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1d
comment Does the Riemann Hypothesis consider mirror symmetry on its non-trivial zeros?
The answer to the question in your title is "Yes." However, I see no relation at all between your post, your picture, and the zeta function. I've downvoted. I also note that this picture was posted in your last question on this topic. I've downvoted.
2d
answered Inquiry on big $O$ notation
Feb
3
answered Asymptotics of “ugly” function elucidate Goldbach's conjecture?
Feb
2
answered How to find proper functions to bound for integral squeeze thereom
Feb
1
answered estimation for n-th prime
Feb
1
comment What is the probability that 3 Heads occur before 2 Tails occur?
One way to do this from first principles would be to find the probability that you get three heads in a row, and add it to (the probability of getting 3 heads and one tail) multiplied by (the number of orders in which you can get 3 heads and one tail). That is, if you mean 3 heads and not 3 heads in a row.
Jan
30
comment Is the Riemann Hypothesis incorrect?
What does this have to do with the Riemann Hypothesis? Actually, let me rephrase. You see to think you can construct a zero of the zeta function that's not on the critical line. So what is it?
Jan
28
comment Proving that if $a|b$ and $b|a$, then $a = \pm b$ for $a,b$ as nonzero integers?
As a brief note: $a \mid b$ is not $a/b$. Instead, $a\mid b$ is either true or false, and means that $a$ divides $b$.
Jan
28
awarded  Nice Answer
Jan
27
revised Product of two sums, one finite and one infinite
edited body
Jan
27
answered Product of two sums, one finite and one infinite
Jan
27
comment Product of two sums, one finite and one infinite
As you've written these, both are infinite. Did you mean something different? Also, you'll likely find yourself confused if you use $k$ as the index of summation for two different summations.
Jan
27
comment Proving convergence for $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$
I'm not sure what you're asking. Do you mean to have $(\ln n)^a$ not in a denominator right now? Regardless, $\ln n \to \infty$. So if you have it to a positive power $a$, then $(\ln n)^a \to \infty$. If you have a negative power $b$, then $(\ln n)^b \to 0$. The first is because positive powers of large numbers are large. The second is because dividing by positive powers of large numbers is like dividing by a large number, and is small.
Jan
27
comment Proving convergence for $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$
Because $\ln n \to \infty$, and dividing $1$ into very very many pieces means those pieces are small (i.e. going to zero).
Jan
27
revised Proving convergence for $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$
deleted 2 characters in body; edited tags
Jan
27
answered Proving convergence for $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$
Jan
27
comment If $a^4 + 4^b$ is prime, then $a$ is odd and $b$ is even.
@CameronWilliams Yep, I sure did. Thank you.
Jan
27
revised If $a^4 + 4^b$ is prime, then $a$ is odd and $b$ is even.
deleted 1 character in body
Jan
27
revised If $a^4 + 4^b$ is prime, then $a$ is odd and $b$ is even.
edited title
Jan
26
answered Using Set Builder Notation on a set that jumps in intervals?