let f(x) =(x+1)^9 e^(−3x) calculstr f^((7)) (0) and f^((5)) (1)


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Question Number 94650 by msup by abdo last updated on 20/May/20

$l e t f (x) = {(x + 1)}^{9} e^{- 3 x}$ $c a l c u l s t r f^{(7)} (0) a n d f^{(5)} (1)$
Commented by i jagooll last updated on 20/May/20
question what number is the integral function of the gamma sir? I do not find it
	Answered by mathmax by abdo last updated on 20/May/20
	$we have f^((n)) (x) =Σ_(k=0) ^n C_n ^k {(x+1)^9 }^((k)) (e^(−3x) )^((n−k)) k>9 ⇒{(x+1)^9 }^((k)) =0 and for k≤9 we have (x+1)^p )^((1)) =p(x+1)^(p−1) ((x+1)^p )^((2)) =p(p−1)(x+1)^(p−1) ⇒((x+1)^p )^((k)) =p(p−1)...(p−k+1)(x+1)^(p−k) ((x+1)^9 )^((k)) =9.8....(10−k)(x+1)^(9−k) ⇒ f^((n)) (x) =(x+1)^9 (−3)^n e^(−3x) +Σ_(k=1) ^n C_n ^k 9.8...(10−k)(x+1)^(9−k) (−3)^(n−k) e^(−3x) ⇒ f^((7)) (0) =1+Σ_(k=1) ^7 C_7 ^k 9.8....(10−k)(−3)^(7−k) f^((5)) (1) =2^9 (−3)^5 e^(−3) +Σ_(k=1) ^5 C_5 ^k 9.8....(10−k)2^(5−k) e^(−3)$
	$we have f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {{(x + 1)}^{9}}^{(k)} {(e^{- 3 x})}^{(n - k)}$ $k > 9 \Rightarrow {{(x + 1)}^{9}}^{(k)} = 0 and for k ⩽ 9 we have$ ${{(x + 1)}^{p})}^{(1)} = p {(x + 1)}^{p - 1}$ ${({(x + 1)}^{p})}^{(2)} = p (p - 1) {(x + 1)}^{p - 1} \Rightarrow {({(x + 1)}^{p})}^{(k)} = p (p - 1) . . . (p - k + 1) {(x + 1)}^{p - k}$ ${({(x + 1)}^{9})}^{(k)} = 9.8 . . . . (10 - k) {(x + 1)}^{9 - k} \Rightarrow$ $f^{(n)} (x) = {(x + 1)}^{9} {(- 3)}^{n} e^{- 3 x} + \sum_{k = 1}^{n} C_{n}^{k} 9.8 . . . (10 - k) {(x + 1)}^{9 - k} {(- 3)}^{n - k} e^{- 3 x} \Rightarrow$ $f^{(7)} (0) = 1 + \sum_{k = 1}^{7} C_{7}^{k} 9.8 . . . . (10 - k) {(- 3)}^{7 - k}$ $f^{(5)} (1) = 2^{9} {(- 3)}^{5} e^{- 3} + \sum_{k = 1}^{5} C_{5}^{k} 9.8 . . . . (10 - k) 2^{5 - k} e^{- 3}$
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