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

f(x)=(x−3)^5 ln(1+x) f^((2020)) (3)=?

$${f}\left({x}\right)=\left({x}−\mathrm{3}\right)^{\mathrm{5}} {ln}\left(\mathrm{1}+{x}\right) \\ $$$${f}^{\left(\mathrm{2020}\right)} \left(\mathrm{3}\right)=? \\ $$

Commented by Tony Lin last updated on 28/Apr/20

f(x+3) =x^5 ln(4+x) =x^5 (x−(x^2 /(2∙4))+(x^3 /(3∙4^2 ))∙∙∙−(x^(2015) /(2015∙4^(2014) ))+∙∙∙) f^((2020)) (3) =−((2020!)/(2015∙4^(2014) ))

$${f}\left({x}+\mathrm{3}\right) \\ $$$$={x}^{\mathrm{5}} {ln}\left(\mathrm{4}+{x}\right) \\ $$$$={x}^{\mathrm{5}} \left({x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\centerdot\mathrm{4}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}\centerdot\mathrm{4}^{\mathrm{2}} }\centerdot\centerdot\centerdot−\frac{{x}^{\mathrm{2015}} }{\mathrm{2015}\centerdot\mathrm{4}^{\mathrm{2014}} }+\centerdot\centerdot\centerdot\right) \\ $$$${f}^{\left(\mathrm{2020}\right)} \left(\mathrm{3}\right) \\ $$$$=−\frac{\mathrm{2020}!}{\mathrm{2015}\centerdot\mathrm{4}^{\mathrm{2014}} } \\ $$

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...

$${f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\:\left\{\left({x}−\mathrm{3}\right)^{\mathrm{5}} \right\}^{\left({k}\right)} \left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}−{k}\right)} \\ $$$$={C}_{{n}} ^{{n}} \:\left({x}−\mathrm{3}\right)^{\mathrm{5}} \left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}\right)} \:+\mathrm{5}\:{C}_{{n}} ^{\mathrm{1}} \left({x}−\mathrm{3}\right)^{\mathrm{4}} \left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}−\mathrm{1}\right)} \\ $$$$+\mathrm{20}\:{C}_{{n}} ^{\mathrm{2}} \:\left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}−\mathrm{2}\right)} \:+\mathrm{60}\:{C}_{{n}} ^{\mathrm{3}} \:\left({x}−\mathrm{3}\right)^{\mathrm{2}} \left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}−\mathrm{3}\right)} \\ $$$$+\mathrm{120}\:{C}_{{n}} ^{\mathrm{4}} \left({x}−\mathrm{3}\right)\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}−\mathrm{4}\right)} \:+\mathrm{120}\:{C}_{{n}} ^{\mathrm{5}} \left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}−\mathrm{5}\right)} \\ $$$${let}\:{determine}\:\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({p}\right)} \:\:\:\:\left({p}>\mathrm{0}\right) \\ $$$${ln}^{\left(\mathrm{1}\right)} \left(\mathrm{1}+{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\Rightarrow\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({p}\right)} =\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\left({p}−\mathrm{1}\right)} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{p}} }\:\Rightarrow\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}−{k}\right)} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−{k}−\mathrm{1}} \left({n}−{k}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}−{k}} }\:\:\left({k}\leqslant{n}\right)\:\:{rest}\:{to}\:{finich} \\ $$$${the}\:{calculus}... \\ $$

Answered by MWSuSon last updated on 28/Apr/20

y=(x−3)^5 log_e (1+x) y_n =[(((−1)^(n−1) (n−1)!)/((1+x)^n ))(x−3)^5 +n(((−1)^(n−2) (n−2)!)/((1+x)^(n−1) ))5(x−3)^4 +((n(n−1))/2)(((−1)^(n−3) (n−3)!)/((1+x)^(n−2) ))20(x−3)^3 +((n(n−1)(n−2))/6)(((−1)^(n−4) (n−4)!)/((1+x)^(n−3) ))60(x−3)^2 +((n(n−1)(n−2)(n−3))/(24))(((−1)^(n−5) (n−5)!)/((x+1)^(n−4) ))120(x−3)+((n(n−1)(n−2)(n−3)(n−4))/(120))(((−1)^(n−6) (n−6)!)/((1+x)^(n−5) ))120] inserting 2020 where you see n and 3 where you see x you will have your answer. y_(2020) ∣_(x=3) =

$${y}=\left({x}−\mathrm{3}\right)^{\mathrm{5}} {log}_{{e}} \left(\mathrm{1}+{x}\right) \\ $$$${y}_{{n}} =\left[\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}} }\left({x}−\mathrm{3}\right)^{\mathrm{5}} +{n}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} \left({n}−\mathrm{2}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} }\mathrm{5}\left({x}−\mathrm{3}\right)^{\mathrm{4}} +\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{3}} \left({n}−\mathrm{3}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} }\mathrm{20}\left({x}−\mathrm{3}\right)^{\mathrm{3}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{6}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{4}} \left({n}−\mathrm{4}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{3}} }\mathrm{60}\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)}{\mathrm{24}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{5}} \left({n}−\mathrm{5}\right)!}{\left({x}+\mathrm{1}\right)^{{n}−\mathrm{4}} }\mathrm{120}\left({x}−\mathrm{3}\right)+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)}{\mathrm{120}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{6}} \left({n}−\mathrm{6}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{5}} }\mathrm{120}\right] \\ $$$${inserting}\:\mathrm{2020}\:{where}\:{you}\:{see}\:{n}\:{and}\:\mathrm{3}\:{where}\:{you}\:{see}\:{x}\:{you}\:{will}\:{have}\:{your}\:{answer}. \\ $$$${y}_{\mathrm{2020}} \mid_{{x}=\mathrm{3}} = \\ $$

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