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Using interpolants to solve a polynomial.

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that a root of the equation x3 - 3x - 5 =0 lies in the interval [2,3], and then find the root using linear interpolation correct to one decimal place.


    2. Relevant equations
    n/a

    3. The attempt at a solution
    This is my first ever time using interpolants ( well at least in the sense of solving a polynomial. I think I've used them before... but instinctively.)

    This is what I did:
    3 [tex]\div[/tex] 13 = (x1 - 2) [tex]\div[/tex] (3-x1)
    to end up with x1 = 2.1875.
    I went on to get the third approximation only to find out that it was outside the range (less than 2).
    I checked what the solution said, and it said:
    3 [tex]\div[/tex] 13 = (3-x1) [tex]\div[/tex] (x1 - 2)

    I'm utterly confused. Please help.
     
  2. jcsd
  3. Sep 14, 2010 #2

    LCKurtz

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    The first thing your problem asked was to show that there was a root of your function between x = 2 and x = 3. Did you do that? And how did you do it?

    Your x1 looks OK, and it is between 2 and 3. Did you check whether you got lucky and x1 is your root? If it isn't, then how do you know whether the root is in [2, x1] or in [x1, 3]? You have to know that in order to know which interval to do the next step with.
     
  4. Sep 14, 2010 #3
    Oh. I forgot to mention that I'd got the first part. Yes. The root is in between 2 and 3.
    I just substituted "x" with 3 and 2 and saw that f(3) / f(2) = -s. i.e., the sign changed, so there had to have had been a number which had f(x) = 0, i.e., the root of the equation.
    My first approximation is different from what the book says, the same with all my further approximations. I thought that it might have been a typo, but then I couldn't do the other questions either, and in the end I checked to see if the answers that the book gave worked, and they did. So theres got to be something wrong in what I'm doing.
     
  5. Sep 14, 2010 #4

    LCKurtz

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    Your first approximation for x1 of 35/16=2.1875 is correct. The value of the function at that point is negative so you want to use x = 2.1875 and x = 3 for your calculation of x2. I get 2.250619230 for it.
     
  6. Sep 15, 2010 #5
    Ok. I'm getting the same thing.
    I understand how this works now. But I'm still having trouble believing that the publishers messed up so bad. Can you just have a look at what they did. I'm really tired right now, and I wont be able to use the book for a week or two so I wont be able to look at it later.
     

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