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Thursday 18 April 2013
Functions
Can we Find any two Distinct Functions
$$ f:\Re \rightarrow \Re $$ such that
$$ f\left ( x \right )=f\left (\frac{x}{2}\right ) $$
It's easy to see that formally speaking $$L(x/2) = L(x)$$. Now that series I gave seems hopeless at first since it consider values $m$ at 0 and infinity. That is fine, let us consider $m(x) = x^3e^{-x^2}$. This rapidly goes to 0 as the $|x|$ gets very large, as well as very small. Then substituting that back into the definition of $L$ gives you a function that is periodic with respect to doubling the period (at least over the reals).
If it is true for all x, then I think it would be f(x)=0.
ReplyDeleteIn general $$f(x)=Const$$ is one such function.
ReplyDeleteWe can engineer a solution pretty easily here. Consider some $m(x)$ (we'll talk about what this is later).
ReplyDeleteConsider
$$L(x)= m(4x) + m(2x) + m(x) + m(\frac{x}{2}) + m(\frac{x}{8}) + ... $$
It's easy to see that formally speaking $$L(x/2) = L(x)$$. Now that series I gave seems hopeless at first since it consider values $m$ at 0 and infinity. That is fine, let us consider $m(x) = x^3e^{-x^2}$. This rapidly goes to 0 as the $|x|$ gets very large, as well as very small. Then substituting that back into the definition of $L$ gives you a function that is periodic with respect to doubling the period (at least over the reals).
The $L(x)$ in my earlier comment should be $L(x) = ... m(4x) ... $ that is, it is infinite in both directions
ReplyDelete