# Bidit Acharya

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bio website biditacharya.wordpress.com location Berkeley, CA age 19 member for 3 years, 3 months seen Sep 13 at 0:48 profile views 674

Freshman at UC Berkeley. Originally from Nepal

The scientist does not study nature because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful. - Henri Poincaré.

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 Aug23 comment Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$. So if $2^{3n}$, when divided by $7$ leaves a remainder of $1$, what remainder must $2^{3n}-1$ (which is, as you may have noticed, $1$ less than $2^{3n}$) leave when divided by 7? Aug23 answered Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$. Aug22 revised Is $\sum \sin{\frac{\pi}{n}}$ convergent? Tex formating Aug22 suggested suggested edit on Is $\sum \sin{\frac{\pi}{n}}$ convergent? Aug20 comment Why isn't math on the sine of angles the same as math on the angles in degrees? @tomasz, "The thing is, before applying a „rule”, you should verify if it is actually true." that is exactly my point. I don't know why you disagree :) Aug20 comment Why isn't math on the sine of angles the same as math on the angles in degrees? (+1) for the 'In maths it's the single most important thing to stick to given rules and not accidentally "invent" new ones.' Aug15 answered A question on iterated sums Aug15 revised Proving that $\mu$ is $\sup S$ added 8 characters in body Aug14 revised Proving that $2^{2^n} + 5$ is always composite by working modulo $3$ deleted 591 characters in body Aug14 comment Proving that $2^{2^n} + 5$ is always composite by working modulo $3$ First of all, please do note that I am not comparing my answers with others. Now that I read my answer, I realize that I badly phrased my little disclaimer :) What I mean to say is I have included parts like "...this means that any even power of 2 is 1 greater than some multiple of 3..." in my answer which are obvious. So I guess I need to edit/ delete the disclaimer, eh? Aug14 comment Proving that $\mu$ is $\sup S$ (+1) Neat answer! Aug14 comment Proving that $\mu$ is $\sup S$ @PeterTamaroff Done! Aug14 revised Proving that $\mu$ is $\sup S$ added 23 characters in body Aug14 comment Proving that $\mu$ is $\sup S$ My bad! I wrote the definition wrong. I did not realize. Aug14 comment Proving that $\mu$ is $\sup S$ but $1.5$ is not an upper bound Aug14 comment Proving that $\mu$ is $\sup S$ I am sorry, I don't follow Aug14 comment Why is $b^x \overset{\mathrm{def}}{=} \sup \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$ for $b > 1$ a sensible definition? oh, okay thanks Aug14 answered Proving that $\mu$ is $\sup S$ Aug14 comment Why is $b^x \overset{\mathrm{def}}{=} \sup \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$ for $b > 1$ a sensible definition? I'm sorry, but what does $B(x)$ mean in general? Aug14 comment Proving that $\mu$ is $\sup S$ Spot on! :) The question says that $\mu$ is the supremum iff there is an element of $S$ in the interval. But what we just did was started out by assuming that there is no element of $S$ in the interval and proved that if this is the case, then $\mu \ne \sup S$. So, for $\mu = \sup S$, there has to be an element of $S$ in the interval