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math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

i need an example in which newtons method converge with order 3
please if some one know the answer ill be greatful
its in numerical analysis
thanx

outermeasure · **Posted:** Fri, 27 Mar 2009 09:28:51 UTC

math 1 wrote:

i need an example in which newtons method converge with order 3
please if some one know the answer ill be greatful
its in numerical analysis
thanx

Hint: something with f''(root)=0.

math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

can you give more details like an example

outermeasure · **Posted:** Fri, 27 Mar 2009 12:58:49 UTC

math 1 wrote:

can you give more details like an example

Please show some work.

math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

i thought of
f(x)=the therd root of any poly.
im not that smart or even have big imagenatin to think of a function that its second derivative and its first =0 and the therd is not ill be greatfull if you could help
thanx

robbwrr · **Posted:** Fri, 27 Mar 2009 14:55:38 UTC

math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

thank you i never thought it is that simple

outermeasure · **Posted:** Fri, 27 Mar 2009 15:07:01 UTC

robbwrr wrote:

$$ f(x) = (x-5)^3 :\ f'=3(x-5)^2, f''=6(x-5), f'''=6, f(5)=f'(5)=f''(5)=0$

... except it doesn't work for the original problem. I'll leave math 1 to figure out how to tweak it.

math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

ok ill think of it if it work and i found it ill tell you but if i didnt ill tell you too maby ill get some thing new

math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

how is this one
f(x)=(cosx-x)^3
am i right or its wrong

math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

thanx to both of your help

outermeasure · **Posted:** Fri, 27 Mar 2009 17:01:06 UTC

math 1 wrote:

how is this one
f(x)=(cosx-x)^3
am i right or its wrong

No, that won't work either. Read my first hint carefully --- there is a subtler hint there.

math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

have you tryed the second derivative on the root is zero if im not mestaken

math 1 · **Joined:** Fri, 27 Mar 2009 08:27:30 UTC **Posts:** 110

by newtons method the root of cosx-x is approximatly 0.739085......
so i think its right what about you what do you think
if it wrong then give me an example
thank you

outermeasure · **Posted:** Fri, 27 Mar 2009 17:36:19 UTC

OK, so spelling that out a bit more

Hint: I have said nothing about f'(root) --- there is a condition that you need to find out for yourself. Note what f'(root) is for robbwrr's and your suggestions.

Once you get that, the proof is just Taylor expand $\alpha+\epsilon-\dfrac{f(\alpha+\epsilon)}{f'(\alpha+\epsilon)}$ , where $\alpha$ is the root and $\lvert\epsilon\rvert\ll 1$

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