Results 1 to 12 of 12

Math Help - Simplify expression, Is it correct?

  1. #1
    Member
    Joined
    May 2008
    Posts
    112

    Simplify expression, Is it correct?

    Attached is the expression and my solution.

    I get a solution of zero, is it correct?
    Attached Thumbnails Attached Thumbnails Simplify expression, Is it correct?-question.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2008
    From
    Singapore
    Posts
    160
    Sorry the attachment is not clear. Can't really tell u from a in the attachment. Can attach a clearer version.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    112
    There is no a? All of the exponents are u (positive and negative).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2008
    Posts
    6
    the correct answer is 4/((e^u+e^(-u))^2)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2008
    From
    Singapore
    Posts
    160
    This is the approach i took.

    (e^u+e^-u)(e^u+e^-u)-(e^u-e^-u)(e^u-e^-u) / (e^u+e^-u)^2
    (e^2u+2+e^-2u)-(e^2u-2+e^-2u)/ (e^u+e^-u)^2
    (e^2u+2+e^-2u-e^2u+2-e^-2u)/ (e^u+e^-u)^2
    4 / (e^u+e^-u)^2

    Quote Originally Posted by nrslmz View Post
    the correct answer is 4/((e^u+e^(-u))^2)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2008
    Posts
    112
    Attached is a better version.
    Attached Thumbnails Attached Thumbnails Simplify expression, Is it correct?-question.jpg  
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2008
    From
    Singapore
    Posts
    160
    Ok noted. Your approach is not quite right. Please look at my previous post for further clarification.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    May 2008
    Posts
    112
    Ahh, I see now. Thanks!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    May 2008
    Posts
    112
    Actually, I noticed that you didn't simplify your denominator.

    The denominator simplified is 3 because (e^2u + e^0 + e^0 + e^-2u).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Oct 2008
    From
    Singapore
    Posts
    160
    You are right. I forgot to simplify the denominator

    In that case it should be

    (e^u+e^-u)(e^u+e^-u)-(e^u-e^-u)(e^u-e^-u) / (e^u+e^-u)^2
    (e^2u+2+e^-2u)-(e^2u-2+e^-2u)/ (e^u+e^-u)^2
    (e^2u+2+e^-2u-e^2u+2-e^-2u)/ (e^u+e^-u)^2
    4 / (e^2u+2+e^-2u)
    4 / 3
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    May 2008
    Posts
    112
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Dec 2008
    From
    NY - USA
    Posts
    28
    Quote Originally Posted by mwok View Post
    Attached is the expression and my solution.

    I get a solution of zero, is it correct?
    Hi,

    If I am not mistaken, this can be solved using hyperbolic functions.
    I'll break this problem into many sections so you can see what I am doing

    Please follow this closely!
    If you consider that the hyperbolic functions are

    \frac{e^{x} - e^{-x}}{2} = sinh(x)

    and that

    \frac{e^{x} + e^{-x}}{2} = cosh(x)

    then we can conclude that

    (e^{x} - e^{-x}) = 2sinh(x)

    e^{x} + e^{-x} = 2cosh(x)

    ------------------------------------------------------
    Which means that if you use them in your problem you will get

    (e^{u} + e^{-u})*(e^{u}+e^{-u}) = 4 cosh^{2}(u)

    (e^{u} - e^{-u})*(e^{u}-e^{-u}) = 4 sinh^{2}(u)


    (e^{u} - e^{-u})^{2} = 4 cosh^{2}(u)



    4 cosh^{2}(u) - 4 sinh^{2}(u) = 4

    \frac{4}{4 cosh^{2}(u)} = \frac{1}{cosh^{2}(u)} = sech^{2}(u)<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Simplify the integral (is my version correct?)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 6th 2011, 09:45 AM
  2. simplify a expression
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 5th 2011, 01:57 PM
  3. Correct Rational Expression?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 26th 2010, 01:49 AM
  4. Replies: 2
    Last Post: May 28th 2009, 12:57 PM
  5. Simplify this expression, is this correct?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 2nd 2008, 05:45 PM

Search Tags


/mathhelpforum @mathhelpforum