# On general expansion of cos nΘ

1. Jun 25, 2014

### julian

So I know

$\cos n \theta + i \sin n \theta = (\cos \theta + i \sin \theta)^n$

and by applying binomial to the RHS and taking the real part gives you:

$\cos n \theta = \sum_{k=0}^{\lfloor {n \over 2} \rfloor} C^n_{2k} (\cos^2 \theta - 1)^k \cos^{n - 2k} \theta$ .

I have come across another expansion:

$\cos n \theta = {n \over 2} \sum_{k=0}^{\lfloor {n \over 2} \rfloor} (-1)^k {(n - k - 1)! \over k! (n - 2 k)!} (2 \cos \theta)^{n-2k} \qquad n > 0$

I'm trying to derive this second expansion for $\cos n \theta$ from the previous expression. You can start by rewriting the first expansion as

$\cos n \theta = \cos^n \theta \sum_{k=0}^{\lfloor {n \over 2} \rfloor} C^n_{2k} (1 - \cos^{-2} \theta)^k$

and then expand the $(1 - \cos^{-2} \theta)^k$ for every value of $k$ and bring together terms of the same power in $(\cos^{-2} \theta)$. 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 $C_0^n + C_2^n + C_4^n + \dots + C_{2 \lfloor {n \over 2} \rfloor}^n = 2^{n-1}$ to obtain the correct coefficient for the $(\cos^{-2} \theta)^0$ term (note this term gets multiplied by the $\cos^n \theta$ outside the sum and so what we have is the coefficient corresponding to the $\cos^n \theta$ term in the expansion of $\cos n \theta$).

2. Jun 25, 2014

### Simon Bridge

Equating coefficients of $cos^{n-2k}\theta$ suggests...
$$C_{2k}^n(\cos^2\theta - 1)^k=(-1)^k2^{n-2k}\frac{(n-k-1)!}{k!(n-2k)!}$$... recalling that:$$C_{2k}^n=\frac{n!}{(2k)!(n-2k)!}$$

BTW: the identities you are looking for in your question can be found by considering how factorials work.

3. Jun 25, 2014

### julian

There are many contributions to the coefficient of $\cos^{n-2k} \theta$ and you have to add them up. Let me be more explicit, I write out the expression:

$\cos^n \theta \; \sum_{k=0}^{\lfloor {n \over 2} \rfloor} C^n_{2k} (1 - \cos^{-2} \theta)^k$
$= cos^n \theta \; [ C^n_{0} + C^n_{2} (1 - \cos^{-2} \theta) + C^n_{4} (1 - \cos^{-2} \theta)^2 + C^n_{6} (1 - \cos^{-2} \theta)^3 + \dots + C_{2 \lfloor {n \over 2} \rfloor}^n (1 - \cos^{-2} \theta)^{\lfloor {n \over 2} \rfloor}]$

and expand every bracket so the total sum is explicitly:

$= \cos^n \theta \; C^n_{0} +$
$+ \cos^n \theta \; C^n_{2} (1 - \cos^{-2} \theta) +$
$+ \cos^n \theta \; C^n_{4} (1 - 2 \cos^{-2} \theta + \cos^{-4} \theta) +$
$+ \cos^n \theta \; C^n_{6} (1 - 3 \cos^{-2} \theta + 3 \cos^{-4} \theta - \cos^{-6} \theta) +$
$+ \dots$
$+ \cos^n \theta \; C^n_{2 \lfloor {n \over 2} \rfloor} (1 - C^{\lfloor {n \over 2} \rfloor}_1 \cos^{-2} \theta + C^{\lfloor {n \over 2} \rfloor}_2 \cos^{-4} \theta + \dots + (-1)^{{\lfloor {n \over 2} \rfloor} - 1} C^{\lfloor {n \over 2} \rfloor}_{{\lfloor {n \over 2} \rfloor} - 1} \cos^{- 2 ({\lfloor {n \over 2} \rfloor} - 1)} \theta + (-1)^{\lfloor {n \over 2} \rfloor} x^{- 2 {\lfloor {n \over 2} \rfloor}} )$

