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    (Original post by Azzer11)
    ...
    Seems to me like you need to first use the given expressions then say

    q(x,1) = -x^3 (1+\frac{1}{1+e^{-1}})\log(2x^2+3x+1)

    Then use Taylor's expansion for it about the point x=4 to determine c_i, and only then approximate the value at x=3
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    (Original post by RDKGames)
    Seems to me like you need to first use the given expressions then say

    q(x,1) = -x^3 (1+\frac{1}{1+e^{-1}})\log(2x^2+3x+1)

    Then use Taylor's expansion for it about the point x=4 to determine c_i, and only then approximate the value at x=3
    Okay I see, I assume I differentiate T(x) to get dT/dx for the equation though?
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    (Original post by Azzer11)
    Okay I see, I assume I differentiate T(x) to get dT/dx for the equation though?
    Oops yeah I didn't notice they put the derivative of it in the equation. Yes you need to differentiate it first.
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    I did the taylor expansion at centre 4, do I then sub in x=3 to get q(3)? I'm a bit confused
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    (Original post by Azzer11)


    I did the taylor expansion at centre 4, do I then sub in x=3 to get q(3)? I'm a bit confused
    What's confusing?

    The first term of q(x) will be q(4) which is -64(1+\frac{1}{1+e^{-1}})\cdot \dfrac{19}{45} which indeed simplifies to -\dfrac{1216(2e+1)}{45(e+1)} as expected.

    To get q(3) you sub in x=3 yes.
 
 
 
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