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Question Number 41627 by alex041103 last updated on 10/Aug/18

Commented by alex041103 last updated on 10/Aug/18

$J u s t f o r f u n! I h a v e a l r e a d y p o s t e d$ $t h i s Q . a r o u n d a y e a r a g o . T h e r e s u l t$ $i s b e a u t i f u l a n d w o r t h s g i v i n g i t a t r y .$ $C a n y o u p r o v e s o m e t h i n g i n t e r e s t i n g .$ $H i n t : C a l c u l a t e t h e r e s u l t .$
Commented by maxmathsup by imad last updated on 10/Aug/18
$we have x^4 (1−x^ )^4 =x^4 (x^2 −2x+1)^2 =x^4 ( (x^2 −2x)^2 +2(x^2 −2x) +1) =x^4 { x^4 −4x^3 +4x^2 +2x^2 −4x +1} =x^4 { x^4 −4x^3 +6x^2 −4x+1} = x^8 −4x^7 +6x^6 −4x^5 +x^4 ⇒ I = ∫_0 ^1 ((x^8 −4x^7 +6x^6 −4x^5 +x^4 )/(1+x^2 )) dx = ∫_0 ^1 (x^8 −4x^7 +6x^6 −4x^5 +x^4 )Σ_(n=0) ^∞ (−1)^n x^(2n) dx = Σ_(n=0) ^∞ (−1)^(n ) ∫_0 ^1 (x^(2n+8) −x^(2n+7) +6 x^(2n+6) −4 x^(2n+5) +x^(2n+4) )dx = Σ_(n=0) ^∞ (((−1)^n )/(2n+9)) −Σ_(n=0) ^∞ (((−1)^n )/(2n+8)) +6 Σ_(n=0) ^∞ (((−1)^n )/(2n+7)) −4 Σ_(n=0) ^∞ (((−1)^n )/(2n+6)) +Σ_(n=0) ^∞ (((−1)^n )/(2n+5)) let w(x) = Σ_(n=0) ^∞ (((−1)^n )/(2n+5)) x^(2n+5) we have w(1) =Σ_(n=0) ^∞ (((−1)^n )/(2n+5)) w^′ (x) =Σ_(n=0) ^∞ (−1)^n x^(2n+4) =x^4 Σ_(n=0) ^∞ (−x^2 )^n =(x^4 /(1+x^2 )) ⇒ w(x) = ∫_0 ^x (t^4 /(1+t^2 )) dt +c but c=w(0) =0 ⇒w(x)=∫_0 ^x (t^4 /(1+t^2 )) dt w(x)= ∫_0 ^x ((t^4 −1+1)/(t^2 +1)) dt = ∫_0 ^x (((t^2 +1)(t^2 −1) +1)/(t^2 +1)) dt =∫_0 ^x (t^2 −1)dt + ∫_0 ^x (dt/(1+t^2 )) = (x^3 /3) −x +arctan(x) ⇒ w(1) =−(2/3) +(π/4) generally let consider w_p (x) =Σ_(n=0) ^∞ (((−1)^n )/(2n+p)) x^(2n+p) we have w_p (1) =Σ_(n=0) ^∞ (((−1)^n )/(2n+p )) and we have w_p ^′ (x) = Σ_(n=0) ^∞ (−1)^n x^(2n+p−1) = x^(p−1) Σ_(n=0) ^∞ (−x^2 )^n = (x^(p−1) /(1+x^2 )) ⇒ w_p (x) = ∫_0 ^x (x^(p−1) /(1+x^2 )) dx +c but c=w_p (1)=0 ⇒w_p (x)= ∫_0 ^x (x^(p−1) /(1+x^2 )) dx ....be continued...$
$w e h a v e x^{4} {(1 - x^{})}^{4} = x^{4} {(x^{2} - 2 x + 1)}^{2} = x^{4} ({(x^{2} - 2 x)}^{2} + 2 (x^{2} - 2 x) + 1)$ $= x^{4} {x^{4} - 4 x^{3} + 4 x^{2} + 2 x^{2} - 4 x + 1} = x^{4} {x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1}$ $= x^{8} - 4 x^{7} + 6 x^{6} - 4 x^{5} + x^{4} \Rightarrow$ $I = \int_{0}^{1} \frac{x^{8} - 4 x^{7} + 6 x^{6} - 4 x^{5} + x^{4}}{1 + x^{2}} d x$ $= \int_{0}^{1} (x^{8} - 4 x^{7} + 6 x^{6} - 4 x^{5} + x^{4}) \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} d x$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} (x^{2 n + 8} - x^{2 n + 7} + 6 x^{2 n + 6} - 4 x^{2 n + 5} + x^{2 n + 4}) d x$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 9} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 8} + 6 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 7} - 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 6} + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 5}$ $l e t w (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 5} x^{2 n + 5} w e h a v e w (1) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 5}$ $w^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n + 4} = x^{4} \sum_{n = 0}^{\infty} {(- x^{2})}^{n} = \frac{x^{4}}{1 + x^{2}} \Rightarrow$ $w (x) = \int_{0}^{x} \frac{t^{4}}{1 + t^{2}} d t + c b u t c = w (0) = 0 \Rightarrow w (x) = \int_{0}^{x} \frac{t^{4}}{1 + t^{2}} d t$ $w (x) = \int_{0}^{x} \frac{t^{4} - 1 + 1}{t^{2} + 1} d t = \int_{0}^{x} \frac{(t^{2} + 1) (t^{2} - 1) + 1}{t^{2} + 1} d t$ $= \int_{0}^{x} (t^{2} - 1) d t + \int_{0}^{x} \frac{d t}{1 + t^{2}} = \frac{x^{3}}{3} - x + a r c t a n (x) \Rightarrow$ $w (1) = - \frac{2}{3} + \frac{π}{4} g e n e r a l l y l e t c o n s i d e r w_{p} (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + p} x^{2 n + p}$ $w e h a v e w_{p} (1) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + p} a n d w e h a v e$ $w_{p}^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n + p - 1} = x^{p - 1} \sum_{n = 0}^{\infty} {(- x^{2})}^{n} = \frac{x^{p - 1}}{1 + x^{2}} \Rightarrow$ $w_{p} (x) = \int_{0}^{x} \frac{x^{p - 1}}{1 + x^{2}} d x + c b u t c = w_{p} (1) = 0 \Rightarrow w_{p} (x) = \int_{0}^{x} \frac{x^{p - 1}}{1 + x^{2}} d x$ $. . . . b e c o n t i n u e d . . .$
	Answered by rahul 19 last updated on 10/Aug/18

	Commented by rahul 19 last updated on 10/Aug/18
	≠ 0 ��
	Commented by alex041103 last updated on 10/Aug/18

	$C a n y o u n o w p r o v e, t h a t \frac{22}{7} > π ?$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

	$b a h d a r u n e x c e l l e n t . . .$
	Commented by rahul 19 last updated on 10/Aug/18

	$Since integrand is non - negative &$ $continous everywhere, it must be strictly$ $positive in [0, 1] .$ $0 < \frac{22}{7} - π$ $\Rightarrow \frac{22}{7} > π .$
	Commented by alex041103 last updated on 10/Aug/18

	$E x a c t l y .$ $A l s o y o u c a n s e e t h a t f o r 0 < x < 1 t h e$ $i n t e g r a n d i s s m a l l . . . t h i s i m p l i e s t h a t$ $\frac{22}{7} i s g o o d r a t i o n a l a p p r o x i m a t i o n$ $f o r π .$
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