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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=…
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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=…
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Jan 8, 2017
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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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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
Bogdan Tushevskyi
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Jan 12, 2017
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