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$$P(c \cdot x) = \cos(x) P(x)$$

For $c=2$, $P(x) = \sin(x)/x$ is a solution to this. I don't know if there's a closed-form solution for $c \ne 2$.

Rather than add my own attempt at solution, which is dubious at best, I'll add context. I've been trying to solve a rather odd line integral here, and I ended up with a very nice infinite product solution. Robert Israel, put this solution into the form of recurrence relation. So, in attempt to continue the wild-goose-chase, I'm pursuing a solution for this recurrence relation.

(I'll take any solution, as long as the "special functions" used are not tautological and have a public-access paper written about them)

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Zach466920
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  • Any restrictions on the range of $c$ ? – Sergio Parreiras Aug 14 '15 at 15:35
  • @SergioParreiras c is any real number greater than 2. However, generalized behavior, for other values, is cool too. Given the nature of the problem however, you could consider c to be any integer greater than or equal to 2 if that helps. – Zach466920 Aug 14 '15 at 15:49
  • Let me see if I understand what you look for: You look for a function (P(cx)), such as that it will have the appearance as you have above, and for c=2 the result is the function as described in the second line? – Moti Aug 14 '15 at 17:01
  • What is wrong with the solution: $P(cx)=\frac{sin(2x)}{2x}$ – Moti Aug 14 '15 at 17:04
  • @Moti what's wrong with that solution? It's a function of both $c$ and $x$ on the left, and then only a function of $x$ on the right. That can't solve the above recurrence relation. – Zach466920 Aug 14 '15 at 18:48
  • $P(c(2x))=cos(2x)/P(2x)$ and also $c=2 leads to P(2(2x))=sin(2x)/2x$ using this two leads to $cos(2x)=cos(x)$, which is solvable. It seems that your original assumptions are questionable. – Moti Aug 14 '15 at 19:49
  • This is not a recurrence relation. It is a functional equation. – Thomas Andrews Sep 05 '15 at 18:16

1 Answers1

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We know that $c \neq 0$, unless $P(x) = 0$ for all $x$. Also $c\neq 1$, since otherwise we have $P(x)=\cos(x)P(x) \implies \cos(x) =1$ for all $x$.

If it is assumed that $P$ can be expressed as a power series centered at the origin (supported by the result for $c=2$) we find:

$$\sum_{n=0}^\infty a_n c^n x^n = \cos(x) \sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!} \sum_{n=0}^\infty a_n x^n.$$

Thus $$a_n c^n = \sum_{i=0}^n a_i b_{n-i}$$ where $$b_{i} = \left\{ \begin{array}{cc} (-1)^{i/2}/(i)! & i \text{ even}\\ 0 & i \text{ odd}\end{array}\right.$$

Writing out each series to compare terms gives:

$$a_0 + a_1cx+a_2c^2x^2+\cdots = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\right) \left( a_0 + a_1x + a_2x^2 +a_3 x^3 + \cdots \right)$$

Thus $a_0 = a_0$, which provides some degree of freedom of choice for $a_0$. (Note that $A\sin(x)/x$ will satisfy $P(2x)=\cos(x)P(x)$ as well).

Then $a_1 c = a_1$. Either this means that $c=1$ or $a_1=0$. Since we have shown that $c\neq 1$ we have $a_1=0$.

Continuing on this line gives us $$a_2 c^2 = a_2 - \frac{a_0}{2!}$$ which yields $$a_2 = \frac{-a_0}{2!(c^2-1)}.$$

If you continue this process for $a_3, a_4,...$, you will find that $a_{2n+1}=0$ for all $n$, and you will find expressions for each $a_{2n}$ in terms of $a_0$. There might be a nice closed form expression for $a_{2n}$, but I haven't committed to finding it.

Joel
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    Ultimately you will wish to show that the power series for $P(x)$ converges in some region. I am guessing that it should have an infinite radius of convergence, based on the decay rate extrapolated from the first few terms I calculated. – Joel Aug 14 '15 at 18:26
  • There is no need to show that the expansion corresponds to a special function. If the question is whether or not the expression $P(cx)=\cos(x)P(x)$ has a solution, then it is sufficient to demonstrate a solution through a power series. The coefficients can be determined independent of a recursion. Each $a_{2n}$ can be expressed as a finite sum of terms, multiplied by $a_0$. I just didn't want to work it out. – Joel Aug 14 '15 at 18:56
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    The title of your question seems to contradict this: "This 1 innocent looking recurrence relation seems to have no solution." I provided a means of determining a solution. Perhaps what you mean to say is that it does not have a solution in terms of special functions. – Joel Aug 14 '15 at 19:30
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    Honestly, I would have been more motivated to help you with your problem, and meeting the specifications that you have added, if it were not for your aggressive attitude. There is a clear way forward to determine the coefficients that is manageable. A pattern will emerge when you work out the coefficients. It's unlikely that it will correspond to a special function, and I wonder why you think it should in the first place. – Joel Aug 14 '15 at 19:49
  • Hello, I've edited the title of my question to better reflect the content in the question. I'd suggest you either delete or complete the answer, as what I'm asking for is a closed form solution. You should also note that the original tag for this question was closed-form... – Zach466920 Sep 02 '15 at 16:11
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    @Joel Very nice approach :). Thank you for sharing it. – MrYouMath Sep 08 '15 at 16:43
  • @MJay1985 I don't mean to be rude, but as the Op, who already has the solution in terms of infinite product, I find that a solution that is incomplete to be unacceptable. In addition, this entire time, I've been asking for a closed-form solution. All this answer does, is make me have to start a new question, and sink another 50 rep on another "goose-chase". It's not that the answer is incorrect, it's that it ignores my needs and prior-knowledge. – Zach466920 Sep 08 '15 at 18:11