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dedylja [7]
2 years ago
14

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

Mathematics
1 answer:
artcher [175]2 years ago
6 0

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

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