^n]\) is converging,
Let \(a_n=(-1)^n\cdot\frac{x^n}{n(x+1)^n}\)
Apply the Ratio Test,
\(L=\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|\)
\(=\lim_{n\rightarrow\infty}|\frac{[\frac{x^{n+1}}{(n+1)(x+1)^{n+1}}\cdot(-1)^{n+1}]}{[\frac{x^n}{n(x+1)^n}\cdot(-1)^n]}|\)
\(=\lim_{n\rightarrow\infty}|\frac{-xn}{(x+1)(n+1)}|\)
\(=|\frac{x}{x+1}|\cdot\lim_{n\rightarrow\infty}|\frac{1}{1+\frac1n}|\)
\(=|\frac{x}{x+1}|\)
The series is converging if,
\( L<1\)
\(|\frac{x}{x+1}|<1\)
\(-1<\frac{x}{x+1}<1\)
The term \(\frac{x}{x+1}<1\) is true for any real value of \(x \in R,\)
Also,
\(\frac{x}{x+1}<1\)
\(x>-x-1\)
\(2x>1\)
\(x>\frac{-1}{2}\)
Thus, the interval of convergence for the given power series is \((\frac{-1}{2},\infty)\)

0

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