Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
For a trivial example take any constant function $f(x)$, any $\delta$ and $a=b=0$. –  Dennis Gulko Sep 12 '12 at 19:00
    
Have you tried looking for one? –  Erick Wong Sep 12 '12 at 19:01
3  
May you have missed some constraint? As it is, the problem is somewhat trivial. Maybe $b \neq 0$? –  Karolis Juodelė Sep 12 '12 at 19:03
    
Oops, yes, I did miss a constraint... $b\neq0$ –  M.B.M. Sep 12 '12 at 20:09

5 Answers 5

up vote 3 down vote accepted

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.

share|improve this answer
    
Hehe, our first two ideas were the simplest ones possible, I think :) –  rschwieb Sep 12 '12 at 19:04
2  
With the same logic any $f(x) = g(x) + kx$ where $g$ has period $\delta$ works. –  Karolis Juodelė Sep 12 '12 at 19:05
    
Good observation, Karolis! –  Cameron Buie Sep 12 '12 at 19:11
    
In fact constant functions are the degenerated special cases of the periodic functions with any period. –  doraemonpaul Sep 13 '12 at 1:11
    
True. As I recall, a function is continuous on $\Bbb R$ and having every period iff it is constant. Probably the same works for any (separable) Banach space. –  Cameron Buie Sep 13 '12 at 1:32

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

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

share|improve this answer
    
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). –  Cameron Buie Sep 13 '12 at 0:37

$f(x)-f(x-\delta)+a+bx^2=0$

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

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 $\delta$

share|improve this answer

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)

share|improve this answer
    
Of course the constant term of $f$ could be arbitrary rather than $0$. –  Erick Wong Sep 12 '12 at 19:35
    
Thanks, I will add it. –  clark Sep 12 '12 at 20:18

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.