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Author: Tinku Tara

e-ipi-1-0-e-ipi-1-e-ipi-2-1-2-e-2ipi-1-

Question Number 9255 by geovane10math last updated on 26/Nov/16 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{{i}\pi} \:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{{i}\pi} \:=\:−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({e}^{{i}\pi} \right)^{\mathrm{2}} \:=\:\left(−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{\mathrm{2}{i}\pi} \:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({e}^{\mathrm{2}{i}\pi} \right)^{{i}} \:=\:\mathrm{1}^{{i}}…

Question-9254

Question Number 9254 by tawakalitu last updated on 25/Nov/16 Answered by mrW last updated on 26/Nov/16 $$\mathrm{2cos}\:\pi\mathrm{t}=\mathrm{x} \\ $$$$\mathrm{y}=\mathrm{1}−\mathrm{4cos}\:\mathrm{2}\pi\mathrm{t}=\mathrm{1}−\mathrm{4}\left(\mathrm{2cos}\:^{\mathrm{2}} \pi\mathrm{t}−\mathrm{1}\right) \\ $$$$\mathrm{y}=\mathrm{5}−\mathrm{2}\left(\mathrm{2cos}\:\pi\mathrm{t}\right)^{\mathrm{2}} \\ $$$$\mathrm{hence}\:\mathrm{y}=\mathrm{5}−\mathrm{2x}^{\mathrm{2}} \\…

Question-74786

Question Number 74786 by chess1 last updated on 30/Nov/19 Commented by abdomathmax last updated on 30/Nov/19 $${S}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{k}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)^{{k}} \:={w}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)\:{with} \\ $$$${w}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{kx}^{{k}} \:\:\:{we}\:{have}\:{for}\:{x}\neq\mathrm{1}…

Question-74782

Question Number 74782 by naka3546 last updated on 30/Nov/19 Commented by abdomathmax last updated on 01/Dec/19 $${let}\:{f}\left({x}\right)=\left\{\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{x}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right\}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow \\ $$$${ln}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left\{\:\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{x}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}}…

Question-74776

Question Number 74776 by chess1 last updated on 30/Nov/19 Answered by mind is power last updated on 01/Dec/19 $$\mathrm{we}\:\mathrm{show} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{tan}^{−} \left(\mathrm{x}\right)}{\mathrm{tan}^{−} \left(\mathrm{x}\right)−\mathrm{x}}\right)^{\mathrm{2}}…