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http://physics.stackexchange.com/questions/59579/how-far-apart-are-galaxies-on-average-if-galaxies-were-the-size-of-peas-how-ma | # How far apart are galaxies on average? If galaxies were the size of peas, how many would be in a cubic meter?
The actual number:
How far apart are galaxies on average?
An attempt to visualize such a thing:
If galaxies were the size of peas, how many would be in a cubic meter?
- | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9563495516777039, "perplexity": 2059.035605568122}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00111-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2138771/gamblers-ruin-probability-of-losing-in-t-steps | # Gambler's Ruin - Probability of Losing in t Steps
I would be surprised if this hasn't been asked before, but I cannot find it anywhere.
Suppose we're given an instance of the gambler's ruin problem where the gambler starts off with $i$ dollars and at every step she wins 1 dollar with probability $p$ and loses a dollar with probability $q = 1 - p$. The gambler stops when she has lost all her money, or when she has $n$ dollars. I am interested in the probability that the gambler loses in $t$ steps.
I know how to find the expected number of steps before reaching either absorbing state, and how to solve the probability that she loses before winning $n$ dollars, but this one is eluding me. Let $P_{i, t}$ be the probability that the gambler goes broke in $t$ steps given that she started with $i$ dollars. I have set up the recurrence: $$P_{i, t} = qP_{i-1, t-1} + pP_{i+1, t-1}$$ and we know that $P_{0, j} = 1$ and $P_{n, j} = 0$for all $j$, and $P_{i, 0} = 0$ for all $i > 0$. I'm struggling to solve this two dimensional recurrence.
If it turns out to be too hard to give closed form solutions for this, can we give tighter bounds than just the probability that the gambler ever loses?
I do not know if I misinterpret your question, but I think the probability of going to ruin in $$t$$ steps is just the probability of losing $$i$$ times more than winning. Let $$m$$ be the number of wins, and $$n$$ be the number of losses, so obviously $$m+n=t$$
To lose means $$n=m+i$$, so $$n=\frac{t-i}{2}$$ and $$m=\frac{t-3i}{2}$$
$$p_0(t) = p^{\frac{t-3i}{2}}\cdot q^{\frac{t-i}{2}}$$
or if you mean "losing in $$\leq t$$ steps", this would change of course to
$$P_0(\leq t) = \sum_{k=i}^{t} p_0(k)$$
• Hi, do we need to consider which state go back and which go forward? – maple Jun 1 '18 at 9:01
• I think when I asked this question I would have been satisfied with answering either exactly t steps or at most t steps. Either way, your solution seems to lose part of the nuance of the question. There are lots of invalid walks that are captured by "losing i times more than winning" such as walks which have the gambler going into negative money, or reaching n dollars and continuing to play. – Andrew S Jun 8 '18 at 23:47
The first (edited) answer is the correct probability for reaching 0 Dollars at exactly step t. That is an incorrect definition of ruin as @Andrew S points out.
It is not the probability of reaching 0 Dollars for the first time without having previously reached n Dollars, during t steps or fewer steps if 0 or n Dollars is reached earlier, which is the correct definition of ruin.
This is a tough problem. I do not know of a closed solution...only a closed approximation and a path-counting summing algorithm which counts and sums only the permitted paths from i Dollars to 0 Dollars for each step from 1 to t.
• I'll settle for a closed approximation if you have it. – Andrew S Jan 18 at 3:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9121060967445374, "perplexity": 231.2551059468032}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00367.warc.gz"} |
https://web2.0calc.com/questions/for-what-real-value-of-is-a-root-of | +0
# For what real value of is a root of?
0
44
1
For what real value of $$k$$ is $$\frac{13-\sqrt{131}}{4}$$ a root of $$2x^2-13x+k$$?
Jul 2, 2021
#1
+234
0
The root given is in the form of the quadratic formula, $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$
The part we are interested in is the discriminant, $$b^2-4ac$$. Using this,
$$b^2 - 4ac = 169 - 8k = 131$$
$$169 = 131 + 8k$$
$$38 = 8k$$
$$k = 4.75$$
Jul 2, 2021 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8346057534217834, "perplexity": 902.7362602010877}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154085.58/warc/CC-MAIN-20210731074335-20210731104335-00362.warc.gz"} |
https://www.physicsforums.com/threads/integration-problem-i-cannot-figure-out.54573/ | # Integration Problem I Cannot Figure Out
1. Nov 29, 2004
### mathemagician
My Professor in my calculus class (1st year) left us with this question at the end of lecture today and told us to think about it. I am baffled as to how to solve it. Anyways, here is what he gave us.
$$\int_{x}^{xy} f(t) dt$$
This is independent of x.
If $$f(2) = 2$$, compute the value of
$$A(x) = \int_{1}^{x} f(t)dt$$
for all $$x > 0$$
He then gave us a hint saying since it is independent of x, the function will be in terms of y.
$$g(y) = \int_{x}^{xy}f(t)dt$$
He also told us the final answer is $$4 \ln x$$
Does this make any sense? I would appreciate it if someone can show me how to solve this.
2. Nov 29, 2004
### arildno
To solve this, differentiate g(y) with respect to x:
$$0=\frac{d}{dx}g(y)=yf(xy)-f(x)\to{f}(xy)=\frac{f(x)}{y}$$
then, differentiate g(y) with respect to y:
$$\frac{dg}{dy}=xf(xy)=\frac{xf(x)}{y}$$
Hope this helps..
3. Nov 29, 2004
### mathemagician
I am a little confused after spending an hour thinking about it. But I think I have something.
Since $$f(2) = 2$$ then $$\frac{dg}{dy} = \frac{2f(2)}{y} = \frac{4}{y}$$
Then we can replace $$f(t)$$ with $$\frac{4}{y}$$
Going back we can now solve for $$A(y) = \int_{1}^{x} \frac{4}{y} dy = 4 \int_{1}^{x} \frac{1}{y} dy = 4[\ln |x| - \ln (1)]$$
and since $$x > 0$$ we finally have:
$$A(y) = 4 \ln x$$
OK, so is this right? I'm a little bit troubled with doing the substitution of f(2) = 2, can you explain to me how that might be justified?
I also have a question about your hint, arildno. Just the first line.
how is it possible that you set $$\frac{d}{dx}g(y) = 0$$? And could you explain $$yf(xy) - f(x)$$ where that came from?
Thanks
4. Nov 29, 2004
### arildno
1) g is solely a function of the variable "y".
Hence, differentiating it with respect to some other variable it does not depend on, yields zero.
2) Using the Leibniz rule for differentiating an integral where the bounds depend on your variable, reads:
$$\frac{d}{dx}\int_{x}^{xy}f(t)dt=f(xy)\frac{d}{dx}xy-f(x)\frac{d}{dx}x=yf(xy)-f(x)$$
3. Since g(y) is independent of x, so is $$\frac{dg}{dy}$$
Hence, we must have:
xf(x)=K, where K is some constant.
We can determine K, with noting 2f(2)=4, that, is,
xf(x)=4 (implying f(x)=\frac{4}{x}), or $$\frac{dg}{dy}=\frac{4}{y}=f(y)$$
5. Nov 29, 2004
### mathemagician
Thank you. I understand. Its much clearer now. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.946662425994873, "perplexity": 526.5881736898283}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171251.27/warc/CC-MAIN-20170219104611-00471-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-r-section-r-2-algebra-essentials-r-2-assess-your-understanding-page-27/37 | ## College Algebra (10th Edition)
$d(D, E) = 2$
$d(D, E)$ represents the distance between the points D and E on a number line. If the coordinates of D and E are $d$ and $e$, respectively, then: $d(D, E) = |d-e|$ Based on the given number line, the coordinates of D and E are 1 and $3$, respectively. Use the formula above to obtain: $d(D, E) = |1-3| = |-2| =2$ Thus, $d(D, E) = 2$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8375102281570435, "perplexity": 149.42688479384137}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160842.79/warc/CC-MAIN-20180924224739-20180925005139-00319.warc.gz"} |
https://www.illustrativemathematics.org/content-standards/6/EE/B | # 6.EE.B
Reason about and solve one-variable equations and inequalities.
## Standards
6.EE.B.5 Understand solving an equation or inequality as a process of answering a question: which values...
6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or...
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p =... 6.EE.B.8 Write an inequality of the form$x > c$or$x < c\$ to represent a constraint or condition...
## Tasks aligned to this cluster
Triangular Tables
Busy Day
Which Goes with Which?
All, Some, or None? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9115992188453674, "perplexity": 4430.411766232061}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00035.warc.gz"} |
https://stacks.math.columbia.edu/tag/0CCG | Remark 33.36.11. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a scheme over $S$. For $n \geq 1$
$X^{(p^ n)} = X^{(p^ n/S)} = X \times _{S, F_ S^ n} S$
viewed as a scheme over $S$. Observe that $X \mapsto X^{(p^ n)}$ is a functor. Applying Lemma 33.36.2 we see $F_{X/S, n} = (F_ X^ n, \text{id}_ S) : X \longrightarrow X^{(p^ n)}$ is a morphism over $S$ fitting into the commutative diagram
$\xymatrix{ X \ar[rr]_{F_{X/S, n}} \ar[rrd] \ar@/^1em/[rrrr]^{F_ X^ n} & & X^{(p^ n)} \ar[rr] \ar[d] & & X \ar[d] \\ & & S \ar[rr]^{F_ S^ n} & & S }$
where the right square is cartesian. The morphism $F_{X/S, n}$ is sometimes called the $n$-fold relative Frobenius morphism of $X/S$. This makes sense because we have the formula
$F_{X/S, n} = F_{X^{(p^{n - 1})}/S} \circ \ldots \circ F_{X^{(p)}/S} \circ F_{X/S}$
which shows that $F_{X/S, n}$ is the composition of $n$ relative Frobenii. Since we have
$F_{X^{(p^ m)}/S} = F_{X^{(p^{m - 1})}/S}^{(p)} = \ldots = F_{X/S}^{(p^ m)}$
(details omitted) we get also that
$F_{X/S, n} = F_{X/S}^{(p^{n - 1})} \circ \ldots \circ F_{X/S}^{(p)} \circ F_{X/S}$
There are also:
• 4 comment(s) on Section 33.36: Frobenii
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.998590350151062, "perplexity": 306.0855979573404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103917192.48/warc/CC-MAIN-20220701004112-20220701034112-00158.warc.gz"} |
http://mathoverflow.net/questions/100146/when-is-the-cohomology-cross-product-square-nonzero | # When is the cohomology cross product square nonzero?
Let $X$ be a finite CW complex, and $A$ an abelian group. Given a nonzero class in singular cohomology $$x\in H^k(X;A),$$ when is its cross product square $$x\times x\in H^{2k}(X\times X;A\otimes A)$$ nonzero?
Remarks: The tensor product is over $\mathbb{Z}$. By the Künneth theorem $x\times x$ is always nonzero if $A$ is a field, or more generally if $A$ is a domain and $H^\ast(X)$ is flat over $A$. Examples where $x\times x=0$ arise in the study of Lusternik-Schnirelmann category, in particular in spaces $X$ for which $\operatorname{cat}(X\times X)<2\operatorname{cat}(X)$.
-
According to the Künneth formula [Spanier, 5.3.10], $x \times x \neq 0$ if $Tor_1^{\mathbb{Z}}(A,A) \neq 0$. This holds for example if $A$ is torsion-free. – Ralph Jun 20 '12 at 17:36
I mean of course $Tor_1^\mathbb{Z}(A,A)=0$. – Ralph Jun 20 '12 at 19:17
Building on Ralph's answer I would think this question reduces completely to algebra. Using the Kunneth theorem and the cohomology cross product, $x\times x$ should correspond precisely to the image of $x\otimes x\in H^k(X;A)\otimes H^k(X;A)$ under the injective cross product. So the question really becomes, given an element of a group $g\in G$, when is it true that $g\otimes g=0\in G\otimes G$. As Ralph notes, this can't happen if $G$ is torsion-free (or even if $g$ generates an infinite cyclic subgroup?), but I guess it could happen if, for example, we have $2\in \mathbb{Z}/4$. – Greg Friedman Jun 21 '12 at 22:53
@Greg Friedman: That's a good point that even $x\otimes x$ may be zero. However, I suppose it might also happen that the cross product fails to be injective (if $\operatorname{Tor}(A,A)$ is non-trivial). – Mark Grant Jun 22 '12 at 9:50
@Ralph: Thanks for pointing out this general formulation of Kunneth, which I had forgotten about (worse is that I forgot to look in Spanier for the most general formulation of a result). – Mark Grant Jun 22 '12 at 9:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9473500847816467, "perplexity": 268.74304244721526}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065828.38/warc/CC-MAIN-20150827025425-00034-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/348968/algebraic-topology | # Algebraic Topology
Can we help me next tags:
1. Let $Y$ vector topological space and $A \subseteq Y$ convex set. Prove that any two continuous mappings $f,g : X \to A$ homotopic, where $X$ is an arbitrary topological space.
2. Prove that every interval $(a, b)$ homotopy equivalent point.
3. Let $X$ be a contractible space. Prove that the space $X$ every two times with the same beginning and end homotopic (rel $\{0,1\}$).
4. Prove that two topological spaces one of which is attached, and the other can not be unlinked homotopy equivalent.
5. Show that the fundamental group of the space $\mathbb R^n$ , $n\geq 1$, trivial.
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Hope this helps.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9965894818305969, "perplexity": 462.8079632256716}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111313.83/warc/CC-MAIN-20160428161511-00061-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://www.math.gatech.edu/node/16135 | ## High dimensional sampling in metabolic networks
Series:
ACO Student Seminar
Friday, March 4, 2016 - 13:05
1 hour (actually 50 minutes)
Location:
Skiles 256
,
Georgia Tech
Organizer:
I will give a tour of high-dimensional sampling algorithms, both from a theoretical and applied perspective, for generating random samples from a convex body. There are many well-studied random walks to choose from, with many of them having rigorous mixing bounds which say when the random walk has converged. We then show that the techniques from theory yield state-of-the-art algorithms in practice, where we analyze various organisms by randomly sampling their metabolic networks.This work is in collaboration with Ronan Fleming, Hulda Haraldsdottir ,and Santosh Vempala. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8201684951782227, "perplexity": 2844.797676372616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889917.49/warc/CC-MAIN-20180121021136-20180121041136-00794.warc.gz"} |
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# Calculate faster - Reciprocal percentage equivalent
Author Message
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Joined: 27 Dec 2013
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Calculate faster - Reciprocal percentage equivalent [#permalink] 06 Jan 2014, 10:18
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Hi All,
If one has to complete the quants section on the GMAT on time, it is imperative that one has to calculate faster. A lot of questions on the GMAT involve division. To divide faster reciprocal percentage equivalent come quite handy.
In this post I will share a basic and an easy method that may help one in memorizing the reciprocal equivalents.
> Reciprocal of 2 (i.e 1/2) is 50%, that of 4 will be half of 50% i.e 25%. Similarly, reciprocal of 8 will be half of 25% = 12.5% and that of 16 will be 6.25%
> Reciprocal of 3 is 33.33%. Thus reciprocal of 6 will be half of 33.33% i.e 16.66% and that of 12 will be 8.33%
> Reciprocal of 9 is 11.11% and reciprocal of 11 is 9.0909%. Reciprocal of 9 is composed of 11's and reciprocal of 11 is composed of 09's.
If any calculation has 9 in the denominator, the decimal part will be only 1111 or 2222 or 3333 or 4444...
ex. 95/9 will be 10.5555
> Reciprocal of 20 is 5% [ you can remember this as 1/5 = 20% so, 1/20 = 5%]
> Reciprocal of 21 is 4.76% and of 19 is 5.26%. Thus we can easily remember reciprocals of 19, 20, 21 as 5.25%, 5 ,4.75% i.e 0.25% more and less than 5%
> Similarly, reciprocal of 25 is 4 % [ Remember this as 1/4 = 25%, so 1/25 = 4%]
> Reciprocal of 24 is 4.16% and of 26 is 3.84%. Thus, we can easily remember reciprocals of 24, 25, 26 as 4.15%, 4, 3.85% i.e 0.15% more and less than 4%.
> Reciprocal of 29 is 3.45% (i.e 345 in order) and reciprocal of 23 is 4.35% ( same digits but order is different. If 1/29 = 3.45% than definitely 1/23 will be more than 3.45%. Reverse the digits and the answer comes to 4.35%)
> Reciprocal of 18 is half of 11.1111% i.e 5.5555% i.e it consists of only 5's.
> Reciprocal of 22 is half of 09.0909%. i.e 4.5454% i.e consists of 45's.
> One can remember 1/8 = 12.5% and tables of 8, and one can easily remember fractions such as 2/8, 3/8, 5/8, 7/8 which are used very regularly.
1/8 is 12.5%, 2/8 is 25% (12.5*2), 3/8 is 37.5% (12.5*3), 5/8 is 62.5% ( 12.5*5), 7/8 = 87.5%
I hope you find this post useful.
Thanks,
Harish
_________________
Kindly consider for kudos if my post was helpful!
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Calculate faster - Reciprocal percentage equivalent [#permalink] 06 Jan 2014, 10:18
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http://mathhelpforum.com/calculus/130658-problem-exponential-integral.html | # Math Help - Problem with exponential integral
1. ## Problem with exponential integral
I'm lost as to how to go about solving this. u = 4x and du = 4 dx but beyond that I'm stumped. That constant in the denominator confuses me.
2. Originally Posted by Archduke01
I'm lost as to how to go about solving this. u = 4x and du = 4 dx but beyond that I'm stumped. That constant in the denominator confuses me.
Hint $e^{4x}=\left( e^{2x}\right)^2$ Then set $u=e^{2x}$
3. Originally Posted by TheEmptySet
Hint $e^{4x}=\left( e^{2x}\right)^2$ Then set $u=e^{2x}$
What would du be? 2?
4. Originally Posted by Archduke01
What would du be? 2?
What is the derivative of $e^{\lambda x}$?
if $u=e^{\lambda x} \implies \frac{du}{dx}=\lambda \cdot e^{\lambda x}$
Then in terms of differentials you get $du=...$
5. Originally Posted by TheEmptySet
What is the derivative of $e^{\lambda x}$?
if $u=e^{\lambda x} \implies \frac{du}{dx}=\lambda \cdot e^{\lambda x}$
Then in terms of differentials you get $du=...$
$2e^{2x}$!
$1/2 \int 2e^{2x}/(e^{4x} + 64)$
$1/2 \int du/(u^2+64)$
... How can I proceed though?
6. Originally Posted by Archduke01
$2e^{2x}$!
$1/2 \int 2e^{2x}/(e^{4x} + 64)$
$1/2 \int du/(u^2+64)$
... How can I proceed though?
Now use a trigonometric substitution $u = 8\tan{\theta}$. Make note that $du = 8\sec^2{\theta}\,d\theta$.
Also note that $\theta = \arctan{\frac{u}{8}}$.
$\int{\frac{1}{u^2 + 64}\,du} = \int{\frac{1}{(8\tan{\theta})^2 + 64}\,8\sec^2{\theta}\,d\theta}$
$= \int{\frac{8\sec^2{\theta}}{64\tan^2{\theta} + 64}\,d\theta}$
$= \int{\frac{8\sec^2{\theta}}{64(\tan^2{\theta} + 1)}\,d\theta}$
$= \int{\frac{\sec^2{\theta}}{8\sec^2{\theta}}\,d\the ta}$
$= \int{\frac{1}{8}\,d\theta}$
$= \frac{1}{8}\theta + C$
$= \frac{1}{8}\arctan{\frac{u}{8}} + C$
$= \frac{1}{8}\arctan{\frac{e^{2x}}{8}} + C$.
7. Originally Posted by Prove It
Now use a trigonometric substitution $u = 8\tan{\theta}$. Make note that $du = 8\sec^2{\theta}\,d\theta$.
Also note that $\theta = \arctan{\frac{u}{8}}$.
Thank you for the detailed solution. But why did you choose 8?
8. Originally Posted by Archduke01
Thank you for the detailed solution. But why did you choose 8?
Because if you have an integral of the form
$\int{\frac{1}{a^2 + x^2}\,dx}$
you make the substitution $x = a\tan{\theta}$.
This is so that you can factor out the $a^2$ and then make use of the identity $1 + \tan^2{\theta} = \sec^2{\theta}$.
And since the derivative of $\tan{\theta}$ is $\sec^2{\theta}$, this also means that the $\sec^2{\theta}$ will be eliminated (as you will end up with them on the top and bottom). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9896054863929749, "perplexity": 445.42712851050254}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443738008122.86/warc/CC-MAIN-20151001222008-00077-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://mathhelpforum.com/number-theory/121989-solving-congruence-equation.html | 1. Solving a congruence equation.
You can solve this since $5$ is invertible $\pmod {26}$ . The following is done using arithmetic $\pmod {26}$ :
$5x+8=22\Longrightarrow 5x=14\Longrightarrow x=\frac{14}{5}=14\cdot 21=294=8$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9920668005943298, "perplexity": 374.66750237621613}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084890187.52/warc/CC-MAIN-20180121040927-20180121060927-00396.warc.gz"} |
https://www.computer.org/csdl/trans/tp/2004/07/i0947-abs.html | Issue No. 07 - July (2004 vol. 26)
ISSN: 0162-8828
pp: 947-951
ABSTRACT
<p><b>Abstract</b>—We consider polarimetric images formed with coherent waves, such as in laser-illuminated imagery or synthetic aperture radar. A definition of the contrast between regions with different polarimetric properties in such images is proposed, and it is shown that the performances of maximum likelihood-based detection and segmentation algorithms are bijective functions of this contrast parameter. This makes it possible to characterize the performance of such algorithms by simply specifying the value of the contrast parameter.</p>
INDEX TERMS
Image processing, contrast definition, detection, segmentation, active contours, polarimetric imaging.
CITATION
P. R?fr?gier and F. Goudail, "Contrast Definition for Optical Coherent Polarimetric Images," in IEEE Transactions on Pattern Analysis & Machine Intelligence, vol. 26, no. , pp. 947-951, 2004.
doi:10.1109/TPAMI.2004.22
CITATIONS
SHARE
88 ms
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http://math.stackexchange.com/questions/110528/characters-being-everywhere-dense-in-the-character-group | Characters being everywhere dense in the character group
Let $k$ be the completion of an algebraic number field at a prime divisor $\mathfrak{p}$. We note that $k$ is locally compact. Let $k^{+}$ be the additive group of $k$ which is a locally compact commutative group.
Tate's Thesis Lemma 2.2.1 states that
If $\xi \rightarrow \chi(\xi)$ is one non-trivial character of $k^{+}$, then for each $\eta \in k^{+}$, $\xi \rightarrow \chi(\eta\xi)$ is also a character. The correspondence $\eta \leftrightarrow \chi(\eta\xi)$ is an isomorphism, both topological and algebraic, between $k^{+}$ and its character group.
The proof of this lemma is divided up into 6 steps, one step is to show that the characters $\chi(\eta\xi)$ are everywhere dense in the character group. Tate writes
$\chi(\eta\xi) = 1$, all $\eta \implies k^{+}\xi \neq k^{+} \implies \xi = 0$. Therefore the characters of the form $\chi(\eta\xi)$ are everywhere dense in the character group.
My question is: How does he get from showing that the $\xi = 0$ to the the result that the $\chi(\eta\xi)$ are everywhere dense?
-
This feels like a weak-$*$ topology thing, but this isn't linear and the topology on the character group is probably the compact-open topology, so I'm getting confused. Nice question! – Dylan Moreland Feb 18 '12 at 2:13
Following up: I think what I said above is on the right track. If you look at Section 4.1 of Folland's book, he shows that the topology on $\widehat G$ coincides with the weak$*$ topology it inherits as a subset of $L^\infty(G)$. I need to sort this out for my own purposes, so I'll try to summarize the argument in the morning, if no one else has done so by then. – Dylan Moreland Feb 18 '12 at 7:26
Denote the character $\xi\rightarrow\chi(\eta\xi)$ by $\chi_\eta$. We want to show that image of the map $f_\chi:\eta\rightarrow\chi_\eta$ is dense in $\hat k$. Take a closed subgroup $H$ of $k$ and set $N_H=\lbrace \xi\in k:\chi_\eta(\xi)=1\ {\rm for\ all}\ \eta\in H\rbrace$. This is also a closed subgroup. We have the short exact sequence $$0\rightarrow N_H\rightarrow k\rightarrow k/N_H\rightarrow 0$$ and the functoriality of Pontryagin duality turns this into $$0\rightarrow \widehat{k/N_H}\rightarrow \widehat{k}\rightarrow \widehat{N_H}\rightarrow 0$$ We have an isomorphism $\widehat{k/N_H}\simeq f_\chi(H)$ (this is basically Theorem 4.39 in Folland's "A Course in Abstract Harmonic Analysis"). Now, setting $H=k$, we see that $N_k=\lbrace 0\rbrace$, so the short exact sequence becomes $$0\rightarrow f_\chi(k)\rightarrow \widehat{k}\rightarrow 0\rightarrow 0$$ Hence $f_\chi(k)\simeq \widehat k$. This is mildly stronger than what Tate has done at this point, but I'm not worried, since we're incorporating the topology directly in the argument. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9635313749313354, "perplexity": 105.37095453875723}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461861848830.49/warc/CC-MAIN-20160428164408-00191-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/20123/problems-concerning-r-and-rx | # Problems concerning R and R[x]
A few questions relevant formally, but quite different in nature:
From now on, let R denote a ring.
1. If R is a UFD , is R[x] also a UFD?
2. If R is Noetherian, is R[x] also Noetherian?
3. If R is a PID, is R[x] also a PID?
4. If R is an Artin ring, is R[x] also an Artin ring?
For 1, we all know it's Gauss's lemma.
For 2, we all know it's Hilbert's basis theorem.
For 3, we all know that in Z[x], the ideal (2,x) is not a principal ideal, so the answer is negative.
But what about 4?
-
The answer to 4 is "no." If $R$ is an Artin ring, then it is Noetherian of Krull dimension zero. It follows from dimension theory that $R[X]$ is Noetherian of dimension one, i.e., not every prime ideal in $R[X]$ is maximal, so $R[X]$ can't be Artin.
-
Another really easy way to see that $R[X]$ fails to be Artin is the descending chain $(X)\supseteq (X^2)\supseteq\cdots$. – Keenan Kidwell Apr 2 '10 at 1:19
...assuming $0 \neq 1$ :) – S. Carnahan Apr 2 '10 at 3:20
Good argument. But let's give a down-to-earth counterexample:
Let $R$ be a field. Then consider $$(x)\supset (x^2)\supset(x^3)\supset\ldots.$$ This is a descending chain of ideals that doesn't become stationary so $R[X]$ is not Artin.
-
Yes, I realized this and posted a comment. – Keenan Kidwell Apr 2 '10 at 1:32
Hm, sorry. I didn't see it early enough. g – Tilemachos Vassias Apr 2 '10 at 1:36
Don't worry about it. – Keenan Kidwell Apr 2 '10 at 2:45
Note that one can give a very similar proof of fact that if integral domain is an artinian ring then it is a field (this also follows from consideration involving dimension). – ifk Apr 2 '10 at 10:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9737840890884399, "perplexity": 625.6060988176081}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122086930.99/warc/CC-MAIN-20150124175446-00018-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/72959-integration-help-riemann-sums.html | # Thread: Integration help. Riemann sums
1. ## Integration help. Riemann sums
Use X*(subscript k) as the left endpoint of each sub interval to find the area under the curve y=F(x) over [a,b]
f(x)= x^3 a=2 b=6 __________________
2. Originally Posted by Khaali91
Use X*(subscript k) as the left endpoint of each sub interval to find the area under the curve y=F(x) over [a,b]
f(x)= x^3 a=2 b=6 __________________
Click the link below for a good explanation of using estimating rectangles. It would be very difficult for someone to teach you how to do this on a forum.
Pauls Online Notes : Calculus I - Area Problem
3. Recall: $\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$
where $\Delta x = \frac{b-a}{n} = \frac{6-2}{n} = \frac{4}{n}$
Edit: Sorry I used right endpoints even though I said I was using left endpoints! I've corrected it:
Since we're using left end points, we consider: $x_i^* = x_{i-1} \ = \ a + (i-1)\Delta x \ = \ 2 + \frac{4(i-1)}{n}$
So, using the formula:
\begin{aligned} \int_2^6 x^3 dx & = \lim_{n \to \infty} \sum_{i=1}^n \left(2 + \frac{4(i-1)}{n}\right)^3 \frac{4}{n} \\ & = \lim_{n \to \infty} \frac{4}{n} \sum_{i=1}^n \left(4 + \frac{48(i-1)}{n} + \frac{96(i-1)^2}{n^2} + \frac{64(i-1)^3}{n^3} \right) \\ & = \lim_{n \to \infty} \frac{4}{n} \left( \sum_{i = 1}^n 4 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right) \end{aligned}
A bit messy but entirely doable. As you can see, right endpoints are usually nicer =|
4. Originally Posted by o_O
Recall: $\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$
where $\Delta x = \frac{b-a}{n} = \frac{6-2}{n} = \frac{4}{n}$
Edit: Sorry I used right endpoints even though I said I was using left endpoints! I've corrected it:
Since we're using left end points, we consider: $x_i^* = x_{i-1} \ = \ a + (i-1)\Delta x \ = \ 2 + \frac{4(i-1)}{n}$
So, using the formula:
\begin{aligned} \int_2^6 x^3 dx & = \lim_{n \to \infty} \sum_{i=1}^n \left(2 + \frac{4(i-1)}{n}\right)^3 \frac{4}{n} \\ & = \lim_{n \to \infty} \frac{4}{n} \sum_{i=1}^n \left(4 + \frac{48(i-1)}{n} + \frac{96(i-1)^2}{n^2} + \frac{64(i-1)^3}{n^3} \right) \\ & = \lim_{n \to \infty} \frac{4}{n} \left( \sum_{i = 1}^n 4 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right) \end{aligned}
A bit messy but entirely doable. As you can see, right endpoints are usually nicer =|
Thank You very much!
But is there a way to eliminate the Limit and the Sigma notation? My teacher briefly swooped over this and moved on so i never fully understood the concept.
From what i remember there are theorems such as $[n(n+1)]/2$
Is that right?
5. Yes. First we have to take care of the sums. Then we have to take the limit and then we finally get a numerical answer.
So, things to note about summations:
• $\sum_{i=1}^n 1 = \overbrace{1 + 1 + 1 + \cdots + 1}^{n \text{ times}} = n$
• $\sum_{i=1}^n c a_i = c \sum_{i=1}^n a_i$ (Basically, you can pull out the constant)
• $\sum_{i=1}^n i \ = \ 1 + 2 + \cdots + n \ = \ \tfrac{1}{2} n(n+1)$
• $\sum_{i=1}^n i^2 \ = \ 1^2 + 2^2 + \cdots + n^2 \ = \ \tfrac{1}{6} n(n+1)(2n+1)$
• $\sum_{i=1}^n i^3 \ = \ 1^3 + 2^3 + \cdots + n^3 \ = \ \tfrac{1}{4}n^2(n+1)^2$
Now, looking at that last expression I gave you: $\lim_{n \to \infty} \frac{4}{n} \left( 4\sum_{i = 1}^n 1 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right)$
Notice how the summations don't exactly look like the bulleted ones. Let's consider $\sum_{i=1}^n (i-1)^2$.
Expand it: $\sum_{i=1}^n (i-1)^2 \ = \ 0^2 + 1^2 + 2^2 + \cdots + n^2 \ = \ 1^2 + 2^2 + \cdots + n^2 \ = \ \sum_{i=1}^n i^2$
which matches our bulleted formula. Basically, we opened up the shorthand notation and rewritten it.
See if you can carry on from here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.998511552810669, "perplexity": 1217.7113370886266}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560285001.96/warc/CC-MAIN-20170116095125-00152-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.zbmath.org/?q=an%3A1184.68118 | ×
# zbMATH — the first resource for mathematics
Transactional contention management as a Non-clairvoyant scheduling problem. (English) Zbl 1184.68118
Summary: The transactional approach to contention management guarantees consistency by making sure that whenever two transactions have a conflict on a resource, only one of them proceeds. A major challenge in implementing this approach lies in guaranteeing progress, since transactions are often restarted.
Inspired by the paradigm of non-clairvoyant job scheduling, we analyze the performance of a contention manager by comparison with an optimal, clairvoyant contention manager that knows the list of resource accesses that will be performed by each transaction, as well as its release time and duration. The realistic, non-clairvoyant contention manager is evaluated by the competitive ratio between the last completion time (makespan) it provides and the makespan provided by an optimal contention manager.
Assuming that the amount of exclusive accesses to the resources is non-negligible, we present a simple proof that every work conserving contention manager guaranteeing the pending commit property achieves an $$O(s)$$ competitive ratio, where $$s$$ is the number of resources. This bound holds for the Greedy contention manager studied by R. Guerraoui et al. [in: Proceedings of the 24th Annual ACM Symposium on Principles of Distributed Computing (PODC), 258–264 (2005)] and is a significant improvement over the $$O(s^2)$$ bound they prove for the competitive ratio of Greedy. We show that this bound is tight for any deterministic contention manager, and under certain assumptions about the transactions, also for randomized contention managers.
When transactions may fail, we show that a simple adaptation of Greedy has a competitive ratio of at most $$O(ks)$$, assuming that a transaction may fail at most $$k$$ times. If a transaction can modify its resource requirements when re-invoked, then any deterministic algorithm has a competitive ratio $$\Omega(ks)$$. For the case of unit length jobs, we give (almost) matching lower and upper bounds.
