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How to solve this functional equation?

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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
A few days ago on MMF the following question was posted with no one showing how to solve it so far:

Given:

$\displaystyle f(f(x))=x^2-x+1\, \forall x\in \mathbb{R}$

find $\displaystyle f(x)$.

I have never known how to solve such equations, except by trial and error, and this one has me stymied.

If someone could give me a nudge in the right direction, I would appreciate it. Thanks!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Check this wiki out under Formulae for Fractional Iteration, especially 6b, since it applies. The (sole) fixed point for the RHS is $x=1$, which is nice. You could at least get a series solution for the answer.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
I don't think there are any closed from in terms of elementary function. Maybe you are interested in a somewhat different bu similar type : \(\displaystyle f(f(x)) = e^x\). \(\displaystyle f\) is called the half-exponential function and usually denoted by \(\displaystyle \mathrm{exp}^{[1/2]}(x)\). Interesting thing is that if we analytically continue this to the complex plane, then super-exponential too gets continued analytically in the same region! If you are interested, here is a link to a forum I am in and here is the Kneser's method which continues this to the complex region by taylor series : Kneser's Super Logarithm

Maybe somewhat similar can be done to continue it in the complex plane.

Ackbach said:
You could at least get a series solution for the answer.
Not necessarily. You have to specify the region of convergence of the series and analytically continue it in the region it can be and where the taylor series diverges. Furthermore, it's not obvious that such an expansion would be analytic!
 
Last edited:

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
[JUSTIFY]Curious. All polynomials of the form:

$f(f(x)) = a_0 x^n + a_1 x^{n - 1} + \cdots + a_n$

Have trivial solutions for $n = m^2$, that is, a perfect square (just expand a generic polynomial of the $m$th degree and solve the resulting system of $n$ equations). Unfortunately, $2$ is not a perfect square, which leads me to believe there is no easy solution to this problem. This is very interesting, in fact, I'll need to look into it deeper. Thanks MarkFL for this![/JUSTIFY]
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Yes, \(\displaystyle f(f(x))=x^2-x+1\) has no easy solution, in fact, not expressible in terms of elementary function as I can see. I strongly believe that we can continue it in the same way we do for half-exponential. I have managed to find a weak approximation which works nicely inside the semi-circle \(\displaystyle |z| = 2\) from the upper-half plane : \(\displaystyle f(x) = x^{\sqrt{2}} - x^{\frac{1}{\sqrt{2}}} + 1\).

Balarka
.
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
A few days ago on MMF the following question was posted with no one showing how to solve it so far:

Given:

$\displaystyle f(f(x))=x^2-x+1\, \forall x\in \mathbb{R}$

find $\displaystyle f(x)$.

I have never known how to solve such equations, except by trial and error, and this one has me stymied.

If someone could give me a nudge in the right direction, I would appreciate it. Thanks!
I propose a solution even if [honestly...] I'm not sure cent per cent on its correctness...

Let's write the unknown function as...

$\displaystyle f(x) = \int_{a}^{x} y(u)\ d u$ (1)

... where a is a constant that we leave undefined, so that is...

$\displaystyle \varphi(x) = f \{f(x)\} = \int_{a}^{x} y(u)\ \int_{a}^{u} y(v)\ d v\ du$ (2)

Deriving (2) we obtain first...

$\displaystyle \varphi^{\ '} (x) = y(x)\ \int_{a}^{x} y(v)\ d v$ (3)

... and after, deriving (3)...

$\displaystyle \varphi^{\ ' '} (x) = 2\ y(x)\ y^{\ '} (x) = 2$ (4)

... so that we arrive to write the ODE...

$\displaystyle y\ y^{\ '}= 1$ (5)

... that, with the condition $\displaystyle y(\frac{1}{2})=0$ supplies...

$\displaystyle y= \sqrt{2 x - 1}$ (6)

Now integrating (6) with the condition $\displaystyle \varphi(\frac{1}{2})= \frac{3}{4}$ we obtain finally...

