What is the connection between Taylor series and Chebyshev polynomials?

Question

Can somebody help me find some historical references for the connection between Chebyshev polynomials and the Taylor series for sine and cosine functions? We know that Chebyshev polynomials are used to represent multiple angle identities for sine and cosine functions. According to Vieta,

$$\cos(nx)=\sum_{k=0}^n \binom{n}{k}\cos^k x\sin^{n-k}x\cos\left(\frac{1}{2}(n-k) \pi \right),$$ which is derived from Euler's formula.

Chebyshev polynomials $T_n(x)$ are those which satisfy $T_n(\cos x)=\cos(nx)$. For example,

$$T_4(x)=1-8x^2+8x^4,$$

$$T_8(x)=1-32x^2+160x^4-256x^6+128x^8.$$

The choice to show multiples of $4$ for $n$ is deliberate because these Chebyshev polynomials most closely resemble the pattern in the Taylor series for cosine, which begins with $1$ and and alternates between positive and negative terms with powers $x^{2k}$.

We know that from elementary trigonometry that any double angle identity can be rewritten to express a half-angle formula. So,

$$\cos(2x)=2\cos^{2}x-1 \implies \cos(x)=2\cos^2(x/2)-1.$$

This can be generalized for any multiple angle identity as well. That is,

$$\cos(x)=\sum_{k=0}^n \binom{n}{k} \cos^{k-1} (x/n)\sin^k (x/n)\cos\left(\frac{1}{2}(n-k) \pi \right).$$

Taking the limit of the above polynomial as $n$ approaches infinity results in small angles for cosine and sine which approach $1$ and $(x/n)^k$ respectively. Also, it can be shown that $\lim_{n \to \infty} \binom{n}{k}(1/n)^k =\frac{1}{k!}$. The result is the familiar Taylor series

$$\sum_{k=0}^{\infty}{(-1)^k \frac{x^{2k}}{(2k!)}}.$$

This can be appreciated by taking a multiple-four Chebyshev polynomial and dividing the coefficient of any $x^k$ term by $n^k$. The higher the $n$, the closer the term approximates $\frac{1}{k!}$. I was surprised to find this because I thought a Taylor series could only be derived by differential calculus.

I can't imagine this has gone unnoticed. Any comments and/or historical references are much appreciated.

Answer 1

I posted this answer more than three years ago. I suspect that I had found it in this book.

Copying from that answer and pasting here:

First recall that $$ \sin(\theta_1+\theta_2+\theta_3+\cdots) = \sum_{\text{odd }k \ge 1} (-1)^{(k-1)/2} \sum_{|A| = k}\ \prod_{i\in A} \sin\theta_i\prod_{i\not\in A} \cos\theta_i. $$ Then let $n$ be an infinitely large integer (that's how Euler phrased it, if I'm not mistaken) and let $$ \theta= \frac \theta n + \cdots + \frac \theta n $$ and apply the formula to find $\sin\theta$. Finally, recall that (as Euler would put it), since $\theta/n$ is infinitely small, $\sin(\theta/n) = \theta/n$ and $\cos(\theta/n) = 1$. Then do a bit of algebra and the series drops out.

The algebra will include things like saying that $$ \frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k} = 1 $$ if $n$ is an infinite integer and $k$ is a finite integer.

Then in the comments below the question, I wrote: $$ \begin{align}\sin(\alpha + \beta + \gamma + \delta + \varepsilon + \cdots) & = \underbrace{\sin\alpha}\;\underbrace{\cos\beta\cos\gamma\cos\delta\cos\varepsilon \cdots} {} + \text{more terms with just one sine} \\ & = {} -\underbrace{\sin\alpha\sin\beta\sin\gamma}\;\underbrace{\cos\delta\cos\varepsilon \cdots} {} + \text{other terms with three sines} \\ & = {} + \text{terms with five sines, etc.} \end{align} $$

. . . and in further comments:

Now imagine a term with three sines and the rest cosines: $$ \begin{align} & = \binom{n}{3}\sin\left(\left(\frac \theta n \right)^3\right) \cos\left(\left(\frac\theta n \right)^{n-3}\right) \\[6pt] & = \frac{n(n-1)(n-2)}{6}\left(\frac{\theta}{n}\right)^3 = \frac{\theta^3} 6 \end{align} $$