let-f-x-x-2-1-3-x-3-2-4-1-3-5-x-5-2-4-6-1-3-5-7-x-7-x-0-1-the-value-of-f-1-2-

Question Number 171484 by infinityaction last updated on 16/Jun/22

l e t f (x) = x + \frac{2}{1.3} x^{3} + \frac{2.4}{1.3.5} x^{5} + \frac{2.4.6}{1.3.5.7} x^{7} + \dots \dots \dots

\forall x \in (0, 1) t h e v a l u e o f f (\frac{1}{\sqrt{2}}) = ?

Commented by mr W last updated on 16/Jun/22

i g o t

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

f (\frac{1}{\sqrt{2}}) = \frac{\sqrt{2} π}{4}

Answered by mr W last updated on 16/Jun/22

f (x) = x + \frac{2}{1.3} x^{3} + \frac{2.4}{1.3.5} x^{5} + \frac{2.4.6}{1.3.5.7} x^{7} + \dots

x f (x) = x^{2} + \frac{2}{1.3} x^{4} + \frac{2.4}{1.3.5} x^{6} + \frac{2.4.6}{1.3.5.7} x^{8} + \dots

\int_{0}^{x} x f (x) d x = \frac{1}{3} x^{3} + \frac{2}{1.3.5} x^{5} + \frac{2.4}{1.3.5.7} x^{7} + \frac{2.4.6}{1.3.5.7.9} x^{9} + \dots

\frac{1}{x} \int_{0}^{x} x f (x) d x = \frac{1}{1.3} x^{2} + \frac{2}{1.3.5} x^{4} + \frac{2.4}{1.3.5.7} x^{6} + \frac{2.4.6}{1.3.5.7.9} x^{8} + \dots

{[\frac{1}{x} \int_{0}^{x} x f (x) d x]}^{'} = \frac{2}{1.3} x + \frac{2.4}{1.3.5} x^{3} + \frac{2.4.6}{1.3.5.7} x^{5} + \frac{2.4.6.8}{1.3.5.7.9} x^{7} + \dots

x^{2} {[\frac{1}{x} \int_{0}^{x} x f (x) d x]}^{'} = \frac{2}{1.3} x^{3} + \frac{2.4}{1.3.5} x^{5} + \frac{2.4.6}{1.3.5.7} x^{7} + \frac{2.4.6.8}{1.3.5.7.9} x^{9} + \dots

x^{2} {[\frac{1}{x} \int_{0}^{x} x f (x) d x]}^{'} = - x + x + \frac{2}{1.3} x^{3} + \frac{2.4}{1.3.5} x^{5} + \frac{2.4.6}{1.3.5.7} x^{7} + \frac{2.4.6.8}{1.3.5.7.9} x^{9} + \dots

x^{2} {[\frac{1}{x} \int_{0}^{x} x f (x) d x]}^{'} = - x + f (x)

x^{2} [- \frac{1}{x^{2}} \int_{0}^{x} x f (x) d x + f (x)] = - x + f (x)

\int_{0}^{x} x f (x) d x = (x^{2} - 1) f (x) + x

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

(x^{2} - 1) f^{'} (x) + x f (x) + 1 = 0

\Rightarrow f^{'} (x) + \frac{x}{x^{2} - 1} f (x) = \frac{1}{1 - x^{2}}

I = e^{\int \frac{x d x}{x^{2} - 1}} = e^{\frac{\ln ∣ 1 - x^{2} ∣}{2}} = \sqrt{1 - x^{2}}

\Rightarrow f (x) = \frac{1}{\sqrt{1 - x^{2}}} \int \frac{\sqrt{1 - x^{2}}}{1 - x^{2}} d x = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}} + C

f (0) = 0 \Rightarrow C = 0

\Rightarrow f (x) = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}}

\Rightarrow f (\frac{1}{\sqrt{2}}) = \frac{\sqrt{2} π}{4}

Commented by infinityaction last updated on 16/Jun/22

t h a n k y o u s i r

n i c e s o l u t i o n

Commented by Tawa11 last updated on 16/Jun/22

Great sir .

Answered by qaz last updated on 16/Jun/22

f (x) = \underset{k = 0}{\sum^{\infty}} \frac{(2 k)!!}{(2 k + 1)!!} x^{2 k + 1} = \underset{k = 0}{\sum^{\infty}} x^{2 k + 1} \int_{0}^{π / 2} s {i n}^{2 k + 1} t d t = \int_{0}^{π / 2} \frac{x \sin t}{1 - {(x \sin t)}^{2}} d t

\Rightarrow f (\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} \int_{0}^{π / 2} \frac{\sin t}{1 - \frac{1}{2} \sin^{2} t} d t

= \sqrt{2} \int_{0}^{π / 2} \frac{\sin t}{1 + \cos^{2} t} d t

= \sqrt{2} \cdot \arctan (\cos t) ∣_{π / 2}^{0}

= \frac{\sqrt{2} π}{4}

Commented by infinityaction last updated on 16/Jun/22

a m a z i n g s o l u t i o n s i r

Answered by infinityaction last updated on 16/Jun/22