Another functional equation

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Another functional equation

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#1 h Jan 29, 2017, 7:44 PM • 2 Y

Y by Adventure10, Mango247

Solve this
$f:\mathbb{R}\longrightarrow\mathbb{R}$ ,
$\forall x, \; f(x+2)-2f(x+1)+f(x)=x^2$

This post has been edited 3 times. Last edited by Hello_Kitty, Jan 29, 2017, 7:48 PM

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#2 h Jan 30, 2017, 2:50 AM • 2 Y

Y by Adventure10, Mango247

This is a forward-difference equation $\Delta^2f(x)=x^2$ so the general solution is a quartic function.

$x^2=\frac{3x^2+3x+1}{3}-x-\frac{1}{3}=\frac{3x^2+3x+1}{3}-\frac{2x+1}{2}+\frac{1}{6}\implies \Delta f(x)=\frac{x^3}{3}-\frac{x^2}{2}+\frac{x}{6}+A$ .

$\frac{x^3}{3}-\frac{x^2}{2}+\frac{x}{6}+A=\frac{4x^3+6x^2+4x+1}{12}-\frac{3x^2+3x+1}{3}+\frac{5(2x+1)}{12}+C_1$ where $C_1=A-\frac{1}{6}$ .

Thus $f(x)=\frac{x^4}{12}-\frac{x^3}{3}+\frac{5x^2}{12}+C_1x+C_2$ .

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pco

#3 h Jan 30, 2017, 6:41 AM • 2 Y

Y by Adventure10, Mango247

rchokler wrote:

This is a forward-difference equation $\Delta^2f(x)=x^2$ so the general solution is a quartic function.

Certainly not.
"Function" does not mean "Polynomial".
There exists a lot of non-polynomial functions.

For example, add to any quartic you gave any additive non continuous function and you get another solution
(and these are not the only solutions).

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pco

#5 h Jan 30, 2017, 7:43 AM • 2 Y

Y by Hello_Kitty, Adventure10

Hello_Kitty wrote:

Solve this
$f:\mathbb{R}\longrightarrow\mathbb{R}$ ,
$\forall x, \; f(x+2)-2f(x+1)+f(x)=x^2$

General solution :
$\boxed{\frac{x^4-4x^3+5x^2}{12}+xg(\{x\})+h(\{x\})}$ whatever are functions $g,h$ from $[0,1)\to\mathbb R$

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#6 h Jan 30, 2017, 2:05 PM • 1 Y

Y by Adventure10

pco wrote:

Hello_Kitty wrote:

Solve this
$f:\mathbb{R}\longrightarrow\mathbb{R}$ ,
$\forall x, \; f(x+2)-2f(x+1)+f(x)=x^2$

General solution :
$\boxed{\frac{x^4-4x^3+5x^2}{12}+xg(\{x\})+h(\{x\})}$ whatever are functions $g,h$ from $[0,1)\to\mathbb R$

Can you explain your solution please. It's seem hard to come up with.

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pco

#7 h Jan 30, 2017, 6:04 PM • 3 Y

Y by Hello_Kitty, Adventure10, Mango247

finn123 wrote:

Can you explain your solution please. It's seem hard to come up with.

In this kind of equation, the simplest way is to find one unique solution $a(x)$ and to write $f(x)=a(x)+b(x)$ and solve the new equation.

1) Finding one solution is rather simple using rchokler's remark and looking only for polynomial solution.
It is easy to find for example $a(x)=\frac{x^4-4x^3+5x^2}{12}$

2) writing then $f(x)=a(x)+b(x)$ , equation becomes
$b(x+2)-2b(x+1)+b(x)=0$

Which is $b(x+2)-b(x+1)=b(x+1)-b(x)$
So $b(x+1)-b(x)$ is periodic with at least period $1$ and one general form for all these periodic functions is $c(\{x\})$

So $b(x+1)=b(x)+c(\{x\})$
Which is $b(x+n)=b(x)+nc(\{x\})$
And so $b(x)=b(\{x\})+\lfloor x\rfloor c(\{x\})$

Or also $b(x)=x c(\{x\})+b(\{x\})-\{x\}c(\{x\})$

And so $b(x)=xg(\{x\})+h(\{x\})$ for some $g,h$
And it is easy to check back that this mandatory form indeed is enough and so is a general solution.

Hence the claim.

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#8 h Mar 27, 2025, 12:42 AM

Y by

Let $g(x)=f(x)-\frac{x^4-4x^3+5x^2-2x}{12}$ (note that this quartic is a solution to the original FE) and $h(x)=g(x+1)-g(x)$ , equation is $h(x+1)=h(x)$ .

We have $h(x+n)=h(x)$ for any $n\in\mathbb Z$ by induction, so $h(x)=h(\{x\}+\lfloor x\rfloor)=h(\{x\})$ , hence the general form $h(x)=b(\{x\})$ for any $b:[0,1)\to\mathbb R$ (which fits).

We have $g(x+1)=g(x)+b(\{x\})$ and so $g(x+n)=g(x)+nb(\{x\})$ (by induction), so $g(x)=g(\{x\}+\lfloor x\rfloor)=g(\{x\})+\lfloor x\rfloor b(\{x\})$ , hence the general form $g(x)=a(\{x\})+\lfloor x\rfloor b(\{x\})$ for any functions $a,b:[0,1)\to\mathbb R$ (which fits).

Then we have:
$\boxed{f(x)=\frac{x^4-4x^3+5x^2-2x}{12}+a(\{x\})+\lfloor x\rfloor b(\{x\})}$ for any functions $a,b:[0,1)\to\mathbb R$ (which fits).

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