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I have been given this to solved:

$$f(x) = \begin{cases}x & \text{if }0<x<\pi/2, \\ \pi-x & \text{if }\pi/2<x<\pi. \end{cases}$$

My attempt:

2l=$\pi$ so l=$\pi$/2

$$a_n=\frac{4}{\pi}\int_0^{\pi}f(x)\cos(2 n x)dx=\frac{4}{\pi}\int_0^{\pi/2}x\cos(2 n x)dx$$

The first integral gives me: $$\frac{-1}{n^2 \pi}{((-1)^n-1)}.$$

Also I get $a_0$ = ${\pi^2}/8$.Are these correct

Also the function is discontinious and I know only to solve for continious function..Pls help

Scáthach
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  • Why do you believe the function is discontinuous? – adfriedman May 20 '20 at 20:54
  • I am guessing that what you mean, perhaps, is that you extend the function's definition in an even way, so that you get a completely continuous periodic function even function, which Fourier series will then contain onle cosines...is this what you meant? Even if not, the function as given is continuous on the given interval and piecewise differentiable there (only at $;x=\frac\pi2;$ it isn't) – DonAntonio May 20 '20 at 21:04
  • the limits of rhs and lhs at $\pi/2$ are equal so it's continious – Scáthach May 20 '20 at 21:15
  • i thought since there's no equality it's not continious... – Scáthach May 20 '20 at 21:15
  • @Scáthach It is very confusing what you write: first you write the limits at $;\pi/2;$ are equal, then you write that "since there's no equality..." ....And you didn't answer my questions above. – DonAntonio May 20 '20 at 22:33
  • @DonAntonio yea i extended limits by extending as an even function – Scáthach May 21 '20 at 03:03

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