newton-raphson failure case

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    I know its a bit random, but does anyone have an example of a failure case for the newton-raphson method of calculating roots of equations? I need one for my cwk, but my graphical calc's started playing up and i've bin trying to find one for about an hour! Thanks!
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    (Original post by JML161)
    I know its a bit random, but does anyone have an example of a failure case for the newton-raphson method of calculating roots of equations? I need one for my cwk, but my graphical calc's started playing up and i've bin trying to find one for about an hour! Thanks!
    How about where the gradient at the point in question (i.e. dy/dx) = 0? Because then the tangent won't intercept the x-axis, which is required for the N-R method to work.

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    GENIUS! Just need to find a line that does that...
    Thanks muchly

    ... now to get that damned calculator to work!
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    (Original post by JML161)
    GENIUS! Just need to find a line that does that...
    Thanks muchly

    ... now to get that damned calculator to work!
    Well that's easy enough. Just find a turning point or point of inflection. That's some of the easy Pure stuff you'll get
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    (Original post by JML161)
    I know its a bit random, but does anyone have an example of a failure case for the newton-raphson method of calculating roots of equations? I need one for my cwk, but my graphical calc's started playing up and i've bin trying to find one for about an hour! Thanks!
    Try the attached part (c)

    It does just what you want!

    Aitch
    Attached Files
  1. File Type: doc nrfail.doc (74.5 KB, 4725 views)
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    (Original post by JohnSPals)
    Well that's easy enough. Just find a turning point or point of inflection. That's some of the easy Pure stuff you'll get
    for my failure can my equation have a fraction eg - 3/4 x^2
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    (Original post by darth_vader05)
    for my failure can my equation have a fraction eg - 3/4 x^2
    Sure. Imagine this curve - a maximum will be at (0,0) hence it fails at x=0.

    Differentiating: d [-0.75x2]/dx = -1.5x

    At turning point, d [-0.75x2]/dx = -1.5x = 0
    Therefore x = 0
    Therefore y = 0 when substituting into the equation of the curve
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    The newton raphson method will break down for f(x) at a if

    f(a)/f'(a)+f(a-f(a)/f'(a))/f'(a-f(a)/f'(a))=0
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    Here's another example from an old Edexcel paper, if you need one:

    Aitch
    Attached Files
  2. File Type: doc nrfail2.doc (18.5 KB, 2314 views)
 
 
 
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