##### MSC:
68M20 Performance evaluation, queueing, and scheduling in the context of computer systems
Full Text:
##### References:
[1] Borodin, A., El-Yaniv, R.: Online Computation and Competitive Analysis. Cambridge University Press, Cambridge (1998) · Zbl 0931.68015 [2] Edmonds, J., Chinn, D.D., Brecht, T., Deng, X.: Non-clairvoyant multiprocessor scheduling of jobs with changing execution characteristics. J. Sched. 6(3), 231–250 (2003) · Zbl 1154.90444 [3] Guerraoui, R., Herlihy, M., Pochon, B.: Toward a theory of transactional contention management. In: Proceedings of the 24th Annual ACM Symposium on Principles of Distributed Computing (PODC), pp. 258–264 (2005) · Zbl 1314.68088 [4] Guerraoui, R., Herlihy, M., Kapałka, M., Pochon, B.: Robust contention management in software transactional memory. In: Synchronization and Concurrency in Object-Oriented Languages (SCOOL). Workshop, in Conjunction with OOPSLA (2005). http://urresearch.rochester.edu/handle/1802/2103 [5] Herlihy, M., Luchangco, V., Moir, M., Scherer III, W.N.: Software transactional memory for dynamic-sized data structures. In: Proceedings of the 22nd Annual ACM Symposium on Principles of Distributed Computing (PODC), pp. 92–101 (2003) [6] Irani, S., Leung, V.: Scheduling with conflicts, and applications to traffic signal control. In: Proceedings of the 7th Annual ACM-SIAM Symposium on Discrete Algorithms (SODA), pp. 85–94 (1996) · Zbl 0845.90072 [7] Kalyanasundaram, B., Pruhs, K.R.: Fault-tolerant scheduling. SIAM J. Comput. 34(3), 697–719 (2005) · Zbl 1075.68004 [8] Motwani, R., Phillips, S., Torng, E.: Non-clairvoyant scheduling. Theor. Comput. Sci. 130(1), 17–47 (1994) · Zbl 0820.90056 [9] Rosenkrantz, D.J., Stearns, R.E., Lewis II, P.M.: System level concurrency control for distributed database systems. ACM Trans. Database Syst. 3(2), 178–198 (1978) [10] Scherer III, W.N., Scott, M.: Contention management in dynamic software transactional memory. In: PODC Workshop on Concurrency and Synchronization in Java Programs, pp. 70–79 (2004) [11] Scherer III, W.N., Scott, M.: Advanced contention management for dynamic software transactional memory. In: Proceedings of the 24th Annual ACM Symposium on Principles of Distributed Computing (PODC), pp. 240–248 (2005) [12] Silberschatz, A., Galvin, P.: Operating Systems Concepts, 5th edn. Wiley, New York (1999) · Zbl 0803.68019 [13] Vossen, G., Weikum, G.: Transactional Information Systems. Morgan Kaufmann, San Mateo (2001) [14] Yao, A.C.C.: Probabilistic computations: towards a unified measure of complexity. In: Proc. 18th Symp. Foundations of Computer Science (FOCS), pp. 222–227. IEEE (1977)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8352521061897278, "perplexity": 4543.95100532905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057504.60/warc/CC-MAIN-20210924050055-20210924080055-00449.warc.gz"} |
http://mathhelpforum.com/algebra/47421-indices-question.html | # Math Help - Indices question
1. ## Indices question
Question:
Attached Thumbnails
2. Originally Posted by madd.ryan
Question:
$\left(\frac{p}{q}\right)^{-2}$
When you have an indice to a negative power, this mean you will need to flip the fraction as you want the reciprocal.
$\left(\frac{q}{p}\right)^{2}$
When a fraction is to a power, then both the numerator and the denominator are to that power too.
$\frac{q^2}{p^2}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9871983528137207, "perplexity": 1133.852782096636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462009.45/warc/CC-MAIN-20150226074102-00052-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/95719-guessing-roots-polynomial.html | # Math Help - Guessing Roots of Polynomial
1. ## Guessing Roots of Polynomial
Is there any way to tell if a polynomial f(x) has some real root for x>c where c is a real constant.
I want to predict this without actually calculating the roots.
2. Originally Posted by hsachdevah
Is there any way to tell if a polynomial f(x) has some real root for x>c where c is a real constant.
I want to predict this without actually calculating the roots.
If f(c) and $\lim_{x\rightarrow \infty} f(x)$ are of different sign, then there must be at least one (in fact an odd number) of roots > c.
Unfortunately, if they are of the same sign, we do not know whether there are any roots > c (if there are any there must be an even number). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9368245005607605, "perplexity": 225.33652992069744}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098685.20/warc/CC-MAIN-20150627031818-00057-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-draw-the-lewis-structure-for-ions | Chemistry
Topics
# How do you draw the lewis structure for ions?
May 13, 2018
Well, what is the ion? Sulfate, chlorate, nitrate....?
#### Explanation:
In all of these cases, we must take the valence electrons of EACH atom in the ion, add the negative charge, (which is usually associated with the most electronegative atom, i.e. oxygen....) and then write the Lewis structure, and then ASSIGN the geometry...
For sulfate we got $5 \times 6 + 2 = 32 \cdot \text{valence electrons}$, i.e. sixteen electron pairs...
And so....${\left(O =\right)}_{2} S {\left(- {O}^{-}\right)}_{2}$...the oxygen atoms are conceived to bear the negative charges given that they own NINE electrons.
For chlorate we got ${\left(O =\right)}_{2} \ddot{C} l - {O}^{-}$...$7 + 3 \times 6 + 1 = 26 \cdot \text{valence electrons}$
For nitrate we got $O = \stackrel{+}{N} {\left(- {O}^{-}\right)}_{2}$...$5 + 3 \times 6 + 1 = 24 \cdot \text{valence electrons}$
And can you assigen the structure of the complex ion on the the basis of VESPER....
##### Impact of this question
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https://stats.stackexchange.com/questions/37761/framing-the-negative-binomial-distribution-for-dna-sequencing | # Framing the negative binomial distribution for DNA sequencing
The negative binomial distribution has become a popular model for count data (specifically the expected number of sequencing reads within a given region of the genome from a given experiment) in bioinformatics. Explanations vary:
• Some explain it as something that works like the Poisson distribution but has an additional parameter, allowing more freedom to model the true distribution, with a variance not necessarily equal to the mean
• Some explain it as a weighted mixture of Poisson distributions (with a gamma mixing distribution on the Poisson parameter)
Is there a way to square these rationales with the traditional definition of a negative binomial distribution as modeling the number of successes of Bernoulli trials before seeing a certain number of failures? Or should I just think of it as a happy coincidence that a weighted mixture of Poisson distributions with a gamma mixing distribution has the same probability mass function as the negative binomial?
• It is also a compound Poisson distribution where you sum a Poisson-distributed number of logarithmic random variables. – Douglas Zare Sep 22 '12 at 0:09
IMOH, I really think that the negative binomial distribution is used for convenience.
So in RNA Seq there is a common assumption that if you take an infinite number of measurements of the same gene in an infinite number of replicates then the true distribution would be lognormal. This distribution is then sampled via a Poisson process (with a count) so the true distribution reads per gene across replicates would be a Poisson-Lognormal distribution.
But in packages that we use such as EdgeR and DESeq this distribution modeled as a negative binomial distribution. This is not because the guys that wrote it didn't know about a Poisson Lognormal distribution.
It is because the Poisson Lognormal distribution is a terrible thing to work with because it requires numerical integration to do the fits etc. so when you actually try to use it sometimes the performance is really bad.
A negative binomial distribution has a closed form so it is a lot easier to work with and the gamma distribution (the underlying distribution) looks a lot like a lognormal distribution in that it sometimes looks kind of normal and sometimes has a tail.
But in this example (if you believe the assumption) it can't possibly be theoretically correct because the theoretically correct distribution is the Poisson lognormal and the two distributions are reasonable approximations of one another but are not equivalent.
But I still think the "incorrect" negative binomial distribution is often the better choice because empirically it will give better results because the integration performs slowly and the fits can perform badly, especially with distributions with long tails.
I looked through a few web pages and couldn't find an explanation, but I came up with one for integer values of $r$. Suppose we have two radioactive sources independently generating alpha and beta particles at the rates $\alpha$ and $\beta$, respectively.
What is the distribution of the number of alpha particles before the $r$th beta particle?
1. Consider the alpha particles as successes, and the beta particles as failures. When a particle is detected, the probability that it is an alpha particle is $\frac{\alpha}{\alpha+\beta}$. So, this is the negative binomial distribution $\text{NB}(r,\frac{\alpha}{\alpha+\beta})$.
2. Consider the time $t_r$ of the $r$th beta particle. This follows a gamma distribution $\Gamma(r,1/\beta).$ If you condition on $t_r = \lambda/\alpha$, then the number of alpha particles before time $t_r$ follows a Poisson distribution $\text{Pois}(\lambda).$ So, the distribution of the number of alpha particles before the $r$th beta particle is a Gamma-mixed Poisson distribution.
That explains why these distributions are equal.
I can only offer intuition, but the gamma distribution itself describes (continuous) waiting times (how long does it take for a rare event to occur). So the fact that a gamma-distributed mixture of discrete poisson distributions would result in a discrete waiting time (trials until N failures) does not seem too surprising. I hope someone has a more formal answer.
Edit: I always justified the negative binomial dist. for sequencing as follows: The actual sequencing step is simply sampling reads from a large library of molecules (poisson). However that library is made from the original sample by PCR. That means that the original molecules are amplified exponentially. And the gamma distribution describes the sum of k independent exponentially distributed random variables, i.e. how many molecules in the library after amplifying k sample molecules for the same number of PCR cycles.
Hence the negative binomial models PCR followed by sequencing.
• That makes sense, but in the context of measuring the number of sequencing reads in a genome is there an intuitive explanation for what the waiting period in the negative binomial distribution represents? In this case there is no waiting period - he's just measuring counts of sequencing reads. – RobertF Sep 21 '12 at 20:31
• See my edit. I don't see how thinking of it in terms of waiting times fits the sequencing setting. The gamma poisson mixture is easier to interpret. But in the end they are the same thing. – Felix Schlesinger Sep 21 '12 at 20:34
• Ok - then perhaps the real question is by what coincidence does modeling k successes + r failures in Bernoulli trials follow a gamma Poisson mixture? Maybe a negative binomial modeling k successes + r failures can be thought of as an overdispersed Poisson dbn due to the many possible permutations of success and failure trials resulting in the exactly k observed successes and r observed failures, which can be described as a collection of separate dbns? – RobertF Sep 21 '12 at 21:02
Assume we have a perfect uniform coverage of the genome before library prep, and we observe $\mu$ reads covering a site on average. Say that sequencing is a process that picks an original DNA fragment, puts it through a stochastic process that goes through PCR, subsampling, etc, and comes up with a base from the fragment at frequency $p$, and a failure otherwise. If sequencing proceeds until $\mu\frac{1-p}{p}$ failures, it can be modeled with a negative binomial distribution, $NB(\mu\frac{1-p}{p}, p)$.
Calculating the moments of this distribution, we get expected number of successes $\mu\frac{1-p}{p}\frac{p}{1-p} = \mu$ as required. For variance of the number of successes, we get $\sigma^2 = \mu(1-p)^{-1}$ - the rate at which the library prep fails for a fragment increases the variance in the observed coverage.
While the above is a slightly artificial description of the sequencing process, and one could make a proper generative model of the PCR steps etc, I think it gives some insight into the origin of the overdispersion parameter $(1-p)^{-1}$ directly from the negative binomial distribution. I do prefer the Poisson model with rate integrated out as an explanation in general. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9415577054023743, "perplexity": 402.52861093378436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317847.79/warc/CC-MAIN-20190823041746-20190823063746-00553.warc.gz"} |
http://www.ecircuitslab.com/2011/06/baud-rate-generator.html | # Baud Rate Generator
In this article, an RC oscillator is used as a baud rate generator. If you can calibrate the frequency of such a circuit sufficiently accurately (within a few percent) using a frequency meter, it will work very well. However, it may well drift a bit after some time, and then…. Consequently, here we present a small crystal-controlled oscillator. If you start with a crystal frequency of 2.45765 MHz and divide it by multiples of 2, you can very nicely obtain the well-known baud rates of 9600, 4800, 2400, 600, 300, 150 and 75. If you look closely at this series, you will see that 1200 baud is missing, since divider in the 4060 has no Q10 output!
If you do not need 1200 baud, this is not a problem. However, seeing that 1200 baud is used in practice more often than 600 baud, we have put a divide-by-two stage in the circuit after the 4060, in the form of a 74HC74 flip-flop. This yields a similar series of baud rates, in which 600 baud is missing. The trimmer is for the calibration purists; a 33 pF capacitor will usually provide sufficient accuracy. The current consumption of this circuit is very low (around 1mA), thanks to the use of CMOS components. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.929248571395874, "perplexity": 883.632133062983}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999642201/warc/CC-MAIN-20140305060722-00094-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/63334/closed-points-on-varieties | Closed points on varieties
I consider a variety over a field $k$, i.e. an integral separated scheme $X$ of finite type over $k$.
One knows by the Nullstellensatz that any closed point on $X$ is a $\bar k-$ rational point (where $\bar k$ denotes the algebraic closure of $k$)as its residue field is finite over $k$.
I know wonder what one can say about the relation between the closedness of a point and its residue field. E.g. it wont hold that any $\bar k-$rational point is closed, but can one say something similar? Or how can one characterize the closed points?
And does the situation change if one assumes the variety furthermore as complete over $k$?
-
Dear Descartes, it doesn't really make sense to say that a closed point on $X$ "is" a $\bar k$- rational point . If you want, I'll write a more detailed answer later , but I must run now! – Georges Elencwajg Sep 10 '11 at 15:59
Every rational point is closed... – Matt Sep 10 '11 at 16:25
Why not, Georges? I thought a $\bar k$-rational point (which is a morphism of $Spec(\bar k)$ to $X$) is equivalent to giving a point with residue field contained in $\bar k$. – Descartes Sep 10 '11 at 17:08
A $\bar{k}$-rational point of $X$ (by definition a morphism of $k$-schemes $\mathrm{Spec}(\bar{k})\rightarrow X$) is equivalent to the data of a point $x$ of $X$ and a $k$-monomorphism $k(x)\rightarrow\bar{k}$. – Keenan Kidwell Sep 10 '11 at 21:25
@Keenan. A perfect description: nothing to add or subtract! – Georges Elencwajg Sep 10 '11 at 21:29
The closed points of a finite type $k$-scheme are precisely the points with residue extension $k(x)/k$ algebraic (equivalently finite). The residue field of a closed point is a domain that is finitely generated as a $k$-algebra, also a field, hence a finite extension of $k$ by (a form of) the Nullstellensatz. For the other implication, assume $X=\mathrm{Spec}(A)$ with $A$ a finitely generated $k$-algebra. If $\mathfrak{p}\in X$ is such that $k(\mathfrak{p})/k$ is algebraic, then $k(\mathfrak{p})$ is integral over the domain $A/\mathfrak{p}$ which is therefore a field, i.e., $\mathfrak{p}$ is maximal.
-
Perfect, thanks a lot for the explanation, Keenan! – Descartes Sep 10 '11 at 18:10
A scheme (over a field $k$, say) really has two sorts of points and much confusion arises from the fact that they are not distinguished linguistically. For clarity's sake I'll call them (just here and now!) physical and functorial points.
The physical points They are elements of the underlying set $|X|$. Such an $x\in |X|$ has a residual field $\kappa (x)$ which is an extension $k \to \kappa (x)$. If that extension is an isomorphism, we say that $x$ is rational or $k$-rational.
The functorial points They are $k$ -morphisms from some $k$-scheme $Y$ to $X$. You are interested in the special case where $Y$ corresponds to a fixed algebraic closure $k\to \bar k$. In that special case, a $\bar k$-point $f:Spec (\bar k) \to X$ of $X$ certainly has an image $x=f(\ast)\in X$.
However the crucial point is that this image does not determine $f$ at all. You also have to give yourself a $k$-algebra morphism $\kappa (x) \to \bar k$ in order to define $f$.
So the same $x$ can correspond to billions of $\bar k$-points, say $7$ billion.
An example Consider $k=\mathbb Q$ and $X=Spec( \mathbb Q[T]/\langle T^{7,000,000,000}-2\rangle)=Spec(K)=\lbrace x\rbrace$.
Although $X$ has only one physical point, namely $x$, there are 7 billion different $\bar {\mathbb Q}$- points in $X$ .
[They correspond -via the affine scheme/ring dictionary- to the $\mathbb Q$-algebra morphisms $K \to \bar {\mathbb Q}$, which in turn are uniquely determined by the choice of a 7,000,000,000-th root of 2 in $\bar {\mathbb Q}$]
The case of varieties In the case of a variety $X$, the closed physical points are exactly the images of the $\bar k$-points of $X$ (see Keenan's answer). Completeness of $X$ is irrelevant.
-
This is a very nice answer Georges (as your answers tend to be)! – Keenan Kidwell Sep 10 '11 at 20:45
Dear Keenan, considering you have written an answer yourself, your comment is a really gracious gesture: bravo and thank you. – Georges Elencwajg Sep 10 '11 at 21:18
One should mention that for a $k$-variety, $|X| = X(\bar{k})/Gal(\bar{k}/k)$: closed points correspond to Galois orbits of $\bar{k}$-points as the Galois group acts transitively on the embeddings $\kappa(x) \hookrightarrow \bar{k}$. In the case $X = \mathbb{A}^1$, this is a generalization of the fact that the Galois orbit of $\alpha \in \bar{k} = \mathbb{A}^1(\bar{k})$ consists of the roots of its minimal polynomial $P$ over $k$, i.e. corresponds to the maximal ideal $(P)$ of $k[X]$. – AFK Sep 10 '11 at 23:22
Dear user8882, thanks for your judicious comment. Just a detail: I would use the notation $Aut(\bar k/k)$ instead of $Gal(\bar k /k)$, since $\bar k /k$ is not Galois unless $k$ is perfect. – Georges Elencwajg Sep 10 '11 at 23:59
I love the clarity of your distinction, Georges. And your example! – Descartes Sep 11 '11 at 8:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9638794660568237, "perplexity": 274.46728839635944}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049278244.51/warc/CC-MAIN-20160524002118-00001-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://space.stackexchange.com/questions/18639/relation-between-air-core-and-ferromagnetic-solenoid-core-magnetorquers-and-mea?noredirect=1 | # Relation between air-core and ferromagnetic solenoid -core magnetorquers and measurement of dipole moment
I am going to be using air coil magnetorquers for a satellite development project. These air coil magnetorquers come embedded in solar panels. I am currently using this magnetorquer , it's a solar panel with a magnetorquer embedded into it. The magnetorquers are going to be used to produce a magnetic moment ($m$) to desaturate the reaction wheels in space. I have a controller that will provide me with the $m$ that I require, but I need a way to experimentally verify that I am getting the correct $m$.
I am pretty familiar with the math behind the ferromagnetic magnetorquers, alot of information is provided by this link.
I mainly have two questions:
Firstly,I am working with the magnetorquers integrated with the solar panels, would the mathematics remain the same? I also know that for normal ferremagnetic magnetorquers , $m=nIA$ , where $m$ is the magnetic dipole moment, $I$ is the current through the magnetorquer and $A$ is the area of the magnetorquer. Are these equations still valid for the air coil magnetorquers?
Secondly, how would I be able to experimentally measure the magnetic dipole moment generated by the air coil? For the ferromagnetic coil magnetoquers, I found this research paper to measure the magnetic dipole moment for a normal ferromagnetic magnetorquer, but I don't know if it would work for an air coil magnetorquer. I am also worried about how the solar panel itself would interfere with the measurements.
Thanks.
• That's a very nice paper about the ferromagnetic rod magnetotorquer testing. I don't think that that the equation $m=nIA$ applies to a ferromagnetic rod - there's no place to but the permeability or dimensions of the rod. However I think is a good equation to use for your flat coil. I believe that the spec of 1.55$m^2$ in the data sheet represents the product $nA$. If the area of one turn of the PCB trace is roughly 9x9 cm${}^2$ then there are roughy 150 to 200 very narrow turns in the trace, which explains the high DC resistance. – uhoh Oct 16 '16 at 1:26
• Yeah, I tried using m=nIA to calculate n, the number of turns, and found that it is indeed 1. The datasheet specifies the max m at the max voltage. But in terms of measuring the actual m, would it be already to think of my air coil as a ferromagnetic rod that is pressed flat, and use the same paper that I linked to find m? – John Oct 16 '16 at 1:29
• No you definitely can't use that equation. It is a (fairly good) approximation for a uniformly magnetized rod of ferromagnetic material, with a permeability $/mu$. The detailed shape of the coil isn't even specified there - it assumes the coil magnetizes the permeable material uniformly. Here you have only a coil and no permeable material. You need math that applies to a flat coil in air. It will also have to be an approximation because this coil has multiple turns of different sizes, and their shape isn't even a perfect square. – uhoh Oct 16 '16 at 1:41
• So again the best way is to measure at large distance where you can call it a point dipole. – uhoh Oct 16 '16 at 1:41
• Ah, alright. I'll edit my question to make it more concise – John Oct 16 '16 at 5:04
As for the equation, the difference between an air core and a magnetic core is reflected in a missing term known as the magnetic permeability, $\mu$, which is a measure of how much the magnetic field is boosted by the presence of the core material.
• Yeah. My main concern is that they are two totally different types of magnetorquers. Unless I can assume that my current embedded magnetorquer to be a compressed version of the torque rod used in the research paper. Which would mean that I'd have to prop up the solar array on it's side and duplicate the same experiment. Which I am not sure if I can do. I can't really apply $m=nIA$ as well, as the no. of turns,$n$, is not provided in the datasheet. – John Oct 16 '16 at 17:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8371826410293579, "perplexity": 451.0145821498876}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655921988.66/warc/CC-MAIN-20200711032932-20200711062932-00112.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/2703/is-the-pauli-group-for-n-qubits-a-basis-for-mathbbc2n-times-2n | # Is the Pauli group for $n$-qubits a basis for $\mathbb{C}^{2^n\times 2^n}$?
The $$n$$-fold Pauli operator set is defined as $$G_n=\{I,X,Y,Z \}^{\otimes n}$$, that is as the set containing all the possible tensor products between $$n$$ Pauli matrices. It is clear that the Pauli matrices form a basis for the $$2\times 2$$ complex matrix vector spaces, that is $$\mathbb{C}^{2\times 2}$$. Apart from it, from the definition of the tensor product, it is known that the $$n$$-qubit Pauli group will form a basis for the tensor product space $$(\mathbb{C}^{2\times 2})^{\otimes n}$$.
I am wondering if the such set forms a basis for the complex vector space where the elements of this tensor product space act, that is $$\mathbb{C}^{2^n\times 2^n}$$. Summarizing, the question would be, is $$(\mathbb{C}^{2\times 2})^{\otimes n}=\mathbb{C}^{2^n\times 2^n}$$ true?
I have been trying to prove it using arguments about the dimensions of both spaces, but I have not been able to get anything yet.
• The set $\{I,X,Y,Z \}^{\otimes n}$ that you describe is only $1/4$ of the Pauli group (in fact it is not a group at all since it fails to be closed under multiplication). The set $\{I,X,Y,Z \}^{\otimes n}$ is mostly famous because it is both a generating set for the Pauli group and, as you point out, an orthogonal basis for the space of $2^n \times 2^n$ complex matrices. It is already clear for $n=1$ that $\{ I,X,Y,Z \}$ fails to be a group. For example just the subgroup generated by $X,Z$ has many extra elements $<X,Z>=I,X,Z,XZ,XZXZ=-I,-X,-Z, ZX=-XZ$ May 13 at 14:17
• Yeah you are right. I was actually referring to the $n$-fold Pauli operator set. This anyway does not change the question nor the answers. Even if the actual Pauli group was considered, it would also form a basis for such complex space. I edited the question for clarity. May 13 at 21:10
• Ya I agree you are totally right the Pauli set is a nice orthogonal basis. And you are right that the actual Pauli group is a spanning set. But the actual Pauli group would not be a basis since not all the vectors are linearly independent (for example $X,-X,iX,-iX$ are all linearly dependent) May 13 at 22:08
• Ok, agreed. I was just thinking about generating the complex space, not about the concept of basis. Nice clarification. 2 days ago
• The Pauli set is the best matrix basis that is also (almost) a group...some details... basically the group $G$ in the linked question is the Pauli group (for $n$ a prime power the Pauli group is called an extraspecial p group, although for $p=2$ it is rather the real Pauli group generated by $I,X,Z,XZ$ that is extraspecial, i.e. drop the global $i$ phase) and $\{ \overline{g_i}: 1 \leq j \leq n^2 \}$ is the Pauli set. When I say this is a group in $PGL$ that just means its a group if you add some phases like $\pm1$. Enjoy! 2 days ago
Yes, the set of tensor products of all possible $$n$$ Pauli operators (including $$I$$) form an orthogonal basis for the vector space of $$2^n \times 2^n$$ complex matrices. To see this first we notice that the space has a dimension of $$4^n$$ and we also have $$4^n$$ vectors ( the vectors are operators in this case). So we only need to show that they are linearly independent.
We can actually show something stronger. It can be easily seen that the members of the Pauli group are orthogonal under the Hilbert-Schmidt inner product. The H-S inner product of two matrices is defined as $$Tr(AB^\dagger)$$. We can easily verify from the definition that the Pauli group is a mutually orthogonal set under this inner product. We simply have to use the elementary property $$Tr(C \otimes D) = Tr(C)Tr(D)$$.
• @biryaniTo prove that the new set you obtained is linearly independent, wouldn't you need to prove that $Tr(AB)=0$ for every $A$ and $B$ of the new set? In that case, I haven't understood how the trace propriety with respect to the tensor product comes into play. When you compute $Tr(A_1 \otimes ... \otimes A_n) = Tr(A_1)...Tr(A_n)$ you are computing the trace of a new element of the set but not the trace of the product of the elements of that set $Tr(AB)$. In the latter case, it's not allowed to write $Tr(AB)=Tr(A)Tr(B)$ and then expand the two traces as you proposed, so I missed a step. Sep 26, 2020 at 10:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9328713417053223, "perplexity": 125.75014762894777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00590.warc.gz"} |
http://mathhelpforum.com/calculus/98439-solved-can-someone-give-me-start-integrating-please.html | # Math Help - [SOLVED] Can someone give me a start integrating this please?
1. ## [SOLVED] Can someone give me a start integrating this please?
$
\int
\frac{1}{10 - kv^2}
\, dv
$
2. Hi Bucephalus
If k is a constant, you can try :
1. Let : $v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta$
or
2. Partial fraction by first factoring the denominator
Originally Posted by songoku
Hi Bucephalus
If k is a constant, you can try :
1. Let : $v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta$
or
2. Partial fraction by first factoring the denominator
Hi Songoku
Two things:
1) I want to understand how you got that expression in your first helper.
2) I have done partial fraction decomposition before, but I don't understand what you're saying "factoring the denominator"
$
\frac{1}{10} - \frac{1}{(kv)v}
$
I'm sorry, I'm really having trouble understanding.
bucephalus
4. Originally Posted by Bucephalus
[snip]
2) I have done partial fraction decomposition before, but I don't understand what you're saying "factoring the denominator"
$
\frac{1}{10} - \frac{1}{(kv)v}
$
I'm sorry, I'm really having trouble understanding.
bucephalus
Factorise $10 - kv^2$.
And if you're studying at the level of integral calculus I would sincerely hope that you know that $\frac{1}{10 - kv^2}$ is certainly NOT equal to $\frac{1}{10} - \frac{1}{(kv)v}$ !!!
5. Hi Bucephalus
I want to understand how you got that expression in your first helper
It's because I want to use trigonometry identity : $1- \sin^2 \theta = \cos^2 \theta$
Using $v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta$ will change the denominator to :
$10-kv^2=10-10 \sin^2 \theta=10(1-\sin^2 \theta)=10 \cos^2 \theta$
In fact, you also can use : $\frac{\sqrt{10}}{\sqrt{k}} \cos \theta$ because it will be relevant to identity : $1- \cos^2 \theta = \sin^2 \theta$
If the question turns out to be $10+kv^2$ , you can use substitution : $v=\frac{\sqrt{10}}{\sqrt{k}} \tan \theta$ because it is relevant to identity : $1+\; tan^2 \theta = \sec^2 \theta$
For the second one, follow Mr. F suggestion :
Factorise $10 - kv^2$
6. ## Factorising
so factorise this.
$10 - kv^2$
The only way I can think of to factorise this is:
$10(1-\frac{k}{10}v^2)$
Is this what you mean?
7. $10-kv^2 = \sqrt{10}^2-\sqrt{k}^2v^2$
now using the difference of two squares $a^2-b^2 = (a-b)(a+b)$
You get
$10-kv^2 = \sqrt{10}^2-\sqrt{k}^2v^2 = \sqrt{10}^2-(\sqrt{k}v)^2 = (\sqrt{10}-\sqrt{k}v)(\sqrt{10}+\sqrt{k}v)$
So now $\frac{1}{10-kv^2} = \frac{A}{\sqrt{10}-\sqrt{k}v}+\frac{B}{\sqrt{10}+\sqrt{k}v}$
Find A and B and the integral will be alot easier in this form.
8. ## Thanks Pickslides
That's an awesome response. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.983392059803009, "perplexity": 720.1982621388477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500829421.59/warc/CC-MAIN-20140820021349-00216-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/dimensions-of-intensity.733286/ | # Dimensions of intensity
1. Jan 17, 2014
### Old Guy
1. The problem statement, all variables and given/known data
I know that intensity is power per unit area. When I do a dimensional analysis, it reduces down to mass per cubic time. Is there any physical significance to that?
2. Relevant equations
I=P/A
3. The attempt at a solution
2. Jan 17, 2014
### lightgrav
Not really, because that form has removed all the geometry of the situation;
Intensity's geometry comes from the energy flow vector piercing an Area, so a flow Volume is indicated.
On the other hand, if you divide the intensity by flow velocity, you end with m v^2 / length^3 ;
light's intensity divided by light speed leaves you with mc^2 / volume, the energy density in the light.
Draft saved Draft deleted
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=207805 | ## ICE
Erik Buetow 1F
Posts: 96
Joined: Fri Aug 30, 2019 12:15 am
### ICE
When should we use the ice table vs when do we just use the normal equilibrium concentration? Is the ICE table only used when a reaction is not in equilibrium and we need to find information from it?
Jessica Booth 2F
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am
### Re: ICE
The ice tables are mainly used when we only know the initial concentration(s) and the equilibrium constant. Basically, we use ice tables when we are unable to just use the equilibrium concentrations and equilibrium constant to find the other equilibrium concentrations.
jisulee1C
Posts: 149
Joined: Thu Jul 25, 2019 12:17 am
### Re: ICE
ICE charts can be used to find equilibrium molar concentration using molar ratios and the change in molar concentration if the initial molar concentration for either reactants or products are given and the equilibrium constant is also given.
DLee_1L
Posts: 103
Joined: Sat Aug 17, 2019 12:17 am
### Re: ICE
Also, ICE tables are used to find the equilibrium concentrations for weak acids and bases that don't fully dissociate.
805373590
Posts: 101
Joined: Wed Sep 11, 2019 12:17 am
### Re: ICE
ICE tables are used when the problem is only providing the initial concentrations and we are trying to find the ph or ooh of weak acids or bases
Tahlia Mullins
Posts: 105
Joined: Thu Jul 25, 2019 12:15 am
### Re: ICE
To determine if an ICE table is necessary, look at the information given. Such as, if the initial concentration(s) are given along with a Ka or Kb value, then an ICE table is the correct method in finding the equilibrium concentrations and other information. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9358290433883667, "perplexity": 2986.2894580975553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655882934.6/warc/CC-MAIN-20200703184459-20200703214459-00154.warc.gz"} |
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[91] L. H. Erbe and Haiyan Wang. On the existence of positive solutions of ordinary differential equations . Proc. Amer. Math. Soc. 120 (1994) 743-748. MR 1204373. Abstract, references, and article information View Article: PDF This article is available free of charge [92] Dariusz Bielawski. Generic properties of the Nicoletti and Floquet boundary value problems . Proc. Amer. Math. Soc. 120 (1994) 831-841. MR 1195712. Abstract, references, and article information View Article: PDF This article is available free of charge [93] Alessandro Fonda. On the existence of periodic solutions for scalar second order differential equations when only the asymptotic behaviour of the potential is known . Proc. Amer. Math. Soc. 119 (1993) 439-445. MR 1154246. Abstract, references, and article information View Article: PDF This article is available free of charge [94] Pierpaolo Omari and Fabio Zanolin. Nonresonance conditions on the potential for a second-order periodic boundary value problem . Proc. Amer. Math. Soc. 117 (1993) 125-135. MR 1143021. Abstract, references, and article information View Article: PDF This article is available free of charge [95] Nicholas D. Alikakos, Peter W. Bates and Giorgio Fusco. Solutions to the nonautonomous bistable equation with specified Morse index. I. Existence . Trans. Amer. Math. Soc. 340 (1993) 641-654. MR 1167183. Abstract, references, and article information View Article: PDF This article is available free of charge [96] Kevin McLeod. Uniqueness of positive radial solutions of $\Delta u+f(u)=0$ in ${\bf R}\sp n$. II . Trans. Amer. Math. Soc. 339 (1993) 495-505. MR 1201323. Abstract, references, and article information View Article: PDF This article is available free of charge [97] Man Kam Kwong and Yi Li. Uniqueness of radial solutions of semilinear elliptic equations . Trans. Amer. Math. Soc. 333 (1992) 339-363. MR 1088021. Abstract, references, and article information View Article: PDF This article is available free of charge [98] Alexander Nabutovsky. Number of solutions with a norm bounded by a given constant of a semilinear elliptic PDE with a generic right-hand side . Trans. Amer. Math. Soc. 332 (1992) 135-166. MR 1066447. Abstract, references, and article information View Article: PDF This article is available free of charge [99] Anna Capietto, Jean Mawhin and Fabio Zanolin. Continuation theorems for periodic perturbations of autonomous systems . Trans. Amer. Math. Soc. 329 (1992) 41-72. MR 1042285. Abstract, references, and article information View Article: PDF This article is available free of charge [100] Jie Jiang and Xue Kong Wang. A singular perturbation nonlinear boundary value problem and the $E$-condition for a scalar conservation law. Quart. Appl. Math. 50 (1992) 547-557. MR MR1178434. 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Results: 91 to 120 of 147 found Go to page: 1 2 3 4 5 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9111024737358093, "perplexity": 1444.9457392155036}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128321309.25/warc/CC-MAIN-20170627101436-20170627121436-00714.warc.gz"} |
http://dev.goldbook.iupac.org/terms/view/M03688 | magnetic moment $$m$$, $$\mu$$
https://doi.org/10.1351/goldbook.M03688
Vector quantity, the vector product of which with the @M03686@ of a homogeneous field is equal to the @T06400@.
Source:
Green Book, 2nd ed., p. 21 [Terms] [Book] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9876159429550171, "perplexity": 2456.7732481545304}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583514162.67/warc/CC-MAIN-20181021161035-20181021182535-00058.warc.gz"} |
http://www.gradesaver.com/textbooks/science/chemistry/chemistry-the-central-science-13th-edition/chapter-15-chemical-equilibrium-exercises-page-663/15-15f | ## Chemistry: The Central Science (13th Edition)
$Kc = [H+][OH-]$ heterogeneous
The expression for Kc is given by the concentration of the products raised to their stoichiometric coefficients divided by the concentration of the reactants raised to their coefficients. $Kc = [H+][OH-]$ There is no denominator because the only reactant is H2O (l) and we cannot have the concentration of a pure liquid. Since there are both liquids and aqueous solutions it is heterogeneous. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8919431567192078, "perplexity": 907.7336722373659}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00626.warc.gz"} |
https://stats.stackexchange.com/questions/412047/the-minimal-sufficient-statistic-of-fx-e-x-thetae-e-x-theta | # The minimal sufficient statistic of $f(x) = e^{-(x-\theta)}e^{-e^{-(x-\theta)}}$
The Casella Berger (2002) solutions manual says that the minimal sufficient statistic for
$$f(x) = e^{-(x-\theta)}e^{-e^{-(x-\theta)}}, \qquad x\in \mathbb{R}$$
are the order statistics $$(X_{(1)},\dots,X_{(n)})$$. This confuses me because it can be written as
$$f(x) = e^{\theta}e^{-x}e^{-e^\theta{e^{-x}}}$$
Which seems to be in the form of a full-rank exponential family, with complete sufficient statistic $$T(X) = \sum_i e^{-X_i}$$, and hence $$T(X)$$ is minimal sufficient by Bahadur's theorem. $$T(X)$$ seems to achieve a much greater reduction in the data aswell, so that the order statistics cannot be minimal sufficient?
I wonder if there is a property of the PDF I am missing that means it's not actually exponential family?
