Question

The expansion of f(x,y), is

(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)-\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]-…\)

This question was posed to me by my college professor while I was bunking the class.

My question comes from Taylor Mclaurin Series in division Differential Calculus of Engineering Mathematics

Answer 1

Right option is (b) f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…\)

The best explanation: By taylor expansion,

f(x,y) = f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…\)