Challenge Math Challenge - April 2020

  • #151
Not anonymous said:
Is there any function that can be expressed using only common standard functions and does not need to be written as a sum of infinite series that is continuous and differentiable in ##(0,1)## which has all rational numbers in its range and no other value as its zeros?
The range of any continuous map ##f:(0,1)\to\mathbb{R}## is connected, so if it contains ##\mathbb{Q}##, then it must be all of ##\mathbb{R}.## There are continuous surjections like this, e.g. ##f(x)=\tan(\pi(x-1/2)).## What do you mean by "no other values as its zeros"?

Not anonymous said:
Thanks for that example! It is nice to be able to prove something using just one simple theorem. But I am a bit confused about this function's relation to @wrobel's. @wrobel's function uses absolute value of ##f(x)## whereas yours does not, so your example appears to be of similar form but not exactly the same as the exponential of @wrobel's function. Am I missing something?
Okay, you're right, the exponential of wrobel's function is ##e^{-x}|f(x)|.## But since he is working on an interval where ##f## has no zeros, this is the same up to a possible minus sign. The point is that ##f## and ##e^f## have the same critical points, and negating doesn't change this.
 

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