Find n for y=ln(cos(x))/cos(1)-x^n to be Zero

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SUMMARY

The discussion focuses on determining the value of n in the equation y=ln(cos(x))/cos(1)-x^n such that it equals zero for all x in the interval [0, 1]. The analysis concludes that no such n exists, as demonstrated by the contradiction arising from the derivative f'(x)=-(tan(x)+nx^(n-1))=0, which implies nx^(n-1) must equal -tan(x). Since both terms are non-negative in the specified interval, this leads to the conclusion that the function cannot be constant and equal to zero across the entire interval.

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Homework Statement


I need to find n such that,

[tex]y=\frac{ln(cos(x))}{cos(1)}-x^n[/tex]

is zero for all [tex]0\leq x\leq 1[/tex].

The Attempt at a Solution


I've already narrowed it down to [tex]2<n<2.5[/tex] 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.
 
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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.
 
I realize that you've withdrawn this, but you should be able to prove that there is no number n satisfying the desired conditions (assuming that I've interpretted your question correctly).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 [itex]f[/itex] such that [itex]f(x)=\log{(\cos{(x)})}-x^n[/itex] and suppose that there is some number [itex]n[/itex] such that [itex]f(x)=0[/itex] for all [itex]x \in [0,1][/itex]. The last condition means that [itex]f[/itex] is constant on an interval; and moreover, [itex]f[/itex] is differentiable on this interval too. Therefore, [itex]f'(x)=-(\tan{(x)}+nx^{n-1})=0[/itex] for all [itex]x \in [0,1][/itex]. It follows from the last equality that [itex]nx^{n-1} = -\tan{(x)}[/itex] for [itex]x \in [0,1][/itex], but this is a contradiction since [itex]nx^{n-1} \geq 0[/itex] and [itex]\tan{(x)} \geq 0[/itex] 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 [itex]n>0[/itex], because otherwise [itex]f[/itex] isn't defined for [itex]x=0[/itex].

** 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).
 

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