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Math Help - Simple Substituting and Rearranging

  1. #1
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    Question Simple Substituting and Rearranging

    Hello,
    may someone be so kind to explain how to arrive, step by step, from equation 23 to 28?

    Most of all I would like to understand the approximation with delta: if I substitute eq26 in 25 I get a different result (e.g. delta^3 terms).

    See the attached image.

    Thank you very much.

    PS
    eq24 may be taken as it is, I mean, phi is simply "(A D Cs / x') - (c D Cs/2)"
    Attached Thumbnails Attached Thumbnails Simple Substituting and Rearranging-formula3.png   Simple Substituting and Rearranging-formula2.png  
    Last edited by Caccioppoli; April 19th 2013 at 04:25 AM.
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  2. #2
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    Re: Simple Substituting and Rearranging

    I've uploaded a bigger version of the image, split in two figures.
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  3. #3
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    Re: Simple Substituting and Rearranging

    I find it a bit hard to see how to go from (23) to (24) without seeing (17)...
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  4. #4
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    Re: Simple Substituting and Rearranging

    eq 23 and 24 are not linked, eq24 does not need to be explained, phi is simply
    Attached Thumbnails Attached Thumbnails Simple Substituting and Rearranging-formula1.png  
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  5. #5
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    Re: Simple Substituting and Rearranging

    I can update the problem since there is some progress:

    I have the following equation

    q=aq^3+b [eq1]

    eq1 is approximated with q=a^{-0.5} + \delta [eq2]

    \delta=b/3

    Where does this approximation come from and why is \delta=b/3?

    Thank you very much.
    Last edited by Caccioppoli; April 22nd 2013 at 01:27 PM.
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  6. #6
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    Re: Simple Substituting and Rearranging

    I've found the solution !!

    Premise the result was not [tex]\delta=b/3[\tex] but [tex]\delta=-b/2[\tex]

    we have two equations, basically:

    aq^3-q+b=0 and q=a^{-0.5}+\delta with a small δ

    Writing those two equations in a system, we have that the δ that satisfy both equations is a δ that satisfy this 3rd order equation:

    a\delta^3+3a^{-0.5}\delta^2+2\delta+b=0

    ignoring third order and second order terms in δ we have simply

    2\delta+b=0 so the result \delta=-b/2

    Last edited by Caccioppoli; April 23rd 2013 at 07:02 AM.
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