Question

The expansion of f(x, y) = e^x ln(1 + y), is

(a) f(x,y) = y + xy – ^y^2⁄2 +…….

(b) f(x,y) = y – xy + ^y^2⁄2 -…….

(c) f(x,y) = y + x – ^y^2⁄2 +……..

(d) f(x,y) = x + y – ^x^2⁄2 +……..

I had been asked this question in a national level competition.

The question is from Taylor Mclaurin Series in portion Differential Calculus of Engineering Mathematics

Answer 1

Right option is (a) f(x,y) = y + xy – ^y^2⁄2 +…….

For explanation: Now, f(x, y) = e^x ln(1 + y) , f(0,0) = 0

Therefore,

\(f_x (x,y)=e^x ln(1+y)\), hence fx (0,0) = 0

\(f_y (x,y)=\frac{e^x}{(1+y)}\), hence fy (0,0) = 1

\(f_{xx} (x,y)=e^x ln(1+y)\), hence fxx (0,0) = 0

\(f_{yy} (x,y)=-\frac{e^x}{(1+y)^2}\), hence fyy (0,0) = -1

\(f_{xy} (x,y)=\frac{e^x}{(1+y)}\), hence fxy (0,0) = 1

By taylor expansion,

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

f(x,y) = y + xy – ^y^2⁄2 +…….