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Category: Number Theory

If-x-and-y-are-integers-prove-that-x-3-7x-divisible-by-3-

Question Number 97823 by john santu last updated on 10/Jun/20 $$\mathrm{If}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{integers}\:,\:\mathrm{prove} \\ $$$$\mathrm{that}\:{x}^{\mathrm{3}} −\mathrm{7}{x}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}\: \\ $$ Commented by bobhans last updated on 10/Jun/20 $$\mathrm{let}\:{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{write} \\…

0-2-f-x-dx-f-a-f-2-a-For-what-values-of-a-is-the-following-formula-accurate-for-polynomials-of-degree-3-

Question Number 163300 by amin96 last updated on 05/Jan/22 $$\int_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}\backsimeq\boldsymbol{{f}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{f}}\left(\mathrm{2}−\boldsymbol{{a}}\right) \\ $$$$ \\ $$For what values ​​of a is the following formula accurate…

Show-that-RE-1-1-z-1-2-where-z-cos-i-sin-

Question Number 97576 by Rio Michael last updated on 08/Jun/20 $$\:\mathrm{Show}\:\mathrm{that}\:{RE}\left[\frac{\mathrm{1}}{\mathrm{1}−{z}}\right]=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{where}\:{z}\:=\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\theta \\ $$$$ \\ $$ Answered by smridha last updated on 08/Jun/20 $$\boldsymbol{{RE}}\left[\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{{e}}^{\boldsymbol{{i}\theta}} }\right]=\boldsymbol{{RE}}\left[\frac{\mathrm{1}+{e}^{{i}\boldsymbol{\theta}} }{\mathrm{1}−{e}^{\mathrm{2}{i}\boldsymbol{\theta}}…

Given-that-e-i-npi-n-N-show-that-1-n-2-n-1-2-e-1-2-in-please-help-me-out-on-this-i-ve-stumbled-on-it-

Question Number 97413 by Rio Michael last updated on 08/Jun/20 $$\mathrm{Given}\:\mathrm{that}\:\omega\:=\:{e}^{{i}\theta} ,\:\theta\neq\:{n}\pi\:,\:{n}\:\in\mathbb{N} \\ $$$$\mathrm{show}\:\mathrm{that}\:\left(\mathrm{1}\:+\:\omega\right)^{{n}} \:=\:\mathrm{2}^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}\left({in}\theta\right)} \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{out}\:\mathrm{on}\:\mathrm{this},\:\mathrm{i}'\mathrm{ve}\:\mathrm{stumbled}\:\mathrm{on}\:\mathrm{it}. \\ $$ Answered by mathmax by abdo…