Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following functional equation:

$$f(a+b)=f^{n-1}(a)f(b)+f^{n-2}(a)f^{1}(b)+...+f(a)f^{n-1}(b)=\sum_{k=0}^{n-1}f^{n-1-k}(a)f^{k}(b)$$

where $a,b$ are complex and the function $f$ is an entire function and $n\in\mathbb N$.

What is the usual way to solve these problems?

share|improve this question
    
Yes, $f^k(x)$ denotes the kth derivative –  Voyage Apr 2 '13 at 15:37
    
Why is $k=0$ wrong? –  Voyage Apr 2 '13 at 15:42
1  
One usually writes $f^k$ for $f\circ f\circ \ldots \circ f$ ($k$ times). And $f^{(k)}$ for the $k$-th derivative. I think... –  1015 Apr 2 '13 at 16:02

2 Answers 2

up vote 2 down vote accepted

Let $\partial_a$ and $\partial_b$ stand for the shorthand of $\frac{\partial}{\partial a}$ and $\frac{\partial}{\partial b}$. The functional equation can be rewritten as:

$$f(a+b) = \left(\partial_a^{n-1} + \partial_a^{n-2}\partial_b + \cdots + \partial_a\partial_b^{n-2} + \partial_b^{n-1}\right)(f(a)f(b))$$

Apply $\partial_a - \partial_b$ on both sides, we get:

$$ 0 = (\partial_a - \partial_b)f(a+b) = \left(\partial_a^n - \partial_b^n\right)(f(a)f(b))$$

which is equivalent to $$(\partial_a^n f(a)) f(b) - f(a)(\partial_b^{n}f(b)) = 0$$ Since $a$ is independent of $b$, this implies the existence of a constant $\lambda$ such that

$$\frac{d^n}{d a^{n}} f(a) = \lambda^{n}f(a)\tag{*1}$$

Let us look at the case $\lambda \ne 0$. Let $\omega$ be a primitive $n^{th}$ root of unity. A general solution for $(*1)$ has the form:

$$f(a) = \sum_{j=0}^{n-1} A_j e^{\lambda\,\omega^j a}$$

where $A_j, j = 0,\ldots,n-1$ are constants. Substitute this back into the functional equation, we get:

$$\sum_{j=0}^{n-1} A_j e^{\lambda\,\omega^j (a+b)} =\sum_{j=0}^{n-1} \sum_{k=0}^{n-1} ( A_j e^{\lambda\,\omega^j a})( A_k e^{\lambda\,\omega^k b})\left( \sum_{l=0}^{n-1} (\lambda\,\omega^j)^l (\lambda\,\omega^k)^{n-1-l} \right) \tag{*2}$$ The last factor in R.H.S of $(*2)$ is given by: $$\sum_{l=0}^{n-1} (\lambda\,\omega^j)^l (\lambda\,\omega^k)^{n-1-l} = \lambda^{n-1}\omega^{-k}\sum_{l=0}^{n-1}\omega^{(j-k)l} = n\lambda^{n-1} \omega^{-k} \delta_{jk}$$

where $\delta_{jk}$ is the Kronecker delta. One can simplify $(*2)$ to:

$$\sum_{j=0}^{n-1} A_j e^{\lambda\,\omega^j (a+b)} = \sum_{j=0}^{n-1} n \lambda^{n-1} \omega^{-j} A_j^2 e^{\lambda\,\omega^j(a + b)}$$

This implies $A_j$ is either $0$ or $\frac{\omega^j}{n\lambda^{n-1}}$. For $\lambda \ne 0$, there are $2^{n}-1$ non-zero solutions for the functional equation:

$$f(a) = \frac{1}{n\lambda^{n-1}} \sum_{j=0}^{n-1} \epsilon_j \omega^j e^{\lambda\,\omega^j a}\tag{*3}$$

where $\epsilon_j \in \{0,1\}$ for $j = 0, \ldots, n-1$ and not all zero.

