Question-132153

Question Number 132153 by Ñï= last updated on 11/Feb/21

Commented by Ar Brandon last updated on 11/Feb/21

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Answered by Olaf last updated on 11/Feb/21

S (x) = \underset{n = 0}{\sum^{\infty}} (2 n - 1)!! \frac{x^{n}}{n!}

Let f (x) = {(1 - 2 x)}^{- \frac{1}{2}}

f^{'} (x) = {(1 - 2 x)}^{- \frac{3}{2}}

f^{″} (x) = 3 {(1 - 2 x)}^{- \frac{5}{2}}

f^{(3)} (x) = 3.5 {(1 - 2 x)}^{- \frac{7}{2}}

f^{(4)} (x) = 3.5.7 {(1 - 2 x)}^{- \frac{9}{2}}

\dots

f^{(n)} (x) = 3.5.7 \dots (2 n - 1) {(1 - 2 x)}^{- \frac{(2 n + 1)}{2}}

f^{(n)} (0) = 3.5.7 \dots (2 n - 1) = \underset{k = 1}{\prod^{n}} (2 k - 1) = (2 n - 1)!!

f (x) = \underset{n = 0}{\sum^{\infty}} f^{(n)} (0) \frac{x^{n}}{n!} = S (x)

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