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Question Number 109574 by mathdave last updated on 24/Aug/20

Commented by mathdave last updated on 24/Aug/20

solution to earlier post

$${solution}\:{to}\:{earlier}\:{post} \\ $$

Commented by Sarah85 last updated on 24/Aug/20

Me and others solved it. Why do you post it  again? Strange behavior...

$$\mathrm{Me}\:\mathrm{and}\:\mathrm{others}\:\mathrm{solved}\:\mathrm{it}.\:\mathrm{Why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{post}\:\mathrm{it} \\ $$$$\mathrm{again}?\:\mathrm{Strange}\:\mathrm{behavior}... \\ $$

Commented by mathdave last updated on 24/Aug/20

because you guys didnt arrived the particular  answer of  ((2π)/(3(√3)))

$${because}\:{you}\:{guys}\:{didnt}\:{arrived}\:{the}\:{particular} \\ $$$${answer}\:{of}\:\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

Commented by Sarah85 last updated on 24/Aug/20

((2π)/(3(√3)))=((2π(√3))/9) as you might know; and I arrived  there and others also.  your path is an unnecessary detour; of course  we can solve 1+1 by starting with  1=Γ(e^(2πi) ×sin^(−1)  ((π/2)−i×ln (2+(√3))) but...

$$\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:\mathrm{as}\:\mathrm{you}\:\mathrm{might}\:\mathrm{know};\:\mathrm{and}\:\mathrm{I}\:\mathrm{arrived} \\ $$$$\mathrm{there}\:\mathrm{and}\:\mathrm{others}\:\mathrm{also}. \\ $$$$\mathrm{your}\:\mathrm{path}\:\mathrm{is}\:\mathrm{an}\:\mathrm{unnecessary}\:\mathrm{detour};\:\mathrm{of}\:\mathrm{course} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{1}+\mathrm{1}\:\mathrm{by}\:\mathrm{starting}\:\mathrm{with} \\ $$$$\mathrm{1}=\Gamma\left(\mathrm{e}^{\mathrm{2}\pi\mathrm{i}} ×\mathrm{sin}^{−\mathrm{1}} \:\left(\frac{\pi}{\mathrm{2}}−\mathrm{i}×\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right)\:\mathrm{but}...\right. \\ $$

Commented by mathdave last updated on 24/Aug/20

yes ma thanks

$${yes}\:{ma}\:{thanks}\: \\ $$

Commented by mnjuly1970 last updated on 24/Aug/20

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