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I'm preparing for a test and i've spent quite some time on this. What I already did was to use the taylor expansion for $\ln(1+x)$ to finally get the sum: $$ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{{2^n} {n^2}} $$ I checked wolfram and it seems this gives the correct result, but how do I calculate this manually? I tried tweaking around with known sums, but to no avail. I also tried calculating the integral without the Taylor expansion approach, but it didn't work out. Any help will be greatly appreciated!

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Note that the terms in your series fall very rapidly to zero. Since it is an alternating sum the error when summing up to and including term $N$ is bounded by the absolute value of term $N+1$. You will not need many terms to get two digits precision.

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  • $\begingroup$ Translating WimC's assertion to a more concrete form, considering $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{{2^n} {n^2}}$, the OP needs to find the smallest value of $n$ such that: $$\left| \frac{(-1)^{n-1}}{2^n n^2} \right| = \frac{1}{2^n n^2} < 0.01$$ That will be the minimum number of terms needed to achieve the desired precision of $0.01$. $\endgroup$ – etothepitimesi Aug 17 '14 at 19:45
  • $\begingroup$ Thank's a lot for the answer. I understand it now :) $\endgroup$ – Xsy Aug 17 '14 at 21:11
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Instead of writing a formula in summation notation of $\ln(1+x)$, I personally find it easier to expand the Taylor series expansion of $\ln (1+x)$, like this: $$\ln (1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5-\cdots,$$ where the expansion is sufficiently long enough to help you integrate and find your answer precisely to at least 2 decimal places. (The more terms in your series expansion, the more precise your answer will be. For me, I cut off after the fifth term upon integration.)

Finally, \begin{align} \int_0^{0.5} \frac{\ln(1+x)}x \, dx &\approx\int_0^{0.5} \frac{x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5}x \, dx \\ &= \int_0^{0.5}1-\frac x2+\frac{x^2}3-\frac{x^3}4+\frac{x^4}5 \, dx \end{align}

Can you take it from here? Your answer should be very well precise to at least two decimal places. If you have more decimal places, simply truncate or round your answer accordingly.

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  • $\begingroup$ Thank you for the answer, that is indeed a much simpler way :) There is a problem though, I did not know the answer prematurely, so if I go into this again "blind" how can I prove that the arbitrary number of terms I picked is enough to be accurate enough? $\endgroup$ – Xsy Aug 17 '14 at 18:11
  • $\begingroup$ I'm sorry if I sound too harsh, but this answer doesn't add anything new to what the OP has done. By specifying a precision, it's clear the teacher wants the students to find the minimum number of terms in the Taylor expansion for which its error is smaller than the precision. WimC's answer is brief, but solves the problem. $\endgroup$ – etothepitimesi Aug 17 '14 at 19:44
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By integrating termwise the Taylor series of $\frac{\log(1+x)}{x}$ around $x=0$ we get: $$\int_{0}^{1/2}\frac{\log(1+x)}{x}\,dx = \sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot2^j }$$ and through partial summation: $$\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot 2^j }=\frac{1}{3}\sum_{j=1}^{+\infty}\frac{2j+1}{j^2(j+1)^2}+\frac{1}{3}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}(2j+1)}{2^j\cdot j^2(j+1)^2}=\frac{1}{3}+\frac{1}{3}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}(2j+1)}{2^j\cdot j^2(j+1)^2}$$ Performing partial summation a second time yields: $$\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot 2^j }=\frac{1}{3}+\frac{1}{12}+\frac{1}{9}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{2^j}\left(\frac{1}{j^2}-\frac{2}{(j+1)^2}+\frac{1}{(j+2)^2}\right)$$ and performing the same trick a third time gives: $$\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot 2^j }=\frac{427}{972}+\frac{1}{27}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{2^j}\left(\frac{1}{j^2}-\frac{3}{(j+1)^2}+\frac{3}{(j+2)^2}-\frac{1}{(j+3)^2}\right)$$ while the fourth time we have: $$\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot 2^j }=\frac{1733}{3888}+\frac{1}{81}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{2^j}\sum_{k=0}^{4}\frac{(-1)^k\binom{4}{k}}{(j+k)^2}.$$ Finally, the first term in the RHS is $0.445\ldots$ while the first term in the last sum, multiplied by $\frac{1}{81}$, is less than $0.003$, hence $0.44$ is the approximation we were looking for.

The Euler acceleration method here slightly improves the Taylor series approximation since the error term goes to zero like $3^{-k}$ instead of $2^{-k}$.

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