Trig++Binomial??

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

NOTE:\textbf{NOTE:}

I am not really sure on the lower limit of the summation, i think it might be 11, but i think, by symmetry it should be 00 .

#Algebra #Trigonometry #BinomialCoefficients #SummationOfSeries

Note by Aritra Jana
6 years, 4 months ago

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Comments

Start taking terms 1 from start and other from end you will see a pattern of sinA.cosB+cosA.sinBsinA.cosB + cosA.sinB

Taking 1st and last term

nC0cos(nx).sin(0)+nCncos(0).sin(nx)=nC0sin(nx)\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

nC1cos((n−1)x).sinx+nC(n−1)cosx.sin((n−1)x)=nC1sin(nx)\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

sin(nx)×[nC0+nC1+nC2⋯ ]\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.

Krishna Sharma - 6 years, 4 months ago

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

U Z - 6 years, 4 months ago

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

Aritra Jana - 6 years, 4 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 (2eix)n \left( 2 e ^ { i x } \right) ^ n . Expand it as (eix+eix)n \left( e^{ix} + e^{ix} \right)^ n .

Calvin Lin Staff - 6 years, 4 months ago
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