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Activity for Snoopy‭

Type On... Excerpt Status Date
Edit Post #287850 Post undeleted about 2 months ago
Edit Post #287849 Post undeleted about 2 months ago
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Comment Post #287849 @#53398 Thank you for your comments. I have edited the post by adding the reference. The sketchy proof in the post is intended as a summary of the standard textbook proof. For the sake of clarity, I add the complete argument explicitly. For the last part of your comment, I think I'm using the def...
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about 2 months ago
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Answer A: Universal property of quotient spaces
I figured out an answer after posting the question for a while. I would like to record it here. The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ ...
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about 2 months ago
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Question Universal property of quotient spaces
A typical textbook theorem about quotient space is as follows: >Theorem (Gamelin-Greene Introduction to Topology p.106): Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equiva...
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about 2 months ago
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Edit Post #287846 Post edited:
The condition (2') edited.
about 2 months ago
Comment Post #287846 Yes, thanks for that. I will edit it into the post.
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about 2 months ago
Edit Post #287846 Initial revision about 2 months ago
Question Criterion in terms of the bases for determining whether one topology is finer than another
When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another: > Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime...
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Question Proving $|{\bf R}^{\bf R}|=|2^{\bf R}|$ using the Schroeder-Bernstein Theorem
Let $A$ be the set of all functions from ${\bf R}$ to ${\bf R}$ and $B$ the power set of ${\bf R}$. Then $|A|=|B|$. This is a well-known result in set theory. A quick search on Google returns answers in various places explicitly using cardinal arithmetic. Question: Can one prove the above res...
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2 months ago
Comment Post #287667 Yes. Adding more text will push the mentioned expression down to the next line. If I change the expression to $f(x,0)$, then the scroll bar disappears.
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3 months ago
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Comment Post #287664 That is the answer to the second question mark in the quoted excerpt.
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Edit Post #287667 Initial revision 3 months ago
Question Extra "bar" in the post
GitHub issue: https://github.com/codidact/qpixel/issues/802 I have experimented with various things but cannot get rid of an extra "bar" in my recent answer (now edited by putting the expression in another separate line to avoid the bar) on the main site: > snapshot of the mentioned post ...
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3 months ago
Comment Post #287625 The answer to your second bullet point is NO already. (See my answer below.) You don't need to write 3 and 4, which are all based on a wrong interpretation of the phrase in 2.
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Answer A: What did James Stewart mean by "the line integral reduces to an ordinary single integral in this case" ?
> How do you symbolize "the line integral reduces to an ordinary single integral in this case"? $\int^ba f(x {\color{goldenrod}{, 0)}} \, dx = \int^ba f(x) \, dx $? NO. For any function with two variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $$x\mapsto f(x,0)$$ gives you a fun...
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Question 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 $
In the Wikipedia article on improper integrals, the function $f(x)=\frac{\sin x}{x}$ gives an example that is improperly integrable: $$ \lim{N\to\infty}\int0^N f(x)dx=\frac{\pi}{2} $$ but not absolutely integrable: $$ \int0^\infty|f(x)|dx=\infty $$ I am looking for such an example for funct...
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4 months ago
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Question Finding the limit $ \lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n} $
> Let $\lfloor x \rfloor$ be the maximum integer $n\le x$. Find the limit $$ \lim{x\to 0^+}e^{1/x}\sum{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n} $$ I do not have an idea how to approach this problem except for a few observations. I don't have much progress from here: - The limit is of the ...
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4 months ago
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Answer A: How can 3/1 ≡ 1/(1/3), when left side features merely integers, but right side features a repetend?
> ... it's impossible to measure and cut anything physical at a repetend. This is incorrect. Given a segment with $1$ unit length, one can easily construct, "physically", a segment with a length of $1/3$ units. In fact, given any line segment, there are many ways to trisect it. See for instance t...
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Question Why can't we conclude the extrema property of a function from its quadratic approximation when the discriminant is zero?
Suppose $f:\mathbf{R}^2\to\mathbf{R}$ is a smooth function and $P=(0,0)$ is a critical point of $f$. The second-derivative test is inconclusive when the discriminant at $P$ is zero: $$f{xx}(0,0)f{yy}(0,0)-(f{xy}(0,0))^2=0\ .$$ For simplicity, assume further that $f(0,0)=0$ and $f{xx}(0,0)\ne 0$....
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Comment Post #287003 Thank you for your answer! I see from your proof that the crucial step is the fact that for prime $p$, one has: $p\mid rs$ if and only if $p\mid r$ or $p\mid s$.
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Edit Post #287006 Initial revision 6 months ago
Answer A: Proving that $p\mid (p+1776)$ if $p$ is a prime and $p(p+1776)$ is a perfect square
Inspired by Derek Elkins's excellent answer, I have the following proof. By the assumption, we have $p(p+1776)=k^2$ for some integer $k$ and thus $p\mid k^2$. Then, Euclid's lemma implies that $p\mid k$. It follows that $k=mp$ for some integer $m$ and thus $(mp)^2=k^2=p(p+1776)$, which implies by...
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6 months ago
Edit Post #287002 Initial revision 6 months ago
Question Proving that $p\mid (p+1776)$ if $p$ is a prime and $p(p+1776)$ is a perfect square
Problem: Suppose $p$ is a prime number and $p(p+1776)$ is a perfect square. Prove that $p\mid (p+1776)$. From the assumption of the problem, $p(p+1776)=k^2$ for some positive integer $k$. This does not help much. Intuitively, one can write $p(p+1776)=p^2m^2$ for some integer $m$ due to the fundam...
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6 months ago
Comment Post #286985 Corrected. Thanks.
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6 months ago
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Question Given two angles of a triangle, finding an angle formed by a median
> Problem: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median) of $\triangle ABC$. Show that $\angle MBC=\angle BAC$. > Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median of $\triangle ABC$....
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7 months ago
Comment Post #286957 Thank you! Do you have a reference for the result in the second paragraph?
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7 months ago
Edit Post #286956 Initial revision 7 months ago
Question If $\mathbf{R}$ is thought of as a vector space over $\mathbf{Q}$, what is its dimension?
It is known that $\mathbf{R}$, as a vector space over the field of real numbers, has the dimension $1$. I know that $\mathbf{Q}$ is also a field. Question: If $\mathbf{R}$ is thought of as a vector space over $\mathbf{Q}$, what is its dimension?
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7 months ago
Comment Post #285036 Here is a meta complain by a former Math SE mod: https://math.meta.stackexchange.com/q/28168.
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8 months ago
Comment Post #285036 @#8046 That's a long story. Basically, on Math SE, there is a small group of people who dominate the "deletion queue" of the site in a chat room. They have a very strong opinions on what kind of questions should *not* be allowed. Whenever they think the posts are not "good questions" using *their* st...
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8 months ago
Comment Post #285036 "What draws newcomers in?" Many good contributors suffer from (a lot of) their valuable highly upvoted answers being deleted frequently by a clique in Math Stack Exchange. Some of them have stopped doing any further contributions on that site. Those are some potential users of codidact.
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8 months ago
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Question The gcd of powers in a gcd domain
Question: $\def\gcd{\operatorname{gcd}}$Let $R$ be a gcd domain. Does it always hold that $\gcd(x^m,y^m)=\gcd(x,y)^m$? Context. If the ring is a Bezout domain, then we can apply this method. However, a gcd domain may not be a Bezout domain. I don't know how I can go on.
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8 months ago