How Does the Inequality (cos(x))^p ≤ cos(px) Work?

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Homework Help Overview

The discussion revolves around the inequality (cos(x))^p ≤ cos(px), with the variable x constrained between 0 and π/2 and p constrained between 0 and 1. Participants are seeking guidance on how to approach proving this inequality.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants express uncertainty about how to start the proof and what methods to use, including references to Taylor's polynomial and the binomial theorem. Some participants question the assumptions regarding the values of p and the implications of differentiating the functions involved.

Discussion Status

There are various hints and suggestions being exchanged, including the use of Taylor's polynomial and differentiation. Participants are actively engaging with each other's ideas, but no consensus or clear resolution has been reached yet.

Contextual Notes

Participants note the lack of clarity regarding the professor's instructions on the values of p and express confusion about the application of mathematical concepts like Taylor's polynomial and the binomial theorem in this context.

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Homework Statement



(cos(x))^p [tex]\leq[/tex] cos(px)

0[tex]\leq[/tex]x[tex]\leq[/tex]pi half

and p, 0[tex]\leq[/tex](not equal) p [tex]\leq[/tex](not equal) 1

i need help, if some one can tell me how to started, what should i used i will really apreciate it! (sorry for my english :confused:)

Homework Equations





The Attempt at a Solution



 
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http://img218.imageshack.us/img218/6451/ine111es8.jpg
 
Last edited by a moderator:
without LaTeX

Nanie said:
http://img218.imageshack.us/img218/6451/ine111es8.jpg
[/URL]

Hi Nanie! :smile:

Or, without LaTeX:

(cosθ)p ≤ cos(pθ)

0 ≤ θ ≤ π/2

0 < p < 1. :wink:
 
Last edited by a moderator:
:biggrin: jeje thanks

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I'm friend of medinap...we really need help with this exercise!

First the professor...gave us (cos[tex]\theta[/tex])p[tex]\leq[/tex] cos(p[tex]\theta[/tex]) but he don't gave us the values of p (we asume that n is for all natural p=2(2n+1)...but yesterday! he made a correction that 0< p< 1

I don't know what to do

and sorry for my english too.
 
Your English is excellent!

Perhaps the simplest way to prove this is to use a Taylor's polynomial. [itex]cos(\theta)= 1- (1/2)\theta^2+ (1/4!)\theta^4-\cdot\cdot\cdot[/itex]
so [itex]cos(\theta)\le 1- (1/2)\theta^2[/itex]. Now use the extended binomial theorem to take that to the p power. You only need the first two terms.
 
i don't know how to used it, explain me please!
 
i know how to used the binomial theorem, but i don't know how to used it with taylor's polynomial i don't even know what that is..
 
Hi medinap! Hi Nanie! :smile:

Hint: differentiate (cosθ)p and cos(pθ), and remember that cos ≤ 1.

(btw, you both need to be more careful about using the past tense …

it's "if some one can tell me how to start, what should i use …"

and "he didn't gave us the values of p (we assumed that …" :wink:)
 
ok thanks... I will try! jeje...I have one hour to think !
 
  • #10
F([tex]\theta[/tex]) = cos ([tex]\theta[/tex]) p- cos (p[tex]\theta[/tex])

How I use this?

iah...I don't Know...

F(theta) = cos (theta)p- cos (p(theta))
 
  • #11
Nanie said:
F([tex]\theta[/tex]) = cos ([tex]\theta[/tex]) p- cos (p[tex]\theta[/tex])

How I use this?

Hi Nanie! :smile:

(what happened to that θ i gave you? :rolleyes:)

What is F'(θ)? Is it positive or negative?
 
  • #12
Thanks!

F([tex]\theta[/tex]) = p (sen[tex]\theta[/tex][tex]/[/tex]cos[tex]^{1-p}[/tex]) - sen(p[tex]\theta[/tex]) > 0 ...however ... cos (p[tex]\theta[/tex]) > cos ([tex]\theta[/tex])[tex]^{p}[/tex]



F'([tex]\theta[/tex]) = p (sen[tex]\theta[/tex][tex]/[/tex]cos[tex]^{1-p}[/tex] - sen p ([tex]\theta[/tex]) > 0 (positive)


We did it...?
 

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