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A function $f(x)$ satisfies the functional equation $x^2{f(x)} +f(1- x) =2x -x^4$ for all real $x$. Then $f(x)$ is given by.

My work $$x^2{f(x)} +f(1- x) =2x -x^4$$ Replacing $x$ $by$ $1- x$ $$(1-x)^2{f(1 - x)} +f(x) =2x -x^4 + 4x^3 -6x^2 + 1$$ Subtracting them $${f(x)}(1-x^2) +{f(1- x)}((1-x)^2 - 1) = 4x^3 -6x^2 + 1$$

What should I do next ?

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  • $\begingroup$ You can notice that $f(\frac{1}{2})=\frac{3}{4}$ by plugging in $x=\frac{1}{2}$. $\endgroup$ – Bérénice Jul 9 '16 at 12:39
  • $\begingroup$ Additionally, put $x=0$ and $x = 1$, to get $f(1) = 0$ and $f(0) = 1$ gives $f(x) = 1 - x^2$ $\endgroup$ – Shailesh Jul 9 '16 at 12:47
  • $\begingroup$ Have you tried solving the first two equations of your work for $f(x)$? $\endgroup$ – Burrrrb Jul 9 '16 at 12:50
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You have the equation \begin{equation} x^2\cdot f(x) + f(1-x) = 2x-x^4. \end{equation} Replacing $x$ by $1-x$ results in \begin{equation} (1-x)^2\cdot f(1-x) + f(x) = 2(1-x)-(1-x)^4. \end{equation} Now we have two equations with two unknowns, namely $f(x)$ and $f(1-x)$. We can subtract $(1-x)^2$ times the first equation from the second one to get \begin{equation} \left(1-(1-x)^2x^2\right)\cdot f(x) = 2(1-x)-(1-x)^4 - (1-x)^2\cdot (2x-x^4). \end{equation} This can be simplified to $$f(x)=1-x^2.$$

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  • $\begingroup$ Very Nice. +1.... $\endgroup$ – Shailesh Jul 9 '16 at 13:06
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    $\begingroup$ Your solution does not work if $x=\frac{1\pm\sqrt{5}}{2}$. In fact, if you define $f\left(\frac{1+\sqrt{5}}{2}\right)=a$ for some $a$, the $f\left(\frac{1-\sqrt{5}}{2}\right)$ depends only on $a$. That is, $f(x)=1-x^2$ holds only for $x\in\mathbb{R}\setminus\left\{\frac{1\pm\sqrt{5}}{2}\right\}$. $\endgroup$ – Batominovski Jul 9 '16 at 13:15
  • $\begingroup$ @Batominovski Why doesn't it work for $x=\frac{1 \pm \sqrt{5}}{2}$? $\endgroup$ – Pjotr5 Jul 9 '16 at 13:20
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    $\begingroup$ @pjotr5 You have at the end that $(1-(1-x)^2x^2)f(x)=(1-(1-x)^2x^2)(1-x^2)$. The numbers $(1\pm\sqrt{5})/2$ are the real roots of $(1-(1-x)^2x^2)=0$. In this case, you have $0f(x)=0$, so you cannot find $f(x)$. $\endgroup$ – Kelenner Jul 9 '16 at 13:27
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    $\begingroup$ @pjotr5 You are obviuouly true if we search for polynomial or rational fraction solution, but the OP has asked for the function solutions, so the remark by Batiminovski is correct. $\endgroup$ – Kelenner Jul 9 '16 at 13:47
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If we assume (the reasoning given in comments) $f(x)$ to be a quadratic $ax^2 + bx + x$, then put $x = 0, \frac12, 1$ in the given equation, you get 3 equations.

$f(0) = 1 = c$

$f(1) = 0 = a + b + c$ and

$f(\frac{1}{2}) = \frac{3}{4} = \frac{a}{4} + \frac{b}{2} + c$

Solving, you get $a = -1$ and $c = 1$ so the function is $f(x) = \color{blue}{1 - x^2}$

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  • $\begingroup$ How you assume it to be quadratic , it may be of higher degree . $\endgroup$ – Aakash Kumar Jul 9 '16 at 12:58
  • $\begingroup$ Suppose it is of a higher degree, then $x^2f(x)$ will be of a degree higher than $4$ which cannot be canceled by $f(1-x)$ since $f(1-x)$ will always be $2$ degrees lower than $x^2f(x)$. The RHS degree is $4$. You can formalize this argument if required $\endgroup$ – Shailesh Jul 9 '16 at 13:02
  • $\begingroup$ @Shailesh That is true, but it is a priori not clear that $f$ has to be a polynomial. If you replace the equation by $$x^2 \cdot f(x) + f(1-x) = x-x^4,$$ then $f$ is given by $$f(x) = \frac{-x^4+x^3+2 x^2-2 x}{x^2-x-1}.$$ $\endgroup$ – Pjotr5 Jul 9 '16 at 13:10
  • $\begingroup$ @Pjotr5 Point taken. Well, Diophantus employed similar methods -- assuming things and then arriving at the answer. So I took the liberty. $\endgroup$ – Shailesh Jul 9 '16 at 13:54
  • $\begingroup$ @Shailesh I didn't mean to bash your method, especially since it worked. I just wanted to point out that some of your assumptions were indeed assumptions. :) $\endgroup$ – Pjotr5 Jul 9 '16 at 14:09
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It is easier to solve a much more general problem: Consider the functional equation

a(x) f(x) + b(x) f (1-x) = g(x).

Call a^* (x) = a(1-x) etc. and the equation becomes the two equations.

a f + b f^* = g

a^* f^* + b^* f = g^*. which is a 2-by-2 linear system:

Solving for f by inverting a 2-by-2 matrix gives

f(x) = (1/ (a a^* - b b^)) ( a^ g - b^* g^*).

This works for all functions a,b,g as long as a a^* - b b^* is not zero.

The given problem has a(x) = x^2, b(x) = 1, and g(x) = 2x-x^4.

Plugging these in gives

f(x) = (1/ (x^2(1-x)^2 -1)) ( (1-x)^2 g(x) - g(1-x) )

 = (1/ (x^2(1-x)^2 -1)) ( (1-x)^2 (2x-x^4) - 2(1-x) + (1-x)^4)

which simplifies to 1-x^2.

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  • $\begingroup$ Welcome to math SE. Have a look at mathjax to improve your mathematical expressions $\endgroup$ – Alain Remillard Jan 28 '20 at 15:59

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