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  1. Jul 11th, 2004, 10:14 PM #1
    MasterJC
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    Newton's Method

    Hi
    Does anyone know how to use Newton's Method to do , say, find the absolute minimum of 1/4*x^4+x^2+5x ?
    thanks
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  2. Jul 14th, 2004, 04:07 AM #2
    Ecniv
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    Isn't that a quadratic equation?

    I thought that the A level method of solving it applies??

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  3. Jul 20th, 2004, 07:35 PM #3
    nicktheman
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    Newton's method is not required.

    y = (x^4)/4 + x^2 + 5x

    dy/dx = x^3 + 2x + 5

    maximum/minimum when dy/dx = 0

    Therefore x^3 + 2x + 5 = 0

    Using graphics calculator :

    x = -1.328268856... (only real root, 2 other roots with imaginary parts)

    If an exact result is required, there is a formula for cubic equations in the same way that for the quadratic equation ax^2 + bx + c = 0,
    x = (-b +/- (b^2 - 4ac)^0.5)/2a

    d2y/dx2 = 3x^2 +2, which is always >0, so point is a minimum.

    PS Yes I know this because i did A Level Maths and Further Maths this year.
    Last edited by nicktheman; Jul 21st, 2004 at 09:21 AM.
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  4. Jul 20th, 2004, 10:28 PM #4
    alkatran
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    Excuse me, but that's not the quadratic formula.

    There's no "^-1" in there, only "^1/2"
    Don't pay attention to this signature, it's contradictory.
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  6. Jul 21st, 2004, 08:00 AM #5
    wossname
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    Originally posted by alkatran
    Excuse me, but that's not the quadratic formula.

    There's no "^-1" in there, only "^1/2"
    Indeed, x^0.5 is the equivalent of taking the square root of x.
    I don't live here any more.
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  7. Jul 21st, 2004, 09:21 AM #6
    nicktheman
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    Originally posted by alkatran
    Excuse me, but that's not the quadratic formula.

    There's no "^-1" in there, only "^1/2"
    Sorry, my mistake

    Used to just writing a square root sign but there isn't one on my keyboard.
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