# Problem Involving Two-variable Taylor's Series

1. Jul 7, 2017

### s3a

1. The problem statement, all variables and given/known data
Write the Taylor's series expansion of the function f(x,y) = e-x2 sin(y) about the point (1,3). (The question doesn't specify how accurate it wants the answer to be, but based on the answer I have, it seems to me that the Taylor's polynomial should be of degree 3.)

e-x2 sin(y) = e-1 sin(3) - 2(x-1) e-1 sin(3) + (y-3) e-1 cos(3) + 1/2 [ 2(x-1)2 e-1 sin(3) - 4(x-1)(y-3) e-1 cos(3) - (y-3)2 e-1 sin(3) ] + . . .

2. Relevant equations
I'm trying to do what the guy in this video is saying to do.:

3. The attempt at a solution
f(x,y) = e-x2 sin(y)

fx = -2x e-x2 sin(y)
fy = e-x2 cos(y)
fxx = -2 sin(y) e-x2 + 4x2 sin(y) e-x2
fyy = -e-x2 sin(y)
fxy = -2x e-x2 cos(y)

fx(1,3) = -2 e-1 sin(3)
fy(1,3) = e-1 cos(3)
fxx(1,3) = 2 sin(3) e-1
fyy(1,3) = -e-1 sin(3)
fxy(1,3) = -2 * e-1 cos(3)

=====================================

f(1,3):
e-1 sin(3)

fx(1,3):
1/1! * -2 e-1 sin(3) x1
= -2 e-1 sin(3) x

fy(1,3):
1/1! * e(-1) cos(3) y1
=e-1 cos(3) y

fxx(1,3):
1/2! sin(3) e-1 x2
=1/2 sin(3) e-1 x2

fyy(1,3):
1/2! * -e-1 sin(3) y2
=1/2 * -e-1 sin(3) y2

fxy(1,3):
1/1! * -2 * e-1 cos(3) x1 y1
=-2 * e-1 cos(3) x y

f(x,y) = sum of the above parts + ...
f(x,y) = [e-1 sin(3)] + [-2 e-1 sin(3) x] + [e-1 cos(3) y] + [1/2 sin(3) e-1 x2] + [1/2 * -e-1 sin(3) y2] + [-2 * e-1 cos(3) x y] + ...

However, checking with software, my answer is not the same as the answer that's apparently the correct answer.

2. Jul 7, 2017

### Ray Vickson

Your series should be about $x=1$ and $y=3$, so must be expressed in powers and products of $x-1$ and $y-3$.

3. Jul 7, 2017

### s3a

Oh, right!

That was the biggest problem I had; for what it's worth, another smaller mistake I made was that I had forgotten to bring the factor of 2 from fxx(x,y) to fxx(1,3), and then to the final answer.

Thanks!