Question

Find the expansion of e^x in terms of x + m, m > 0.

(a) \(e^m [1+(x+m)+\frac{(x+m)^2}{2!}+\frac{(x+m)^3}{3!}+….]\)

(b) \(e^{-m} [1+(x-m)+\frac{(x-m)^2}{2!}+\frac{(x-m)^3}{3!}+….]\)

(c) \(e^m [1+(x-m)+\frac{(x-m)^2}{2!}+\frac{(x-m)^3}{3!}+….]\)

(d) \(e^{-m} [1+(x+m)+\frac{(x+m)^2}{2!}+\frac{(x+m)^3}{3!}+….]\)

I got this question during an interview for a job.

This intriguing question comes from Taylor Mclaurin Series topic in division Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (d) \(e^{-m} [1+(x+m)+\frac{(x+m)^2}{2!}+\frac{(x+m)^3}{3!}+….]\)

For explanation: Let, h = x + m = > f(x) = f(h-m) = e^(h-m)

By taylor theorem, putting a = -m, we get,

f(a+h) = \(f(a)+\frac{h}{1!} f’ (a)+\frac{h^2}{2!} f” (a)…\)

f(h-m) = \(f(-m)+h/1! f'(-m)+\frac{h^2}{2!} f” (-m)….(1)\)

now, f(-m) = f’(-m) = f’’(-m)=e^-m

hence,

f(x)=e^x=f(h-m)=\(e^{-m} [1+(x+m)+\frac{(x+m)^2}{2!}+\frac{(x+m)^3}{3!}+….]\)