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# Trig$$+$$Binomial??

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 $$0$$ .

Note by Aritra Jana
2 years, 11 months ago

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Start 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.

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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 $$\left( e^{ix} + e^{ix} \right)^ n$$.

Staff - 2 years, 11 months ago