Answer to Question #345830 in Calculus for Roy

Question

if f(x) = sin^-1x , show that ( 1-x²) f^'' (x) -x f^'(x) = 0 . Hence prove that , f⁽ⁿ⁺²⁾(0) = n² fⁿ (0)

Answer 1

$$f(x)=\sin^{-1}(x)$$

$$f'(x)=\dfrac{1}{\sqrt{1-x^2}}$$

$$f''(x)=-\dfrac{1}{2(1-x^2)^{3/2}}(-2x)=\dfrac{x}{(1-x^2)^{3/2}}$$

Then

$$( 1-x^2) f''(x) -xf'(x)$$

$$=(1-x^2)(\dfrac{x}{(1-x^2)^{3/2}})-x(\dfrac{1}{\sqrt{1-x^2}})$$

$$=\dfrac{x}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}=0, -1<x<1$$

$$( 1-x^2) f''(x) -xf'(x)=0$$

$$( 1-x^2) f''(x)=xf'(x)$$

Differentiate with respect to $x$

$$-2xf''(x)+( 1-x^2) f'''(x) =f'(x)+xf''(x)$$

$$( 1-x^2) f'''(x) =f'(x)+3xf''(x)$$

Differentiate with respect to $x$

$$( 1-x^2)f^{(4)}(x)-2xf'''(x)=f''(x)+3f''(x)$$

$$+3xf'''(x)$$

$$( 1-x^2)f^{(4)}(x)=4f''(x)+5xf'''(x)$$

Differentiate with respect to $x$

$$( 1-x^2)f^{(5)}(x)-2xf^{(4)}(x)=4f'''(x)+5f'''(x)$$

$$+5xf^{(4)}(x)$$

$$( 1-x^2)f^{(5)}(x)=9f'''(x)+7xf^{(4)}(x)$$

Differentiate with respect to $x$

$$( 1-x^2)f^{(6)}(x)-2xf^{(5)}(x)=9f^{(4)}(x)+7f^{(4)}(x)$$

$$+7xf^{(5)}(x)$$

$$( 1-x^2)f^{(6)}(x)=16f^{(4)}(x)+9xf^{(5)}(x)$$

Let $P(n)$ be the proposition that

$$( 1-x^2)f^{(n+2)}(x)=n^2f^{(n)}(x)+(2n+1)xf^{(n+1)}(x)$$

$$n=1,2,...$$

Basic Step

$P(1)$ is true because

$$( 1-x^2) f^{(1+2)}(x) =(1)^2f^{(1)}(x)+(2(1)+1)xf^{(1+1)}(x)$$

Inductive Step

We assume that $P(k)$ holds for an arbitrary positive integer $k$

That is, we assume that

$$( 1-x^2)f^{(k+2)}(x)=k^2f^{(k)}(x)+(2k+1)xf^{(k+1)}(x)$$

Differentiate with respect to $x$

$$( 1-x^2)f^{(k+2+1)}(x)-2xf^{(k+2)}(x)=k^2f^{(k+1)}(x)$$

$$+(2k+1)f^{(k+1)}(x)+(2k+1)xf^{(k+1+1)}(x)$$

Then

$$( 1-x^2)f^{((k+1)+2)}(x)=(k+1)^2f^{(k+1)}(x)$$

$$+(2(k+1)+1)xf^{((k+1)+1)}(x)$$

This last equation shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n.$ That is

$$( 1-x^2)f^{(n+2)}(x)=n^2f^{(n)}(x)+(2n+1)xf^{(n+1)}(x)$$

$$n=1,2,...$$

Substitute $x=0$

$$f^{(n+2)}(0)=n^2f^{(n)}(0)$$