Skip terms in power series of $\cosh$ and $\sinh$

Question

Is there a way to skip every second term in the power series representation of $\sinh{x}$ and $\cosh{x}$ and adjust the other terms accordingly (approx.)?

So, instead of

$$\sinh{x} \approx x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \frac{x^9}{9!} + ... + \frac{x^{2n+1}}{(2n+1)!}$$ the following $$\sinh{x} \approx a_1 x + a_5\frac{x^5}{5!} + a_9 \frac{x^9}{9!} + ... + a_{4n+1}\frac{x^{4n+1}}{(4n+1)!}$$ and accordingly, instead of $$\cosh{x} \approx 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \frac{x^8}{8!} + ... + \frac{x^{2n}}{(2n)!}$$ the following $$\cosh{x} \approx a_0 + a_4\frac{x^4}{4!} + a_8 \frac{x^8}{8!} + ... + a_{4n}\frac{x^{4n}}{(4n)!}$$

The approximations do not need to be steady at $x = 0$

Answer 1

At a particular value of $x$, yes you can. For instance, to approximate $\sinh{(\frac12)}$ by the series $\sum_{n=0}^{N}a_{4n+1}\frac{x^{4n+1}}{(4n+1)!}$, choose $a_1=1$, $a_5=\frac{4\cdot 5}{(1/2)^2}$, $a_{4n+1}=\frac{(4n+1)!}{(4n-1)!(1/2)^2}$, etc. so that the resulting sum is the McLauren series! The problem here is that every time you change the value of $x$ where you are approximating the function, you're going to need calculate the new values for the coefficients $a_k$. The marvelous thing about approximating a function by its McLauren/Taylor series though, is that you only have to find one set of values for coefficients and you can reuse them for any value of $x$ at all (so long as it's within the radius of convergence).

Answer 2

In your representation, how would you represent the third derivative of $\sinh x$ at $x=0$? The value of this derivative at $x=0$ needs to be 1 because it is $\sinh x$, but your representation gives the value 0. The Maclaurin expansion is the way it is because it has to be that way in order to get the right derivatives -- you can't just pick and choose terms to include in the power series.

Answer 3

Series expansions around any particular x=a are unique. Your two examples would represent different expansions around x=0, so this would not work.

If you'd like a bit more of a hands-on explanation, consider that the difference between two different representations of the same function must be the zero function (f(x) = 0 for all x), yet the difference between these two series will be a series with non-zero coefficients, which (pardon my lack of rigour in this statement) will itself be a non-zero function.