let-f-x-e-x-2-0-x-e-t-2-dt-1-find-a-d-e-verified-by-f-2-developpf-at-integr-serie-

Question Number 34699 by abdo imad last updated on 10/May/18

l e t f (x) = e^{- x^{2}} \int_{0}^{x} e^{t^{2}} d t

1) f i n d a d . e v e r i f i e d b y f

2) d e v e l o p p f a t i n t e g r s e r i e .

Commented by math khazana by abdo last updated on 11/May/18

w e h a v e f^{^{'}} (x) = - 2 x e^{- x^{2}} \int_{0}^{x} e^{t^{2}} d t + e^{- x^{2}} e^{x^{2}}

= - 2 x f (x) + 1 s o f i s s o l u t i o n o f t h e d . e .

y^{^{'}} = - 2 x y + 1

2) l e t p u t f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow f^{^{'}} (x) = \sum_{n = 1}^{\infty} {n a}_{n} x^{n - 1}

f^{^{'}} + 2 x f - 1 = 0 \Leftrightarrow \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}

+ 2 x \sum_{n = 0}^{\infty} a_{n} x^{n} - 1 = 0 \Leftrightarrow

\sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n} + 2 \sum_{n = 0}^{\infty} a_{n} x^{n + 1} - 1 = 0 \Leftrightarrow

\sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n} + 2 \sum_{n = 1}^{\infty} a_{n - 1} x^{n} - 1 = 0 \Leftrightarrow

a_{1} + \sum_{n = 1}^{\infty} {(n + 1) a_{n + 1} + 2 a_{n - 1}} x^{n} - 1 = 0 \Leftrightarrow

a_{1} = 1 a n d \forall n ⩾ 1 (n + 1) a_{n + 1} + 2 a_{n - 1} = 0 \Rightarrow

(n + 1) a_{n + 1} = - 2 a_{n - 1} \Rightarrow a_{n + 1} = \frac{- 2}{n + 1} a_{n - 1} \Rightarrow

a_{2 n + 1} = \frac{- 2}{2 n + 1} a_{2 n - 1} a n d a_{2 n} = - \frac{2}{2 n} a_{2 n - 2} = \frac{- 1}{n} a_{2 n - 2}

\prod_{k = 1}^{n} a_{2 k + 1} = {(- 2)}^{n + 1} \frac{\prod_{k = 1}^{n} a_{2 k - 1}}{\prod_{k = 0}^{n} (2 k + 1)}

. a_{3} . a_{5} \dots . . a_{2 n + 1} = \frac{{(- 2)}^{n + 1}}{1.3.5 \dots . (2 n + 1)} a_{1} . a_{3} \dots a_{2 n - 1}

\Rightarrow a_{2 n + 1} = \frac{{(- 2)}^{n + 1}}{1.3.5 \dots . . (2 n + 1)} a l s o w e h a v e

\prod_{k = 1}^{n} a_{2 k} = \frac{{(- 1)}^{n}}{n!} \prod_{k = 1}^{n} a_{2 k - 2} \Rightarrow

a_{2} . a_{4} \dots . . a_{2 n} = \frac{{(- 1)}^{n}}{n!} a_{0} a_{2} \dots . a_{2 n - 2} \Rightarrow

a_{2 n} = \frac{{(- 1)}^{n}}{n!} a_{0}

f (x) = \sum_{n = 0}^{\infty} a_{2 n} x^{2 n} + \sum_{n = 0}^{\infty} a_{2 n + 1} x^{2 n + 1}

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} a_{0} x^{2 n} + \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n + 1}}{1.3.5 \dots (2 n + 1)} x^{2 n + 1}

Commented by math khazana by abdo last updated on 11/May/18

w e h a v e f (x) = e^{- x^{2}} \int_{0}^{x} e^{t^{2}} d t \Rightarrow

f (- x) = e^{- x^{2}} \int_{0}^{- x} e^{t^{2}} d t

=_{t = - u} e^{- x^{2}} \int_{o}^{x} e^{u^{2}} (- d u) = - e^{x^{2}} \int_{0}^{x} e^{u^{2}} d u = - f (x)

s o f i s o d d \Rightarrow a_{2 n} = 0 \Rightarrow

f (x) = \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n + 1}}{1.3.5 \dots . (2 n + 1)} x^{2 n + 1} .