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Context:

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The question is :

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This is my working:

for 0$\leqslant$ x $\leqslant$p, $A_{m}$ = $2\over\pi$$\int^\pi_0 {xh\over p} sin(mx)\,dx$ = $2h\over\pi p$$\int^\pi_0 xsin(mx)\,dx$ = $2h\over\pi p$[$\left.\ -xcos(mx)\over m\right|_0^\pi$ + $1\over m$$\int^\pi_0 cos(mx)\,dx$] = $2h\over\pi p$ [$-\pi cos(m\pi)\over m$ + $1\over m^{2}$ $\left.\sin (mx)\right|_0^\pi$] = $2h\over\pi p$[$-(-1)^{m}\pi \over m $]

similarly, for p$\leqslant$ x $\leqslant$$\pi$, $A_{m}$ = $2\over\pi$$\int^\pi_0 {h(\pi-x)\over (\pi-p)} sin(mx)\,dx$ = $2h\over\pi(\pi-p)$$\int^\pi_0 (\pi-x)sin(mx)\,dx$ = $2h\over\pi(\pi-p)$[$\left.\ -(\pi-x)cos(mx)\over m\right|_0^\pi$ - $1\over m$$\int^\pi_0 cos(mx)\,dx$] = $2h\over\pi(\pi-p)$ [$\pi cos(m\pi)\over m$ - $1\over m^{2}$ $\left.\sin (mx)\right|_0^\pi$] = $2h\over\pi(\pi-p)$[$(-1)^{m}\pi \over m $]

Hope there is no calculation mistake. As there is only one $A_{m}$ expected, should I sum the two coefficients? But I failed to get that desired one, Google finds me similar questions (plucked string), and in some of them there is an issue of "coefficients vanishes when m is even", but I can't see that here, and some use initial conditions http://www.math.hmc.edu/~ajb/PCMI/lecture7.pdf, but in this question what should I use?

I believe quite some people has worked on or seen there exercises before, and I admits my working is long and ugly, so, can people just tell me am I in the right track? And what has been wrong? Any hints are welcomed and would be appreciated! My self-study is lack of guidelines and there is no classmates I can talk to.

Bob
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1 Answers1

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Your limits of integration are wrong. The first integral should be $\int_0^p$, and the second $\int_p^{\pi}$. (Then you just add them.)

TonyK
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  • Thanks, but do you know why we can directly sum the coefficients derived from different parts of the function? If I go to extreme, put 3 functions in one graph, each defined on adjacent intervals,if I sum the 3 fs, I got a piecewise defined function, why the Fourier coefficient of it is the sum of 3 coefficients from 3 originally not related functions? As the Fourier expansion is a representation of the function, on any 1 of the 3 intervals, the Fourier expansion using the summed coefficient is going to be different with the one using coefficient from the function defined on it – Bob Dec 08 '13 at 02:46
  • The separate integrals are not Fourier coefficients. You are computing the Fourier coefficient of the function in Figure 8 by integrating it over $[0,\pi]$ $-$ but you need to calculate the two parts of the integral separately, because of the way the function is defined. – TonyK Dec 08 '13 at 08:59
  • ok, got it,thanks man~ – Bob Dec 08 '13 at 13:35