Question

Expand ln(x) in the power of (x-m).

(a) ln⁡(m)+\(\frac{h}{m}-\frac{1}{2!} (h/m)^2+\frac{2}{3!} (h/m)^3-……\)

(b) ln⁡(m)-\(\frac{h}{m}-\frac{1}{2!} (h/m)^2-\frac{2}{3!} (h/m)^3-……\)

(c) ln⁡(m)-\(\frac{1}{2!} (h/m)^2+\frac{2}{3!} (h/m)^4-……\)

(d) ln⁡(m)+\(\frac{h}{m}+\frac{2}{3!} (h/m)^3-……\)

I have been asked this question in an interview.

I would like to ask this question from Taylor Mclaurin Series topic in portion Differential Calculus of Engineering Mathematics

Answer 1

Right option is (a) ln⁡(m)+\(\frac{h}{m}-\frac{1}{2!} (h/m)^2+\frac{2}{3!} (h/m)^3-……\)

Easiest explanation: where, h = x-m

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

now,f(m) = ln(m), f’(m)=1/m, f” (m)=-1/m^2, f”’ (m)=2/m^3,……

hence,

f(x)=ln(x)=f(h+m)=\(ln⁡(m)+\frac{h}{m}-\frac{h^2}{2!}\frac{1}{m^2}+\frac{h^3}{3!} \frac{2}{m^3}-……\)

f(x)=ln(x)=ln⁡(m)+\(\frac{h}{m}-\frac{1}{2!} (h/m)^2+\frac{2}{3!} (h/m)^3-……\)

where, h = x-m