Question

A function f(x) which is continuous and differentiable over the real domain exists such that f^(n) (x) = [f^(n + 1) (x)]^2, f(0) = a and f^(1)(0) = 1.

(a) True

(b) False

I had been asked this question by my school principal while I was bunking the class.

My question is from Taylor Mclaurin Series in division Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (a) True

Easiest explanation: Writing out the Mclaurin series we have

f(x)=f(0)+\(\frac{f^{(1)}(0).x}{1!}+\frac{f^{(2)}(0).x^2}{2!}+…\infty\)

Now sustituting f^(n+1)(0) = \(\sqrt{f^{(n)}(0)}\)we have

f(x) = f(0) + \(\frac{f^{(1)}(0).x}{1!}+\frac{\sqrt{f^{(1)}(0)}.x^2}{2!}+\frac{\sqrt{\sqrt{f^{(1)}(0)}.x^3}}{3!}…\infty\)

Now f^(1)(0)=1 and f(0)=a

Substituting this we have

f(x)=a+\(\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+…\infty\)

f(x)=a-1+e^x

This is a well defined function.