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Tayor series with truncation error

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider dy/dx=x+y, a function of both x and y subject to initial condition, y(x0)=y0.
    Use Taylor series to determine y(x0+[itex]\Delta[/itex]x) to 4th order accuracy.

    Initial condition: x0=0, y(x0)=1
    step size: [itex]\Delta[/itex]x=0.1

    Show 5 significant digits in the answer.

    2. Relevant equations
    [itex]\epsilon[/itex]=O([itex]\Delta[/itex]x5)
    Do the calculations for only one step.


    3. The attempt at a solution
    dy/dx=f(x,y)

    Taylor series:
    y(x0+[itex]\Delta[/itex]x)=y(x0)+[itex]\Delta[/itex]xf(x0,y(x0))+[itex]\epsilon[/itex]


    My solution:

    f(x0,y(0))=f(0,1)=0+1=1

    y(0+0.1)=1+0.1(1)+.00001=1.10001

    Does this seem correct. It feels like I missed something. On the other hand it makes sense. Did I miss something or mess up a step?
     
  2. jcsd
  3. Aug 27, 2011 #2

    Ray Vickson

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    To 4th order in h (= Delta x), y(x0 + h) = y(x0) + sum{D^k y(x0) * h^k/k!: k=1..4}, where D^k y(x0) = (d/dx)^k y(x)|_{x = x0} = kth derivative of y(x) at x0. Can you figure out (d/dx)^2 y, (d/dx)^3 y, etc?

    RGV
     
  4. Aug 27, 2011 #3
    I thought that doing calculations for one step meant that I didn't need to take any derivatives. For example, a calculation using 3 steps would mean the Taylor series would have the original function, 1st and 2nd derivatives. Wouldn't I just get zeros beyond the first derivative of x+y anyways?
     
  5. Aug 28, 2011 #4
    Can anyone provide some insight on this. This is review and I don't ever remember doing a Taylor series on a differential equation.
     
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