• I think you're right. The full location-scale family Gumbel is not exponential family but the location-family with scale=1 as here would be. Jun 8, 2019 at 3:26
Indeed it is clear from the density of $$X_1,\ldots,X_n$$ that $$\sum\limits_{i=1}^n e^{-X_i}$$ is a minimal complete sufficient statistic for $$\theta$$ as the pdf is a member of a regular full-rank exponential family as you say:
\begin{align} f_{\theta}(x_1,\cdots,x_n)&=e^{-\sum\limits_{i=1}^n x_i+n\theta}\exp\left(-\sum_{i=1}^n e^{-(x_i-\theta)}\right) \\&=\exp\left(-e^{\theta}\sum_{i=1}^n e^{-x_i}+n\theta\right)e^{-\sum\limits_{i=1}^nx_i}\quad\small\text{ for all }(x_1,\ldots,x_n)\in\mathbb R^n\,,\,\theta\in\mathbb R \end{align} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8440777063369751, "perplexity": 521.2258406891397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00642.warc.gz"} |
http://www.enotes.com/homework-help/find-equation-line-passes-through-point-5-2-216901 | # Find the equation of the line passes through the point ( 5,-2) and parallel to the line : 2y-x = 5
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
First we will write the equation for the line in the standard format:
y-y1 = m(x-x1) where (x1,y1) any point passes through the line and m is the slope.
==> y--2 = m(x-5)
==> y+ 2 = m(x-5)
But given the the line 2y-x = 5 is perpendicular to the line.
Then the products of the slopes should be -1.
2y-x = 5
We will rewrite in slope form:
2y= x+ 5
==> y= (1/2)x + 5/2
Then the slpe = 1/2
==> 1/2* m = -1
==> m= -2
==> y+ 2 = -2(x-5)
==> y= -2x + 10 -2
==> y= -2x +8
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
Two lines are parallel if and only if their slopes are equal or if the ratio of their correspondent coefficients are also equal.
We'll put the given equation into the standard form. For this reason, we'll isolate 2y to the left side:
2y = x + 5
We'll divide by 2:
y = x/2 + 5/2
We'll write the standard form of the equation of the parallel line and we'll identify the value of the slope for both lines.
y = mx + n
m1 = 1/2 and m2 = m
The slopes have like values.
m = 1/2
Now, we'll write the equtaion of the line that has the slope m=1/2 and it passes through the point (5,-2).
y - (-2) = (1/2)(x - 5)
y + 2 = x/2 - 5/2
y = x/2 - 5/2 - 2
y = x/2 - 9/2
2y = x - 9
2y - x + 9 = 0
The equation of the parallel line whose slope is m = 1/2 and it is passing through the point (5,-2) is 2y - x + 9 = 0. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8028510212898254, "perplexity": 1663.5092080508246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661155.56/warc/CC-MAIN-20160924173741-00142-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://api.philpapers.org/rec/BECKCF | # Kolmogorov complexity for possibly infinite computations
Abstract In this paper we study the Kolmogorov complexity for non-effective computations, that is, either halting or non-halting computations on Turing machines. This complexity function is defined as the length of the shortest input that produce a desired output via a possibly non-halting computation. Clearly this function gives a lower bound of the classical Kolmogorov complexity. In particular, if the machine is allowed to overwrite its output, this complexity coincides with the classical Kolmogorov complexity for halting computations relative to the first jump of the halting problem. However, on machines that cannot erase their output –called monotone machines–, we prove that our complexity for non effective computations and the classical Kolmogorov complexity separate as much as we want. We also consider the prefix-free complexity for possibly infinite computations. We study several properties of the graph of these complexity functions and specially their oscillations with respect to the complexities for effective computations Keywords infinite computations Kolmogorov complexity monotone machines non-effective computations program-size complexity Turing machines Categories (categorize this paper) DOI 10.1007/s10849-005-2255-6 Options Mark as duplicate Export citation Request removal from index
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## References found in this work BETA
On Degrees of Unsolvability.J. R. Shoenfield - 1964 - Journal of Symbolic Logic 29 (4):203-204. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9709477424621582, "perplexity": 2170.829376160615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103324665.17/warc/CC-MAIN-20220627012807-20220627042807-00608.warc.gz"} |
https://science.sciencemag.org/content/343/6172/754.abstract | Report
# Designing Collective Behavior in a Termite-Inspired Robot Construction Team
See allHide authors and affiliations
Science 14 Feb 2014:
Vol. 343, Issue 6172, pp. 754-758
DOI: 10.1126/science.1245842
## Abstract
Complex systems are characterized by many independent components whose low-level actions produce collective high-level results. Predicting high-level results given low-level rules is a key open challenge; the inverse problem, finding low-level rules that give specific outcomes, is in general still less understood. We present a multi-agent construction system inspired by mound-building termites, solving such an inverse problem. A user specifies a desired structure, and the system automatically generates low-level rules for independent climbing robots that guarantee production of that structure. Robots use only local sensing and coordinate their activity via the shared environment. We demonstrate the approach via a physical realization with three autonomous climbing robots limited to onboard sensing. This work advances the aim of engineering complex systems that achieve specific human-designed goals.
## Robot Rules
In the case of mound-building termites, colonies comprising thousands of independently behaving insects build intricate structures, orders of magnitude larger than themselves, using indirect communication methods. In this process, known as stigmergy, local cues in the structure itself help to direct the workers. Werfel et al. (p. 754; see the Perspective by Korb) wanted to construct complex predetermined structures using autonomous robots. A successful system was designed so that for a given final structure, the robots followed basic rules or “structpaths” in order to complete the task.
View Full Text | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8228097558021545, "perplexity": 4600.251062577938}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046155458.35/warc/CC-MAIN-20210805063730-20210805093730-00464.warc.gz"} |
http://mathhelpforum.com/calculus/47804-limit-derivative.html | 1. ## limit & derivative
hey guys having trouble with this question. any help?
Use f '(x) = lim h->0 f(x + h) - f(x)/h
to find dy/dx when y = 3x + 4.
much appreciated!
2. Originally Posted by jvignacio
hey guys having trouble with this question. any help?
Use f '(x) = lim h->0 f(x + h) - f(x)/h
to find dy/dx when y = 3x + 4.
much appreciated!
$f(x) = 3x + 4$.
$f(x + h) = 3(x + h) + 4$.
$f(x + h) - f(x) = 3h$.
$\frac{f(x+h)-f(x)}{h} = 3$.
$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \rightarrow 0} 3 = 3$.
3. Originally Posted by mr fantastic
$f(x) = 3x + 4$.
$f(x + h) = 3(x + h) + 4$.
$f(x + h) - f(x) = 3h$.
$\frac{f(x+h)-f(x)}{h} = 3$.
$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \rightarrow 0} 3 = 3$.
haha any way of explaining what u did ??? so i can do the others myself.
4. Originally Posted by jvignacio
haha any way of explaining what u did ??? so i can do the others myself.
That's about as plain as it will ever get! What don't you understand?
5. Originally Posted by mr fantastic
That's about as plain as it will ever get! What don't you understand?
actually its okay. i understand now. thank u | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9534005522727966, "perplexity": 2365.5582893613782}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320863.60/warc/CC-MAIN-20170626184725-20170626204725-00524.warc.gz"} |
http://link.springer.com/article/10.1007%2FBF00334039 | , Volume 77, Issue 2, pp 231-240
# On the shape of the convex hull of random points
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## Summary
Denote by E n the convex hull of n points chosen uniformly and independently from the d-dimensional ball. Let Prob(d, n) denote the probability that E n has exactly n vertices. It is proved here that Prob(d, 2 d/2 d )→1 and Prob(d, 2 d/2 d (3/4)+ɛ)→0 for every fixed ɛ>0 when d→∞. The question whether E n is a k-neighbourly polytope is also investigated. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9412215352058411, "perplexity": 2001.2657739366675}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637898629.32/warc/CC-MAIN-20141030025818-00032-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://mathoverflow.net/feeds/question/103561 | On finding A-polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:57:07Z http://mathoverflow.net/feeds/question/103561 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103561/on-finding-a-polynomials On finding A-polynomials Satoshi Nawata 2012-07-30T23:38:32Z 2012-08-02T13:36:40Z <p>I have two questions to obtain the explicit forms of A-polynomials.</p> <p><a href="http://arxiv.org/abs/math/0401068" rel="nofollow">Takata</a> used the mathematica pacage qMultisum.m to obtain the recursion relation of the colored Jones polynomials for twist knots. As Stavros Garoufalidis and Xinyu Sun pointed out in <a href="http://de.arxiv.org/abs/0802.4074" rel="nofollow">this paper</a>, the simple use of the mathematica pacage qZeil.m, qMultisum.m does not give the recursion relation of minimal order. They made use of the method, so-called creative telescoping, to get the recursion relation of minimal order by using the certificat function.</p> <blockquote> <blockquote> <ol> <li>How do you implement this method in Mathematica, say, to get the recursion relation for $5_2$ and $6_1$ knots as in p.4 of <a href="http://de.arxiv.org/abs/0802.4074" rel="nofollow">the paper</a>?</li> </ol> </blockquote> </blockquote> <p>Recently, <a href="http://arxiv.org/abs/1205.1515v2" rel="nofollow">Gukov, Sulkowski and Fuji</a> conjecture that, in the limit, $$q = e^{\hbar} \to 1 \,, \qquad a = \text{fixed} \,, \qquad t = \text{fixed} \,, \qquad x = q^n = \text{fixed}$$ the $n$-colored superpolynomials $P_n (K;a,q,t)$ exhibit the following large color'' behavior: $$P_n (K;a,q,t) \;\overset{{n \to \infty \atop \hbar \to 0}}{\sim}\; \exp\left( \frac{1}{\hbar} \int \log y \frac{dx}{x} \,+\, \ldots \right)$$ where ellipsis stand for regular terms (as $\hbar \to 0$) and the leading term is given by the integral on the zero locus of the super-$A$-polynomial: $$A^{\text{super}} (x,y;a,t) \; = \; 0 \ .$$</p> <p>For example, the critical points of the leading term of colored superpolynomials of torus knots $T^{2,2p+1}$ are give by \begin{eqnarray} 1 \; &=& \; -\frac{t^{-2-2p}(x-z_0)z_0^{-1-2p}(-1+t^2z_0)(1+ at^3 xz_0)}{(-1+z_0)(atx+z_0)(-1 + t^2 x z_0)} \cr y(x,t,a)&=& \frac{a^p t^{2 + 2 p} (-1 + x) x^{1 + 2 p} (atx + z_0) (1 + a t^3 x z_0)}{(1 + a t^3 x) (x - z_0) (-1 + t^2 x z_0)} , \end{eqnarray} which is written in Eq.(2.35) and (2.36). By eliminating $z_0$, you will obtain the super-$A$-polynomials for torus knots $T^{2,2p+1}$. Off course, it should be doable in principle, but</p> <blockquote> <blockquote> <p>$2$. how can it be implemented explicitly to obtain the super-$A$-polynomials as in Table 5 of <a href="http://arxiv.org/abs/1205.1515v2" rel="nofollow">this paper</a>? In other words, how do you explicitly eliminate $z_0$ in such a way that you will obtain the super-$A$-polynomials?</p> </blockquote> </blockquote> <p>I have the same problem to obtain the $Q$-deformed $A$-polynomials from Eq.(A.21) in <a href="http://arxiv.org/abs/1203.2182v1" rel="nofollow">this paper</a>.</p> | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0, "math_score": 0.9179399013519287, "perplexity": 4164.473129952219}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368702298845/warc/CC-MAIN-20130516110458-00010-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://mathhelpforum.com/math-topics/167790-boundaries-print.html | # Boundaries
• Jan 8th 2011, 11:17 AM
Natasha1
Boundaries
I never understand boundaries and would need some help on this question
In a race, Paula runs 25 laps of a track.
Each lap of the track is 400m, correct to the nearest metre.
Paula's average speed is 5.0 m/s, correct to one decimal place.
Calculate the upper bound for the time that Paula takes to run the race.
Give your answer in mins and seconds, correct ot the nearest second!
• Jan 8th 2011, 11:34 AM
snowtea
Upper bound for time is found using the lower bound on average speed and upper bound on track length.
I'm not sure what correct to one decimal place means.
If this means the digit is correct, then the average speed lower bound is 5.0m/s.
If it means correct rounded to 1 decimal place, then the average speed lower bound is 4.95 m/s.
If it means with an error of 0.1 m/s, then 4.9 m/s.
Upper bound on track distance is 401m.
• Jan 8th 2011, 11:38 AM
Natasha1
so I do 401 divide by 4.95 which gives me 81.01010101 mins/sec is this correct? And are the units correct too?
• Jan 8th 2011, 11:41 AM
snowtea
I think you have the right idea.
$time = distance / (avg\, speed)$, so you want to maximize distance and minimize average speed to get maximum time.
Units for time should be seconds.
• Jan 8th 2011, 11:43 AM
Natasha1
so time = 81 seconds
• Jan 8th 2011, 11:48 AM
snowtea
Quote:
Originally Posted by Natasha1
so time = 81 seconds
Close. Is 81 an upper bound of 81.010101... ?
• Jan 8th 2011, 11:50 AM
Natasha1
I don't know
• Jan 8th 2011, 12:09 PM
snowtea
Is $81 \geq 81.010101...$?
• Jan 8th 2011, 12:11 PM
Natasha1
strickly smaller
• Jan 8th 2011, 12:25 PM
snowtea
The question says compute an upper bound for time.
How can 81 seconds be an upper bound, when you computed something larger than 81 seconds for your answer.
• Jan 8th 2011, 12:27 PM
Natasha1
Got it! It's 82 dooooooooohhhhhh! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9794625043869019, "perplexity": 3283.597932304674}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720000.45/warc/CC-MAIN-20161020183840-00248-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://socratic.org/questions/when-n2-g-reacts-with-h2-g-to-form-nh3-g-92-2-kj-of-energy-are-evolved-for-each- | Chemistry
Topics
# When N2(g) reacts with H2(g) to form NH3(g) , 92.2 kJ of energy are evolved for each mole of N2(g) that reacts.?
## Write a balanced thermochemical equation for the reaction with an energy term in kJ as part of the equation. Note that the answer box for the energy term is case sensitive. Use the SMALLEST INTEGER coefficients possible and put the energy term (including the units) in the last box on the appropriate side of the equation. If a box is not needed, leave it blank. ++__------+____
Mar 7, 2018
$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right) + 46.1 \cdot k J$
#### Explanation:
And according to the terms of the question...
${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \rightarrow 2 N {H}_{3} \left(g\right) + 92.2 \cdot k J$
Enthalpy change is usually quoted per mole of reaction as written. Here it was specified that $92.2 \cdot k J$ is evolved per mole of dinitrogen...
For the lower reaction we could also write $\Delta {H}_{\text{rxn}}^{\circ} = - 92.2 \cdot k J \cdot m o {l}^{-} 1$...
##### Impact of this question
919 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9321096539497375, "perplexity": 1325.4639325963906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232258620.81/warc/CC-MAIN-20190526004917-20190526030917-00142.warc.gz"} |
https://www.newton.ac.uk/seminar/20150318113012301 | We consider a model of a population of fixed size $N$ in which each individual acquires beneficial mutations at rate $\mu$. Each individual dies at rate one, and when a death occurs, an individual is chosen with probability proportional to the individual's fitness to give birth. We obtain rigorous results for the rate at which mutations accumulate in the population, the distribution of the fitnesses of individuals in the population at a given time, and the genealogy of the population. Our results confirm predictions of Desai and Fisher (2007), Desai, Walczak, and Fisher (2013), and Neher and Hallatschek (2013). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9843673706054688, "perplexity": 476.711512176471}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00459.warc.gz"} |
https://www.physicsforums.com/threads/surface-area-surface-of-revolution-discrepancy.61887/ | # Surface Area (Surface of Revolution) - Discrepancy
1. Jan 30, 2005
### DoubleMike
So this question has been bothering me for a very long time... but only recently have I mustered up courage to register to ask it.
Before though, let me draw a parallel analogy, so you can see where I'm coming from. When you take the volume of a region enclosed by one or more functions revolved about a line, either the method of diks or shells is used.
Let us examine the volume of a sphere. The simplest way to calculate the volume of a sphere of radius R is to use the disk method with limits of integration -R to R, in respect to either variable (due to the symmetry).
So in essence the Riemann sum is the addition of the volume of infinitely many cylinders. These have parallel sides to the axis of the sphere/integration. Though strictly speaking we are adding very small volumes, I'd rather think of it as adding an infinite number of areas. This goes well with my understanding of infinities (as I've entreated in the General Math Forum). If this is flawed however, please feel free to elucidate.
Anyway.... When asked to take the surface area of a sphere, my book goes about it a very strange method indeed. It breaks the sphere up into an infinite number of partial cones, and uses the formula of their lateral surface area. I understand why this works... it makes sense. But why won't this work:
Breaking the sphere up into an infinite number of cylinders, and using the surface area formula for those (translational surface of a circle).
Here I have supplied a picture of what I mean:
http://www.geocities.com/boeclan/sphere.jpg
That picture goes for both the surface area and volume integrals. In the case of volume, that is the correct cross-section, whereas for surface area it is not.
Why is it that the function's slope is ignored when calculating volume, but for surface area, one must use the circuitous method described by my book?
*edit = typo
2. Jan 30, 2005
### HallsofIvy
So what you are saying is that you are dividing the sphere into very thin disks, of height dh and radius "r" where, since the sphere is given by x2+ y2+ z2= R2, r= $\sqrt{R^2- y^2}$. The lateral area of such a cylinder is given by $2\pir dy= 2\pi\sqrt{R^2-y^2}dy. Integrate [itex]2\pi\int_{-R}^R\sqrt{R^2-y^2}dy$ and see what you get.
(If you are right, it will be 4πr2.)
3. Jan 31, 2005
### DoubleMike
Is the integral $2\pi\int_{-R}^R\sqrt{R^2-x^2}dx$?
How would go integrate that without resorting to trigonometric substitution?
Well anyway, I checked the above integral for a random radius and it differs from the actual surface area!
4. Jan 31, 2005
### dextercioby
Write it (leave the 2pi for the moment):
$$\int_{-R}^{+R} \frac{R^{2}-x^{2}}{\sqrt{R^{2}-x^{2}}} dx$$
and then use the fact that
$$\int \frac{du}{\sqrt{1-u^{2}}}=\arcsin u +C$$
and partial integration to get the result...
Daniel.
P.S.Use the fact that the integrand is even and change the lower integration limit correspondingly.
5. Jan 31, 2005
### mathwonk
a smarter way to do the surface area of a sphere is by analogy with the method you used for the volume, i.e. by finding the derivative of the volume formula.
i.e. if one grows the volume of a sphere outwards it follows that the derivative of the volume function is the area of a cylindrical slice. Thus the volume itself is the integral of the area formula for this cylindrical slice, the so called "cylindrical shells" method.
On the other hand if one grows the volume of a sphere radially, it follows that the derivative of the volume function, wrt the radius is the area function of the sphere. hence the derivative of volume wrt radius is area for a sphere.
now this does not help compute the volume since we are not given the area, but since we already know the volume it does let us compute area, i.e. the area formula for a sphere, is the derivative of (4/3)pi r^3, wrt r, namely 4pi r^2.
you might like to read up on pappus theorem as well, for another analog of computing area via slices.
Last edited: Feb 1, 2005
6. Jan 31, 2005
### DoubleMike
I'm not entirely sure I know what you mean by even integrand. Also, I think trigonometric substitution is simpler in this case, but I keep messing it up:
$2\pi\int_{-R}^R\sqrt{R^2-x^2}dx$
I constructed a triangle with hypothenuse R and legs x and $\sqrt{R^2-x^2}$
so $\sin\theta = \frac{\sqrt{R^2-x^2}}{R}$ and $\cos\theta = \frac{x}{R}$
hence $-R\sin\theta d\theta = dx$ and $R\sin\theta = \sqrt{R^2-x^2}$
after the appropriate substitutions I get
$2\pi R^2 \int_{0}^\pi \sin^2 \theta d\theta$
using a power reducing formula
$\pi R^2 \int_{0}^\pi 1+\cos 2\theta$
$\pi R^2 [\theta + \frac{1}{2}\sin 2\theta]_{0}^\pi$
$\pi R^2 [\pi + 0]$
$\pi^2 R^2$
I'm not entirely sure of my math there, I didn't get the correct answer before, and working it out on itex was even worse... I might very well be wrong.
Regardless, integrating on my calculator, the method of disks differs largely from the actual formula derived using the slope and lateral surface of cones. What gives?
Last edited: Jan 31, 2005
7. Jan 31, 2005
### Justin Lazear
You're integrating just fine. Your cylinders model is incorrect, though.
Your model doesn't account for the fact that the length element ds of the cylinder must be longer, since it's at an angle. Compare the arclength traced out by the circle in element dx at x = 0 and x = near R. The first is just about straight, but the second is at a large angle, so it also goes down a good bit in the length dx, and therefore it should be longer.
--J
8. Feb 1, 2005
### DoubleMike
Yeah I realize that, but why do you have to account for the angle when finding the surface area, whereas for volume you can pretty much ignore it?
I don't understand that, besides... As the differential of x approaches 0, the significance of slope should be negligible.
9. Feb 1, 2005
### Justin Lazear
You do have to account for it. That's why when you change to cylindrical coordinates you get an $r d\theta$ term or an $r^2 \sin{\theta} dr d\theta$ term in spherical coordinates.
--J
10. Feb 1, 2005
### mathwonk
i am trying to actually answer the question of how to view the calculation of surface area as analogous to that of volume, but i do not seem to be making any impression on the questioner. oh well. cest la vie. people would rather calculate than think.
11. Feb 1, 2005
### Justin Lazear
Thinking is hard. Calculating is easy. Or at least that's what's taught in high school.
--J
12. Feb 2, 2005
### DoubleMike
I'm not sure whether to be insulted or not... Perhaps a better explanation is in order...
As it stands, I have to yet see a reason why I have to account for the curvature (besides the fact that the numbers don't match up). Yeah I understand that the curvature becomes more and more severe... But this never seemed to be a problem for calculating volume... And even if it is, why is it that $\pi \int_{-R}^R R^2-X^2 dx$ is a valid integral for volume? I don't see any reference to the slope of line tangent to the sphere's surface...
13. Feb 2, 2005
### mathwonk
Last edited: Feb 2, 2005
14. Feb 2, 2005
### DoubleMike
Ok, slow down.. Let's make up. I'm sorry I didn't read your post as carefully as I should have.
I understand what you're saying though, I had thought about it before (I fancied it suspicious that the derivative of volume happened to be the surface area.) I haven't done anything with spherical coordinates yet, but I follow the reasoning.
But still, that doesn't explain why breaking the sphere into small cylindrical segments and taking their lateral surface area doesn't work.
I know that the curvature is important, but there is nothing in the volume's Riemann sum that refers to it... So why should surface area be any different?
15. Feb 2, 2005
### Justin Lazear
The slope of the lines in the Cartesian coordinate system is always 1! dxdydz is always just a big box! As I said before, the case is not the same for coordinate systems where your dV element is not always just a big (edit: little) box, and in those coordinate systems, you do have to account for curvature.
--J
16. Feb 2, 2005
### mathwonk
what i am saying is there is no reason why what you suggested should work, since it is not really analogous to the method of calculating volume. in my post i give two methods which ARE analogous. i.e. if one works so should the other.
in your post, you give two rather different methods and ask why one works and the other does not. i ask you to think about why the area computation you suggest should work. i.e. what does it have in common with the shells method of computing volume?
and i apologize for getting my nose in a snit. i tried to edit but too slowly.
ill be back later but right now duke / wake is on.
Last edited: Feb 2, 2005
17. Feb 2, 2005
### DoubleMike
Ok, so perhaps a better question to ask it why the volume of sphere doesn't require that argument.
as it stands the Riemann sum is
$\pi\sum (R^2 - x^2) \delta x$ (not sure how to make the triangle on itex)
why not a sum of some function that takes the arguments radius and slope to calculate the volume of infinitely small cone sections?
$\pi\sum f(x, \frac{dy}{dx}) \delta x$
though it came to my mind that the function for the volume of a cone was derived using disks, so it too would be compromised.
Anyway, the point being that for a sphere, as you sum up the disks, shouldn't the fact that they have parallel sides skew the integral? (Because they don't follow the sphere's curvature.)
I don't know how to express my question!
Last edited: Feb 2, 2005
18. Feb 2, 2005
### mathwonk
what i mean is that calculating voluimes has nothing to do with approximations by cylinders or disks, it has to do with defining a growing volume function along some axis or other, and finding the derivative. if you consider a family of discs with enlarging radii, centered along the z axis, and let V(r) be the volume of your figure intersected with the solid cylinder of radius r, then the derivative is the area of the intersection of your figure intersected with the shell of radius r. this is called the method of cylindrical shells.
On the other hand if you consider a growing cylinder of fixed radius and growing height, intersected with your figure, then the derivative wrt height is the area of your figure intersected with the top circle of the cylinder of than height. then it is called the method of discs.
If you consider a famliy of solid spheres growing radially outward, intersected with your figure, then the derivative is the area of you figure with the surface area of the sphere of that radius. this method is not usually taught hence has no name.
But basically these methods for finding volumes have nothing to do with the shape of the family of figures that are growing, only with the relationship between their volume and the area of their leading face. I.e. these methods of calculating volumes are based on the fact that the derivative of a growing family of volumes is the surface area of the leading face of the family of figures. (provided the leading face is perpendicular to the direction of growth.)
In Pappus method for instance, we grow the volume in a circle by revolving a disk about the z axis, and consider the intersection of our figure with the solid partial torus generated by revolving the disc through a given angle. the derivative is the intersection of our figure with the area of the leading face, i.e. the moving circle.
Thus this method is much smarter for computing the volume of a torus generated by revolving a circle, since the derivative, the area of the moving circle is constant. thus the volume is that area times the circumference of the circle of centers of the moving circles.
so the whole point is to come up with a moving volume function, actually compute its derivative, usually an area, and then integrate back to get the volume.
the fact that the usual moving volume functions have derivative either the area of a cyliunder or a disc, is due to lack of imagination of textbook writers.
In particualr, it is not a consequence of that accident that one should do all calculations by approximating things with discs and cylinders.
rather one should try to compute ones desired quantity by some kind of appropriate moving function to the given problem, and then try to figure out what its derivative is.
so for example to compute the area of a torus, generate by revolving a circle, the derivative is delta area/ delta circumference of path of revolution, = circumference of the circle being revolved, again a constant. so the area of a torus equals the length of the circle being revoolved times the circumference of the path of revolution.
this is pappus method from thousands of years before calculus, explained by virtue of newton's ideas from calculus.
these things are not explained well in textbooks because most textbook writers just copy what other writers have said, without thinking about it first.
the methjod of computing areas using tangential cones is based on another entirely different idea they do not choose to explain. the key point is that since nbow the approximating regions no longer lie inside the given region, they mkust at elast be tangent to it.
tyhere is abetter way to view it absed on aparmetrizations, and change of area formulas under paramewtrizations, but enoiugh for now.
your question by the way is very intelligent and shows wonderful curiosity.
Last edited: Feb 2, 2005
19. Feb 2, 2005
### mathwonk
so ultimately the point is to understand why things are true. as presented in some books, the volume calculation is just, well these cylinders approximate that volume so in the limit we get the actual volume. oh yeah? why? whats the proof?
if you actually understand the proof of why the cylinder calculation does give the volume, then try to make that same proof go through for area, you should see that something crucial is then missing in the second case.
of coure it helps if you have a definition of volume and area.
but the method i am giving comes equipped with an argument. i.e. define a volume function V(r) or V(z) or V(something), and then compoute dV/dsomething as an area. then integrate back to get volume.
in the books they are skipping the explanation of why the area of that cylinder or of that disc, is really dV/dr or dV/dz, for some volujme function.
But if it were not, then you not get volume by integrating it.
so in the case of surface area, you are trying to compute surface area as a limit of areas of a different type.
what if you approximated your surface by discs that stuck straight out of your surface? would the areas of those discs have any relation to the area of your surface?
the better analogy for your problem, is why does the computation of arclength work?
i.e. the computation of arclength does work by approximating the curve by straight lines.
now if you look under the radical of the integrand for arclength you will see that what is being integrated is the length of approximating tangent line segments to your curve.
this is the analogy that does generalize to surface area. i.e. if you approximate your aurface by tangential pieces of surface, then their areas will approximate the surface area.
best wishes.
sorry if these explanations are inadequate, but they are the best i have at the moment.
20. Feb 2, 2005
### mathwonk
actually the arclength anal;ogy completely explains the surface area method, since if a piece of tangential segment is a good approximation to the curve then the cone obtained by revolving it is a good approximation to the surface area of revolution.
i.e. if the arclength is the integral of ds then the surface area is the integral of 2piRds. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9100978374481201, "perplexity": 503.5912390858479}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156192.24/warc/CC-MAIN-20180919102700-20180919122700-00116.warc.gz"} |
http://op.salesautopilot.com/9fbjj/dipole-moment-calculator-b67c07 | Therefore the CGS unit of dipole moment = … Methane; Benzene; Water dimer; Optimize; Atoms. 5 $\begingroup$ Question: The dipole moment of $\ce{HBr}$ is $2.60 \times 10^{-30}$ and the interatomic spacing is $1.41$. Hello, I would like to know how Avogadro calculates the dipole moment of molecule when I optimise the moelcule geometry by UFF (where no eletron distribution is taken into account). Use 75 or 50 ohm cable to feed the centre of the dipole. Dipole Moment Calculation Plugin calculates the total dipole moment of a molecule as a vector expressed in the principal axis frame. Dipole Moment Conversion. Magnetic Dipole moment- The magnetic field, B due to a current loop carrying a current i of radius, R at a distance l along its axis is given by: B = $$\frac {μ_0 i R^2}{2(R^2~+~l^2 )^{\frac32}}$$ Now if we consider a point very far from the current loop such that l>>R, then we can approximate the field as: Derivation of Magnetic Dipole Moment Formula. dipole moment calculation example: how to find the dipole moment of a molecule: how to calculate dipole moment of a molecule: dipole moment unit conversion: how to find zero dipole moment: how to calculate dipole moment chemistry: formula for finding magnetic moment: Molecular dipole moments can be obtained from microwave spectra. Electric Dipole Moment calculator uses Electric Dipole Moment=Charge*Radius Charge to calculate the Electric Dipole Moment, The electric dipole moment is a measure of the separation of positive and negative electrical charges within a system. You can also see the magnitude of the dipole moment increases in the second calculation. This is very common in excited states. Continue reading below if you want to learn more about the moment of force, its formula, or how to find the moment of force with examples? 9 4 A ˚, determine magnitude of the charge on the oxygen atom in the water molecule. In presence of an electric field, the rotational energy levels of a gaseous polar molecule are split (The Stark effect) according to the square of the permanent electric dipole moment. General Discussion. Being vectors, these can reinforce or cancel each other, depending on the geometry of the molecule; it is therefore not uncommon for molecules Enter the desired operating frequency in megahertz to get a good starting length for a dipole in both feet and meters. If the net charge of the species to which you are computing the dipole moment is zero, then the origin will not matter. Permanent Dipole Moment Curves Of The Lowest 1 + , 1 -and Molecular Modelling Analysis Of The Metabolism Of Caffeine Dipole-bound Anions And Relations Simulations Of The Physical Origin Of Far-field. In general the transition dipole moment is a complex vector quantity that includes the phase factors associated with the two states. Molecular polarity depends on both individual bond polarities and molecular geometry, the latter of which we can predict using VSEPR theory. Leticia_Pasetto April 4, 2019, 1:47pm #1. Dipole moment calculation. From the bond angle and vector moment μ = e × l. Thus, e (charge) can be determined in e s u. This organic chemistry video tutorial provides a basic introduction into dipole moment and molecular polarity. Therefore, 1Debye = 10-18 esu cm = 3.336 × 10-30 coulomb meter. Dipole moments as vector sums In molecules containing more than one polar bond, the molecular dipole moment is just the vector combination of what can be regarded as individual "bond dipole moments". Dipole Antenna Calculator. Unit of µ = unit of charge × unit of length. The SI unit of dipole moment is the coulomb meter, which is much too large for use with molecules, so dipole moments are measured in debye (symbol D) where 1 D = 3.335641×103-30 C m. Chlorobenzene has a dipole moment of magnitude 1.5 D due to the polarized C-Cl bond. The net dipole points through oxygen down the y-axis in the negative direction. It is a measure of the system's overall polarity. 3D; 2D; Starting structure. Dipole moment ( μ) is the measure of net molecular polarity, and describes the charge separation in a molecule, where electron density is shared unequally between atoms. The Wire Size can range from 16 AWG to 12 AWG. Dipole Moment. Place a 1:1 Balun on the Antenna end of the Feedline. The ORCA documentation also seems to state that *** CIS RELAXED DENSITY *** indicates this is the TD-DFT dipole moment. of Structural Biology Weizmann Institute, 761000 Rehovot, Israel With this server you can discover if your protein might have an unusually large net charge or dipole moment, and how this might relate to specific structural features of the protein, and thereby its function. Enter the desired frequency and select the desired calculation from the drop box. The electric dipole moment lies at the heart of a widely used experimental method for probing the vibrational dynamics of a system. Dipole Moment Calculation presents the overall dipole moment of a molecule as a vector expressed in the principal axis frame. This calculator is designed to give the horizontal length of a particular dipole (including Tees) antenna, or one side of it, for the frequency chosen. In a polar molecule, electron density is unevenly distributed throughout the molecule, resulting in regions of partial negative charge and regions of partial positive charge. Support. 1) In order to compute the dipole moment, you need to first choose an origin. Mind: the calculation of the dipole (IDIPOL=1-4) requires a definition of the center of the cell, and results might differ for different positions. 5 o and O - H bond length is 0. The molecular dipole moment can be represented as the sum of the individual atomic dipole moments and the pairwise atomic dipole contributions. Menu. Let's suppose we put water on the xy-plane like so: The dipole moment is calculated by looking up the dipole moment contributions from each "O"-"H" bond, which are polar, and summing them to get the net dipole. Download Image. Dipole moment of H 2 O is 1.85 D. If bond angle is 1. The half-wave dipole is very simple to construct. The dipole moment is defined much more broadly than just (point or finite) dipoles consisting of oppositely-charged pairs ─ there's plenty of other distributions with the same or similar properties , and the dipole moment is still an important quantity even if the shape differs from that. Hence the calculation of dipole moment order = 10-10 × 10-8 = 10-18 esu = 1 Debye. Protein Dipole Moments Server Clifford Felder and Joel Sussman, Dept. Electric dipole moments calculated from atom charges compared to dipoles from the wavefunction. The larger the wire, the wider the bandwidth. Dipoles antennas are easy to build and can be very effective when placed half a wavelenth or more above ground. The electric dipole moment for a pair of opposite charges of magnitude q is defined as the magnitude of the charge times the distance between them and the defined direction is toward the positive charge. Dipole moment (μ) is the measure of net molecular polarity, and describes the charge separation in a molecule, where electron density is shared unequally between atoms. Each contribution is "1.5 D" (debyes). Notice that the magnitude of the electric dipole moment is completely dependent on the displacement such that the dipole gains more polarity as the two charges move further and further from one another. Viewed 83k times 9. Physikalisch-chemisches Praktikum I Dipole Moment { 2016 Assuming that the vectors P~, E~and E~ 0 are parallel, we can insert4into8to obtain the scalar relation E= ˙ 0 P 0 def= ˙ 0 (9) The electric eld Ein a dielectric can thus be treated the same way as a eld in vacuum. Dimension of the Dipole Moment. $\endgroup$ – Emilio Pisanty Apr 12 '18 at 16:27 The electric dipole moment is a measure of the charge distribution in a molecule. The measurements below are for building a simple Dipole Antenna. One debye corresponds to a dipole moment occurring between two charges of one ten billionth franklin (10-10 Fr) separated by distance of one agstrem (1 Å). Unit of dipole moment derived from the centimeter-gram-second system (CGS). For covalent molecules, calculation of dipole moment requires knowledge of partial charge on all atoms and their locations. 1 Coulomb Meter = 2.997920E+029 debye = 1.000000E+000 C.m = 2.997920E+011 Eu Active 2 years, 3 months ago. Like bonds, molecules can also be polar. A physical dipole consists of two equal and opposite point charges An electric dipole is a separation of positive and negative charges.The direction of an electric field is defined as the direction of the force on a positive charge. The antenna is designed to be fed with 50 or 75 Ohm Coax Cable of most any length with a Balun. Ask Question Asked 6 years, 5 months ago. 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https://istopdeath.com/solve-for-x-log-base-4-of-8x/ | # Solve for x log base 4 of 8=x
Rewrite the equation as .
