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Hi could anyone help me solve this problem.(without expanding the function). I cant seem to get it.

Express the function as a power series

$$\ln\left(\frac{1+x}{1-x}\right)$$

I managed to split the $\ln(1+x)-\ln(1-x)$ apart but i do not know how to proceed from here.

ys wong
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    Can you write $\ln (1+x)$ as a power series? – Daniel Fischer Mar 26 '14 at 11:53
  • Yes but the question i do not know how to combine the Ln(1+x) and ln(1-x) together without expanding them. – ys wong Mar 26 '14 at 11:56
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    If both series do not diverge, you can sum them term by term – 7raiden7 Mar 26 '14 at 11:59
  • One way to do it "without expanding the function" is: (1) Take derivative of the original function, (2) You will get a rational function which you can expand in series by decomposing in partial fractions and then expanding each of them in series, (3) integrate the series you obtained. (5) But is too much work that wasn't needed. – OR. Mar 26 '14 at 12:00

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Based on $$ln(1+x)=\sum _{ n=1 }^{ \infty }{ { \left( -1 \right) }^{ n-1 } } \frac { { x }^{ n } }{ n } ,\quad \left| x \right| <1$$ we get $$ln(1-x)=-\sum _{ n=1 }^{ \infty }{ \frac { { x }^{ n } }{ n } } ,\quad \left| x \right| <1$$

Therefore,

$$ln\frac { 1+x }{ 1-x } =2\sum _{ k=0 }^{ \infty }{ \frac { x^{ 2k+1 } }{ 2k+1 } } ,\quad \left| x \right| <1$$

ZYX
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