let-L-n-x-e-x-e-x-x-n-n-1-prove-that-L-n-is-a-polynomial-2-find-degL-n-and-the-leading-coefficient-

Question Number 31500 by abdo imad last updated on 09/Mar/18

l e t L_{n} (x) = e^{x} {(e^{- x} x^{n})}^{(n)}

1) p r o v e t h a t L_{n} i s a p o l y n o m i a l

2) f i n d {d e g L}_{n} a n d t h e l e a d i n g c o e f f i c i e n t .

Commented by abdo imad last updated on 13/Mar/18

w e h a v e b y L e n i z f o r m u l a e

{(e^{- x} x^{n})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} {(x^{n})}^{(k)} {(e^{- x})}^{(n - k)} b u t

{(x^{n})}^{(k)} = n (n - 1) (n - 2) \dots (n - k + 1) x^{n - k}

= \frac{n!}{(n - k)!} x^{n - k} a n d {(e^{- x})}^{(n - k)} = {(- 1)}^{n - k} e^{- x} \Rightarrow

{(e^{- x} x^{n})}^{(n)} = e^{- x} \sum_{k = 0}^{n} {(- 1)}^{n - k} C_{n}^{k} \frac{n!}{(n - k)!} x^{n - k} \Rightarrow

L_{n} (x) = {(- 1)}^{n} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} \frac{n!}{(n - k)!} x^{n - k} c h . o f i n d i c e

n - k = p g i v e L_{n} (x) = {(- 1)}^{n} \sum_{p = 0}^{n} {(- 1)}^{n - p} C_{n}^{n - p} \frac{n!}{p!} x^{p}

L_{n} (x) = \sum_{p = 0}^{n} {(- 1)}^{p} C_{n}^{p} \frac{n!}{p!} x^{p} a n d i s a p o l y n o m i a l

2) i t s c l e a r t h a t d e g L_{n} = n a n d t h e l e a d i n g c o e f f i c i e n t i s

{(- 1)}^{n} x^{n} .

Commented by abdo imad last updated on 13/Mar/18

L e i b n i z f o r m u l a e \dots

Commented by abdo imad last updated on 13/Mar/18

L_{n} a r e n a m e d p o l y n o m i a l s o f L a g u e r r e .