Find a power series for f (x) = xln (1+x). / [f (x) = xln (1+x) / ]. / [/frac{d}{dx}xln (1+x) / ]. / [=/frac{x}{x+1}+ln (x+1) / ]. / [=/sum_{n=0}^{/infty} (-1) ^nx^n+/frac{d}{dx}ln (x+1) / ]. / [/sum_{n=0}^{/infty} (-1) ^nx^n+/sum_{n=0}^{/infty} (-1) ^n/frac{x^{n+1}}{n+1}/]

Question

Shiloh Bender

Mathematics

25 July, 22:01

Find a power series for f (x) = xln (1+x). / [f (x) = xln (1+x) / ]. / [/frac{d}{dx}xln (1+x) / ]. / [=/frac{x}{x+1}+ln (x+1) / ]. / [=/sum_{n=0}^{/infty} (-1) ^nx^n+/frac{d}{dx}ln (x+1) / ]. / [/sum_{n=0}^{/infty} (-1) ^nx^n+/sum_{n=0}^{/infty} (-1) ^n/frac{x^{n+1}}{n+1}/]

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Alivia Hayes · Answer 1

In your question that first im confused but i arrange if correctly though, in my calculation the possible power of the series is infinity ∑ n-0 times the - 1 one to the power of n over n+1 times x to the power of n+2