How do I find the closed form of the sum: \[\large{\sum\limits_{r=0}^{n} \binom{n}{r}cos[(n-r)x].sin(rx)}\]

\(\textbf{NOTE:}\)

I am not really sure on the lower limit of the summation, i think it might be \(1\), but i think, by symmetry it should be \(0\) .

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TopNewestStart taking terms 1 from start and other from end you will see a pattern of \(sinA.cosB + cosA.sinB \)

Taking 1st and last term

\(\displaystyle ^{n}C_0 cos(nx).sin(0) + ^{n}C_n cos(0).sin(nx) = ^{n}C_0 sin(nx)\)

Now 2nd and 2nd last

\(\displaystyle ^{n}C_1 cos((n-1)x).sinx + ^{n}C_{(n-1)} cosx.sin((n-1)x) = ^{n}C_1 sin(nx)\)

And so on

Finally you will get

\(\displaystyle sin(nx) \times [ ^{n}C_0 + ^{n}C_1 + ^{n}C_2 \cdots]\)

Closed form depends on number of terms of it is even or odd, I guess this you can do by yourself :).

Note:- when n is odd middle term should be written seperately.

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Very similar , my method - writing the whole series from end and adding the term with equal binomial coefficients

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thanks.. i don't know how this escaped me, it was a fairly simple one, but then again, thanks! :D

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Given the binomial coefficients, and trigonometric multiple angles, it is possible that there is a complex interpretation.

Hint:Consider the imaginary part of \( \left( 2 e ^ { i x } \right) ^ n \). Expand it as \( \left( e^{ix} + e^{ix} \right)^ n \).Log in to reply