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Activity for r~~‭

Type On... Excerpt Status Date
Edit Post #292189 Post edited:
Be more precise about the n = 0 case
3 months ago
Edit Post #292189 Post edited:
Yikes, missed an important minus sign!
3 months ago
Comment Post #292189 Done, thanks for the tip!
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4 months ago
Edit Post #292189 Post edited:
Update log notation
4 months ago
Edit Post #292189 Post edited:
4 months ago
Edit Post #292189 Post edited:
Simplify sigma formulas
4 months ago
Edit Post #292189 Initial revision 4 months ago
Answer A: Is there a closed formula for multiplication of imaginary units in the direct limit of the Cayley-Dickson construction?
You might find this paper helpful: An Alternate Cayley-Dickson Product. The author observes that while the standard product formula for the Cayley-Dickson construction leads to precisely this difficulty with finding a closed form, there are alternatives for the product that produce isomorphic al...
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4 months ago
Edit Post #290772 Initial revision 9 months ago
Answer A: Minimal non-standard number in non-standard models of PA
Quite the opposite; in no non-standard model of Peano arithmetic is there a minimal non-standard number. Consider the formula \\(\phi(x) = \left(x = 0\right) \vee \exists y \left(x = S(y)\right)\\). The first-order induction axiom for \(\phi\) is \\[ \phi(0) \wedge \forall x \bigl(\phi(x) \Rig...
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9 months ago
Edit Post #290744 Initial revision 9 months ago
Answer A: Cyclical or “loop” fractals?
Does the Cantor set qualify? This animation loops infinitely, and shows that you get back to where you started every time you zoom in by a factor of 3. Show animated GIF An animation of the Cantor set
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9 months ago
Comment Post #290318 Your claim is correct as written, I think. It is the case that for most numbers, it is unknown if the number is normal. (That is, for a real number between 0 and 1 chosen uniformly, the probability that we know whether it is normal or not is 0.) It is also the case that we know that most number...
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12 months ago
Comment Post #290305 But from the Atlas link it seems that unshackling yourself from a specific geometry unlocks *thousands* of new regular 4-polytopes. I don't see how this answers the question. You assert in your answer that most of those linked polytopes aren't regular but can you support that claim at all? The Atlas ...
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12 months ago
Comment Post #290196 You probably mean to specify that M is a *connected* n-dimensional compact manifold; the nth Betti number is in fact the number of connected components if M is orientable.
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12 months ago
Comment Post #290305 > They are the only ones aside from the 6 convex Euclidean 4-polytopes. This can't be true; there are also 5 regular projective 4-polytopes, for example (hemi-tesseract through hemi-600-cell). I don't have deep expertise here but I don't know of any reason why the linked list from the Atlas of Sma...
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12 months ago
Comment Post #289826 > because what the code does is pick a random number with random.randint() for each candidate, and not have a fixed total population size. So instead of simulating each voter making a random choice, you're choosing the number of voters for each candidate from a uniform distribution? This has a sim...
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about 1 year ago
Comment Post #289826 Does your code vary the population size at all? You should definitely see a change in your observed probability for the same $n$ if so, which means that a formula expressed only in terms of $n$ can't be right.
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about 1 year ago
Edit Post #289826 Post edited:
about 1 year ago
Edit Post #289826 Initial revision about 1 year ago
Answer A: Strange behavior in elections and pie charts
> with randomized positions in the chart Implicit in this description of your model is the notion that every location for a separator line is equally likely. This doesn't describe the actual distribution of vote totals; with random voters, lines are much less likely to appear near other lines than...
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about 1 year ago
Comment Post #289671 Wow, you seem pretty twisted up over the purpose of this site. This isn't a social media site. We are collaboratively building a public repository of knowledge here. Your submissions aren't primarily acts of personal expression; they are primarily bricks in an edifice of which we are all custodians. ...
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about 1 year ago
Edit Post #289644 Post edited:
Embarassing; it's the chi-square distribution for stddev, not t
about 1 year ago
Edit Post #289644 Post edited:
about 1 year ago
Edit Post #289644 Initial revision about 1 year ago
Answer A: What is the formula for sample standard deviation of a small sample size?
The sample standard deviation (with Bessel's correction) is defined to be the first formula in your post. It doesn't ‘become’ anything else. You were possibly remembering using the sample standard deviation in an estimator for the population mean. The \(t\)-value is multiplied by the sample standa...
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about 1 year ago
Comment Post #288820 > One obvious criterion is that all the open squares be connected. If a set of squares are "walled off" with mines, then there is no way to get clues to the mines inside from the outside. The total number of mines is known, so if there are multiple disconnected open regions but all but one of them...
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over 1 year ago
Comment Post #288168 Do you intend the mod to be outside the sum or inside it—i.e., are you expecting the final result to necessarily be less than M?
