Crazy Calculus 5

Calculus Level 2

$\large f\left( \dfrac xy \right) = f( x) - f( y)$

Let $f(x)$ be defined for all real $x > 1$ such that it satisfies the above functional equation for all real $x$ and $y$ and that $f(e) =1$ . Which is of the following options is correct?

$P: f \left( x \right)$ is bounded
$Q: f\left( \dfrac 1x \right) \to 0$ as $x\to 0$
$R: xf\left( x \right) \to 1$ as $x \to 0$
$S: f\left( x \right) =\ln x$

P

S

Q

R

1 solution

Tom Engelsman
Sep 2, 2016

If one differentiates the above functional equation with respect to x and y each, then one obtains:

(1/y) f' (x/y) = f' (x) (i), (-x/y^2) f' (x/y) = f' (y) (ii).

Equating (i) with (ii) yields y f' (x) = (-y^2 / x) f' (y) => f' (x) = [-y f' (y)]/x => f' (x) = A/x (iii), where A is an arbitrary real constant. Integrating (iii) will produce f(x) = A ln(x) + B and substituting this expression back into the original functional equation yields:

A ln(x/y) + B = [A ln(x) + B] - [A ln(y) + B] => A ln(x/y) + B = A*ln(x/y)

which forces B = 0. Taking the initial value f(e) = 1 now gives: 1 = A*ln(e) => A = 1, or f(x) = ln(x). Thus, choice S is correct.

Crazy Calculus 5

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