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This is what I've atempted so far in solving $\lambda^3 - 3.250\lambda^2 + \lambda - 0.063 = 0$. The following are the steps:

step 1: $f(\lambda) = \lambda^3 - 3.250\lambda^2 + \lambda - 0.063 $

step 2a: $f(0) = -0.063$

step 2b: $f(1) = -1.313$

step 3: $f(2) = -4.063$

step 4: $f(3) = -17.313$

step 5: $f(4) = 14.937$

This shows that the value of the root is close to $\lambda = 3$. Now we use the iterative formula: $$r_{4} = r_{3} - \frac{f(r_{3})}{f^\prime(r_{3})}$$

where $f(r_{3}) = -17.313$ and $f^\prime(r_{3}) = 8.5$. Then we compute $$r_{4} = 3-\frac{(-17.313)}{8.5} = 5.0368$$

Question: Have I followed the preliminary steps correctly?

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Your value for $f(3)$ is incorrect, I think. It should be 0.687, according to W|A? wolframalpha.com/input/… –  James Fennell Aug 16 '12 at 11:05
    
I relabeled the two step2's so they can be referred to without confusion. –  rschwieb Aug 16 '12 at 12:31
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Small notational point. Your first estimate is $3$. So call it $r_0$ or $r_1$. As James Fennell points out, $f(3)=0.687$, so your next estimate is about $2.9191765$. –  André Nicolas Aug 16 '12 at 13:29
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Additional comment: Note that our function is positive at $\lambda=0.1$. So there is a root between $0$ and $0.1$, and another between $0.1$ and $1$. There is also a root near $3$. Where you start on your Newton Method calculation depends on which root you want to get with high accuracy. –  André Nicolas Aug 16 '12 at 15:00
    
I had a calculator do this, until the precision was 110 digits pastebin.com/0U3Lu1FG –  Darksonn May 17 at 13:16
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1 Answer 1

Your iteration seems to be incorrect with a guess of 3, I have calculated the root using newtons method to 110 digits here.
The next value is about 2.919 but your value is about 5.037.

Also the polynomial has two more roots at 0.24885... and 0.0868705...

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