This is actually a very simple problem, but I am going to type everything out in case I really overlooked something

I am trying to find a recurrence relation in the series solution I got. The ODE is $y'' + xy'- y = 0$.

$\begin{align} y'' + xy' - y & = \sum_{n=2}^{\infty} n(n - 1)a_nx^{n-2} + x\sum_{n=1}^{\infty} na_nx^{n-1} - \sum_{n=0}^{\infty} a_nx^n \\ & = \sum_{n=2}^{\infty} n(n - 1)a_nx^{n-2} + \sum_{n=1}^{\infty} na_nx^n - \sum_{n=0}^{\infty} a_nx^n \\ & = \sum_{n=0}^{\infty} (n+2)(n+1)a_nx^n + \sum_{n=0}^{\infty} na_nx^n - \sum_{n=0}^{\infty} a_nx^n\\ & = \sum_{n=0}^{\infty} x^n[(n+1)(n+2)a_{n+2} - a_n + na_n ] \\ & = 0 \end{align}$

The power series is 0 iff $(n+1)(n+2)a_{n+2} - a_n(1- n) = 0 \iff a_{n+2} = \dfrac{a_n (1 - n)}{(n+1)(n+2)}$

So testing out various values of n = 1, 2, 3... led me to

$a_2 = \dfrac{a_0}{2}$

$a_3 = 0$

$a_4 = \dfrac{-a_2}{12} = \dfrac{a_0}{24}$

$a_5 = 0$

$a_6 = -\dfrac{a_4}{10} = \dfrac{a_0}{240}$

Okay so clearly the odd terms all go to 0 and there will be no odd terms. I know that the even terms alternate back and forth in series (so I will get an alternating series). I have no idea how the denominator can jump from 2 to 24 and then to 240 and then to a stunning 240*56. I see no pattern in how the denominator jumps at all, it jumps too fast if you ask me

  • $\begingroup$ What's the ODE here? is it $y''+xy'+y=0$ or $y''+xy'-y=0$? $\endgroup$ – Mercy King Jun 21 '12 at 1:31
  • $\begingroup$ @Mercy, sorry it is $y'' +xy' - y= 0$ $\endgroup$ – Hawk Jun 21 '12 at 1:35
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    $\begingroup$ Don't be too hasty with "there will be no odd terms": $a_1$ is arbitrary (and in fact $y=x$ is a solution). Remember that for a second-order differential equation there must be two arbitrary constants in the general solution. $\endgroup$ – Robert Israel Jun 21 '12 at 2:08
  • $\begingroup$ Oh wow, I had completely forgotten that $a_1$ existed $\endgroup$ – Hawk Jun 21 '12 at 2:20

Rewriting the recurrence $$a_{n+2}=\frac{a_n(1-n)}{(n+1)(n+2)}=-\frac{a_n(n-1)}{(n+1)(n+2)}$$ as $$a_{n+2}(n+1)=-\frac{a_n(n-1)}{n+2}$$ suggests the substitution $b_n=a_n(n-1)$, $b_0=-a_0$. Then we have simply $$b_{n+2}=-\dfrac{b_n}{n+2}\;.$$

Sticking to the even-numbered terms, we get $b_0,b_2=-\dfrac{b_0}2,b_4=\dfrac{b_0}{2\cdot4},b_6=-\dfrac{b_0}{2\cdot4\cdot6}$, and it’s apparent (and easily proved by induction) that $$b_{2n}=\frac{(-1)^nb_0}{2^nn!}\;.$$

