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Calculus and Beyond Homework Help
Generate the Fourier transform of the function
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[QUOTE="jbunniii, post: 5442979, member: 81553"] I think the formula your lecturer meant is $$z^{-N} + z^{-(N-1)} + z^{-(N-2)} + \cdots + z^{-1} + 1 + z + \cdots + z^{N-1} + z^{N} = \frac{z^{-(N+1/2)} - z^{N + 1/2}} {z^{-1/2}-z^{1/2}}$$ This can be derived as follows. We can write the left hand side more compactly as $$\sum_{k=-N}^{N}z^{k}$$ Now introduce the change of variable ##m = k + N## to rewrite the sum as $$\begin{aligned} \sum_{m=0}^{2N}z^{m-N} &= z^{-N}\sum_{m=0}^{2N}z^m \\ \end{aligned}$$ Do you see what to do next? Then, assuming you have established the formula $$\sum_{k=-N}^{N}z^{k} = \frac{z^{-(N+1/2)} - z^{N + 1/2}} {z^{-1/2}-z^{1/2}}$$ you can apply it to your situation by setting ##z = e^{-ipt}## and simplifying the resulting right hand side. [/QUOTE]
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Generate the Fourier transform of the function
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