Logarithm base of is .
Rewrite as an equation.
Rewrite in exponential form using the definition of a logarithm. If and are positive real numbers and does not equal , then is equivalent to .
Create expressions in the equation that all have equal bases.
Rewrite as .
Since the bases are the same, then two expressions are only equal if the exponents are also equal.
Solve for .
The variable is equal to .
The result can be shown in multiple forms.
Exact Form:
Decimal Form:
Mixed Number Form:
Solve for x log base 4 of 8=x | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9454767107963562, "perplexity": 1011.2463474439924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00283.warc.gz"} |
http://mathhelpforum.com/geometry/217852-point-infinity-elliptic-curve.html | ## Point of infinity on an elliptic curve?
How to find the point of infinity on an elliptic curve y^2=X^3-5 ? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9480263590812683, "perplexity": 842.5947404525251}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663467.44/warc/CC-MAIN-20140930004103-00261-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://www.math.wsu.edu/kcooper/M300/essential/node11.html | Next: Quotations Up: Essential LATEX Previous: Document Styles and Style
# Environments
We mentioned earlier the idea of identifying a quotation to LATEX so that it could arrange to typeset it correctly. To do this you enclose the quotation between the commands \begin{quotation} and \end{quotation}. This is an example of a LATEX construction called an environment. A number of special effects are obtained by putting text into particular environments.
Subsections
Kevin Cooper 2002-03-01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9985822439193726, "perplexity": 3196.08992610698}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823442.17/warc/CC-MAIN-20181210191406-20181210212906-00003.warc.gz"} |
http://allcharters.info/finding-the-equation-of-a-line-in-standard-form | # Finding The Equation Of A Line In Standard Form
## Finding Slope From The Standard Form Of A Linear Equation
Writing The Equation Of A Line In Standard Form Youtube. Solved Write The Equation Of Each Line In Standard Form . Finding Slope From The Standard Form Of A Linear Equation . Straight Lines. How To Find The Equation Of A Tangent Line 8 Steps. How To Find The Equation Of A Tangent Line 8 Steps. Y Y1 Mx X1 Find The Equation Of A Line Using Point Slope Form . Finding Equations Of Lines Given A Parallel Or Perpendicular Line . Standard And Vertex Form Of The Equation Of Parabola And How It . Standard And Vertex Form Of The Equation Of Parabola And How It . How To Find The Equation Of A Tangent Line 8 Steps. Write A Slope Intercept Equation Given An X Y Table Youtube. How To Find A Slope Of A Straight Line With Ax By C 0 Math . Equation Of A Line Passing Through Two Points Worksheets. Finding The Slope And Y Intercept. How To Find The Vertex Of A Quadratic Equation 10 Steps. Finding The Equation Of A Line Given A Point And The Slope Youtube. Linear Equation Of A Line Worksheets. Equation Of A Circle In Standard Form Formula Practice Problems . | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8640519976615906, "perplexity": 217.75989426693255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00249.warc.gz"} |
https://terrytao.wordpress.com/2014/11/13/254a-announcement-analytic-prime-number-theory/ | In the winter quarter (starting January 5) I will be teaching a graduate topics course entitled “An introduction to analytic prime number theory“. As the name suggests, this is a course covering many of the analytic number theory techniques used to study the distribution of the prime numbers ${{\mathcal P} = \{2,3,5,7,11,\dots\}}$. I will list the topics I intend to cover in this course below the fold. As with my previous courses, I will place lecture notes online on my blog in advance of the physical lectures.
The type of results about primes that one aspires to prove here is well captured by Landau’s classical list of problems:
1. Even Goldbach conjecture: every even number ${N}$ greater than two is expressible as the sum of two primes.
2. Twin prime conjecture: there are infinitely many pairs ${n,n+2}$ which are simultaneously prime.
3. Legendre’s conjecture: for every natural number ${N}$, there is a prime between ${N^2}$ and ${(N+1)^2}$.
4. There are infinitely many primes of the form ${n^2+1}$.
All four of Landau’s problems remain open, but we have convincing heuristic evidence that they are all true, and in each of the four cases we have some highly non-trivial partial results, some of which will be covered in this course. We also now have some understanding of the barriers we are facing to fully resolving each of these problems, such as the parity problem; this will also be discussed in the course.
One of the main reasons that the prime numbers ${{\mathcal P}}$ are so difficult to deal with rigorously is that they have very little usable algebraic or geometric structure that we know how to exploit; for instance, we do not have any useful prime generating functions. One of course can create non-useful functions of this form, such as the ordered parameterisation ${n \mapsto p_n}$ that maps each natural number ${n}$ to the ${n^{th}}$ prime ${p_n}$, or one could invoke Matiyasevich’s theorem to produce a polynomial of many variables whose only positive values are prime, but these sorts of functions have no usable structure to exploit (for instance, they give no insight into any of the Landau problems listed above; see also Remark 2 below). The various primality tests in the literature, while useful for practical applications (e.g. cryptography) involving primes, have also proven to be of little utility for these sorts of problems; again, see Remark 2. In fact, in order to make plausible heuristic predictions about the primes, it is best to take almost the opposite point of view to the structured viewpoint, using as a starting point the belief that the primes exhibit strong pseudorandomness properties that are largely incompatible with the presence of rigid algebraic or geometric structure. We will discuss such heuristics later in this course.
It may be in the future that some usable structure to the primes (or related objects) will eventually be located (this is for instance one of the motivations in developing a rigorous theory of the “field with one element“, although this theory is far from being fully realised at present). For now, though, analytic and combinatorial methods have proven to be the most effective way forward, as they can often be used even in the near-complete absence of structure.
In this course, we will not discuss combinatorial approaches (such as the deployment of tools from additive combinatorics) in depth, but instead focus on the analytic methods. The basic principles of this approach can be summarised as follows:
1. Rather than try to isolate individual primes ${p}$ in ${{\mathcal P}}$, one works with the set of primes ${{\mathcal P}}$ in aggregate, focusing in particular on asymptotic statistics of this set. For instance, rather than try to find a single pair ${n,n+2}$ of twin primes, one can focus instead on the count ${|\{ n \leq x: n,n+2 \in {\mathcal P} \}|}$ of twin primes up to some threshold ${x}$. Similarly, one can focus on counts such as ${|\{ n \leq N: n, N-n \in {\mathcal P} \}|}$, ${|\{ p \in {\mathcal P}: N^2 < p < (N+1)^2 \}|}$, or ${|\{ n \leq x: n^2 + 1 \in {\mathcal P} \}|}$, which are the natural counts associated to the other three Landau problems. In all four of Landau’s problems, the basic task is now to obtain a non-trivial lower bounds on these counts.
2. If one wishes to proceed analytically rather than combinatorially, one should convert all these counts into sums, using the fundamental identity
$\displaystyle |A| = \sum_n 1_A(n),$
(or variants thereof) for the cardinality ${|A|}$ of subsets ${A}$ of the natural numbers ${{\bf N}}$, where ${1_A}$ is the indicator function of ${A}$ (and ${n}$ ranges over ${{\bf N}}$). Thus we are now interested in estimating (and particularly in lower bounding) sums such as
$\displaystyle \sum_{n \leq N} 1_{{\mathcal P}}(n) 1_{{\mathcal P}}(N-n),$
$\displaystyle \sum_{n \leq x} 1_{{\mathcal P}}(n) 1_{{\mathcal P}}(n+2),$
$\displaystyle \sum_{N^2 < n < (N+1)^2} 1_{{\mathcal P}}(n),$
or
$\displaystyle \sum_{n \leq x} 1_{{\mathcal P}}(n^2+1).$
3. Once one expresses number-theoretic problems in this fashion, we are naturally led to the more general question of how to accurately estimate (or, less ambitiously, to lower bound or upper bound) sums such as
$\displaystyle \sum_n f(n)$
or more generally bilinear or multilinear sums such as
$\displaystyle \sum_n \sum_m f(n,m)$
or
$\displaystyle \sum_{n_1,\dots,n_k} f(n_1,\dots,n_k)$
for various functions ${f}$ of arithmetic interest. (Importantly, one should also generalise to include integrals as well as sums, particularly contour integrals or integrals over the unit circle or real line, but we postpone discussion of these generalisations to later in the course.) Indeed, a huge portion of modern analytic number theory is devoted to precisely this sort of question. In many cases, we can predict an expected main term for such sums, and then the task is to control the error term between the true sum and its expected main term. It is often convenient to normalise the expected main term to be zero or negligible (e.g. by subtracting a suitable constant from ${f}$), so that one is now trying to show that a sum of signed real numbers (or perhaps complex numbers) is small. In other words, the question becomes one of rigorously establishing a significant amount of cancellation in one’s sums (also referred to as a gain or savings over a benchmark “trivial bound”). Or to phrase it negatively, the task is to rigorously prevent a conspiracy of non-cancellation, caused for instance by two factors in the summand ${f(n)}$ exhibiting an unexpectedly large correlation with each other.
4. It is often difficult to discern cancellation (or to prevent conspiracy) directly for a given sum (such as ${\sum_n f(n)}$) of interest. However, analytic number theory has developed a large number of techniques to relate one sum to another, and then the strategy is to keep transforming the sum into more and more analytically tractable expressions, until one arrives at a sum for which cancellation can be directly exhibited. (Note though that there is often a short-term tradeoff between analytic tractability and algebraic simplicity; in a typical analytic number theory argument, the sums will get expanded and decomposed into many quite messy-looking sub-sums, until at some point one applies some crude estimation to replace these messy sub-sums by tractable ones again.) There are many transformations available, ranging such basic tools as the triangle inequality, pointwise domination, or the Cauchy-Schwarz inequality to key identities such as multiplicative number theory identities (such as the Vaughan identity and the Heath-Brown identity), Fourier-analytic identities (e.g. Fourier inversion, Poisson summation, or more advanced trace formulae), or complex analytic identities (e.g. the residue theorem, Perron’s formula, or Jensen’s formula). The sheer range of transformations available can be intimidating at first; there is no shortage of transformations and identities in this subject, and if one applies them randomly then one will typically just transform a difficult sum into an even more difficult and intractable expression. However, one can make progress if one is guided by the strategy of isolating and enhancing a desired cancellation (or conspiracy) to the point where it can be easily established (or dispelled), or alternatively to reach the point where no deep cancellation is needed for the application at hand (or equivalently, that no deep conspiracy can disrupt the application).
5. One particularly powerful technique (albeit one which, ironically, can be highly “ineffective” in a certain technical sense to be discussed later) is to use one potential conspiracy to defeat another, a technique I refer to as the “dueling conspiracies” method. This technique may be unable to prevent a single strong conspiracy, but it can sometimes be used to prevent two or more such conspiracies from occurring, which is particularly useful if conspiracies come in pairs (e.g. through complex conjugation symmetry, or a functional equation). A related (but more “effective”) strategy is to try to “disperse” a single conspiracy into several distinct conspiracies, which can then be used to defeat each other.
As stated before, the above strategy has not been able to establish any of the four Landau problems as stated. However, they can come close to such problems (and we now have some understanding as to why these problems remain out of reach of current methods). For instance, by using these techniques (and a lot of additional effort) one can obtain the following sample partial results in the Landau problems:
1. Chen’s theorem: every sufficiently large even number ${N}$ is expressible as the sum of a prime and an almost prime (the product of at most two primes). The proof proceeds by finding a nontrivial lower bound on ${\sum_{n \leq N} 1_{\mathcal P}(n) 1_{{\mathcal E}_2}(N-n)}$, where ${{\mathcal E}_2}$ is the set of almost primes.
2. Zhang’s theorem: There exist infinitely many pairs ${p_n, p_{n+1}}$ of consecutive primes with ${p_{n+1} - p_n \leq 7 \times 10^7}$. The proof proceeds by giving a non-negative lower bound on the quantity ${\sum_{x \leq n \leq 2x} (\sum_{i=1}^k 1_{\mathcal P}(n+h_i) - 1)}$ for large ${x}$ and certain distinct integers ${h_1,\dots,h_k}$ between ${0}$ and ${7 \times 10^7}$. (The bound ${7 \times 10^7}$ has since been lowered to ${246}$.)
3. The Baker-Harman-Pintz theorem: for sufficiently large ${x}$, there is a prime between ${x}$ and ${x + x^{0.525}}$. Proven by finding a nontrivial lower bound on ${\sum_{x \leq n \leq x+x^{0.525}} 1_{\mathcal P}(n)}$.
4. The Friedlander-Iwaniec theorem: There are infinitely many primes of the form ${n^2+m^4}$. Proven by finding a nontrivial lower bound on ${\sum_{n,m: n^2+m^4 \leq x} 1_{{\mathcal P}}(n^2+m^4)}$.
We will discuss (simpler versions of) several of these results in this course.
Of course, for the above general strategy to have any chance of succeeding, one must at some point use some information about the set ${{\mathcal P}}$ of primes. As stated previously, usefully structured parametric descriptions of ${{\mathcal P}}$ do not appear to be available. However, we do have two other fundamental and useful ways to describe ${{\mathcal P}}$:
1. (Sieve theory description) The primes ${{\mathcal P}}$ consist of those numbers greater than one, that are not divisible by any smaller prime.
2. (Multiplicative number theory description) The primes ${{\mathcal P}}$ are the multiplicative generators of the natural numbers ${{\bf N}}$: every natural number is uniquely factorisable (up to permutation) into the product of primes (the fundamental theorem of arithmetic).
The sieve-theoretic description and its variants lead one to a good understanding of the almost primes, which turn out to be excellent tools for controlling the primes themselves, although there are known limitations as to how much information on the primes one can extract from sieve-theoretic methods alone, which we will discuss later in this course. The multiplicative number theory methods lead one (after some complex or Fourier analysis) to the Riemann zeta function (and other L-functions, particularly the Dirichlet L-functions), with the distribution of zeroes (and poles) of these functions playing a particularly decisive role in the multiplicative methods.
Many of our strongest results in analytic prime number theory are ultimately obtained by incorporating some combination of the above two fundamental descriptions of ${{\mathcal P}}$ (or variants thereof) into the general strategy described above. In contrast, more advanced descriptions of ${{\mathcal P}}$, such as those coming from the various primality tests available, have (until now, at least) been surprisingly ineffective in practice for attacking problems such as Landau’s problems. One reason for this is that such tests generally involve operations such as exponentiation ${a \mapsto a^n}$ or the factorial function ${n \mapsto n!}$, which grow too quickly to be amenable to the analytic techniques discussed above.
To give a simple illustration of these two basic approaches to the primes, let us first give two variants of the usual proof of Euclid’s theorem:
Theorem 1 (Euclid’s theorem) There are infinitely many primes.
Proof: (Multiplicative number theory proof) Suppose for contradiction that there were only finitely many primes ${p_1,\dots,p_n}$. Then, by the fundamental theorem of arithmetic, every natural number is expressible as the product of the primes ${p_1,\dots,p_n}$. But the natural number ${p_1 \dots p_n + 1}$ is larger than one, but not divisible by any of the primes ${p_1,\dots,p_n}$, a contradiction.
(Sieve-theoretic proof) Suppose for contradiction that there were only finitely many primes ${p_1,\dots,p_n}$. Then, by the Chinese remainder theorem, the set of natural numbers ${A}$ that is not divisible by any of the ${p_1,\dots,p_n}$ has density ${\prod_{i=1}^n (1-\frac{1}{p_i})}$, that is to say
$\displaystyle \lim_{N \rightarrow \infty} \frac{1}{N} | A \cap \{1,\dots,N\} | = \prod_{i=1}^n (1-\frac{1}{p_i}).$
In particular, ${A}$ has positive density and thus contains an element larger than ${1}$. But the least such element is one further prime in addition to ${p_1,\dots,p_n}$, a contradiction. $\Box$
Remark 1 One can also phrase the proof of Euclid’s theorem in a fashion that largely avoids the use of contradiction; see this previous blog post for more discussion.
Both proofs in fact extend to give a stronger result:
Theorem 2 (Euler’s theorem) The sum ${\sum_{p \in {\mathcal P}} \frac{1}{p}}$ is divergent.
Proof: (Multiplicative number theory proof) By the fundamental theorem of arithmetic, every natural number is expressible uniquely as the product ${p_1^{a_1} \dots p_n^{a_n}}$ of primes in increasing order. In particular, we have the identity
$\displaystyle \sum_{n=1}^\infty \frac{1}{n} = \prod_{p \in {\mathcal P}} ( 1 + \frac{1}{p} + \frac{1}{p^2} + \dots )$
(both sides make sense in ${[0,+\infty]}$ as everything is unsigned). Since the left-hand side is divergent, the right-hand side is as well. But
$\displaystyle ( 1 + \frac{1}{p} + \frac{1}{p^2} + \dots ) = \exp( \frac{1}{p} + O( \frac{1}{p^2} ) )$
and ${\sum_{p \in {\mathcal P}} \frac{1}{p^2}\leq \sum_{n=1}^\infty \frac{1}{n^2} < \infty}$, so ${\sum_{p \in {\mathcal P}} \frac{1}{p}}$ must be divergent.
(Sieve-theoretic proof) Suppose for contradiction that the sum ${\sum_{p \in {\mathcal P}} \frac{1}{p}}$ is convergent. For each natural number ${k}$, let ${A_k}$ be the set of natural numbers not divisible by the first ${k}$ primes ${p_1,\dots,p_k}$, and let ${A}$ be the set of numbers not divisible by any prime in ${{\mathcal P}}$. As in the previous proof, each ${A_k}$ has density ${\prod_{i=1}^k (1-\frac{1}{p_i})}$. Also, since ${\{1,\dots,N\}}$ contains at most ${\frac{N}{p}}$ multiples of ${p}$, we have from the union bound that
$\displaystyle | A \cap \{1,\dots,N \}| = |A_k \cap \{1,\dots,N\}| - O( N \sum_{i > k} \frac{1}{p_i} ).$
Since ${\sum_{i=1}^\infty \frac{1}{p_i}}$ is assumed to be convergent, we conclude that the density of ${A_k}$ converges to the density of ${A}$; thus ${A}$ has density ${\prod_{i=1}^\infty (1-\frac{1}{p_i})}$, which is non-zero by the hypothesis that ${\sum_{i=1}^\infty \frac{1}{p_i}}$ converges. On the other hand, since the primes are the only numbers greater than one not divisible by smaller primes, ${A}$ is just ${\{1\}}$, which has density zero, giving the desired contradiction. $\Box$
Remark 2 We have seen how easy it is to prove Euler’s theorem by analytic methods. In contrast, there does not seem to be any known proof of this theorem that proceeds by using any sort of prime-generating formula or a primality test, which is further evidence that such tools are not the most effective way to make progress on problems such as Landau’s problems. (But the weaker theorem of Euclid, Theorem 1, can sometimes be proven by such devices.)
The two proofs of Theorem 2 given above are essentially the same proof, as is hinted at by the geometric series identity
$\displaystyle 1 + \frac{1}{p} + \frac{1}{p^2} + \dots = (1 - \frac{1}{p})^{-1}.$
One can also see the Riemann zeta function begin to make an appearance in both proofs. Once one goes beyond Euler’s theorem, though, the sieve-theoretic and multiplicative methods begin to diverge significantly. On one hand, sieve theory can still handle to some extent sets such as twin primes, despite the lack of multiplicative structure (one simply has to sieve out two residue classes per prime, rather than one); on the other, multiplicative number theory can attain results such as the prime number theorem for which purely sieve theoretic techniques have not been able to establish. The deepest results in analytic number theory will typically require a combination of both sieve-theoretic methods and multiplicative methods in conjunction with the many transforms discussed earlier (and, in many cases, additional inputs from other fields of mathematics such as arithmetic geometry, ergodic theory, or additive combinatorics).
— 1. Topics covered —
Analytic prime number theory is a vast subject (the 615-page text of Iwaniec and Kowalski, for instance, gives a good indication as to its scope). I will therefore have to be somewhat selective in deciding what subset of this field to cover. I have chosen the following “core” topics to focus on:
• Elementary multiplicative number theory.
• Heuristic random models for the primes.
• The basic theory of the Riemann zeta function and Dirichlet L-functions, and their relationship with the primes.
• Zero-free regions for the zeta function and the Dirichet L-function, including Siegel’s theorem.
• The prime number theorem, the Siegel-Walfisz theorem, and the Bombieri-Vinogradov theorem.
• Sieve theory, small and large gaps between the primes, and the parity problem.
• Exponential sum estimates over the integers, and the Vinogradov-Korobov zero-free region.
• Zero density estimates, Hohiesel’s theorem, and Linnik’s theorem.
• Exponential sum estimates over finite fields, and improved distribution estimates for the primes.
• (If time permits) Exponential sum estimates over the primes, the circle method, and Vinogradov’s three-primes theorem.
In order to cover all this material, I will focus on more qualitative results, as opposed to the strongest quantitative results, in particular I will not attempt to optimise many of the numerical constants and exponents appearing in various estimates. This also allows me to downplay the role of some key components of the field which are not essential for establishing the core results of this course at such a qualitative level:
• I will minimise the use of algebraic number theory tools (such as the class number formula).
• I will avoid deploying the functional equation (or related identities, such as Poisson summation) if they are unnecessary at a qualitative level (though I will note when the functional equation can be used to improve the quantitative results). As it turns out, all of the core results mentioned above can in fact be derived without ever invoking the functional equation, although one usually gets poorer numerical exponents as a consequence.
• Somewhat related to this, I will reduce the reliance on complex analytic methods as compared to more traditional presentations of the material, relying in some places instead on Fourier-analytic substitutes, or on results about harmonic functions. (But I will not go as far as deploying the primarily real-variable “pretentious” approach to analytic number theory currently in development by Granville and Soundararajan, although my approach here does align in spirit with that approach.)
• The discussion on sieve methods will be somewhat abridged, focusing primarily on the Selberg sieve, which is a good general-purpose sieve for qualitative applications at least.
• I will almost certainly avoid any discussion of automorphic forms methods.
• Similarly, I will not cover methods that rely on additive combinatorics or ergodic theory.
Of course, many of these additional topics are well covered in existing textbooks, such as the above-mentioned text of Iwaniec and Kowalski (or, for the finer points of sieve theory, the text of Friedlander and Iwaniec). Other good texts that can be used for supplementary reading are Davenport’s “Multiplicative number theory” and Montgomery-Vaughan’s “Multiplicative number theory I.”. As for prerequisites: some exposure to complex analysis, Fourier analysis, and real analysis will be particularly helpful, although we will review some of this material as needed (particularly with regard to complex analysis and the theory of harmonic functions). Experience with other quantitative areas of mathematics in which lower bounds, upper bounds, and other forms of estimation are emphasised (e.g. asymptotic combinatorics or theoretical computer science) will also be useful. Knowledge of algebraic number theory or arithmetic geometry will add a valuable additional perspective to the course, but will not be necessary to follow most of the material.
— 2. Notation —
In this course, all sums will be understood to be over the natural numbers unless otherwise specified, with the exception of sums over the variable ${p}$ (or variants such as ${p_1}$, ${p_2}$, etc.), which will be understood to be over primes.
We will use asymptotic notation in two contexts, one in which there is no asymptotic parameter present, and one in which there is an asymptotic parameter (such as ${x}$) that is going to infinity. In the non-asymptotic setting (which is the default context if no asymptotic parameter is explicitly specified), we use ${X = O(Y)}$, ${X \ll Y}$, or ${Y \gg X}$ to denote an estimate of the form ${|X| \leq CY}$, where ${C}$ is an absolute constant. In some cases we would like the implied constant ${C}$ to depend on some additional parameters such as ${k}$, in which case we will denote this by subscripts, for instance ${X = O_k(Y)}$ denotes the claim that ${|X| \leq C_k Y}$ for some ${C_k}$ depending on ${k}$.
In some cases it will instead be convenient to work in an asymptotic setting, in which there is an explicitly designated asymptotic parameter (such as ${x}$) going to infinity. In that case, all mathematical objects will be permitted to depend on this asymptotic parameter, unless they are explicitly referred to as being fixed. We then use ${X = O(Y)}$, ${X \ll Y}$, or ${Y \gg X}$ to denote the claim that ${|X| \leq CY}$ for some fixed ${C}$. Note that in slight contrast to the non-asymptotic setting, the implied constant ${C}$ here is allowed to depend on other parameters, so long as these parameters are also fixed. As such, the asymptotic setting can be a convenient way to manage dependencies of various implied constants on parameters. In the asymptotic setting we also use ${X = o(Y)}$ to denote the claim that ${|X| \leq c Y}$, where ${c}$ is a quantity which goes to zero as the asymptotic parameter goes to infinity.
Remark 3 In later posts we will make a distinction between implied constants ${C}$ that are effective (they can be computed, at least in principle, by some explicit method) and those at are ineffective (they can be proven to be finite, but there is no algorithm known to compute them in finite time).
We use ${d|n}$ to denote the assertion that ${d}$ divides ${n}$, and ${a\ (q)}$ to denote the residue class of ${a}$ modulo ${q}$.
We use ${1_E}$ to denote the indicator function of a set ${E}$, thus ${1_E(x) = 1}$ when ${x \in E}$ and ${1_E(x) = 0}$ otherwise. Similarly, for any mathematical statement ${S}$, we use ${1_S}$ to denote the value ${1}$ when ${S}$ is true and ${0}$ when ${S}$ is false. Thus for instance ${1_{2|n} = 1_{n\ (2) = 0}}$ is the indicator function of the even numbers.
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https://www.jobilize.com/course/section/verbal-inverse-trigonometric-functions-by-openstax?qcr=www.quizover.com | # 5.6 Inverse trigonometric functions (Page 6/15)
Page 6 / 15
Find a simplified expression for $\text{\hspace{0.17em}}\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(4x\right)\right)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}-\frac{1}{4}\le x\le \frac{1}{4}.$
$\frac{4x}{\sqrt{16{x}^{2}+1}}$
Access this online resource for additional instruction and practice with inverse trigonometric functions.
Visit this website for additional practice questions from Learningpod.
## Key concepts
• An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.
• Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.
• For any trigonometric function $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x={f}^{-1}\left(y\right),\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}f\left(x\right)=y.\text{\hspace{0.17em}}$ However, $\text{\hspace{0.17em}}f\left(x\right)=y\text{\hspace{0.17em}}$ only implies $\text{\hspace{0.17em}}x={f}^{-1}\left(y\right)\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in the restricted domain of $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ See [link] .
• Special angles are the outputs of inverse trigonometric functions for special input values; for example, $\text{\hspace{0.17em}}\frac{\pi }{4}={\mathrm{tan}}^{-1}\left(1\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{\pi }{6}={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right).$ See [link] .
• A calculator will return an angle within the restricted domain of the original trigonometric function. See [link] .
• Inverse functions allow us to find an angle when given two sides of a right triangle. See [link] .
• In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, $\text{\hspace{0.17em}}\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(x\right)\right)=\sqrt{1-{x}^{2}}.\text{\hspace{0.17em}}$ See [link] .
• If the inside function is a trigonometric function, then the only possible combinations are $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-x\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}0\le x\le \pi \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-x\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}-\frac{\pi }{2}\le x\le \frac{\pi }{2}.$ See [link] and [link] .
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function. See [link] .
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides. See [link] .
## Verbal
Why do the functions $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{sin}}^{-1}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{cos}}^{-1}x\text{\hspace{0.17em}}$ have different ranges?
The function $\text{\hspace{0.17em}}y=\mathrm{sin}x\text{\hspace{0.17em}}$ is one-to-one on $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right];\text{\hspace{0.17em}}$ thus, this interval is the range of the inverse function of $\text{\hspace{0.17em}}y=\mathrm{sin}x,$ $f\left(x\right)={\mathrm{sin}}^{-1}x.\text{\hspace{0.17em}}$ The function $\text{\hspace{0.17em}}y=\mathrm{cos}x\text{\hspace{0.17em}}$ is one-to-one on $\text{\hspace{0.17em}}\left[0,\pi \right];\text{\hspace{0.17em}}$ thus, this interval is the range of the inverse function of $\text{\hspace{0.17em}}y=\mathrm{cos}x,f\left(x\right)={\mathrm{cos}}^{-1}x.\text{\hspace{0.17em}}$
Since the functions $\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y={\mathrm{cos}}^{-1}x\text{\hspace{0.17em}}$ are inverse functions, why is $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{6}\right)\right)\text{\hspace{0.17em}}$ not equal to $\text{\hspace{0.17em}}-\frac{\pi }{6}?$
Explain the meaning of $\text{\hspace{0.17em}}\frac{\pi }{6}=\mathrm{arcsin}\left(0.5\right).$
$\frac{\pi }{6}\text{\hspace{0.17em}}$ is the radian measure of an angle between $\text{\hspace{0.17em}}-\frac{\pi }{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\frac{\pi }{2}$ whose sine is 0.5.
Most calculators do not have a key to evaluate $\text{\hspace{0.17em}}{\mathrm{sec}}^{-1}\left(2\right).\text{\hspace{0.17em}}$ Explain how this can be done using the cosine function or the inverse cosine function.
Why must the domain of the sine function, $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ be restricted to $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\text{\hspace{0.17em}}$ for the inverse sine function to exist?
In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\text{\hspace{0.17em}}$ so that it is one-to-one and possesses an inverse.
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
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https://www.jiskha.com/questions/405128/suppose-that-the-proportions-of-blood-phenotypes-in-a-particular-population-are-a-follows | # statistics
Suppose that the proportions of blood phenotypes in a particular population are a follows:
A = .42
B = .10
AB = .04
O = .44
Assuming that the phenotypes of two randomly selected indiviuals are independent of one another, what is the probablility that both phenotypes are O? What is the probablility that the phenotypes of two randomly selected individuals match?
I'm really stuck on how to approach and solve this problem. From what I see, it says they are independent so would we need to use a formula of the kind
P(O Phenotype) = P(A) * P(B)
where A and B are the events of two randomly selected individuals?
1. 👍
2. 👎
3. 👁
1. Right!
1. 👍
2. 👎
2. now i don't know what to do frrom there
1. 👍
2. 👎
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4. ### Stats
Suppose a random sample of size 58 is selected from a population with σ = 9. Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate). A) The | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8271366953849792, "perplexity": 1060.9653150636173}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00312.warc.gz"} |
https://www.physicsforums.com/threads/how-do-we-measure-particle-momentum.227477/ | # How do we measure particle momentum?
1. Apr 8, 2008
### pellman
How do we measure a quantum particle's momentum?
2. Apr 8, 2008
### lbrits
The field of spectroscopy is devoted to that issue.
3. Apr 8, 2008
### pam
Momentum p is most commonly measured by curvature in a magnetic field B.
R=p/qB.
4. Apr 8, 2008
### ZapperZ
Staff Emeritus
There are no unique methods to do this. In angle-resolved photoemission spectroscopy (ARPES), the detector has a finite rectangular slit in which one dimension is the energy, while the other is the momentum of the emitted photoelectrons. So you would get a raw image of something like my avatar. For certain types of material (such as 2D, layered material), the transverse momentum of the photoelectrons corresponds to the transverse momentum of such electrons while it is in the material.
Believe it or not, the diffraction pattern that you see on the screen corresponds to the transverse momentum of whatever particle that passed through the single slit.
Zz.
5. Apr 8, 2008
### pellman
So do all the methods amount to making a position measurement from which we infer the momentum? How does that square with the uncertainty principle?
6. May 12, 2010
### sweet springs
Hi.
Measurement of momentum is always achieved by measurement of position, so I assume that observation of momentum is not how it IS but how it WAS . After measurement of momentum value p, the state cannot keep on |p>. It is a kind of destructive observation.
Regards.
7. May 12, 2010
### SpectraCat
This is a very important point that IMO is *way* under-emphasized in discussions of QM. We talk rather blithely about measurement of observables in an experimental sense, however the reality is that every measurement we can actually *do* involves either an explicit or implicit measurement of position to one extent or another. The closest thing to a counter-example that I can think of right away might be a direct absorption spectroscopy experiment on a gas confined to a large-volume sample cell. In that case the "uncertainty" in the position of the gas molecules giving rise to the observed spectral lines is fairly large, however you can still say with certainty that they must have been somewhere inside the cell at the time they absorbed a photon(s). So even there there is an implicit measurement of position involved.
At least in my own academic arc, which included a significant amount of formal QM training, this aspect was "pushed under the rug" to the extent that I am not even sure how to address it! I have thought of it on my own several times, and then pushed it back under the rug with a rationalization such as, "well, it must not really be important, or it would be addressed explicitly in QM texts". However, in light of many of the discussions I have participated in here, I think this deserves a closer look. For example, if we look at Zz's answer above, you see that the momentum is inferred from a position measurement of displacement along a given spatial axis. Thus the momentum observable was never really measured at all! Yet we consistently talk as though it had been directly measured. Couldn't this lead to fundamental mis-interpretations of such measurements?
8. May 16, 2010
### DaTario
I would say that neither do position observable. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.934503436088562, "perplexity": 726.2673720025296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170864.16/warc/CC-MAIN-20170219104610-00346-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://wias-berlin.de/publications/wias-publ/run.jsp?lang=0&template=abstract&type=Preprint&year=2002&number=758 | WIAS Preprint No. 758, (2002)
Eigen mode computation of microwave and laser structures including PML
Authors
• Hebermehl, Georg
• Hübner, Friedrich-Karl
• Schlundt, Rainer
ORCID: 0000-0002-4424-4301
• Tischler, Thorsten
• Zscheile, Horst
• Heinrich, Wolfgang
2010 Mathematics Subject Classification
• 35Q60 65F15 65N22
Keywords
• Microwave device, Optoelectronic device, Simulation, Maxwell's equations, PML boundary condition, Eigenvalue problem
DOI
10.20347/WIAS.PREPRINT.758
Abstract
The field distribution at the ports of the transmission line structure is computed by applying Maxwell's equations to the structure. Assuming longitudinal homogeneity an eigenvalue problem can be derived, whose solutions correspond to the propagation constants of the modes. The nonsymmetric sparse system matrix is complex in the presence of losses and Perfectly Matched Layer. The propagation constants are found solving a sequence of eigenvalue problems of modified matrices with the aid of the invert mode of the Arnoldi method. Using coarse and fine grids, and a new parallel sparse linear solver, the method, first developed for microwave structures, can be applied also to high dimensional problems of optoelectronics.
