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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

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 .