$f \colon \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous , and $f(x+1)+f(x)=x^2$

Question

I would appreciate if somebody could help me with the following problem:

Find $f(x)$ ($f(x)$ is not Polynomial function), given that:

$f \colon \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous, and $f(x+1)+f(x)=x^2$

I tried but couldn't get it that way.

Answer 1

It might be helpful to first find $f:\Bbb N \to \Bbb R$, then extend the result.

We can solve the recurrence relation $f(n) + f(n+1) = n^2$ via generating functions: define $F(x) = \sum_{n=0}^\infty f(n) x^n$. Then $$ f(n) + f(n+1) = n^2\\ \sum_{n=0}^\infty [f(n) + f(n+1)]x^n = \sum_{n=0}^\infty n^2 x^n \\ \sum_{n=0}^\infty f(n)x^n + \sum_{n=0}^\infty f(n+1)x^n = \sum_{n=0}^\infty n^2 x^n\\ \sum_{n=0}^\infty f(n)x^n + \sum_{n=1}^\infty f(n)x^{n-1} = \sum_{n=0}^\infty n^2 x^n\\ F(x) + \frac 1x \left(F(x) - f(0)\right) = \frac{x(x+1)}{(x-1)^3}\\ xF(x) + F(x) - f(0) = \frac{x^2(x+1)}{(x-1)^3}\\ (x+1)F(x) = \frac{x^2(x+1)}{(x-1)^3} + f(0)\\ F(x) = \frac{x^2}{(x-1)^3} + \frac{f(0)}{x+1}\\ \sum_{n=0}^\infty f(n) x^n = \sum_{n=0}^\infty \frac{n(n+1)}{2} x^n + f(0)\sum_{n=0}^\infty (-1)^n x^n $$ So setting $f(0) = a$ for any $a$, we find $$ f(n) = \frac{n(n+1)}{2} + a(-1)^n = \frac{n(n+1)}{2} + a\cos(\pi n) $$ if a solution to the problem over $\Bbb R$ exists, it must be an extension of this function for some value of $a$.

In fact, $f(x) = \frac{x(x+1)}{2} + a\cos(\pi x)$ seems to work for any $a \in \Bbb R$.

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A more general solution (taken from Salomo's work) is $$ f(x) = \frac{x(x+1)}{2} + g(x) $$ where $g(x)$ satisfies $g(x+1) + g(x) = 0$.

Answer 2

Let $f(x)=\frac{x^2-x}{2}+\sin(\pi x)$. Then, $f(x+1)+f(x)=\frac{x^2+2x+1-(x+1)}{2}+\sin(\pi x + \pi)+\frac{x^2-x}{2}+\sin(\pi x)=x^2$.

As $\frac{x^2-x}{2}$ can be found easily, $sin(\pi x)$ can be replaced by any other non-polynomial continuous function $g$ with $g(x+1)+g(x)=0$.

Answer 3

Let $f$ be a solution of the equation $f(x+1)+f(x)=x^2$. Consider $h(x)=f(x) - \frac{x(x-1)}{2}$, i.e. $f(x)=h(x) + \frac{x(x-1)}{2}$. Then the equation simply becomes: $$h(x+1) + h(x) = 0$$ It is clear that such a function $h$ is completely determined by what it does on $[0,1[$, where $h$ can be defined arbitrarily. Conversely, every $h$ gives a solution $f$. Finally, $f : \mathbb{R} \to \mathbb{R}$ is continuous if and only if $h : \mathbb{R} \to \mathbb{R}$ is continuous.

Answer 4

Let's first consider the function on $\mathbb{Z}$. $$ \begin{align} \sum_{k=0}^{n-1}(-1)^k(f(k+1)+f(k)) &=\sum_{k=1}^n(-1)^{k-1}f(k)+\sum_{k=0}^{n-1}(-1)^kf(k)\\ &=(-1)^{n-1}f(n)+f(0)\tag{1} \end{align} $$ Furthermore, $$ \begin{align} \sum_{k=0}^{n-1}(-1)^kk^2\tag{2} &=(-1)^{n-1}\frac{n(n-1)}2 \end{align} $$ Combining $(1)$ and $(2)$ yields $$ f(n)=\frac{n(n-1)}2+(-1)^nf(0)\tag{3} $$ If we set $f(0)=0$, we get $$ \bbox[5px,border:2px solid #FFC000]{f(x)=\frac{x(x-1)}2}\tag{4} $$ Checking, we get $$ f(x+1)+f(x)=x^2\tag{5} $$ This gives us one solution. Suppose we have two solutions, $f$ and $g$. Then $$ (f-g)(x+1)+(f-g)(x)=0\tag{6} $$ That is, $(f-g)(x+1)=-(f-g)(x)$. Thus, the general solution is $$ \bbox[5px,border:2px solid #FFC000]{f(x)=\frac{x(x-1)}2+h(x)}\tag{7} $$ where $h$ is any continuous function where $h(x+1)=-h(x)$.

An example of such an $h(x)$ would be $a\cos(\pi x)+b\sin(\pi x)$.