derivativesfunctional-equationsfunctions
Find all differentiable functions with $f'(0)=1$, such that they satisfy the following functional equation:
$$ f(2x) = 2f^2(x) -4f(x) + 3$$
So, here's what I tried:
And now, I'm stuck because I have no clue how to simplify the final limit.
Plus, I think there must be a more general method to solve functional equations of the kind $f(nx) = g(f(x))$ where $g(x)$ is a polynomial of degree $n$, or at least the case $f(2x) = quadratic in f(x)$.
So, what can I do ahead of this?
As for iterating polynomials, there are two general forms of polynomials that we currently know how to iterate. They are polynomials of the form $$P(x)=a(x-b)^d+b$$ and $$T(x)=\cos(k\arccos(x))$$ and yes, I know the second one doesn't look like a polynomial, but for integer values of $k$, it is a polynomial where the inverse cosine is defined, and can be extended to other values (accurately, for our purposes) using the hyperbolic cosine function. For more info, see Chebyshev Polynomials.
The iteration formulas for each of these is given by $$P^n(x)=a^{\frac{b^n-1}{b-1}}(x-b)^{d^n}+b$$ and $$T^n(x)=\cos(k^n\arccos(x))$$
In case you were wondering how this applies to your problem of finding $f$ given that $$f(1+n)=(Q\circ f)(n)$$ where $Q$ is a polynomial, here's how: if you assign a value for $f(0)$, say $y_0$, then you can say that $$f(n)=(Q^n\circ f)(0)=Q^n(y_0)$$ which allows you to find $f:\mathbb Z\to\mathbb Z$.
For every $x$,
$$a[f(x)]^2+bf(x)+c-g(x)=0$$
is an ordinary quadratic equation and
$$f(x)=\frac{-b\pm\sqrt{b^2-4ac+4ag(x)}}{2a}$$ indeed holds.
There are real roots when
$$b^2-4ac+4ag(x)\ge0,$$ a quartic inequation.
If there are no constraints on $f$, except that it must be a function, then for any $x$ such that there are distinct real roots, you can choose either of them. So there are infinitely, non-countably many solutions.
If $f$ is restricted to be continuous, then the choice of the sign must be consistent across the intervals of the domain.
For the case $(1,1,0)$, the roots are
$$\frac{-1\pm\sqrt{1+4(x^4+2x^3+2x^2+x)}}{2}=\frac{-1\pm(2x^2+2x+1)}2=x^2+x,-x^2-x-1.$$
A solution, among many many many others:
Best Answer
First note that differentiating the relation $$ f(2x) = 2f(x)^2 - 4f(x)+3 $$ at $x=0$ we obtain $$ 2f'(0) = 4f'(0)f(0) - 4f'(0) $$ and since $f'(0)=1$ this gives $f(0)=3/2$.
Instead of your choice of $g$, take $$ g(x) = 2(f(x)-1), $$ and thus $g(0)=1$. Then you obtain the functional equation $$ g(2x) = g(x)^2, $$ where $g'(0)=2$. Now, we have that $$ g(x) = g\left( \frac{x}{2} \right)^2 \geq 0 $$ for all $x\in\mathbb{R}$. If $g(x_0)=0$ for some $x_0\in\mathbb{R}$, then by induction we can prove that $g(2^{-n}x_0)=0$ for all positive integer $n$, so taking the limit as $n\to \infty$ we obtain that $g(0)=0$ (note that $g$ is continuous at $0$ since it is differentiable at $0$), but this contradicts $g(0)=1$. Thus we have that $g(x)>0$ for all $x\in\mathbb{R}$.
Now, define $$ h(x) = \log(g(x)). $$ Here $\log$ is the natural logarithm. Then you obtain the functional equation $$ h(2x) = 2h(x) $$ with $h(0)=0$ and $h$ differentiable at $0$. This equation is well known and is solved for example here. Note that by the first answer to that question, you only need the differentiability at $0$, not on $\mathbb{R}$.
Thus by the answer to the above mentioned question, we have that $h(x)=ax$ for some $a\in\mathbb{R}$. Thus $g(x)=e^{ax}$ and hence $$ f(x) = 1 + \frac{e^{ax}}{2}. $$ From the condition $f'(0)=1$ you can show that $a=2$ so $$ f(x) = 1 + \frac{e^{2x}}{2} $$ is the only solution to the given functional equation.