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$$a_0\bigg[1-\frac{\lambda}{2!}x^2-\frac{(4-\lambda)\lambda}{4!}x^4-\frac{(8-\lambda)(4-\lambda)\lambda}{6!}x^6-\cdots \bigg]$$

$$a_1\bigg[x+\frac{2-\lambda}{3!}x^3+\frac{(6-\lambda)(2-\lambda)}{5!}x^5+\cdots \bigg].$$

Hey guys, can someone help me put these into a power series with respect to $x$. $a_0$ and $a_1$ on the outside are just constants which you may ignore. They are answers I've received from solving a differential equation but I'm having trouble putting each one into a series.

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    $\begingroup$ They are power series in $x$. If what you want is a formula for the general term the ask for that. The answer probably won't be pretty, and probably won't help in further computations. The form you have is often the most informative. $\endgroup$ – Ethan Bolker Aug 18 '16 at 13:26
  • $\begingroup$ are these separate terms or is it a sum of $a_0 [\cdots]+a_1 [\cdots]+a_2[\cdots]+\cdots$? $\endgroup$ – hypergeometric Aug 18 '16 at 14:46
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Both are power series expanded at $x_0=0$ since they have the form \begin{align*} \sum_{n=0}^\infty \lambda_nx^n \end{align*}

In fact both series are Maclaurin series and we could look for an expression using sigma notation.

Here we take a look at the first series.

  • Since the exponents are even, we have an even power series $A(x)$, i.e. $A(x)=A(-x)$. \begin{align*} A(x)=\sum_{n=0}^\infty b_{2n}x^{2n} \end{align*}

  • There is a factor $(2n)!$ in the denominator of each term. So, we have in fact an exponential power series

\begin{align*} \sum_{n=0}^\infty \lambda_{2n}\frac{x^{2n}}{(2n)!} \end{align*}

  • There are $n$ factors in the numerator of $x^{2n}$ of the form $$\lambda(4-\lambda)\cdots(4(n-1)-\lambda)=-\prod_{j=0}^{n-1}\left(4j-\lambda\right)$$

  • We assume only the first term $1$ has positive sign and we obtain \begin{align*} A(x)=a_0\left(1-\sum_{n=1}^\infty\prod_{j=0}^{n-1}\left(4j-\lambda\right)\frac{x^{2n}}{(2n)!}\right) \end{align*}

With similar reasoning we can represent the second series in sigma notation.

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Assuming this is a sum which goes up to $a_n[\cdots]$, the expression required is $$\color{red}{\sum_{r=0}^na_r\sum_{j=0}^\infty \frac {(2(r+2j-2)-\lambda)!!!!\;\; x^{r+2j}}{(r+2j)!}}$$ where $(m-\lambda)!!!!$ is a specifically defined quadruple factorial, i.e. $$(m-\lambda)!!!!=(m-\lambda)(m-4-\lambda)(m-8-\lambda)\cdots ((m-1)\mod 4+1-\lambda)$$ and $(m-\lambda)!!!!=1$ for $m<0$.


NB: For the coefficient of $a_0$, it's helpful to write $-\lambda$ as $(0-\lambda)$, in which case the expression shown in the original question becomes $$a_0\bigg[1+\frac{(0-\lambda)}{2!}x^2+\frac{(4-\lambda)(0-\lambda)}{4!}x^4+\frac{(8-\lambda)(4-\lambda)(0-\lambda)}{6!}x^6+\cdots \bigg]$$

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