I am doing some practice problems for solving second order ODEs, and I am a bit stuck on this one.

Here is what I have:


Let $y = \sum_{n=0}^{\infty} C_nx^n \implies y' = \sum_{n=0}^{\infty} nC_nx^{n-1} \implies y'' = \sum_{n=0}^{\infty} n(n-1)C_nx^{n-2} $

Substituting this into the ODE, and I get:

$$ \sum_{n=0}^{\infty} n(n-1)C_nx^{n-2} -2\sum_{n=0}^{\infty} nC_nx^{n}+ \sum_{n=0}^{\infty} C_nx^n = 0$$

Then getting each term to $x^n$ and starting each sum at $n=0$, I have:

$$ \sum_{n=0}^{\infty} [(n+2)(n+1)C_{n+2}-2 nC_n+ C_n]x^n = 0 $$ $$ \implies C_{n+2} = \frac{(2n-1)C_n}{(n+2)(n+1)}$$

I notice that this decouples into two series' for odd and even terms, but I am having trouble with determining the general formula for $C_n$ for each series:

For $n$ even:

When $n=0: C_2 = \frac{-C_0}{2} $

When $n=2: C_4 = \frac{3C_2}{4 \cdot 3} = \frac{-3C_0}{4!} $

When $n=4: C_6 = \frac{7C_4}{6 \cdot 5} = \frac{-7 \cdot 3C_0}{6!} $

When $n=6: C_8 = \frac{11C_6}{8 \cdot 7} = \frac{-11 \cdot 7 \cdot 3C_0}{8!} $

For $n$ odd:

When $n=1: C_3 = \frac{C_1}{3 \cdot 2} $

When $n=3: C_5 = \frac{5C_3}{5 \cdot 4} = \frac{5C_3}{5!} $

When $n=5: C_7 = \frac{9C_5}{7 \cdot 6} = \frac{9 \cdot 5C_1}{7!} $

When $n=7: C_9 = \frac{13C_7}{9 \cdot 8} = \frac{13 \cdot 9 \cdot 5C_1}{9!} $

I am mainly finding it difficult to determine the closed formula for the numerator in each series, so that I can calculate the radius of convergence of each one.

Thanks so much, any help is greatly appreciated.

  • $\begingroup$ Try writing $C_{n}$ in terms of $C_{n-2}$ using your relation, then replace $C_{n-2}$ in terms of $C_{n-4}$, etc... and deduce a pattern for $C_{n}$ in terms of $C_0$ or $C_1$ (even v. odd). $\endgroup$ – Semiclassical Aug 30 '14 at 0:54
  • $\begingroup$ thanks, will give that a go. $\endgroup$ – Terrence J Aug 30 '14 at 1:40
  • $\begingroup$ Just in case you were wondering or wanted to check your answer...mathworld.wolfram.com/HermiteDifferentialEquation.html $\endgroup$ – ClassicStyle Aug 30 '14 at 1:53
  • $\begingroup$ @Semiclassical - I did as you suggested, and it didn't really help me. I'm still stuck with the same issue. $\endgroup$ – Terrence J Aug 30 '14 at 3:01
  • $\begingroup$ Is the second term $-2xy$ or $-2xy'$? Your substitution seems to be for $-2xy'$ but the equations say $-2xy$. $\endgroup$ – JiK Aug 30 '14 at 23:07

Basically what you need is a notation for things like $1\cdot5\cdot9\cdot13\cdots$. These are generalizations of the factorial, where numbers with increment more than $1$ are multiplied. When the increment is $2$ this is called double factorial $n!!:=n(n-2)(n-4)\cdots$, and when the increment is $4$ it is quadruple factorial $n!!\,!!:=n(n-4)(n-8)\cdots$, so $1\cdot5\cdot9\cdot13=13!!\,!!$, $3\cdot7\cdot11=11!!\,!!$. These are all particular cases of the multifactorial. With this notation $$C_{n}=-\frac{(2n-5)!!\,!!}{n!}C_0\text{ and } C_{n}=\frac{(2n-5)!!\,!!}{n!}C_1$$ depending on whether $n$ is even or odd. This gives two linearly independent solutions, one even one odd.

As for computing the radius of convergence you don't actually need an explicit formula. Since both solutions have only even or only odd powers the ratio formula for the radius can not be used, but applying the ratio test directly gives $\lim_{n\to\infty}\frac{|C_{n+2}x^{n+2}|}{|C_{n}x^n|}<1$ since the consecutive powers in each series differ by $2$. This only requires knowing the limit of $\frac{|C_{n+2}|}{|C_{n}|}|x|^{2}=\frac{(2n-1)}{(n+2)(n+1)}|x|^{2}$, which is $0$ for any $x$. Therefore both series converge everywhere and both radii are $\infty$.

  • $\begingroup$ Thanks, is there anywhere that defines what $3!!!!$ is? Also, for computing the radius of convergence, don't you have to do each series separately, as they are different solutions? $\endgroup$ – Terrence J Aug 30 '14 at 3:00
  • $\begingroup$ apologies, you're right, in this case, since the coefficients only differ by the negative for the two series' I see that their radius of convergence will be the same. $\endgroup$ – Terrence J Aug 30 '14 at 3:28
  • $\begingroup$ @user153663 Definition is usually done recursively en.wikipedia.org/wiki/Factorial#Multifactorials. You are right actually, these solutions should be tested separately, and the ratio test has to be applied directly since half of the powers are missing in each series, see the edit. $\endgroup$ – Conifold Aug 30 '14 at 21:35

Well, may be I am wrong but it seems for me a general formula can be easily deduced from your recurrent one: $$ C_{2n}=\frac{(2n-2))!}{(2n)!} \prod_{i=1}^n (4i-5) C_0\\ C_{2n+1}=\frac{(2n-1))!}{(2n+1)!} \prod_{i=1}^n (4i-3) C_1\\ $$ The product also can expressed in factorials but I let it do for somebody else :). $C_0$ and $C_1$ are free constant.


Starting from your last relation $$ C_{n+2} = \frac{(2n-1)C_n}{(n+2)(n+1)}$$ it seems that $$C_n=\frac{2^{n-\frac{9}{2}} \Gamma \left(\frac{n}{2}-\frac{1}{4}\right) \left({C_1} \left((-1)^n-1\right) \Gamma \left(-\frac{1}{4}\right)-2 {C_0} \left((-1)^n+1\right) \Gamma \left(\frac{1}{4}\right)\right)}{\pi \Gamma (n+1)}$$ which effectively separates the $C$'s for odd and even values because of the $\Big((-1)^n\pm 1\Big)$ terms which multiply the $C_0$ and $C_1$ terms.

This so gives $$C_{2n}=-\frac{2^{2 n-\frac{5}{2}} \Gamma \left(\frac{1}{4}\right) \Gamma \left(n-\frac{1}{4}\right)}{\pi \Gamma (2 n+1)}C_0$$ $$C_{2n+1}=-\frac{ 2^{2 n-\frac{5}{2}} \Gamma \left(-\frac{1}{4}\right) \Gamma \left(n+\frac{1}{4}\right)}{\pi \Gamma (2 n+2)}C_1$$


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