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Activity for TheCodidacter, or rather ACodidacter‭

Type On... Excerpt Status Date
Comment Post #292112 That's a good point. I'll look more into this. 🙂
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5 months ago
Edit Post #292113 Post edited:
5 months ago
Edit Post #292112 Post edited:
Removed edit message
5 months ago
Comment Post #292113 Oh, that's right. As a LaTeX user I instinctively use `amsmath` for that black square symbol (turns out you don't need to here). Thanks for pointing this out!
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5 months ago
Edit Post #292112 Post edited:
Removed amsmath
5 months ago
Edit Post #292113 Post edited:
Overflow
5 months ago
Edit Post #292112 Post edited:
Edit comment
5 months ago
Edit Post #292113 Post edited:
Question link
5 months ago
Edit Post #292113 Initial revision 5 months ago
Question MathJax not rendered correctly when using package
(Edit: the issue has been solved, as pointed out in this thread.) I have just posted a question) using packages in MathJax, and the math rendering seems to only work starting from the line where the packages are first used. Screenshot: Screenshot (The rendering only works starting from the ...
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5 months ago
Edit Post #292112 Post edited:
Neater formatting
5 months ago
Edit Post #292112 Post edited:
Rendering
5 months ago
Edit Post #292112 Initial revision 5 months ago
Question Find all functions \(f:\mathbb N\to\mathbb N\) such that for all primes \(p\), \(p\mid f(a)^{f(p)}-f(b)^p\iff p\mid a-b\)
\(\require{centernot}\) > Find all functions \(f:\mathbb N\to\mathbb N\) such that for all primes \(p\), \[p\mid f(a)^{f(p)}-f(b)^p\] if and only if \(p\mid a-b\). My proposed solution: Notice that \(p\mid a-b\iff a\equiv b\pmod p\) and by Euler's theorem, \(p\mid f(a)^{f(p)}-f(b)^p\iff ...
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5 months ago
Edit Post #291711 Initial revision 6 months ago
Answer A: How to find a point on a vector equation with another vector equation and perpendicular distance?
Well, with my not-so-advanced vector knowledge, I've got a simpler approach in mind. We name the angle bisector $n$. Let $A(3,0)$ and $P$ be a point in $n$ such that the distance from $P$ to $m$ is $\sqrt{10}$. Drawing the projection of $\vec{AP}$ onto $m$, we get a right triangle and plugging in ...
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6 months ago
Comment Post #291255 > Since floor is monotonic, we also have $x^2 + 1 \geq rx$ implies $\lfloor x^2 + 1 \rfloor \geq \lfloor rx \rfloor$. Oh goodness, I never thought of that. 2 hours in 2 minutes. Thank you very much!
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8 months ago
Edit Post #291254 Initial revision 8 months ago
Question Prove that $\forall x\in\Bbb R:\lfloor x^2\rfloor-\lfloor rx\rfloor\ge-1\iff|r|\le2$.
> Prove that the inequality $\lfloor x^2\rfloor-\lfloor rx\rfloor\ge-1$ holds true for all real numbers $x$ if and only if $|r|\le2$. My (proposed) solution, which took me around 2 hours: > First, we reframe the inequality to $\lfloor x^2\rfloor-\lfloor rx\rfloor+1\ge0$.Suppose $r=a$ satisfies th...
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8 months ago
Comment Post #291104 Update: looked for a while, didn't find anything
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9 months ago
Comment Post #291104 Indeed, the only properties I've used are relating to the angles. I suspected nothing will come out from most other properties. Interesting observation though, there might be something lurking from that $CD\perp MD$. I'll check.
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9 months ago
Edit Post #291104 Post edited:
Geo*****G******ebra.
9 months ago
Edit Post #291104 Post edited:
Geo**G**ebra. The more you know.
9 months ago
Edit Post #291104 Post edited:
Length
9 months ago
Edit Post #291104 Initial revision 9 months ago
Question Find length $MP$ in trapezoid
Trapezoid $ABCD$ is given such that $AB\parallel CD$ and $AD=BD$. Let $M$ be the midpoint of $AB$ and $P$ be the intersection of diagonal $AC$ with the circumcircle of $\triangle BCD$. Given that $BC=27$, $CD=25$, and $AP=10$, find $MP$. Here's some visualization that I made in GeoGebra: image ...
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9 months ago
Comment Post #289826 My code runs on an online IDE, which makes it _much_ slower 🙂 I have tested just 3 candidates on 10,000 simulations, and admittedly to my surprise, you were **absolutely right!** So that's what I've been confusing this whole time–randomly choosing the number of votes per candidates makes it mor...
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about 1 year ago
Comment Post #289826 Continued: ``` Sample probability [average population 2999153595322]: 0.008285 Sample probability [average population 29998082716178]: 0.008359 Sample probability [average population 300075885268942]: 0.008453 Sample probability [average population 2999350982695126]: 0.008428 Sample probability...
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about 1 year ago
Comment Post #289826 Initially my code only relied on `random.random()` for population size. I then tweaked it to make the population size vary (as you said) by using `random.randint()`, and it does converge around $\frac1{(n-1)!}$. The code starts at a small population size then increases it by powers of 10. Interesting...
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about 1 year ago
Comment Post #289826 If I understand your reasoning correctly: the standard deviation of a binomial distribution is proportional to the square root of the population, which in turn makes it more likely for the votes cast between participants to be roughly equal as the population grows. I'd say I agree with this... The...
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about 1 year ago
Edit Post #289823 Post edited:
Clearer wording
about 1 year ago
Edit Post #289823 Initial revision about 1 year ago
Question Strange behavior in elections and pie charts
So, a friend asked me the probability for a candidate to get at least 50% of the total votes in an election consisting of 5 candidates (let's pretend everyone picks at random 🙂). I thought for a bit and then presented an answer: > Let's first generalize so that the election has $n$ candidates. > ...
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about 1 year ago
Edit Post #288411 Post edited:
Slight notational mistake
over 1 year ago
Suggested Edit Post #288411 Suggested edit:
Slight notational mistake
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helpful over 1 year ago
Comment Post #288411 Very clear explanation, thank you 🙂 I'll rephrase your explanation about using $r^2$ on the integral (to make sure I understand correctly): dividing $C_1$ into concentric rings (like [this](https://upload.wikimedia.org/wikipedia/commons/1/17/WA_80_cm_archery_target.svg)), we can visualize that the...
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over 1 year ago
Edit Post #288409 Initial revision over 1 year ago
Question Average distance from circle's center to a point
> What is the average distance from a point inside a circle to the circle's center? I came across this problem, and I've heard the solution is $\frac23R$ for the radius $R$. So, I tried to tackle this myself: Let there be two concentric circles $C1$ and $C2$ with radii $R$ and $r$ ($R\gt r$)....
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over 1 year ago