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$$f(x)=\sum_{k=0}^\infty \frac{2^{-k}}{k+1}(x-1)^k$$

Find the Taylor series for this function around $x=1$.

I don't understand this, isn't that summation already the Taylor polynomial of $f(x)$? If I follow the procedure of Taylor's theorem and differentiate I get the exact same result again.

Also it's a power series and by Taylor's theorem the coefficients $a_k$ should be equal to $f^{(k)}(1)$, which, non-surprisingly, they are.

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  • $\begingroup$ Perhaps they want you to evaluate the sum? $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:21
  • $\begingroup$ Find the $f(x)$ it represents? $\endgroup$ – QuestionMaker Jan 25 '17 at 1:22
  • $\begingroup$ Perhaps. In closed form or something. That might be what they really want, though it seems unclear. $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:24
  • $\begingroup$ Any idea on how to do that though? I suppose it's convergence radius is 1 but I'd have to check that. These exercises are really hard, they basically explain you nothing but Taylor's theorem and I'm expected to be able to reverse that process and all.. Don't really understand what's going on here. $\endgroup$ – QuestionMaker Jan 25 '17 at 1:25
  • $\begingroup$ Radius of convergence should be $2$. And, if I may, I'd point to the polylogarithm. $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:27
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Taking upon the semi-agreed question as per the comments, your function is:

$$f(x)=\frac{2\ln\left(\frac12(3-x)\right)}{1-x}$$

As per the Taylor series of the natural logarithm:

$$\ln(1-x)=\sum_{k=1}^\infty\frac{x^k}k$$

and the given Taylor series has radius of convergence $R=2$ by the ratio test.

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  • $\begingroup$ I'm sorry but I don't understand that. I get that the standardpolynomial can be applied to $ln(2-x)$ by using $x=x-1$, but how does the rest fit in? And how to get the sum to start at $k=0$? $\endgroup$ – QuestionMaker Jan 25 '17 at 1:39
  • $\begingroup$ @QuestionMaker $$\sum_{k=0}^\infty a_k=\sum_{k=1}^\infty a_{k-1}$$The rest should fit in quite nicely :-) $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:43
  • $\begingroup$ But wouldn't you get: $$\ln(1-(x-1))=\ln(2-x)=\sum_{k=0}^\infty\frac{(x-1)^{k+1}}{k+1}$$ $\endgroup$ – QuestionMaker Jan 25 '17 at 1:46
  • $\begingroup$ @QuestionMaker Oh, right, I forgot the $1/2$. Thanks for the catch! XD $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:47
  • $\begingroup$ Ok but how would you use the standard polynomial of the natural logarithm for something like $(3-x)/2$ $\endgroup$ – QuestionMaker Jan 25 '17 at 1:49

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