# Trying to solve $\lambda^3 - 3.250\lambda^2 + \lambda - 0.063 = 0$ using Newton-Raphson method

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
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
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 – Kristoffer Ryhl May 17 '14 at 13:16