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Question Number 145338 by puissant last updated on 04/Jul/21
$$\mathrm{d}\acute {\mathrm{e}riv}\acute {\mathrm{e}e}\:\:\:\mathrm{n}-\mathrm{i}\grave {\mathrm{e}me}\:\:\:\mathrm{de}\:\:\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{1}+\mathrm{x}^{\mathrm{6}} } \\ $$
Answered by mathmax by abdo last updated on 04/Jul/21
![f(x)=(x^3 /(1+x^6 )) x^6 +1=0 ⇒x^6 =e^(i(2k+1)π) ⇒roots are x_k =e^(i(2k+1)(π/6)) k∈[[0,5]] ⇒f(x)=(x^3 /(Π_(k=0) ^5 (x−x_k )))=Σ_(k=0) ^5 (a_k /(x−x_k )) a_k =(x_k ^3 /(6z_k ^5 ))=−(x_k ^4 /6) ⇒f(x)=−(1/6)Σ_(k=0) ^5 (x_k ^4 /(x−x_k )) =−(1/6)Σ_(k=0) ^5 (e^(i(2k+1)((2π)/3)) /(x−x_k )) ⇒f^((n)) (x)=−(1/6)Σ_(k=0) ^5 e^(i(2k+1)((2π)/3)) ×(((−1)^n n!)/((x−x_k )^(n+1) )) ⇒f^((n)) (x)=(((−1)^(n+1) n!)/6)Σ_(k=0) ^5 (e^(i(2k+1)((2π)/3)) /((x−x_k )^(n+1) ))](Q145342.png)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{1}+\mathrm{x}^{\mathrm{6}} } \\ $$$$\mathrm{x}^{\mathrm{6}} +\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{x}^{\mathrm{6}} \:=\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi} \:\Rightarrow\mathrm{roots}\:\mathrm{are}\:\mathrm{x}_{\mathrm{k}} =\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\frac{\pi}{\mathrm{6}}} \\ $$$$\mathrm{k}\in\left[\left[\mathrm{0},\mathrm{5}\right]\right]\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{3}} }{\prod_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \left(\mathrm{x}−\mathrm{x}_{\mathrm{k}} \right)}=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{x}−\mathrm{x}_{\mathrm{k}} } \\ $$$$\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{3}} }{\mathrm{6z}_{\mathrm{k}} ^{\mathrm{5}} }=−\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{4}} }{\mathrm{6}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{4}} }{\mathrm{x}−\mathrm{x}_{\mathrm{k}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\frac{\mathrm{2}\pi}{\mathrm{3}}} }{\mathrm{x}−\mathrm{x}_{\mathrm{k}} }\:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\frac{\mathrm{2}\pi}{\mathrm{3}}} ×\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!}{\left(\mathrm{x}−\mathrm{x}_{\mathrm{k}} \right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{n}!}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\frac{\mathrm{2}\pi}{\mathrm{3}}} }{\left(\mathrm{x}−\mathrm{x}_{\mathrm{k}} \right)^{\mathrm{n}+\mathrm{1}} } \\ $$
Commented by puissant last updated on 04/Jul/21
$$\mathrm{Thanks} \\ $$