Question

Expand (1 + x)^^1⁄x, gives ___________

(a) e[1 + ^x⁄2 + ^11x^2⁄24 -…..]

(b) e[1 – ^x⁄2 + ^11x^2⁄24 -…..]

(c) e[^x⁄2 – ^11x^2⁄24 -…..]

(d) e[^x⁄2 + ^11x^2⁄24 -…..]

I had been asked this question in homework.

This key question is from Taylor Mclaurin Series topic in division Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (b) e[1 – ^x⁄2 + ^11x^2⁄24 -…..]

To explain: Given, y = (1 + x)^^1⁄x

Hence,

ln⁡(y)=\(\frac{ln⁡(1+x)}{x}\)

Hence, ln⁡(y)=\(\frac{1}{x} [x-\frac{x^2}{2}+\frac{x^3}{3}-……]=1-\frac{x}{2}+\frac{x^2}{3}\)-……

Hence, \(y=e^{1-\frac{x}{2}+\frac{x^2}{3}-……}=e^{1+z}\), where, \(z=\frac{-x}{2}+\frac{x^2}{3}-……\)

Hence, \(y=e.e^z=e(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+…)\)

y=\(e[1+(\frac{-x}{2}+\frac{x^2}{3}-……)+\frac{1}{2!}(\frac{-x}{2}+\frac{x^2}{3}-…)^2+…]\)

y=\(e[1-x/2+x^2/e+x^2/8+…]\)

y = e[1 – ^x⁄2 + ^11x^2⁄24 -…..].