ece205: fix bug
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@ -395,7 +395,7 @@ Thus if a Fourier series on $(0,L)$ exists, it can be expressed as either a **Fo
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We have $L=\pi,a=\sqrt 2$.
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We have $L=\pi,a=\sqrt 2$.
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\begin{align*}
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\begin{align*}
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u(x,t)&=\sum^\infty_{n=1}\alpha_ne^{left(\frac{n\pi\sqrt 2}{\pi}\right)^2t}\sin(\frac{n\pi x}{\pi})
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u(x,t)&=\sum^\infty_{n=1}\alpha_ne^{\left(\frac{n\pi\sqrt 2}{\pi}\right)^2t}\sin(\frac{n\pi x}{\pi})
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&=\sum^\infty_{n=1}\alpha_ne^{-2n^2t}\sin(nx) \\
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&=\sum^\infty_{n=1}\alpha_ne^{-2n^2t}\sin(nx) \\
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\alpha_n&=\frac 2 L\int^L_0f(x)\sin(\frac{n\pi x}{L})dx \\
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\alpha_n&=\frac 2 L\int^L_0f(x)\sin(\frac{n\pi x}{L})dx \\
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&=\frac2\pi\int^{\pi/2}_0\frac\pi 2\sin(nx)dx+\frac2\pi\int^\pi_{\pi/2}(x-\frac\pi2\sin(nx)dx \\
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&=\frac2\pi\int^{\pi/2}_0\frac\pi 2\sin(nx)dx+\frac2\pi\int^\pi_{\pi/2}(x-\frac\pi2\sin(nx)dx \\
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