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3h
answered The divisibility of $a^p-1$ by $a-1$ and by $(a-1)^2$
3h
comment Simple complex analysis inverse
Try this method math.stackexchange.com/questions/1121059/…
15h
comment Complex hyperbolic Trigonometry
@columbus8myhw, Agreed & the current version is also technically correct
16h
comment Complex hyperbolic Trigonometry
@phandaman, How about this?
16h
answered Complex hyperbolic Trigonometry
20h
comment Is there an obvious reason why $4^n+n^4$ cannot be prime for $n\ge 2$?
@barakmanos, Please find the edited version
20h
answered Is there an obvious reason why $4^n+n^4$ cannot be prime for $n\ge 2$?
21h
comment For how many integers is this a prime number?
@Amad27, Think of the reverse condition.
21h
answered For how many integers is this a prime number?
23h
revised how old are Inee and Imee now?
added 55 characters in body
23h
answered how old are Inee and Imee now?
23h
comment Limit find $f(x)$ $ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $
@Parhs, Yes as we can use $\lim_{h\to0}\dfrac{\sin h}h=1$
23h
answered Limit find $f(x)$ $ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $
1d
answered Antiderivative of $\frac{\sqrt{4-x}}{x\sqrt{x}}$
1d
comment Imaginary part of $ln(\sqrt{i})?$
@Did, How to increase the legibility of the exponent?
1d
revised Imaginary part of $ln(\sqrt{i})?$
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1d
revised Finding the limit of the following expression
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1d
answered Finding the limit of the following expression
1d
answered Imaginary part of $ln(\sqrt{i})?$
1d
comment Number of real roots of $2 \cos\left(\frac{x^2+x}{6}\right)=2^x+2^{-x}$
@User786, The 1st one is AM-GM. We have $2\cos\dfrac{x^2+x}6\le2$ and $2^x+2^{-x}\ge2$. For the existence of the solution, what do we need?