let f(x) =e^(−2x) ln(1+2x) 1) find f^((n)) (x) and f^((n)) (0) 2)developp f at integr serie


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Question Number 83245 by mathmax by abdo last updated on 29/Feb/20

$l e t f (x) = e^{- 2 x} l n (1 + 2 x)$ $1) f i n d f^{(n)} (x) a n d f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e$
Commented by mathmax by abdo last updated on 03/Mar/20
$1) f^((n)) (x)=Σ_(k=0) ^n C_n ^k (ln(2x+1))^((k)) (e^(−2x) )^((n−k)) =ln(2x+1)(−2)^n e^(−2x) +Σ_(k=1) ^n C_n ^k (ln(2x+1))^((k)) (−2)^(n−k) e^(−2x) we have (ln(2x+1))^((1)) =(2/(2x+1)) =(1/(x+(1/2))) ⇒ (ln(2x+1))^((k)) =((1/(x+(1/2))))^((k−1)) =(((−1)^(k−1) (k−1)!)/((x+(1/2))^k )) ⇒ f^((n)) (x)=ln(2x+1)(−2)^n e^(−2x) +Σ_(k=1) ^n C_n ^k ((2^k (−1)^(k−1) (k−1)!)/((2x+1)^k ))(−2)^(n−k) e^(−2x) =ln(2x+1)(−2)^n e^(−2x) +2^n Σ_(k=1) ^n C_n ^k (((−1)^(k−1+n−k) (k−1)!)/((2x+1)^k )) e^(−2x) f^((n)) (x)=(−2)^n e^(−2x) { ln(2x+1)+Σ_(k=1) ^n (k−1)! ×(C_n ^k /((2x+1)^k ))} f^((n)) (0) =Σ_(k=1) ^n (k−1)!C_n ^k =Σ_(k=1) ^n (k−1)!×((n!)/(k!(n−k)!)) =n!Σ_(k=1) ^n (1/(k(n−k)!))$
$1) f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(l n (2 x + 1))}^{(k)} {(e^{- 2 x})}^{(n - k)}$ $= l n (2 x + 1) {(- 2)}^{n} e^{- 2 x} + \sum_{k = 1}^{n} C_{n}^{k} {(l n (2 x + 1))}^{(k)} {(- 2)}^{n - k} e^{- 2 x}$ $w e h a v e {(l n (2 x + 1))}^{(1)} = \frac{2}{2 x + 1} = \frac{1}{x + \frac{1}{2}} \Rightarrow$ ${(l n (2 x + 1))}^{(k)} = {(\frac{1}{x + \frac{1}{2}})}^{(k - 1)} = \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + \frac{1}{2})}^{k}} \Rightarrow$ $f^{(n)} (x) = l n (2 x + 1) {(- 2)}^{n} e^{- 2 x} + \sum_{k = 1}^{n} C_{n}^{k} \frac{2^{k} {(- 1)}^{k - 1} (k - 1)!}{{(2 x + 1)}^{k}} {(- 2)}^{n - k} e^{- 2 x}$ $= l n (2 x + 1) {(- 2)}^{n} e^{- 2 x} + 2^{n} \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1 + n - k} (k - 1)!}{{(2 x + 1)}^{k}} e^{- 2 x}$ $f^{(n)} (x) = {(- 2)}^{n} e^{- 2 x} {l n (2 x + 1) + \sum_{k = 1}^{n} (k - 1)! \times \frac{C_{n}^{k}}{{(2 x + 1)}^{k}}}$ $f^{(n)} (0) = \sum_{k = 1}^{n} (k - 1)! C_{n}^{k} = \sum_{k = 1}^{n} (k - 1)! \times \frac{n!}{k! (n - k)!}$ $= n! \sum_{k = 1}^{n} \frac{1}{k (n - k)!}$
Commented by mathmax by abdo last updated on 03/Mar/20

$2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = f (0) + \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$ $= \sum_{n = 1}^{\infty} (\sum_{k = 1}^{n} \frac{1}{k (n - k)!}) x^{n}$
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