Proof that $\frac{d^a}{dx^a}\sin(x) = \sin(x+\pi a/4)$ iff these two infinite series are equivalent?

Question

I am interested in analytically continuing the differentiation operation on functions, and I am currently focusing on trigonometric functions. Any high schooler can conjecture that $\frac{d^a}{dx^a}\sin(x) = \sin(x+\pi a/2)$ applies not only for integers, but for all complex numbers; I, however, am trying to prove the statement. By manipulating the power rule, repeating it with factorials, and analytically continuing with the gamma function as so: $$\frac{d^a}{dx^a}x^b=\frac{\Gamma(b+1)x^{b-a}}{\Gamma(b-a+1)}$$ Which can easily be applied to the Taylor series of any analytic function. Plugging in this formula to the Taylor series of a sine wave we get: $$\frac{d^a}{dx^a}\sin(x)=\sum_{n=0}^\inf \frac{(-1)^nx^{2n+1-a}}{\Gamma(2n+2-a)}$$ The Taylor series for $\sin(x+\pi a/2)$ is: $$\sum_{n=0}^\inf \sin^{(n)}(\pi a/2)x^n/n!$$ Where $\sin^{(n)}$ denotes the nth derivative of sine. If you zoned out during all that, that's fine. Here's the important part: $$\frac{d^a}{dx^a}\sin(x) = \sin(x+\pi a/2) \iff \sum_{n=0}^\inf \frac{(-1)^nx^{2n+1-a}}{\Gamma(2n+2-a)} = \sum_{n=0}^\inf \sin^{(n)}(\pi a/2)x^n/n!$$ How can I prove either one of these statements? Or has one of them already been proved, therefore proving the other?

Answer 1

Your operator is linear and homogeneous, i.e. : $$ \frac{d^a}{dx^a}f(\lambda x)=\lambda^a \frac{d^af}{dx^a}(\lambda x) $$ Therefore, your claim is equivalent to proving that $\exp$ is fixed by your generalized derivative for non integer $a$. It’s clear that it is not the case for positive non integer $a$: $$ \begin{align} \exp_a(x) &:= \frac{d^a}{dx^a}e^x \\ &= \sum_{n=0}^\infty \frac{x^{n-a}}{\Gamma(n-a+1)}\\ &\sim \frac{x^{-a}}{\Gamma(-a+1)} && x\to0\\ &\to \infty \end{align} $$

So you don’t even have the correct asymptotic at $0$. You can even apply this reasoning directly to $\sin$. In general, by using non integer powers, you are generating a branch cut so your function is not entire anymore.

Hope this helps.