How do I solve $f(x) - f(x - 1) = x$? Is it possible to solve the functional equation $f(x) - f(x - 1) = x$ by algebraic means?
 A: First note that we can add a constant to $f$ and it will still be a solution, so we might as well define $f(0)=0$  Then we can find $f(1)=1, f(2)=3, f(3)=6$ and so on.  We can take finite differences and find that a quadratic polynomial fits all of these, then find that $\frac 12x(x+1)$ satisfies our equation.  The full solution is then $f(x)=\frac 12x(x+1)+f(0)$
A: $\require{cancel}$
A logical guess is a quadratic because we know that taking differences of polynomials of the same degree and leading term will eliminate the top level term, so we try
$$f(x)=ax^2+bx+c$$
then
$$f(x)-f(x-1) = \cancel{ax^2}+\cancel{bx}+\cancel{c}-(\cancel{ax^2}-2ax+a+\cancel{bx}-b+\cancel{c})$$
leaving us with
$$\color{blue}{2a}\cdot x + \color{red}{(b-a)}\cdot 1 = \color{blue}1 \cdot x + \color{red}0 \cdot 1$$
i.e.
$$\begin{cases} 2a = 1 \\ a-b=0\end{cases}$$
i.e. $a=b={1\over 2}$ and yielding $f(x)={x(x+1)\over 2}+c$ since there are no restrictions on $c$ whatsoever.
A: Note that a particular solution is $\frac12(x^2+x)$. Let $g(x) = f(x)-\frac12(x^2+x)$, the given equation becomes
$$g(x)=g(x-1)$$
which means $g(x)$ is periodic with period $1$. 
So $$f(x) = \frac{x^2+x}{2}+g(x)$$ where $g(x)$ is periodic with period $1$. 
I think that is the best one can get out of the given equation without any other assumptions.
A: I don't quite get your question, but a possible solution can be obtained by
$$f(x) = \frac{x(x+1)}{2} + C$$
Then
$$f(x)-f(x-1) = \frac{x(x+1) - (x-1)x}2 + C - C = \frac12 x(x-x+2) = x$$
To find it, set $f(0) = C$ and see
$$f(n) = C + \sum_{i=1}^n i = \frac{n(n+1)}2$$
for $n\in\mathbb N$. Finally extend canonically to $x\in\mathbb R$

A more general solution if $f$ is allowed to have discontinuities is given $C : [0,1) \to\mathbb R$ let
$$f(x) = C(\{x\}) + \frac{x(x+1)}2$$
Where $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part. $f$ should be continuous as long as $C(0) = \lim_{t\to1} C(1)$ and $C$ is continuous.
A: $$f(x)=\frac{x(x+1)}{2}$$
is a solution, but I have not yet found a way to get it.
A: Yes, just take any $f(0)$, and $f(n) = f(0)$ plus some summation.
