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a) Show the partial sum $$S = \frac{4}{\pi} \sum_{n=1}^N \frac{\sin((2n-1)t)}{2n-1}$$ which may also be written as $$ \frac{2}{\pi}\int_0^x\frac{\sin(2Nt)}{\sin(t)}dt$$ has extrema at $x= \frac{m\pi}{2N} $

where $m$ is any positive integer except m=2kN, k also integer.

Solution: The derivative of $$S = \frac{2}{\pi}\frac{\sin(2Nx)}{\sin(x)} = 0 $$ $$\text{where }\sin(2Nx)=0,$$ $\sin(x)$ cannot equal zero.

$\sin(x) = 0$ where $x$ is a multiple of $\pi$.

Therefore, $$\sin(2Nx)=0 $$ where $ x=\frac{m\pi}{2N}$

however $\sin(x)$ cannot equal zero, $$\sin(\frac{m\pi}{2N})\neq 0 $$ so$ $m is any positive integer except $m=2kN$, $k$ also integer. Is this complete?!

b) Consider the first extrema to the right of the discontinuity, located at $x=\frac{\pi}{2N}$. By considering a suitable small angle formula show that the value of the sum at this point $$S(\frac{\pi}{2N})≈\frac{2}{\pi}\int_0^{\pi} \frac{\sin(u)}{u}du$$

Solution: I'm not sure which small angle formula i'm meant to considering?! Taylor series of sin? or how to consider it? I see that $$ S(\frac{\pi}{2N}) = \frac{4}{\pi} (\sin(\frac{\pi}{2N})+\frac{\sin(\frac{3\pi}{2N})}{3}+\frac{\frac{\sin(5\pi)}{2N}}{5}+\ldots)$$ $$=\frac{2}{\pi}(\frac{\pi}{N}(\frac{\sin(\frac{\pi}{2N})}{\frac{\pi}{2N}}+\frac{\sin(\frac{3\pi}{2N})}{\frac{3\pi}{2N}}+\frac{\sin(\frac{5\pi}{2N})}{\frac{5\pi}{2N}}+\cdots)$$

c) and by getting a computer to evaluate this numerically show that $$S(\frac{\pi}{2N})≈1.1790$$ independently of $N$.

Not really sure how I could show this?

Hence comment on the accuracy of Fourier series at discontinuities (also known as Gibbs phenomenon). Given that the error at $\frac{π}{2N}$ is nearly constant explain why the Fourier Convergence theorem is, or is not, valid for this problem?

Where a function has a jump discontinuity, the fourier series will overshoot as it approaches the discontinuity. As the number of terms in the fouler series increases, the amount of overshoot will converge to a constant percentage (around 17.9) of the amount of the jump

could someone explain this to me please? -sorry for the long winded question!

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In this answer, it is shown that the overshoot, on each side, is approximately $$ \frac1\pi\int_0^\pi\frac{\sin(t)}{t}\mathrm{d}t-\frac12=0.089489872236 $$ of the total jump. Thus, the overshoot you mention is twice that.

This answer tries to explain the phenomenon in a bit more intuitive manner.

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