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I have the function $$f(x) :=\begin{cases}1,& \frac{\pi}{2}< x < \pi \\0, &-\frac{\pi}{2}< x < \frac{\pi}{2} \\-1, &-\pi < x < -\frac{\pi}{2}\end{cases}$$

and I compute $a_0, a_n$ and $b_n$, $$a_0=a_n=0,\quad\text{and}\quad b_n = \frac{1}{\pi}\left(\frac{2}{k}\cos\big(\frac{\pi k}{2}\big) - \frac{2}{k}\cos(\pi k)\right)$$

Now I have a problem with the calculation of the next sum: $$\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n-1)}$$

I do not know how to compute this sum with $a_0,a_n$ and $b_n$. Thank you for your help!

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  • $\begingroup$ Hint: What's the taylor series for $\arctan(x)$? $\endgroup$ – Alex R. Jul 5 '17 at 20:42
  • $\begingroup$ arctan(x) = x - (1/3)x^3 + (1/5)x^5 - (1/7)x^7...? $\endgroup$ – J.Doe Jul 5 '17 at 20:49
  • $\begingroup$ And what if $x=1$? $\endgroup$ – Alex R. Jul 5 '17 at 20:57
  • $\begingroup$ Solution is arctan(1). Thank you :) $\endgroup$ – J.Doe Jul 5 '17 at 21:26
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over 2n - 1} & = \ic\sum_{n = 1}^{\infty}{\ic^{2n - 1} \over 2n - 1} = \ic\sum_{n = 1}^{\infty}{\ic^{n} \over n} - \ic\sum_{n = 1}^{\infty}{\ic^{2n} \over 2n} = \ic\sum_{n = 1}^{\infty}{\ic^{n} \over n} - {1 \over 2}\,\ic\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n} \\[5mm] & = -\ic\ln\pars{1 - \ic} - {1 \over 2}\,\ic\braces{-\ln\pars{1 - \bracks{-1}}} = -\ic\bracks{{1 \over 2}\,\ln\pars{2} - {\pi \over 4}\,\ic} + {1 \over 2}\,\ic\ln\pars{2} \\[5mm] & = \bbx{-\,{\pi \over 4}} \end{align}

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