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Newton's Methodstuck on a simple problem

  1. Apr 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Find x(sub3) when x(sub1) is 3 and x^3+3x-5 = 0
    Give your answer to 4 correct decimal places.

    2. Relevant equations
    Mean Value Theoren Equation with Newton's Method.

    x(sub n+1) = x(sub n) - [(f(x)/f'(x)]


    3. The attempt at a solution

    Okay, so we have to find x(sub2) before x(sub3).

    x(sub2) = 3 - [(3^3 + 3(3) - 5)/(3(3)^2 + 3)

    I get x(sub 2) to be ~ 1.966666667

    Substituting that in the equation, I keep calculating x(sub3) to be ~1.010865585, which would be 1.0109 when rounded to 4 decimal places.

    The homework program (WebAssign) that I am using is telling me that answer is wrong. I thought it might be a rounding error, so I tried 1.0108 and 1.0110, both of which were also wrong.

    I don't know what I am doing wrong. I have solved this the same way as the ones prior to it, and they came out correct.

    Thanks for any and all help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 7, 2008 #2

    HallsofIvy

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    Okay, that's what I get.

    That's not what I get. Check your arithmetic again. It's more like 2.5.

     
  4. Apr 7, 2008 #3
    Halls' - can you check that second iterate again? - I get something more like 1.38
     
  5. Apr 7, 2008 #4

    Integral

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    I get 1.38 for x3 also.
     
  6. Apr 7, 2008 #5

    HallsofIvy

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    Oh, of course, I subtracted from the original 3 instead of x1! Still be a good idea to check his arithmetic though, wouldn't it?
     
  7. Apr 7, 2008 #6

    Integral

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    Yep, he has a mistake, it is not ~1.01
     
  8. Apr 7, 2008 #7
    Yes, absolutely.
     
  9. Apr 7, 2008 #8
    Thanks everybody. It did turn out to be ~1.38...

    Not sure what I was doing wrong before, but oh well. I guess I will just have to be more careful when test day comes.

    thanks again.
     
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