Recurrence relation in a recurrence relation?

[MevsEinstein]

TL;DR

I tried to solve a functional equation by using a recurrence relation, but then I decided to do a recurrence relation in a recurrence relation.

There's a famous functional equation that was asked in the 2019 IMO. It looks like this: find all f: Z -> Z where f(2a)+2f(b)=f(f(a+b)).

I thought of solving it using a recurrence relation where a_n=f(nx). But when I substituted values in the functional equation (after setting a and b equal, and then changing the variable to x), I got this: a_2 + 2a_1 = f(a_2). Is it allowed in Mathematics to make a second relation (where in this case a_n_i=f(...f(i times)(nx)...), so I get this recurrence relation: a_2_1 + 2a_1_1 = a_2_2?

 

[anuttarasammyak]

I do not think recurrence is an appropriate approach for this problem. I rewrote the formula for any x and t f(f(x))=2f(t)+f(2x-2t) and applied differentiation.

 

[fresh_42]

I do not think recurrence is an appropriate approach for this problem. I rewrote the formula for any x and t f(f(x))=2f(t)+f(2x-2t) and applied differentiation.

You differentiate a function that is defined on integers only?

 

[fresh_42]

Summary:: I tried to solve a functional equation by using a recurrence relation, but then I decided to do a recurrence relation in a recurrence relation.

There's a famous functional equation that was asked in the 2019 IMO. It looks like this: find all f: Z -> Z where f(2a)+2f(b)=f(f(a+b)).

I thought of solving it using a recurrence relation where a_n=f(nx). But when I substituted values in the functional equation (after setting a and b equal, and then changing the variable to x), I got this: a_2 + 2a_1 = f(a_2). Is it allowed in Mathematics to make a second relation (where in this case a_n_i=f(...f(i times)(nx)...), so I get this recurrence relation: a_2_1 + 2a_1_1 = a_2_2?

It usually helps to consider small integers first: $f(0)\, , \,f(-1)\, , \,f(1)$ or generally $f(a)=f(b)$. There is no one method fits all solution.

Is it allowed in Mathematics to make a second relation

Of course, it is. I don't think that it will be of great help here, but you can define whatever you want as long as it is unambiguously. You can sometimes consider properties of the function without knowing the function itself, e.g. monotony, fixed points, or injectivity.

 

[MevsEinstein]

It usually helps to consider small integers first: $f(0)\, , \,f(-1)\, , \,f(1)$ or generally $f(a)=f(b)$. There is no one method fits all solution.

Of course, it is. I don't think that it will be of great help here, but you can define whatever you want as long as it is unambiguously. You can sometimes consider properties of the function without knowing the function itself, e.g. monotony, fixed points, or injectivity.

I was trying to make my own solution to this functional equation. I'm almost done with my solution using recurrence relations (I now have $f(x)\, \, = 2c_1\, - \,c_2,$ and I want to prove that $c_1$ is $x$).

 

[anuttarasammyak]

You differentiate a function that is defined on integers only?

Yea, I extend f(x) for continuous x, apply $\frac{d}{dx}\frac{d}{dt}$ to the formula in post #2, and get $f"(x)=0$ for any x. At least it is one of all the solutions of the original problem.

 

[fresh_42]

Yea, I extend f(x) for continuous x, apply $\frac{d}{dx}\frac{d}{dt}$ to the formula in post #2, and get $f"(x)=0$ for any x. At least it is one of all the solutions of the original problem.

It could be a method to get an educated guess. The difficulty is then uniqueness. There should be more real solutions than there are integer solutions. But what if the real solution doesn't give an integer solution? However, such questions are primarily an invitation to play with them. Btw., there is a discrete analogon to differentiation, namely investigating the differences $f(n+1)-f(n).$

 

[martinbn]

Yea, I extend f(x) for continuous x, apply $\frac{d}{dx}\frac{d}{dt}$ to the formula in post #2, and get $f"(x)=0$ for any x. At least it is one of all the solutions of the original problem.

$f(x)=x$ satisfies $f''(x)=0$, but is not a solution to the original problem. In fact linear functions $f(x)=mx+n$ are solutions only if $m=2$. You can check after substituting in the equation.

Btw., there is a discrete analogon to differentiation, namely investigating the differences $f(n+1)-f(n).$

This actually helps. Set $a=1$ and $b=n$, then $a=0$ and $b=n+1$, you get
$f(2)+2f(n)=f(f(n+1))$
and
$f(0)+2f(n+1)=f(f(n+1))$
subtract and rearrange, you get
$f(n+1)-f(n)=\frac12 (f(0)-f(2))$
So this is independent of $n$, hence the function is linear.

 

[anuttarasammyak]

f(x)=x satisfies f″(x)=0, but is not a solution to the original problem. In fact linear functions f(x)=mx+n are solutions only if m=2. You can check after substituting in the equation.

Thanks for full writing. I would just add obvious m=n=0.