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over 1 year ago
Edit Post #288123 Initial revision over 1 year ago
Answer A: 2 construals "of 100 patients presenting with a lump like the claimant’s in Gregg v Scott, 42 will be ‘cured’ if they are treated immediately."
The author of this passage is proposing a very simple hidden variable model. This model has two variables: $A$ — whether the patient has certain unknown genes (this variable is hidden) $B$ — whether the treatment succeeds (this variable is observed) And the model proposes that $A$ influences $...
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over 1 year ago
Comment Post #288087 I don't think being akin to a statement in predicate logic is sufficient to make a question on topic for math. Contemplating the mortality of Socrates would also be off topic, unless it was specifically asking about applying modus ponens or something. Likewise, I think it would be okay if OP were usi...
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over 1 year ago
Comment Post #288087 Not a math question. OP should perhaps ask this in a community for English language learners.
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over 1 year ago
Edit Post #288039 Post edited:
Not a bug, but a support question.
over 1 year ago
Suggested Edit Post #288039 Suggested edit:
Not a bug, but a support question.
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helpful over 1 year ago
Edit Post #287975 Initial revision over 1 year ago
Answer A: The meaning of $\pm$
I think I would describe that as ‘the two claims $|x| = \pm x$’, not a single claim. It's comparable to saying ‘the two roots of $x^2 - 4$ are $x = \pm2$’ and not ‘the root of $x^2 - 4$ is $x = \pm2$’. Whether those two claims are meant to be and-ed or or-ed would depend on context. In a vacuum, I...
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over 1 year ago
Edit Post #287870 Initial revision almost 2 years ago
Question Generalization of categorical product
I'm only a dabbler in category theory; this might be a basic concept just outside of my sphere of exposure. I'm looking for references to the following universal construction, a generalization of the categorical product. Given a category $\mathcal{C}$ and a functor $F$ from $\mathcal{C}$ to som...
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almost 2 years ago
Comment Post #287674 Arguably, this numeral system is better thought of as a mixed radix system alternating between bases 5 and 4 than as a simple base-20 system. There's a valid mathematical (or maybe mathematics-educational?) question about the pros and cons of mixed radix systems versus common radix systems (I think i...
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almost 2 years ago
Edit Post #287494 Initial revision almost 2 years ago
Answer A: Example of $f:[0,1]\to\mathbf{R}$ with $\lim_{a\to 0^+}\int_a^1f(x)dx=L $ for some real number $L$ but $\int_0^1|f(x)|dx=\infty $
Since you haven't specified that such a function needs to be continuous or well-behaved in any way, it's quite easy to describe one. The integral $\int0^1 \frac1x\,dx$ diverges, so there is an infinite amount of area to work with. Measure off the section of curve with area 1 starting at $x = 1$ an...
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almost 2 years ago
Edit Post #287490 Post edited:
almost 2 years ago
Edit Post #287490 Initial revision almost 2 years ago
Answer A: Finding the limit $ \lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n} $
First, let's make things simpler and do a change of variable, replacing $x$ with $\frac{1}{x}$: $$ \lim{x\to \infty}e^x\sum{n=\lfloor x\rfloor}^\infty\frac{1}{nx^n} $$ Massage into an explicit L'Hôpital indeterminate form: $$ \lim{x\to \infty}\frac{\sum{n=\lfloor x \rfloor}^\infty\frac{1}...
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almost 2 years ago
Edit Post #287437 Initial revision about 2 years ago
Answer A: What is the probability that the convex hull of $n$ randomly distributed points has $l$ of the points on its boundary?
Your conjecture is incorrect. Consider the case of four points. Place three of these points randomly; these points form a triangle, and each pair of points defines a line. Consider the three half-planes bounded by each of these lines and containing the centroid of the triangle. If and only if a fo...
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about 2 years ago
Edit Post #287308 Initial revision about 2 years ago
Answer A: Do linear and group invariant functions allowed to go inside(?) integral operators?
If $f$ is a linear map (between finite-dimensional spaces), then it can be represented by multiplying by some matrix. The fact that $f$ is $\mathfrak G$-invariant presumably means that this matrix is constant with respect to the $\mathfrak g$ used as the variable of integration. It's easy to se...
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about 2 years ago
Edit Post #287165 Initial revision about 2 years ago
Answer A: Why can't we conclude the extrema property of a function from its quadratic approximation when the discriminant is zero?
You've basically walked right up to the explanation. The second-order approximation you describe, when $D = 0$, is $$g(x, y) = a\left(x + \frac{b}{2a}y\right)^2$$ Define $z = x + \frac{b}{2a}y$, and we have $g(x, y) = g(z) = az^2$. This means that $g$ describes a function that really only ha...
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about 2 years ago