How to find a function $f$ that satisfy the equation $f(x)=f(x-1)+x^2$ How to find a continuous function $f$ that satisfy the equation $f(x)=f(x-1)+x^2$
Where both $x$ and $f$ are members of the set of real numbers.
There’s probably more than one $f$ that satisfies the equation. So is there like a general form for all $f$ that do so?
 A: That is similar to the function that gives the sum of the first $n$ perfect squares. It actually follows from how that formula is proved by induction.
In this case then $f(x)=\dfrac{x(x+1)(2x+1)}{6}$
This is because for example: $f(3)-f(2)=3^2$ which should fit that sum of squares description.
However, it doesn't hurt to be more detailed. Let's say the $x^2$ defines a polynomial function. I think this is the only kind of function possible but I can't prove it.
Let's also pick one of degree 3. I'll explain why in the end.
Define it generally as $f(x)=ax^3+bx^2+cx+d$
Then:
$ax^3+bx^2+cx+d=f(x)=a(x-1)^3+b(x-1)^2+c(x-1)+d+x^2$
Let me just do the messy expansion for you.
$ax^3+bx^2+cx+d=ax^3+(b-3a+1)x^2+(3a-2b+c)x+(b+d-c-a)$
Equating coefficients we have:
$c=3a-2b+c \implies 3a=2b$
$b=b-3a+1 \implies a=\dfrac{1}{3}$
$b=3 \cdot \dfrac{1}{3} \cdot \dfrac{1}{2}=\dfrac{1}{2}$
$d=b+d-c-a \implies c=\dfrac{1}{6}$
$d$ is not determined so in the function it can be anything.
Thus all functions satisfying this equation are:
$f(x)=\dfrac{1}{3}x^3+\dfrac{1}{2}x^2+\dfrac{1}{6}x+d$
Or when you factorize and add elegance:
$\boxed{f(x)=\dfrac{x(x+1)(2x+1)}{6}+k \quad k \in \mathbb{R}}$
The reason why I picked a cubic is because upon expansion, the $ax^3$ terms cancel out as the working shows. That's why I didn't consider them. In that case the next lowest power which is $bx^2$ needs to match the highest power on the other side. So we don't need to go any higher than 3.
$\text{Edit}^2$
I hear that in this case $k$ can also be a function with a period of one e.g. $\sin(2\pi x)$ @ WE Tutorial School. Why didn't I think of that? Anyway, the other backbone is legitimate reasoning.
I also tried adding a higher power $ex^4$ on $f(x)$ but I got $a=a-4e$ which gives me a stronger statement.