It is easy to see now how the coefficient of $\cos^n \theta$ term is equal to

$C_0^n + C_2^n + C_4^n + \dots C_{2 {\lfloor {n \over 2} \rfloor}}^n$

so you have many contributions and we have to add them up. We can prove this sum of binomial coefficients is equal to $2^{n-1}$ by noting:

$2^n = (1+x)^n |_{x=1} + (1+x)^n |_{x=-1} = [C_0^n + C_1^n + C_2^n + C_3^n + \dots + (-1)^n C_n^n] + [C_0^n - C_1^n + C_2^n - C_3^n + \dots + (-1)^n C_n^n]$
$= 2 [C_0^n + C_2^n + C_4^n + \dots + C_{2 {\lfloor {n \over 2} \rfloor}}^n]$.

I think I might know how to find coefficient of $\cos^{n-2} \theta$ we need to add

$-[C_2^n + 2 C_4^n + 3 C_6^n + \dots ]$

which can be found from considering:

$n 2^{n-1} = {d \over dx} (1+x)^n |_{x=1} - {d \over dx} (1+x)^n |_{x=-1} = 2^2 [C_2^n + 2 C_4^n + 3 C_6^n + \dots]$

These are the kind of identities I'm talking about.

The coefficient of $\cos^{n-4} \theta$ we need to add

$C_4^n + {3 \times 2 \over 2} C_6^n + {4 \times 3 \over 2} C_8^n + \dots$

not sure how to do this sum.

Last edited: Jun 25, 2014
4. Jun 25, 2014

### julian

Not sure what that equation is supposed to mean.

5. Jun 26, 2014

### Simon Bridge

Exactly what I said. (Oh I think I missed out an n/2)

You don't really need to derive one from the other - just demonstrate that they are the same.
They aught to ... but they may each involve different circumstances so the two expressions you are trying to relate may not actually be equal.

Have you seen how the second expression is usually derived?

Last edited: Jun 26, 2014
6. Jun 26, 2014

### julian

They are equal, I'm simply taking

$\sum_{k=0}^{\lfloor {n \over 2} \rfloor} C^n_{2k} (\cos^2 \theta - 1)^k \cos^{n-2k} \theta$

multiplying everything out and grouping the terms into the form

$\sum_{k=0}^{\lfloor {n \over 2} \rfloor} A_k \cos^{n-2k} \theta$

where I need to determine the coefficients $A_k$ (this is what I'm calling deriving the second expression). The $k$ in the first expression isn't to be identified with the $k$ in the second expression, I'm just using it as a dummy variable, maybe I should write for the second expression

$\sum_{k'=0}^{\lfloor {n \over 2} \rfloor} A_{k'} \cos^{n-2k'} \theta$.

In other words, I'm taking the original expression multiplying out and grouping the terms into a power-series form involving descending powers of $\cos \theta$. Finding $A_{k'}$ isn't as straightforward as you seem to be making it out to be. You have to do certain summations of binomial coefficients and that is where I get stuck! I dont know how to do them in general. And this is what my question is about!

Have I seen how the second expression is usually derived?

No. But I know how all this is related to Chebyshev polynomials $T_n (x)$ ($T_n = \cos n \theta$ and $x = \cos \theta$), and I have read that the expression

$T_n (x) = {n \over 2} \sum_{k=0}^{\lfloor {n \over 2} \rfloor} (-1)^k {(n-k-1) \over k! (n - 2m)!} (2x)^{n-2k}$

can be found from the generating function, or from the differential equation for Chebyshev polynomials, but I haven't tried doing it that way yet. I wanted to do it this way first.

Last edited: Jun 26, 2014
7. Jun 26, 2014

### julian

I still dont know what to make of this - there is no summation over $k$ on the left hand side, there is no $\cos^{n - 2k} \theta$ there either and there are no $\cos \theta$'s on the right hand side.

Sorry I'm a bit confused. You realise that the $k$'s in the two expressions are not to be identified with each other? They are just being used as dummy variables.

