Question

Calculate second-degree Taylor's approximation for f(x,y)=2ysin(4x), at the point (π/2,1)

Answer 1

\[f \left(a,b \right) \approx f \left(0,0 \right)+\frac{\partial f}{\partial x}(0,0)(x-a)+\frac{\partial f}{\partial y}(0,0)(y-b)+2\frac{\partial }{\partial x} \left(\frac{\partial f}{\partial y}\right)(x-a)(y-b)\]
\[+2\frac{\partial ^2 f}{\partial x^2}(0,0)(x-a)^2+2\frac{\partial ^2 f}{\partial y^2}(0,0)(y-a)^2\]

Answer 2

Sorry, all the \[(0,0)\] should be \[(a,b)=(\frac{\pi}{2},1) \]

Answer 3

\[ \frac{\partial f}{\partial x}(\frac{\pi}{2},1)=8(1)\sin(4\cdot \frac{\pi}{2})=8(1)\sin(2 \pi)=0\]

Answer 4

Continue on like that, and then substitute into the first equation (remembering to ignore my error).

Answer 5

Thanks a lot ! But \[df/dx = 8y*\cos(4x) => df/dx (π/2,1)=8 \]

Answer 6

Of course, sorry.