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Find the Maclaurin series for the function $$f(x)=x\ln(x+1)$$

So finding the derivatives is the first step. How many derivatives I need to find is explicitly said so I'll just go till the $4^{th}$ derivative.

$$\begin{align} f'(x)&=\frac{x}{x+1}+\ln(x+1) & f'(0) &=0 \\ f''(x)&=\frac{1}{(x+1)^2}+\frac{1}{x+1} & f''(0)&=2 \\ f^{(3)}(x)&=\frac{-x-3}{(x+1)^3} & f^{(3)}(0)&=0 \\ f^{(4)}(x)&=\frac{2x+8}{(x+1)^4} & f^{(4)}(0)&=8 \end{align}$$

So now plugging in for the Maclaurin form

$$P_n(x)=0+x^2-\frac{1}{2}x^3+\frac{1}{3}x^4+....$$

Is this correct?

Blue
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Eric Brown
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    You have not written a function, but instead an integral with a fixed value. – David G. Stork Dec 03 '19 at 22:54
  • @DavidG.Stork So does the integral being definite mean I don't have to take all those derivatives? – Eric Brown Dec 03 '19 at 22:56
  • Did you integrate the series term by term? – AnyAD Dec 03 '19 at 23:05
  • @AnyAD No I didn't, could you explain what you mean by that? – Eric Brown Dec 03 '19 at 23:06
  • Perhaps was the question about approximating that integral with a power series? However it wouldn't make much sense either, since you can explicitly find the primitive of $x ln(x+1)$ and you usually work with the approximating power series when the integral is not expressible in terms of elementary functions. – Federico Clerici Dec 03 '19 at 23:07
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    I guess you can simply use the Mac Laurin series of $\log(1+x)$ and multply through by $x$...? – dfnu Dec 03 '19 at 23:16
  • Looks good. You can check against the answer here https://math.stackexchange.com/questions/878374/taylor-series-of-ln1x – AnyAD Dec 03 '19 at 23:25
  • Using the comment above re multiplying by $x$ the series for $log(1+x)$ – AnyAD Dec 03 '19 at 23:26

2 Answers2

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Note that $$\ln(1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}x ^k}{k}$$ So the Mclaurin series for $x\ln(1+x)$ is $$ x\ln(1+x)= \sum_{k=1}^{n} \frac{(-1)^{k+1} x^{k+1}}{k}=x^2-\frac{x^3}{2}+\frac{x^4}{3}-\frac{x^5}{4}+....$$

Z Ahmed
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You may use

$$\frac1{1-t} = 1 + t + t^2 + \>...$$

to obtain,

$$x\ln(1+x) = -x\int_0^{-x} \frac{dt}{1-t}=-x\int_0^{-x}\left(1+t+t^2+\>...\right)dt = x^2 - \frac{x^3}2 + \frac{x^4}3+\>...$$

Quanto
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