Results 1 to 8 of 8

Math Help - How to prove?

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    22

    How to prove?

    1. { x= sin t ; y= sin kt} <=> (1-x)*[(d^2*y)/(dx^2)] - x*[(dy)/(dx)]+ k^2*y=0
    ?

    And how to solve this limit? ->

    2. Lim[x->Pi/2][tan x]^(2x-Pi) ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,347
    Thanks
    30
    I don't believe that

    x = \sin t, y = \sin kt is a solution to

    (1-x) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0

    for general k. It works for k = 1 though. Check both the equation and solution.

    EDIT. I got it to work for

    (1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0

    Is this the equation?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,347
    Thanks
    30
    Interesting variant to this thread. Introduce the new independent variable

    x = \sin t

    in the equation

    (1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0

    Spoiler:
    \dfrac{d^2y}{dt^2} + k^2 y = 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294
    \displaystyle \lim_{x \to \frac{\pi}{2}}(\tan{x})^{2x-\pi} = \lim_{x \to \frac{\pi}{2}}e^{\ln{[(\tan{x})^{2x-\pi}]}}

    \displaystyle = \lim_{x \to \frac{\pi}{2}}e^{(2x-\pi)\ln{(\tan{x})}}

    \displaystyle = e^{\lim_{x \to \frac{\pi}{2}}(2x-\pi)\ln{(\tan{x})}}

    \displaystyle = e^{\lim_{x \to \frac{\pi}{2}}\frac{\ln{(\tan{x})}}{(2x-\pi)^{-1}}}

    \displaystyle = e^{\lim_{x \to \frac{\pi}{2}}\frac{\frac{\sec^2{x}}{\tan{x}}}{-2(2x-\pi)^{-2}}} by L'Hospital's Rule

    \displaystyle = e^{-\frac{1}{2}\lim_{x \to \frac{\pi}{2}}\frac{(2x-\pi)^2\sec^2{x}}{\tan{x}}}

    \displaystyle = e^{-\frac{1}{2}\lim_{x \to \frac{\pi}{2}}\frac{(2x-\pi)^2}{\sin{x}\cos{x}}}

    \displaystyle = e^{-\frac{1}{2}\lim_{x \to \frac{\pi}{2}}\frac{(2x - \pi)^2}{\frac{1}{2}\sin{2x}}}

    \displaystyle = e^{-\lim_{x \to \frac{\pi}{2}}\frac{(2x-\pi)^2}{\sin{2x}}}

    \displaystyle = e^{-\lim_{x \to \frac{\pi}{2}}\frac{4(2x-\pi)}{2\cos{2x}}} by L'Hospital's Rule

    \displaystyle = e^{-2\lim_{x \to \frac{\pi}{2}}\frac{2x-\pi}{\cos{2x}}

    \displaystyle = e^{-2\left(\frac{0}{-1}\right)}

    \displaystyle = e^0

    \displaystyle = 1.



    PHEW!!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, RCola!

    \displaystyle 2.\;\lim_{x\to\frac{\pi}{2}}(\tan x)^{2x-\pi}

    Let: \tan x)^{2x-\pi}" alt="y \:=\\tan x)^{2x-\pi}" />

    Take logs: . \ln y \:=\:\ln(\tan x)^{2x-\pi} \;=\;(2x-\pi)\ln(\tan x)

    . . . . . . . . . . . . =\;\dfrac{\ln(\tan x)}{(2x-\pi)^{-1}}\quad\Rightarrow\quad \dfrac{\infty}{\infty}


    Apply L'Hopital: . \dfrac{\frac{\sec^2\!x}{\tan x}}{-(2x-\pi)^{-2}\cdot2} \;=\;\dfrac{\frac{1}{\cos^2\!x}\cdot\frac{\cos x}{\sin x}}{-2(2x-\pi)^{-2}}

    . . . . . . . . =\;\dfrac{\frac{1}{\sin x\cos x}}{-2(2x-\pi)^{-2}} \;=\;\dfrac{\frac{2}{\sin2x}}{\frac{-2}{(2x-\pi)^2}} \;=\;-\dfrac{(2x-\pi)^2}{\sin2x}\quad\Rightarrow\quad \dfrac{0}{0}


    Apply l'Hopital: . -\dfrac{2(2x-\pi)\cdot2}{2\cos2x} \;=\;\dfrac{2(\pi-2x)}{\cos 2x}

