find-the-polynial-p-wich-verify-p-x-p-x-x-n-then-calculate-0-1-p-x-dx-

Question Number 31499 by abdo imad last updated on 09/Mar/18

f i n d t h e p o l y n i a l p w i c h v e r i f y p (x) - p^{^{'}} (x) = x^{n} t h e n

c a l c u l a t e \int_{0}^{1} p (x) d x .

Commented by abdo imad last updated on 14/Mar/18

l e t p u t p (x) = \sum_{k = 0}^{n} a_{k} x^{k} w e h a v e p^{^{'}} (x) = \sum_{k = 1}^{n} k a_{k} x^{k - 1}

p (x) - p^{^{'}} (x) = x^{n} \Leftrightarrow \sum_{k = 0}^{n} a_{k} x^{k} - \sum_{k = 1}^{n} k a_{k} x^{k - 1} = x^{n}

\Leftrightarrow \sum_{k = 0}^{n} a_{k} x^{k} - \sum_{k = 0}^{n - 1} (k + 1) a_{k + 1} x^{k} = x^{n} \Leftrightarrow

\sum_{k = 0}^{n - 1} (a_{k} - (k + 1) a_{k + 1}) x^{k} + a_{n} x^{n} = x^{n} \Leftrightarrow

a_{k} - (k + 1) a_{k + 1} = 0 \forall k \in [[0, n - 1]] a n d a_{n} = 1 \Leftrightarrow a_{n} = 1 a n d

a_{k + 1} = \frac{a_{k}}{k + 1} \Rightarrow \prod_{k = 0}^{n - 1} a_{k + 1} = \frac{\prod_{k = 0}^{n - 1} a_{k}}{\prod_{k = 0}^{n - 1} (k + 1)} \Rightarrow

a_{1} . a_{2} \dots a_{n} = \frac{1}{n!} a_{0} . a_{1} . a_{2} \dots a_{n - 1} \Rightarrow a_{n} = \frac{a_{0}}{n!} \Rightarrow

p (x) = \sum_{k = 0}^{n - 1} \frac{a_{0}}{k!} x^{k} + x^{n}

p (o) - p^{^{'}} (0) = 0 \Rightarrow a_{0} = 0 \Rightarrow p (x) = \sum_{k = 1}^{n - 1} \frac{x^{k}}{k!} + x^{n}

2) \int_{0}^{1} p (x) d x = \sum_{k = 1}^{n - 1} \frac{1}{k!} \int_{0}^{1} x^{k} d x + \frac{1}{n + 1}

= \sum_{k = 1}^{n - 1} \frac{1}{(k + 1)!} + \frac{1}{n + 1}