Taylor series with 3 variables

[ozharu]

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

Homework Statement

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)

Homework Equations

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

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 *\frac{f'(f)}{f'(x)} + dy *\frac{f'(f)}{f'(y)} + dz *\frac{f'(f)}{f'(z)} + dxdy *\frac{f''(f)}{f'(x)f'(y) } + dxdz *\frac{f'(f)}{f''(x)f'(z)} + dydz *\frac{f''(f)}{f'(y)f'(z)} + dxdydz \frac{f'''(f)}{f'(x)f'(y)f'(z)}

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

 

[DJ HAN]

sin (.3) is not small enough to be approximated by first order.