Question

Given f (x,y)=e^x cos⁡y, what is the value of the fifth term in Taylor’s series near (1,\(\frac{π}{4}\)) where it is expanded in increasing order of degree & by following algebraic identity rule?

(a) \(\frac{-e(x-1)(y-\frac{π}{4})}{\sqrt{2}}\)

(b) \(-\sqrt{2}  e(x-1)(y-\frac{π}{4})\)

(c) \(\frac{e(x-1)^2}{\sqrt{2}}\)

(d) \(\frac{e(y-\frac{π}{4})^2}{\sqrt{2}}\)

The question was asked during a job interview.

My question is taken from Taylor’s Theorem Two Variables topic in portion Maxima and Minima of Engineering Mathematics

Answer 1

Correct option is (a) \(\frac{-e(x-1)(y-\frac{π}{4})}{\sqrt{2}}\)

Best explanation: Taylor’s series expansion is given by

\(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\

+2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)\)

Thus fifth term is given by \(2 \frac{(x-a)(x-b)}{2!} f_{xy} (a,b)\)..(1) where a=1, b=π/4 &                                          \(f_{xy}=\frac{∂}{∂x}(\frac{∂f(x,y)}{∂x}) = \frac{∂}{∂x}(\frac{∂e^x  cos⁡y}{∂y}) =- e^x \,sin⁡y \) at (1,  \(\frac{π}{4}), f_{xy}=\frac{-e}{\sqrt{2}}\) substituting in (1)

We get fifth term as \(2 \frac{(x-1)(x-π/4)}{2!} \frac{-e}{\sqrt{2}} = \frac{-e(x-1)(y-\frac{π}{4})}{\sqrt{2}}\).