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What are the analytic solutions to $f(3x)=\dfrac{f(x)f(4x)}{2f(2x)}+\dfrac{f^2(2x)}{2f(x)}$?

Are there solutions that do not satisfy a linear first or second ODE ?

See also : About the addition formula $f(x+y) = f(x)g(y)+f(y)g(x)$

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    $\begingroup$ $f(x)=x$ is a solution. $\endgroup$ – Rohan Dec 24 '16 at 12:27
  • $\begingroup$ The functions $x\mapsto e^{cx}$, $c$ constant, are solutions as well. $\endgroup$ – Christian Blatter Dec 24 '16 at 13:10
  • $\begingroup$ Any polynomial $f$ must have degree $0$ or $1$. $\endgroup$ – Michael Hoppe Dec 24 '16 at 13:12
  • $\begingroup$ Those examples all satisfy a first or second ODE. Just saying. $\endgroup$ – mick Dec 26 '16 at 11:49
  • $\begingroup$ Does "a first or second ODE" mean any first order or second order ODE? $\endgroup$ – Yuriy S Apr 20 '18 at 20:04
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This is not a full answer, just some experimental results.

As we are asked for analytic solutions, we can simply substitute a general Taylor series:

$$f(x)=a_0+a_1 x+a_2 x^2+\dots$$

into the equation:

$$2f(x)f(2x)f(3x)-f(x)^2 f(4x)-f(2x)^3=0$$

Discounting the extra solution $f(x)=0$, which doesn't satisfy the original equation.

1) First case $a_0 \neq 0$. Then with some help from Mathematica, we obtain:

$$a_1 \in \mathbb{C}$$

$$a_{ n \geq 2}=\frac{a_1^n}{n! ~a_0^{n-1}}$$

Which makes the general solution for this case:

$$f(x)=a_0 \exp \left( \frac{a_1}{a_0}x \right)$$

Which satisfies 1st order linear ODE with constant coefficients. This fits both the solutions proposed by Christian Blatter.

2) Second case $a_0=0$. Here it becomes more interesting.

First we notice that $a_1 \neq 0$. Because experimentally from $a_1=0$ follows $a_{n \geq 2}=0$ as well, which makes $f(x)=0$, the solution we have already discounted.

If $a_1 \neq 0$ then we have:

$$a_2, a_3 \in \mathbb{C}$$

And the following recurrence relations:

$$a_4=\frac{3 a_1 a_2 a_3-a_2^3}{3 a_1^2}$$

$$ a_5=\frac{4 a_1 a_2 a_4+3 a_1 a_3^2-2 a_2^2 a_3}{10 a_1^2}$$

$$a_6= \frac{5 a_1 a_2 a_5+23 a_1 a_3 a_4-2 a_2^2 a_4-5 a_2 a_3^2}{65 a_1^2}$$

Not sure if I see the pattern in the general case, however experimentally:

2a) If we take $a_1=1$, $a_2=\frac{1}{2}$ and $a_3=\frac{1}{6}$ then we get the exponential function again.

2b) If we take $a_2=0$, it then follows that $a_{2n}=0$ for any $n$, so we obtain an odd function (note that an even function is not a solution, as the above shows, $a_1 \neq 0$).

The first few coefficients are given by:

$$a_3 \in \mathbb{C}$$

$$a_5=\frac{3a_3^2}{10 a_1}$$

$$a_7=\frac{3a_3^3}{70 a_1^2}$$

$$a_9=\frac{a_3^4}{280 a_1^3}$$

$$a_{11}=\frac{3a_3^5}{15400 a_1^4}$$

$$a_{13}=\frac{3a_3^6}{400400 a_1^5}$$

$$a_{15}=\frac{3a_3^7}{14014000 a_1^6}$$

$$a_{17}=\frac{9a_3^8}{1905904000 a_1^7}$$

The denominators are partial products in the sequence $$10 \cdot 7 \cdot 4 \cdot 55 \cdot 26 \cdot 35 \cdot 136 \cdot \cdots$$

This seems promising as there's no apparent pattern.

Taking $a_3=0$ leads us to the solution $f(x)=a_1 x$, which a more general case of Rohan's comment.


I'll look into this further at a later date, but the question itself is not clear to me because I don't believe it's possible to prove that some function given by its Taylor series doesn't satisfy any first order or second order ODE, as there's so many nonlinear equations possible. If we only consider linear ODEs with constant coefficients, the problem becomes more simple.

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  • $\begingroup$ Thank you. A followup question will come. $\endgroup$ – mick Jun 26 '18 at 19:46

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