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Taylor series homework check

  1. Aug 30, 2011 #1
    1. The problem statement, all variables and given/known data
    I posted this already but decided to revive this thread since I re-worked the problem.

    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.
    Do the calculations for only one step.

    2. Relevant equations
    [itex]\epsilon[/itex]=O([itex]\Delta[/itex]x5)

    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))+
    \Delta[/itex]x21/2!f''(x0,y(x0))+[itex]\Delta[/itex]x31/3!f'''(x0,y(x0))+[itex]\Delta[/itex]x41/4!f''''(x0,y(x0))+[itex]\epsilon[/itex]+[itex]

    Is this correct?

    My solution:

    The derivatives:
    f'(x,y)=dy/dx=y+x=1+0=1
    f''(x,y)=d2y/dx2=dy/dx+1=1+1=2
    f'''(x,y)=d3y/dx3=dy2/dx2=2
    f''''(x,y)=d4y/dx4=d3y/dx3=2

    y(0+.1)=1+(.1)(1)+1/2!(.1)2(2)+1/3!(.1)3(2)+1/4!(.1)4(2)+(.1)5

    y(.1)=1+.1+.01+.001/3+.0001/12+.00001=1.11035

    Did I solve this correctly? I want to be able to have something decent when I meet with the professor tomorrow.
     
  2. jcsd
  3. Aug 30, 2011 #2

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Everything looks right. I am not willing to double check all your arithmetic.
     
  4. Aug 30, 2011 #3
    That's fine. I just wanted to know if I was on the right track. Thanks for taking a look.
     
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