Question

If y = cos^-1x, then

(a) (yn+2)(0) = -n(2n-1) yn(0)

(b) (yn+2)(0) = n(2n-1) yn(0)

(c) (yn+2)(0) = n(n-2) yn(0)

(d) (yn+2)(0) = n(n-3) yn(0)

This question was addressed to me in a job interview.

This intriguing question originated from Leibniz Rule in chapter Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (b) (yn+2)(0) = n(2n-1) yn(0)

To elaborate: y = cos^-1x

Differentiating it

Y1 = \(\frac{1}{\sqrt{1-x^2}}\)

(1-x^2)(y1)^2= 1

Again Differentiating we get

(1-x^2)2y1y2 – 2x(y1)^2 = 0

(1-x^2)y2 = xy1

By Leibniz Rule, Diff it n times,

(1-x^2)yn+2 – 2xnyn+1 – 2n(n-1)yn = xyn+1 + nyn

(1-x^2) yn+2 – xyn+1(2n-1) = nyn(2(n-1)+1)

(1-x^2) yn+2 – xyn+1(2n-1) = nyn(2n-1)

At x=0, we get

 (yn+2)(0) = n(2n-1) yn(0).