Appeared in
• Scientific Computing in Electrical Engineering, Eds. W. H. A. Schilders, E. J. W. ter Maten, St. H. M. J. Houben, Mathematics in Industry, Springer Verlag, Vol. 4, pp. 196--205, 2004 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8611308932304382, "perplexity": 4223.32467070074}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655887377.70/warc/CC-MAIN-20200705152852-20200705182852-00355.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/105621-finding-spanning-set-null-space-print.html | # Finding a spanning set of a null space
• Oct 2nd 2009, 02:38 AM
Mathmaticious
Finding a spanning set of a null space
Let U be the subspace of R4 given by:
U = nullspace of the matrix
[0 0 2 3 ]
[0 -3 -2 -2 ]
Find a spanning set for U.
Any and all help will be awesome.
Not sure how to start at all (Worried)
Cheers
• Oct 2nd 2009, 05:41 AM
Swlabr
Quote:
Originally Posted by Mathmaticious
Let U be the subspace of R4 given by:
U = nullspace of the matrix
[0 0 2 3 ]
[0 -3 -2 -2 ]
Find a spanning set for U.
Any and all help will be awesome.
Not sure how to start at all (Worried)
Cheers
The first thing to do is to find out what the nullspace of your matrix actually is - can you find a general form that these vectors take?
Can you think of a spanning set for this vector space? For instance, if they were of the form $\{(2a,5b, 6c,d): a, b, c, d \in \mathbb{F} \}$ then your space would be spanned by the vectors $(2,0,0,0)$, $(0,5,0,0)$, $(0,0,6,0)$ and $(0,0,0,1)$.
I hope that that helps...
• Oct 2nd 2009, 07:47 AM
aman_cc
Quote:
Originally Posted by Swlabr
The first thing to do is to find out what the nullspace of your matrix actually is - can you find a general form that these vectors take?
Can you think of a spanning set for this vector space? For instance, if they were of the form $\{(2a,5b, 6c,d): a, b, c, d \in \mathbb{F} \}$ then your space would be spanned by the vectors $(2,0,0,0)$, $(0,5,0,0)$, $(0,0,6,0)$ and $(0,0,0,d)$.
I hope that that helps...
1. In your example $\{(2a,5b, 6c,d): a, b, c, d \in \mathbb{F} \}$ what is the relevance of 2,5,6? Isn't it just same as $\{a,b,c,d): a, b, c, d \in \mathbb{F} \}$
2. Nevertheless, is there a good/structured approach to find dimension and basis of a space given by something like $\{a,2a,a+b,b): a, b \in \mathbb{F} \}$
• Oct 2nd 2009, 08:17 AM
Swlabr
Quote:
Originally Posted by aman_cc
1. In your example $\{(2a,5b, 6c,d): a, b, c, d \in \mathbb{F} \}$ what is the relevance of 2,5,6? Isn't it just same as $\{a,b,c,d): a, b, c, d \in \mathbb{F} \}$
Yes. Yes it is. No excuses, other than to point that I was still correct...I mean, 4+2=80712/13452...
Quote:
2. Nevertheless, is there a good/structured approach to find dimension and basis of a space given by something like $\{a,2a,a+b,b): a, b \in \mathbb{F} \}$
Yes. The answer to your problem is either one or two, but is clearly two as the entries in the 1st and 4th positions are independent of one another. It is no more than the number of variables given, but can be less if they cancel out with one another.
$(1, 2, 1, 0)$ and $(0, 0, 1, 1)$ span the space you gave.
• Oct 2nd 2009, 08:24 AM
aman_cc
Quote:
Originally Posted by Swlabr
Yes. Yes it is.
Yes. The answer to your problem is either one or two, but is clearly two as the entries in the 1st and 4th positions are independent of one another. It is no more than the number of variables given, but can be less if they cancel out with one another.
$(a, 2a, a, 0)$ and $(0, 0, b, b)$ span the space you gave.
Thanks Swalbr
If you won't mind can you explain why you say - "It is no more than the number of variables given"?
Also is there is a formal way to solve such a problem - I am looking at something equivalent to row-reduction to find the rank of matrix. Thanks
• Oct 2nd 2009, 08:32 AM
Swlabr
Quote:
Originally Posted by aman_cc
Thanks Swalbr
If you won't mind can you explain why you say - "It is no more than the number of variables given"?
Also is there is a formal way to solve such a problem - I am looking at something equivalent to row-reduction to find the rank of matrix. Thanks
In the vectors you gave you had an $a$ and a $b$, so two variables. Say you had a subspace of vectors where you can write them like you did but with $n>0$ variables. Then you can easily split your general form into $n$ vectors each with precisely one variable. In each vector you can take out this variable as a common factor, and the set of all of these vectors (the ones with the variables removed) forms a spanning set. You may be able to reduce this spanning set, if not then it is a basis.
Does that make sense?
Note also that a vector space cannot have a basis of order greater than the length of the vector...
• Oct 2nd 2009, 08:35 AM
aman_cc
Thanks. I need to think this carefully though. I was also wondering if it has to do something with the fact that each component is a linear combination of the variables. What happens if we relax that - for e.g. {a^2,b^3,a+b^-1,b} (I have not checked if it is a sub-space.)
• Oct 2nd 2009, 03:48 PM
HallsofIvy
I presume you mean the matrix
$\begin{bmatrix}0 & 0 & 2 & 3 \\ 0 & -3 & -2 & -2\end{bmatrix}$
The kernel is, by definition, the set of vectors $\begin{bmatrix}x \\ y \\ z \\ t\end{bmatrix}$ such that
$\begin{bmatrix}0 & 0 & 2 & 3 \\ 0 & -3 & -2 & -2\end{bmatrix}\begin{bmatrix}x \\ y \\ z \\ t\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$
which means we must have 2z+ 3t= 0 and -3y- 2z- 2t= 0. Those two equations drop the dimension from 4 to 2. We can write z= (-3/2)t and then y= (-1/3)(2z+ 2t)= (-1/3)(-3t+ 2t)= (1/3)t. That is, y and z depend on t while x, since it does not appear in the equations can be anything. Use x and t as "free variables". Any vector in the kernel are of the form $\begin{bmatrix}x \\ (1/3)t \\ (-3/2)t \\ t\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}+ t\begin{bmatrix}0 \\ 1/3 \\ -3/2 \\ 1\end{bmatrix}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 28, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9212437868118286, "perplexity": 366.7069385524639}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891791.95/warc/CC-MAIN-20180123072105-20180123092105-00180.warc.gz"} |
http://mathhelpforum.com/calculus/198320-volume-body-obtained-rotating-around-x-axis-y-axis.html | # Math Help - Volume of body obtained by rotating around the x axis and y axis.
1. ## Volume of body obtained by rotating around the x axis and y axis.
Hi,
If y= 4x^3+3x^2-4x-3, calculate the volume of the body obtained by rotating y around 1) the x axis 2) the y axis.
I'm at a bit of a loss here. I know the formula for rotating about the x axis is the integral of pi.y^2 and integral of pi.x^2 for rotation about the y axis.
For the x axis rotation, i presume I have to square out the whole function and then integrate bit by bit. However, for rotation about the y axis, how do I express x^2 in terms of y? Bit confused, would appreciate help!
2. ## Re: Volume of body obtained by rotating around the x axis and y axis.
First, find the roots $4x^3+3x^2-4x-3=0$.
3. ## Re: Volume of body obtained by rotating around the x axis and y axis.
Hi.
I've factorised to get(x-1)(x+1)(4x+3). So the roots are x=1, x=-1 and x=-3/4
4. ## Re: Volume of body obtained by rotating around the x axis and y axis.
around the x-axis:
$V_x=\pi\int_{-1}^{1}(4x^3+3x^2-4x-3)^2\,dx=\ldots=\frac{1264}{105}\,\pi$
5. ## Re: Volume of body obtained by rotating around the x axis and y axis.
Thanks but do you know how to do it about the y axis? Can i say at y=0 that the above function squared then equals zero and then bring the x squared term over to the left and side...
6. ## Re: Volume of body obtained by rotating around the x axis and y axis.
Sorry, Just realised , can I use the shell method for rotating about the y axis? i.e that the volume of a solid rotated about the y axis can be written as the integral of 2pixf(x) between the limits a and b?
7. ## Re: Volume of body obtained by rotating around the x axis and y axis.
Originally Posted by orlacoon
Hi,
If y= 4x^3+3x^2-4x-3, calculate the volume of the body obtained by rotating y around 1) the x axis 2) the y axis.
I'm at a bit of a loss here. I know the formula for rotating about the x axis is the integral of pi.y^2 and integral of pi.x^2 for rotation about the y axis.
For the x axis rotation, i presume I have to square out the whole function and then integrate bit by bit. However, for rotation about the y axis, how do I express x^2 in terms of y? Bit confused, would appreciate help!
I see your problem. are you sure you have given us the complete question. For example it does not tell us between what values of x they are asking. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8147098422050476, "perplexity": 376.2343913488921}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246650671.76/warc/CC-MAIN-20150417045730-00024-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/tags/gauge-theory/new | # Tag Info
## New answers tagged gauge-theory
5
The gauge connection is not unique, and this has nothing to do with the presence of matter fields. Let $\Sigma$ be our space-time, $P$ a principal $G$-bundle, and $\mathcal{A}$ the space of connections on $P$. Then, gauge transformations $t : P \to G$, forming the group of gauge transformations $\mathcal{G}$ have an action on $\mathcal{A}$ given by $$A ... 5 Yes, one traditional alternative to the path integral formalism is the operator formalism. For QED with abelian gauge group, the old quantization formulation is the Gupta-Bleuler formulation. For QCD/Yang-Mills theory with non-abelian gauge group, the Gupta-Bleuler formulation is replaced by the BRST formulation. The BRST formulation exists in at least 3 ... 3 There exists an extensive literature for discretization of the abelian and the non-abelian gauge theories, known as lattice QED and lattice QCD, respectively. Here we will only sketch the main idea. Let us for simplicity use Euclidean signature (+,+,+,+). A small Wilson-loop$$\tag{1} W~=~{\rm Tr}{\cal P}e^{ig\int_{\gamma}A}$$lies approximately in a ... 2 The key point in all of this is that general relativity is a gauge theory, and, as the saying goes, "the gauge always hits twice" (apparently attributed to Claudio Teitelboim). What this means is that (1) you have an arbitrary freedom in defining your evolution, corresponding to the ability to make gauge transformations, and (2) some of the evolution ... 2 It is interesting to look at a linearized version of gravity, with g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} If you choose the Lorentz gauge :$$\partial^\mu \bar h_{\mu\nu}=0 \quad\quad \bar h_{\mu\nu} = h_{\mu\nu} - \frac{1}{2} h^i_i \,\eta_{\mu\nu} \tag{0}$$the equations of movement in the vaccuum are simply :$$\square \bar h_{\mu\nu}=0 \tag{1}$$... 1 It isn't classical field theory, but there are a few features of using of 4-potential in QFT. The first one is that 4-potential as 4-vector can't be used for describing massless photons. It is because the fact that it must describe massless particles leads to its transformations not as 4-vector under the Lorentz group. Specifically,$$ A^{\mu} \to ...
2
If your question is asking whether the four-potential is more useful in classical electromagnetism from a purely computational standpoint, the answer would be no. It's not to say that it isn't useful, it's just that it only groups together two equations in the Lorentz gauge that are already useful themselves. The Lorentz gauge, $$\Box\phi = ... 5 Why do we gauge-fix the path integral in the first place? If we were doing lattice gauge theory, we didn't need to gauge-fix. But in the continuum case, (the Hessian of) the action for a generalized^1 gauge theory has zero-directions that lead to infinite factors when performing the path integral over gauge orbits. In a BRST formulation (such as, e.g., the ... 4 We only have one contribution from each gauge-equivalent matter field configuration: Let P be the principal G-bundle associated to our gauge theory on the spacetime \mathcal{M} (for simplicity, assume it is \mathcal{M} \times G. The matter fields are constructed as sections of an associated vector bundle P \times_G V_\rho, where V_\rho is a ... 2 We have no choice. Let G be our gauge group and \Sigma our spacetime. Then, for the theory to actually be gauge invariant, every field must have a defined action of the gauge group upon it, i.e. every field must transform in a representation of this group:$$\phi : \Sigma \to V_\rho \text{ where there is a group morphism } \rho : G \to ...
3
Feynman diagrams are more than just the Lagrangian. They can be acquired by expanding the path integral of the theory into a perturbative series. There is a priori no reason to assume that all quantities needed in order to produce sensible results are consistent with gauge invariance. One possible issue is the problem of regularization: the way your ...
2
You may always promote "couplings constants" (charge, mass, etc...) to fields. Now, as a physicist, you need to make some contact with reality. So you have to tell why and which field you are using (for instance the Higgs field (up to a constant), which has a $SU(2)$ charge, is used to replace a constant mass coupling in the interaction \$m (\bar e_R e_L + ...
1
The Einstein equivalence principle states : The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime. Emphasis added. Note that this principle has done well in explaining quite a few things about gravity. So there is no a priori reason why you ...
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https://nyuscholars.nyu.edu/en/publications/scenario-submodular-cover | # Scenario submodular cover
Nathaniel Grammel, Lisa Hellerstein, Devorah Kletenik, Patrick Lin
Research output: Chapter in Book/Report/Conference proceedingConference contribution
## Abstract
We introduce the Scenario Submodular Cover problem. In this problem, the goal is to produce a cover with minimum expected cost, with respect to an empirical joint probability distribution, given as input by a weighted sample of realizations. The problem is a counterpart to the Stochastic Submodular Cover problem studied by Golovin and Krause [6], which assumes independent variables. We give two approximation algorithms for Scenario Submodular Cover. Assuming an integervalued utility function and integer weights, the first achieves an approximation factor of O(logQm), where m is the sample size and Q is the goal utility. The second, simpler algorithm achieves an approximation factor of O(logQW), where W is the sum of the weights. We achieve our bounds by building on previous related work (in [4,6,15]) and by exploiting a technique we call the Scenario-OR modification. We apply these algorithms to a new problem, Scenario Boolean Function Evaluation. Our results have applciations to other problems involving distributions that are explicitly specified by their support.
Original language English (US) Approximation and Online Algorithms - 14th International Workshop, WAOA 2016, Revised Selected Papers Monaldo Mastrolilli, Klaus Jansen Springer Verlag 116-128 13 9783319517407 https://doi.org/10.1007/978-3-319-51741-4_10 Published - 2017 14th International Workshop on Approximation and Online Algorithms, WAOA 2016 - Aarhus, DenmarkDuration: Aug 25 2016 → Aug 26 2016
### Publication series
Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) 10138 LNCS 0302-9743 1611-3349
### Other
Other 14th International Workshop on Approximation and Online Algorithms, WAOA 2016 Denmark Aarhus 8/25/16 → 8/26/16
## ASJC Scopus subject areas
• Theoretical Computer Science
• Computer Science(all)
## Fingerprint
Dive into the research topics of 'Scenario submodular cover'. Together they form a unique fingerprint. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9050565958023071, "perplexity": 3456.607309541227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00571.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/194879-continuum-mechanics-elasticity-help-re-polar-decomposition-theorem.html | # Math Help - Continuum Mechanics/Elasticity help re: Polar Decomposition Theorem
1. ## Continuum Mechanics/Elasticity help re: Polar Decomposition Theorem
That is a model question. b) is just stating the theorem, which is as follows in my notes:
If a linear transformation F is invertible with det F > 0, then there exists unique symmetric positive-definite linear transformations U and V, and a unique proper orthogonal transofrmation R, such that:
RU = F = VR.
I'm just having a bit of trouble applying the theorem, specifically "Show that the deformation can be considered to be the result of three simple stretches followed by a rotation. Explain the precise nature of the stretches and rotation.
Looking through the printed and my written notes, it revolves around the homogeneous deformation:
x = A + H(X - A) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9697060585021973, "perplexity": 1054.6068817742498}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500829754.11/warc/CC-MAIN-20140820021349-00251-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://mathoverflow.net/users/38468/lev-borisov?tab=topactivity | Lev Borisov
33 Advice for pure-math Phd students 32 The sum of squared logarithms conjecture 17 Interesting examples of vacuous / void entities 15 Die hard nilpotent spaces 10 Eigenvalues of a matrix with entries involving combinatorics
### Reputation (4,796)
+10 Crepant resolutions of toric varieties +10 What is the meaning of $(h^{11},h^{21})\to (h^{11}-240,h^{21}+240)$ in Calabi-Yau threefolds? +10 which varieties can appear as exceptional divisors? +10 Proof for a Rank-One Decomposition Theorem of Positive (semi) Definite Matrices
### Questions (27)
36 What is the meaning of $(h^{11},h^{21})\to (h^{11}-240,h^{21}+240)$ in Calabi-Yau threefolds? 24 Real square roots of symmetric matrices 15 Derived categories of arithmetic schemes? 11 What is Koszul dual of a curve? 11 Determinant and eigenvalues of a specific matrix
### Tags (112)
69 ag.algebraic-geometry × 35 32 symmetric-polynomials 51 polynomials × 6 32 matrix-theory 35 matrices × 7 29 co.combinatorics × 9 33 reference-request × 9 27 linear-algebra × 7 32 inequalities 22 convex-polytopes × 6
### Bookmarks (25)
64 A bestiary of topologies on Sch 57 “Gross-Zagier” formulae outside of number theory 39 The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic 38 The sum of squared logarithms conjecture 37 Is there a finite family of functions such that the max of any two functions can be dominated by a third? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8649719953536987, "perplexity": 1696.8310854983151}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00312.warc.gz"} |
https://www.physicsforums.com/threads/3-dimension-expectation-values-qm.145495/ | # 3-Dimension Expectation Values (QM)
1. Nov 26, 2006
### moo5003
Hello, I have a problem that wants me to find the expectation value of <r> <r^2> for the ground state of hydrogen (part a.). My friend and I already completed the exercise but i'm concerned about how we found the expectation value. Since the ground state of hydrogen is only dependent on r do we only integrate over r? I notice that if we integrate over psi and phi we will add an extra 2pi^2 multiplied with what we had previously. Any help would be appreciated:
Recap- Do you integrate over all three dimensions if the wave function is only dependent on one?
2. Nov 26, 2006
### OlderDan
You need to make sure the wave functions are normalized and that you are accounting for the variation of spatial volume in the vecinity of any given r. If you are using the full wave function with its normalization constant then you need the angular integrals to get the normalization correct. There are also factors of r in the volume element dV that are important. People often look at the function u(r) = rR(r) as a better representation of the radial wave function because the amount of 3-D space associated with any given dr is proportional to r². This all comes together naturally if you use the full normalized wave function integrated over all space. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.965848445892334, "perplexity": 339.6861737614728}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171932.64/warc/CC-MAIN-20170219104611-00580-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.theguardian.com/sustainable-business/2015/may/28/switching-to-biofuels-would-place-unsustainable-demands-on-water-use | # Switching to biofuels could place unsustainable demands on water use
Transition away from fossil fuels is underway, but without assessing demands on land and water resources we put the future health of our planet at risk
# Switching to biofuels could place unsustainable demands on water use
Transition away from fossil fuels is underway, but without assessing demands on land and water resources we put the future health of our planet at risk
As the world moves towards renewable sources of energy, it faces an accompanying challenge: water scarcity. The intensive water use in the coal, oil, gas and nuclear industries is well-documented, but if we want to encourage a faster transition to renewables we must also contemplate the water use of the alternatives.
It is a great challenge to limit the drain on land and water resources now the transition has taken off. Bioenergy, hydropower, and wind, solar and geothermal energy all require substantial amounts of land and water resources. Given limitations to the availability of land and water, what energy scenarios are feasible in the long run?
With fossil fuels we have learned to worry about energy scarcity as a major concern for economic development and national security. In contrast, renewable energy seems inexhaustible: incoming solar radiation, for example, is far beyond what we need. The fact that renewable energy is available into infinity reinforces this idea of limitlessness. This, however, is a misunderstanding: we will replace energy scarcity by land and water scarcity.
## Biofuels
Bioenergy production in particular requires vast amounts of land and water. Besides, with current energy-intensive agricultural practices, net energy output is far lower than gross energy production, sometimes even near zero. If only 10% of fossil fuels in the global transport sector were replaced by bioethanol from relatively efficient crops, global water demand would increase by 6-7%.
The production of biofuels at the rate we are used to consuming fossil fuels will require more land and water than sustainably available. Already today we have land and water footprints beyond maximum sustainable levels and bioenergy increasingly competing with food.
## Hydropower and the dam debate
Hydropower, accounting for 16% of the world’s electricity supply, is regarded as a clean form of energy. However, we cannot simply increase hydroelectric capacity. Dams can heavily impact on riparian ecosystems and societies, and any further damming of rivers should be subject to careful consideration.
Building new dams and reservoirs is often difficult because the required land is generally already in use for other purposes. For the Three Gorges Dam in China, over one million people were displaced. Besides, hydropower can be a large water consumer because of the additional evaporation from the reservoir created, which affects downstream water availability for other purposes. Damming rivers has therefore become a contentious topic.
## Solar, wind and geothermal energy
Per unit of energy, the water footprint of bioenergy and hydroelectricity is two to three orders of magnitude larger than that of fossil fuels and nuclear. The water footprint of photovoltaic (PV) and wind energy is one to two orders of magnitude smaller.
Electricity from concentrated solar power has a similar water footprint to fossil fuels, while geothermal can be an order of magnitude smaller or even less. From a water consumption and scarcity perspective, it matters greatly whether we shift from fossil energy to bio and hydro or to solar, wind and geothermal energy.
All existing “green” energy scenarios, called “green” because of their considerable fractions of renewable energy, are based on considerable growth of bio and hydro in the mix, which means that the water footprint of the energy sector will grow sky-high if we follow such scenarios. True green scenarios, with a declining rather than increasing water footprint, must be primarily based on solar, wind and geothermal energy.
## The transition to electricity
Solar energy is more efficient than biomass from a land use perspective because PV panels and concentrated solar power systems are more efficient at capturing incoming solar radiation than photosynthesis, thus generating more energy per square metre.
Photosynthesis, however, has the advantage that it results in storable bioenergy and can be turned into energy-dense biofuels, while PV results in non-storable electricity. Concentrated solar power systems can store energy by use of thermal energy storage, but the final product will still be electricity, not fuel.
Since substantial growth of bioenergy – beyond using rest streams of organic material – is impossible, our economies need to be further electrified: electric transport, but also electric heating, at least where surplus heat from industrial processes or geothermal energy doesn’t offer a solution. We need to find ways to store energy and design electrical grids that can handle the large variability of both electricity demand and supply.
Solar and wind power and earth’s heat offer possibilities to achieve energy self-sufficiency at much smaller scales than we are used to in our globalised fossil fuel economy. The time is ripe for a transition away from fossil fuels. Let’s be smart enough to invest in real sustainable solutions, which excludes biofuels that have been so much at the centre of attention in government policies. Decarbonising our economy can be combined with lowering our water footprint, let’s go for that choice.
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Topics | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8196578025817871, "perplexity": 2536.0781870005285}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463612399.20/warc/CC-MAIN-20170529145935-20170529165935-00260.warc.gz"} |
http://banubula.blogspot.com/2007_07_01_archive.html | ## Sunday, July 01, 2007
### The Broken Mirror: III
Building better Mirrors
The fall of Parity triggered the question: If our Universe does not obey simple mirror-symmetry, then what kind of symmetry does it obey? In other words, perhaps we were looking in the wrong mirror.
P-symmetry is a mirror of parity, reversing left and right. We can think of it also as a transformation: It is an exchange of every Left-handed interaction with a Right-handed interaction.
C-Symmetry
Let's consider other types of mirrors. Physicists also examined another mirror called C-symmetry, or Charge symmetry. C Symmetry is a mirror which exchanges every positively charged particle with a negatively charged particle. In the world of physics this essentially means that the C mirror transforms particles into anti-particles and vice-versa. A proton and electron become an anti-proton and a positron.
So, what happens when we use the C Mirror, when we swap particles for anti-particles? Do the anti-particles obey the same laws of physics? The answer is No. C-symmetry is not a property of this Universe.
T-symmetry
Ok, let's consider another abstract mirror. What if we take all particles and interactions in the Universe and move them backwards? Shown a film of a particle interaction, can we tell whether the film had been reversed? (Note to all clever folks: This is not the same as the question addressed by the Second Law since that applies only to macroscopic systems.)
Is our Universe symmetric in Time? This question is known as that of T-symmetry. The answer is...mostly yes. That is, we cannot detect a reversed film unless it contains some particular time-asymmetric interactions. This all comes down to the interactions of one particle: The Kaon. I summarize here from a physics abstract:
Most fundamental physical processes are symmetric in time. The motion of the planets in the gravity field of the sun is reversable- a film of the motion of a planet around its sun can be shown backwards without anyone being able to tell. Similarly to gravity, the strong nuclear and electromagnetic forces are also time-symmetric. Only the weak nuclear force appears to violate this symmetry, and this so far only in the behavior of the neutral kaon.
Combining symmetries
If I reflect a mirror image of a mirror image, I get the original. Mirrors can be combined to produce more mirrors.
The failure of P and C and T symmetry in our Universe is confounding. Recall that P symmetry is closest to what we consider a classic mirror, reversing left and right. What if we combine two mirrors (or more), combining the C mirror with the P mirror?
What this means in practice is that the mirror on the wall does show a valid Universe if we also exchange particles for anti-particles. The resulting symmetry is known as CP symmetry.
In 1980, James Cronin and Val Fitch earned the Nobel prize in physics for demonstrating that CP symmetry is not conserved. It was mostly conserved but violated in some interactions of the Kaon.
CPT: the Ultimate Mirror
The CP violation of the Kaon and the T violation of the Kaon are in fact related. They can be canceled out by throwing togther CP and T to make CPT: antiparticles in a mirror going backwards in time.
There is every reason to believe that CPT symmetry holds for our Universe. Even the Kaon (so far) obeys CPT symmetry. But what exactly does CPT symmetry imply? And what makes the Kaon such an anomaly? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8194789290428162, "perplexity": 767.7834187898986}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645323734.78/warc/CC-MAIN-20150827031523-00060-ip-10-171-96-226.ec2.internal.warc.gz"} |
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http://mathhelpforum.com/algebra/197898-understanding-factoring.html | # Math Help - Understanding this for factoring
1. ## Understanding this for factoring
Factor completly:
(3n-1)2 - (3n-1) - 72
What do you use to process this? If there is nothing to work with the -72? You can factor out (3n-1) correct? But still left with the -72?
2. ## Re: Understanding this for factoring
Originally Posted by Kibbygirl
Factor completly:
(3n-1)2 - (3n-1) - 72
$x^2-x-72=(x-9)(x+8)$. Let $x=3n-1.$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8989927768707275, "perplexity": 4605.860891310493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375099173.16/warc/CC-MAIN-20150627031819-00168-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://open.library.ubc.ca/cIRcle/collections/48630/items/1.0347363 | # Open Collections
## BIRS Workshop Lecture Videos
### Clustering Implies Geometry in Networks Krioukov, Dmitri
#### Description
Two common features of many large real networks are that they are sparse and that they have strong clustering, i.e., large number of triangles homogeneously distributed across all nodes. In many growing real networks for which historical data is available, the average degree and clus- tering are roughly independent of the growing network size. Recently, (soft) random geometric graphs, also known as latent-space network models, with hyperbolic and de Sitter latent geome- tries have been used successfully to model these features of real networks, to predict missing and future links in them, and to study their navigability, with applications ranging from designing optimal routing in the Internet, to identification of the information-transmission skeleton in the human brain. Yet it remains unclear if latent-space models are indeed adequate models of real networks, as random graphs in these models may have structural properties that real networks do not have, or vice versa. We show that the canonical maximum-entropy ensemble of random graphs in which the expected numbers of edges and triangles at every node are fixed to constants, are approximately soft random geometric graphs on the real line. The approximation is exact in the limit of standard random geometric graphs with a sharp connectivity threshold and strongest clustering. This result implies that a large number of triangles homogeneously distributed across all vertices is not only necessary but also a sufficient condition for the presence of a latent/effective metric space in large sparse networks. Strong clustering, ubiquitously observed in real networks, is thus a reflection of their latent geometry. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9098852872848511, "perplexity": 694.3367314602388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00151.warc.gz"} |
http://sci-gems.math.bas.bg/jspui/browse?type=subject&order=ASC&rpp=20&offset=3358 | ## Browsing by Subject
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http://www.analyzemath.com/calculus_questions/analytical/calculus_questions_3.html | # Calculus Questions with Answers (3)
Approximate graphically the first derivative of a function f from its graph. Questions are presented along with solutions.
Question 1: Below is shown the graph of function f. a) Assuming that the only extrema of function f are the ones shown in the graph, for what values of x is f '(x) = 0? b) Assuming that the graph of function f rises indefinitely on the right and on the left, for what values of x is f '(x) negative? For what values of x is f '(x) positive? Solution to Question 1: a) The graph of f has two minimums, one at x = -2 and one at x = 4, and one maximum at x = 1. Therefore f '(x) = 0 for x = -2, x = 1 and x = 4. b) Function f is decreasing on the intervals (- ∞ , -2) and (1 , 4). Hence f' (x) will be negative on these same intervals. Function f is increasing on the intervals (-2 , 1) and (4 , + ∞). Hence f' (x) will be positive on these same intervals. Question 2: The graph of function f is shown below. Assuming that function f is odd and has horizontal asymptotes, approximate graphically the graph of the first derivative f ' of f. Solution to Question 2: We first note that f is an increasing function. Hence f'(x) is positive for all values of and its graph is above the x axis. The value of the first derivative f '(a) at a given value of x = a is equal to the slope of the tangent line at the point (a , f(a)) . Hence, a possible approximation of f '(x) would be an approximation of the slope of the tangent line at the graph of f. For the given graph, it seems that the slope of the tangent line to the graph of f will be maximum around the origin (0,0). Let us use three points A, B and C, close to the origin 0 to approximate the slope mo of the tangent line at x = 0. If points A and C have the coordinates A(xA,yA) and C(xC,yC), then mo may be approximated by mo = (yC - yA) / (xC - xA) = (0.5 - (-0.5))/ (0.5 - (-0.5)) = 1 We can also use points C and E to approximate the slope m1 of the tangent line at point D. m1 = (1 - 0.5) / (1.5 - 0.5) = 0.5 Since f has horizontal asymptotes, we would expect f'(x) to be close to zero as x increases indefinitely (+ ∞) or decreases indefinitely (- ∞). Since f has horizontal asymptotes, we would expect the tangent line to the graph of to be horizontal and its slope close to zero as x increases indefinitely (+ ∞) or decreases indefinitely (- ∞). Using all the above information, one possible approximation of f'(x) would be as shown below in blue. Question 3: Approximate the graph of the first derivative f ' of function f when given the graph of f below. Assume that the graph of f is symmetric with respect to the vertical line x = -0.5 and that f has y = 0 as a horizontal asymptote. Solution to Question 3: Using similar ideas to those used in questions 1 and 2 above, note that f'(x) = 0 at x = -2, -0.5 and 1 since these are the locations of the extrema of f unction f. The groups of three (green) points are used to approximate the slope of the tangent line at the middle point of each group. For example using points D, E and F one can approximate the slope mo of the tangent line at point E as follows mo = (0.8 - 0.4) / (-2.4 - (-2.9)) = 0.8 We can use the next group of the three (green) points, to the right of point A, to approximate the slope at the middle point which will be close to -0.8. The above information may easily be used to plot points on the graph of f' as shown below (blue points). The sign of f'(x) is determined by the increase and decrease of f. Hence f'(x) is positive for values of on the intervals of increase of f given by: (- ∞ , -2) and (-0.5 , 1) f'(x) is negative for values of x in the intervals of decrease of f and are defined by: (-2 , -0.5) and (1 , + ∞) Using all the above information, we may approximate f'(x) as follows (blue graph). | 1 | 2 | 3 | 4 | 5 | More references on calculus questions with answers, tutorials and problems . | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9007874727249146, "perplexity": 255.191570641014}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376825363.58/warc/CC-MAIN-20181214044833-20181214070333-00317.warc.gz"} |
https://whatmaster.com/electrostatics/ | PHYSICS
Electrostatics
Basic Theoretical Information
Electric charge and its properties
The electric charge is a physical quantity that characterizes the ability of particles or bodies to enter into electromagnetic interactions. The electric charge is usually designated by the letters q and Q . In the SI system, the electric charge is measured in Pendants (C). A free charge of 1 C is a huge amount of charge that is practically not found in nature. As a rule, you will have to deal with microcolourants (1 μC = 10 –6 C), nanocolons (1 nC = 10 –9 C) and picoculons (1 p = 10 –12 C). Electric charge has the following properties:
1. Electric charge is a kind of matter.
2. The electric charge does not depend on the motion of the particle and on its speed.
3. Charges can be transferred (for example, by direct contact) from one body to another. Unlike body mass, electric charge is not an integral characteristic of a given body. The same body under different conditions can have a different charge.
4. There are two kinds of electric charges, conventionally called positive and negative .
5. All charges interact with each other. In this case, like charges repel each other, opposite charges attract each other. The forces of interaction of charges are central, that is, they lie on the straight line connecting the centers of charges.
6. There is a minimum possible (modulo) electric charge, called an elementary charge . Its meaning is:
e = 1.602177 · 10 –19 Cl ≈ 1.6 · 10 –19 Cl.
The electric charge of any body is always a multiple of the elementary charge:
where: N is an integer. Please note that the existence of a charge equal to 0.5 e ; 1.7 e ; 22.7 e and so on. Physical quantities that can take only a discrete (non-continuous) series of values are called quantized . Elementary charge e is a quantum (the smallest portion) of electric charge.
7. The law of conservation of electric charge. In an isolated system, the algebraic sum of the charges of all bodies remains constant:
The law of conservation of electric charge states that in a closed system of bodies the processes of creation or disappearance of charges of only one sign cannot be observed. It also follows from the charge conservation law, if two bodies of the same size and shape, having charges 1 and 2 (no matter what sign of charge), come into contact, and then dissolve back, then the charge of each of the bodies becomes equal:
From the modern point of view, charge carriers are elementary particles. All ordinary bodies are composed of atoms, which include positively charged protons , negatively charged electrons and neutral particles – neutrons . Protons and neutrons are part of atomic nuclei, electrons form the electron shell of atoms. The electric charges of the proton and the electron are exactly the same in modulus and equal to the elementary (that is, the minimum possible) charge e .
In a neutral atom, the number of protons in the nucleus is equal to the number of electrons in the shell. This number is called the atomic number. An atom of a given substance may lose one or more electrons, or acquire an extra electron. In these cases, the neutral atom becomes a positively or negatively charged ion. Please note that positive protons are part of the nucleus of an atom, so their number can only change during nuclear reactions. Obviously, when electrifying bodies of nuclear reactions does not occur. Therefore, in any electrical phenomena the number of protons does not change, only the number of electrons changes. So, sending a negative charge to the body means passing extra electrons to it. And the message of a positive charge, in spite of a common error, means not the addition of protons, but the taking of electrons.
