Problem Involving Two-variable Taylor's Series

[s3a]

Homework Statement

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.)

Apparently, the correct answer is:
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) ] + . . .

Homework Equations

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

The Attempt at a Solution

f(x,y) = e^-x2 sin(y)

f_x = -2x e^-x2 sin(y)
f_y = e^-x2 cos(y)
f_xx = -2 sin(y) e^-x2 + 4x² sin(y) e^-x2
f_yy = -e^-x2 sin(y)
f_xy = -2x e^-x2 cos(y)

f_x(1,3) = -2 e^-1 sin(3)
f_y(1,3) = e^-1 cos(3)
f_xx(1,3) = 2 sin(3) e^-1
f_yy(1,3) = -e^-1 sin(3)
f_xy(1,3) = -2 * e^-1 cos(3)

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

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

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

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

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

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

f_xy(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.

Could someone please help me figure out what it is that I am doing wrong?

 

[Ray Vickson]

Homework Statement

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.)

Apparently, the correct answer is:
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) ] + . . .

Homework Equations

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

The Attempt at a Solution

f(x,y) = e^-x2 sin(y)

f_x = -2x e^-x2 sin(y)
f_y = e^-x2 cos(y)
f_xx = -2 sin(y) e^-x2 + 4x² sin(y) e^-x2
f_yy = -e^-x2 sin(y)
f_xy = -2x e^-x2 cos(y)

f_x(1,3) = -2 e^-1 sin(3)
f_y(1,3) = e^-1 cos(3)
f_xx(1,3) = 2 sin(3) e^-1
f_yy(1,3) = -e^-1 sin(3)
f_xy(1,3) = -2 * e^-1 cos(3)

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

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

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

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

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

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

f_xy(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.

Could someone please help me figure out what it is that I am doing wrong?

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

 

[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 f_xx(x,y) to f_xx(1,3), and then to the final answer.

Thanks!