expand function f(x,y)=x2 + xy +y2 in powers of (x-2) and (y-3)

Question

i just want solution.

Answer 1

(x-2)²=x²-4x+4; (y-3)²=y²-6y+9; (x-2)(y-3)=xy-3x-2y+6.

Let x²+xy+y²=A(y²-6y+9)+B(x²-4x+4)+C(xy-3x-2y+6)+D(x-2)+E(y-3)+F, then we need to match the coefficients of x², x, y², y, xy and constant.

x²: B=1

y²: A=1

x: -4B-3C+D=0

y: -6A-2C+E=0

xy: C=1

constant: 9A+4B+6C-2D-3E+F=0.

Since we know A=B=C=1, substitute in the other equations:

x: -4-3+D=0, D=7

y: -6-2+E=0, E=8

constant: 9+4+6-14-24+F=0, F=19.

f(x,y)=(y²-6y+9)+(x²-4x+4)+(xy-3x-2y+6)+7(x-2)+8(y-3)+19=

x²+y²+(-4-3+7)x+(-6-2+8)y+xy+(9+4+6-14-24+19)=x²+y²+xy.

So f(x-2,y-3)=(x-2)²+(y-3)²+(x-2)(y-3)+7(x-2)+8(y-3)+19.

f(x,y)=x²+y²+xy+7x+8y+19.