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Question Number 73487 by abdomathmax last updated on 13/Nov/19

$$let\alpha and\beta rootsoftheequation{x}^2-x+2=0$$$$simplify{A}_p={\alpha }^p+{\beta }^{p}andcalculate$$$$\sum _{p=0}^{n-1}{A}_{p}and\sum _{p=0}^{n-1}{A}_p^2$$

Commented by mathmax by abdo last updated on 15/Nov/19

$$letsolve{x}^2-x+2=0\to \mathrm{\Delta }=1-8=-7\Rightarrow$$$$\alpha =\frac{1+i\sqrt{7}}{2}and\beta =\frac{1-i\sqrt{7}}{2}$$$$\mid \alpha \mid =\frac{1}{2}\sqrt{1+7}=\frac{2\sqrt{2}}{2}=\sqrt{2}\Rightarrow \alpha =\sqrt{2}{e}^{iarctan\left(\sqrt{7}\right)}and\beta =\sqrt{2}{e}^{-iarctan\left(\sqrt{7}\right)}$$$$A_p={\alpha }^p+{\beta }^p={\left(\sqrt{2}\right)}^p{e}^{iparctan\left(\sqrt{7}\right)}+{\left(\sqrt{2}\right)}^p{e}^{-iparctan\left(\sqrt{7}\right)}$$$$=2^{\frac{p}{2}}\times 2Re\left(e^{iparctan\left(\sqrt{7}\right)}\right)=2^{1+\frac{p}{2}}cos\left(parctan\left(\sqrt{7}\right)\right)$$$$\sum _{p=0}^{n-1}{A}_p=\sum _{p=0}^{n-1}{2}^{1+\frac{p}{2}}cos\left(parctan\left(\sqrt{7}\right)\right)$$$$=Re\left(2\sum _{p=0}^{n-1}{\left(\sqrt{2}\right)}^p{e}^{iparctan\left(\sqrt{7}\right)}\right)wehave$$$$\sum _{p=0}^{n-1}{\left(\sqrt{2}\right)}^p{e}^{iparctan\left(\sqrt{7}\right)})=\sum _{p=0}^{n-1}{\left(\sqrt{2}{e}^{iarctan\left(\sqrt{7}\right)}\right)}^p$$$$=\frac{1-{\left(\sqrt{2}{e}^{iarctan\left(\sqrt{7}\right)}\right)}^n}{1-\sqrt{2}{e}^{iarctan\left(\sqrt{7}\right)}}=\frac{1-{\left(\sqrt{2}\right)}^n{e}^{inarctan\left(\sqrt{7}\right)}}{1-\sqrt{2}{e}^{isrctan\left(\sqrt{7}\right)}}$$$$=\frac{(1-{\left(\sqrt{2}\right)}^n(cos(narctan\left(\sqrt{7}\right)+isin\left(narctan\left(\sqrt{7}\right)\right)}{1-\sqrt{2}(cos(arctan\left(\sqrt{7}\right)+isin\left(arctan\left(\sqrt{7}\right)\right)}$$$$resttoextractRe(ofthisquantity...)$$

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