Does $x^2+ x^4/(3\cdot4) + x^6/(3\cdot4\cdot5\cdot6) + \cdots$ have any compact form?

Question

Is there any compact form for the following series

$$F_1(x) = x^2+ \frac{x^4}{3\cdot4} + \frac{x^6}{3\cdot4\cdot5\cdot6} + \cdots$$ $$F_2(x) = x+ \frac{x^3}{2\cdot3} + \frac{x^5}{2\cdot3\cdot4\cdot5} + \cdots$$ $$F_3(x) = F_2(x) - F_1(x)$$

Answer 1

$F_2$ is the sum of the odd powers of $x$ divided by factorials, it should remind you of the Taylor development of $e^x$. Indeed, you get the series with only odd terms by combining $e^x$ and $e^{-x}$: $$F_2(x)=\frac{e^x-e^{-x}}2=\sinh x.$$ This is known as the hyperbolic sine. Similarly, $F_1$ is the sum of even powers, the hyperbolic cosine, but all factorials are missing the factor $2$, and the constant term is absent. $$F_1(x)=e^x+e^{-x}-2=2\cosh x-2.$$ By difference, $$F_3(x)=2-\frac{e^x+3e^{-x}}2.$$

Answer 2

Yes,

$F_1(x) = 2(\cosh x - 1)$;

$F_2(x) = \sinh x$.

(see Taylor series).

Answer 3

First of all, as yoyo says, these can be solved through differential equations. For example, if $y=2+F_1(x)$, then we have

$$y''=y,y(0)=2,y'(0)=0$$

However, given both $F_1(x)$ and $F_2(x)$ to work with, we can use $e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$ to see that

$$\frac{F_1(x)}2+F_2(x)+1=e^x$$ $$\frac{F_1(x)}2-F_2(x)+1=e^{-x}$$

From here, it's simple algebra.