hard question about functional equations (1 Viewer)

Paradoxica · Mar 6, 2017

By inspection, the only answer is:

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

omegadot · Mar 8, 2017

Mahan1 said:
$\textrm{1.If} \ f(\frac{x-1}{x+1})+f(\frac{-1}{x}) + f(\frac{1+x}{1-x}) = x.$

$\textrm{Find f(x)}$

$$\noindent Note that $x \neq 0,\pm 1.$

$\noindent Let $u = \frac{x -1}{x + 1}$. Thus $x = \frac{1 + u}{1 - u}$ and the functional equation becomes$

$f(u) + f \left (\frac{u - 1}{u + 1} \right ) + f\left (-\frac{1}{u} \right ) = \frac{1 + u}{1 - u} \qquad (1)$

$$\noindent And let $u = \frac{1 + x}{1 - x}$. Thus $x = \frac{u - 1}{u + 1}$ and the functional equation becomes$

$f\left (-\frac{1}{u} \right ) + f\left (\frac{1 + u}{1 - u} \right ) + f(u) = \frac{u - 1}{u + 1} \qquad (2)$

$\noindent Adding (1) to (2) gives$

$\begin{align*}2 f(u) + f\left (\frac{u - 1}{u + 1} \right ) + f \left (\frac{1 + u}{1 - u} \right ) + 2 f \left (-\frac{1}{u} \right ) &= \frac{1 + u}{1 - u} + \frac{u - 1}{u + 1}\\2 f(u) + f\left (-\frac{1}{u} \right ) + \left [f \left (\frac{u - 1}{u + 1} \right ) + f \left (-\frac{1}{u} \right ) + f \left (\frac{1 + u}{1 - u} \right ) \right ] &= \frac{4u}{1 - u^2}\\2f(u) + f \left (-\frac{1}{u} \right ) + u &= -\frac{4u}{u^2 - 1}\\\Rightarrow 2 f(u) + f \left (-\frac{1}{u} \right ) &= -\frac{4u}{u^2 - 1} - u \qquad (3)\end{align*}$

$\noindent Now in (3) let $u = -1/t$. Thus $t = -1/u$ and (3) becomes$

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

$\noindent Writing (3) and (4) back in terms of $x$, solving for $f(x)$ by subtracting (4) from twice of (3) one has$

$3 f(x) = -\frac{8x}{x^2 - 1} - 2x + \frac{4x}{x^2 - 1} - \frac{1}{x}$

$\noindent or$

$f(x) = -\frac{1}{3} \left (2x + \frac{4x}{x^2 -1} + \frac{1}{x} \right ),$

$\noindent in agreement with Paradoxica's result found by inspection!$

$\noindent \textbf{Note} - Functional equations are not really part of the 4 Unit syllabus.$

Shadowdude · Mar 9, 2017

Paradoxica said:
By inspection, the only answer is:

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

"by inspection"

not sure if srs

Sure you might be able to pull an f that works, but uniqueness by inspection...?

hard question about functional equations (1 Viewer)

Mahan1

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Paradoxica

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omegadot

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Shadowdude

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