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Math Help - Taylor Expansion on d/d

  1. #1
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    Taylor Expansion on d/d(f(x))[sqrt(x+f(x))]

    I need to find the first 2 orders of the taylor series on the expression below.

    sqrt(Psi_s0 + Phi_t*exp((Psi_s0-2*Phi_F - V_SB)/Phi_t))

    where Phi_t*exp((Psi_s0-2*Phi_F - V_SB)/Phi_t) is defined as Xi, and the taylor series is around Xi = 0.
    Last edited by was1984; June 15th 2008 at 10:53 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by was1984 View Post
    I need to find the first 2 orders of the taylor series on the expression below.

    sqrt(Psi_s0 + Phi_t*exp((Psi_s0-2*Phi_F - V_SB)/Phi_t))

    where Phi_t*exp((Psi_s0-2*Phi_F - V_SB)/Phi_t) is defined as Xi, and the taylor series is around Xi = 0.
    This is unreadable,

    \sqrt{\psi_{s0}+\phi_{t}e^{\frac{\psi_{s0}-2\cdot\phi_f-V_{SB}}{\phi_t}}}??
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  3. #3
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    Yes, thank you, that is correct.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by was1984 View Post
    Yes, thank you, that is correct.
    What is the variable in which we are differentiating in respect to? I will assume it is \psi

    Then let f(\psi) be equal to the above expression, then the second order polynomial would be

    f(0)+f'(0)x+\frac{f''(0)x^2}{2}
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  5. #5
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    The variable we are differentiating with respect to is Xi, which is the second term under the square root.
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  6. #6
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    I'll try to elaborate. We are setting \xi = \phi_{t}e^{\frac{\psi_{s0}-2\cdot\phi_f-V_{SB}}{\phi_t}}

    Then we are solving the series around \xi = 0

    So I actually want expand an equation of the form \sqrt{\psi_{s0}+\xi(\psi_{s0})} for \xi(\psi_{s0}), and I only need the first two terms, fortunately.
    Last edited by was1984; June 15th 2008 at 10:37 PM.
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