About $\cos(\sqrt{-x})$

Question

By Euler's Formula $e^{ix}=\cos{x}+i\sin{x}$ we can deduce that:

$\cos{\sqrt{-x}}=\cosh {\sqrt{x}}$

My question is the following true:

$\cos{\sqrt{-x}}=\begin {cases} \cos{\sqrt{-x}} & ,x \text{ is negative real number} \\ \cosh {\sqrt{x}} & ,x \text{ is positive real number} \end{cases}$

and it is differentiable and continuous at zero.

If this is true...is it useful?

Read the following to know my level in mathematics:

I am second year student of mathematics I know calculus 1+2+3 ,ODES,logic and

writing proofs . at my current semester I am studying Abstract Algebra 01

,Elementary Number Theory ,Introduction to Real Analysis ,PDES 01 and Linear

Algebra 01. This question comes to my mind because I love mathematics and

I am curious about this idea about $\cos{\sqrt{-x}}$ whether it is true or false ,whether it is useful

or useless.

Answer 1

In a word, "yes". In fact, the function you've identified is entire, complex differentiable on the whole complex plane, and given by the power series $$ C(x) = \sum_{k=0}^{\infty} \frac{x^{k}}{(2k)!} = 1 + \frac{x}{2!} + \frac{x^{2}}{4!} + \frac{x^{3}}{6!} + \cdots. $$ (So, $C(x^{2}) = \cosh x$ and $C(-x^{2}) = \cos x$, as your piecewise formula says.)

There's a companion, $$ S(x) = \sum_{k=0}^{\infty} \frac{x^{k}}{(2k+1)!} = 1 + \frac{x}{3!} + \frac{x^{2}}{5!} + \frac{x^{3}}{7!} + \cdots, $$ satisfying $S(x^{2}) = \dfrac{\sinh x}{x}$ and $S(-x^{2}) = \dfrac{\sin x}{x}$.

Both functions do occur "in the wild", for example when expressing the exponential map for the Lie group $SL(2, \mathbf{R})$. I'm not aware of any name for them.