Functional Equations

You're told a function $f$ satisfies $f(x+y) = f(x) + f(y)$ for all real $x,y$, and that $f(3) = 12$. Find $f(1)$ and $f(10)$. There's no formula for $f$ written down anywhere — just a rule it must obey for every pair of inputs. That "for all" is the whole game: since the rule holds for every possible $x$ and $y$, you're free to plug in whatever specific values expose the most information, and a few well-chosen substitutions pin the function down completely.

The idea: plug in the values that break the equation open

A functional equation is really an infinite family of ordinary equations, one for each choice of $x$ and $y$. Most of those choices are useless, but a few specific ones — zero, equal inputs, one input replacing another — tend to make terms cancel or repeat in a way that reveals structure. The standard menu:

• Set $x=y=0$ to find $f(0)$.
• Set $y=x$ to relate $f(2x)$ to $f(x)$.
• Set $y=-x$ to relate $f(0)$ to $f(x)$ and $f(-x)$.
• Look for a pattern and guess the general form, then verify it satisfies the original equation for all inputs, not just the ones you tested.

In plain English: you're using the universality of the rule against itself, extracting a normal, solvable equation from a statement that otherwise has no single unknown to isolate.

Worked example 1: additive functions

With $f(x+y)=f(x)+f(y)$ and $f(3)=12$, set $x=y=1.5$: $f(3) = f(1.5)+f(1.5) = 2f(1.5)$, so $f(1.5) = 6$. More directly, set $x=1,y=1,\text{and }1$ again — i.e., $f(3) = f(1)+f(1)+f(1) = 3f(1)$ by applying the rule twice, so $3f(1) = 12 \Rightarrow f(1) = 4$. Once $f(1)=4$ is known, $f(10) = f(1)+f(1)+\cdots+f(1)$ (ten times) $= 10f(1) = 40$ — repeated application of the same rule. This is the signature of additive functional equations: $f(n) = n \cdot f(1)$ for integer $n$, i.e. $f$ behaves like multiplication by a constant, $f(x) = 4x$.

Worked example 2: finding f(0) and checking a guess

Suppose $f(x)f(y) = f(x+y)$ for all real $x,y$, with $f(1) = 3$. Setting $x=y=0$: $f(0)^2 = f(0)$, so $f(0)$ is $0$ or $1$; if $f(0)=0$ then setting $y=0$ generally forces $f\equiv 0$ everywhere, a degenerate solution, so take $f(0)=1$. Guessing $f(x) = 3^x$ (since $3^x \cdot 3^y = 3^{x+y}$ matches the rule exactly, and $3^1=3$ matches the given value), check it satisfies both the equation and $f(0)=3^0=1$ — it does. So $f(2) = 9$, $f(-1) = 1/3$, read straight off the verified formula rather than solved case by case.