Is it possible to solve the functional equation $f(x) - f(x - 1) = x$ by algebraic means?
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1Do we know that $f$ is continuous? Otherwise $f$ may be defined differently on each set of the form $r+\mathbb{N}$, $r\in\mathbb{R}$. – Reinstate Monica Nov 12 '14 at 22:10
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@jef $r\in [0,1)$ for disambiguation. – AlexR Nov 12 '14 at 22:12
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@AlexR Right, sorry, of course I meant each element of $\mathbb{R}/\mathbb{N}$ but I wasn't sure that OP was familiar with the terminology. – Reinstate Monica Nov 12 '14 at 22:14
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@jef No problem ;) Note that $\mathbb R/\mathbb N \simeq [0,1)$ for these matters, wich may be easier to grasp. – AlexR Nov 12 '14 at 22:17
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@jef I think you mean $\Bbb R/\Bbb Z$. – Adam Hughes Nov 12 '14 at 22:19
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@AdamHughes The equivalence relation induced on $\mathbb R$ by $\mathbb Z$ and $\mathbb N$ is the same due to transitivity. – AlexR Nov 12 '14 at 22:23
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@AlexR Yes, but it seems silly to say $\Bbb N$ when $\Bbb Z$ is really at the heart of it, especially since periodic behavior is fundamentally related to the circle with is $\mod \Bbb Z$. – Adam Hughes Nov 12 '14 at 22:46
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@AdamHughes Yes, we can agree on that. At least it's not wrong in a mathematical sense though. – AlexR Nov 12 '14 at 22:49
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@AdamHughes: You are right, I should have said $\mathbb{Z}$. I'm dumb. – Reinstate Monica Nov 12 '14 at 22:55
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Oddly, the accepted answer does not provide all the solutions, which the question seems to be asking for. – Did Nov 15 '14 at 17:58
6 Answers
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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 $f(x)=\frac 12x(x+1)+f(0)$
Ross Millikan
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This does not provide all the solutions, which the question seems to be asking for. – Did Nov 15 '14 at 17:59
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$\require{cancel}$
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.
Adam Hughes
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It's a formal equivalence of polynomials, so that's not necessary to mention. It's the same with partial fractions, when you clear denominators you cannot technically substitute the values in which make some terms $0$ because it doesn't converge in the original sense, but the formal identity needs to hold. – Adam Hughes Nov 12 '14 at 22:15
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@AlexR yeah, sorry about that, but they help to show what we're going for by matching coefficients on both sides. – Adam Hughes Nov 12 '14 at 22:17
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I suggest
\color{blue}for that? $$\color{blue}{2a}\cdot x + \color{blue}{(b-a)}\cdot 1 = \color{blue}1 \cdot x + \color{blue}0 \cdot 1$$Copyable:\color{blue}{2a}\cdot x + \color{blue}{(b-a)}\cdot 1 = \color{blue}1 \cdot x + \color{blue}0 \cdot 1– AlexR Nov 12 '14 at 22:21 -
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This does not provide all the solutions, which the question seems to be asking for. – Did Nov 15 '14 at 17:58
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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.
Quang Hoang
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This does provide all the solutions, which the question seems to be asking for. – Did Nov 15 '14 at 17:59
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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.
AlexR
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