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Taylor series with 3 variables

  1. Apr 5, 2008 #1
    Hi am trying to solve this Taylor series with 3 variables but my result is not equal to the solution- So i think i might be wrong expanding the taylor series, or the solution is not correct

    1. The problem statement, all variables and given/known data
    Find an a approximated value for the function f(x,y,z) = 2x + ( 1 + y) * sin z at the point
    x= 0.1 y = 0.2 z = 0.3 [asnwer = 0.5546]

    Use only linear terms ( dx dy dz has only 1 power)


    2. Relevant equations

    df = dx*[tex]\frac{f'(f)}{f'(x)}[/tex] + dy*[tex]\frac{f'(f)}{f'(y)}[/tex] + dz *[tex]\frac{f'(f)}{f'(z)}[/tex] + ...

    3. The attempt at a solution
    So i started evaluating the function at the point (0,0,0) which gives me the value 0. Then i expand the taylor series like this

    df = dx *[tex]\frac{f'(f)}{f'(x)}[/tex] + dy *[tex]\frac{f'(f)}{f'(y)}[/tex] + dz *[tex]\frac{f'(f)}{f'(z)}[/tex] + dxdy *[tex]\frac{f''(f)}{f'(x)f'(y) }[/tex] + dxdz *[tex]\frac{f'(f)}{f''(x)f'(z)}[/tex] + dydz *[tex]\frac{f''(f)}{f'(y)f'(z)}[/tex] + dxdydz [tex]\frac{f'''(f)}{f'(x)f'(y)f'(z)}[/tex]

    Then i calculate the differential equations
    [tex]\frac{f'(f)}{f'(x)}[/tex] = 2
    [tex]\frac{f'(f)}{f'(y)}[/tex] = sin z
    [tex]\frac{f'(f)}{f'(z)}[/tex] = (1 + y) cos z
    [tex]\frac{f''(f)}{f'(x)f'(y) }[/tex] = 0
    [tex]\frac{f'(f)}{f''(x)f'(z)}[/tex] = 0
    [tex]\frac{f''(f)}{f'(y)f'(z)}[/tex] = cos z
    [tex]\frac{f'''(f)}{f'(x)f'(y)f'(z)}[/tex] = 0

    So my final formula is

    df = 2 dx + dz * cos(0) + dydz * cos (0)

    An evaluating that formula in (0.1,0.2,0.3)

    df(0.1,0.2,0.3) = 2 *0.1 + 0.3 + 0.06 = 0.56

    But the solution given by the exercise is 0.5546
     
    Last edited: Apr 5, 2008
  2. jcsd
  3. Mar 9, 2009 #2
    sin (.3) is not small enough to be approximated by first order.
     
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