Activity for Snoopy
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #287664 |
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— | about 2 years ago |
| Edit | Post #287720 | Initial revision | — | about 2 years 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) |
— | about 2 years 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) |
— | about 2 years ago |
| Edit | Post #287664 |
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| Edit | Post #287664 |
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| Edit | Post #287664 |
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| Edit | Post #287664 |
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— | about 2 years ago |
| Comment | Post #287664 |
That is the answer to the second question mark in the quoted excerpt. (more) |
— | about 2 years ago |
| Edit | Post #287667 |
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— | about 2 years ago |
| Edit | Post #287667 | Initial revision | — | about 2 years 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) |
— | about 2 years 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) |
— | about 2 years ago |
| Edit | Post #287664 |
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| Edit | Post #287664 |
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| Edit | Post #287664 |
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| Edit | Post #287664 |
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— | about 2 years ago |
| Edit | Post #287664 | Initial revision | — | about 2 years ago |
| 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... (more) |
— | about 2 years ago |
| Edit | Post #287492 |
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— | over 2 years ago |
| Edit | Post #287492 |
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— | over 2 years ago |
| Edit | Post #287492 | Initial revision | — | over 2 years 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 2 years ago |
| Edit | Post #287484 |
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— | over 2 years ago |
| Edit | Post #287484 |
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— | over 2 years ago |
| Edit | Post #287484 | Initial revision | — | over 2 years 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) |
— | over 2 years ago |
| Edit | Post #287201 |
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— | over 2 years ago |
| Edit | Post #287201 |
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— | over 2 years 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 | Initial revision | — | over 2 years 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 2 years ago |
| Edit | Post #287160 |
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| Edit | Post #287160 |
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— | over 2 years ago |
| Edit | Post #287160 | Initial revision | — | over 2 years 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) |
— | over 2 years ago |
| Edit | Post #287006 |
Post edited: |
— | over 2 years 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 2 years ago |
| Edit | Post #287006 | Initial revision | — | over 2 years 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 2 years ago |
| Edit | Post #287002 | Initial revision | — | over 2 years 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 2 years ago |
| Comment | Post #286985 |
Corrected. Thanks. (more) |
— | over 2 years ago |
| Edit | Post #286985 |
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| Edit | Post #286985 |
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| Edit | Post #286985 |
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| Edit | Post #286985 | Initial revision | — | over 2 years 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) |
— | over 2 years ago |