EDIT: I think you've just equated the $k$'th term in the first expression with the $k$'th term in the second expression, you shouldn't have done that, you've misunderstood what I'm doing, and you need to understand that the $k$ is just a dummy variable.

Last edited: Jun 26, 2014
8. Jun 26, 2014

### julian

Example:

$\cos 3 \theta = \cos^3 \theta + (\cos^2 \theta - 1) \cos \theta$

comes from the first expression, gets rearranged to

$\cos 3 \theta = 2 \cos^3 \theta - \cos \theta$

now it is written as a series in descending powers of $\cos \theta$.

So in the first expression the terms are a mixture of $(\cos^2 \theta - 1)$ and $\cos \theta$, whereas in the second expression the terms are just powers of $\cos \theta$.

I want to do this for general $\cos n \theta$.

Last edited: Jun 26, 2014
9. Jun 26, 2014

### julian

Next example:

$\cos 4 \theta = \cos^4 \theta + 6 (\cos^2 \theta - 1) \cos^2 \theta + (\cos^2 \theta - 1)^2$

comes from the first expression and the terms are a mixture of $(\cos^2 \theta - 1)$ and $\cos \theta$, multiplying out and collecting terms in the same power of $\cos \theta$ we get

$\cos 4 \theta = 8 \cos^4 \theta - 8 \cos^2 \theta + 1$,

and so on...

And this multiplied-out/grouped-together answer is what you are supposed to get from the second expression.

Last edited: Jun 26, 2014
10. Jun 26, 2014

### AlephZero

Google for Chebyshev polynomials. They can be defined as

\begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x). \end{align}

and the property you want is $T_n(\cos\theta)=\cos n\theta$.

11. Jun 26, 2014

### julian

So in

$\cos 4 \theta = \cos^4 \theta + 6 (\cos^2 \theta - 1) \cos^2 \theta + (\cos^2 \theta - 1)^2$, $\cos^4 \theta$ appears three times and they add up to $8 \cos^4 \theta$, $\cos^2 \theta$ appears twice times and they add up to $- 8 \cos^2 \theta$, and a constant appears once.

I want to do this multiplying out and adding up of like terms for arbitrary $n$. In

$\cos n \theta = \cos^n \theta + C_2^n (\cos^2 \theta - 1) \cos^{n-2} \theta + C_4^n (\cos^2 \theta - 1)^2 \cos^{n-4} \theta + C_6^n (\cos^2 \theta - 1)^3 \cos^{n-6} \theta \dots$

the $\cos^n \theta$ will appear multiple times and I need to know how to add them up, $\cos^{n-2} \theta$ will appear multiple times and I need to know how to add them up, and so on.

In my second post I established this adding up of like terms is equivalent to certain sums over binomial coefficients, namely

For $\cos^n \theta \;$: $C_0^n + C_2^n + C_4^n + C_6^n + C_8^n + C_{10}^n + \dots$

For $\cos^{n-2} \theta \;$: $-[C_2^n + 2 C_4^n + 3 C_6^n + 4 C_8^n + 5 C_{10}^n + \dots]$

For $\cos^{n-4} \theta \;$: $C_4^n + {3 \times 2 \over 2} C_6^n + {4 \times 3 \over 2} C_8^n + {5 \times 4 \over 2} C_{10}^n + \dots$

For $\cos^{n-6} \theta \;$: $-[C_6^n + {4 \times 3 \times 2 \over 3!} C_8^n + {5 \times 4 \times 3 \over 3!} C_{10}^n + \dots]$

and so on...

My question was how to do these summations...hoping someone might have a reference or something.

12. Jun 26, 2014

### julian

Yeah, it might probably easier to derive the formula I want from a recursive relation.

