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Question Number 90960 by mathmax by abdo last updated on 27/Apr/20

let f(x) =x^3 cos(2x) calculate f^((n)) (x) and f^((n)) (0)

$${let}\:{f}\left({x}\right)\:={x}^{\mathrm{3}} {cos}\left(\mathrm{2}{x}\right) \\ $$$${calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$

Commented by mathmax by abdo last updated on 27/Apr/20

leibniz give f^((n)) (x) =Σ_(k=0) ^n C_n ^k (x^3 )^((k)) ( cos(2x))^((n−k)) first let find (cos(2x))^((p)) (cos(2x))^((1)) =−2sin(2x) =2 cos(2x+(π/2)) (cos(2x))^((2)) =4cos(2x+2(π/2)) let suppose (cos(2x))^((p)) =2^p cos(2x+((pπ)/2)) ⇒(cos(2x))^((p+1)) =−2^(p+1) sin(2x+((pπ)/2)) =2^(p+1) cos(2x+(((p+1)π)/2)) ⇒ f^((n)) (x) =C_n ^0 x^3 2^n cos(2x+((nπ)/2))+C_n ^1 3x^2 2^(n−1) cos(2x+(((n−1)π)/2)) +C_n ^2 (6x) 2^(n−2) cos(2x+(((n−2)π)/2)) +C_n ^3 6 2^(n−3) cos(2x+(((n−3)π)/2)) f^((n)) (x)=2^n x^3 cos(2x+((nπ)/2))+3n2^(n−1) x^2 cos(2x+(((n−1)π)/2)) +3n(n−1) 2^(n−2) x cos(2x+(((n−2)π)/2))+n(n−1)(n−2)2^(n−3) cos(2x+(((n−3)π)/2))

$${leibniz}\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \:\left({x}^{\mathrm{3}} \right)^{\left({k}\right)} \left(\:{cos}\left(\mathrm{2}{x}\right)\right)^{\left({n}−{k}\right)} \\ $$$${first}\:{let}\:{find}\:\:\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left({p}\right)} \\ $$$$\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left(\mathrm{1}\right)} \:=−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)\:=\mathrm{2}\:{cos}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left(\mathrm{2}\right)} \:=\mathrm{4}{cos}\left(\mathrm{2}{x}+\mathrm{2}\frac{\pi}{\mathrm{2}}\right)\:{let}\:{suppose} \\ $$$$\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left({p}\right)} \:=\mathrm{2}^{{p}} \:{cos}\left(\mathrm{2}{x}+\frac{{p}\pi}{\mathrm{2}}\right)\:\Rightarrow\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left({p}+\mathrm{1}\right)} =−\mathrm{2}^{{p}+\mathrm{1}} \:{sin}\left(\mathrm{2}{x}+\frac{{p}\pi}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}^{{p}+\mathrm{1}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({p}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:={C}_{{n}} ^{\mathrm{0}} {x}^{\mathrm{3}} \:\mathrm{2}^{{n}} \:{cos}\left(\mathrm{2}{x}+\frac{{n}\pi}{\mathrm{2}}\right)+{C}_{{n}} ^{\mathrm{1}} \mathrm{3}{x}^{\mathrm{2}} \:\mathrm{2}^{{n}−\mathrm{1}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$$+{C}_{{n}} ^{\mathrm{2}} \:\left(\mathrm{6}{x}\right)\:\mathrm{2}^{{n}−\mathrm{2}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{2}\right)\pi}{\mathrm{2}}\right)\:+{C}_{{n}} ^{\mathrm{3}} \mathrm{6}\:\mathrm{2}^{{n}−\mathrm{3}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right) \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\mathrm{2}^{{n}} \:{x}^{\mathrm{3}} \:{cos}\left(\mathrm{2}{x}+\frac{{n}\pi}{\mathrm{2}}\right)+\mathrm{3}{n}\mathrm{2}^{{n}−\mathrm{1}} {x}^{\mathrm{2}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$$+\mathrm{3}{n}\left({n}−\mathrm{1}\right)\:\mathrm{2}^{{n}−\mathrm{2}} \:{x}\:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{2}\right)\pi}{\mathrm{2}}\right)+{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{3}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right) \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 27/Apr/20

f^((n)) (0) =n(n−1)(n−2)2^(n−3) cos((((n−3)π)/2))

$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{3}} \:{cos}\left(\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right) \\ $$

Answered by MWSuSon last updated on 27/Apr/20

f^((n)) (x)=[2^n cos (2x+((nπ)/2))x^3 +n2^(n−1) cos (2x+(((n−1)π)/2))3x^2 +((n(n−1)2^(n−2) cos (2x+(((n−2)π)/2))6x)/2)+((n(n−1)(n−2)2^(n−3) cos (2x+(((n−3)π)/2))6)/6). f^((n)) (0)=[n(n−1)(n−2)2^(n−3) cos ((((n−3)π)/2))]

$${f}^{\left({n}\right)} \left({x}\right)=\left[\mathrm{2}^{{n}} \mathrm{cos}\:\left(\mathrm{2}{x}+\frac{{n}\pi}{\mathrm{2}}\right){x}^{\mathrm{3}} +{n}\mathrm{2}^{{n}−\mathrm{1}} \mathrm{cos}\:\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right)\mathrm{3}{x}^{\mathrm{2}} \right. \\ $$$$+\frac{{n}\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}−\mathrm{2}} \mathrm{cos}\:\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{2}\right)\pi}{\mathrm{2}}\right)\mathrm{6}{x}}{\mathrm{2}}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{3}} \mathrm{cos}\:\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right)\mathrm{6}}{\mathrm{6}}. \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)=\left[{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{3}} \mathrm{cos}\:\left(\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right)\right] \\ $$

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