f(x)=(x−3)^5 ln(1+x) f^((2020)) (3)=?


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Question Number 91185 by Tony Lin last updated on 28/Apr/20

$f (x) = {(x - 3)}^{5} l n (1 + x)$ $f^{(2020)} (3) = ?$
Commented by Tony Lin last updated on 28/Apr/20

$f (x + 3)$ $= x^{5} l n (4 + x)$ $= x^{5} (x - \frac{x^{2}}{2 \cdot 4} + \frac{x^{3}}{3 \cdot 4^{2}} \cdot \cdot \cdot - \frac{x^{2015}}{2015 \cdot 4^{2014}} + \cdot \cdot \cdot)$ $f^{(2020)} (3)$ $= - \frac{2020!}{2015 \cdot 4^{2014}}$
Commented by abdomathmax last updated on 28/Apr/20
$f^((n)) (x) =Σ_(k=0) ^n C_n ^k {(x−3)^5 }^((k)) (ln(1+x))^((n−k)) =C_n ^n (x−3)^5 (ln(1+x))^((n)) +5 C_n ^1 (x−3)^4 (ln(1+x))^((n−1)) +20 C_n ^2 (x−3)^3 (ln(1+x))^((n−2)) +60 C_n ^3 (x−3)^2 (ln(1+x))^((n−3)) +120 C_n ^4 (x−3)(ln(1+x))^((n−4)) +120 C_n ^5 (ln(1+x))^((n−5)) let determine (ln(1+x))^((p)) (p>0) ln^((1)) (1+x) =(1/(1+x)) ⇒(ln(1+x))^((p)) =((1/(1+x)))^((p−1)) =(((−1)^(p−1) (p−1)!)/((1+x)^p )) ⇒(ln(1+x))^((n−k)) =(((−1)^(n−k−1) (n−k−1)!)/((1+x)^(n−k) )) (k≤n) rest to finich the calculus...$
$f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {{(x - 3)}^{5}}^{(k)} {(l n (1 + x))}^{(n - k)}$ $= C_{n}^{n} {(x - 3)}^{5} {(l n (1 + x))}^{(n)} + 5 C_{n}^{1} {(x - 3)}^{4} {(l n (1 + x))}^{(n - 1)}$ $+ 20 C_{n}^{2} {(x - 3)}^{3} {(l n (1 + x))}^{(n - 2)} + 60 C_{n}^{3} {(x - 3)}^{2} {(l n (1 + x))}^{(n - 3)}$ $+ 120 C_{n}^{4} (x - 3) {(l n (1 + x))}^{(n - 4)} + 120 C_{n}^{5} {(l n (1 + x))}^{(n - 5)}$ $l e t d e t e r m i n e {(l n (1 + x))}^{(p)} (p > 0)$ ${l n}^{(1)} (1 + x) = \frac{1}{1 + x} \Rightarrow {(l n (1 + x))}^{(p)} = {(\frac{1}{1 + x})}^{(p - 1)}$ $= \frac{{(- 1)}^{p - 1} (p - 1)!}{{(1 + x)}^{p}} \Rightarrow {(l n (1 + x))}^{(n - k)}$ $= \frac{{(- 1)}^{n - k - 1} (n - k - 1)!}{{(1 + x)}^{n - k}} (k ⩽ n) r e s t t o f i n i c h$ $t h e c a l c u l u s . . .$
	Answered by MWSuSon last updated on 28/Apr/20

	$y = {(x - 3)}^{5} {l o g}_{e} (1 + x)$ $y_{n} = [\frac{{(- 1)}^{n - 1} (n - 1)!}{{(1 + x)}^{n}} {(x - 3)}^{5} + n \frac{{(- 1)}^{n - 2} (n - 2)!}{{(1 + x)}^{n - 1}} 5 {(x - 3)}^{4} + \frac{n (n - 1)}{2} \frac{{(- 1)}^{n - 3} (n - 3)!}{{(1 + x)}^{n - 2}} 20 {(x - 3)}^{3} + \frac{n (n - 1) (n - 2)}{6} \frac{{(- 1)}^{n - 4} (n - 4)!}{{(1 + x)}^{n - 3}} 60 {(x - 3)}^{2} + \frac{n (n - 1) (n - 2) (n - 3)}{24} \frac{{(- 1)}^{n - 5} (n - 5)!}{{(x + 1)}^{n - 4}} 120 (x - 3) + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{120} \frac{{(- 1)}^{n - 6} (n - 6)!}{{(1 + x)}^{n - 5}} 120]$ $i n s e r t i n g 2020 w h e r e y o u s e e n a n d 3 w h e r e y o u s e e x y o u w i l l h a v e y o u r a n s w e r .$ $y_{2020} ∣_{x = 3} =$
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