$\displaystyle f(x) = \frac{1}{3}\ (2 x -1)^{\frac{3}{2}} + \frac{3}{4}$ (7)

We have to observe that the solution is valid for $\displaystyle x \ge \frac{1}{2}$ and not for all x...

Kind regards

$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
chisigma said:
I propose a solution even if [honestly...] I'm not sure cent per cent on its correctness...
1. First thing is that I don't understand your derivation of (2). It seems to be a wrong step.

2. You solution doesn't seemed to be giving the right solution. For fo2(1), your solution gives approximately 1.17 but the answer is 1. Your form seems to be an overestimate of f(x) which I think is because you used f(x)*f(x) instead of f(f(x)) in your calculation at step 2.
 

chisigma

Well-known member
Feb 13, 2012
1,704
1. First thing is that I don't understand your derivation of (2). It seems to be a wrong step.

2. You solution doesn't seemed to be giving the right solution. For fo2(1), your solution gives approximately 1.17 but the answer is 1. Your form seems to be an overestimate of f(x) which I think is because you used f(x)*f(x) instead of f(f(x)) in your calculation at step 2.
If the step (2) is wrong [perfecly possible of course..] then someone has to answer to that question: if $\displaystyle f(x)= \int_{a}^{x} y(u)\ du$ then what is $\displaystyle f\{f(x)\}$?...


Kind regards


$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
if $\displaystyle f(x)= \int_{a}^{x} y(u)\ du$ then what is $\displaystyle f\{f(x)\}$?...
\(\displaystyle \int_{a}^{f(x)} y(u) du\), of course.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
\(\displaystyle \int_{a}^{f(x)} y(u) du\), of course.
Which would equal

$$\int_{a}^{\int_{a}^{x}y(v)\,dv}y(u)\,du.$$

I've never seen that construction before, though there's certainly nothing wrong with the expression.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
The main problem, and what makes it more interesting, is that it's only fixed point, \(\displaystyle z = 1\) is nor attractive neither repelling. It's neutral : \(\displaystyle |f'(z)| = 1\)! If there was an attracting fixed point, one would directly transform this functional form into the Schröder's equation and use modified Koening's method to solve it.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Your half-iterate formally diverges in the complex plane when continued by taylor. Hence, in one sense, your function actually don't exists. But, instead, we can use Ecalle's method to numerically approximate f(x) to several digits of accuracy. Take a look at here : half-iterates of x^2-x+1.

Balarka
.
 

jacobi

Active member
May 22, 2013
58
The function f(x) is an iterative square root of \(\displaystyle h(x)=x^2-x+1\). If we find a solution g(x) of Schroder's equation, \(\displaystyle g(h(x))=s g(x)\), then the iterative square root would be given by \(\displaystyle h_{1/2}(x)=g^{-1}(s^{1/2} (g(x)))\), where \(\displaystyle s=f'(a)\), where a is a fixed point of f.
 

jacobi

Active member
May 22, 2013
58
The substitution x=1 into Schroder's equation gives \(\displaystyle g(1)=s g(1)\), which implies that s=1. However, substitution of this value into the half-iterate equation gives \(\displaystyle h_{1/2}(x)=g^{-1}(g(x))=x\), which is not true.
Substituting values into the original functional equation gives \(\displaystyle f(0)=f^{-1}(1)\) and \(\displaystyle f(1)=f^{-1}(1)\), which suggests that \(\displaystyle f(0)=f(1)\), and the inverse function is not one-to-one. This is why the half-iterate formula does not work.
Alternatively, one could take \(\displaystyle 1^{1/2}=-1\), and then \(\displaystyle h_{1/2}=g^{-1}(-g(x))\)...
 

jacobi

Active member
May 22, 2013
58
There might be no function at all that solves this equation. The form is similar to Problem 7 on this page. One might use the same method used to solve Problem 7 on this functional equation. I don't know how to do this, but maybe someone else does.