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)=−1(1+x)2 using f(x)=11+x

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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)=−1(1+x)2 using f(x)=11+x

Viktor Wiley · Accepted Answer

Given the series f(x)=11+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: 1.11+x=∑n=0∞xn 2.ddx(xn)=nxn−1 3. If ∑n=1∞an(x−c)n is a power series with radius of convergence R, then the interval of convergence is c−R 4.The series ∑anxn is convergent then limn→∞|anan+1|&amp;gt;1 Now, Power series by differentiating f, 11+x=11−(−x)=∑n=0∞(−x)n=∑n=0∞(−1)nxn 11+x=∑n=0∞(−1)nxn diff. both sides, ddx(11+x)=ddx(∑n=0∞(−1)nxn) −1(1+x)2=∑n=0∞(−1)nnxn−1 −1(1+x)2=0+∑n=1∞(−1)nnxn−1 −1(1+x)2=∑n=1∞(−1)nnxn−1 Now to find the interval of convergence: We have an=(−1)nnxn−1 Now, limn→∞|anan+1|=limn→∞|(−1)nnxn−1(−1)n+1(n+1)xn|=limn→∞|n−1(n+1)x|=limn→∞|1(1+1n)x|=|1x| now limn→∞|anan+1|=|1x|&amp;gt;1 1|x|&amp;gt;1 |x|&amp;lt;1 i.e. −1 Now check at boundary points of the interval: At x=1, ∑n=1∞(−1)nn(1)n−1=∑n=1∞(−1)nn does not converge. At x=−1,∑n=1∞(−1)nn(−1)n−1=∑n=1∞(−1)2n−1n does not converge. Hence the interval of convergence is (-1, 1).

Jeffrey Jordon · Accepted Answer

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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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