Let $f:[-1,1]\to\Bbb R$ be a continuous function such that $$f(x)=\frac{2-x^2}{2} f(\frac{x^2}{2-x^2})$$ for every x in $[-1,1]$, $f(0)=1$, find $f(x)$.
Seeing the expression I think that a recurrence relation may be created by some suitable substitution but couldn’t proceed like that. I tried substitution $x\to\frac{x^2}{2-x^2} $ but then expression became worse. Or a trigonometric substitution like $x=\sqrt2$ $cos\theta$ but then LHS becomes unworkable. Please provide an approach to continue.
Solving the functional equation $f(x)=\frac{2-x^2}{2} f(\frac{x^2}{2-x^2})$
functional-equationsreal-analysis
Related Solutions
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:
I have a problem but, for sure, I may be mistaken.
Starting from scratch, we need to solve $$y\,y''+y'(y'-1)-y=0 \qquad \text{with} \qquad y(0)=0 \,,\,\, \,y'(0)=1$$ with $$y=\sum_{n=1}^\infty c_n\,x^n$$ "My" first coefficients are $$\left\{1,\frac{1}{4},-\frac{1}{72},\frac{1}{288},-\frac{11}{10800 },\frac{337}{1036800},-\frac{16657}{152409600},\frac{92879}{24 38553600},-\frac{8971901}{658409472000},\frac{32752889}{658409 4720000},-\frac{17642030237}{9560105533440000},\frac{106092907 271}{152961688535040000},-\frac{330158692960783}{1252756229101 977600000},\frac{590867235322373}{5846195735809228800000},\cdots\right\}$$ which, for $2 \leq n \leq 200$ at least, are alternating in sign.
Using the coefficients for $10 \leq n \leq 200$, a linear regression gives, with $R^2>0.999999$, $$\log(|c_n|)=-0.2815-0.845614\, n-1.51134\, \log (n)$$
$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & -0.2815005 & 0.0009018 & \{-0.2832795,-0.2797215\} \\ b & -0.8456144 & 3.7\times 10^{-6} & \{-0.8456216,-0.8456071\} \\ c & -1.5113390 & 0.0002813 & \{-1.5118940,-1.5107841\} \\ \end{array}$$
Concerning the absolute errors, the mean is $0.00065$ and the maximum is $0.00645$.
The sum of all coefficients (from $n=1$ to $n=200$) is $1.23880865563816$ which already atteined for $n=31$.
Edit
What I did by hand before this answer $$y=x+\sum_{n=2}^5 c_n\,x^n$$ leads to $$0=(4 a_2-1)x+(6 a_2^2-a_2+9 a_3)x^2+(20 a_2 a_3-a_3+16 a_4)x^3+(15 a_3^2+30 a_2 a_4-a_4+25 a_5)x^4+(42 a_3 a_4+42 a_2 a_5-a_5)x^5+ (28 a_4^2+56 a_3 a_5)x^6+72 a_4 a_5x^7+45 a_5^2x^8$$ giving $$a_2=\frac 14 \qquad a_3=-\frac{1}{72} \qquad a_4=\frac{1}{288} \qquad a_5=-\frac{11}{10800}$$ which are correct. What remains is $$-\frac{337 x^5}{28800}+\frac{1757 x^6}{1555200}-\frac{11 x^7}{43200}+\frac{121 x^8}{2592000}$$
Adding $a_6$ to the list gives $a_6=\frac{337}{1036800}$ (which is correct) and what remains is $$\frac{16657 x^6}{3110400}-\frac{601 x^7}{1036800}+\frac{4099 x^8}{27648000}-\frac{40777 x^9}{1119744000}+\frac{1249259 x^{10}}{179159040000}$$
Best Answer
By trial and error I found the following solution: $$ f(x)=\sqrt{1-x^2}. $$