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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 approved 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
Aug
4
comment Prove that $e^{\sum 1/p_k^2} > \pi/2$
Yes, realized that a while ago. Proving things without sleep is a bad idea.
Aug
4
answered Prove that $e^{\sum 1/p_k^2} > \pi/2$
Aug
1
comment Prove: $\int_0^{\infty}\left(\frac{\sin x}{x}\right)^2dx=\pi/2$
@PeterTamaroff is correct (+1). Put $\delta = \frac{1}{m}$ for some $m$. What does that give you?
Jul
31
comment Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$
@BarryCipra Agreed.
Jul
31
comment Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$
+1 Shouldn't that be <= because it is equal for n = 1?
Jul
29
comment A minimization problem
Taking derivative gives $w - u + 2\beta\frac{w}{x^{\circ 2}}$ and second derivative is semi-definite. Therefore, setting the derivative to zero should give you a local minimum.
Jul
29
comment Find the sum : $\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$
+1 This is a very beautiful answer!
Jul
29
revised Show if $\|\cdot\|$ is a norm then $\|f(\cdot)\|$ is a norm where $f$ is linear and invertible
Minor correction to notation
Jul
29
suggested approved edit on Show if $\|\cdot\|$ is a norm then $\|f(\cdot)\|$ is a norm where $f$ is linear and invertible
Jul
27
revised show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$
added 95 characters in body
Jul
27
answered show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$
Jul
26
comment How to show $\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$?
@EugeneBulkin True. I apologize for the confusion caused.
Jul
25
comment How to show $\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$?
Your definition of f(x) should not contain that n.
Jul
23
comment Given $\{\log_ab \mid a,b\in \mathbb N, \mathrm{gcd}(a,b)=1,a,b≥3\}$ does the sum of any two $\log_ab$ form an irrational or rational number?
Try verifying the conditions for the group. Is it closed? Is it associative? Does it have an identity? What about inverses?