1,526 reputation
410
bio website cse.iitb.ac.in/~aruniyer
location India
age 30
visits member for 2 years, 3 months
seen 8 hours ago

Aug
25
suggested suggested edit on solving equations by the method of elimination
Aug
24
comment Metric spaces and openness
a] math.stackexchange.com/questions/203728/… , b] Given any set X, consider all open sets in the metric space that contain X (call them $G_\alpha, \alpha \in A$ where A is some index set) and let $Y = \cap_{\alpha \in A} G_\alpha$. Show that $Y = X$.
Aug
20
suggested suggested edit on Ratio problem to find the woman weekly salary
Aug
15
revised Relationship between lagrange multiplier and constraint
added 1 characters in body
Aug
15
comment Inequality $(1+a_2)^{2}(1+a_3)^{3}… (1+a_n)^{n}>n^{n}$
Your title and description look different.
Aug
15
revised Relationship between lagrange multiplier and constraint
added 3 characters in body
Aug
15
answered Relationship between lagrange multiplier and constraint
Aug
14
comment What is the meaning of $E[g(x)]=Pr(|x| >\varepsilon)E[g(x)|g(x) > g(\varepsilon)]+Pr(g(x) \leq\varepsilon)E[g(x)|g(x)\leq g(\varepsilon)]$
Where did you come across this? Why is there a $|x|$ in there?
Aug
10
comment If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$.
+1 This is very clean.
Aug
10
comment Finding the $n^{\text{th}}$ term of $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots$
Yes, that is a lot more cleaner. :-)
Aug
10
comment Probability of $P(A \cup B \cup C)$.
@user2213654 subtract 1 from both sides, what do you get?
Aug
10
comment Finding the $n^{\text{th}}$ term of $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots$
Consider the numerator: $1\times3\times5\times7 = \frac{1\times2\times3\times4\times5\times6\times7}{2\times4\times6} = \frac{1\times2\times3\times4\times5\times6\times7}{2^3(1\times2\times3)} = \frac{7!}{2^3 3!}$. Can you generalize this? Can you do something similar for the denominator?
Aug
10
answered Probability of $P(A \cup B \cup C)$.
Aug
10
comment Why is there never a proof that extending the reals to the complex numbers will not cause contradictions?
:-) You have asked some really nice questions here. I am sure someone qualified will give an awesome answer soon enough. Regardless, you should take this chance and read about the formal definition of complex numbers and surreal numbers. Have fun! :-)
Aug
5
revised Prove that $e^{\sum 1/p_k^2} > \pi/2$
added 216 characters in body
Aug
5
comment Prove that $e^{\sum 1/p_k^2} > \pi/2$
@TheoJohnson-Freyd I was bounding the series $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} \ln(\zeta(2k))$ below with just the first term in the series. The first term gives $\ln(\pi^2/6)$ which as observed in the comments is slightly larger the final sum and hence the bound as used clearly isn't true. Maybe there is a way to salvage it yet by considering few more terms from the series and showing that it does bound the series below, but my math-fu skill are lacking for that :-).
Aug
4
revised show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$
Minor typo in the final result of the second technique
Aug
4
comment show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$
(+1) The second technique is amazing and makes the whole integral brilliantly simple!
Aug
4
suggested suggested edit on show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$
Aug
4
comment Prove that $e^{\sum 1/p_k^2} > \pi/2$
(+1) for your answer. And it was my answer that used Prime Zeta Function and it wasn't correct, so I retracted it before it got any undeserved attention -_-; sheepish grin