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, 2015 at 0:26

1 Answer 1


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

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

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.

  • $\begingroup$ Hi Daniel. When you do $$\begin{align} a_{n} &= \frac {3(n-3)}{(n-1)(n-0)} \times \frac {3(n-5)}{(n-3)(n-2)} \times \frac {3(n-7)}{(n-5)(n-4)} \\ & \times \dots \\ & \times \frac {3(3)}{(6)(5)} \times \frac{3(1)}{(4)(3)}\times \frac{3(-1)}{(2)(1)} \times a_{0} \end{align}$$ how do you know what the last term in the multiplication is? $\endgroup$
    – user56202
    Sep 4, 2021 at 0:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.