Question

Find the expansion of f(x) = ln⁡(1+e^x)?

(a) \(ln(2)+x/2+x^2/8-x^4/192+….\)

(b) \(ln⁡(2)+x/2+x^2/8+x^4/192+….\)

(c) \(ln⁡(2)+x/2+x^3/8-x^5/192+….\)

(d) \(ln⁡(2)+x/2+x^3/8+x^5/192+….\)

The question was asked at a job interview.

Asked question is from Taylor Mclaurin Series topic in chapter Differential Calculus of Engineering Mathematics

Answer 1

Right option is (a) \(ln(2)+x/2+x^2/8-x^4/192+….\)

To explain: Given, f(x) = ln⁡(1+e^x), f(0) = ln⁡(2)

Differentiating it we get

\(f{‘}(x)=\frac{e^{x}}{1+e^x}=1-1/(1+e^x), f{‘}(0)=1/2\)

Again differentiating we get

\(f”(x)=e^x/(1+e^x)^2 =(f'(x))/(1+e^x),f”(0)=1/4\)

\(f”'(0)=((1+e^x) f”(x)-f'(x) e^x)/(1+e^x)^2 =(f”(x))/(1+e^x)-f'(x) f”(x)\), hence f”'(0)=0

\(f””(0)=(1(1+e^x) f”'(x)-f”(x) e^x)/(1+e^x)^2 -(f”(x))^2-f'(x) f”'(x)\), hence f””(0)=1/8

Hence, by mclaurin’s series,

\(f(x)=ln⁡(1+e^x)=ln⁡(2)+x/2+x^2/8-x^4/192+…\)