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given the equation

$$ 1= f(x)+f(x/2)+f(x/3)+f(x/4) $$

how could i solve it ?? or the most general equation

$$ 1= f(x)+f(x/2)+f(x/3)+f(x/4)+....+f(x/N) $$

for a given 'N' number, where could i use Wolfram math online to solve it ? thanks.

for the case $$ 1= f(x)+f(x-2)+f(x-3)+f(x-4) $$

i could find a generating function since it is just a recurrence equation

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A trivial solution is the constant function $f(x) = 1/N$ – FiveLemon May 10 '13 at 10:01
You might as well consider $g(x)=f(x)-1/N$ which satisfies the homogeneous equation with 0 instead of 1. The functions would have to oscillate rapidly near zero if they are nonzero. – Sharkos May 10 '13 at 10:05
up vote 3 down vote accepted

If we assume that the function is analytic, then the first observation to make is that $$ \lim_{x\to0} \sum_{i=1}^N f(x/i)=N\lim_{x\to0}f(x) = 1\\ \lim_{x\to0}f(x) = \frac1N $$ Now, if you take the derivative, you have $$ \sum_{i=1}^N \frac1if'(x/i) = 0 $$ In the limit as $x\to0$, we have $$ \lim_{x\to0}\sum_{i=1}^N \frac1if'(x/i) = \sum_{i=1}^N \frac1i \cdot \lim_{x\to0}f'(x) = 0\\ \lim_{x\to0}f'(x) = 0 $$ And this process can be repeated for all derivatives. As such, as it's analytic, we find that $f(x)=\frac1N$ is the only analytic solution.

If we don't require analyticity, there may be many more solutions - at the very least, I expect some interesting everywhere-discontinuous functions.

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+1, but we already expected funky behaviour at the origin, so analytic solutions there are not what we're really interested in (: – Sharkos May 10 '13 at 10:54

Let $h(x)=f(e^x)-1/N$.

Then $$h(x)+h(x-\log 2) + h(x-\log 3) +\cdots + h(x-\log N)=0$$

This brings it close to the form of a recurrence relation, but is of a type I do not know how to even attempt to solve. I suspect that very little is actually known about such defining equations. (The only thing I could find was Recurrence relations on a continuous domain in which nothing was actually deduced in the end anyway!)

As FiveLemon points out, in fact one can seek $$h(x)=a^x$$ type solutions to this equation just as for recurrence relations. One finds a nonpolynomial equation in $a$ which is the natural generalization of the recurrence relation case. It's not immediately clear whether there are such solutions or whether these are in any sense whatsoever complete. (Yes? Probably not? resp.) The equation should be investigated! I don't have a computer atm...

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Here is a somewhat unrigorous attempt. I have no actually theory for this; I was just messing around. Following Sharkos's comment we can let $g(x) = f(x) - 1/N$ and solve the equation

$$g(x) + g(x/2) + g(x/3) + g(x/4) = 0.$$

This is the same as solving

$$g(12x) + g(6x) + g(4x) + g(3x) = 0.$$

Let $y$ be a root of $$y^{\log(12)} + y^{\log(6)} + y^{\log(4)} + y^{\log(3)} = 0.$$ Then

$$g(x) = y^{\log(x)}$$

is a possible solution for $g$. We can check this as follows

\begin{eqnarray} g(12x) + g(6x) + g(4x) + g(3x) &=& y^{\log(12x)} + y^{\log(6x)} + y^{\log(4x)} + y^{\log(3x)}\\ &=& (y^{\log(12)} + y^{\log(6)} + y^{\log(4)} + y^{\log(3)})\cdot y^{\log(x)}\\ &=& 0 \cdot y^{\log(x)} \end{eqnarray}

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I'm thinking about this; could multivaluedness be a problem? Also +1 – Sharkos May 10 '13 at 11:02
It maybe the case that $0$ is the only possible value for $y$ above anyhow. However for $N=2$, we have $y=-1$. – FiveLemon May 10 '13 at 11:05
what does 'mathematica softwary' say ?? – Jose Garcia May 10 '13 at 12:08

Let $r$ be an non-negative integer. Let $\lambda_k = \alpha_k + \beta_k i$ be any set of roots of the equation:

$$ 1 + 2^{-\lambda} + 3^{-\lambda} + 4^{-\lambda} + \cdots + N^{-\lambda} = 0\tag{*1}$$ subject to the constraint $\alpha_i > r$ and $\beta_i > 0$.

For any real numbers $A_k^{\pm}$ and $\delta_k^{\pm}$, consider the function $g$ defined by:

$$g(x) = \begin{cases} \sum_{k} A_k^{+} \Re\left(e^{i\delta_k^{+}} x^{\lambda_k} \right)\\ \sum_{k} A_k^{-} \Re\left(e^{i\delta_k^{-}} |x|^{\lambda_k} \right) \end{cases} = \begin{cases} \sum_{k} A_k^{+} x^{\alpha_k}\cos(\beta_k \log x + \delta_k^{+}), &\quad\text{ for } x \ge 0\\ \sum_{k} A_k^{-} |x|^{\alpha_k}\cos(\beta_k \log |x| + \delta_k^{-}),&\quad\text{ for } x \le 0 \end{cases} $$

It is easy to check $f(x) = g(x) + \frac{1}{N}$ satisfy the functional equation

$$1 = f(x) + f(\frac{x}{2}) + \cdots + f(\frac{x}{N})\tag{*2}$$

provided the sum over $k$ converges. Furthermore, if $A_k^{\pm}$ decrease fast enough, the constraint $\alpha_i > r$ will force $g$ and its first $r^{th}$ derivatives vanishes and continuous at $0$.

In short,

if the set of roots of $(*1)$ is non-empty, then there are non-trivial $C^{r}$ solution for $(*2)$.

For an example, consider the case $N = 4$ and $r = 0$. It seems there are infinite many solution for $(*1)$. The one with smallest $\beta$ is give by:

$$\lambda_1 = \alpha_1 + \beta_1 i \sim 0.62597108186373 + 3.127120203586539 i$$

Which leads to a non-trivial continuous solution for $(*2)$: $$f(x) = \frac{1}{4} + |x|^{\alpha_1} \cos(\beta_1 \log|x|) \sim \frac14 + |x|^{0.62597108186373} \cos( 3.127120203586539 \log |x|)$$ Image of a non-trivial solution Please note that for $N = 4$, it seems all the complex roots of $(*1)$ has $\Re(\lambda) < 1$. It is not clear whether $(*2)$ has any non-trivial $C^{1}$ solution at all.

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Very nice! I started to think in this direction by looking at the Mellin transform of $f$ but didn't get far. (+1) – Start wearing purple May 10 '13 at 15:30
@O.L. This is the third? time I answered question of similar nature. Every time I understand the problem a little bit more and every time I fail to prove the set of solutions is complete :-( – achille hui May 10 '13 at 15:52

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