Assuming the solution to my differential equation is of the form $y=\sum_{n=0}^\infty a_nx^n$, I was able to get to the recurrence relation.

The recurrence relation is $$a_{n+2} = \dfrac {3n-3}{(n+2)(n+1)}a_n$$

From this I get $$a_2 = \frac {-3}{2}a_0$$ $$a_3 = 0 \implies a_{\text{odd}} = 0 \tag{except maybe $a_1$} \\ a_4 = \frac 14a_2 = \frac {-3}8a_0 \\ a_6 = \frac 3{10}a_4 = \frac {-9}{80}a_0 \\ \vdots$$

Thus my solution should be of the form $$y= a_1(x) + a_0 (1-\frac 32x^2 -\frac 38x^4 -\frac 9{80}x^6 + \cdots)$$

How do I simplify this -- that is write the infinite series for the even terms? I've been trying for a while, but I'm not sure what the standard technique is for this.

  • 1
    $\begingroup$ What is the differential equation that $y(x)$ solves? $\endgroup$ – DVD May 2 '15 at 0:26

$$y(x) = \sum_{n=0}^{\infty} a_nx^n$$ $$a_{n+2} = \dfrac {3n-3}{(n+2)(n+1)}a_n$$ $$a_{2n+1} = 0$$

$$\begin{align} % a_{n} &= \frac {3(n-3)}{(n-1)(n-0)} \times \frac {3(n-5)}{(n-3)(n-2)} \times \frac {3(n-7)}{(n-5)(n-4)} \\ & \times \dots \\ & \times \frac {3(3)}{(6)(5)} \times \frac{3(1)}{(4)(3)}\times \frac{3(-1)}{(2)(1)} \times a_{0} \end{align}$$

And canceling out terms:

$$\begin{align} % a_{n} &= a_0~3^{n/2}~\frac {1}{(n-1)(n-0)} \times \frac {1}{(n-2)} \times \frac {1}{(n-4)} \\ & \times \dots \\ & \times \frac {1}{6} \times \frac{1}{4}\times \frac{(-1)}{2} \end{align}$$

$$a_{n} = a_0~3^{n/2}~\frac{-1}{n-1}~\frac{1}{(n-0)(n-2)(n-4) \dots (4)(2)}$$

And rewriting it in various ways:

$$a_{n} = a_0~3^{n/2}~\frac{-1}{2(n/2)-1}~\frac{2^{-n/2}}{(n/2-0)(n/2-1)(n/2-2) \dots (2)(1)}$$

$$f_n(x) = a_{2n}x^{2n} = a_0~(3/2)^{n}~\frac{-1}{2n - 1}~\frac{x^{2n}}{(n-0)(n-1)(n-2) \dots (2)(1)}$$

$$f_n(x) = \frac{-a_0}{2n - 1}~\frac{\left(3x^2/2\right)^n}{n!}$$

$$-a_0~g_n\left( \frac{3x^2}{2} \right) = f_n(x) \iff \color{darkred}{g_n(x) = \frac{1}{2n - 1}~\frac{x^n}{n!}}$$

This $g$ doesn't look like any common taylor series, so checking wolfram online:

$$\sum_{n=0}^{\infty} \frac{1}{2n - 1}~\frac{x^n}{n!} = \sqrt{\pi x}~~ {\rm erfi}(\sqrt{x}) - e^{x}$$

where erfi is the imaginary error function. So while your recursion has the above closed form, your series does not have an elementary closed form.


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