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Can anyone prove how the two equations are equal? Thanks

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$$=\frac1\pi \int_0^{2\pi} f(x) \left\{\frac12+\sum_{n=1}^N \cos [n(t-x)] \right\} \, dx$$

$$=\frac1{2\pi} \int_0^{2\pi} f(x) \frac{(N+\frac12)(x-t)}{\sin\frac12(x-t)}\, dx$$

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  • $\begingroup$ Hint : try to work with the sum of $e^{in(t-x)} = (e^{i(t-x) \text{ }}) ^ n$ … $\endgroup$
    – Watson
    Feb 8, 2016 at 14:33
  • $\begingroup$ Have a look here and on some other similar posts. $\endgroup$ Feb 8, 2016 at 17:24

1 Answer 1

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\begin{align*}\sin\left[\frac{1}{2}(x-t)\right]\left[\frac{1}{2}+\sum_{n=1}^{N}\cos[n(t-x)]\right]=\frac{1}{2}\sin\left[\frac{1}{2}(x-t)\right]+\sum_{n=1}^N\sin\left[\frac{1}{2}(x-t)\right]\cos\left[n(x-t)\right]\\=\frac{1}{2}\sin\left[\frac{1}{2}(x-t)\right]+\frac{1}{2}\sum_{n=1}^{N}\sin\left[\left(n+\frac{1}{2}\right)(x-t)\right]-\sin\left[\left(n-\frac{1}{2}\right)(x-t)\right]\end{align*}

Now, this is a telescoping sum, hence we can write $$\frac{1}{2}\sin\left[\frac{1}{2}(x-t)\right]+\frac{1}{2}\left[\sin\left[\left(N+\frac{1}{2}\right)(x-t)\right]-\sin\left[\frac{1}{2}(x-t)\right]\right]=\frac{1}{2}\sin\left[\left(N+\frac{1}{2}\right)(x-t)\right]$$

Eventually $$\boxed{\frac{1}{2}+\sum_{n=1}^{N}\cos\left[n(t-x)\right]=\frac{\sin\left[\left(N+\frac{1}{2}\right)(x-t)\right]}{2\sin\left[\frac{1}{2}(x-t)\right]}} \hspace{3cm}\blacksquare$$

BTW, it is called Dirichlet kernel (you might want to google it).

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  • $\begingroup$ @MBM, You are welcome. Please upvote and accept answers if you found them helpful. $\endgroup$
    – Galc127
    Feb 8, 2016 at 14:59

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