Consider the function $f(x) = \frac{x}{2}$, defined over the interval $[0, 2\pi]$. Show that $\frac{\pi}{4} = 1 − \frac13 +\frac15 −\frac17 + \cdots$.
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$\begingroup$ I think you might have typoed. The function $f(x)=\frac{x}{2}$ does not have a period. $\endgroup$– Ian MillerCommented Dec 18, 2015 at 12:47
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$\begingroup$ Assuming the case that it has a period 2pi $\endgroup$– Zaryab SaeedCommented Dec 18, 2015 at 12:50
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$\begingroup$ So do you mean: $$f(x)=\pi\left(\frac{x}{2\pi}-\bigg\lfloor\frac{x}{2\pi}\bigg\rfloor\right)$$ $\endgroup$– Ian MillerCommented Dec 18, 2015 at 12:54
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$\begingroup$ You can keep the function to be f(x) = x/2. This is one of the questions asked in my quiz yesterday. I think you would know better what we can do with this and how this can be shown. $\endgroup$– Zaryab SaeedCommented Dec 18, 2015 at 12:57
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$\begingroup$ Hint: You must have done Fourier series. $\endgroup$– Geoff RobinsonCommented Dec 18, 2015 at 13:00
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1 Answer
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You just have to expand $y=x/2$ as a Fourier series: $$ {x\over2}=\sin (x)-\frac{1}{2} \sin (2 x)+\frac{1}{3} \sin (3 x)-\frac{1}{4} \sin (4x)+\frac{1}{5} \sin (5 x)+\ldots $$ and put here $x=\pi/2$.
answered Dec 18, 2015 at 13:01
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$\begingroup$ Keeping the period to be 2pi? Can you explain it a bit? $\endgroup$ Commented Dec 18, 2015 at 13:04
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$\begingroup$ As a matter of fact the question is not well formulated: $f(x)$ should be defined as $x/2$ on $[-\pi,\pi]$ and extended by periodicity for the other values of $x$. $\endgroup$ Commented Dec 18, 2015 at 14:43