if y = cos(x^2 ) then y^((n)) = .........


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Question Number 105488 by 175mohamed last updated on 29/Jul/20

$i f y = c o s (x^{2}) t h e n$ $y^{(n)} = . . . . . . . . .$
	Answered by mathmax by abdo last updated on 29/Jul/20

	$y (x) = \cos (x^{2}) = \frac{e^{{ix}^{2}} + e^{- {ix}^{2}}}{2} \Rightarrow y^{(n)} (x) = \frac{1}{2} {(e^{{ix}^{2}})}^{(n)} + \frac{1}{2} {(e^{- {ix}^{2}})}^{) n)}$ $let w (x) = (e^{{ix}^{2}}) we have w^{(1)} (x) = 2 ix e^{{ix}^{2}}$ $w^{(2)} (x) = 2 i e^{{ix}^{2}} + 2 ix (2 ix) e^{{ix}^{2}} = (- {4 x}^{2} + 2 i) e^{{ix}^{2}} \Rightarrow$ $w^{(n)} = p_{n} (x) e^{{ix}^{2}} with p_{n} \in C [x] we have$ $w^{(n + 1)} (x) = p_{n}^{^{'}} (x) e^{{ix}^{2}} + 2 ix p_{n} (x) e^{{ix}^{2}} = ({2 ip}_{n} (x) + p_{n}^{^{'}} (x)) e^{{ix}^{2}} \Rightarrow$ $\Rightarrow p_{n + 1} (x) = {2 ip}_{n} (x) + p_{n}^{^{'}} (x)$ $let φ (x) = e^{- {ix}^{2}} \Rightarrow φ^{(1)} (x) = - 2 ix e^{- {ix}^{2}} and$ $φ^{(2)} (x) = - 2 i e^{- {ix}^{2}} + (- 2 ix) (- 2 ix) e^{- {ix}^{2}} = (- {4 x}^{2} - 2 i) e^{- {ix}^{2}} \Rightarrow$ $φ^{(n)} (x) = q_{n} (x) e^{- {ix}^{2}} (q_{n} \in C [x]) we have φ^{(n + 1)} (x) = q_{n}^{^{'}} (x) e^{- {ix}^{2}} - {2 ixq}_{n} (x) e^{- {ix}^{2}}$ $= (q_{n}^{^{'}} (x) - {2 iq}_{n} (x)) e^{- {ix}^{2}} \Rightarrow q_{n + 1} = q_{n}^{^{'}} - {2 iq}_{n} (x) and we get$ $y^{(n)} (x) = w^{(n)} (x) + φ^{(n)} (x)$
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