2

Inspired by my alternate answer to this question, I was investigating powers of cosine when turned into frequencies. The rule was: take $(2\cos(X))^n$, turn it into a sum of frequencies (that is, things of the form $\cos(kX)$ instead of $\cos(X)^k$), and (for clarity) replace $\cos(kX)$ with $y^k$. The results were: $$\begin{array}{l}2y\\2y^2+2\\2y^3+6y\\2y^4+8y^2+6\\2y^5+10y^3+20y\\2y^6+12y^4+30y^2+20\\2y^7+14y^5+42y^3+70y\\2y^8+16y^6+56y^4+112y^2+70\\2y^9+18y^7+72y^5+168y^3+252y\\2y^{10}+20y^8+90y^6+240y^4+420y^2+252\\2y^{11}+22y^9+110y^7+330y^5+660y^3+924y\\2y^{12}+24y^{10}+132y^8+440y^6+990y^4+1584y^2+924\\\end{array}$$ I stared at this for a while, and was able to see a recursion that was nearly identical to the recursion for the binomial coefficients in Pascal's triangle. Let $c_{n,k}$ denote the coefficient of $y^k$ in row $n$. Then, with appropriate initial conditions, $$ \begin{cases} c_{n,k} = c_{n-1,k-1}+c_{n-1,k+1},&k\ne 1\\ c_{n,1} = 2c_{n-1,0} + c_{n-1,2} \end{cases} $$For instance, this says $72=c_{9,5}=c_{8,4}+c_{8,6}=56+16$ and $252=c_{9,1}=2c_{8,0}+c_{8,2}=2\cdot 70+112$.

What struck me is that, aside from the presence of $2$ in the $c_{n-1,0}$ term, this is very similar to the binomial recurrence: each term is found from the terms to the upper right and upper left, so to speak. However, though I am reasonably familiar with the trig functions and the Chebyshev polynomials, I have not seen this recurrence given so explicitly in any literature. I would be shocked if this was not known, so I am requesting a source giving this recursion.

Integrand
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  • On further thought, the $2$ isn't really exceptional: it's the same formula, but uses the parity of cosine to get $\cos(X)=\cos(-X)$. I think the recurrence then follows directly from angle-addition, so it's not a big surprise. I will leave this up, but I don't think a reference is super necessary. – Integrand Oct 18 '24 at 13:59
  • Noting that $2\cos(\theta) = e^{i\theta} + e^{-i\theta}$, it is the variance from Pascal's triangle that is more surprising. – Paul Sinclair Oct 19 '24 at 20:54

1 Answers1

-1

As observed by Paul, it's Pascal's triangle with a 'folding' or combination of terms ash $\cos(x)=\cos(-x)$:

Table[ExpToTrig[Expand[(Exp[I x]+Exp[-I x])^k]],{k,0,7}]//MatrixForm

results in

(1
2 Cos[x]
2+2 Cos[2 x]
6 Cos[x]+2 Cos[3 x]
6+8 Cos[2 x]+2 Cos[4 x]
20 Cos[x]+10 Cos[3 x]+2 Cos[5 x]
20+30 Cos[2 x]+12 Cos[4 x]+2 Cos[6 x]
70 Cos[x]+42 Cos[3 x]+14 Cos[5 x]+2 Cos[7 x]
)
Michael T
  • 1,742