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unscientific
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Homework Statement
Part (a): State inverse Fourier transform. Show Fourier transform is:
Part (b): Show Fourier transform is:
Part (c): By transforming LHS and RHS, show the solution is:
Part(d): Using inverse Fourier transform, find an expression for T(x,t)
Homework Equations
The Attempt at a Solution
Part(d)
[tex]T_{(x,t)} = \int_{-\infty}^{\infty} T e^{-Dk^2t}e^{ikx} dk [/tex]
= [tex] \int_{-\infty}^{\infty} (T_0 + \sum_{m=1}^{\infty} Tm cos(\frac {m\pi x}{L}) e^{-Dk^2t + ikx} dk [/tex]
= [tex]\int_{-\infty}^{\infty} T_0 e^{-Dk^2t + ikx} dk + \sum_{m=1}^{\infty} T_m cos(\frac {m \pi x}{L}) \int_{-\infty}^{\infty} e^{-Dk^2t + ikx} dk [/tex]
Attempt at evaluating the integral, letting [tex] a^2 = \frac {1}{Dt} [/tex]
[tex] \int_{-\infty}^{\infty} e^{-(\frac{k^2}{a^2} - ikx)} dx [/tex]
[tex] = \int_{-\infty}^{\infty} e^{-\frac{({k - \frac{ixa^2}{2}})^2}{a^2}} e^{-\frac {x^2a^2}{4}} dk [/tex]
[tex] = e^-{\frac{x^2a^2}{4}} \int_{-\infty - i\frac{x^2a^2}{4}}^{\infty - i\frac {x^2a^2}{4}} e^{- \frac{k^2}{a^2}} dk [/tex]
[tex] = e^{\frac {-x^2a^2}{4}} \sqrt{\pi a^2} = \sqrt {\frac{\pi}{Dt}} e^{\frac {-x^2a^2}{4}}[/tex]
But this appears to be wrong as their final expression do not have the factor of √(1/t) in their coefficients..
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