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Question Number 90960 by mathmax by abdo last updated on 27/Apr/20 | ||
$${let}\:{f}\left({x}\right)\:={x}^{\mathrm{3}} {cos}\left(\mathrm{2}{x}\right) \\ $$$${calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$ | ||
Commented by mathmax by abdo last updated on 27/Apr/20 | ||
$${leibniz}\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \:\left({x}^{\mathrm{3}} \right)^{\left({k}\right)} \left(\:{cos}\left(\mathrm{2}{x}\right)\right)^{\left({n}−{k}\right)} \\ $$$${first}\:{let}\:{find}\:\:\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left({p}\right)} \\ $$$$\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left(\mathrm{1}\right)} \:=−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)\:=\mathrm{2}\:{cos}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left(\mathrm{2}\right)} \:=\mathrm{4}{cos}\left(\mathrm{2}{x}+\mathrm{2}\frac{\pi}{\mathrm{2}}\right)\:{let}\:{suppose} \\ $$$$\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left({p}\right)} \:=\mathrm{2}^{{p}} \:{cos}\left(\mathrm{2}{x}+\frac{{p}\pi}{\mathrm{2}}\right)\:\Rightarrow\left({cos}\left(\mathrm{2}{x}\right)\right)^{\left({p}+\mathrm{1}\right)} =−\mathrm{2}^{{p}+\mathrm{1}} \:{sin}\left(\mathrm{2}{x}+\frac{{p}\pi}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}^{{p}+\mathrm{1}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({p}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:={C}_{{n}} ^{\mathrm{0}} {x}^{\mathrm{3}} \:\mathrm{2}^{{n}} \:{cos}\left(\mathrm{2}{x}+\frac{{n}\pi}{\mathrm{2}}\right)+{C}_{{n}} ^{\mathrm{1}} \mathrm{3}{x}^{\mathrm{2}} \:\mathrm{2}^{{n}−\mathrm{1}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$$+{C}_{{n}} ^{\mathrm{2}} \:\left(\mathrm{6}{x}\right)\:\mathrm{2}^{{n}−\mathrm{2}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{2}\right)\pi}{\mathrm{2}}\right)\:+{C}_{{n}} ^{\mathrm{3}} \mathrm{6}\:\mathrm{2}^{{n}−\mathrm{3}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right) \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\mathrm{2}^{{n}} \:{x}^{\mathrm{3}} \:{cos}\left(\mathrm{2}{x}+\frac{{n}\pi}{\mathrm{2}}\right)+\mathrm{3}{n}\mathrm{2}^{{n}−\mathrm{1}} {x}^{\mathrm{2}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$$+\mathrm{3}{n}\left({n}−\mathrm{1}\right)\:\mathrm{2}^{{n}−\mathrm{2}} \:{x}\:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{2}\right)\pi}{\mathrm{2}}\right)+{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{3}} \:{cos}\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right) \\ $$$$ \\ $$ | ||
Commented by mathmax by abdo last updated on 27/Apr/20 | ||
$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{3}} \:{cos}\left(\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right) \\ $$ | ||
Answered by MWSuSon last updated on 27/Apr/20 | ||
$${f}^{\left({n}\right)} \left({x}\right)=\left[\mathrm{2}^{{n}} \mathrm{cos}\:\left(\mathrm{2}{x}+\frac{{n}\pi}{\mathrm{2}}\right){x}^{\mathrm{3}} +{n}\mathrm{2}^{{n}−\mathrm{1}} \mathrm{cos}\:\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right)\mathrm{3}{x}^{\mathrm{2}} \right. \\ $$$$+\frac{{n}\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}−\mathrm{2}} \mathrm{cos}\:\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{2}\right)\pi}{\mathrm{2}}\right)\mathrm{6}{x}}{\mathrm{2}}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{3}} \mathrm{cos}\:\left(\mathrm{2}{x}+\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right)\mathrm{6}}{\mathrm{6}}. \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)=\left[{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{3}} \mathrm{cos}\:\left(\frac{\left({n}−\mathrm{3}\right)\pi}{\mathrm{2}}\right)\right] \\ $$ | ||