let f(x) = e^(−x^2 ) 1) prove that f^((n)) (x)=p_n (x)e^(−x^2 ) with p_n is a polynom 2) find a relation of recurrence between the p_n 3) calculate p_1 ,p_2 ,p_3 ,p


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 36926 by maxmathsup by imad last updated on 07/Jun/18

$l e t f (x) = e^{- x^{2}}$ $1) p r o v e t h a t f^{(n)} (x) = p_{n} (x) e^{- x^{2}} w i t h p_{n} i s a p o l y n o m$ $2) f i n d a r e l a t i o n o f r e c u r r e n c e b e t w e e n t h e p_{n}$ $3) c a l c u l a t e p_{1}, p_{2}, p_{3}, p_{4}$
	Answered by math khazana by abdo last updated on 10/Jun/18
	$we have f^′ (x) =−2x e^(−x^2 ) f^((2)) (x) =−2 e^(−x^2 ) +4x^2 e^(−x^2 ) =(4x^2 −2)e^(−x^2 ) let suppose that f^((n)) (x) =p_n (x) e^(−x^2 ) with p_n a polynome ⇒ f^((n+1)) (x)= (p_n (x)e^(−x^2 ) )^′ =p_n ^′ (x) e^(−x^2 ) −2x p_n (x) e^(−x^2 ) ={ p_n ^′ (x) −2x p_n (x)}e^(−x^2 ) =p_(n+1) (x)e^(−x^2 ) with p_(n+1) (x)= p_n ^′ (x) −2x p_n (x) and its clear that deg(p_n ) =n 2) p_(n+1) (x)= −2xp_n (x) +p_n ^′ (x) 3) p_1 (x) =−2x and p_2 (x) =−2xp_1 (x) +p_1 ^′ (x) =4x^2 −2 p_3 (x) =−2xp_2 (x) +p_2 ^′ (x) =−2x(4x^2 −2) +8x=−8x^3 +4x +8x =−8x^3 +12x p_4 (x) =−2xp_3 (x) +p_3 ^′ (x) =−2x(−8x^3 +12x) −24x^2 +12 =16 x^4 −24x^2 −24x^2 +12 =16 x^4 −48x^2 +12 .$
	$w e h a v e f^{^{'}} (x) = - 2 x e^{- x^{2}}$ $f^{(2)} (x) = - 2 e^{- x^{2}} + 4 x^{2} e^{- x^{2}} = (4 x^{2} - 2) e^{- x^{2}}$ $l e t s u p p o s e t h a t f^{(n)} (x) = p_{n} (x) e^{- x^{2}} w i t h p_{n}$ $a p o l y n o m e \Rightarrow f^{(n + 1)} (x) = {(p_{n} (x) e^{- x^{2}})}^{^{'}}$ $= p_{n}^{^{'}} (x) e^{- x^{2}} - 2 x p_{n} (x) e^{- x^{2}}$ $= {p_{n}^{^{'}} (x) - 2 x p_{n} (x)} e^{- x^{2}} = p_{n + 1} (x) e^{- x^{2}}$ $w i t h p_{n + 1} (x) = p_{n}^{^{'}} (x) - 2 x p_{n} (x) a n d i t s c l e a r$ $t h a t d e g (p_{n}) = n$ $2) p_{n + 1} (x) = - 2 {x p}_{n} (x) + p_{n}^{^{'}} (x)$ $3) p_{1} (x) = - 2 x a n d p_{2} (x) = - 2 {x p}_{1} (x) + p_{1}^{^{'}} (x)$ $= 4 x^{2} - 2$ $p_{3} (x) = - 2 {x p}_{2} (x) + p_{2}^{^{'}} (x)$ $= - 2 x (4 x^{2} - 2) + 8 x = - 8 x^{3} + 4 x + 8 x$ $= - 8 x^{3} + 12 x$ $p_{4} (x) = - 2 {x p}_{3} (x) + p_{3}^{^{'}} (x)$ $= - 2 x (- 8 x^{3} + 12 x) - 24 x^{2} + 12$ $= 16 x^{4} - 24 x^{2} - 24 x^{2} + 12$ $= 16 x^{4} - 48 x^{2} + 12 .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum