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Newtons method

  1. Nov 12, 2004 #1
    i have the answer to this problem i just see why i'm not getting the same answer as my solution manuel. i Have two functtions f(x) = x and g(x) = tan(x)
    and i have to find where these two functions are equal using newtons method.
    i subtracted the two functions to get this new function H(x) = x -tan(x) then i found the differential of this new function to be 1 - sec^2 (x). now by looking at the given graph of the first two functions i estimated the soulution to be about 4.1. then by unising newtons method i calculated two iterations and got an answer of 8.5987, which is wrong the answer is 4.493. what could i be doing wrong? i know how to use newtons method.
  2. jcsd
  3. Nov 12, 2004 #2
    You need to use a better first approximation.
  4. Nov 12, 2004 #3


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    Are you sure that the two functions are only equal in one place? x and tan x intersect in *lots* of places.
  5. Nov 12, 2004 #4
    x has to be between pi/2 and 3pi/2 these are the asmyspotes.
  6. Nov 13, 2004 #5


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    You need a real good first approximation, otherwise it never converges. If you know that the tangent of 3 pi/2 is undefined and start a little lower than that, you're okay. If your guess is less than around 4.29, you won't converge.
  7. Nov 13, 2004 #6
    thanks guys the last reply helped me out.
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