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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?

  • 3
    Your value for $f(3)$ is incorrect, I think. It should be 0.687, according to W|A? http://www.wolframalpha.com/input/?i=%CE%BB3%E2%88%923.250%CE%BB2%2B%CE%BB%E2%88%920.063+at+32012-08-16
  • 0
    I relabeled the two step2's so they can be referred to without confusion.2012-08-16
  • 1
    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 $2.9191765$.2012-08-16
  • 2
    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.2012-08-16
  • 0
    I had a calculator do this, until the precision was 110 digits http://pastebin.com/0U3Lu1FG2014-05-17

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