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Is there a function $f(x)$ on the real domain and real constants $a$ and $b\neq 0$ for which the following is true:

$f(x)-f(x-\delta)+a+bx^2=0$ for some real $\delta\neq 0$?

EDIT: I missed a very important constraint in the original posting of this question: $b\neq 0$. Sorry!

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    Oops, yes, I did miss a constraint... $b\neq0$2012-09-12

5 Answers 5

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Edit

Let's develop a concrete solution, shall we? Observe that for any polynomial $p(x)$ of degree $n>0$ and any $\delta$, we have that $p(x)-p(x-\delta)$ is again a polynomial, having degree at most $n-1$. This suggests that we try for a degree $3$ polynomial. For the moment, we may as well assume the constant term is $0$, since if not, there'd be cancellation in our subtraction, anyhow. Let $f(x)=c_1x+c_2x^2+c_3x^3$, so we need $-a-bx^2=f(x)-f(x-\delta)=(c_1\delta-c_2\delta^2+c_3\delta^3)+(2c_2\delta-3c_3\delta^2)x+3c_3\delta x^2.$ Equating the coefficients of the terms of same degree, we see that: $-a=c_1\delta-c_2\delta^2+c_3\delta^3$ $0=2c_2\delta-3c_3\delta^2\quad(\text{so}\: 0=2c_2-3c_3\delta,\:\text{since}\:\delta\neq 0)$ $-b=3c_3\delta$ The third of these equations gives us $c_3=-\frac{b}{3\delta}$ since $\delta\neq 0$, and substituting $-b$ in for $3c_3\delta$ in the adapted second equation yields $c_2=-\frac{b}2.$ Finally, the first equation becomes $-a=c_1\delta-c_2\delta^2+c_3\delta^3=c_1\delta+\frac{b\delta^2}2-\frac{b\delta^2}3=c_1\delta+\frac{b\delta^2}6\!,$ so $c_1=-\frac{a}\delta-\frac{b\delta}6=\frac{6a+b\delta^2}6.$

Thus, we have in fact that for any $a,b$ and any $\delta\neq 0$, the polynomial family $f(x)=d-\frac{6a+b\delta^2}6x-\frac{b}{2}x^2-\frac{b}{3\delta}x^3$ satisfies the desired functional equation for all $d$.


Addendum It occurred to me (once I processed Karolis's comment below) that instead of the arbitrary $d$, and instead of requiring $f$ to be a polynomial, we can take any function $g$ with period $\delta$ (of which constant functions are merely an uninteresting family of examples), and then let $f(x)=g(x)-\frac{6a+b\delta^2}6x-\frac{b}{2}x^2-\frac{b}{3\delta}x^3.$ I'm not sure if this describes all $f$ that fit this equation, as I've never done any serious functional analysis, but it certainly gives a fairly general family of solutions.

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    Thank you all for the great answers! I accept this answer because it's the most intuitive to me (and points out how to approach other problems of the similar form...) However, answers by doraemonpaul and clark are also very good (I would accept those if I could accept more than one answer.)2012-09-13
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Take the function $f$ be a polynomial of degree $3$ and solve for the coefficients. $f(x)=a_3x^3+a_2 x^2 +a_1x+a_0$ let us take the difference in respect of the powers $a_3x^3-a_3(x-\delta)^3=3a_3\delta x^2-3a_3x \delta ^2+\delta ^3 a_3$ $a_2x^2-a_2(x-\delta)^2=2a_2\delta x-\delta ^2 a_2$ $a_1x-a_1(x-\delta)=a_1 \delta$ Hence $f(x)-f(x-\delta)=3a_3\delta x^2+(2a_2\delta-3a_3 \delta ^2)x+\delta ^3 a_3-\delta ^2 a_2+a_1 \delta$ Now since $\delta$ is not zero you can solve this system $3a_3\delta=b$ $2a_2\delta-3a_3 \delta ^2=0$ $\delta ^3 a_3-\delta ^2 a_2+a_1 \delta=a$ which is linear with unique solution. Also notice that $a_0$ can be arbitary (as will Erick Wong notified me)

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    Of course the constant term of $f$ could be arbitrary rather than $0$.2012-09-12
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$f(x)-f(x-\delta)+a+bx^2=0$

$f(x)-f(x-\delta)=-bx^2-a$

In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe1108.pdf.

The general solution of this functional equation is $f(x)=\Theta(x)+f_p(x)$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$

Luckily we can find $f_p(x)$ by method of undetermined coefficients:

Let $f_p(x)=Ax^3+Bx^2+Cx$ ,

Then $f_p(x-\delta)=A(x-\delta)^3+B(x-\delta)^2+C(x-\delta)=Ax^3-3A\delta x^2+3A\delta^2x-A\delta^3+Bx^2-2B\delta x+B\delta^2+Cx-C\delta=Ax^3+(B-3\delta A)x^2+(3\delta^2A-2\delta B+C)x-\delta^3A+\delta^2B-\delta C$

$\therefore Ax^3+Bx^2+Cx-(Ax^3+(B-3\delta A)x^2+(3\delta^2A-2\delta B+C)x-\delta^3A+\delta^2B-\delta C)\equiv-bx^2-a$

$3\delta Ax^2+(2\delta B-3\delta^2A)x+\delta^3A-\delta^2B+\delta C\equiv-bx^2-a$

$\therefore\begin{cases}3\delta A=-b\\2\delta B-3\delta^2A=0\\\delta^3A-\delta^2B+\delta C=-a\end{cases}$

$\begin{cases}A=-\dfrac{b}{3\delta}\\B=-\dfrac{b}{2}\\C=-\dfrac{b\delta}{6}-\dfrac{a}{\delta}\end{cases}$

$\therefore f(x)=\Theta(x)-\dfrac{bx^3}{3\delta}-\dfrac{bx^2}{2}-\dfrac{b\delta x}{6}-\dfrac{ax}{\delta}$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$

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The restrictions are extremely mild in this question: if $g_x(y)$ is any family of bijections on $\mathbb R$, then for any $\delta > 0$ there are infinitely many solutions to the functional equation $f(x-\delta) = g_x(f(x))$: just define $f$ arbitrarily on $[0,\delta)$ and use the functional equation to extend it in either direction ($g_x$ has an inverse for any $x$).

In your case you want $g_x(y) = y + a + bx^2$. No matter what $a$ and $b$ are, this is a bijection.

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Let $f(x)=x$, $b=0$ and $a=-\delta\neq 0$.

Edit: This was posted before the OP introduced the $b\neq 0$ condition.

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    Uh oh! I guess you missed the restriction $b\neq 0$ that the OP added later, and someone downvoted you. I've upvoted you to compensate (as it was a fine answer based on the initial post).2012-09-13