Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for ƒ (perhaps more than once). Give the interval of convergence for the resulting series. g(x)=-frac{1}{(1+x)^2}text{ using }f(x)=frac{1}{1+x}

Kye 2021-03-09 Answered
Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for ƒ (perhaps more than once). Give the interval of convergence for the resulting series.
\(g(x)=-\frac{1}{(1+x)^2}\text{ using }f(x)=\frac{1}{1+x}\)

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Expert Answer

Viktor Wiley
Answered 2021-03-10 Author has 17497 answers

Given the series
\(f(x)=\frac{1}{1+x}\)
We have to find the power series of g(x) by differentiating or integrating f(x). Also we have to find the interval of convergence.
Formula used:
1.\(\frac{1}{1+x}=\sum_{n=0}^\infty x^n\)
2.\(\frac{d}{dx}(x^n)=nx^{n-1}\)
3. If \(\sum_{n=1}^\infty a_n(x-c)^n\) is a power series with radius of convergence R, then the interval of convergence is \(c-R\)
4.The series \(\sum a_nx^n\) is convergent then \(\lim_{n\rightarrow\infty}|\frac{a_n}{a_{n+1}}|>1\)
Now, Power series by differentiating f,
\(\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum_{n=0}^\infty(-x)^n=\sum_{n=0}^\infty(-1)^nx^n\)
\(\frac{1}{1+x}=\sum_{n=0}^\infty(-1)^nx^n\)
diff. both sides,
\(\frac{d}{dx}(\frac{1}{1+x})=\frac{d}{dx}(\sum_{n=0}^\infty(-1)^nx^n)\)
\(\frac{-1}{(1+x)^2}=\sum_{n=0}^\infty(-1)^nnx^{n-1}\)
\(\frac{-1}{(1+x)^2}=0+\sum_{n=1}^\infty(-1)^nnx^{n-1}\)
\(\frac{-1}{(1+x)^2}=\sum_{n=1}^\infty(-1)^nnx^{n-1}\)
Now to find the interval of convergence:
We have
\(a_n=(-1)^nnx^{n-1}\)
Now, \(\lim_{n\rightarrow\infty}|\frac{a_n}{a_{n+1}}|=\lim_{n\rightarrow\infty}|\frac{(-1)^nnx^{n-1}}{(-1)^{n+1}(n+1)x^n}|=\lim_{n\rightarrow\infty}|\frac{n}{-1(n+1)x}|=\lim_{n\rightarrow\infty}|\frac{1}{(1+\frac1n)x}|=|\frac1x|\)
now \(\lim_{n\rightarrow\infty}|\frac{a_n}{a_{n+1}}|=|\frac1x|>1\)
\(\frac{1}{|x|}>1\)
\(|x|<1\ i.e.\ -1\)
Now check at boundary points of the interval:
At \(x=1,\ \sum_{n=1}^\infty(-1)^nn(1)^{n-1}=\sum_{n=1}^\infty(-1)^nn\)
does not converge.
At \(x=-1,\sum_{n=1}^\infty(-1)^nn(-1)^{n-1}=\sum_{n=1}^\infty(-1)^{2n-1}n\)
does not converge.
Hence the interval of convergence is (-1, 1).

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Answered 2021-12-17 Author has 10829 answers

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