Sometimes in problems the electric charge is distributed over a certain body. To describe this distribution, the following values are entered:
1. Linear charge density. Used to describe the charge distribution over a thread:
where: L is the length of the thread. Measured in C / m.
2. Surface charge density. Used to describe the charge distribution over the body surface:
where: S is the surface area of the body. Measured in C / m 2 .
3. Bulk charge density. Used to describe the distribution of charge over body volume:
where: V – body volume. Measured in C / m 3 .
Note that the electron mass is equal to:
e = 9.11 ∙ 10–31 kg.
Coulomb’s law
A point charge is a charged body whose dimensions under the conditions of this task can be neglected. Based on numerous experiments, Coulomb established the following law:
The interaction forces of fixed point charges are directly proportional to the product of charge modules and inversely proportional to the square of the distance between them:
where: ε is the dielectric constant of a medium — a dimensionless physical quantity that indicates how many times the force of electrostatic interaction in a given medium will be less than in a vacuum (that is, how many times the medium weakens the interaction). Here k is the coefficient in the Coulomb law, the value that determines the numerical value of the interaction force of the charges. In the SI system, its value is assumed to be:
k = 9 ∙ 10 9 m / F.
The interaction forces of the pointless stationary charges obey the third law of Newton, and are repulsive forces from each other with the same signs of charges and forces of attraction to each other with different signs. The interaction of fixed electric charges is called electrostatic or Coulomb interaction. The section of electrodynamics that studies the Coulomb interaction is called electrostatics .
Coulomb’s law is valid for point charged bodies, uniformly charged spheres and balls. In this case, for distances r take the distance between the centers of spheres or balls. In practice, the Coulomb’s law is well satisfied if the size of charged bodies is much smaller than the distance between them. The coefficient k in the SI system is sometimes written in the form:
where: ε 0 = 8.85 ∙ 10 –12 F / m is the electric constant.
Experience shows that the forces of the Coulomb interaction obey the superposition principle: if a charged body interacts simultaneously with several charged bodies, then the resultant force acting on a given body is equal to the vector sum of the forces acting on this body from all other charged bodies.
Remember also two important definitions:
Conductors – substances containing free charge carriers. Inside the conductor, free movement of electrons – charge carriers is possible (an electric current can flow through the conductors). The conductors include metals, solutions and melts of electrolytes, ionized gases, plasma.
Dielectrics (insulators) are substances in which there are no free charge carriers. The free movement of electrons inside dielectrics is impossible (no electric current can flow through them). It is dielectrics that have some non-unit dielectric constant ε .
For the dielectric constant of a substance, the following is true (that is, an electric field is slightly lower):
Electric field and its intensity
According to modern concepts, electric charges do not act on each other directly. Each charged body creates an electric field in the surrounding space . This field has a powerful effect on other charged bodies. The main property of the electric field is the effect on electric charges with some force. Thus, the interaction of charged bodies is carried out not by their direct action against each other, but through the electric fields surrounding the charged bodies.
The electric field surrounding a charged body can be investigated with the help of the so-called trial charge — a small point charge that does not introduce a noticeable redistribution of the charges under study. To quantify the electric field force characteristic is introduced – the electric field strength E .
The strength of the electric field is called a physical quantity equal to the ratio of the force with which the field acts on the trial charge placed at a given point of the field to the magnitude of this charge:
Electric field strength is a vector physical quantity. The direction of the intensity vector coincides at each point in space with the direction of the force acting on the positive test charge. The electric field of the fixed and unchanging with time charges is called electrostatic.
For a visual representation of the electric field using lines of force . These lines are drawn so that the direction of the intensity vector at each point coincides with the direction of the tangent to the field line. Power lines have the following properties.
• The power lines of the electrostatic field never intersect.
• The electrostatic field lines are always directed from positive to negative charges.
• When depicting an electric field with the help of lines of force, their density should be proportional to the modulus of the field intensity vector.
• Power lines start at a positive charge or infinity, and end at a negative or infinity. The density of the lines is greater, the greater the tension.
• At this point in space, only one line of force can pass; The electric field strength at a given point in space is uniquely defined.
An electric field is called homogeneous if the intensity vector is the same at all points of the field. For example, a uniform field creates a flat capacitor — two plates, charged by an equal in magnitude and opposite in sign charge, separated by a dielectric layer, and the distance between the plates is much smaller than the sizes of the plates.
At all points of a uniform field, a charge q introduced into a uniform field with a strength E acts on the same in magnitude and direction force, equal to F = Eq . Moreover, if the charge q is positive, then the direction of the force coincides with the direction of the intensity vector, and if the charge is negative, then the strength and intensity vectors are oppositely directed.
The force lines of the Coulomb fields of positive and negative point charges are shown in the figure:
Principle of superposition
If using a trial charge, the electric field created by several charged bodies is investigated, then the resulting force is equal to the geometric sum of the forces acting on the trial charge from each charged body separately. Consequently, the intensity of the electric field created by the charge system at a given point in space is equal to the vector sum of the strengths of the electric fields created at the same point by the charges separately:
This property of the electric field means that the field obeys the superposition principle . In accordance with the Coulomb’s law, the strength of the electrostatic field created by a point charge Q at a distance r from it is equal in absolute value:
This field is called Coulomb. In the Coulomb field, the direction of the intensity vector depends on the sign of the charge Q: if Q > 0, then the intensity vector is directed from the charge, if Q <0, then the intensity vector is directed towards the charge. The magnitude of the strength depends on the magnitude of the charge, the medium in which the charge is located, and decreases with increasing distance.
The electric field strength, which creates a charged plane near its surface:
So, if in the task it is required to determine the field strength of the charge system, then we must act according to the following algorithm :
1. Draw a picture.
2. Draw the field strength of each charge separately at the desired point. Remember that the intensity is directed towards the negative charge and from the positive charge.
3. Calculate each of the tensions using the appropriate formula.
4. Add the stress vector geometrically (i.e., vector).
Potential charge interaction energy
Electric charges interact with each other and with the electric field. Any interaction describes the potential energy. The potential energy of interaction of two point electric charges is calculated by the formula:
Pay attention to the absence of modules in charges. For opposite charges, the interaction energy is negative. The same formula is valid for the interaction energy of uniformly charged spheres and balls. As usual, in this case the distance r is measured between the centers of the balls or spheres. If the charges are not two, but more, then the energy of their interaction should be considered as follows: break the system of charges into all possible pairs, calculate the interaction energy of each pair and sum up all the energies for all pairs.
Tasks on this topic are solved, as well as tasks on the law of conservation of mechanical energy: first, the initial interaction energy is found, then the final one. If the task is asked to find work on the movement of charges, then it will be equal to the difference between the initial and final total interaction energy of the charges. The interaction energy can also transfer to kinetic energy or to other forms of energy. If the bodies are at a very large distance, then the energy of their interaction is set to 0.
Note: if the task requires to find the minimum or maximum distance between the bodies (particles) while moving, then this condition will be fulfilled at that moment of time when the particles move in one direction with the same speed. Therefore, the solution must begin with the recording of the law of conservation of momentum, from which this identical velocity is found. And then you should write the law of energy conservation taking into account the kinetic energy of the particles in the second case.
Potential. Potential difference. Voltage
The electrostatic field has an important property: the work of the electrostatic field forces when a charge moves from one point of the field to another does not depend on the shape of the trajectory, but is determined only by the position of the initial and final points and the charge value.
The consequence of the independence of the work from the shape of the trajectory is the following statement: the work of the electrostatic field forces when the charge moves along any closed trajectory is zero.
The property of potentiality (independence of work from the shape of the trajectory) of an electrostatic field allows one to introduce the concept of potential energy of a charge in an electric field. A physical quantity equal to the ratio of the potential energy of an electric charge in an electrostatic field to the magnitude of this charge is called the potential φ of the electric field:
The potential φ is the energy characteristic of the electrostatic field. In the International System of Units (SI), the potential unit (and hence the potential difference, i.e. voltage) is the volt [V]. Potential is a scalar quantity.
In many problems of electrostatics when calculating the potentials for the reference point, where the values of potential energy and potential turn to zero, it is convenient to take an infinitely distant point. In this case, the concept of potential can be defined as follows: the potential of the field at a given point in space is equal to the work that the electric forces do when a single positive charge is removed from a given point to infinity.
Recalling the formula for the potential energy of the interaction of two point charges and dividing it by the value of one of the charges in accordance with the definition of potential, we obtain that the potential φ of the point charge field Q at a distance r from it relative to an infinitely remote point is calculated as follows
The potential calculated by this formula can be positive and negative, depending on the sign of the charge that created it. The same formula expresses the potential of the field of a uniformly charged ball (or sphere) with r ≥ R (outside the ball or sphere), where R is the radius of the ball, and the distance r is measured from the center of the ball.
For a visual representation of the electric field, along with the lines of force, equipotential surfaces are used . A surface, at all points of which the potential of the electric field has the same value, is called an equipotential surface or a surface of equal potential. The electric field lines are always perpendicular to the equipotential surfaces. The equipotential surfaces of the Coulomb field of a point charge are concentric spheres.
Electrical voltage is simply a voltage difference, i.e. The definition of electrical voltage can be given by:
In a uniform electric field, there is a relationship between field strength and voltage:
The work of the electric field can be calculated as the difference between the initial and final potential energy of the charge system:
The work of the electric field in the general case can also be calculated by one of the formulas:
In a uniform field when the charge moves along its lines of force, the field operation can also be calculated using the following formula:
In these formulas:
• φ is the electric field potential.
• ∆ φ is the potential difference.
• W is the potential energy of a charge in an external electric field.
• A – the work of the electric field on the movement of charge (charges).
• q – charge, which is moved in an external electric field.
• U is the voltage.
• E is the electric field strength.
• d or ∆ l is the distance over which the charge moves along the lines of force.
All the previous formulas dealt specifically with the work of the electrostatic field, but if the task states that “the work must be done,” or it is referred to the “work of external forces,” then this work should be considered the same as the work of the field, but with opposite sign.
Principle of superposition of potential
The principle of superposition for potentials follows from the principle of superposition of the strengths of fields generated by electric charges (the sign of the field potential depends on the sign of the charge that created the field):
Notice how much easier it is to apply the principle of superposition of potential than tension. Potential is a scalar quantity that has no direction. Adding potentials is simply summing up the numerical values.
Electric capacity Flat capacitor
When communicating to the charge conductor, there is always a certain limit, over which the body cannot be charged. To characterize the body’s ability to accumulate electrical charge, the concept of electrical capacitance is introduced . The capacity of a solitary conductor is the ratio of its charge to potential:
In the SI system, capacitance is measured in Farads [F]. 1 Farad – extremely large capacity. For comparison, the capacity of the entire globe is significantly less than one farad. The capacity of the conductor does not depend on its charge, nor on the potential of the body. Similarly, density does not depend on mass or on the volume of the body. Capacity depends only on the shape of the body, its size and the properties of its environment.
The electrical capacity of a system of two conductors is a physical quantity, defined as the ratio of the charge q of one of the conductors to the potential difference Δ φ between them:
The amount of electrical capacity of conductors depends on the shape and size of the conductors and on the properties of the dielectric separating the conductors. There are configurations of conductors in which the electric field is concentrated (localized) only in a certain region of space. Such systems are called capacitors , and the conductors that make up the capacitor are called plates .
The simplest capacitor is a system of two flat conducting plates arranged parallel to each other at a small distance compared to the dimensions of the plates and separated by a dielectric layer. Such a capacitor is called flat . The electric field of a plane capacitor is mainly localized between the plates.
Each of the charged plates of a flat capacitor creates an electric field near its surface, the strength of which is expressed by the ratio already mentioned above. Then the modulus of the strength of the final field inside a capacitor created by two plates is equal to:
Outside of the capacitor, the electric fields of the two plates are directed in different directions, and therefore the resulting electrostatic field is E = 0. The capacitance of the flat capacitor can be calculated by the formula
Thus, the electrical capacitance of a flat capacitor is directly proportional to the area of the plates (plates) and inversely proportional to the distance between them. If the space between the plates is filled with a dielectric, the capacitance of the capacitor increases by a factor of ε . Note that S in this formula is the area of only one capacitor plate. When the problem is referred to as “area of the plates”, they mean precisely this quantity. It is never necessary to multiply or divide by 2.
Once again we give the formula for charging the capacitor . Under the charge of the capacitor understand only the charge of its positive lining:
The force of attraction of the capacitor plates. The force acting on each plate is determined not by the full field of the capacitor, but by the field created by the opposite plate (the plate itself does not act). The strength of this field is equal to half the strength of the total field, and the strength of the interaction of the plates:
Condenser energy. It is also called the energy of the electric field inside the capacitor. Experience shows that a charged capacitor contains a supply of energy. The energy of a charged capacitor is equal to the work of external forces, which must be expended to charge the capacitor. There are three equivalent forms for writing a formula for the energy of a capacitor (they follow one of the other if we use the relation q = CU ):
Pay special attention to the phrase: “The capacitor is connected to the source.” This means that the voltage across the capacitor does not change. And the phrase “The capacitor was charged and disconnected from the source” means that the charge of the capacitor will not change.
Electric field energy
Electrical energy should be considered as potential energy stored in a charged capacitor. According to modern concepts, the electric energy of a capacitor is localized in the space between the capacitor plates, that is, in an electric field. Therefore, it is called the electric field energy. The energy of charged bodies is concentrated in space, in which there is an electric field, i.e. You can talk about the energy of the electric field. For example, for a capacitor, energy is concentrated in the space between its plates. Thus, it makes sense to introduce a new physical characteristic – the bulk density of the electric field energy. Using the example of a flat capacitor, one can obtain the following formula for the bulk energy density (or energy per unit volume of the electric field):
Capacitor connections
A Parallel connection of capacitors – to increase capacity. Capacitors are connected by the same charged plates, as if increasing the area of equally charged plates. The voltage on all capacitors is the same, the total charge is equal to the sum of the charges of each of the capacitors, and the total capacity is also equal to the sum of the capacitors of all the capacitors connected in parallel. We write the formulas for parallel connection of capacitors:
When connecting capacitors in series, the total capacitance of the capacitor bank is always less than the capacity of the smallest capacitor entering the battery. A series connection is used to increase the breakdown voltage of capacitors. We write the formulas for the serial connection of capacitors. The total capacitance of series-connected capacitors is found from the relationship:
From the law of conservation of charge, it follows that the charges on the adjacent plates are equal:
The voltage is equal to the sum of the voltages on the individual capacitors.
For two series-connected capacitors, the formula above will give us the following expression for the total capacity:
For N identical series-connected capacitors:
Conductive sphere
The field strength inside the charged conductor is zero. Otherwise, the free charges inside the conductor would be acted upon by an electric force, which would force these charges to move inside the conductor. This movement, in turn, would lead to heating of the charged conductor, which in fact does not occur.
The fact that there is no electric field inside the conductor can be understood in another way: if it were, then the charged particles would move again, and they would move precisely so as to reduce this field to zero by its own field, since in general, they would not want to move, because any system tends to equilibrium. Sooner or later, all moving charges would stop at exactly that place so that the field inside the conductor becomes equal to zero.
On the surface of the conductor the electric field strength is maximum. The magnitude of the electric field of a charged ball outside it decreases with distance from the conductor and is calculated using a formula similar to the formulas for the field strength of a point charge, in which the distances are measured from the center of the ball.
Since the field strength inside a charged conductor is zero, the potential at all points inside and on the surface of the conductor is the same (only in this case, the potential difference, and hence the strength is zero). The potential inside a charged ball is equal to the potential on the surface. The potential outside the ball is calculated by a formula similar to the formulas for the point charge potential, in which the distances are measured from the center of the ball.
Electrical capacity of the ball of radius R :
If the ball is surrounded by a dielectric, then:
Properties of the conductor in the electric field
1. Inside the conductor, the field strength is always zero.
2. The potential inside the conductor at all points is the same and equal to the potential of the surface of the conductor. When the problem says that “the conductor is charged to the potential … B”, then it is the surface potential that is meant.
3. Outside of the conductor near its surface, the field strength is always perpendicular to the surface.
4. If the conductor is charged, it is distributed over a very thin layer near the surface of the conductor (they usually say that the entire charge of the conductor is distributed on its surface). This is easily explained: the fact is that when we communicate the charge to the body, we give it charge carriers of the same sign, i.e. charges of the same name that repel each other. So they will strive to scatter from each other to the maximum distance of all possible, i.e. accumulate at the very edges of the conductor. As a result, if the core is removed from the conductor, its electrostatic properties will not change at all.
5. Outside the conductor, the field strength is greater, the more curved the surface of the conductor. The maximum value of tension is reached near the tips and sharp kinks of the conductor surface.
Tips for solving complex problems
1. Grounding something means connecting the conductor of a given object to the Earth. In this case, the potentials of the Earth and the existing object are aligned, and the necessary charges for this run across the conductor from the Earth to the object or vice versa. In this case, it is necessary to take into account several factors that follow from the fact that the Earth is disproportionately larger than any object that is not located there:
• The total charge of the Earth is conventionally equal to zero, therefore its potential is also equal to zero, and it will remain equal to zero after the connection of the object with the Earth. In short, grounding means zeroing the potential of the object.
• To reset the potential (and hence the object’s own charge, which could have been both positive and negative), the object will have to either accept or give the Earth some (possibly very large) charge, and the Earth will always be able to provide such an opportunity.
2. We repeat once again: the distance between repulsive bodies is minimal at the moment when their velocities become equal in magnitude and are directed in one direction (the relative velocity of the charges is zero). At this moment, the potential energy of the interaction of charges is maximum. The distance between the attracting bodies is maximal, also at the moment of equality of speeds directed in one direction.
3. If the problem has a system consisting of a large number of charges, then it is necessary to consider and describe the forces acting on a charge that is not in the center of symmetry. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9433480501174927, "perplexity": 270.91213085239815}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737883.59/warc/CC-MAIN-20200808135620-20200808165620-00336.warc.gz"} |
https://socratic.org/questions/how-do-you-rationalize-sqrt-5-8 | Algebra
Topics
# How do you rationalize sqrt(5/8)?
Write it as: $\frac{\sqrt{5}}{\sqrt{8}}$
then multiply and divide by $\sqrt{8}$ to get:
$\frac{\sqrt{5}}{\sqrt{8}} \cdot \frac{\sqrt{8}}{\sqrt{8}} = \frac{\sqrt{5} \sqrt{8}}{8} =$
$= \frac{\sqrt{5} \sqrt{4 \cdot 2}}{8} = 2 \frac{\sqrt{5} \sqrt{2}}{8} = \frac{\sqrt{5} \sqrt{2}}{4} = \frac{\sqrt{10}}{4}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9821943044662476, "perplexity": 1244.6123969814355}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107896778.71/warc/CC-MAIN-20201028044037-20201028074037-00106.warc.gz"} |
https://www.physicsforums.com/threads/recommand-a-nice-qm-textbook-on-perturbation-and-scattering.421192/ | # Recommand a nice qm textbook on perturbation and scattering
1. Aug 9, 2010
### kof9595995
I just finished the first 4 chapters of Sakurai's Modern qm, and now I'm begining to learn purterbation method and scattering theory, but from the feedback it seems that many people are quite unsatisfied with Modern qm on these parts. Could you guys recommand a nice book on perturbation and scattering?
2. Aug 9, 2010
### humanino
I like "Scattering theory : the quantum theory of non-relativistic collisions" by John R. Taylor.
3. Aug 10, 2010
### kof9595995
Thanks for the information, btw does it also include perturbation theory ?
4. Aug 11, 2010
### humanino
Yes it does. It is not the fastest shortcut to perturbation and perturbation technologies are not applied to many examples. Rather, it is a formal text justifying rigorously the perturbation formalism, pointing out the subtleties and caveats.
5. Aug 23, 2010
### kof9595995
Thanks, just had a look at the content, it seems a bit too andvanced for me.
6. Aug 23, 2010
### Daverz
For scattering theory, Quantum Mechanics II: A Second Course in Quantum Theory by Rubin Landau is good.
Although, maybe what you need is a second QM text. Shankar, perhaps. Or at a more sophisticated level, the book by Arno Bohm.
Last edited: Aug 23, 2010
Similar Discussions: Recommand a nice qm textbook on perturbation and scattering | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8389682173728943, "perplexity": 2994.4033372632957}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886117874.26/warc/CC-MAIN-20170823055231-20170823075231-00385.warc.gz"} |
https://educatingphysics.com/a-level/module-4-electrons-waves-and-photons/electricity-charge-and-current/5-the-current-equation/ | Objectives:
• To state what is meant by the term mean drift velocity of charge carriers;
• To be able to select and use the equation $I = Anev$;
• To describe the difference between conductors, semiconductors and insulators in terms of the number density n.
Mean Drift Velocity
In order for a current to flow in a wire the charge carriers, i.e. electrons, must be moving. How fast they are moving depends on several factors, some of which include the potential difference across the wire, the resistance of the wire and the number of electrons present.
When there is no potential difference across a particular wire, each individual charge carrier (in most cases electrons) will be moving in its own direction with a speed dependent on the amount of energy it has previously gained; this energy could depend on the thermal energy it has gained from its surroundings. So a circuit like the following; with a wire, as located in the circuit, will have electrons in it which are moving with their own internal energy;
When a potential difference (voltage) is supplied across the end of a wire, each electron is forced to move in a certain direction. The potential difference forces the electrons along – work is done on the charged particles;
Here we will be discussing how fast the electrons need to be travelling in order to produce an ‘adequate’ current. Since $I = \frac{\Delta Q}{\Delta t}$; a current can only flow if the charges move some distance in a time t seconds.
Deriving the current equation
For this derivation we will be making the assumption that all the electrons will be travelling at the same speed, this speed will actually be the average speed, or mean drift velocity, of the electrons.
If a detector is placed at position and is capable of timing a single charge flowing past this point;
Suppose over a time $t$ second, the highlighted electron will travel a distance $l$.
Assuming all the electrons are travelling at the same speed, all the electrons in the cylinder of length $l$ will have in fact flowed past the detector;
In order to work out how much current has flowed, we first need to work out how much charge has passed the detector. The total amount of charge that will have passed the detector would fill the volume, and this volume can be calculated;
Volume of the cylinder, $V = l \times A$, where $A$ is the cross sectional area of the wire.
This cylinder will not be entirely made up of electrons, there will be ions for example as well, however for a specific material there will be a certain concentration of electrons per metre cubed $n$.
The number of electrons in this cylinder can then be written as;
$number\ of\ electrons\ in\ cylinder = n \times A \times l$
If each electron carries a charge $q$ then;
The charge charge by the cylinder, $Q = n \times A \times l \times q$
Since this cylinder is dependent on how long the detector is timing for, an element of time should be in the equation for the $Q$. If all the electrons are travelling at a mean drift velocity $v$ for a time $t$ seconds, and $v = \frac{l}{t}$ we can write $l = v \times t$
therefore; $Q = n \times A \times v \times t \times q$
since $I = \frac{Q}{t}$
we can write; $I = \frac{Q}{t} = \frac{n \times A \times v \times t \times q}{t}$
The $t$‘s on the right hand side of this equation cancel to give:
$I = n \times A \times v \times q$
Since the charge of an electron, in this case $q$, is $1.6 \times 10^{-19}$ or also denoted with an $e$, we can write;
$I = n \times A \times v \times e$
$I = nAve$
where;
$I$ is the current, measured in $A$
$n$ is the number of electrons per metre cubed, therefore measured in $m^{-3}$
$A$ is the cross sectional area, measured in $m^{2}$
$v$ is the mean drift velocity of the electrons, measured in $ms^{-1}$
$e$ is the charge of an electron, $1.6 \times 10^{-19}$, measure in $C$
One method of checking whether this equation is valid is to check the units are conserved on each side;
$I = nAve$
$A = m^{-3} \times m^{2} \times ms^{-1} \times C$
$A = s^{-1} \times C$ Since, $I = \frac{Q}{t}$ and therefore $A = \frac {C}{s}$, the equation holds true.
So, to address the original question how fast do electrons need to be travelling in order to produce an ‘adequate’ current?
Lets plug in some realistic numbers to determine the mean drift velocity:
If a current of $0.50 \ A$ flows through a copper wire measured to be $0.2019 \ mm$ in thickness, using a micrometer, and copper has $8.5 \times 10^{28}$ electrons per metre cubed:
$I = 0.50 \ A$
$n = 8.5 \times 10^{28} \ m^{-3}$
$d = 0.2019 \ mm = 0.2019 \times 10^{-3} \ m$
A ‘non-deformed’ wire can be assumed to be cylindrical and therefore the cross sectional area can be said to be circular. So $A = \pi r^{2}$
$A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.2019 \times 10^{-3}}{2})^{2} = 3.202 \times 10^{-8} m^{2}$
Rearranging $I = nAve$ to give $v = \frac{I}{nAe}$ and substituting everything in gives:
$v = \frac{0.50}{8.5 \times 10^{28} \times 3.202 \times 10^{-8} \times 1.6 \times 10^{-19}}$
$v = \frac{0.50}{435.47}$
$v = 1.148 \times 10^{-3}$
$v = 1.1 \times 10^{-3} ms^{-1} \ ({2s.f})$
This shows that the mean drift velocity of electrons flowing down a standard gauge wire with a current of $0.50 \ A$ is very slow, approximately $1 mms^{-1}$. Many people think that electrons travel down wires extremely quickly, which would explain why a light bulb would turn on almost immediately after a switch has been flicked). However, they actually travel very slowly and the reason a bulb still turns on as if instantaneously is because all electrons repel one another, as soon as one electrons is propelled along by a power source it repels its neighbouring electrons, which does the same to its neighbouring electrons, and so on and so forth until the bulb (this repulsive force os what causes a light bulb to light up so quickly.
Electrons as incredibly small, they have a mass of $9.11 \times 10^{-31} kg$, they are not visible to any human instrument (they can only be detected so far), so the fact that they can travel even $1 mm$ in just one second is actually quite impressive. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 60, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9011223316192627, "perplexity": 524.2476809415851}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141191692.20/warc/CC-MAIN-20201127103102-20201127133102-00059.warc.gz"} |
https://math.stackexchange.com/questions/1689013/the-fundamental-theorem-of-symmetric-polynomials | # The Fundamental Theorem of Symmetric Polynomials
This theorem stays that any symmetric polynomial can be expressed as a polynomial of elementary polynomials. So let's suppose I have a symmetric polynomial $f(x_1,x_2,...,x_n)$ in $R[\mathbb{X}]$. I can find a unique polynomial $g(s_1,s_2,...,s_n)$ in $R[\mathbb{X}]$ so I can express f in term of g, where $s_1,s_2,...,s_n$ are elementary polynomials. My question to you is: Is there a algorithm/set of stepts to determine g? Thanks.
Let $c x_1^{a_1} x_2^{a_2} \dots x_n^{a_n}$ be the lexicographically largest monomial of $f$, that is there are no monomials with strictly larger exponent of $x_1$, and no monomials with $x_1$ exponent ${a_1}$ that have a higher power of $x_2$, and so on. We'll think of this as being the leading term of the polynomial.
Now the key thing to notice is that $e_n^{a_n}e_{n-1}^{(a_{n-1}-{a_n})} \dots e_1^{(a_1-a_2)}$ contains the monomial $x_1^{a_1} x_2^{a_2} \dots x_n^{a_n}$ with coefficient $1$ and all other monomials it contains are smaller lexicographically.
Now the point is you can consider the leading term of $f - c e_n^{a_n}e_{n-1}^{(a_{n-1}-{a_n})} \dots e_1^{(a_1-a_2)}$, and so on. Each step reduces the leading term (in lexicographic order), so this process must eventually terminate, at which point you have written $f$ in terms of the elementary symmetric functions.
• Ok, you kinda lost me at the last paragraph. Can you please show me how this work on a example. Let's say I have the following polynomial: $$P(x_1,x_2,x_3) = x_1^4+x_2^4+x_3^4-2x_1^2x_2^2-2x_2^2x_3^2-2x_1^2x_3^2$$. Thanks a lot for your help. – Raducu Mihai Mar 9 '16 at 22:35
• The lexicographically largest term is $x_1^4$. So according to the algorithm described you should subtract off $e_1^4 = (x_1+x_2+x_3)^4$. The point being that what remains, while having more terms, will have a smaller leading term namely $-4x_1^3x_2$. Then we apply the same operation again to this polynomial $P - e_1^4$, this time adding $4e_2e_1^2$ to cancel the leading term. Next we would cancel the leading term $P - e_1^4 + 4e_2e_1^2$, and so on. Eventually we are left with $P - g(e_1,e_2,e_3)$ having no leading term and hence $P - g(e_1,e_2,e_3) = 0$ as desired. – Nate Mar 10 '16 at 15:14
You can use divided difference operators to express the polynomial in terms of Schur polynomials, then you can apply the second Jacobi-Trudi formula, which expresses a Schur polynomial as the determinant of a matrix whose entries are elementary symmetric polynomials. Computationally this is rather inefficient, but it is nonrecursive, which is sometimes advantageous.
• Hey, thank you for your answare. Can you please show me how this would work on a real example. See the polynomial in the comment section for @Nate 's answare. – Raducu Mihai Mar 10 '16 at 15:10
• @Raducu I don't have a computer algebra system here, but my hand calculations resulted in $\partial^{4123}(p)=3$, $\partial^{5423}(p)=-1$, $\partial^{3423}(p)=-2$, so $p=3s_{(31)}-s_{(211)}-2s_{(22)}$, and you can go to the Wikipedia article on the Jacobi-Trudi formula and substitute what it gives you. – Matt Samuel Mar 10 '16 at 16:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8914452791213989, "perplexity": 170.7191693871135}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317516.88/warc/CC-MAIN-20190822215308-20190823001308-00308.warc.gz"} |
http://math.stackexchange.com/questions/160626/prove-that-g-abelian-if-g-pq2 | # Prove that $G$ abelian if $|G|= pq^2$.
Let $G$ be a group of order $pq^2$, where $p \neq q$ prime and $p$ does not divide $| Aut (G) |$. Show that $G$ is abelian.
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If this is homework, have you at least tried something? Can you show some work? Note that this is proved in some algebra books exercises ; I know that in Dummit & Foote's Abstract Algebra, there's a sketch of proof in the exercises. – Patrick Da Silva Jun 20 '12 at 4:40
It would be nice if you didn't phrase your question as a command, if you told us what you know about the problem, what theorems you are aware of that might be relevant, what work you have put into it yourself, etc., etc., etc. – Gerry Myerson Jun 20 '12 at 4:40
Isn't it important to know that $p$>$q$ or $q$>$p$ in this problem? – Babak S. Jun 20 '12 at 6:23
Firstly,we can consider a homomorphism $f:G\rightarrow Aut(G)$ such that: $f(x)=t_x$,where $t_x:G\rightarrow G$ is defined by $t_x(g)=xgx^{-1}$.
Note that $ker f=Z(G)$,then by first isomorphism theorm, we get: $G/Z(G)\cong f(G) \le Aut(G)$.So $|G|$can be divided by $|Z(G)||Aut(G)|$.
Hence, $Z(G)$ is divided by $p$.By Cauchy Theorem,$Z(G)$ contains a element of order $p$. We find that the cyclic subgroup generated by that element is of order $p$ and hence is Sylow $p$ subgroup and since it is in the center, we can conclude that it is the unique Sylow $p$ subgroup,which is also normal.
Let $Q$ be a Sylow $q$ subgroup in $G$.Since $p$ and $q$ are relative prime,we have $P\cap Q=\{1\}$.Besides, $PQ$ is a subgroup of order $pq^2$ since $P$ is normal and $|PQ|=|P||Q|/|P\cap Q|$.We have $PQ=G$.
Now, we only focus on $Q$.From class equation ,we get that $Z(Q)$ is nontrivial.If $Z(Q)$ is of order $q^2$,then $Q$ commutes with its elements, otherwise $|Z(Q)|=q$. In this case,$Q/Z(Q)$ which is of order $q$ and hence is cyclic, so $Q$ is abelian. In both case, $Q$ always commutes with its elements.
Since $G=PQ$,$P$ is contained in the centre and $Q$ commutes with its elements,we can conclude that $G$ is abelian.
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I don't have that reference noted by Patrick, but I think you can use the following hint.
Hint: Use the fact that $|\frac{G}{Z(G)}|$ divides the order of $Aut(G)$.
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So $|G/Z(G)|$ is not divisible by p which implies $|Z(G)|$ is $p, pq$ or $pq^2$, the case $Z(G)=pq$ is ruled out as otherwise $G/Z(G)$ is cyclic and thus $Z(G)=G$. It remains to rule out the case $Z(G)=p$ ? – user31899 Jun 20 '12 at 8:29
@user31899: If $Z(G)=p$ then $G≅Z(G)×\mathbb Z_{q^2}$? – Babak S. Jun 20 '12 at 8:39
Good work, nice hint! +1 – amWhy Mar 6 '13 at 0:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9689511656761169, "perplexity": 226.40655081031971}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065464.19/warc/CC-MAIN-20150827025425-00272-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/92190/formula-of-sum-the-nimbers?answertab=oldest | # formula of sum the nimbers
If have well-known formula $(n + 1) n / 2 = 1 + 2 +\cdots+ n$. If the difference between the closest numbers smaller, will obtain, for example $(n + 0,1) n / (2 \cdot 0,1) = 0,1 + 0,2 +\cdots+ n$. Now if the difference between the closest numbers the smallest possible, will obtain $(n + 0,0\ldots1) n / (2 \cdot 0,0\ldots 1) = 0,0\ldots1 + 0,0\ldots2 + \ldots + n$, so can conclude $n ^ 2 / 2 = (0,0\ldots1 + 0,0\ldots2 + \cdots + n) / 0,0\ldots1$ whether conclude is correct?
EDITED VERSION:
If have well-known formula $\frac{(n + 1)n}2 = 1 + 2 +\dots+ n$.
If the difference between the closest numbers smaller, will obtain, for example $\frac{(n + 0,1) n}{2.0,1} = 0,1 + 0,2 +\dots + n$.
Now if the difference between the closest numbers the smallest possible, will obtain $\frac{(n + 0,0\dots1) n}{2 . 0,0\dots 1} = 0,0\dots 1 + 0,0\dots 2 + \dots + n$ , so can conclude $\frac{n ^ 2}2 = \frac{0,0\dots1 + 0,0\dots2 + \dots + n}{0,0\dots1}$ whether conclude is correct?
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By comma symbol, you mean decimal point? – user18325 Dec 17 '11 at 7:40
@ZeeshanMahmud Comma is used as decimal separator in many countries. – Martin Sleziak Dec 17 '11 at 7:48
Marko: I guess both these account belong to you: math.stackexchange.com/users/20189/marko and math.stackexchange.com/users/21380/marko; Maybe it would be good for you to register, so that you can better follow all questions you posted. (After you do it, you can even ask moderators to merge you account with the older unregistered ones.) – Martin Sleziak Dec 17 '11 at 7:52
shiver me nimbers – jspecter Dec 17 '11 at 8:00
Marko: I've tried to edit your question using TeX for better readability. (And maybe some other users will improve it a little more.) You should check whether I did not change the meaning of your question, unintentionally. If you're satisfied with the edited version, you can remove the original one. – Martin Sleziak Dec 17 '11 at 8:07
If we denote $k=n/\alpha$ then we get $$\sum\limits_{i=1}^k i\alpha=\frac{k(k+1)}2 \alpha = \frac{k\alpha(k\alpha+\alpha)}{2\alpha} = \frac{n(n+\alpha)}{2\alpha}.$$ For $\alpha$ of the form $10^{-s}$, i.e. $0,00\dots01$, this is precisely the first part of your question.
