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Need help with a nooby question

  1. Sep 18, 2006 #1
    alright there is this question which uses NR method to find intersect but i kept getting the wrong ans can anybody help me check which part i went wrong?:biggrin:

    f(x)=2x + ln4x , f'(x)=2+ 1/x

    btw,differentiate ln4x and lnx get the same result but the arbitary constant is different right?so do we consider the constant when finding the intersect?
     
    Last edited: Sep 18, 2006
  2. jcsd
  3. Sep 18, 2006 #2

    radou

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    If you integrate 1/x, you get lnx + C. ln4x = ln4 + lnx, so C = ln4 in this case.
     
  4. Sep 18, 2006 #3
    hmm...i see now so the constant changes although theres 1/x in both cases.

    regarding the question, if i were to find the 2nd approx. , i am suppose to use 0.1 as 1st approx. however, 0.1-f(0.1)/f'(0.1) is not the ans:confused:
     
  5. Sep 19, 2006 #4

    HallsofIvy

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    Why not? It won't be the exact solution to the equation, of course, but it certainly is a "second approximation" with first approximation 0.1. I get 0.16 as the second approximation.

    By the way, the reason ln(4x) and ln(x) have the same derivative (1/x) is that ln(4x)= ln(x)+ ln(4).
     
  6. Sep 19, 2006 #5
    hmm....i also get 0.16 guess the ans is wrong :biggrin: Thanks
     
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