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Math Help - Series expansion when y is small

  1. #1
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    Exclamation Series expansion when y is small

    I'm quite rusty when it comes to these, but I need to do this to finish a uni assignment which is due tomorrow!

    I have

    (y^0.5)(1+y)^0.5 - loge[y^0.5 + (1+y)^0.5] = T + c

    Now, the question is:

    By series expansion, show that
    when y << 1, T + c = (2/3)y^(3/2)

    Make sure you expand to include terms up to y^(3/2).



    They've been nice enough to give a hint, which is:

    You will need the results

    (1+y)^0.5 = 1 + 0.5y - (1/8)y^2 + .....

    and

    loge[y^0.5 + (1+y)^0.5] = loge[1 + y^0.5 + 0.5y - (1/8)y^2]



    It also tells me that the expression 'loge[1 + y^0.5 + 0.5y - (1/8)y^2]' as above can be expanded further by writing

    Q = y^0.5 +0.5y - (1/8)y^2

    and recalling that loge[1+Q] = Q - (1/2)Q^2 + (1/3)Q^3 - ....





    Yeah, if anyone can help, please do, and QUICKLY!

    Thanks in advance!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Shoelet View Post
    I'm quite rusty when it comes to these, but I need to do this to finish a uni assignment which is due tomorrow!

    I have

    (y^0.5)(1+y)^0.5 - loge[y^0.5 + (1+y)^0.5] = T + c

    Now, the question is:

    By series expansion, show that
    when y << 1, T + c = (2/3)y^(3/2)

    Make sure you expand to include terms up to y^(3/2).



    They've been nice enough to give a hint, which is:

    You will need the results

    (1+y)^0.5 = 1 + 0.5y - (1/8)y^2 + .....

    and

    loge[y^0.5 + (1+y)^0.5] = loge[1 + y^0.5 + 0.5y - (1/8)y^2]



    It also tells me that the expression 'loge[1 + y^0.5 + 0.5y - (1/8)y^2]' as above can be expanded further by writing

    Q = y^0.5 +0.5y - (1/8)y^2

    and recalling that loge[1+Q] = Q - (1/2)Q^2 + (1/3)Q^3 - ....





    Yeah, if anyone can help, please do, and QUICKLY!

    Thanks in advance!
    From the hints you have been given you have:

    \log_e(1 + y^{0.5} + 0.5y - (1/8)y^2) = (y^{0.5} + 0.5y - (1/8)y^2) -(1/2)(y^{0.5} + 0.5y - (1/8)y^2) ^2+(1/3)(y^{0.5} + 0.5y - (1/8)y^2) ^3

    Now expand the right hand side as far as the term in y^{3/2}

    \log_e(1 + y^{0.5} + 0.5y - (1/8)y^2) =  (y^{1/2}+0.5y) - (1/2) (y+y^{3/2}+(1/3)(y^{3/2})+ ..=y^{1/2}-(1/6)y^{3/2}+..

    Check this for errors!

    Doing the same for the other term on the left gives:

    (y^{1/2})(1+y)^{1/2}=y^{1/2}(1+(1/2)y-(1/8)y^2+..)=y^{1/2}+(1/2)y^{3/2}+ ..

    and you should be able to take it from there.

    RonL
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