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Question Number 147467 by mathmax by abdo last updated on 21/Jul/21

f (x) = x^{n} e^{- x}

1) calculate f^{(n)} (0) and f^{(n)} (1)

2) developp f at integr serie

3) calculate \int_{0}^{1} f (x) dx

Answered by mathmax by abdo last updated on 22/Jul/21

1) leibniz give f^{(p)} (x) = \sum_{k = 0}^{p} C_{p}^{k} {(x^{n})}^{(k)} {(e^{- x})}^{(p - k)} \Rightarrow

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

if k > n \Rightarrow {(x^{n})}^{(k)} = 0

if k ⩽ n \Rightarrow {(x^{n})}^{(k)} = n (n - 1) \dots (n - k + 1) x^{n - k}

= \frac{n!}{(n - k)!} x^{n - k} we have

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

= \sum_{k = 0}^{n - 1} (\dots) x^{n - k} + e^{- x} n! \Rightarrow f^{(n)} (0) = n!

Commented by mathmax by abdo last updated on 22/Jul/21

f^{(n)} (1) = e^{- 1} \sum_{k = 0}^{n} {(- 1)}^{n - k} \frac{n!}{k! (n - k)!} \times \frac{n!}{(n - k)!}

= {(- 1)}^{n} e^{- 1} \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k!} \times \frac{{(n!)}^{2}}{{((n - k)!)}^{2}}

Answered by mathmax by abdo last updated on 22/Jul/21

2) we have f (x) = x^{n} e^{- x} = x^{n} \sum_{p = 0}^{\infty} \frac{{(- x)}^{p}}{p!}

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

Answered by mathmax by abdo last updated on 22/Jul/21

3) A_{n} = \int_{0}^{1} x^{n} e^{- x} dx by parts we get

A_{n} = {[\frac{x^{n + 1}}{n + 1} e^{- x}]}_{0}^{1} + \int_{0}^{1} \frac{x^{n + 1}}{n + 1} e^{- x} dx = \frac{e^{- 1}}{n + 1} + \frac{1}{n + 1} A_{n + 1} \Rightarrow

(n + 1) A_{n} = \frac{1}{e} + A_{n + 1} \Rightarrow A_{n + 1} = (n + 1) A_{n} - \frac{1}{e} \Rightarrow

A_{n} = {nA}_{n - 1} - \frac{1}{e} let u_{n} = \frac{A_{n}}{n!} \Rightarrow

u_{n + 1} = \frac{A_{n + 1}}{(n + 1)!} = \frac{(n + 1) A_{n} - \frac{1}{e}}{(n + 1)!} = \frac{A_{n}}{n!} - \frac{1}{e (n + 1)!}

= u_{n} - \frac{1}{e (n + 1)!} \Rightarrow u_{n + 1} - u_{n} = - \frac{1}{e (n + 1)!} \Rightarrow

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

u_{n} - u_{0} = - \frac{1}{e} \sum_{k = 1}^{n + 1} \frac{1}{k!} \Rightarrow u_{n} = u_{o} - \frac{1}{e} \sum_{k = 1}^{n + 1} \frac{1}{k!}

u_{o} = A_{0} = \int_{0}^{1} e^{- x} dx = {[- e^{- x}]}_{0}^{1} = 1 - \frac{1}{e} \Rightarrow

u_{n} = 1 - \frac{1}{e} \sum_{k = 0}^{n + 1} \frac{1}{k!}

A_{n} = n! u_{n} = n! (1 - \frac{1}{e} \sum_{k = 0}^{n + 1} \frac{1}{k!})