Question #345830

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)


1
Expert's answer
2022-05-31T10:56:49-0400
f(x)=sin1(x)f(x)=\sin^{-1}(x)

f(x)=11x2f'(x)=\dfrac{1}{\sqrt{1-x^2}}

f(x)=12(1x2)3/2(2x)=x(1x2)3/2f''(x)=-\dfrac{1}{2(1-x^2)^{3/2}}(-2x)=\dfrac{x}{(1-x^2)^{3/2}}

Then

(1x2)f(x)xf(x)( 1-x^2) f''(x) -xf'(x)

=(1x2)(x(1x2)3/2)x(11x2)=(1-x^2)(\dfrac{x}{(1-x^2)^{3/2}})-x(\dfrac{1}{\sqrt{1-x^2}})

=x1x2x1x2=0,1<x<1=\dfrac{x}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}=0, -1<x<1


(1x2)f(x)xf(x)=0( 1-x^2) f''(x) -xf'(x)=0

(1x2)f(x)=xf(x)( 1-x^2) f''(x)=xf'(x)

Differentiate with respect to xx

2xf(x)+(1x2)f(x)=f(x)+xf(x)-2xf''(x)+( 1-x^2) f'''(x) =f'(x)+xf''(x)

(1x2)f(x)=f(x)+3xf(x)( 1-x^2) f'''(x) =f'(x)+3xf''(x)

Differentiate with respect to xx


(1x2)f(4)(x)2xf(x)=f(x)+3f(x)( 1-x^2)f^{(4)}(x)-2xf'''(x)=f''(x)+3f''(x)

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

(1x2)f(4)(x)=4f(x)+5xf(x)( 1-x^2)f^{(4)}(x)=4f''(x)+5xf'''(x)

Differentiate with respect to xx

(1x2)f(5)(x)2xf(4)(x)=4f(x)+5f(x)( 1-x^2)f^{(5)}(x)-2xf^{(4)}(x)=4f'''(x)+5f'''(x)

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


(1x2)f(5)(x)=9f(x)+7xf(4)(x)( 1-x^2)f^{(5)}(x)=9f'''(x)+7xf^{(4)}(x)

Differentiate with respect to xx

(1x2)f(6)(x)2xf(5)(x)=9f(4)(x)+7f(4)(x)( 1-x^2)f^{(6)}(x)-2xf^{(5)}(x)=9f^{(4)}(x)+7f^{(4)}(x)

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

(1x2)f(6)(x)=16f(4)(x)+9xf(5)(x)( 1-x^2)f^{(6)}(x)=16f^{(4)}(x)+9xf^{(5)}(x)

Let P(n)P(n) be the proposition that


(1x2)f(n+2)(x)=n2f(n)(x)+(2n+1)xf(n+1)(x)( 1-x^2)f^{(n+2)}(x)=n^2f^{(n)}(x)+(2n+1)xf^{(n+1)}(x)

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

Basic Step

P(1)P(1) is true because

(1x2)f(1+2)(x)=(1)2f(1)(x)+(2(1)+1)xf(1+1)(x)( 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)P(k) holds for an arbitrary positive integer kk

That is, we assume that


(1x2)f(k+2)(x)=k2f(k)(x)+(2k+1)xf(k+1)(x)( 1-x^2)f^{(k+2)}(x)=k^2f^{(k)}(x)+(2k+1)xf^{(k+1)}(x)

Differentiate with respect to xx


(1x2)f(k+2+1)(x)2xf(k+2)(x)=k2f(k+1)(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)+(2k+1)f^{(k+1)}(x)+(2k+1)xf^{(k+1+1)}(x)

Then

(1x2)f((k+1)+2)(x)=(k+1)2f(k+1)(x)( 1-x^2)f^{((k+1)+2)}(x)=(k+1)^2f^{(k+1)}(x)

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

This last equation shows that P(k+1)P(k + 1) is true under the assumption that P(k)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)P(n) is true for all positive integers n.n. That is


(1x2)f(n+2)(x)=n2f(n)(x)+(2n+1)xf(n+1)(x)( 1-x^2)f^{(n+2)}(x)=n^2f^{(n)}(x)+(2n+1)xf^{(n+1)}(x)

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

Substitute x=0x=0


f(n+2)(0)=n2f(n)(0)f^{(n+2)}(0)=n^2f^{(n)}(0)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS