prove that x^4 divide (1+x^2 )^n −nx^2 −1 without use binomial formula.


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Question Number 72026 by mathmax by abdo last updated on 23/Oct/19

$p r o v e t h a t x^{4} d i v i d e {(1 + x^{2})}^{n} - {n x}^{2} - 1 w i t h o u t u s e b i n o m i a l$ $f o r m u l a .$
Commented by mathmax by abdo last updated on 24/Oct/19
$let P_n (x)=(1+x^2 )^n −nx^2 −1 let prove that x^4 divide P_n (x) by recurrence n=0 ⇒P_n (0)=1−1=0 and x^4 divide 0 let suppose x^4 divide P_n (x) ⇒ ∃ Q ∈R[x] /P_n (x)=x^4 Q(x) we have P_(n+1) (x)=(1+x^2 )^(n+1) −(n+1)x^2 −1 =(1+x^2 )(1+x^2 )^n −(n+1)x^2 −1 =(1+x^2 )(P_n (x)+nx^2 +1)−(n+1)x^2 −1 =(1+x^2 )P_n (x)+nx^2 (1+x^2 )+1+x^2 −nx^2 −x^2 −1 =(1+x^2 )P_n (x)+nx^4 =(1+x^2 )(x^4 Q(x)) +nx^4 =x^4 { (1+x^2 )Q(x) +n} ⇒x^4 divide P_(n+1) (x) so the result is proved.$
$l e t P_{n} (x) = {(1 + x^{2})}^{n} - {n x}^{2} - 1 l e t p r o v e t h a t x^{4} d i v i d e P_{n} (x)$ $b y r e c u r r e n c e n = 0 \Rightarrow P_{n} (0) = 1 - 1 = 0 a n d x^{4} d i v i d e 0$ $l e t s u p p o s e x^{4} d i v i d e P_{n} (x) \Rightarrow \exists Q \in R [x] / P_{n} (x) = x^{4} Q (x)$ $w e h a v e P_{n + 1} (x) = {(1 + x^{2})}^{n + 1} - (n + 1) x^{2} - 1$ $= (1 + x^{2}) {(1 + x^{2})}^{n} - (n + 1) x^{2} - 1$ $= (1 + x^{2}) (P_{n} (x) + {n x}^{2} + 1) - (n + 1) x^{2} - 1$ $= (1 + x^{2}) P_{n} (x) + {n x}^{2} (1 + x^{2}) + 1 + x^{2} - {n x}^{2} - x^{2} - 1$ $= (1 + x^{2}) P_{n} (x) + {n x}^{4} = (1 + x^{2}) (x^{4} Q (x)) + {n x}^{4}$ $= x^{4} {(1 + x^{2}) Q (x) + n} \Rightarrow x^{4} d i v i d e P_{n + 1} (x) s o t h e r e s u l t i s$ $p r o v e d .$
	Answered by mind is power last updated on 23/Oct/19

	${(1 + x^{2})}^{n} - {nx}^{2} - 1$ $for n = 0, n = 1, n = 2 juste evaluate$ $for n ⩾ 2$ $p (x) = {(1 + x^{2})}^{n} - {nx}^{2} - 1$ $p (0) = 0$ $p^{'} (x) = 2 xn {(1 + x^{2})}^{n - 1} - 2 nx$ $p^{'} (0) = 0$ $p^{″} (x) = 2 n {(1 + x^{2})}^{n - 1} + {4 x}^{2} n (n - 1) {(1 + x^{2})}^{n - 2} - 2 n$ $p^{″} (0) = 2 n - 2 n = 0$ $p^{‴} (x) = 4 nx (n - 1) {(1 + x^{2})}^{n - 2} + 8 xn (n - 1) {(1 + x^{2})}^{n - 2} + {8 x}^{3} n (n - 1) (n - 2) {(1 + x^{2})}^{n - 3}$ $p^{‴} (0) = 0$ $taylor formula$ $p (x) = \underset{k = 1}{\sum^{2 n}} p^{(k)} (0) . \frac{x^{k}}{k!} = \frac{x^{4} p^{(4)} (0)}{4!} + . . . .$ $= x^{4} (. Q (x))$ $Q (x) = \underset{k = 4}{\sum^{2 n}} \frac{p^{(k)} x^{k - 4}}{k!}$ $\Rightarrow X^{4} ∣ {(1 + x^{2})}^{n} - {nx}^{2} - 1$
	Commented by mathmax by abdo last updated on 24/Oct/19

	$t h a n k s s i r .$
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