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Math Help - Why won't think equation work with the newton-raphson method?

  1. #1
    Newbie XShmalX's Avatar
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    Why won't think equation work with the newton-raphson method?

    (x^3+e^2x+5)^(1/3) It just ends up oscillating between 0.854 & -2.884
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    Quote Originally Posted by XShmalX View Post
    (x^3+e^2x+5)^(1/3) It just ends up oscillating between 0.854 & -2.884
    what you posted is an expression, not an equation.

    I assume you mean to find the value of x where the given expression equals zero ... a look at the graph will show you why the iterations fail to converge.
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    Newbie XShmalX's Avatar
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    Quote Originally Posted by skeeter View Post
    what you posted is an expression, not an equation.

    I assume you mean to find the value of x where the given expression equals zero ... a look at the graph will show you why the iterations fail to converge.
    Yes I meant to set it equal to zero, sorry! I tried looking at the graph myself first & I assumed It was to do with the gradients of the linebut didn't understand exactly what... :S
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    Quote Originally Posted by XShmalX View Post
    Yes I meant to set it equal to zero, sorry! I tried looking at the graph myself first & I assumed It was to do with the gradients of the linebut didn't understand exactly what... :S
    Hence you could solve for x^3+e^{2x}+5=0
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    Very good point! a^{1/3}= 0 if and only if a= 0. That will not have the vertical tangent at the solution point. Skeeter's point was that the derivative of y= (x^3+ e^x+ 5)^{1/3}, y= (1/3)(x^3+ e^x+ 5)^{-2/3}(3x^2+ e^x) is infinite where x^2+ e^x+ 5= 0, exactly the point you are looking for.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by XShmalX View Post
    (x^3+e^2x+5)^(1/3) It just ends up oscillating between 0.854 & -2.884
    If the equation to be solved is x^{3} + e^{2x} + 5 =0 the Newton-Raphson iterations are given by the difference equation...

    \displaystyle \Delta_{n} = x_{n+1} - x_{n} = - \frac{x^{3} + e^{2x} + 5}{3 x^{2} + 2 e^{2 x}} = \varphi (x_{n}) (1)

    The function \varphi(x) is illustrated here...



    There is only one 'attractive fixed point' in x_{0} = -1.71366985564... and for any initial value the sequence (1) converges at x_{0}...

    Kind regards

    \chi \sigma
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    MHF Contributor chisigma's Avatar
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    It is interesting to observe that the search of the solution of the equation \displaystyle (x^{3} + e^{2x} +5)^{\frac{1}{3}} with the Newton-Raphson method conducts to the difference equation...

    \displaystyle \Delta_{n} = x_{n+1} - x_{n} = -\frac{(x^{3} + e^{2x} +5)^{\frac{1}{3}}}{\frac{1}{3}\ {(x^{3} + e^{2x} +5)^{-\frac{2}{3}} (3 x^{2} + 2 e^{2 x})}} = - 3\ \frac{x^{3} + e^{2 x} + 5}{3 x^{2}+2 e^{2x}} = 3\ \varphi (x_{n}) (1)

    ... where \varphi(x) is the function described in my previous post. At first it seems that nothing changes but it isn't!... The function 3\ \varphi(x) is illustrated here...



    The 'attractive fixed point' is [of course...] again at x_{0} = -1.71366985564... but [taht is the difference...] around x=x_{0} is |\varphi(x)|> |2\ (x_{0} - x)| [the 'red line' in figure...] and that means that the sequence obtained from (1) will never converge to x_{0} ...

    Kind regards

    \chi \sigma
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  8. #8
    Newbie XShmalX's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Very good point! a^{1/3}= 0 if and only if a= 0. That will not have the vertical tangent at the solution point. Skeeter's point was that the derivative of y= (x^3+ e^x+ 5)^{1/3}, y= (1/3)(x^3+ e^x+ 5)^{-2/3}(3x^2+ e^x) is infinite where x^2+ e^x+ 5= 0, exactly the point you are looking for.
    Ok, but why does the gradient being infinite mean I can't yet the answer through the normal NR forumula. I want to understand why...
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  9. #9
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    Newton's method requires that you continually evaluate the derivative to find more points. How can you use \displaystyle \infty to find the next point?

    The Bisection Method is guaranteed to converge, though it will converge slowly.
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    Newbie XShmalX's Avatar
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    Quote Originally Posted by Prove It View Post
    Newton's method requires that you continually evaluate the derivative to find more points. How can you use \displaystyle \infty to find the next point?

    The Bisection Method is guaranteed to converge, though it will converge slowly.
    Ahh ok, thankyou!
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  11. #11
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    Quote Originally Posted by XShmalX View Post
    Ok, but why does the gradient being infinite mean I can't yet the answer through the normal NR forumula. I want to understand why...
    The graph is positive-going.
    When you pick an initial approximation to the root,
    calculate the slope of the tangent,
    find your next approximation and continue with the iterations,
    your "next approximation" keeps going either side of the root.

    There is one value of x that would cause you to "land on" the root.
    But that would be pure luck!!
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