let f(x) =(√(1+x^2 )) 1) let U_n =f^((n)) (x) prove that Σ_(k=0) ^n C_n ^k U_k U_(n+1−k) =0 for n≥2 2) developp f at integr serie .


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Question Number 60706 by maxmathsup by imad last updated on 24/May/19

$l e t f (x) = \sqrt{1 + x^{2}}$ $1) l e t U_{n} = f^{(n)} (x) p r o v e t h a t \sum_{k = 0}^{n} C_{n}^{k} U_{k} U_{n + 1 - k} = 0 f o r n ⩾ 2$ $2) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by maxmathsup by imad last updated on 03/Jun/19
$1) we have f(x)=(√(1+x^2 )) ⇒f^2 (x) =1+x^2 ⇒2f(x)f^′ (x)=2x ⇒ ∀n≥2 {f(x)f^′ (x)}^((n)) =0 ⇒Σ_(k=0) ^n C_n ^k (f(x))^((k)) (f^′ (x))^((n−k)) =0 ⇒ Σ_(k=0) ^n C_n ^k f^((k)) (x) f^((n−k+1)) (x) =0 ⇒Σ_(k=0) ^n C_n ^k U_k U_(n+1−k) =0 for n≥2$
$1) w e h a v e f (x) = \sqrt{1 + x^{2}} \Rightarrow f^{2} (x) = 1 + x^{2} \Rightarrow 2 f (x) f^{^{'}} (x) = 2 x \Rightarrow$ $\forall n ⩾ 2 {f (x) f^{^{'}} (x)}^{(n)} = 0 \Rightarrow \sum_{k = 0}^{n} C_{n}^{k} {(f (x))}^{(k)} {(f^{^{'}} (x))}^{(n - k)} = 0 \Rightarrow$ ${\sum^{n}}_{k = 0} C_{n}^{k} f^{(k)} (x) f^{(n - k + 1)} (x) = 0 \Rightarrow \sum_{k = 0}^{n} C_{n}^{k} U_{k} U_{n + 1 - k} = 0 f o r n ⩾ 2$
Commented by maxmathsup by imad last updated on 03/Jun/19

${(1 + u)}^{α} = 1 + \frac{α}{1!} u + \frac{α (α - 1)}{2!} u^{2} + . . . . \frac{α (α - 1) . . . (α - n + 1)}{n!} u^{n} + . . . \Rightarrow f o r α = \frac{1}{2}$ $a n d u = x^{2} w e g e t f (x) = 1 + \frac{\frac{1}{2}}{1!} x^{2} + \frac{\frac{1}{2} (\frac{1}{2} - 1)}{2!} x^{4} + . . . \frac{\frac{1}{2} (\frac{1}{2} - 1) . . (\frac{1}{2} - n + 1)}{n!} x^{2 n} + . . .$ $= 1 + \frac{1}{2} x^{2} - \frac{1}{8} x^{4} + \frac{\frac{1}{2} (- \frac{1}{2}) (\frac{- 3}{2}) . . . . (\frac{3 - 2 n}{2})}{n!} x^{2 n} + . . . .$
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