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The forward difference operator (discrete derivative) $\Delta$ is defined as $\Delta f(x) = f(x+1) - f(x)$.

The "discrete $e^x$" / eigenfunction of $\Delta$ is $2^x$.

Since $2^{x+1} - 2^x = (2-1)2^x = 2^x$. The analogue for $e^{\lambda x}$ is $(1+\lambda)^x$ since $$\Delta (1+\lambda)^x = (1+\lambda)^{x+1} - (1+\lambda)^x = (1+\lambda -1)(1+\lambda)^x = \lambda(1+\lambda)^x$$

A Newton series representation for $2^x$ is $\sum\limits_{n=0}^{\infty} \frac{x^{\underline{n}}}{n!}$, using $x^{\underline{n}}$ = $n! \displaystyle {x \choose {n}}$ to denote the falling power. This holds because \begin{equation*}\sum_{n=0}^{\infty} \frac{x^{\underline{n}}}{n!} =\sum_{n=0}^{\infty} {x\choose{k}} = 2^x\end{equation*}

And using the nice analogue of the power rule that $\Delta x^{\underline{n}} = nx^{\underline{n-1}}$ we have

\begin{equation*}\Delta 2^x = \Delta \left( \sum_{n=0}^{\infty} \frac{x^{\underline{n}}}{n!} \right) = \sum_{n=0}^{\infty} \frac{\Delta x^{\underline{n}}}{n!} = \sum_{n=0}^{\infty} \frac{x^{\underline{n}}}{n!} = 2^x\end{equation*}

If I define a "discrete cosine" as \begin{equation*}c(x) = \frac{(1+i)^x + (1-i)^x}{2} = \frac{1}{2}\sum_{n=0}^{\infty}{x \choose {n}}i^n + \frac{1}{2}\sum_{n=0}^{\infty}{x \choose {n}}(-i)^n\end{equation*}

I want to show that this is equal to $\displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{x^{\underline{2n}}}{(2n)!}$ and likewise show

$s(x) := \displaystyle \frac{(1+i)^x - (1-i)^x}{2i} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{\underline{2n+1}}}{(2n+1)!}$

The proof is complete if can get an expression of the form $\displaystyle \sum_{n=0}^{\infty}{x \choose {2n}}(i)^{2n}$ for $c(x)$ which is similar to the above expression which I get by a binomial expansion, and the same for $s(x)$

I also recall to have seen other functions $s^*(x), c^*(x)$ with the same property of solving $\Delta^2 f = -f$ in a pdf I can no longer find, but I remember them looking extremely different, involving a $2^x$ term and the respective "normal" trig functions. The closed form I think was roughly in the form $2^{kx}$ $\cos (\alpha x + \beta)$ for $c^*(x)$ and $2^{kx}$ $\sin(\alpha x + \beta)$ for $s^*(x)$ (same constants if I remember correctly).

I would like to know how this was obtained (and what the formula actually is), though this is more of a side question I would be grateful if anyone else knows of this and its proof.

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    $\begingroup$ For the first question, do the same as for the regular sine, cosine and derivative: Take $\Delta$ on both sides repeatedly and evaluate at $x=0$. $\endgroup$ – user517969 Jan 2 '18 at 21:57
  • $\begingroup$ @plumbus you are indeed correct, although I am still curious about other proofs. $\endgroup$ – Theo Diamantakis Jan 3 '18 at 1:58
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    $\begingroup$ As a sanity check... you do realize that in your "sum of two summations" formula for cosine, the odd indexed terms cancel out and the even index terms are the same? $\endgroup$ – user14972 Jan 3 '18 at 2:50
  • $\begingroup$ Ah I see what you mean, the proof follows from that simple observation, and you can do the same trick with sine. If you post a few words as an answer I'll give you the reputation and accept it haha. Thanks. $\endgroup$ – Theo Diamantakis Jan 3 '18 at 2:52
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Given a power series, there is a standard trick to isolate its even or its odd indexed terms. If

$$ f(x) = \sum_{n=0}^{\infty} a_n x^n $$

then

$$ f(-x) = \sum_{n=0}^{\infty} a_n (-x)^n = \sum_{n=0}^{\infty} (-1)^n a_n x^n $$

so that both series have the same even terms and negated opposite terms. Thus

$$ \frac{f(x) + f(-x)}{2} = \sum_{\substack{n=0 \\ n \text{ even}}}^\infty a_n x^n = \sum_{m=0}^{\infty} a_{2m} x^{2m}$$

$$ \frac{f(x) - f(-x)}{2} = \sum_{\substack{n=0 \\ n \text{ odd}}}^\infty a_n x^n = \sum_{m=0}^{\infty} a_{2m+1} x^{2m+1}$$

The combination of sums you're asking about happens to be precisely of this form:

$$ \frac{1}{2}\left( \sum_{n=0}^{\infty}{x \choose {n}}i^n + \sum_{n=0}^{\infty}{x \choose {n}}(-i)^n\right) = \sum_{m=0}^{\infty}{x \choose {2m}} i^{2m} $$

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After lots of fruitless searching I managed to guess the mysterious $c^*(x), s^*(x)$ via trial and error.

The functions $$2^{\frac{x}{2}} \sin(\frac{\pi}{4}x + 2) $$ and $$2^{\frac{x}{2}} \cos(\frac{\pi}{4}x + 2) $$

Satisfy $\Delta s^*(x) = c^*(x)$ and $\Delta c^*(x) = -s^*(x)$, but $s^*(x) \neq s(x)$ and $c^*(x) \neq c(x)$. Since $\Delta f = \Delta (f+g)$ when $g$ is a function which has period $1$ i.e $g(x+1) = g(x)$ (because $\Delta g = 0$ and $\Delta$ is linear) $s(x), s^*(x)$ must differ by a $1$ periodic function.

$s(x)$ and $c(x)$ can be still considered the "true" discrete trigonometric functions because of the power series analogy, but this is just a matter of taste.

For another interesting choice see this question and answer, where the periodic property is kept but not the similar looking power series https://math.stackexchange.com/a/1439726/462531

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