Functional equation $ f(x)+f(x+1)=x$ What functions satisfy $f(x)+f(x+1)=x$?
I tried but I do not know if my answer is correct.
$f(x)=y$
$y+f(x+1)=x$
$f(x+1)=x-y$
$f(x)=x-1-y$
$2y=x-1$
$f(x)=(x-1)/2$
 A: Your conclusion isn't quite right, and isn't found correctly either; your definition of $f(x)=y$ is a bit misleading because we have to interpret it as applying to a particular $x$ and not to all $x$. So when you step from
$$f(x+1)=x-y$$
to
$$f(x)=x-1-y$$
by "shifting" the argument by one (which be allowable if $x$ was "free"), you're making a misstep, since, if we wanted to be sure, the proper way to make this shift would be to set $u=x+1$ and then
$$f(u)=u-1-y$$
which is true, but we can't say $f(u)=y$.
One way to approach this sort of problem is to try to find some function related to $f$ which satisfies a simpler relation. In particular, suppose we let
$$g(x)=f(x)-\frac{x}2-\frac{1}4$$
which we choose to "undo" the $x$ on the left hand side. Then we can show
$$g(x)+g(x+1)=f(x)+f(x+1)-\frac{x}2-\frac{x+1}2-\frac{1}4-\frac{1}4=x-\frac{x}2-\frac{x+1}2-\frac{1}4-\frac{1}4=0.$$
So, we can say that we can build solutions by choosing any $g$ satisfying
$$g(x)=-g(x+1)$$
and adding the quantity $\frac{x}2+\frac{1}4$ to them. So, for instance, in a vein similar to your solution, we get $f(x)=\frac{x}2+\frac{1}4$ by setting $g$ to be zero everywhere. More generally, we can choose the values of $g$ arbitrarily in $[0,1)$ and then use the relation $g(x+n)=(-1)^n g(x)$ for integer $n$ to find its value everywhere else. A simple function we could use would be $g(x)=\sin(\pi x)$, yielding the solution $f(x)=\sin(\pi x)+\frac{x}2+\frac{1}4$, however there are infinitely many possible solutions.
A: Replacing $x$ by $x+1$ you get
$$f(x+1)+f(x+2)=x+1=f(x)+f(x+1)+1$$
This shows that 
$$f(x+2)=f(x)+1$$
Let $g(x)=f(x)-\frac{x}{2}$. Then $g(x)$ is periodic with period $2$, and 
$$g(x)+g(x+1)=-\frac{1}{2}$$
This gives your answer:
Define $g(x)$ to be any function on $[0,1)$. Define $g:[1,2) \to \mathbb R$ by
$$g(x):=-\frac{1}{2}-g(x-1)$$
This defines $g:[0,2) \to \mathbb R$. Extend it to $\mathbb R$ by asking that $g$ is 2-periodic.
Now, let $f(x)=g(x)+\frac{x}{2}$.
It is easy to check that $f$ satisfies the given condition, and the above shows that any function which satisfies the given condition has this form.
Now, the answer can change if one adds extra conditions like continuity, differentiability, but even so I think there is enough freedom to generate uncountably many examples.
A: $f(x) = x - 1 - y$ is wrong. $y$ is a function of $x$. In my opinion, it was such a good try. Wouldn't have thought of that approach. Anyway, is $f(x)$ supposed to be linear? Write out: $f(x) = ax + b$ if so.
Then plug in:
$a(x + 1) + b + ax + b = 1$
Hint: $1 = 0x + 1$.
A: Concluding that from $f(x+1)=x-y$ it follows $f(x)=x-1-y$ is wrong, because the variable $x$ was not changed in the Argument $y$. 
Try the Ansatz $f(x)=ax+b$ for some coefficients $a,b$ to solve this equation.
A: This is a linear inhomogeneous problem. The general philosophy for such problems is the following:
(a) Find the general solution $f_h$ of the associated homogeneous problem
$$f(x)+f(x+1)=0\ .$$
To this end write $f_h(x):=e^{i\pi x} g(x)$ with a new unknown function $x\mapsto g(x)$. You will obtain a huge set of solutions.
(b) Find a single particular solution $x\mapsto f_p(x)$ of the given equation. To this end make the "Ansatz" $f_p(x):=a x+b$ with undetermined coefficients $a$ and $b$, and fix $a$ and $b$ such that $f_p$ fulfills the given functional equation.
(c) The general solution of the given functional equation is then 
$$f(x)=f_h(x)+f_p(x)\ .$$
A: The simplest answer is:
$$f(x)=\frac x2-\frac14$$
If you have any function $g$ of period 2, such that $g(x)=-g(x+1)$ for all $g$ ($\sin(\pi x)$ is a good example), then $f(x)=g(x)+\frac x2-\frac14$ also works. For example, the function $\sin(\pi x)+\frac x2-\frac14$ also satisfies it.
