What's the original function

Question

I have this Maclaurin series: $\sum_{n=0}^\infty \cfrac{(n+1)(2n^2+10n+3)}{3(2n+3)!}x^{2n}$. Can someone help me getting its original function? I have tried to get it from $sin(x)$ and $cos(x)$ but got no result so far. Cheers!

Answer 1

Wolfram alpha gives

$\frac{1}{180x^3}\left(x^5 \ _3F_4(2,2,2;1,1,3,\frac{7}{2};\frac{x^2}{4}) + 180 x^2 \sinh(x)-120x +630 \sinh(x) -510x\cosh(x) \right)$

but this may not be the form you are after

Answer 2

Hint. Recall that $$\cosh(x)=\sum_{n=0}^\infty \cfrac{x^{2n}}{(2n)!},\quad \frac{\sinh(x)}{x}=\sum_{n=0}^\infty \cfrac{x^{2n}}{(2n+1)!},\\ \frac{\cosh(x)-1}{x^2}=\sum_{n=0}^\infty \cfrac{x^{2n}}{(2n+2)!},\quad \frac{\sinh(x)-x}{x^3}=\sum_{n=0}^\infty \cfrac{x^{2n}}{(2n+3)!}. $$ Now if $P$ is a third degree polynomial and $$f(x):=\sum_{n=0}^\infty \cfrac{P(n)}{(2n+3)!}x^{2n}$$ then $f$ will be a linear combination of the above functions.

Just find the constants $A$, $B$, $C$ and $D$ such that
$$P(n)= A(2n+3)(2n+2)(2n+1)+B(2n+3)(2n+2)+C(2n+3)+D$$ then $$f=A\cdot\frac{\sinh(x)-x}{x^3}+B\cdot\frac{\cosh(x)-1}{x^2}+C\cdot\frac{\sinh(x)}{x}+D\cdot\cosh(x).$$

Answer 3

The $2n$ is a clue: these are the terms of

$$ g(x) = \sum_{k=0}^\infty \dfrac{(k/2+1)(k^2/2 + 5k+3)}{3(k+3)!} x^k$$ for odd $k$, i.e. $(g(x) + g(-x))/2$. Now relate $g(x)$ to an exponential...

Answer 4

Hint. Alternatively, on setting $$ f(x)=\sum_{n=0}^\infty \cfrac{(n+1)(2n^2+10n+3)}{3(2n+3)!}x^{2n} $$ one may multiply $f(x)$ by $x^3$ then one may differentiate as follows obtaining $$ \begin{align} \frac{d}{dx}\left( x \frac{d}{dx} \left(x^3f(x)\right)\right)=\frac1{12} x^2(x^2+24) \cosh x+\frac1{12}(13x^3-x^2)\sinh x \end{align} $$ then integrating reversely one gets

$$ f(x)=\left(\frac1{12}-\frac{5}{4x^2}\right)\cosh x+\left(\frac1{2x}+\frac{5}{4x^3}\right)\sinh x. $$