Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    2 more questions to put to you lot!!

    review exercise q49
    show that the curve with the equation y=sechx has 2nd differental =0
    at points where x= +/- lnp and state value of p


    the next one... Q50

    I[n]=integral of sin^n x dx between pi/2 and 0
    a) show that (n+2)I[n+2]=(n+1)I[n] this bit was fine

    show that I[2n]=(2n)!pi / 2^(2n+1)(n!)^2 which i really can't do!

    and show further that I[2n-1]I[2n]=pi/4n


    sorry people!!! thankyou
    Offline

    0
    ReputationRep:
    (Original post by ogs)
    2 more questions to put to you lot!!

    review exercise q49
    show that the curve with the equation y=sechx has 2nd differental =0
    at points where x= +/- lnp and state value of p


    the next one... Q50

    I[n]=integral of sin^n x dx between pi/2 and 0
    a) show that (n+2)I[n+2]=(n+1)I[n] this bit was fine

    show that I[2n]=(2n)!pi / 2^(2n+1)(n!)^2 which i really can't do!

    and show further that I[2n-1]I[2n]=pi/4n


    sorry people!!! thankyou
    Attached Images
     
    Offline

    2
    ReputationRep:
    y = sech(x)
    dy/dx = -tanh(x)sech(x)
    d^2y/dx^2
    = -sech^3(x) + tanh^2(x)sech(x)
    = sech^3(x) (-1 + sinh^2(x))

    When d^2y/dx^2 = 0 we have sinh(x) = 1 or -1. If sinh(x) = 1 then

    e^x - e^(-x) = 2
    e^2x - 2e^x - 1 = 0
    e^x = [2 + sqrt(4 + 4)] / 2 = 1 + sqrt(2)
    x = ln(1 + sqrt(2)).

    Similarly (or using the fact that sinh is odd) if sinh(x) = -1 then x = -ln(1 + sqrt(2))
    Offline

    0
    ReputationRep:
    For your second question see here: http://www.nrich.maths.org/discus/me...tml?1081469967.
    Offline

    2
    ReputationRep:
    I(2n)
    = [(2n - 1)/2n] * I(2n - 2)
    = [(2n - 1)/2n] * [(2n - 3)/(2n - 2)] * I(2n - 4)
    = [(2n - 1)/2n] * [(2n - 3)/(2n - 2)] * [(2n - 5)/(2n - 4)] * I(2n - 6)
    = ...
    = [(2n - 1)(2n - 3) ... 1] I(0) / [2n (2n-2) ... 2]
    = [(2n - 1)(2n - 3) ... 1] pi / {2[2n (2n-2) ... 2]}
    = (2n)! pi / {2[2n (2n-2) ... 2]^2}
    = (2n)! pi / {2^(2n + 1) (n!)^2}
    Offline

    0
    ReputationRep:
    (Original post by mikesgt2)
    For your second question see here: http://www.nrich.maths.org/discus/me...tml?1081469967.
    hah i remeber this!
    • Thread Starter
    Offline

    0
    ReputationRep:
    Thankyou every one!
 
 
 
Turn on thread page Beta
Updated: June 20, 2004
Poll
Do you think parents should charge rent?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.