13. Jun 30, 2014

### julian

RECURSIVE RELATION:

We can easily derive the (Chebyshev) recursive relation directly as a trig identity:

$2 \cos \theta \cos n \theta = \dfrac{e^{i \theta} + e^{-i \theta}}{2} \dfrac{e^{i n \theta} + e^{-i n \theta}}{2} = \dfrac{e^{i (n+1) \theta} + e^{-i (n+1) \theta}}{2} + \dfrac{e^{i (n-1) \theta} + e^{-i (n-1) \theta}}{2} = \cos (n+1) \theta + \cos (n-1) \theta$

and so

$\cos (n+1) \theta = 2 \cos \theta \cos n \theta - \cos (n-1) \theta \qquad Eq. 1$.

PROOF BY INDUCTION:

We can use this recursive relation to prove

$\cos n \theta = {n \over 2} \sum_{k=0}^{\lfloor {n \over 2} \rfloor} (-1)^k {(n - k - 1)! \over k! (n - 2k)!} (2 \cos \theta)^{n-2k} \qquad n > 0 \qquad Eq 2.$

by induction.

First we assume it is true for $n-1$ and $n$ and substitute this assumption into the right hand side of the recursive relation Eq. 1:

$\cos (n+1) \theta = {n \over 2} \sum_{k=0}^{\lfloor {n \over 2} \rfloor} (-1)^k {(n - k - 1)! \over k! (n - 2k)!} (2 \cos \theta)^{n+1-2k} - {n-1 \over 2} \sum_{k=0}^{\lfloor {n - 1 \over 2} \rfloor} (-1)^k {(n - k - 2)! \over k! (n - 1 - 2k)!} (2 \cos \theta)^{n - 1 -2k}$

write $k = k' - 1$ in the second term, and we get

${n \over 2} \sum_{k=0}^{\lfloor {n \over 2} \rfloor} (-1)^k {(n - k - 1)! \over k! (n - 2k)!} (2 \cos \theta)^{n+1-2k} - {n-1 \over 2} \sum_{k'=1}^{\lfloor {n - 1 \over 2} \rfloor + 1} (-1)^{k' - 1} {(n - k' - 1)! \over (k' - 1)! (n + 1 - 2k')!} (2 \cos \theta)^{n + 1 -2k'} \qquad Eq.3$

It is easy to check that for $n$ even $\lfloor {n - 1 \over 2} \rfloor + 1 = \lfloor {n \over 2} \rfloor$ and for $n$ odd $\lfloor {n - 1 \over 2} \rfloor + 1 = {n+1 \over 2} = \lfloor {n + 1 \over 2} \rfloor$.

$n \; {\bf even:}$

Separating out the ($k=0$ term of the first part, and then replacing $k'$ by $k$ we obtain

${n \over 2} {(n-1)! \over 0! n!} (2 \cos \theta)^{n+1} + \sum_{k=1}^{\lfloor {n \over 2} \rfloor} (-1)^k \Big[ {n \over 2} {(n-k-1)! \over k! (n-2k)!} + {n-1 \over 2} {(n-k-1)! \over (k-1)! (n -2k +1)!} \Big] (2 \cos \theta)^{n+1-2k}$

The content of the square brackets simplify:

${1 \over 2} {(n-k-1)! \over k! (n + 1 -2k)!} \Big\{ n (n + 1 - 2k) + k (n - 1) \Big\} = {1 \over 2} {(n-k-1)! \over k! (n + 1 -2k)!} \Big\{ (n + 1) (n-k) \Big\} = {n+1 \over 2} {(n-k)! \over k! (n + 1 -2k)!}$

and also

${n \over 2} {(n-1)! \over 0! n!} (2 \cos \theta)^{n+1} = {1 \over 2} (2 \cos \theta)^{n+1} = {n+1 \over 2} {(n-0)! \over 0! (n + 1 -0)!} (2 \cos \theta)^{n+1-0}$

and so then we have

$\cos (n+1) \theta = {n+1 \over 2} \sum_{k=0}^{\lfloor {n + 1 \over 2} \rfloor} (-1)^k {(n+1-k-1)! \over k! (n + 1 -2k)!} (2 \cos \theta)^{n+1-2k}$

where we have used $\lfloor {n + 1 \over 2} \rfloor = {n \over 2} = \lfloor {n \over 2} \rfloor$ for $n$ even.

