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Math Help - Series expansion

  1. #1
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    Series expansion

    Hi,
    Just wondering if anybody might be able to guide me through this problem, I'm not really sure where to start.

    I wish to show that when y<<1,

    τ + c ≈ [2y^(2/3)]/3 where c is a constant.

    I already know the results
    √(1+y) = 1+y/2 - (y^2)/8 + ....

    and

    ln[√(y)+√(1+y)] = ln[1 + √(y) + y/2 -(y^2)/8 + ....]
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sk1001 View Post
    Hi,
    Just wondering if anybody might be able to guide me through this problem, I'm not really sure where to start.

    I wish to show that when y<<1,

    τ + c ≈ [2y^(2/3)]/3 where c is a constant.

    I already know the results
    √(1+y) = 1+y/2 - (y^2)/8 + ....

    and

    ln[√(y)+√(1+y)] = ln[1 + √(y) + y/2 -(y^2)/8 + ....]
    What is t and what is c?
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  3. #3
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    I originally obtained t and c from the DE

    dy/dt = √(1+[1/y])

    Seperation of variables and various other change of variables led me to

    √(y)√(1+y) - ln[√(y)+√(1+y)] = t + c

    Obviously, t and c come from the RHS (where t is a variable and c is a constant obtained from integration)
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sk1001 View Post
    I originally obtained t and c from the DE

    dy/dt = √(1+[1/y])

    Seperation of variables and various other change of variables led me to

    √(y)√(1+y) - ln[√(y)+√(1+y)] = t + c

    Obviously, t and c come from the RHS (where t is a variable and c is a constant obtained from integration)
    I do not understand why you are using power series







    Now



    So let



    Now go from there
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  5. #5
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    sorry, I think I'm being unclear here.

    I'll start again.
    Basically given dy/dt = √(1+[1/y]), I needed to show the solution was given by √(y)√(1+y) - ln[√(y)+√(1+y)] = t + c

    To get there I used seperation of variables and 2 change of variables (y=q
    and then q=sinh(θ))

    the next lead on question then asks me to show by series expansion that when y<<1,

    t+c ≈ [2y^(2/3)]/3

    I hope this has cleared things up
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  6. #6
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    basically, I wish to show that

    √(y)√(1+y) - ln[√(y)+√(1+y)] = 2y^(3/2)/3 when y<<1 by series expansion.

    Can anybody start me off?
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