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Question

f(x) = ln$\frac {1+x^2}{1-x}$

How to find the 5th degree Taylor polynomial of f(x) about x = 0 without finding all derivatives of ln$\frac {1+x^2}{1-x}$ ? Is there a better way to find this?

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1 Answer

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What you could do is to consider first that$$\frac{x^2+1}{1-x}=1+x+2\sum_{n=2}^\infty x^n$$ and then take the logarithm of$$\log(1+t)=\sum_{n=1}^\infty (-1)^{n+1}\frac{t^n}{n} $$ with $t=x+2\sum_{n=2}^\infty x^n$ and use the binomial expansion stopping as soon as you find an exponent greater than $5$.

In fact, you could show that, if $$\log \left(\frac{x^2+1}{1-x}\right)=\sum_{n=1}^\infty a_n x^n$$ the coefficients are simply$$a_{2n}=\frac{1-2 \cos (\pi n)}{2 n}\qquad \text{and}\qquad a_{2n+1}=\frac{1}{2 n+1}$$

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