Jtdylktuy

Give the forth order taylor polynomial for the function $f(x, y) = cos(xy)$ around the point $(x, y) = (0,0)$

$f(x, y) = cos(x, y)|_{0,0} = 1$

$f_{x}(x, y) = -xsin(xy)|_{0,0} = 0$

$f_{y}(x, y) = -xsin(x, y)|_{0,0} = 0$

$f_{xx}(x, y) = -x^2cos(x, y)|_{0,0} = 0$

$f_{yy}(x, y) = -x^2cos(x, y)|_{0,0} = 0$

$f_{xy} = -[y^2cos(xy) + sin(xy)]|_{0,0} = 0$

$f_{xxx} = y^3(sin(x, y))|_{0,0} = 0$

$f_{xxy} = [-y^3sin(xy) + 2ycos(xy)]|_{0,0} = 0$

$f_{xyy} = [-y^3sin(xy) + 2ycos(xy) + cos(xy)]|_{0,0} = 0$

$f_{xxxx} = y^4cos(x, y)|_{0,0} = 0$

$f_{yyyy} = x^4cos(x, y)|_{0,0} = 0$

$f_{xxxy} = [y^4cos(x, y) + 3y^2sin(xy)]|_{0,0} = 0$

$f_{xxyy} = [-(y^4cos(xy) + 3y^2sin(xy))]|_{0,0} = -2$

$f_{xyyy} = -[-(y^4cos(xy) + 3y^2sin(xy)) + 3[-y^2sin(xy) + cos(xy)]]|_{0,0} = -3$

using the equation

$f(x, y) = 1 + [0 + 0] + 1/2 [0 + 0 + 0] + 1/6 [0 + 0 + 0] + 1/24 [0 + 0] + 6x^2y^2(-2) + 4xy^5 -3 + 0$

$f(x, y) = 1 + 1/24(-12x^2y^2 - 12xy^3) + cdots$

$= 1 - 1/2(x^2y^2 - xy^3) + cdots$

is this right?

Tried a simpler approach:

$f(xy) = cos(xy)$ let $xy = t$

$f(t) = 1 - frac{t^2}{2} + frac{t^4}{4} - frac{t^6}{6}$

$f(x, y) = 1 - frac{x^2y^2}{2} + frac{x^4y^4}{4} + cdots$

4th order polynomial is

$p_4(x) = 1 - frac{x^2y^2}{2} + frac{x^4y^4}{2}$