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lemon14 · Feb 11, 2012, 03:26 AM

I'm not sure how to deal with this type of exercises. Here is what I solved so far:

1.
$f(x)=f(x^2), x \geq 0$
$f(x)=f(x^2)=f(x^4)=...=f(x^{2n})$
$f(x)=f(x^{2n})\Rightarrow f(x)= \lim_{n \to \infty} f(x^{2n})= f(\lim_{x \to \infty} x^{2n})$
$f(x)=0$ if x=0 and $f(x)= \infty$ if x>0
Is this possible? I don't think it's correct.

2.
$f(2^x)=f(3^x), x \epsilon R$
$3^x = t \Rightarrow \log_3 {t} = x$
$f(2^{\log_3{t}})=f(t) \Rightarrow f(x)=f(2^{\log_3{x}})$
What should I do next?

ebaines · Feb 12, 2012, 12:50 PM

For question 1, it seems to me that if x = 1 it would also satisfy the equations. But in general the conclusion that f(x) = f(x^4) applies only if f(x) = f(x^2) throughout the entire domain of x. That leads me to believe that this only works if f(x) = C (a constant).

There are pleanty fop functions where f(x) = f(x^2) for some values of x, but f(x^2) does not equal f(x^4). For example: if f(x) = sin(x), then $\sin(x) = \sin(x^2)$ for $x^2 = x +2n\pi$ . For n = 1 this results in

$<br /> x = 3.056 \\<br /> \sin(x) = \sin(x^2) = 0.0855<br />$

but
$\sin(3.056^4) =-0.6773<br />$

For 2: if you take log base 2 of both sides then x = x log3, so x = 0. Or again, perhaps f(x) = C.