Activity for Snoopy
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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... (more) |
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Edit | Post #287849 |
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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 $f$ ... (more) |
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Edit | Post #287849 |
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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... (more) |
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Edit | Post #287846 |
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Post edited: The condition (2') edited. |
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Comment | Post #287846 |
Yes, thanks for that. I will edit it into the post. (more) |
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Edit | Post #287846 | Initial revision | — | about 2 months ago |
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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... (more) |
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Edit | Post #287667 |
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Edit | Post #287720 | Initial revision | — | 2 months ago |
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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... (more) |
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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. (more) |
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Edit | Post #287664 |
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Comment | Post #287664 |
That is the answer to the second question mark in the quoted excerpt. (more) |
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Edit | Post #287667 |
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Edit | Post #287667 | Initial revision | — | 3 months ago |
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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 ... (more) |
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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. (more) |
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Edit | Post #287664 |
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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... (more) |
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Edit | Post #287492 |
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Edit | Post #287492 | Initial revision | — | 4 months ago |
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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... (more) |
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Edit | Post #287484 |
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Edit | Post #287484 | Initial revision | — | 4 months ago |
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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 ... (more) |
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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... (more) |
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Edit | Post #287160 |
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Edit | Post #287160 | Initial revision | — | 6 months ago |
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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 $f{xx}(0,0)\ne 0$.... (more) |
— | 6 months ago |
Edit | Post #287006 |
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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 $p\mid s$. (more) |
— | 6 months ago |
Edit | Post #287006 | Initial revision | — | 6 months ago |
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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... (more) |
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Edit | Post #287002 | Initial revision | — | 6 months ago |
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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... (more) |
— | 6 months ago |
Comment | Post #286985 |
Corrected. Thanks. (more) |
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Edit | Post #286985 | Initial revision | — | 7 months ago |
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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 $\triangle ABC$.... (more) |
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Comment | Post #286957 |
Thank you! Do you have a reference for the result in the second paragraph? (more) |
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Edit | Post #286956 | Initial revision | — | 7 months ago |
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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? (more) |
— | 7 months ago |
Comment | Post #285036 |
Here is a meta complain by a former Math SE mod: https://math.meta.stackexchange.com/q/28168. (more) |
— | 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... (more) |
— | 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. (more) |
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Edit | Post #286788 |
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Edit | Post #286788 | Initial revision | — | 8 months ago |
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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. (more) |
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