The expansion of f(x,y), is
Posted: Thu Jul 14, 2022 9:29 am
a) f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+….\)
b) f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…\)
c) f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}-y^2 \frac{∂^2 f}{∂y^2}]+…\)
d) f(0,0)-\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]-…\)
b) f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…\)
c) f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}-y^2 \frac{∂^2 f}{∂y^2}]+…\)
d) f(0,0)-\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]-…\)