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Consider $$ f(t)= \begin{cases} 1 \mbox{ ; } 0<t<1\\ 2-t \mbox{ ; } 1<t<2 \end{cases}$$

Let $f_1(t)$ be the Fourier sine series and $f_2(t)$ be the Fourier cosine series of $f$, $f_1(t)=f_2(t), 0<t<2$. Write the form of the series (without computing the coefficients) and graph $f_1$ and $f_2$ on [-4,4] (including the endpoints $\pm 4$) using *'s to identify the value of the series at points of discontinuity.

I think we have:

$f_1(t)=\sum \limits_{n=1}^{\infty} b_n \sin \frac{n \pi t}{2}$
$f_2(t)=\frac{a_0}{2}+\sum \limits_{n=1}^{\infty} a_n \cos \frac{n \pi t}{2}$

I think we have $f_2=1$ and for $0<t<2, f_1=f_2=1$

Can we do anything else? Can someone help me with the end?

Thank you

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I could be mistaken, but shouldn't the angular frequency be just $n\pi$, not $n\pi / 2$, seeing as $2\pi / 2 = \pi$? – Sam Apr 6 '13 at 16:34
I don't think so. If we have a periodic function of period $P=2L$ then in the fourier series we have $\cos \frac{n \pi t}{L}$ – Carpediem Apr 6 '13 at 16:43
I agree with that, but isn't the function assumed to have a period $P = 2$ (that is, $L = 1$)? – Sam Apr 6 '13 at 16:46
yeah chill i said i gonna do it ... – Dominic Michaelis Apr 6 '13 at 17:35
up vote 2 down vote accepted

Ok at first we gonna plot our function

enter image description here

We know that on jump discontinuities it will converge to the arithmetic mean of them, so the first approximation is just taking $\frac{1}{2}$. This gonna look like

The Cos terms gonna look like

enter image description here

The Sin terms are looking like

enter image description here

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It's only a partial answer i will add the rest after eating – Dominic Michaelis Apr 6 '13 at 17:57
Can you edit please ? – Carpediem Apr 6 '13 at 19:58
On the cosine graph, the horizontal line the constant function ? – Carpediem Apr 6 '13 at 20:23
yeah it is, it should be on 1/2 but it isn't i found that strange – Dominic Michaelis Apr 6 '13 at 20:24
So the overlaps are normal right? Since those are then other terms – Carpediem Apr 6 '13 at 20:24

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