I am not sure I understand what you mean by the last part of your question, but if you want to say that $\sum\limits_{i=1}^k i\alpha=\frac{k(k+1)}2$ is approximately $\frac{n^2}{2\alpha}$, in some sense it is true, since the error is: $$\frac{n(n+\alpha)}{2\alpha} - \frac{n^2}{2\alpha} = \frac {n\alpha}{2\alpha}=\frac n2,$$ so the error has smaller order than the sum. (The error is $o(n^2)$, if you're familiar with this notation.)
If you know something about Riemann integral, you may notice that the sum $$\alpha \sum\limits_{i=1}^k i\alpha$$ is in fact upper Riemann sum for the function $f(x)=x$ on the interval $[0,n]$ (and for uniform partition of this interval). To see this just notice that the lengths of the intervals of partitions is $\alpha$ and the value at the end of the interval $[(i-1)\alpha,i\alpha]$ is $i\alpha$.
So it is expected that this sum is approximately $\frac{n^2}2$ for small values of $\alpha$. Your sum is $\alpha$-times smaller. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9662213325500488, "perplexity": 325.979333289402}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999653645/warc/CC-MAIN-20140305060733-00011-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/215297/suppose-i-said-x-spans-w | # Suppose I said “$X$ spans $W$”…
So I've seen two definitions of this:
Let $V$ be a vector space with subspace $W$. We say that $X \subseteq V$ spans $W$ if and only if
(Definition 1): Every $\vec{w} \in W$ can be written as a linear combination of vectors in $X$.
(Definition 2): $span(X) = W$.
I don't think the two definitions are equivalent are they? Clearly the condition in Def. 2 implies the condition in Def. 1, but not the other way round.
For example take $V$ to be the vector space of ordered pairs over $\mathbb{Z}_2$, then if $W = \{(0,0),(1,1)\}$ and $X=\{(0,1),(1,0)\}$, $X$ spans $W$ according to the first definition, but not the second.
What then would you have understood if I were to tell you that "$X$ spans $W$"? Would you have agreed or disagreed with me?
I know the usual drill is "clarify your definitions from the start" in whatever work you're doing, and that as long as you do that it will be fine, but I'm just wanting to know what peoples' immediate interpretation would be. Because it does impact a little on the reading and understanding process.
(For the record I'm a fan of Definition 1.)
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One usually understands $X$ spans $W$ to include the condition that $X$ is a subset of $W$. – Gerry Myerson Oct 17 '12 at 1:22
I would vote for definition 2. – Bitwise Oct 17 '12 at 1:57
My feeling for this terminology is that the phrase "$X$ spans $W$" should be consistent with the phrase "the span of $X$ is $W$". It is standard to define "the span" of a set, and so the first definition, which implies that $X$ could span distinct subspaces, is inconsistent with "the span".
Probably a good definition is: if $V$ is a vector space and $X\subseteq V$ is a set then we say that $X$ spans $V$ if every $v\in V$ can be written as a linear combination of elements of $X$. This way you have to refer to a set of vectors within a vector space.
So in this case to say that $X$ spans $W$ only makes sense in the case that $X\subseteq W$. It is as if $V$ ceased to exist for a brief moment of mathematical thought. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.96741783618927, "perplexity": 98.54204476529746}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928780.77/warc/CC-MAIN-20150521113208-00041-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.theochem.ru.nl/~pwormer/Knowino/knowino.org/wiki/Rotation_matrix.html | # Rotation matrix
In mathematics and physics a rotation matrix is synonymous with a 3×3 orthogonal matrix, which is a real 3×3 matrix R satisfying
$\mathbf{R}^\mathrm{T} = \mathbf{R}^{-1},$
where T stands for the transposed matrix and R−1 is the inverse of R.
## Connection of an orthogonal matrix to a rotation
In general a motion of a rigid body (which is equivalent to an angle and distance preserving transformation of affine space) can be described as a translation of the body followed by a rotation. By a translation all points of the body are displaced, while under a rotation at least one point of the body stays in place. Let the the fixed point be O. By Euler's theorem follows that then not only the point is fixed but also an axis—the rotation axis— through the fixed point. Write $\hat{n}$ for the unit vector along the rotation axis and φ for the angle over which the body is rotated, then the rotation operator on ℝ3 is written as ℛ(φ, ).
Erect three Cartesian coordinate axes with the origin in the fixed point O and take unit vectors $\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z$ along the axes, then the 3×3 rotation matrix $\mathbf{R}(\varphi, \hat{n})$ is defined by its elements $R_{ji}(\varphi, \hat{n})$ :
$\mathcal{R}(\varphi, \hat{n})(\hat{e}_i) = \sum_{j=x,y,x} \hat{e}_j R_{ji}(\varphi, \hat{n}) \quad\hbox{for}\quad i=x,y,z.$
In a more condensed notation this equation can be written as
$\mathcal{R}(\varphi, \hat{n})\left(\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z\right) = \left(\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z\right) \; \mathbf{R}(\varphi, \hat{n}).$
Given a basis of the linear space ℝ3, the association between a linear map and its matrix is one-to-one.
A rotation $\mathcal{R}$ (for convenience sake the rotation axis and angle are suppressed in the notation) leaves the shape of a rotated rigid body intact, so that all distances within the body are invariant. If the body is 3-dimensional—it contains three linearly independent vectors with origins in the invariant point—it holds that for any pair of vectors $\vec{a}$ and $\vec{b}$ in ℝ3 the inner product is invariant, that is,
$\left(\mathcal{R}(\vec{a}),\;\mathcal{R}(\vec{b}) \right) = \left(\vec{a},\;\vec{b}\right).$
A linear map with this property is called orthogonal. It is easily shown that a similar vector-matrix relation holds. First we define column vectors (stacked triplets of real numbers given in bold face):
$\vec{a} =\left(\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z\right)\begin{pmatrix}a_x\\a_y\\a_z\end{pmatrix} \;\stackrel{\mathrm{def}}{=}\; \left(\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z\right) \mathbf{a} \quad\hbox{and}\quad \vec{b} =\left(\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z\right)\begin{pmatrix}b_x\\b_y\\b_z\end{pmatrix} \;\stackrel{\mathrm{def}}{=}\; \left(\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z\right) \mathbf{b}$
and observe that the inner product becomes by virtue of the orthonormality of the basis vectors
$\left( \vec{a},\; \vec{b} \right) = \mathbf{a}^\mathrm{T} \mathbf{b}\equiv \left(a_x,\;a_y,\;a_z\right) \begin{pmatrix}b_x\\b_y\\b_z\end{pmatrix} \equiv a_xb_x+a_yb_y+a_zb_z.$
The invariance of the inner product under the rotation operator $\mathcal{R}$ leads to
$\mathbf{a}^\mathrm{T}\; \mathbf{b} =\big(\mathbf{R}\mathbf{a}\big)^\mathrm{T}\; \mathbf{R}\mathbf{b} = \mathbf{a}^\mathrm{T} \mathbf{R}^\mathrm{T}\; \mathbf{R}\mathbf{b},$
since this holds for any pair a and b it follows that a rotation matrix satisfies
$\mathbf{R}^\mathrm{T} \mathbf{R} = \mathbf{E},$
where E is the 3×3 identity matrix. For finite-dimensional matrices one shows easily
$\mathbf{R}^\mathrm{T} \mathbf{R} = \mathbf{E} \quad \Longleftrightarrow\quad\mathbf{R}\mathbf{R}^\mathrm{T} = \mathbf{E}.$
A matrix with this property is called orthogonal. So, a rotation gives rise to a unique orthogonal matrix.
Conversely, consider an arbitrary point P in the body and let the vector $\overrightarrow{OP}$ connect the fixed point O with P. Expressing this vector with respect to a Cartesian frame in O gives the column vector p (three stacked real numbers). Multiply p by the orthogonal matrix R, then p′ = Rp represents the rotated point P′ (or, more precisely, the vector $\overrightarrow{OP'}$ is represented by column vector p′ with respect to the same Cartesian frame). If we map all points P of the body by the same matrix R in this manner, we have rotated the body. Thus, an orthogonal matrix leads to a unique rotation.
Note that the Cartesian frame is fixed here and that points of the body are rotated, this is known as an active rotation. Instead, the rigid body could have been left invariant and the Cartesian frame could have been rotated, this also leads to new column vectors of the form p′ ≡ Rp, such rotations are referred to as passive.
## Properties of an orthogonal matrix
Writing out matrix products it follows that both the rows and the columns of the matrix are orthonormal (normalized and orthogonal). Indeed,
\begin{align} \mathbf{R}^\mathrm{T} \mathbf{R} &= \mathbf{E} \quad\Longleftrightarrow\quad \sum_{k=1}^{3} R_{ki}\, R_{kj} =\delta_{ij} \quad\hbox{(columns)} \\ \mathbf{R} \mathbf{R}^\mathrm{T} &= \mathbf{E} \quad\Longleftrightarrow\quad \sum_{k=1}^{3} R_{ik}\, R_{jk} =\delta_{ij} \quad\hbox{(rows)} \\ \end{align}
where δij is the Kronecker delta.
Orthogonal matrices come in two flavors: proper (det = 1) and improper (det = −1) rotations. Indeed, invoking some properties of determinants, one can prove
$1=\det(\mathbf{E})=\det(\mathbf{R}^\mathrm{T}\mathbf{R}) = \det(\mathbf{R}^\mathrm{T})\det(\mathbf{R}) = \det(\mathbf{R})^2 \quad\Longrightarrow \quad \det(\mathbf{R}) = \pm 1.$
### Compact notation
A compact way of presenting the same results is the following. Designate the columns of R by r1, r2, r3, i.e.,
$\mathbf{R} = \left(\mathbf{r}_1,\, \mathbf{r}_2,\, \mathbf{r}_3 \right)$.
The matrix R is orthogonal if
$\mathbf{r}_i^\mathrm{T} \mathbf{r}_j \equiv \mathbf{r}_i \cdot \mathbf{r}_j = \delta_{ij}, \quad i,j = 1,2,3 .$
The matrix R is a proper rotation matrix, if it is orthogonal and if r1, r2, r3 form a right-handed set, i.e.,
$\mathbf{r}_i \times \mathbf{r}_j = \sum_{k=1}^3 \, \varepsilon_{ijk} \mathbf{r}_k .$
Here the symbol × indicates a cross product and $\varepsilon_{ijk}$ is the antisymmetric Levi-Civita symbol,
\begin{align} \varepsilon_{123} =&\; \varepsilon_{312} = \varepsilon_{231} = 1 \\ \varepsilon_{213} =&\; \varepsilon_{321} = \varepsilon_{132} = -1 \end{align}
and $\varepsilon_{ijk} = 0$ if two or more indices are equal.
The matrix R is an improper rotation matrix if its column vectors form a left-handed set, i.e.,
$\mathbf{r}_i \times \mathbf{r}_j = - \sum_{k=1}^3 \, \varepsilon_{ijk} \mathbf{r}_k \; .$
The last two equations can be condensed into one equation
$\mathbf{r}_i \times \mathbf{r}_j = \det(\mathbf{R}) \sum_{k=1}^3 \; \varepsilon_{ijk} \mathbf{r}_k$
by virtue of the the fact that the determinant of a proper rotation matrix is 1 and of an improper rotation −1. This was proved above, an alternative proof is the following: The determinant of a 3×3 matrix with column vectors a, b, and c can be written as scalar triple product
$\det\left(\mathbf{a},\,\mathbf{b},\, \mathbf{c}\right) = \mathbf{a} \cdot (\mathbf{b}\times\mathbf{c})$.
It was just shown that for a proper rotation the columns of R are orthonormal and satisfy,
$\mathbf{r}_1 \cdot (\mathbf{r}_2 \times \mathbf{r}_3 ) = \mathbf{r}_1 \cdot\left(\sum_{k=1}^3 \, \varepsilon_{23k} \, \mathbf{r}_k \right) = \varepsilon_{231} = 1 .$
Likewise the determinant is −1 for an improper rotation.
## Explicit expression of rotation operator
Rotation of vector $\scriptstyle \vec{r}$ around over φ.
(i) The red vectors and black axis are in the plane of the screen.
(ii) The blue vectors are obtained by rotation into the screen.
(iii) The green cross product in x-y plane is perpendicular to the screen, pointing away from the reader.
Let $\overrightarrow{OP} \equiv \vec{r}$ be a vector pointing from the fixed point O of a rigid body to an arbitrary point P of the body. A rotation of this arbitrary vector around the unit vector over an angle φ can be written as
\begin{align} \mathcal{R}(\varphi, \hat{n})(\vec{r}\,)&= \vec{r}\,' = \\ & \vec{r}\;\cos\varphi + \hat{n} (\hat{n}\cdot\vec{r}\,)\; (1- \cos\varphi) + (\hat{n} \times \vec{r}\,) \sin\varphi . \\ \end{align}
where • indicates an inner product and the symbol × a cross product.
It is easy to derive this result. Indeed, decompose the vector to be rotated into two components, one along the rotation axis and one perpendicular to it (see the figure on the right)
$\vec{r} = \vec{r}_\parallel + \vec{r}_\perp\quad\hbox{with}\quad \vec{r}_\parallel = \hat{n} (\hat{n}\cdot\vec{r}\,),$
and
$\vec{r}_\perp = \vec{r} - \vec{r}_\parallel =\vec{r}- \hat{n} (\hat{n}\cdot\vec{r}\,),$
so that, in view of || = 1,
$|\vec{r}_\perp|^2 = r^2 - (\hat{n}\cdot\vec{r}\,)^2$
and from |a×b|2 = a2b2 − (ab)2 follows
$|\hat{n} \times \vec{r}\,|^2 = r^2 - (\hat{n}\cdot\vec{r}\,)^2.$
Upon rotation, the component of the rotated vector along the rotation axis is invariant, only its component orthogonal to the rotation axis changes. By virtue of the following fact (in words: length of basis vector along x-axis is length of basis vector along y-axis):
$|\vec{r}_\perp| = |\hat{n} \times \vec{r}\,| = \sqrt{|\vec{r}\,|^2 - (\hat{n}\cdot\vec{r}\,)^2},$
the rotation property simply is
$\vec{r}_\perp \mapsto \vec{r}\,'_\perp = \cos\varphi\;\vec{r}_\perp + \sin\varphi\;(\hat{n}\times\vec{r}\,).$
Hence
$\vec{r}\,' = \hat{n} (\hat{n}\cdot\vec{r}\,) + \left[\vec{r}- (\hat{n}\cdot\vec{r}\,) \hat{n}\right]\cos\varphi+(\hat{n}\times\vec{r}\,)\sin\varphi.$
Some reshuffling of the terms gives the required result.
## Explicit expression of rotation matrix
It will be shown that
$\mathbf{R}(\varphi, \hat{n}) = \mathbf{E} + \sin\varphi \mathbf{N} +(1-\cos\varphi)\mathbf{N}^2,$
where (see cross product for more details)
$\mathbf{N} \equiv \begin{pmatrix} 0 & -n_z & n_y \\ n_z& 0 & -n_x \\ -n_y& n_x & 0 \end{pmatrix}\quad\hbox{and}\quad \hat{n} \equiv (\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z\,) \begin{pmatrix} n_x \\ n_y \\ n_z \\ \end{pmatrix} \equiv (\hat{e}_x,\;\hat{e}_y,\;\hat{e}_z\,) \; \hat{\mathbf{n}}$
Note further that the dyadic product satisfies (as can be shown by squaring N),
$\hat{\mathbf{n}}\otimes\hat{\mathbf{n}} = \mathbf{N}^2 + \mathbf{E}, \quad\hbox{with}\quad |\hat{\mathbf{n}}| = 1$
and that
$\hat{n} (\hat{n}\cdot\vec{r}\,) \leftrightarrow (\hat{\mathbf{n}}\otimes\hat{\mathbf{n}})\;\mathbf{r} = \left(\mathbf{N}^2 + \mathbf{E}\right) \mathbf{r}.$
Translating the result of the previous section to coordinate vectors and substituting these results gives
\begin{align} \mathbf{r}' &= \left[\cos\varphi\; \mathbf{E} +(1-\cos\varphi)(\mathbf{N}^2 + \mathbf{E}) + \sin\varphi \mathbf{N} \right] \mathbf{r} \\ &= \left[\mathbf{E} + (1-\cos\varphi) \mathbf{N}^2 + \sin\varphi \mathbf{N}\right] \mathbf{r}, \qquad\qquad\qquad\qquad\qquad(1) \end{align}
which gives the desired result for the rotation matrix. This equation is identical to Eq. (2.6) of Biedenharn and Louck[1], who give credit to Leonhard Euler (ca. 1770) for this result.
For the special case φ = π (rotation over 180°), equation (1) can be simplified to
$\mathbf{R}(\pi, \hat{n}) = \mathbf{E} + 2 \mathbf{N}^2 = -\mathbf{E} +2\, \hat{\mathbf{n}}\otimes\hat{\mathbf{n}}$
so that
$\mathbf{R}(\pi, \hat{n})\, \mathbf{r} = -\mathbf{r} +2\; \hat{\mathbf{n}}\; (\hat{\mathbf{n}} \cdot\mathbf{r}),$
which sometimes[2] is referred to as reflection over a line, although it is not a reflection.
Equation (1) is a special case of the transformation properties given on p. 7 of the classical work (first edition 1904) of Whittaker[3]. To see the correspondence with Whittaker's formula, which includes also a translation over a displacement d and who rotates the vector (xa, yb, zc), we must put equal to zero: a, b, c, and d in Whittaker's equation. Furthermore Whittaker uses that the components of the unit vector n are the direction cosines of the rotation axis:
$n_x \equiv \cos\alpha,\quad n_y \equiv\cos\beta,\quad n_z\equiv\cos\gamma.$
Under these conditions the rotation becomes
\begin{align} x' &= x-(1-\cos\varphi)\left[x\;\sin^2\alpha - y\;\cos\alpha\cos\beta - z\;\cos\alpha\cos\gamma\right] +\left[z\;\cos\beta- y\;\cos\gamma\right]\sin\varphi \\ y' &= y-(1-\cos\varphi)\left[y\;\sin^2\beta - z\;\cos\beta\cos\gamma - x\;\cos\beta\cos\alpha\right] +\left[x\;\cos\gamma- z\;\cos\alpha\right]\sin\varphi \\ z' &= z-(1-\cos\varphi)\left[z\;\sin^2\gamma - x\;\cos\gamma\cos\alpha - y\;\cos\gamma\cos\beta\right] +\left[y\;\cos\alpha- x\;\cos\beta\right]\sin\varphi \\ \end{align}
## Vector rotation
Sometimes it is necessary to rotate a rigid body so that a given vector in the body (for instance an oriented chemical bond in a molecule) is lined-up with an external vector (for instance a vector along a coordinate axis).
Let the vector in the body be f (the "from" vector) and the vector to which f must be rotated be t (the "to" vector). Let both vectors be unit vectors (have length 1). From the definition of the inner product and the cross product follows that
$\mathbf{f} \cdot \mathbf{t} = \cos\varphi,\quad |\mathbf{f}\times\mathbf{t}| = \sin\varphi,$
where φ is the angle (< 180°) between the two vectors. Since the cross product f × t is a vector perpendicular to the plane of both vectors, it is easily seen that a rotation around the cross product vector over an angle φ moves f to t.
Write
$\mathbf{u} = \mathbf{f}\times\mathbf{t}\quad \Longrightarrow\quad \mathbf{u} = \sin\varphi\hat{\mathbf{n}},$
where $\hat{\mathbf{n}}$ is a unit vector. Define accordingly
$\mathbf{U} \; \stackrel{\mathrm{def}}{=} \; \sin\varphi \mathbf{N}= \sin\varphi \begin{pmatrix} 0 & -n_z & n_y \\ n_z& 0 & -n_x \\ -n_y& n_x & 0 \end{pmatrix}= \begin{pmatrix} 0 & -u_z & u_y \\ u_z& 0 & -u_x \\ -u_y& u_x & 0 \end{pmatrix} ,$
so that
\begin{align} \mathbf{R}(\varphi, \hat{n})&= \mathbf{E} + \mathbf{U} + \frac{1-\cos\varphi}{\sin^2\varphi} \mathbf{U}^2 \\ &= \mathbf{E} + \mathbf{U} + \frac{1}{1+\cos\varphi} \mathbf{U}^2 \qquad\qquad\qquad\qquad\qquad\qquad\qquad(2)\\ &= \cos\varphi\,\mathbf{E}+ \mathbf{U} + \frac{1}{1+\cos\varphi} \mathbf{u}\otimes\mathbf{u}\\ \end{align}
The last equation, which follows from U2 = uu − sin2φ E, is Eq. (3) of Ref. [4] after substitution of
$\frac{1-\cos\varphi}{\sin^2\varphi} =\frac{1-\cos\varphi}{1-\cos^2\varphi} = \frac{1}{1+\cos\varphi} .$
Indeed, write the rotation matrix in full, using the short-hand notations c = cos φ and h = (1-c)/(1-c2)
$\mathbf{R}(\varphi, \hat{n}) = \begin{pmatrix} c+hu_x^2 & h u_xu_y - u_z & h u_x u_z + u_y \\ h u_x u_y + u_z & c + h u_y^2 & h u_y u_z - u_x \\ h u_x u_z -u_y & h u_y u_z + u_x & c + h u_z^2 \\ \end{pmatrix},$
which is the matrix in Eq. (3) of Ref. [4]. If the vectors are parallel, f = t, then u = 0, U = 0 and ft = cosφ = 1, so that the equation is well-defined—it gives R(φ, n) = E, as expected. Provided f and t are not anti-parallel, the matrix $\mathbf{R}(\varphi, \hat{n})$ can be computed easily and quickly. It requires not much more than the computation of the inner and cross product of f and t.
### Case that "from" and "to" vectors are anti-parallel
If the vectors f and t are nearly anti-parallel, f ≈ − t, then ft = cosφ ≈ − 1, and the denominator in Eq. (2) becomes zero, so that Eq. (2) is not applicable. The vector defining the rotation axis is nearly zero: uf × tf × (−f) ≈ 0.
Clearly, a two-fold rotation (angle 180°) around any rotation axis perpendicular to f will map f onto −f. This freedom in choice of the two-fold axis is consistent with the indeterminacy in the equation that occurs for cosφ = −1.
Since −E (inversion) sends f to −f, one could naively assume that all position vectors of the rigid body may be inverted by −E. However, if the rigid body is not symmetric under inversion (does not have a symmetry center), inversion turns the body into a non-equivalent one. For instance, if the body is a right-hand glove, inversion sends it into a left-hand glove; it is well-known that a right-hand and a left-hand glove are different.
Recall that f is a unit vector. If f is on the z-axis, |fz| = 1, then the following rotation around the y-axis turns f into −f:
$\mathbf{R}(\pi, \hat{e}_{y}) \mathbf{f} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\\pm 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\\mp 1 \end{pmatrix}$
If f is not on the z-axis, fz ≠ 1, then the following orthogonal matrix sends f to −f and hence gives a 180° rotation of the rigid body
$\mathbf{R}(\pi, \hat{e}_{y'}) \mathbf{f} = \frac{1}{1-f^2_z} \begin{pmatrix} -(f^2_x-f^2_y) & -2f_xf_y & 0 \\ -2f_xf_y &(f^2_x-f^2_y) & 0 \\ 0 & 0 & -(1-f^2_z) \\ \end{pmatrix} \begin{pmatrix} f_x \\ f_y \\ f_z \end{pmatrix} = \begin{pmatrix} -f_x \\ -f_y \\ -f_z \end{pmatrix}$
To show this we insert
$\hat{\mathbf{n}} = \frac{1}{1-f^2_z} \begin{pmatrix} f_y \\-f_x \\0 \end{pmatrix}$
into the expression for rotation over 180° introduced earlier:
$\mathbf{R}(\pi, \hat{e}_{y'}) = -\mathbf{E} +2 \hat{\mathbf{n}}\otimes \hat{\mathbf{n}} .$
Note that the vector $\hat{\mathbf{n}}$ is normalized, because f is normalized, and that it lies in the x-y plane normal to the plane spanned by the z-axis and f; it is in fact a unit vector along a rotated y-axis obtained by rotation around the z-axis.
Given f = (fx, fy, fz), the matrix $\mathbf{R}(\pi, \hat{e}_{y'})$ is easily and quickly calculated. The matrix rotates the body over 180° sending f to −f exactly. When −f and t are close, but not exactly equal, the earlier rotation formula Eq. (2), may be used to perform the final (small) rotation of the body that lets −f and t coincide exactly.
## Change of rotation axis
Euler's theorem states that a rotation is characterized by a unique axis (an eigenvector with unit eigenvalue) and a unique angle. As a corollary of the theorem follows that the trace of a rotation matrix is equal to
$\mathrm{Tr}[\mathbf{R}(\varphi, \hat{n})] = 2\cos\varphi +1,$
and hence the trace depends only on the rotation angle and is independent of the axis. Let A be an orthogonal 3×3 matrix (not equal to E), then, because cyclic permutation of matrices under the trace is allowed (i.e., leaves the trace invariant),
$\mathrm{Tr}[\mathbf{A} \mathbf{R}(\varphi, \hat{n}) \mathbf{A}^\mathrm{T}] = \mathrm{Tr}[ \mathbf{R}(\varphi, \hat{n}) \mathbf{A}^\mathrm{T}\mathbf{A}] = \mathrm{Tr}[\mathbf{R}(\varphi, \hat{n})] .$
As a consequence it follows that the two matrices
$\mathbf{R}(\varphi, \hat{n})\quad \hbox{and}\quad \mathbf{A} \mathbf{R}(\varphi, \hat{n}) \mathbf{A}^\mathrm{T}$
describe a rotation over the same angle φ but around different axes. The eigenvalue equation can be transformed
$\mathbf{R}(\varphi, \hat{n})\;\hat{n} = \hat{n} \;\Longrightarrow\; \left(\mathbf{A}\,\mathbf{R}(\varphi, \hat{n})\,\mathbf{A}^\mathrm{T}\right) \;\mathbf{A}\hat{n} = \mathbf{A}\hat{n}$
which shows that the transformed eigenvector is the rotation axis of the transformed matrix, or
$\mathbf{R}(\varphi, \mathbf{A}\hat{n})= \mathbf{A}\,\mathbf{R}(\varphi, \hat{n})\,\mathbf{A}^\mathrm{T} .$
## References
1. L. C. Biedenharn and J. D. Louck, Angular Momentum in Quantum Physics, Addison-Wesley, Reading, Mass. (1981) ISBN 0-201-13507-8
2. Wikipedia Retrieved July 20, 2009
3. E. T. Whittaker, A Treatise on the Dynamics of Particles and Rigid Bodies, Cambridge University Press (1965)
4. 4.0 4.1 T. Möller and J. F. Hughes, J. Graphics Tools, 4 pp. 1–4 (1999). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 69, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9915676116943359, "perplexity": 741.7744924385461}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500058.1/warc/CC-MAIN-20230203154140-20230203184140-00673.warc.gz"} |
http://mathhelpforum.com/algebra/67656-urgent-help-needed-solve-equation-print.html | # Urgent Help needed to solve equation.
• January 11th 2009, 02:49 AM
den
1 Attachment(s)
Urgent Help needed to solve equation.
Urgent Help needed to solve equation.
• January 11th 2009, 03:37 AM
Prove It
Quote:
Originally Posted by den
Urgent Help needed to solve equation.
Complete the square on the bottom.
• January 11th 2009, 04:02 AM
den
What I cannot figure out is how we get
0.46y2 + 1.84y + 1.84 = 2.87
• January 11th 2009, 04:06 AM
Prove It
Quote:
Originally Posted by den
What I cannot figure out is how we get
0.46y2 + 1.84y + 1.84 = 2.87
Are you trying to solve this? Just get everything on one side and use the Quadratic formula...
• January 11th 2009, 04:16 AM
den
When I multiply I get different figures than 0.46y2 + 1.84y + 1.84 = 2.87.
• January 11th 2009, 05:31 AM
running-gag
Hi
Double posting is forbidden on MHF
http://www.mathhelpforum.com/math-he...lp-needed.html
What is your problem exactly ?
• January 11th 2009, 05:41 AM
den
Missing line/calulation
What I cannot figure out is how we get
0.46y2 + 1.84y + 1.84 = 2.87
• January 11th 2009, 06:05 AM
running-gag
Starting from
$0.080 = 0.051 \,\frac{56.25}{5.76 y^2 + 23.04 y + 23.04}$
You just have to multiply 0.080 par the denominator
$0.080 \,(5.76 y^2 + 23.04 y + 23.04) = 0.051 \,\, 56.25 = 2.87$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8692230582237244, "perplexity": 4955.385457579164}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395679.92/warc/CC-MAIN-20160624154955-00174-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/finding-a-formula-for-a-sequence-with-recurring-digits.623732/ | # Finding a formula for a sequence with recurring digits
1. Jul 26, 2012
### spiritzavior
how do i determine the formula for the sequence below?
1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,...,n,n,n,n,n,...,n,...
need some instructions. thanks in advance
2. Jul 26, 2012
### Dickfore
Ok, there are subsequences of equal entries (increasing by one) in a row, and the numbers of equal entries in each of them form the following sequence:
2, 4, 5, 8, ...
I can't see any regularity. Sorry.
3. Jul 26, 2012
### haruspex
I can't see a pattern here: you have two 1s, four 2s, five 3s and eight 4s. Can you confirm that you've written it out correctly?
4. Jul 27, 2012
### ramsey2879
Seems like one too many 2's with n-1 repeated F(n+1) times for n = 2,3,4,5,... .
Last edited: Jul 27, 2012
5. Jul 28, 2012
### spiritzavior
sorry, i've typed it incorrectly, it should be:
1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,...
two 1s, four 2s, six 3s, eight 4s,...
6. Jul 28, 2012
### haruspex
If you don't mind the use of [] (integer part of) it's [1/2 +√(n-3/4)]
7. Jul 28, 2012
### Dickfore
So, there are $2 n$ n's. The total number of elements no smaller than n is:
$$\sum_{k = 1}^{n}{2 k} = n(n + 1)$$
Thus, all the elements with an index k that satisfies:
$$(n - 1) n < k \le n (n + 1)$$
are equal to $x_k = n$ (why?). Can you find n, given k from the above inequalities?
$$n^2 + n - k \ge 0 \Rightarrow n \ge -\frac{1}{2} + \sqrt{k + \frac{1}{4}}$$
$$n^2 - n - k < 0 \Rightarrow n < \frac{1}{2} + \sqrt{k + \frac{1}{4}}$$
If you know the [STRIKE]floor[/STRIKE] floor function, which is the same as integer part for positive integers, you should be able to deduce:
$$x_k = \mathrm{Ceiling} \left[ \sqrt{k + \frac{1}{4}} - \frac{1}{2}\right]$$
Last edited: Jul 28, 2012
8. Jul 28, 2012
### acabus
$a(n) = round(\sqrt{n})$ seems to work, not that I could prove it.
This is where a(1) is the first entry in the sequence.
9. Jul 28, 2012
### Anti-Crackpot
Good intuition abacus...
A000194 n appears 2n times; also nearest integer to square root of n
http://oeis.org/A000194
If one desires that digits are repeated 2n - 1 times, that's an easy one: floor [sqrt n]
10. Jul 28, 2012
### Dickfore
The definition of ceiling and round functions is:
$$\mathrm{Ceiling}(x) \equiv n, \ n \le x < n + 1$$
$$\mathrm{Round}(x) \equiv m, \ \vert x - m \vert < 1/2$$
(the last definition is ambiguous if the argument is half-integer, but there can never happen for acabus's argument)
We want to know for what x, and y, $\mathrm{Ceiling}(x) \stackrel{?}{=} \mathrm{Round}(y) = n$. From the above definitions, this happens when:
$$n \le x < n + 1, \ n - 1/2 < y < n + 1/2$$
or, if we rewrite these inequalities for n, we get:
$$x - 1 < n \le x, \ y - 1/2 < n < y + 1/2$$
Then, a necessary and sufficient (why?) condition is:
$$x - 1 < y + 1/2, \ y - 1/2 < x$$
In our case, $x = \sqrt{k + 1/4} - 1/2$, and $y= \sqrt{k}$, so we have:
$$\sqrt{k + 1/4} - 1/2 - 1 < \sqrt{k} + 1/2 \Leftrightarrow \sqrt{k + 1/4} - \sqrt{k} < 2$$
$$\sqrt{k} - 1/2 < \sqrt{k + 1/4} - 1/2 \Leftrightarrow \sqrt{k + 1/4} - \sqrt{k} > 0$$
The second inequality is surely true, but the first is proven as follows:
$$\begin{array}{l} \sqrt{k + 1/4} - \sqrt{k} \stackrel{?}{<} 2 \\ 0 < \sqrt{k + 1/4} \stackrel{?}{<} \sqrt{k} + 2 \\ k + 1/4 \stackrel{?}{<} k + 4 \sqrt{k} + 4 \\ 0 \stackrel{\surd}{<} \sqrt{k} + 15/16 \end{array}$$
This is surely true. Therefore, both the formulas give the same output for all positive integers.
11. Jul 28, 2012
### spiritzavior
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https://ftp.aimsciences.org/article/doi/10.3934/cpaa.2014.13.1337 | # American Institute of Mathematical Sciences
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On the blow-up criterion of smooth solutions for Hall-magnetohydrodynamics system with partial viscosity
May 2014, 13(3): 1337-1345. doi: 10.3934/cpaa.2014.13.1337
## Global existence of strong solutions to incompressible MHD
1 The Institute of Mathematical Sciences, University of Science and Technology of China, Anhui, 230026 2 The Institute of Mathematical Sciences, The Chinese University of Hong Kong
Received September 2013 Revised November 2013 Published December 2013
We establish the global existence and uniqueness of strong solutions to the initial boundary value problem for the incompressible MHD equations in bounded smooth domains of $\mathbb R^3$ under some suitable smallness conditions. The initial density is allowed to have vacuum, in particular, it can vanish in a set of positive Lebessgue measure. More precisely, under the assumption that the production of the quantities $\|\sqrt\rho_0u_0\|_{L^2(\Omega)}^2+\|H_0\|_{L^2(\Omega)}^2$ and $\|\nabla u_0\|_{L^2(\Omega)}^2+\|\nabla H_0\|_{L^2(\Omega)}^2$ is suitably small, with the smallness depending only on the bound of the initial density and the domain, we prove that there is a unique strong solution to the Dirichlet problem of the incompressible MHD system.