Consider the special case all $\epsilon_j = 1$. If one expand R.H.S of $(*3)$ as a Laurent series at $\lambda = 0$. It is not hard to see all the pole in $\lambda$ cancel out. This give us a solution of the functional equation for $\lambda = 0$, namely:

$$f(a) = \lim_{\lambda\to 0}\frac{1}{n\lambda^{n-1}} \sum_{j=0}^{n-1} \omega^j e^{\lambda\,\omega^j a} = \frac{a^{n-1}}{(n-1)!}\tag{*4}$$

EDIT

In fact, this is the only non-zero solution for $\lambda = 0$. Apply $\partial_a$ to both sides of the functional equation, we get:

$$\begin{align}f'(a+b) &= \partial_a f(a+b)\\ &= \partial_a^n f(a)f(b) + (\partial_a^{n-1}\partial_b + \cdots + \partial_a\partial_b^{n-1})(f(a)f(b))\\ &= (\partial_a^{n-2} + \partial_a^{n-3}\partial_b + \cdots + \partial_b^{n-2})(f'(a)f'(b)) \end{align}$$

Notice $\partial_a^n f(a) = 0 \implies \partial_a^{n-1} f'(a) = 0$. This means for any $\lambda = 0$ solution $f(\cdot)$ of the functional equation for $n-1$. $f'(\cdot)$ is a $\lambda = 0$ solution for the functional equation for $n-2$.

If we have shown $(*4)$ is the only non-zero $\lambda = 0$ solution for $n < N$, then for $n = N$,

$$f'(a) = \frac{a^{N-2}}{(N-2)!} \;\;\text{ or }\;\; 0 \implies f(a) = \frac{a^{N-1}}{(N-1)!} + c\;\;\text{ or }\;\;c$$ for some constant $c$. Plug this into the functional equation for $n = N$, $c$ need to satisfy:

$$c = (\partial_{a}^{N-1}f(a)) c + c(\partial_b^{N-1}f(b)) = 2 c\;\;\text{ or }\;\;0$$

In both cases, $c = 0$ and $(*4)$ is indeed the only non-zero $\lambda = 0$ solution.

share|improve this answer
    
Thanks, I have two questions. 1) In your term $(\sum_{l=0}^{n-1} (\lambda\,\omega^j)^l (\lambda\,\omega^k)^{n-1-l})$ , does it refer to the (n-1-l) derivative? 2) Why are there exactly $2^n-1$ solutions? –  Voyage Apr 2 '13 at 18:50
    
1) Yes, it refer to the $(n-1-l)$-th derivative. $$\partial_b^{n-1-l} e^{\lambda\,\omega^k b} = (\lambda\,\omega^k)^{n-1-l} e^{\lambda\,\omega^k b}$$ 2) When $\lambda \ne 0$, there are $n$ constants $\epsilon_0, \epsilon_1, \ldots, \epsilon_{n-1}$ one can pick. Since each $\epsilon_j$ is either 0 or 1, there are totally $2^{n}$ solutions. $2^{n} - 1$ of them are non-zero. –  achille hui Apr 3 '13 at 0:32

If $n=1$, we have $f(a+b) = f(a)f(b)$. If $f$ is assume to be continuous, then we have $f(x) = e^{kx}$.

If $n=2$, we have $f(a+b) = f'(a)f(b) + f(a) f'(b)$. Taking a cue from the above, if we let $f(x) = ce^{kx}$, we get $$ce^{k(a+b)} = 2c^2ke^{k(a+b)}$$ This gives us $2ck = 1 \implies c = \dfrac1{2k}$.

If $n=3$, we have $f(a+b) = f^2(a)f(b) + f'(a) f'(b) + f(a) f^2(b)$. As above, if we let $f(x) = ce^{kx}$, we get $$ce^{k(a+b)} = (c^2k^2 + c^2k^2+c^2k^2)e^{k(a+b)}$$ This gives us $3ck^2 = 1 \implies c = \dfrac1{3k^2}$.

Hence, in general, $$f(x) = \dfrac{e^{kx}}{nk^{n-1}}$$ satisfies the recurrence. This is one possible solution. You could try $f(x) = P_{n-1}(x)e^{kx}$ where $P_{n-1}(x)$ is a polynomial of degree $n-1$ and try to obtain some constraint on its coefficients.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.