This translates back to $$a_{2n}=\frac{(-1)^{n+1}a_0}{(2n-1)2^nn!}\;.$$

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  • $\begingroup$ The second equation should be $$a_{n+2}(n+1)=-\dfrac{a_n(n-1)}{n+2}. $$ $\endgroup$ – the_fox Jun 21 '12 at 1:48
  • $\begingroup$ @Stefanos: Thanks! $\endgroup$ – Brian M. Scott Jun 21 '12 at 1:49
  • $\begingroup$ @BrianM.Scott, what is motivation behind writing it as $a_{n+2}(n+1)=-\frac{a_n(n-1)}{n+2}$? I was taught to always solve for the highest subscript $\endgroup$ – Hawk Jun 21 '12 at 1:57
  • $\begingroup$ @jak: I was looking for some way to simplify it and noticed that if I pulled the $n+1$ over to the other side, I had (bar the minus sign) something of the form $a_k(k-1)$ on each side. That meant that I could substitute and get a much simpler recurrence. If you’ve never seen the trick of substituting to get a simpler series, it probably isn’t at all obvious. $\endgroup$ – Brian M. Scott Jun 21 '12 at 2:06
  • $\begingroup$ @BrianM.Scott, to find the new $b_{n+2}$, is this what you did? $b_n = a_n(n-1) \implies b_{n+2} = a_{n+2}(n + 2 - 1) = a_{n+2}(n + 1)$ $\endgroup$ – Hawk Jun 21 '12 at 2:13

A start: Everything will go more nicely if we avoid premature cancellation.

Let's forget about $a_0$, it is a constant multiplier, we can put it in later. (If I were doing this for myself, I would also forget temporarily about minus signs, they are under control. And $1-n$ really wants to be written as $-(n-1)$.)

When we calculate without cancellation we get $a_2=-\frac{1}{1\cdot 2}$, $\,a_4=\frac{1}{1\cdot 2\cdot 3\cdot 4}$, $\,a_6=-\frac{1\cdot 3}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6}$. The next term has $1\cdot 3\cdot 5$ on top.

That should be roughly enough to see what's going on. If not, one or two more terms are decisive.

We get the obvious factorials at the bottom. At the top, we have the product of successive odd numbers. It may be useful to simplify this. For example, look at $1\cdot 3\cdot 5\cdot 7$, which comes up a little later. This is $8!$ with the even terms missing. So it is $\frac{8!}{2\cdot 4\cdot 6\cdot 8}$, which is $\frac{8!}{(2^4)(4!)}$.

Once we know the answer, we can, if we are in the mood, provide a formal proof by induction.

Remarks: $1.$ Products such as $7\cdot 5\cdot 3\cdot 7$ or $10\cdot 8\cdot 6\cdot 4\cdot 2$ come up often enough that a special notation has been devised for them. the first is denoted by $7!!$, the second by $10!!$. This is a somewhat strange notation, since $7!!$ is definitely not the same as $(7!)!$.

$2.$ Premature cancellation can hide structure. There is an associated more modern affliction called digititis. Everything gets pushed into a calculator. The result is usually a structureless jumble of digits.

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It is obvious that $a_{2n+1}=0$ for every $n \ge 1$. Set $b_n=a_{2n}$ for $n \ge 0$. Then $$ b_k=a_{2(k-1)+2}=-\frac{2k-3}{(2k-1)(2k)}b_{k-1}=-\frac{1}{2}\cdot\frac{1}{k}\cdot\frac{2k-3}{2k-1}b_{k-1} \quad \forall k \ge 0. $$ Taking the product on both sides for $1 \le k \le n$ we get $$ b_nb_{n-1}\cdots b_1=\left(-\frac{1}{2}\right)^{n}\cdot\frac{1}{n!}\cdot\frac{-1.3.\cdots(2n-3)}{1.3.\cdots(2n-3)(2n-1)}b_{n-1}b_{n-2}\cdots b_1b_0. $$ After simplication we get $$ b_n=\left(-\frac{1}{2}\right)^{n}\cdot\frac{-1}{n!(2n-1)}b_0, $$ i.e. $$ a_{2n}=\frac{(-1)^{n+1}}{n!2^n(2n-1)}a_0. $$

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  • $\begingroup$ how did you change the subscript in the first line there? I am a little confused. $\endgroup$ – Hawk Jun 21 '12 at 2:06
  • $\begingroup$ Do you mind explaining again how you got from $b_n = a_{2n}$ to $b_k = a_{2(k-1)+2}$? $\endgroup$ – Hawk Jun 22 '12 at 19:06
  • $\begingroup$ $2(k-1)+2=2k-2+2=2k$ $\endgroup$ – Mercy King Jun 22 '12 at 19:33

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