Question

Compute the Taylor expansion of order n=2 of the function sin(xy) at x=0 and y=0

Answer 1

Answer:

f(x, y) = Sin(x*y)

We want the second order taylor expansion around x = 0, y = 0.

This will be:

$f(x,y) = f(0,0) + \frac{df(0,0)}{dx} x + \frac{df(0,0)}{dy} y + \frac{1}{2} \frac{d^2f(0,0)}{dx^2} x^2 +\frac{1}{2} \frac{d^2f(0,0)}{dy^2}y^2 + \frac{d^2f(0,0)}{dydx} x*y$

So let's find all the terms:

Remember that:

$\frac{dsin(ax)}{dx} = a*cos(ax)$

$\frac{dcos(ax)}{dx} = -a*cos(ax)$

f(0,0) = sin(0*0) = 1.

$\frac{df(0,0)}{dx}*x = y*cos(0*0)*x = x*y$

$\frac{df(0,0)}{dy} *y = x*cos(00)*y = x*y$

$\frac{1}{2} \frac{d^2f(0,0)}{dx^2}*x^2 = -\frac{1}{2} *y^2*sin(0*0)*x^2 = 0$

$\frac{1}{2} \frac{d^2f(0,0)}{dy^2}*y^2 = -\frac{1}{2} *x^2*sin(0*0)*y^2 = 0$

$\frac{d^2f(0,0)}{dxdy} x*y = (cos(0*0) -x*y*sin(0*0))*x*y = x*y$

Then we have that the taylor expansion of second order around x = 0 and y = 0 is:

sin(x,y) = x*y + x*y + x*y = 3*x*y