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    This may be a pretty basic question for some of you so bear with me.
    If I have a exponential graph with 3 solutions and I use iteration
    xt+1=xt-[f(x)/f'(x)] to find the middle solution starting at 1 how do I find the other solutions?
    I was thinking that starting near the other solutions would give different results (i've graphed it out so I know roughly the solutions).
    Also I was thinking by rearranging the equation.
    say for example
    exp(x)=2x^2
    instead of f(x)=exp(x)-2x^2 f(x)=2x^2-exp(x)
    Thanks
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    (Original post by Lay-Z)
    This may be a pretty basic question for some of you so bear with me.
    If I have a exponential graph with 3 solutions and I use iteration
    xt+1=xt-[f(x)/f'(x)] to find the middle solution starting at 1 how do I find the other solutions?
    I was thinking that starting near the other solutions would give different results (i've graphed it out so I know roughly the solutions).
    Also I was thinking by rearranging the equation.
    say for example
    exp(x)=2x^2
    instead of f(x)=exp(x)-2x^2 f(x)=2x^2-exp(x)
    Thanks
    Use the same iteration with starting points close to your other solutions
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    If the onlly info you`re given is that you have 3 solutions, then plotting the individual curves (exp(x), 2x^2) on the same axes is sensible, so you can "see" what the solutions will be, as you`ve already done.

    The simpler iteration formula is given by f(x)=2x^{2}-e^{x}

    Viz:

    \displaystyle  N(x) = \frac{e^{x}(x-1)-2x^{2}}{e^{x}-4x}

    Takes aroud 5 iterations if you start the negative root at the highest integer value (from your graph estimate), the positive one at it`s respective next highest (i.e. -1 and 2)
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    (Original post by Hasufel)
    If the onlly info you`re given is that you have 3 solutions, then plotting the individual curves (exp(x), 2x^2) on the same axes is sensible, so you can "see" what the solutions will be, as you`ve already done.

    The simpler iteration formula is given by f(x)=2x^{2}-e^{x}

    Viz:

    \displaystyle  N(x) = \frac{e^{x}(x-1)-2x^{2}}{e^{x}-4x}

    Takes aroud 5 iterations if you start the negative root at the highest integer value (from your graph estimate), the positive one at it`s respective next highest (i.e. -1 and 2)
    Thanks for the reply.
    Ive iterated from 1 and 3 to get 2 solutions 1.488 and 2.618 which coincide with the graph but I cant seem to get the final solution which looking at the graph is about -0.5.
    If I start at 0 it the 2nd iteration of the algortihm goes to 1 and I get 1.488 again. The other numbers i've tried dont converge.

    I've been using the f(x)=exp(x)-2x^2/exp(x)-4x. Not sure what the 2nd equation you posted is
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    have you tried starting from -1?

    this is what i get:

    {-0.626336, -0.544081, -0.539846, -0.539835, -0.539835, -0.539835}

    (the problem with the other starting points you have (and others) is that they`re in intervals of choas - sometimes within a given interval a "guess" converges and sometimes not - sometimes it cycles between a few numbers and other times it goes all over the place (chaos theory, but that`s too deep for here)

    try x = -1...

    My N(x) equation is just the newton iteration for y(x)=e^x-2x^2 , where it`s common denominator is e^x-4x

    \displaystyle N(x)= x - \frac{e^{x}-2x^{2}}{e^{x}-4x}= \frac{e^{x}(x-1)-2x^{2}}{e^{x}-4x}

    (EDITED)
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    (Original post by Hasufel)
    have you tried starting from -1?

    this is what i get:

    {-0.626336, -0.544081, -0.539846, -0.539835, -0.539835, -0.539835}

    (the problem with the other starting points you have (and others) is that they`re in intervals of choas - sometimes within a given interval a "guess" converges and sometimes not - sometimes it cycles between a few numbers and other times it goes all over the place (chaos theory, but that`s too deep for here)

    try x = -1...

    My N(x) equation is just the newton iteration for y(x)=e^x-2x^2 , where it`s common denominator is e^x-4x

    \displaystyle N(x)= x - \frac{e^{x}-2x^{2}}{e^{x}-4x}= \frac{e^{x}(x-1)-2x^{2}}{e^{x}-4x}

    (EDITED)
    What values did you get for f and f'.
    this is what I get for -1

    Attached Images
     
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    i get these:
    Attached Images
  1. File Type: pdf TUESDAY.pdf (62.0 KB, 82 views)
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    (Original post by Hasufel)
    i get these:
    Thanks mate
 
 
 
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