# Function help.

1. Aug 30, 2010

### Mentallic

1. The problem statement, all variables and given/known data
I need to find n such that,

$$y=\frac{ln(cos(x))}{cos(1)}-x^n$$

is zero for all $$0\leq x\leq 1$$.

3. The attempt at a solution
I've already narrowed it down to $$2<n<2.5$$ and I understand that the answer will probably be an approximation. I'm hoping for an exact solution though, however ugly it may be.

Any ideas?

p.s. this isn't a homework problem, so it may well be that you can't have the function be zero for all x between 0 to 1 for any n. It seems as though for some n I've chosen, the function is always going to be under the x axis or above the x axis depending on my n. If it so happens that this is always the case even for n approaching very close to my desired value, then my question should have a valid solution. I'm curious as to how I could show it is always above/below the axis for some n or if it isn't.

Last edited: Aug 30, 2010
2. Aug 30, 2010

### Mentallic

Oh even for n=2.5 it cuts the axis in between 0 and 1, what a let down... in that case, I withdraw my question.

3. Aug 31, 2010

### jgens

I realize that you've withdrawn this, but you should be able to prove that there is no number n satisfying the desired conditions (assuming that I've interpretted your question correctly).

Define the function $f$ such that $f(x)=\log{(\cos{(x)})}-x^n$ and suppose that there is some number $n$ such that $f(x)=0$ for all $x \in [0,1]$. The last condition means that $f$ is constant on an interval; and moreover, $f$ is differentiable on this interval too. Therefore, $f'(x)=-(\tan{(x)}+nx^{n-1})=0$ for all $x \in [0,1]$. It follows from the last equality that $nx^{n-1} = -\tan{(x)}$ for $x \in [0,1]$, but this is a contradiction since $nx^{n-1} \geq 0$ and $\tan{(x)} \geq 0$ on the desired interval.* This essentially completes the proof.**

I'm sorry if there's some fundamental error in the 'proof' above or if I misunderstood your question. Hopefully you'll find something in the above post useful.

* I'm assuming $n>0$, because otherwise $f$ isn't defined for $x=0$.

** I used a modification of your function, but the proof for the function that you gave should be analogous to this one (assuming that I didn't screw up somewhere).