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The well-known trig identity

$$2\sin(x)\cos(x) = \sin(2x)$$

gives use the differential equation (which I'm not even sure qualifies as an ODE, given the factor in the argument?)

$$2f(x)f'(x) = f(2x)$$

Assuming one had no idea that $\sin(x)$ is a solution to this equation, I was wondering if there was a way to derive the solution anyway.

My attempt

Assume that $f(x)$ has a power series expansion valid for all values of $x$ that we are interested in, that is

$$f(x) = \sum_{n=0}^{\infty}a_nx^n$$

for some coefficients $a_n$.

Plugging this in gives us

$$\sum_{n=0}^\infty 2^na_nx^n=2\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty (n+1)a_{n+1}x^n\right) \tag1$$

Assuming absolute convergence, we can rewrite the RHS using the Cauchy product formula:

\begin{align} \left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty (n+1)a_{n+1}x^n\right)&=\sum_{n=0}^\infty\sum_{k=0}^n(k+1)a_{k+1}x^ka_{n-k}x^{n-k}\\ &=\sum_{n=0}^\infty x^n\sum_{k=0}^n(k+1)a_{k+1}a_{n-k} \end{align}

Plugging this into $(1)$ and comparing coefficients of powers of $x$, we get:

$$a_n=2^{1-n}\sum_{k=0}^n(k+1)a_{k+1}a_{n-k}$$

However, I have no idea how to proceed with this recurrence relation for the coefficients of the power series that I am searching for.

Therefore, my two questions are:

$1.$ Is my approach correct and is there a way to solve this recurrence relation for the known coefficients of the power series expansion of $\sin(x)?$

$2.$ Is there a general approach that is maybe less cumbersome to this kind of problem?

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    Power series methods are only standard for linear equations. There's no guarantee you'll get anything useful here. You might be able to get a few terms though. – 2017-01-09
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    @Kaynex: Absolutely not: think, for instance, of the [Cauchy-Kowaleski theorem](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem), and this is not the only example. On the contrary, power series may prove to be helpful in situations where "linear" techniques fail. – 2017-01-09

2 Answers 2

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One could imagine the solution to be an exponential of the following form:

$$f(x)=ae^{bx}$$

Substituting values in, we have

$$2a^2be^{2bx}=ae^{2bx}$$

$$\implies2ab=1$$

$$\implies a=1/2b$$

$$\implies f(x)=\frac1{2b}e^{bx}$$

which is another more general solution to your differential equation. If $b$ is complex, we might end up with $f(x)=\sin(x)$ due to Euler's formula.

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    True, but then you'd need a uniqueness theorem of some sort, guaranteeing that solutions of this form are the only ones. Plus, your approach uses guessing (even if educated), which some people might not appreciate. – 2017-01-09
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    @AlexM. Also true, but it is a good start most likely. – 2017-01-09
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    I like this approach - it makes the calculations of both the derivatives and $f(2x)$ easy. I'd still be interested to know if there is a way to proceed with the coefficients that I have found and if there is some way to manipulate that expression to somehow arrive at this formula or something similar, but thanks all the same. If no answer comes along during the next 24 hours that expands on that, I'll accept this. :) – 2017-01-10
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    @Tom Uh, sure then. I did notice an oddity, that my solution does not yield $f(x)=\sin(x)$ since at $x=0$, it is never $0$... Unless some amazing Euler's formula comes through of course. – 2017-01-10
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    @SimpleArt Apart from some constant, it arises as the real part of $f(x)$ if $b=i$, but I'm not seeing how we are justified separating it from its imaginary part, as the equation is not linear and therefore the superposition principle is not applicable, I think, is that correct? – 2017-01-11
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    @Tom Yes, I was having the same problem! :-( – 2017-01-11
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Entering the "Ansatz" $$f(x):=\sum_{k=0}^n a_k x^k$$ with, e.g., $n=8$ into Mathematica and putting the coefficients of $$\psi:=2f(x)f'(x)-f(2x)+O[x]^{n+1}$$ to $0$ produces successive numerical values $a_k$ $(0\leq k\leq n)$ that can be attributed to the following solution functions: $$f(x)=\qquad x\>,\qquad{1\over2\lambda}e^{\lambda x}\>,\qquad {1\over\lambda}\sin(\lambda x)$$ with $\lambda\ne0$ an arbitrary complex constant.

Note that a solution which is $\ne0$ in a punctured neighborhood $\dot U$ of $0$ is automatically $C^\infty$ in $\dot U$. But there might exist solutions that have an essential singularity at $0$.