Question

I don't understand this unit on Newton Method:
x^3 - 6x^2 + 10x - 4 = 0
Root between 0 and 1. Take X1 = 1.5 as initial guess.
Where do I begin?

Answer 1

you can guess as x=0 and then iterate

Answer 2

In the question he's just asking for f(x) and f'(x)

Answer 3

\[f(x) = x ^{3} - 6x^{2} + 10x - 4\]
\[f'(x) = 3x ^{2} - 12x + 10\]
Is this correct?

Answer 4

Should I sub in the 1.5?

Answer 5

the method says

\[x _{1} = x _{0} - f(x _{0})/f'(x _{0})\]

in your question

\[x _{0} = 1.5, f(1.5) = 0.875, f'(1.5) = -1.25\]

so the 1st approximation is

\[x _{1} = 1.5 - 0.875/-1.25\]

\[x _{1} =2.2\]

the only problem with the question is that the polynomial has 2 roots

1 between 0 and 1 while the other is at x = 2.

Because of the initial estimate the 1st interation is at 2.2

so you will converge on the root at x = 2

To find the root between 0 and 1... start with a value between 0 and 1... say 0.5 and use the above formula.

Answer 6

Your f(1.5) and f'(1.5) are different from what I got...

Answer 7

f(r) = (4/3)(pi)(r^3) + 4(pi)(r^2) - 50
f'(r) = (4)(pi)(r^2) + 8(pi)(r)
For now I just need to know if that's the correct equation for each, because if so then I think I can handle the subbing in of each new r and using that in the equation. But if those are wrong then I'm gonna waste a lot of time.

Answer 8

the equation for this question is

f(x) = x^3 - 6x^2 + 10x -4

you seem to be working on a different question

Answer 9

Oh wow lmfao, my mistake it's very late. The question I was working on is this:
http://openstudy.com/study#/updates/4f55b88be4b0862cfd070d63

Answer 10

Lemme look this one over now though.

Answer 11

the differentiation is correct...

Answer 12

okay lol thank you :D I'll do this one first I think it's a bit easier.

Answer 13

I think the problem with this question is with the initial estimate of the root.

Answer 14

Our teacher is telling us to use that. However, our teacher is one of those madman professors and not very good. Our entire class complains about his teaching methods and how he constantly interrupts and doesn't explain well. So it's possible he's giving us poor instructions.

Answer 15

well if you use 1.5 with 2 iterations you will get close to 2... and not the desired root between 0 and 1

Answer 16

:\ So our teacher is giving us a heck of a time, would you say?

Answer 17

If so, I might just do the first few iterations and write a note saying I refuse to continue as I've demonstrated the ability to find the desired root but doing so would be too tedious and time consuming. Sound legit?

Answer 18

Anyways, when I do my second iteration do I do 2.2 - (0.875/-1.25) or start all over for new f(2.2) f'(2.2)?

Answer 19

it will be

\[x _{2} = 2.2 - f(2.2)/f'(2.2)\]

Answer 20

then \[x _{2} = 1.99149\]

Answer 21

Oh snap, that's so tedious :(

Answer 22

Uhh, so apparently if I use 1.5 as my initial guess I just get stuck on 2 as my answer.

Answer 23

Oh right, that's what you said. I'll discuss this with my teacher.

Answer 24

Thanks for your help, sorry if I derped a bit, it's been a rough night.

Answer 25

thats ok... on early evening here in Australia

Answer 26

3:58am here :D Anyways yeah even if I kept 4 decimal places, I still ended up getting 2, so I'm going to take it up with the teacher (and by take it up I mean give him an earful, he truly should not be teaching, although I don't have the heart to take it to higher ups because I'm worried he'll lose his job). Lol just had to get that out there, thanks for your help, again.