- #1
julian
Gold Member
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So I know
[itex]\cos n \theta + i \sin n \theta = (\cos \theta + i \sin \theta)^n[/itex]
and by applying binomial to the RHS and taking the real part gives you:
[itex]\cos n \theta = \sum_{k=0}^{\lfloor {n \over 2} \rfloor} C^n_{2k} (\cos^2 \theta - 1)^k \cos^{n - 2k} \theta[/itex] .
I have come across another expansion:
[itex]\cos n \theta = {n \over 2} \sum_{k=0}^{\lfloor {n \over 2} \rfloor} (-1)^k {(n - k - 1)! \over k! (n - 2 k)!} (2 \cos \theta)^{n-2k} \qquad n > 0[/itex]
I'm trying to derive this second expansion for [itex]\cos n \theta[/itex] from the previous expression. You can start by rewriting the first expansion as
[itex]\cos n \theta = \cos^n \theta \sum_{k=0}^{\lfloor {n \over 2} \rfloor} C^n_{2k} (1 - \cos^{-2} \theta)^k[/itex]
and then expand the [itex](1 - \cos^{-2} \theta)^k[/itex] for every value of [itex]k[/itex] and bring together terms of the same power in [itex](\cos^{-2} \theta)[/itex]. This is where I get a bit stuck.
I think you might need to use some identity for the sum of the product of binomial coefficients. I have used the identity [itex]C_0^n + C_2^n + C_4^n + \dots + C_{2 \lfloor {n \over 2} \rfloor}^n = 2^{n-1}[/itex] to obtain the correct coefficient for the [itex](\cos^{-2} \theta)^0[/itex] term (note this term gets multiplied by the [itex]\cos^n \theta[/itex] outside the sum and so what we have is the coefficient corresponding to the [itex]\cos^n \theta[/itex] term in the expansion of [itex]\cos n \theta[/itex]).
[itex]\cos n \theta + i \sin n \theta = (\cos \theta + i \sin \theta)^n[/itex]
and by applying binomial to the RHS and taking the real part gives you:
[itex]\cos n \theta = \sum_{k=0}^{\lfloor {n \over 2} \rfloor} C^n_{2k} (\cos^2 \theta - 1)^k \cos^{n - 2k} \theta[/itex] .
I have come across another expansion:
[itex]\cos n \theta = {n \over 2} \sum_{k=0}^{\lfloor {n \over 2} \rfloor} (-1)^k {(n - k - 1)! \over k! (n - 2 k)!} (2 \cos \theta)^{n-2k} \qquad n > 0[/itex]
I'm trying to derive this second expansion for [itex]\cos n \theta[/itex] from the previous expression. You can start by rewriting the first expansion as
[itex]\cos n \theta = \cos^n \theta \sum_{k=0}^{\lfloor {n \over 2} \rfloor} C^n_{2k} (1 - \cos^{-2} \theta)^k[/itex]
and then expand the [itex](1 - \cos^{-2} \theta)^k[/itex] for every value of [itex]k[/itex] and bring together terms of the same power in [itex](\cos^{-2} \theta)[/itex]. This is where I get a bit stuck.
I think you might need to use some identity for the sum of the product of binomial coefficients. I have used the identity [itex]C_0^n + C_2^n + C_4^n + \dots + C_{2 \lfloor {n \over 2} \rfloor}^n = 2^{n-1}[/itex] to obtain the correct coefficient for the [itex](\cos^{-2} \theta)^0[/itex] term (note this term gets multiplied by the [itex]\cos^n \theta[/itex] outside the sum and so what we have is the coefficient corresponding to the [itex]\cos^n \theta[/itex] term in the expansion of [itex]\cos n \theta[/itex]).