prove-that-x-4-divide-1-x-2-n-nx-2-1-without-use-binomial-formula-

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

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!} + \dots .

= 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 .