3,455 reputation
11344
bio website
location Brazil
age 17
visits member for 2 years, 10 months
seen 1 hour ago

3h
comment Is mathematics invented or discovered?
This was the very first question in the philosophy site.
12h
comment Challenging Infinite summation involving the zeta function
@DavidH I see, thanks! It is indeed a duplicate, then.
13h
comment Challenging Infinite summation involving the zeta function
@DavidH, it seems to me that the proposed duplicate talks about the derivative of the trigamma function $\psi'$, not about $\psi$ itself.
2d
comment Why is there a 'missing' $1$ in the Euler–Mascheroni constant?
Yes, they are identical (in the limit). Compute $\lim\limits_{n\to\infty} \log(n+1)-\log(n)$.
Aug
26
comment How to show the convergence of this infinite series: $\frac{x}{1+x}- \frac{x^2}{1+x^2}+ \frac{x^3}{1+x^3}\dots$
Just show the terms are going to zero and apply the alternating series test.
Aug
26
comment Heuristic for Dirichlet's Theorem on Arithemtic Progression
Mastrel, the heuristic is that the primes seem are essentially random and that there are easy cases to prove. Of course, one could ask how do we know primes are random. We don't because they aren't random, but experience shows they are quite. This lack of understanding in the behaviour of primes keeps this heuristic as heuristic, and not as method of proof.
Aug
26
comment Countable and uncountable sets in Riemann integration
We either have to give up the fact that the union of the intervals is a partition or that there are only countably many intervals. In my opinion, the second option is much more sensible.
Aug
26
comment Countable and uncountable sets in Riemann integration
Hafej, just to make it clear: what happens when $\delta=0$ does not matter in any way for the limit. As we will see, the case $\delta=0$ is ill-defined and basically can do what we want it to do. If we insist that it is still a partition of the original interval, it cannot generate only countably many intervals, and for any $p\in[a,b]$ there is a set $\{p\}$ now. If we insist there are only countably many intervals, they cannot cover $[a,b]$ (and be a partition of it) because it would generate a countable set.
Aug
26
comment Tractable indefinite integral of the exponentiation of some function
I deleted my answer because Antonio Vargas has already said the same thing.
Aug
26
answered Countable and uncountable sets in Riemann integration
Aug
26
answered Heuristic for Dirichlet's Theorem on Arithemtic Progression
Aug
26
comment Suppose f(x) + 2f(1/x) = x . Evaluate f(5) in simplest form.
Study the cases $x=5$ and $x=1/5$
Aug
23
revised Where did the negative answer come from in $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}$?
added 5 characters in body
Aug
23
revised Where did the negative answer come from in $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}$?
added 22 characters in body; edited title
Aug
15
comment Find all non-trivial triplets $(a,b,c)$ such that $ a+b^{d}\equiv0\pmod c $ for all $d>0$
To get nice spacing for $\pmod p$, use \pmod{c}; for example, compare $\mod p^k$, given by \mod p^k, with $\pmod{p^k}$, given by \pmod{p^k}.
Aug
15
comment Find all non-trivial triplets $(a,b,c)$ such that $ a+b^{d}\equiv0\pmod c $ for all $d>0$
Why don't you just write $a+b^d$, for $d\gt 0$?
Aug
14
comment Convergence of $\sum_{n=1}^{\infty} n$ and integral test
To make the situation clearer, you probably know $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}.$$ This means $$\sum_{k=0}^\infty 2^k=1+2+4+8+\cdots = \frac{1}{1-2}=-1,$$ right? Err, no. This does not converge to $-1$, we are outside the area of convergence $|x|<1$. But I can abuse the notation and write $1+2+4+\cdots=-1$ if I want, even though this is outside the area of convergence. This is very, very, similar to what is happening: the $-1$ is outside the area of convergence of the sum. I suggest you take a look at the many questions previously asked and try to digest the information.
Aug
14
comment Convergence of $\sum_{n=1}^{\infty} n$ and integral test
Short answer: you are right, this does not converge and the value $-1/12$ is instead associated to $\zeta(-1)$, a "extended version" of $p$-series. This abuse of notation was exhaustively addressed on his site before. As dylan7 noted, the integral test cannot be applied to this test blindly (as this series is not monotone decreasing) but you can easily show by area comparison that the sum is divergent.
Aug
14
revised Prove that expression is never an perfect square.
edited body
Aug
14
answered Prove that expression is never an perfect square.