Question

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

Answer 1

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) + (df(0,0))/(dx) x + (df(0,0))/(dy) y + (1)/(2) (d^2f(0,0))/(dx^2) x^2 +(1)/(2) (d^2f(0,0))/(dy^2)y^2 + (d^2f(0,0))/(dydx) x*y`

So let's find all the terms:

Remember that:

`(dsin(ax))/(dx) = a*cos(ax)`

`(dcos(ax))/(dx) = -a*cos(ax)`

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

`(df(0,0))/(dx)*x = y*cos(0*0)*x = x*y`

`(df(0,0))/(dy) *y = x*cos(00)*y = x*y`

`(1)/(2) (d^2f(0,0))/(dx^2)*x^2 = -(1)/(2) *y^2*sin(0*0)*x^2 = 0`

`(1)/(2) (d^2f(0,0))/(dy^2)*y^2 = -(1)/(2) *x^2*sin(0*0)*y^2 = 0`

`(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