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# Newtons algorithm watch

1. 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
Thanks
2. (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
Thanks
Use the same iteration with starting points close to your other solutions
3. If the onlly info youre 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 youve already done.

The simpler iteration formula is given by

Viz:

Takes aroud 5 iterations if you start the negative root at the highest integer value (from your graph estimate), the positive one at its respective next highest (i.e. -1 and 2)
4. (Original post by Hasufel)
If the onlly info youre 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 youve already done.

The simpler iteration formula is given by

Viz:

Takes aroud 5 iterations if you start the negative root at the highest integer value (from your graph estimate), the positive one at its respective next highest (i.e. -1 and 2)
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
5. 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 theyre 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 thats too deep for here)

try x = -1...

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

(EDITED)
6. (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 theyre 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 thats too deep for here)

try x = -1...

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

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

Attached Images

7. i get these:
Attached Images
8. TUESDAY.pdf (62.0 KB, 82 views)
9. (Original post by Hasufel)
i get these:
Thanks mate

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