Activity for Snoopy
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #287846 | Initial revision | — | about 1 year 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... (more) |
— | about 1 year ago |
| Edit | Post #287667 |
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| Edit | Post #287664 |
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| Edit | Post #287720 | Initial revision | — | over 1 year ago |
| 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... (more) |
— | over 1 year 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. (more) |
— | over 1 year ago |
| Edit | Post #287664 |
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| Edit | Post #287664 |
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— | over 1 year ago |
| Edit | Post #287664 |
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| Edit | Post #287664 |
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— | over 1 year ago |
| Comment | Post #287664 |
That is the answer to the second question mark in the quoted excerpt. (more) |
— | over 1 year ago |
| Edit | Post #287667 |
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— | over 1 year ago |
| Edit | Post #287667 | Initial revision | — | over 1 year 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 ... (more) |
— | over 1 year 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. (more) |
— | over 1 year ago |
| Edit | Post #287664 |
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— | over 1 year ago |
| Edit | Post #287664 |
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— | over 1 year ago |
| Edit | Post #287664 |
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| Edit | Post #287664 |
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| Edit | Post #287664 | Initial revision | — | over 1 year ago |
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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) |
— | over 1 year ago |
| Edit | Post #287492 |
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| Edit | Post #287492 |
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| Edit | Post #287492 | Initial revision | — | over 1 year ago |
| 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... (more) |
— | over 1 year ago |
| Edit | Post #287484 |
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— | over 1 year ago |
| Edit | Post #287484 |
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— | over 1 year ago |
| Edit | Post #287484 | Initial revision | — | over 1 year ago |
| 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 ... (more) |
— | over 1 year ago |
| Edit | Post #287201 |
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| Edit | Post #287201 |
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| Edit | Post #287201 |
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| Edit | Post #287201 |
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| Edit | Post #287201 |
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| Edit | Post #287201 | Initial revision | — | over 1 year ago |
| 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... (more) |
— | over 1 year ago |
| Edit | Post #287160 |
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| Edit | Post #287160 |
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— | over 1 year ago |
| Edit | Post #287160 | Initial revision | — | over 1 year ago |
| 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 $f{xx}(0,0)\ne 0$.... (more) |
— | over 1 year ago |
| Edit | Post #287006 |
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— | over 1 year ago |
| 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) |
— | over 1 year ago |
| Edit | Post #287006 | Initial revision | — | over 1 year 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... (more) |
— | over 1 year ago |
| Edit | Post #287002 | Initial revision | — | over 1 year 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) |
— | over 1 year ago |
| Comment | Post #286985 |
Corrected. Thanks. (more) |
— | over 1 year ago |
| Edit | Post #286985 |
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— | over 1 year ago |
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