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Let $f:\Bbb R\to\Bbb R$ such that $f(1)=3$ and $f$ satisfies the functional equation $$f(x)f(y) = f(x+y) + f(x-y)$$ Find the value of $f(7)$.

Attempt:

If $x=1$ and $y=0$, we find

$$f(1) f(0) = 2f(1) \implies f(0) = 2$$

If we fix $y=1$, we get the recurrence relation

$$\begin{cases} f(0) = 2 \\ f(1) = 3 \\ f(x + 1) - 3f(x) + f(x - 1) = 0 & \text{for } x\ge1 \end{cases}$$

From here I can solve for $f(x)$ or sequentially compute $f(2),f(3),f(4),\ldots$ to arrive at $f(7) = 843$.

Question:

Is there a more elegant way of finding $f(7)$ directly from the functional equation?

user170231
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  • well you can solve the linear induction relation $u_n=f(n)$ satisfying $u_{n+2}-3u_{n+1}+u_n=0\iff u_n=(1+\varphi)^n+(1-\frac 1{\varphi})^n$ and calculate $u_7$ but not sure if this really is easier considering that $f(7)$ is quite accessible quickly. – zwim Sep 21 '22 at 20:29
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    I believe this is $f(n)=L_{2n}$ where $L$ is the Lucas numbers, defined by $L_0=2$, $L_1=1$, and $L_n=L_{n-1}+L_{n-2}$ (the same recurrence as the Fibonacci numbers). Wikipedia lists some identities, one of which implies $L_{2n}=F_{2n-1}+F_{2n+1}$. – Akiva Weinberger Sep 21 '22 at 20:43
  • (only for integer inputs) – Akiva Weinberger Sep 21 '22 at 20:49
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    For complete resolution note that if you set $f(x)=2g(x)$ then $g$ is solution of d'Alembert functional equation of the form $\cos(\alpha x)$ or $\cosh(\alpha x)$ depending on initial conditions. See for instance here https://math.stackexchange.com/q/1570977/399263 – zwim Sep 21 '22 at 21:22
  • @zwim Thanks for the reference. I had a feeling this equation was named. – user170231 Sep 21 '22 at 21:51

2 Answers2

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Well, is not too fancy, but you could do

$f(7)=f(4+3)=f(4)f(3)-f(1),$

and using the fact that $f(4)=f(3)f(1)-f(2),$

you could compute $f(7)$ by knowing only $f(1),\;f(2),$ and $f(3)$.

Fernando
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3

Evaluate $\pmb{f(0)}$

$f(1)=3$ and $$ f(x)f(y)=f(x+y)+f(x-y)\tag1 $$ Setting $x=1$ and $y=0$, $(1)$ gives $f(0)=2$.

Determine and Solve a Recursion

Setting $y=1$, $(1)$ gives $$ 3f(x)=f(x+1)+f(x-1)\tag2 $$ Solving the second order linear recurrence in $(2)$ and using $f(0)=2$ and $f(1)=3$ gives, for $x\in\mathbb{Z}$, $$ \begin{align} f(x)&=\left(\frac{3+\sqrt5}2\right)^{\large\!x}+\left(\frac{3-\sqrt5}2\right)^{\large\!x}\tag{3a}\\ &=\phi^{2x}+\phi^{-2x}\tag{3b}\\[6pt] &=L_{2x}\tag{3c} \end{align} $$ where $L_n$ is the $n^\text{th}$ Lucas Number (as Akiva Weinberger mentioned in a comment).

Check the Functional Equation $\pmb{(1)}$

If $(1)$ has a solution, then it is given by $(3)$. However, to show that $(1)$ has a solution, we need to verify that $(3)$ satisfies $(1)$. $$ \begin{align} f(x)f(y) &=\left(\phi^{2x}+\phi^{-2x}\right)\left(\phi^{2y}+\phi^{-2y}\right)\tag{4a}\\[4pt] &=\phi^{2(x+y)}+\phi^{-2(x+y)}+\phi^{2(x-y)}+\phi^{-2(x-y)}\tag{4b}\\[6pt] &=f(x+y)+f(x-y)\tag{4c} \end{align} $$ So, $(1)$ is satisfied.

Apply to the Question

Thus, $$ \begin{align} f(7) &=L_{14}\tag{5a}\\[4pt] &=843\tag{5b} \end{align} $$

robjohn
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