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Taylor Series Approximation Help

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data

    Use the "Three Term" Taylor's approximation to find approximate values y_1 through y_20 with h=.1 for this Initial Value Problem:
    y'= cosh(4x^2-2y^2)

    And write a computer program to do the grunt work approximation

    2. Relevant equations

    3. The attempt at a solution

    y''= (cosh(4x^2-2y^2)(4/3x^3-2/3y^3)
    I'm pretty sure I'm doing this wrong, and I also have no idea how to go about writing a computer program to do this work
  2. jcsd
  3. Oct 31, 2007 #2
    In the original problem statement, is that the second derivative, ie,

    y'' = cosh(4x^2-2y^2)

    or is it the first derivative?
  4. Oct 31, 2007 #3
    wow, my professor changed the IVP to:
    y'=xy^3 - cos(x)sin(y)
    y(0)= -1
    still using h= .1 and looking for y_1 to y_20, as well as writing the program. Apparently the original blew up too quickly.
    Last edited: Oct 31, 2007
  5. Nov 1, 2007 #4
    OK. Now to start, what were the various expressions that your prof gave you for approximations to first derivatives. Hint: a two term approx might be

    [tex]y'_{+} = \frac{y_{n+1}-y_{n}}{h} + O(h)[/tex]

    then again, it might be

    [tex]y'_{-} = \frac{y_{n}-y_{n-1}}{h} + O(h)[/tex]

    Do you remember how your prof established the three point approx?
  6. Nov 1, 2007 #5
    Thanks for the help Theo, but my problem now lies in the programming bit of the problem. I think I've got the steps before it figured out. Please correct me if I'm wrong.

    y''= x3(y^2)y' + y^3 -[(-sin(x)siny) + (cos(x)cos(y)y')]
    = 3x(y^2)y' + y^3 +[cos(x)cos(y)y' - sin(x)sin(y)]
    y'= (0)(-1)^3- [cos(0)sin(-1)] = sin(-1)
    y''= 3(0) + (-1)^3 = [(cos(0)cos(-1)sin(-1)) - (sin(0)sin(-1))]
    = -1 + sin(-1)
    y_1= (x_0-y_0) = 0-(-1) = 1
    y_2= y_1 + (x_1+y_1)h + (x_1+y_1+1)h^2/2
    = 1 + (.1+1)(.1) + (.1+1+1).005 = 1.1205
    y_3=y_2 + (x_2+y_2)h + (x_2+y_2+1)h^2/2
    =1.1205 +(.2+1.1205)(.1) + (.2+1.1205+1).005 = 1.2641525

    You see, now I have no idea how to go about writing a program for this approximation. What sort of program do I use, and where can I find it? Can I just download one? Is it very complicated to program this?
    Last edited: Nov 2, 2007
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