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What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution.

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    $\begingroup$ Have you tried exploting the symmetry? $\endgroup$ – Pedro Tamaroff Jul 26 '12 at 23:58
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We start from $$x^2f(x)+f(1-x)=2x-x^4.$$ Replace $x$ by $1-x$. Then $1-x$ gets replaced by $x$. So $$(1-x)^2f(1-x)+f(x)=2(1-x)-(1-x)^4.$$ Two linear equations in two unknowns, $f(x)$ and $f(1-x)$. Solve for $f(x)$.

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Another, more mechanical approach: First of all, just by counting degrees it's clear that any polynomial solution $f$ must be quadratic, so we can write $f(x)=ax^2+bx+c$. What's more, plugging in $x=0$ gives $0^2\cdot f(0)+f(1) = 2\cdot0-0^4$, or $f(1)=0$, and plugging in $x=1$ gives $1^2\cdot f(1)+f(0) = 2\cdot1-1^4$, or $f(0) = 1$, so we know that $c=1$ and $a+b+c=0$, or $b=-(1+a)$. Using these, we can rewrite the quadratic as $f(x) = ax^2-(a+1)x+1$. Now, it's obvious that $x^2f(x) = ax^4+\mathrm{smaller\ terms}$, and $f(1-x)$ will also be only quadratic — so for the LHS to equal $2x-x^4$, it must be that $a=-1$ and $f(x)=1-x^2$; all that's left is to compute the left hand side in full and see that it solves the equation. Note that this approach doesn't prove that such an $f$ is unique, only that it exists and that it's unique among polynomial solutions — you need one of the other approaches to show that this is the only function that works.

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As I commented, you can exploit the symmetry.

$$\eqalign{ & f(1 - x) + {x^2}f(x) = 2x - {x^4} \cr & f(x) + {\left( {1 - x} \right)^2}f(1 - x) = 2\left( {1 - x} \right) - {\left( {1 - x} \right)^4} \cr} $$

Now you get

$$\eqalign{ & f(1 - x) + {x^2}f(x) = 2x - {x^4} \cr & f(1 - x) = \frac{{2\left( {1 - x} \right) - {{\left( {1 - x} \right)}^4} - f\left( x \right)}}{{{{\left( {1 - x} \right)}^2}}} \cr} $$

So all you have to do is solve for $f(x)$ in

$$2x - {x^4} - {x^2}f(x) = \frac{{2\left( {1 - x} \right) - {{\left( {1 - x} \right)}^4} - f\left( x \right)}}{{{{\left( {1 - x} \right)}^2}}}$$

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There's an algebraic way (which may be a bit messy without using a CAS). It involves exploiting the symmetry of the problem, as Peter Tamaroff suggested.

$$ \begin{eqnarray} x^2 f(x) + f(1-x) &=& 2x - x^4 \\ (1-x)^2f(1-x) + f(x) &=& 2(1-x) - (1-x)^4 \quad \textrm{subs. $1-x$ for $x$}\\ \\ -(1-x)^2x^2f(x) - (1-x)^2f(1-x) &=& -(1-x)^2(2x - x^4) \quad \textrm{Eqn.1 times $-(1-x)^2$}\\ \textrm{Now add to Eqn.2:} \quad f(x) - (1-x)^2x^2f(x) &=& 2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)\\ f(x)\left[1 - (1-x)^2x^2\right] &=& 2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)\quad \textrm{factoring out $f(x)$}\\ f(x) &=& \frac{2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)}{1 - (1-x)^2x^2}. \end{eqnarray} $$

This is an unreduced final answer, which (according to sage) cancels out to leave $f(x) = 1 - x^2$.

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Given, $x^2f(x)+f(1−x)=2x−x^4$

$f(-1) = 0$

$f(0) = 1$

$f(1) = 0$

$f(2) = -3$

$f(3) = - 8$

$f(4) = -15$

By careful observation one can see that the difference between the terms vanishes after the second stage

i.e. $f(x)$ is of the form $a{x^2} + b{x} +c$

i.e. $f(x) = a{x^2} + b{x} + c$ -------(1)

i.e. $f(0) = c = 1$

i.e. $c = 1$ -------(2)

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

i.e. $f(1) = a + b + 1= 0$

i.e. $a + b = -1$ -------(3)

$f(2) = 4a + 2b + c = -3$

i.e. $4a + 2b + 1 = -3$

i.e. $4a + 2b = -2$

i.e. $2a + b = -2$ -------(4)

Solve equations $(3)$ and $(4)$ to get $a = -1$ and $b = 0$

i.e. $f(x) = {-1}{x^2} + {0}{x} + 1$

i.e. $f(x) = {-x^2} + 1$

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  • $\begingroup$ This derivation is based on algebra $\endgroup$ – HOLYBIBLETHE Jul 27 '12 at 3:53
  • $\begingroup$ NOTE : a, b, c are constant $\endgroup$ – HOLYBIBLETHE Jul 27 '12 at 3:56
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$f(x)=1-x^2$ inserting that gives:

$$ \begin{eqnarray} x^2 f(x) + f(1-x) &=&x^2(1-x^2) + (1 - (1-x)^2)\\ &=&x^2-x^4 +(1-(1-2x+x^2))\\ &=&x^2-x^4 +(1-1+2x-x^2)\\ &=&x^2-x^4+2x-x^2 \\ &=& 2x -x^4 \end{eqnarray} $$ as required. But I admit: I had some help...

But you can also write it as

$$ x^4+f(x) x^2 -2x-f(1-x)=0 $$ and try to solve the depressed quartic equation...

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