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Question Number 72026 by mathmax by abdo last updated on 23/Oct/19
prove that x^4 divide (1+x^2 )^n −nx^2 −1 without use binomial formula.
$${prove}\:{that}\:{x}^{\mathrm{4}} \:{divide}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} −{nx}^{\mathrm{2}} −\mathrm{1}\:\:{without}\:{use}\:\:{binomial} \\ $$$${formula}. \\ $$
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.
$${let}\:{P}_{{n}} \left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} −{nx}^{\mathrm{2}} −\mathrm{1}\:{let}\:{prove}\:{that}\:{x}^{\mathrm{4}} \:{divide}\:{P}_{{n}} \left({x}\right) \\ $$$${by}\:{recurrence}\:\:{n}=\mathrm{0}\:\Rightarrow{P}_{{n}} \left(\mathrm{0}\right)=\mathrm{1}−\mathrm{1}=\mathrm{0}\:\:{and}\:{x}^{\mathrm{4}} \:{divide}\:\mathrm{0} \\ $$$${let}\:{suppose}\:{x}^{\mathrm{4}} \:{divide}\:{P}_{{n}} \left({x}\right)\:\Rightarrow\:\exists\:{Q}\:\in{R}\left[{x}\right]\:/{P}_{{n}} \left({x}\right)={x}^{\mathrm{4}} {Q}\left({x}\right) \\ $$$${we}\:{have}\:{P}_{{n}+\mathrm{1}} \left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{\mathrm{2}} −\mathrm{1} \\ $$$$=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} −\left({n}+\mathrm{1}\right){x}^{\mathrm{2}} −\mathrm{1} \\ $$$$=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({P}_{{n}} \left({x}\right)+{nx}^{\mathrm{2}} \:+\mathrm{1}\right)−\left({n}+\mathrm{1}\right){x}^{\mathrm{2}} −\mathrm{1} \\ $$$$=\left(\mathrm{1}+{x}^{\mathrm{2}} \right){P}_{{n}} \left({x}\right)+{nx}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\mathrm{1}+{x}^{\mathrm{2}} −{nx}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$=\left(\mathrm{1}+{x}^{\mathrm{2}} \right){P}_{{n}} \left({x}\right)+{nx}^{\mathrm{4}} \:\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{4}} {Q}\left({x}\right)\right)\:+{nx}^{\mathrm{4}} \\ $$$$={x}^{\mathrm{4}} \left\{\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){Q}\left({x}\right)\:+{n}\right\}\:\Rightarrow{x}^{\mathrm{4}} \:{divide}\:{P}_{{n}+\mathrm{1}} \left({x}\right)\:{so}\:{the}\:{result}\:{is} \\ $$$${proved}. \\ $$
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)=2xn(1+x^2 )^(n−1) −2nx p′(0)=0 p′′(x)=2n(1+x^2 )^(n−1) +4x^2 n(n−1)(1+x^2 )^(n−2) −2n p′′(0)=2n−2n=0 p′′′(x)=4nx(n−1)(1+x^2 )^(n−2) +8xn(n−1)(1+x^2 )^(n−2) +8x^3 n(n−1)(n−2)(1+x^2 )^(n−3) p′′′(0)=0 taylor formula p(x)=Σ_(k=1) ^(2n) p^((k)) (0).(x^k /(k!))=((x^4 p^((4)) (0))/(4!))+.... =x^4 (.Q(x)) Q(x)=Σ_(k=4) ^(2n) ((p^((k)) x^(k−4) )/(k!)) ⇒X^4 ∣(1+x^2 )^n −nx^2 −1
$$\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} −\mathrm{nx}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{0}\:,\mathrm{n}=\mathrm{1},\mathrm{n}=\mathrm{2}\:\mathrm{juste}\:\mathrm{evaluate} \\ $$$$\mathrm{for}\:\mathrm{n}\geqslant\mathrm{2} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} −\mathrm{nx}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{p}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{p}'\left(\mathrm{x}\right)=\mathrm{2xn}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}−\mathrm{1}} −\mathrm{2nx} \\ $$$$\mathrm{p}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{p}''\left(\mathrm{x}\right)=\mathrm{2n}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}−\mathrm{1}} +\mathrm{4x}^{\mathrm{2}} \mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}−\mathrm{2}} −\mathrm{2n} \\ $$$$\mathrm{p}''\left(\mathrm{0}\right)=\mathrm{2n}−\mathrm{2n}=\mathrm{0} \\ $$$$\mathrm{p}'''\left(\mathrm{x}\right)=\mathrm{4nx}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}−\mathrm{2}} +\mathrm{8xn}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}−\mathrm{2}} +\mathrm{8x}^{\mathrm{3}} \mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}−\mathrm{3}} \\ $$$$\mathrm{p}'''\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{taylor}\:\mathrm{formula} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2n}} {\sum}}\mathrm{p}^{\left(\mathrm{k}\right)} \left(\mathrm{0}\right).\frac{\mathrm{x}^{\mathrm{k}} }{\mathrm{k}!}=\frac{\mathrm{x}^{\mathrm{4}} \mathrm{p}^{\left(\mathrm{4}\right)} \left(\mathrm{0}\right)}{\mathrm{4}!}+…. \\ $$$$=\mathrm{x}^{\mathrm{4}} \left(.\mathrm{Q}\left(\mathrm{x}\right)\right) \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\underset{\mathrm{k}=\mathrm{4}} {\overset{\mathrm{2n}} {\sum}}\frac{\mathrm{p}^{\left(\mathrm{k}\right)} \mathrm{x}^{\mathrm{k}−\mathrm{4}} }{\mathrm{k}!} \\ $$$$\Rightarrow\mathrm{X}^{\mathrm{4}} \mid\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} −\mathrm{nx}^{\mathrm{2}} −\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 24/Oct/19
thanks sir.
$${thanks}\:{sir}. \\ $$

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