    . . Then: . \displaystyle \lim_{x\to\frac{\pi}{2}} \frac{2(\pi-2x)}{\cos2x} \;=\;\frac{0}{-1} \;=\;0


    \displaystyle \text{Since }\lim_{x\to\frac{\pi}{2}} \ln y \;=\;0,\:\text{ then: }\:\lim_{x\to\frac{\pi}{2}} y \;=\;e^0 \;=\;1

    . . \displaystyle \text{Therefore: }\;\lim_{x\to\frac{\pi}{2}} (\tan x)^{2x-\pi} \;=\;1


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Edit: Ah, Prove It beat me to it! .And did an excellent job!
    . . . .(Oh well, I did my work at "ground level" . . .)

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294
    Quote Originally Posted by Soroban View Post
    Hello, RCola!


    Let: \tan x)^{2x-\pi}" alt="y \:=\\tan x)^{2x-\pi}" />

    Take logs: . \ln y \:=\:\ln(\tan x)^{2x-\pi} \;=\;(2x-\pi)\ln(\tan x)

    . . . . . . . . . . . . =\;\dfrac{\ln(\tan x)}{(2x-\pi)^{-1}}\quad\Rightarrow\quad \dfrac{\infty}{\infty}


    Apply L'Hopital: . \dfrac{\frac{\sec^2\!x}{\tan x}}{-(2x-\pi)^{-2}\cdot2} \;=\;\dfrac{\frac{1}{\cos^2\!x}\cdot\frac{\cos x}{\sin x}}{-2(2x-\pi)^{-2}}

    . . . . . . . . =\;\dfrac{\frac{1}{\sin x\cos x}}{-2(2x-\pi)^{-2}} \;=\;\dfrac{\frac{2}{\sin2x}}{\frac{-2}{(2x-\pi)^2}} \;=\;-\dfrac{(2x-\pi)^2}{\sin2x}\quad\Rightarrow\quad \dfrac{0}{0}


    Apply l'Hopital: . -\dfrac{2(2x-\pi)\cdot2}{2\cos2x} \;=\;\dfrac{2(\pi-2x)}{\cos 2x}

    . . Then: . \displaystyle \lim_{x\to\frac{\pi}{2}} \frac{2(\pi-2x)}{\cos2x} \;=\;\frac{0}{-1} \;=\;0


    \displaystyle \text{Since }\lim_{x\to\frac{\pi}{2}} \ln y \;=\;0,\:\text{ then: }\:\lim_{x\to\frac{\pi}{2}} y \;=\;e^0 \;=\;1

    . . \displaystyle \text{Therefore: }\;\lim_{x\to\frac{\pi}{2}} (\tan x)^{2x-\pi} \;=\;1


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Edit: Ah, Prove It beat me to it! .And did an excellent job!
    . . . .(Oh well, I did my work at "ground level" . . .)

    Well it's nice to know I didn't make any mistakes
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2010
    Posts
    22
    Quote Originally Posted by Danny View Post
    I don't believe that

    x = \sin t, y = \sin kt is a solution to

    (1-x) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0

    for general k. It works for k = 1 though. Check both the equation and solution.

    EDIT. I got it to work for

    (1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0

    Is this the equation?
    Quote Originally Posted by Danny View Post
    Interesting variant to this thread. Introduce the new independent variable

    x = \sin t

    in the equation

    (1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0

    Spoiler:
    \dfrac{d^2y}{dt^2} + k^2 y = 0
    \dfrac{dy}{dx} = smth...
    This function is given in parametric form. Also there was given some definition, something like \dfrac{d^2y}{dx^2} =  some reference to t variable; and ; . So it need to be proven that taking derivatives up to 2nd power/ level will be equal to (1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0

    Is there any tool to easily write math formulas between MATH quotes?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294
    Quote Originally Posted by RCola View Post
    Is there any tool to easily write math formulas between MATH quotes?
    Yes, improving your touch typing :P
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove that
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 21st 2010, 05:48 AM
  2. Prove n^2<= ......
    Posted in the Advanced Algebra Forum
    Replies: 12
    Last Post: November 17th 2009, 05:52 AM
  3. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  4. prove that
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 7th 2008, 05:14 PM
  5. prove
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 7th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum