Series Expansion by differentiation

Mathematics Asked by dodgevipert56 on December 12, 2020

We know that 1/(1-x) = 1+x+x^2+x^3+….x^n

Say we need to find a suitable function for the expansion x+x^2+x^3+x^4+….x^(n+1)

We would differentiate 1/(1-x) = 1+x+x^2+x^3+….x^n

This would yield 1/(1-x)^2 = 1 + 2x+3x^2+4x^3 +…+nx^(n-1)**

Multiplying it by x would yield x/(1-x)^2 = x + 2x^2 + 3x^3 + 4x^4 +…nx^n Hence the respective function would be x/(1 – x)^2 = x + 2x^2 + 3x^3 + 4x^4 +…nx^n

But how would we obtain the function g[x] for the expansion 1+ 2x^2 + 3x^3 + 4x^4 +…

One Answer

Your work is fine, now we just have to subtract $x$ and add $1$ that is


Answered by user on December 12, 2020

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