$n \; {\bf odd:}$

We can go through the same steps as before, but this time there is an extra term. So from Eq.3 we have

${n+1 \over 2} \sum_{k=0}^{\lfloor {n \over 2} \rfloor} (-1)^k {(n+1-k-1)! \over k! (n + 1 -2k)!} (2 \cos \theta)^{n+1-2k} + \Big( k' = {n+1 \over 2} \; term \; in \; second \; part \; of \; Eq.3 \Big)$
$\qquad \qquad \qquad = \; \qquad \dots \qquad + {n-1 \over 2} (-1)^{n + 1 \over 2} {(n - {n + 1 \over 2} - 1)! \over ({n + 1 \over 2} - 1)! (n + 1 - 2{n + 1 \over 2})!} (2 \cos \theta)^{n + 1 -2{n + 1 \over 2}}$
$\qquad \qquad \qquad = \; \qquad \dots \qquad + {n-1 \over 2} {2 \over n-1} (-1)^{n + 1 \over 2}$
$\qquad \qquad \qquad = \; \qquad \dots \qquad + (-1)^{n + 1 \over 2}$

But we get from:

${n+1 \over 2} (-1)^{n+1 \over 2} {(n + 1 - {n+1 \over 2} -1)! \over ({n+1 \over 2})! (n+1 - 2{n+1 \over 2})!} (2 \cos \theta)^{n+1 - 2{n+1 \over 2}} = (-1)^{n+1 \over 2}{n+1 \over 2} {({n-1 \over 2})! \over ({n+1 \over 2})! (0)!} = (-1)^{n+1 \over 2}$

and so then we have

$\cos (n+1) \theta = {n+1 \over 2} \sum_{k=0}^{\lfloor {n + 1 \over 2} \rfloor} (-1)^k {(n+1-k-1)! \over k! (n + 1 -2k)!} (2 \cos \theta)^{n+1-2k}$

So we have obtained the correct formula, i.e. Eq.2 with $n \mapsto n+1$, for both $n$ even and $n$ odd.

Now we can argue that if it works for $n = 1$ and for $n = 2$ then it works for $n = 3$. We then know it works for $n = 2$ and for $n = 3$ therefore it works for $n = 4$, and so on...

So for $n = 1$:

${1 \over 2} \sum_{k=0}^{0} (-1)^0 {(0)! \over 0! (1)!} (2 \cos \theta)^{1} = \cos \theta$

and for $n = 2$:

${2 \over 2} \sum_{k=0}^1 (-1)^k {(1 - k)! \over k! (2 - 2k)!} (2 \cos \theta)^{2-2k} = 2 \cos^2 \theta - 1 = \cos 2 \theta.$

BINOMIAL COEFFICIENTS SUMMATION IDENTITIES.

Inadvertently from this result we have the following Binomial coefficient summation identities:

$C_0^n + C_2^n + C_4^n + C_6^n + C_8^n + C_{10}^n + \dots + C_{2 \lfloor {n \over 2} \rfloor}^n = 2^{n-1}$

$C_2^n + 2 C_4^n + 3 C_6^n + 4 C_8^n + 5 C_{10}^n + C_{2 \lfloor {n \over 2} \rfloor}^n = n 2^{n-3}$

$C_4^n + {3 \times 2 \over 2} C_6^n + {4 \times 3 \over 2} C_8^n + {5 \times 4 \over 2} C_{10}^n + \dots + C_{2 \lfloor {n \over 2} \rfloor}^n = {n (n-3) \over 2} 2^{n-5}$

$C_6^n + {4 \times 3 \times 2 \over 3!} C_8^n + {5 \times 4 \times 3 \over 3!} C_{10}^n + \dots + C_{2 \lfloor {n \over 2} \rfloor}^n= {n (n-4) (n-5) \over 3!} 2^{n-7}$

and so on...

I got from wiki (Chebyshev Polynomials page) how to order the terms properly in my second post.

Last edited: Jun 30, 2014