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I was wondering if you could help me find the exact function that this Taylor series is meant to represent as I have been stuck for a couple hours: $$f(x) = \frac12(x - 1) - \frac18(x - 1)^2 + \frac1{24}(x - 1)^3 - \frac1{64}(x - 1)^4 + \cdots + \frac{(-1)^{n + 1}}{n 2^n}(x - 1)^n + \cdots$$

Thanks!

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  • $\begingroup$ @Bladewood I tried writing down the first couple derivatives, but I'm not sure how it works out with the factorial cancelling the factorial in the generic Taylor series expression $\endgroup$ – Maglor May 5 '18 at 1:13
  • $\begingroup$ what series do you get if you take the first derivative? $\endgroup$ – Will Jagy May 5 '18 at 1:13
  • $\begingroup$ $(1.) \quad f(x+1)$ would be easier to work with. $(2.) \quad f'(x+1)$ should look familiar. $\endgroup$ – steven gregory May 5 '18 at 1:13
  • $\begingroup$ indeed, start with $x-1 = 2 w.$ Then take the $w$ derivative to see what function of $w$ you have....If you prefer, you can start with $x-1 = -2t$ $\endgroup$ – Will Jagy May 5 '18 at 1:16
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Let's rewrite this as a summation: $$f(x) = \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n 2^n}(x - 1)^n.$$ We note that there is a power of $n$ in the denominator and numerator.

\begin{align} f(x) & = \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}n \left(\frac{x - 1}2 \right)^n \\ & = \sum_{n = 1}^\infty \frac{-(-1)^n}n \left(\frac{x - 1}2 \right)^n \\ & = - \sum_{n = 1}^\infty \frac1n \left(\frac{1 - x}2 \right)^n. \end{align} This series should look familiar. Indeed, setting $u = \frac{1 - x}2$, this becomes the well known Taylor series for the natural logarithm:

$$f(x) = - \sum_{n = 1}^\infty \frac{u^n}n = -\left( - \ln(1 - u) \right) = \ln(1 - u) = \ln \frac{1 + x}2.$$ This series is absolutely convergent for all $\lvert u \rvert \lt 1 \rightarrow \lvert \frac{1 - x}2 \rvert \lt 1 \rightarrow \lvert 1 - x \rvert < 2.$

Therefore, our series is $f(x) = \ln(1 + x) - \ln 2$ for $\lvert 1 - x \rvert \lt 2$. Put differently, it is the expansion of $\ln(1 + x) - \ln 2$ around $x = 1$.

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  • $\begingroup$ Thank you! I knew that I was sort of heading in the right direction, but this helped me make that last jump from series to ln which for some reason I couldn't see. $\endgroup$ – Maglor May 5 '18 at 13:09

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