Compute y' and y"?

Question

Given that y=sigma of n=0, infinity of nx^n, compute y' and y" and write out the first four terms of each series as well as the coefficient of x^n in the general term.

Answer: y'=1+2²x+3²x²+4²x³+…+(n+1)²xⁿ+…

y"=2^2+3^2*2x+4^2*3x^2+5^2*4x^3+…+(n+2)^2*(n+1)xⁿ+…

Please help me step by step.

Answer 1

Sun,

 your problem is pretty straightforward      you have a power series. Σ_n=0^∞ nx^n.

First thing you need to find out if the series is convergent and, if so, for what value of x. I think you should be able to prove that it converges for -1<x<1. Note that for x=1 you would get the series  Σ_n=0^∞ n which does not converge.  Show what happens for x=-1....  After that, when x is inside the range (-1, 1) .. the derivative of the sum is equal to the sum of the derivatives of its terms. Now, the first derivative of the generic term of the series, nxⁿ, is   n*nx^n-1 =n²x^n-1.  So with simple algebra you get your answer for y', i.e.,

y' = Σ_n=1^∞ n²x^n-1 =1x⁰ +2²x¹+3²x²+....+(n+1)²xⁿ+ .. 

Repeat the same argument starting from y' this time and its defining series to compute y''. Do not forget to discuss the convergence issues or you could not be doing the derivation term by term.