My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 30th, 2007, 01:05 AM   #1
Site Founder
 
julien's Avatar
 
Joined: Nov 2006
From: France

Posts: 824
Thanks: 7

Trigonometric equation / Differential inequality (12/18/06)

A solution received for the trigonometric equation (characters edited):

"Find all the solutions of the equation cos4(x) + sin4(x) = 1/4.

cos(4x) ≡ 2cos2(2x) _ 1
≡ 2(2cos2x _ 1)2 _ 1
≡ 4(2cos4x _ 2cos2x + 1) _ 3
≡ 4(cos4x + (cos2x _ 1)2) _ 3
≡ 4(cos4x + sin4x) _ 3

(Or this result can be quoted as a standard identity.)

Hence the equation to be solved is equivalent to cos(4x) = -2.
Hence x =1/4 Cos-1(-2) (where Cos-1 is multi-valued).
You probably would prefer this in explicit complex form, involving a logarithm.

cos x ≡ (eix + e-ix)/2, where i = sqrt(-1)
and eiπ = -1

Solving the quadratic equation obtainable by rearranging the identity and taking
natural logarithms (or using a standard logarithmic formula for Cos-1), leads to

1/4 Cos-1(-2) = (m π +/- i ln(2 + sqrt(3)))/4, where m is any odd integer.

The plus sign can be replaced by a minus sign without affecting the result.
I chose plus so that the logarithm is positive." [Skipjack]

A solution to the differential inequality problem (characters edited):

"Let g : R -> R be a real function of class C2 (meaning g is twice differentiable everywhere).
Assume also that for all x in R, 0 <= g(x) + g´´(x).
Show that for all x in R, 0 <= g(π + x) + g(x).

To avoid confusion between the two uses of x, I shall prove

0 <= g(π) + g(0),

which implies the general result (by a simple translation).

Let f(x) = g(x) + g´´(x)


Let h(x) = int( f(x) sin x dx, 0..x) = int( (sin x g(x) + sin x g´´(x)) dx, 0..x)

= [- cos x g(x) + sin x g´(x)]x=x - [- cos x g(x) + sin x g´(x)]x=0

= g(0) - cos x g(x) + sin x g´(x)

Now assume 0 <= x <= π, so that 0 <= f(x) sin x and hence 0 <= h(x) also.


0 <= int(h(x) / sin^2x dx, t..n-t) (where 0 < t < Pi/2)



=Int (g(0) / sin^2x - (cos x / sin^2x) g(x) + g´(x) / sin x) dx, t..n-t)


= [-(cos x / sin x))g(0) + g(x) / sin x]x=Pi-t - [-(cos x / sin x))g(0) + g(x) / sin x]x=t

= (cos t g(0) + g(p - t) - (-cos t g(0) + g(t)) / sin t (since sin(p - t) = sin t, etc.)

Multiplying by sin t and then taking the limit as t -> 0+ gives

0 <= g(Pi) + g(0)." [Skipjack]
julien is offline  
 
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
differential, equation, inequality, trigonometric



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
A Trigonometric Inequality fahad nasir Trigonometry 2 July 21st, 2013 05:40 AM
trigonometric inequality bboy114crew Trigonometry 3 July 30th, 2011 12:07 PM
Trigonometric Inequality wkhur Calculus 4 August 18th, 2010 07:52 PM
Trigonometric inequality Tomate Algebra 4 June 18th, 2010 11:51 AM
trigonometric inequality mikeportnoy Algebra 7 October 4th, 2009 06:47 AM





Copyright © 2019 My Math Forum. All rights reserved.