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Show an approximation

  1. Jun 9, 2009 #1
    1. The problem statement, all variables and given/known data
    This comes from a book on relativity but it basically comes down to a math problem. The problem is to prove that if
    [tex]T^2 \ll\frac{c^2}{\alpha^2}[/tex]
    then
    [tex]{t}\approx{T}(1-\frac{\alpha^2{T^2}}{6c^2})[/tex]
    given
    [tex]\frac{\alpha{T}}{c}=sinh(\frac{\alpha{t}}{c})[/tex]

    2. Relevant equations
    See above.


    3. The attempt at a solution
    I've tried solving for t from the equation
    [tex]\frac{\alpha{T}}{c}=sinh(\frac{\alpha{t}}{c})[/tex]
    which gives
    [tex]t=\frac{c}{\alpha}\log(\frac{T\alpha}{c}+\sqrt{1+\frac{T^2\alpha^2}{c^2}})[/tex]
    I thought I'd be able to use maclaurin expansion at this point becuase of how the approximation looks but I keep making mistakes and I'm not getting anywhere at the moment so I'd really appreciate some help.
     
  2. jcsd
  3. Jun 9, 2009 #2

    cepheid

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    What if you tried writing the hyperbolic sine function in terms of exponential functions first? Their Taylor expansions are easy...
     
  4. Jun 9, 2009 #3
    Thanks for the reply!

    Using the first four terms of each of the expansions I end up with what looks like the some kind of opposite to what I want
    [tex]
    {T}\approx{t}(1+\frac{\alpha^2{t^2}}{6c^2})
    [/tex]
    I can't see that I can get where I want from this and I haven't used all information in the problem to get here, maybe I'm missing something.
     
  5. Jun 9, 2009 #4

    Dick

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    You need to reverse the roles of t and T. You could also write at/c=arcsinh(aT/c) and use arcsinh(x)=x-x^3/3!+...
     
  6. Jun 9, 2009 #5

    HallsofIvy

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    No reason in the world to be always writing hyperbolic functions as exponentials!

    sihh(0)= 0.
    (sinh(x))'= cosh(x) and cosh(0)= 1.
    (sinh(x))"= (cosh(x))'= sinh(x) and sinh(0)= 0.
    (sinh(x))'''= (sinh(x))'= cosh(x) and cosh(0)= 1, etc.
     
  7. Jun 9, 2009 #6

    Hurkyl

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    Well, have you at least substituted that approximation into [itex]T(1-\frac{\alpha^2{T^2}}{6c^2})[/itex]?
     
  8. Jun 9, 2009 #7
    Thanks a lot that does it! Also gives me another question and that is how to find that arcsinh(x)=x-x^3/3!+...? My thought of simply solving for t and then looking at what I got didn't really get me there.

    How can I do that substitution? I'm feeling very rusty here, I'm sorry.
     
  9. Jun 9, 2009 #8
    Here's one way. The derivative of arcsinh x is [tex](1+x^2)^{-1/2}[/tex] so use the binomial series to find the series for that, then integrate term by term. You only need a few terms anyway, not the whole series.
     
  10. Jun 10, 2009 #9

    Dick

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    A physicist would just take
    [tex]

    {T}\approx{t}(1+\frac{\alpha^2{t^2}}{6c^2})

    [/tex]
    and say, hey, that means t=T with correction of higher order. So I can just replace t^2 with T^2 without making any errors I would care about. So I can change that into
    [tex]

    {T}\approx{t}(1+\frac{\alpha^2{T^2}}{6c^2})

    [/tex]
    Now move all the T's to one side and use 1/(1+x) is approximately equal to 1-x for x<<1.
     
    Last edited: Jun 10, 2009
  11. Jun 10, 2009 #10
    Thank you everyone!
     
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