1,427 reputation
827
bio website biditacharya.wordpress.com
location Berkeley, CA
age 19
visits member for 3 years, 3 months
seen Sep 13 at 0:48

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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Aug
23
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?
Aug
23
answered Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.
Aug
22
revised Is $\sum \sin{\frac{\pi}{n}}$ convergent?
Tex formating
Aug
22
suggested suggested edit on Is $\sum \sin{\frac{\pi}{n}}$ convergent?
Aug
20
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 :)
Aug
20
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.'
Aug
15
answered A question on iterated sums
Aug
15
revised Proving that $\mu$ is $\sup S$
added 8 characters in body
Aug
14
revised Proving that $2^{2^n} + 5$ is always composite by working modulo $3$
deleted 591 characters in body
Aug
14
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?
Aug
14
comment Proving that $\mu$ is $\sup S$
(+1) Neat answer!
Aug
14
comment Proving that $\mu$ is $\sup S$
@PeterTamaroff Done!
Aug
14
revised Proving that $\mu$ is $\sup S$
added 23 characters in body
Aug
14
comment Proving that $\mu$ is $\sup S$
My bad! I wrote the definition wrong. I did not realize.
Aug
14
comment Proving that $\mu$ is $\sup S$
but $1.5$ is not an upper bound
Aug
14
comment Proving that $\mu$ is $\sup S$
I am sorry, I don't follow
Aug
14
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
Aug
14
answered Proving that $\mu$ is $\sup S$
Aug
14
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?
Aug
14
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