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8h
comment Has there ever been an application of dividing by zero?
no, actually you reach infinity when $v = c$, and there are particles that do what (they have $m_0 = 0$). Going faster then $c$ would mean that the square root becomes complex. I don't know about physics but my guess is that in that case this equation can no longer be used. (and the existence of this equation per se does not imply that $v > c$ is impossible, only that this equation makes sense only when $v < c$)
20h
comment How to debug math?
+1 for the last line. It really does teach you cool math stuff! When I was in high school I wrote a program to calculate a certain probability, but I made (several) mistakes and my program was always outputting $2.7182818$ which I recognized as $e$. After inspection I found out I was actually calculating $1 + 1/2 + 1/6 + 1/24 + \dots$, and the only logical explanation was that $\sum_{n=1}^\infty \frac 1{n!} = e$ ! I did not know that at the time and I was absolutely thrilled :-)
21h
revised What is the probability that a pokemon gets frozen within 3 turns using ice beam?
added 457 characters in body
21h
answered What is the probability that a pokemon gets frozen within 3 turns using ice beam?
1d
comment Complex integration problem via Cauchy's integral formula
@mrf Ah, right :-) Thanks for the clarification ;)
1d
comment How is Mathematics and Space related?
you work your way up from general principles, experience with the real world and previous theories. You don't just dream up one night what to write down
1d
comment How to debug math?
Also worth noting that the quote is a little misleading: You didn't use "used right calculations, reasoning"; they were in fact wrong. It just seems to be true, but it is not (and this is why proofs are important!). As @barakmanos says, you should be dividing by $3! = 3 \cdot 2$ instead of only $3$.This is because you're not only counting $TTTH$ three times (hence divide by $3$) but you're also counting $TTHT$ two times (and should divide by $2$). This is just to get some intuition, is by no means rigorous
1d
comment How to debug math?
You didn't find anything because "debug math" is a funny way to put it :-) Much better would be "finding the error" or "how to be sure a proof is correct". In a software, you make a mistake somewhere. You keep looking at every part of the code until you find the problem. In math you keep looking at every part of an argument until you find the problem. There are no shortcuts or methodologies: Keep looking until you find the error.If you wrote "debug" with the idea that you can find some parallels like unit testing or whatnot, well there aren't
1d
comment Complex integration problem via Cauchy's integral formula
This works but it's completely useless. Cauchy integral formula works directly on the function $\frac 1{z^2-1}$.. Why do you want to split it?
2d
comment Expected value of a product with indicator function, using brownian motion
Informations are missing. Who is $X$? Are the $X(t)$ iid ?
Apr
30
revised Mathematical Notation of Sequence of Functions
added 141 characters in body
Apr
30
answered Mathematical Notation of Sequence of Functions
Apr
28
comment Need to prove continuity for Intermediate Value Theorem
You're welcome! :-)
Apr
28
comment Need to prove continuity for Intermediate Value Theorem
Indeed, you have to prove continuity. Since $x^a$ is continuos for every $a\ge 0$ and sum of continuos functions is continuos, you are done. If you want to prove those properties from first principle, try to work it out using definitions of continuity and properties of limit
Apr
28
comment Let $A = (0,1]$. Then $\inf(A) = 0$
@Meer remember, $\inf A$ is the biggest number with that property. So you found one number that works i.e. $0$, which is smaller than any element of $A$. But there are many such numbers (like $-1$, $-2$, etc.) you want the biggest of them. Since $0$ works, the $\inf A$ (which is the biggest such number) must be $\ge 0$
Apr
28
comment Let $A = (0,1]$. Then$\text{ inf}(A) = 0$
Indeed it is not. You proof seems a little confused; the fact that $b \in A$ does not mean that it cannot be a lower bound for $A$. What confused you in the other post? I posted an answer, if you have trouble understanding it comment on it! :-) Doing a very similar question is not encouraged :-)
Apr
28
comment For $T:\mathbb{R}^6 \to \mathbb{R}^6$ and $T^5 \neq 0, \; T^6 = 0,$ prove there exists no $S$ such that $S^2 = T.$
how do you go from $S^{12} = 0 \implies S^6 = 0$? How is that different then doing $T^6 = 0 \implies T^3 = 0$, which is false?
Apr
28
comment You have to estimate $\binom{63}{19}$ in $2$ minutes to save your life.
+1 but I am not following the steps when you approximate the factorial :) Could you explain it better?
Apr
28
comment You have to estimate $\binom{63}{19}$ in $2$ minutes to save your life.
@imranfat was that a joke or you can actually write with two hands? :P
Apr
28
revised Can the boy escape the teacher for a regular $n$-gon?
added 6 characters in body