Citation: Huajun Gong, Jinkai Li. Global existence of strong solutions to incompressible MHD. Communications on Pure & Applied Analysis, 2014, 13 (3) : 1337-1345. doi: 10.3934/cpaa.2014.13.1337
##### References:
[1] A. G. Kulikovskiy and G. A. Lyubimov, Magnetohydrodynamics, Addison–Wesley, Reading, MA, 1965. Google Scholar [2] L. D. Landau and E. M. Lifchitz, Electrodynamics of Continuous Media, 2nd ed., Pergamon, New York, 1984. Google Scholar [3] M. Sermange and R. Temam, Some mathematical questions related to the MHD equations, Comm. Pure Appl. Math., 36 (1983), 635-664. doi: 10.1002/cpa.3160360506. Google Scholar [4] G. Duvaut and J. L. Lions, Inequations en thermoelasticite et magnetohydrodynamique, Ach.Rational Mech. Anal., 46 (1972), 241-279. Google Scholar [5] J. F. Gerbeau and C. Le Bris, Existence of solution for a density-dependent magnetohydrodynamic equation, Adv. Differential Equations, 2 (1997), 427-452. Google Scholar [6] P. L. Lions, Mathematical Topics in Fluid Mechanics. Vol. 1. Incompressible Models, Oxford Lecture Series in Mathematics and its Applications, 3. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1996. Google Scholar [7] P. L. Lions, Mathematical topics in fluid mechanics. Vol. 2. Compressible models, Oxford Lecture Series in Mathematics and its Applications, 10. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1998. Google Scholar [8] X. P. Hu and D. H. Wang, Global existence and large-time behavior of solutions to the three-dimensional equations of compressible Magnetohydrodynamic flows, Arch. Rational Mech. Anal., 197 (2010), 203-238. doi: 10.1007/s00205-010-0295-9. Google Scholar [9] X. P. Hu and D. H. Wang, Global solutions to the three-dimensional full compressible Magnetohydrodynamic flows, Commun. Math. Phys., 283 (2008), 255-284. doi: 10.1007/s00220-008-0497-2. Google Scholar [10] J. S. Fan and W. H. Yu, Global variational solutions to the compressible magnetohydrodynamic equations, Nonlinear Analysis, 69 (2008), 3637-3660. doi: 10.1016/j.na.2007.10.005. Google Scholar [11] E. Feireisl, Dynamics of viscous compressible fluids, Oxford Lecture Series in Mathematics and its Applications, 26. Oxford University Press, Oxford, 2004. Google Scholar [12] E. Feireisl, A. Novotný and H. Petzeltová, On the existence of globally defined weak solutions to the Navier-Stokes equations, J. Math. Fluid Mech., 3 (2001), 358-392. doi: 10.1007/PL00000976. Google Scholar [13] B. Ducomet and E. Feireisl, The equation of Magnetohydrodynamics: on the interaction between matter and ration in the evolution of gaseous stars, Commun. Math. Phys., 266 (2006), 595-629. doi: 10.1007/s00220-006-0052-y. Google Scholar [14] Q. Chen, Z. Tan and Y. J. Wang, Strong solutions to the incompressible magnetohydrodynamic equations, Math. Methods Appl. Sci., 34 (2011), 94-107. doi: 10.1002/mma.1338. Google Scholar [15] H. W. Wu, Strong solutions to the incompressible magnetohydrodynamic equations with vacuum, Comput. Math. Appl., 61 (2011), 2742-2753. doi: 10.1016/j.camwa.2011.03.033. Google Scholar [16] X. D. Huang and Y. Wang, Global strong solution to the 2D nonhomogeneous incompressible MHD system, J. Differential Equations, 254 (2013), 511-527. doi: 10.1016/j.jde.2012.08.029. Google Scholar [17] J. S. Fan and W. H. Yu, Strong solution to the compressible magnetohydrodynamic equations with vacuum, Nonlinear Anal. Real World Appl., 10 (2009), 392-409. doi: 10.1016/j.nonrwa.2007.10.001. Google Scholar [18] X. L. Li, N. Su, and D. H. Wang, Local strong solution to the compressible magnetohydrodynamic flow with large data, J. Hyperbolic Differ. Equ., 8 (2011), 415-436. doi: 10.1142/S0219891611002457. Google Scholar [19] X. L. Li and D. H. Wang, Global strong solution to the three-dimensional density-dependent incompressible magnetohydrodynamic flows, J. Differential Equations, 251 (2011), 1580-1615. doi: 10.1016/j.jde.2011.06.004. Google Scholar [20] W. Von Wahl, Estimating $\nabla u$ by $\text{div} u$ and $\text{curl}u$, Math. Methods Appl. Sci., 15 (1992), 123-143. doi: 10.1002/mma.1670150206. Google Scholar [21] Y. Zhou, Remarks on regularities for the 3D MHD equations, Discrete Contin. Dyn. Syst., 12 (2005), 881-886. doi: 10.3934/dcds.2005.12.881. Google Scholar [22] Y. Zhou, Regularity criteria for the 3D MHD equations in terms of the pressure, Internat. J. Non-Linear Mech., 41 (2006), 1174-1180. doi: 10.1016/j.ijnonlinmec.2006.12.001. Google Scholar [23] Y. Zhou, Regularity criteria for the generalized viscous MHD equations, Ann. Inst. H. Poincaré Anal. Non Linéaire, 24 (2007), 491-505. doi: 10.1016/j.anihpc.2006.03.014. Google Scholar
show all references
##### References:
[1] A. G. Kulikovskiy and G. A. Lyubimov, Magnetohydrodynamics, Addison–Wesley, Reading, MA, 1965. Google Scholar [2] L. D. Landau and E. M. Lifchitz, Electrodynamics of Continuous Media, 2nd ed., Pergamon, New York, 1984. Google Scholar [3] M. Sermange and R. Temam, Some mathematical questions related to the MHD equations, Comm. Pure Appl. Math., 36 (1983), 635-664. doi: 10.1002/cpa.3160360506. Google Scholar [4] G. Duvaut and J. L. Lions, Inequations en thermoelasticite et magnetohydrodynamique, Ach.Rational Mech. Anal., 46 (1972), 241-279. Google Scholar [5] J. F. Gerbeau and C. Le Bris, Existence of solution for a density-dependent magnetohydrodynamic equation, Adv. Differential Equations, 2 (1997), 427-452. Google Scholar [6] P. L. Lions, Mathematical Topics in Fluid Mechanics. Vol. 1. Incompressible Models, Oxford Lecture Series in Mathematics and its Applications, 3. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1996. Google Scholar [7] P. L. Lions, Mathematical topics in fluid mechanics. Vol. 2. Compressible models, Oxford Lecture Series in Mathematics and its Applications, 10. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1998. Google Scholar [8] X. P. Hu and D. H. Wang, Global existence and large-time behavior of solutions to the three-dimensional equations of compressible Magnetohydrodynamic flows, Arch. Rational Mech. Anal., 197 (2010), 203-238. doi: 10.1007/s00205-010-0295-9. Google Scholar [9] X. P. Hu and D. H. Wang, Global solutions to the three-dimensional full compressible Magnetohydrodynamic flows, Commun. Math. Phys., 283 (2008), 255-284. doi: 10.1007/s00220-008-0497-2. Google Scholar [10] J. S. Fan and W. H. Yu, Global variational solutions to the compressible magnetohydrodynamic equations, Nonlinear Analysis, 69 (2008), 3637-3660. doi: 10.1016/j.na.2007.10.005. Google Scholar [11] E. Feireisl, Dynamics of viscous compressible fluids, Oxford Lecture Series in Mathematics and its Applications, 26. Oxford University Press, Oxford, 2004. Google Scholar [12] E. Feireisl, A. Novotný and H. Petzeltová, On the existence of globally defined weak solutions to the Navier-Stokes equations, J. Math. Fluid Mech., 3 (2001), 358-392. doi: 10.1007/PL00000976. Google Scholar [13] B. Ducomet and E. Feireisl, The equation of Magnetohydrodynamics: on the interaction between matter and ration in the evolution of gaseous stars, Commun. Math. Phys., 266 (2006), 595-629. doi: 10.1007/s00220-006-0052-y. Google Scholar [14] Q. Chen, Z. Tan and Y. J. Wang, Strong solutions to the incompressible magnetohydrodynamic equations, Math. Methods Appl. Sci., 34 (2011), 94-107. doi: 10.1002/mma.1338. Google Scholar [15] H. W. Wu, Strong solutions to the incompressible magnetohydrodynamic equations with vacuum, Comput. Math. Appl., 61 (2011), 2742-2753. doi: 10.1016/j.camwa.2011.03.033. Google Scholar [16] X. D. Huang and Y. Wang, Global strong solution to the 2D nonhomogeneous incompressible MHD system, J. Differential Equations, 254 (2013), 511-527. doi: 10.1016/j.jde.2012.08.029. Google Scholar [17] J. S. Fan and W. H. Yu, Strong solution to the compressible magnetohydrodynamic equations with vacuum, Nonlinear Anal. Real World Appl., 10 (2009), 392-409. doi: 10.1016/j.nonrwa.2007.10.001. Google Scholar [18] X. L. Li, N. Su, and D. H. Wang, Local strong solution to the compressible magnetohydrodynamic flow with large data, J. Hyperbolic Differ. Equ., 8 (2011), 415-436. doi: 10.1142/S0219891611002457. Google Scholar [19] X. L. Li and D. H. Wang, Global strong solution to the three-dimensional density-dependent incompressible magnetohydrodynamic flows, J. Differential Equations, 251 (2011), 1580-1615. doi: 10.1016/j.jde.2011.06.004. Google Scholar [20] W. Von Wahl, Estimating $\nabla u$ by $\text{div} u$ and $\text{curl}u$, Math. Methods Appl. Sci., 15 (1992), 123-143. doi: 10.1002/mma.1670150206. Google Scholar [21] Y. Zhou, Remarks on regularities for the 3D MHD equations, Discrete Contin. Dyn. Syst., 12 (2005), 881-886. doi: 10.3934/dcds.2005.12.881. Google Scholar [22] Y. Zhou, Regularity criteria for the 3D MHD equations in terms of the pressure, Internat. J. Non-Linear Mech., 41 (2006), 1174-1180. doi: 10.1016/j.ijnonlinmec.2006.12.001. Google Scholar [23] Y. Zhou, Regularity criteria for the generalized viscous MHD equations, Ann. Inst. H. Poincaré Anal. Non Linéaire, 24 (2007), 491-505. doi: 10.1016/j.anihpc.2006.03.014. Google Scholar
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2020 Impact Factor: 1.916 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8311110734939575, "perplexity": 2526.8751958651437}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00498.warc.gz"} |
https://blog.jungsoo.kim/laser-population-inversion-2state/ | I reviewed some basic laser physics from a class. Laser requires a gain medium with state > 2. Here's a quick analysis behind why.
Following Einstein's conventions, 3 things can happen in the system: spontaneous emission, stimulated emission, and photon absorption.
Let's look into each component.
1.Spontaneous emission
$$\Bigg(\frac{dN_2}{dt}\Bigg)_{spontaneous} = -A_{21} N_{2}$$
$$\Bigg(\frac{dN_1}{dt}\Bigg)_{spontaneous} = A_{21} N_{2}$$
2.Stimulated emission
$$\Bigg(\frac{dN_1}{dt}\Bigg)_{- absorption} = B_{21} N_{2} \rho(v)$$
3.Photon absorption
$$\Bigg(\frac{dN_1}{dt}\Bigg)_{+ absorption} = -B_{12} N_{1} \rho(v)$$
Combining the above and since Einstein showed that $$B_{12} = B_{21}$$ because of thermodynamics,
$$\frac{dN_1}{dt} = A_{21} N_2 + B N_2 \rho(v) - B N_1 \rho(v)$$
$$\frac{dN_2}{dt} = - A_{21} N_1 + B N_1 \rho(v) - B N_2 \rho(v)$$
Simplifying,
$$\frac{dN_1}{dt} = A_{21} N_2 + \rho(v) B (N_2-N_1)$$
$$\frac{dN_2}{dt} = - A_{21} N_2 + \rho(v) B (N_1-N_2)$$
At steady state $$\frac{dN_1}{dt} = \frac{dN_2}{dt} = 0$$,
$$0 = 2 \rho(v) B (N_1 - N_2) - 2 A_{21} N_2$$
$$\frac{N_1 - N_2}{N_2} = \frac{A_{21}}{\rho(v)B}$$
$$\frac{N_1}{N_2} = \frac{A_{21}}{\rho(v) B} + 1$$
$$\frac{N_2}{N_1} = \frac{\rho(v) B}{A_{21} +\rho(v) B}$$
$$\frac{N_2 + N_2}{N_1 + N_2} = \frac{\rho(v) B}{A_{21} +\rho(v) B}$$
$$\frac{N_2}{N_1 + N_2} = \frac{\rho(v) B}{2(A_{21} +\rho(v) B)}$$
Now $$\frac{N_2}{N_1 + N_2}$$ is the fraction of state 2 out of total (state 1 + state 2). In order for the population inversion to happen, this ratio needs to be greater than 1/2.
Now let's take a limit where $$\frac{B}{A_{21}}\to\infty$$ i.e. pumping the gain medium:
$$\lim_{\frac{B}{A_{21}}\to\infty} \frac{\rho(v) B}{2(A_{21} +\rho(v) B)} = 1/2$$
The result of the limit shows that the maximum achievable ratio is 1/2, which means that population inversion is not theoretically possible in a 2-state gain medium. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9886326193809509, "perplexity": 1230.8100543913192}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703521139.30/warc/CC-MAIN-20210120151257-20210120181257-00371.warc.gz"} |
https://math.stackexchange.com/questions/3008406/show-that-a-given-measure-is-equal-to-the-lebesgue-measure-on-borel-subsets-on | # Show that a given measure is equal to the Lebesgue measure on Borel subsets on $\mathbb{R}$
Suppose we have a measure $$\mu$$ on $$\left( \mathbb{R},\mathcal{B}(\mathbb{R}) \right)$$ with $$\mu((0,1])=1$$, and $$\mu$$ invariant under translations i.e. $$\mu(A) = \mu(A+c)$$ for every $$c \in \mathbb{R}$$ and Borel set $$A \subseteq \mathbb{R}$$.
Now letting $$\lambda$$ be the Lebesgue measure, in the first part of the question I was able to show that for any interval $$A=(a,b]$$ with $$b-a \in \mathbb{Q}$$ that $$\mu(A) = \lambda(A)$$.
The second part of the question asks to extend this to all Borel sets, i.e. $$\forall A \in \mathcal{B}(\mathbb{R})$$ we get that$$\mu(A) = \lambda(A)$$, but I am really struggling to think of a practical way of achieving this.
Clearly using the first part of the question is the correct approach, so I was attempting to come up with some ways of using this.
The first thing I thought of was maybe to approximate all of the intervals in $$\mathcal{B}(\mathbb{R})$$ by these rational intervals in the first part of the question, but I could not think of a rigorous way of doing this and I do not think this is the correct approach.
My next thought was to potentially show that the intervals from the first part form a $$\pi$$-system of some sort and maybe generate the Borel sets and so agreeing only on these intervals is sufficient to show they agree on the whole space, but I am not too confident in arguing this so any help would be appreciated thanks.
Having $$\mu (a,b]=b-a$$ for rational $$a,b$$ gives $$\mu (a,b)=b-a$$ for all real $$a Proof: Write $$(a,b)$$ as the increasing union of $$(a_n,b_n]$$ for appropritate rational $$a_n,b_n.$$ Standard measure theory with your result for rationals then gives the result.
Since every open set in $$\mathbb R$$ is the disjoint union of open intervals, we see $$\mu(U) = \lambda (U)$$ for all open $$U\subset \mathbb R.$$
• Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work? – UsernameInvalid Nov 21 '18 at 22:09
• But if both $a_n,b_n$ are rational, you have it covered. – zhw. Nov 21 '18 at 22:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9722693562507629, "perplexity": 75.91484974852006}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526401.41/warc/CC-MAIN-20190720004131-20190720030131-00500.warc.gz"} |
https://www.physicsforums.com/threads/injective-proof.408789/ | # Homework Help: Injective proof
1. Jun 8, 2010
### rallycar18
Prove that:
If f : X → Y is injective, g, h : W → X, and f ∘ g = f ∘ h, then g = h.
2. Jun 8, 2010
### Tedjn
What is your progress on this problem?
3. Jun 8, 2010
### rallycar18
Definition of injective for If f : X → Y :
For all y $$\in$$ Y, there exists at most one x $$\in$$ X such that f(x) = y
Because f : X → Y and g, h : W → X,
f ∘ g : W → X → Y and f ∘ h : W → X → Y
so f ∘ g, f ∘ h : W → Y
that's where I get stuck.
4. Jun 8, 2010
### lanedance
note that an injective function is a function that preserves distinctness... so you can consider the inverse of f
5. Jun 8, 2010
### rallycar18
Then f^(-1) : Y → X is also injective.. but I don't see what I can do from that.
6. Jun 8, 2010
### Tedjn
It may be easier to see via contradiction. If there is w in W such that g(w) is not equal to h(w), what happens to (f ∘ g)(w) and (f ∘ h)(w)?
7. Jun 8, 2010
### lanedance
or if f o g = f o h, then apply f-1 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.90260249376297, "perplexity": 1526.5868762867528}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156513.14/warc/CC-MAIN-20180920140359-20180920160759-00179.warc.gz"} |
http://physics.stackexchange.com/questions/16939/capacitance-of-non-concentric-spheres/33165 | # Capacitance of Non-concentric Spheres
We all know how to obtain the capacitance $C=\frac{ab}{b-a}$ (ignoring constants) for two concentric spheres of radii $a,b$.
I was just thinking to myself, what would happen to the capacitance for non-concentric spheres?
Suppose we perturbed the inner sphere off the origin. Does capacitance increase? Ideally I would be looking for an explicit calculation (under sufficiently nice assumptions) but I am also guessing it would be a rather ugly formula.
-
We know that the system is unstable. After you perturb the inner sphere, it'll move all the way until it touches the outer sphere. Since the potential energy is $\frac{Q^2}{2C}$, and it's decreasing along the way -> the capacitance increases – pcr Nov 14 '11 at 6:09
Here is my estimate for the change in potential energy $\approx U_0 (\frac{\delta}{b})^2$ (I got this from the change in $\frac12 E^2 2\pi b^2 \delta$). This gives us $\frac{\delta C}{C}\approx (\frac{\delta}{b})^2$. Not sure – pcr Nov 14 '11 at 6:38
@pcr: I do not see why the system is unstable... Irregardless, I am defining my region, with boundaries of the conductors! So everything is held fixed! – Chris Gerig Nov 14 '11 at 20:57
I cheated by using Earnshaw theorem =) (the system is unstable if you don't nail down the conductors). Alternatively, you can also argue by drawing the E-field line: the lines in the "forward pole" region (using Georg's convention) are denser than those in the "backward pole" region, which suggests an attraction. – pcr Nov 14 '11 at 21:14
Check section III and figures 5 to 9 of this note. – mmc Nov 14 '11 at 23:25
Important notice: My previous result was a little bit incorrect. I found the factor $1/2$ by comparison with the textbook V.V. Batygin, I.N. Toptygin «Problems in electrodynamics».
Let's denote the radius of the inner sphere $S_1$ as $a$, the radius of the outer sphere $S_2$ as $b$ and the displacement as $c$, so that $c\ll a,b$. We choose the origin of the coordinate system to be in the center of the inner sphere. Then, up to the second order in $c$ the distance from the origin to the outer sphere has the form: $$R\left( \theta\right) =b+c\cos\theta.$$ Therefore, the potential in the space between them can be found as $$\phi=\left( \alpha_{1}+\frac{\beta_{1}}{r}\right) +c\left( \alpha _{2}r+\frac{\beta_{2}}{r^{2}}\right) \cos\theta,$$ where $\alpha_{i}$ and $\beta_{i}$ are constant which should be found from the boundary conditions: $$\left. \phi\right\vert _{S_{1}}=const,\quad\left. \phi\right\vert _{S_{2} }=0,\quad \oint_{S_{1}} \mathrm{d}S\,\mathbf{n}\cdot\mathbf{\nabla}\phi=-4\pi Q,\,$$ where $\mathbf{n}=\mathbf{r}/r$. Hence the potential reads as follows: $$\phi=Q\left( \frac{1}{r}-\frac{1}{b}\right) +\frac{Qc}{b^{3}-c^{3}}\left( r-\frac{a^{3}}{r^{2}}\right) \cos\theta.$$ Therefore the potential on the inner sphere doesn't depend on $c$ up to the second order: $$\left. \phi\right\vert _{S_{1}}=Q\,\frac{b-a}{ab}.$$ The charge distribution on the inner sphere can be found as follows: $$\sigma=-\frac{1}{4\pi}\left( \mathbf{n}\cdot\mathbf{\nabla}\phi\right) =-\frac{1}{4\pi}\left. \frac{\partial}{\partial r}\,\phi\right\vert _{r=a}=\frac{Q}{4\pi a^{2}}\left( 1-\frac{3a^{2}c}{b^{3}-c^{3}}\cos \theta\right) .$$ Hence, the force acting on the inner sphere has the form: $$\mathbf{F}=-\frac{1}{2}\oint_{S_{1}} \mathrm{d}S\,\sigma\mathbf{\nabla}\phi,$$ $$F =-\frac{Q^{2}}{4}\int_{-1}^{1}\mathrm{d}\cos\theta\,\left( 1-\frac{3ca^{2}}{b^{3}-c^{3}}\cos\theta\right) \left. \frac{\partial }{\partial z}\left[ \frac{1}{r}+\frac{c}{b^{3}-c^{3}}\left( 1-\frac{a^{3} }{r^{3}}\right) z\right] \right\vert _{r=a}\\ =-\,\frac{Q^{2}c}{b^{3}-c^{3}},$$ where I use the trivial identity: $$\frac{\partial}{\partial z}\frac{1}{r^{n}}=-\frac{nz}{r^{3}}.$$ The capacity $C$ can be found from the potential energy: $$U=\frac{CV^{2}}{2}\quad\Rightarrow\quad F=-\frac{\Delta U}{\Delta c} =-\frac{\phi^{2}}{2}\frac{\Delta C}{\Delta c},$$ thus $$\frac{\Delta C}{\Delta c}=-\frac{2F}{\phi^{2}}=\frac{2ca^{2}b^{2}}{\left( b^{3}-a^{3}\right) \left( a-b\right) ^{2}}.$$ Finally, we obtain $$C=\frac{ab}{b-a}+\frac{a^{2}b^{2}c^{2}}{\left( b^{3}-a^{3}\right) \left( a-b\right) ^{2}}\quad\quad (1)$$
UPDATE: The comments given above give the reference to the article of «Capacitance Bounds for Geometries Corresponding to an Advanced Simulator Design» by M.I. Sancer and A.D. Varvatsis. The article in turn contains the reference to the book:
W. R. Smythe, Static and Dynamic Electricity, McGraw-Hill, New York, 1950
where the following exact result for the capacitance is presented: $$C=ab\sinh\alpha\sum_{n=1}^{\infty}\frac{1}{b\sinh n\alpha-a\sinh\left( n-1\right) \alpha},\quad\quad\left( 2\right)$$ so that $$\quad\cosh\alpha=\frac{a^{2}+b^{2}-c^{2}}{2ab}.$$ Sancer and Varvatsis claim that they found the approximation of the exact result in the $c\rightarrow0$ limit: $$C=\frac{ab}{b-a}\left[ \frac{1}{2}\left( \sqrt{\frac{1-y^{2}/\left( 1+x\right) ^{2}}{1-y^{2}/\left( 1-x\right)^{2}}}+1\right) + \frac{x}{2}\left( \sqrt{\frac{1-y^{2}/\left( 1+x\right) ^{2}}{1-y^{2}/\left( 1-x\right) ^{2}}}-1\right) \right] ,\quad\quad\left( 3\right)$$ where $$x=\frac{a}{b},\quad y=\frac{c}{b}.$$ It is easy to see that the expansion of the result (3) doesn't coincide with mine result (1). The numerical comparison of all three results presented in the figure below:
One can see that the result (3) of Sancer and Varvatsis is incorrect.
-
In one of the above comments, a paper was referenced, and it shows that: Letting $d$ denote the distance between the spheres' centers, $C\sim 4\pi\varepsilon\frac{ab}{b-a}\cdot F$, where $F=\frac{1}{2}[\sqrt{\frac{1-(\frac{d}{a+b})^2}{1-(\frac{d}{b-a})^2}}+1]+\frac{1}{2}\frac{a}{b}[\sqrt{\frac{1-(\frac{d}{a+b})^2}{1-(\frac{d}{b-a})^2}}-1]$. Does your approximation fit with this? Sorry I'm being lazy. – Chris Gerig Jul 31 '12 at 1:11
@ChrisGerig I added the comparison with the article you mentioned. – Grisha Kirilin Aug 1 '12 at 14:27
AWESOME, thanks for the comparison! – Chris Gerig Aug 1 '12 at 17:02
+1, nice answer! Although you leave the limit of large c undiscussed. – Ron Maimon Aug 11 '12 at 10:06
The whole system is unstable (you can convince yourself by Earnshaw theorem or by drawing the field lines). After you perturb the inner sphere, it'll move all the way until it touches the outer sphere. Since the potential energy is $\frac{Q^2}{2C}$, and it's decreasing along the way -> the capacitance increases.
Here is my estimate for the change in potential energy $\approx -U_0 \frac{b-a}{a}(\frac{\delta}{b})^2$ .
I got this from the change in bulk electrostatic energy $\frac12 E^2 2\pi b^2\delta$
Here I shift the outer conductor because it is easier to follow my calculation this way. The shift is exaggerated for clarity's sake.
Initially, the electrostatic energy comes from region 1 and 2. After the shift, it comes from region 2 and 3. We can roughly see that the energy in region 3 is less than the one in region 1: region 1 is nearer to the inner conductor than region 3.
Clearly the change in energy is
$$U_3 - U_1 = \frac12 (E_3^2 -E_1^2) 2\pi b^2 \delta$$ with $$E_3^2 - E_1^2 \approx 2E_1 \frac{dE}{dr} \delta \approx -4 E_1^2 \frac{\delta}b$$
we get $$U_3 - U_1 = - \frac{e^2}{4\pi b} (\frac{\delta}b)^2= U_0 \frac{b-a}a (\frac{\delta}b)^2$$
This gives us $-\frac{\Delta U}{U}=\frac{\Delta C}C\approx\frac{b-a}{a}(\frac{\delta}{b})^2$.
-
See my comment above, I do not see how Earnshaw is sufficiently applied, because there are not only electrostatic forces here... And the field lines argument is not rigorous so it could be incorrect. But in addition, I would like to see your energy change calculation as well. – Chris Gerig Nov 14 '11 at 21:40
@pcr, I erased my comment when I realized that Chris knows better than any answer, especially when he wrote ""especially when these are rings moving in weird ways (or just rings moving towards the other sphere?) and not planes moving closer to each other..."". He lacks imagination of the one-dimensional nature of the problem. – Georg Nov 15 '11 at 11:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9350438117980957, "perplexity": 436.5579401375068}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049274985.2/warc/CC-MAIN-20160524002114-00037-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://mathhelpforum.com/new-users/220978-integrating-trignometric-functions-print.html | # Integrating Trignometric Functions
• August 1st 2013, 12:57 AM
reindeer7
Integrating Trignometric Functions
I know that to integrate functions like cos(x) you search for the expression which gave cos(x) as its derivative in the first place so the integral of cos(x) will be sin(x).So i should think that the integral of cos(2x) should be sin(2x) but is that correct? Afterall maybe z=2x and the integarl of cos(z) will be sin(z).but i have seen a different expression for the integrals of those trignometric functions which have a coefficient>1 with x,in their domain.i did not really understand what they did and please also give some simple reasoning as to why sin(2x) will be incorrect.
And similarly how will I integrate functions like cos(3x+1) , cos(3x^2) , cos(x^(2x+4)) , cos((3x+4)^99).What is the underling principle behind all of them?
• August 1st 2013, 01:08 AM
MarkFL
Re: Integrating Trignometric Functions
Consider:
$\int\cos(ax+b)\,dx$ where $a,b$ are constants.
If we use the substitution:
$u=ax+b\,\therefore\,du=a\,dx$
then we have:
$\frac{1}{a}\int\cos(u)\,du=\frac{1}{a}\sin(u)+C= \frac{1}{a}\sin(ax+b)+C$
If the argument of the cosine function is non-linear, that is, its derivative is not constant, then you will need a function that is a constant times this derivative as a factor to the cosine function within the integrand in order to use this substitution technique. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9849477410316467, "perplexity": 1149.427940566444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164582561/warc/CC-MAIN-20131204134302-00069-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/3539273/topology-induced-by-a-norm | # Topology induced by a norm
I came across the notion of a $$\textit{topology induced by a norm}$$.
If $$(X,\Vert\ . \Vert)$$ is a normed space w.r.t a norm $$\Vert\ . \Vert: X \to \mathbb{R}$$. Most sources define the topology $$\tau$$ on $$X$$ induced by $$\Vert\ . \Vert$$ as the sets $$U \subset X$$ open w.r.t. the metric $$d: X \times X \to \mathbb{R}$$ given by $$d(x,y) = \Vert x - y \Vert$$.
But would I be correct in assuming that an equivalent definition would be $$\tau = \{\Vert\ . \Vert^{-1}(U) \mid U \subset \mathbb{R}\ \textrm{open} \}$$?
• Note that the preimage of U would be a subset of $X\times X$, not $X$ – Ottavio Bartenor Feb 8 at 17:42
• Wouldnt you be applying the norm preimage to a real number (representing the norm) and trying to get the original set which has that value as a norm back out? Or am I misreading? – SquishyRhode Feb 8 at 17:46
• We know topologies are induced by metrics. But norms induce metrics too. So a norm should imply a metric, which then implies a topology. Assuming all other criteria are met. – SquishyRhode Feb 8 at 17:48
• @BrianMoehring My bad, didn't pay enough attention – Ottavio Bartenor Feb 8 at 17:48
• I think you only get a neighborhood basis at $0$ with this definition, but that uniquely determines the topology, so in a sense it is sufficient. – TSU Feb 8 at 17:48
With your version and $$\Bbb R$$ with norm $$|\cdot|$$, the set $$(-3,-2)\cup(2,3)$$ would be open, but $$(2,3)$$ not.
Note that your sets $$U'=\|\cdot\|^{-1}[U]$$ are always symmetrical in that $$x \in U$$ implies $$-x \in U'$$. Not all open sets will obey that. These sets will give the open balls around the origin for $$U$$ that are symmetric around $$0$$, but not all other translated balls.
In fact $$\tau$$ is the topology induced by $$\{ \lVert. \rVert^{-1}(U): U\subseteq \mathbb{R} \ \ \text{is Open} \ \}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9386056065559387, "perplexity": 361.6103044476968}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347432237.67/warc/CC-MAIN-20200603050448-20200603080448-00153.warc.gz"} |
http://itknowledgeexchange.techtarget.com/itanswers/outlook-express-compacting/?watch=22199 | 15 pts.
Outlook Express Compacting
Hi All, I have a peculire problem here, I am using outlook express to access our emails, A while before one of my colleage complained that the emails for last three months are removed from deleted folder of outlook express after the system finished the compacting folders ??? I checked the maintinance tab, there isn't any option seleted to delete the email to compact the folders, however the compacting is set run if there is 5% of wasted space. Now the question is how the emails are deleted? and Can I restore them? Any help will be highly appriciated. Thanking You and Best Regards.
Software/Hardware used:
ASKED: December 15, 2005 7:49 AM
UPDATED: February 8, 2006 10:49 AM
Hi !! Check out the following tool. I installed it after i read ur post. It works. http://www.office-addins.com/-outlook-express-addins/restore-deleted-from-outlook-express.html Let me know if it helped you. Regards, Nilesh.
Last Wiki Answer Submitted: December 15, 2005 8:48 am by Nileshroy 0 pts.
All Answer Wiki Contributors: Nileshroy 0 pts.
Discuss This Question:
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The first reply is a good way to get the email messages back. You might want to think about moving to Outlook (not outlook express) you get the option to backup your folders, to a .pst file (example Dec2005.pst) if something goes wrong, you can restore the individual folders, and there is an option for no duplicates, I would look into this option, or create a folder called ‘holdtemp’, and then only send messages to delete that you are sure you want deleted.
0 pts.
Oulook Express store messages into a *.DBX files. Each folder has its DBX file(ie Inbox.dbx, sent.DBX, ect) search for all *.DBX files and make comparison to see.
Another thing, its possible that the autmatic Archiving has been enabled and after some certains days, the process will automatically archive the items to another DBX file… in Outlook 2000 and greater version, this option exist(for PST files).
0 pts. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8299304246902466, "perplexity": 1775.4213957034972}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368703001356/warc/CC-MAIN-20130516111641-00062-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://proceedings.neurips.cc/paper/2019/file/3a0844cee4fcf57de0c71e9ad3035478-MetaReview.html | NeurIPS 2019
Sun Dec 8th through Sat the 14th, 2019 at Vancouver Convention Center
This paper proposes extending temporal matrix factorization to incorporate neural network regularization for time series forecasting. The intuition is to capture global and local structure to make forecasts. The work is interesting because it bridges more classical forecasting ideas with new state of the art deep learning approaches. The ideas presented in this paper seem novel as the authors take existing building blocks for deep learning and combine them in a creative way to capture interesting structure of time series. This is in contrast to simply applying an existing deep learning approach such as seq2seq. There were some criticisms of the experimental evaluation which the authors seem to have addressed in their response and will include new comparisons that the reviewers asked for. The outstanding concern is in regards to understanding what each component of the proposed model is contributing to the increased performance as the current experiments do not elucidate this. However, I think that the ideas presented in this paper are very interesting and that the improved forecasting abilities of the model are good enough for publication now with a deeper understanding being followup work. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9215236902236938, "perplexity": 495.3266498830361}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487622234.42/warc/CC-MAIN-20210616063154-20210616093154-00067.warc.gz"} |
https://www.varsitytutors.com/ap_calculus_bc-help/geometric-series | # AP Calculus BC : Geometric Series
## Example Questions
### Example Question #1 : Series Of Constants
Consider: . Will the series converge or diverge? If converges, where does this coverge to?
Explanation:
This is a geometric series. Use the following formula, where is the first term of the series, and is the ratio that must be less than 1. If is greater than 1, the series diverges.
Rationalize the denominator.
### Example Question #2 : Series Of Constants
Consider the following summation: . Does this converge or diverge? If it converges, where does it approach?
Explanation:
The problem can be reconverted using a summation symbol, and it can be seen that this is geometric.
Since the ratio is less than 1, this series will converge. The formula for geometric series is:
where is the first term, and is the common ratio. Substitute these values and solve.
### Example Question #3 : Series Of Constants
A worm crawls up a wall during the day and slides down slowly during the night. The first day the worm crawls one meter up the wall. The first night the worm slides down a third of a meter. The second day the worm regains one third of the lost progress and slides down one third of that distance regained on the second night. This pattern of motion continues...
Which of the following is a geometric sum representing the distance the worm has travelled after 12-hour periods of motion? (Assuming day and night are both 12 hour periods).
Explanation:
The sum must be alternating, and after one period you should have the worm at 1m. After two periods, the worm should be at 2/3m. There is only one sum for which that is true.
### Example Question #1 : Geometric Series
Determine whether the following series converges or diverges. If it converges, what does it converge to?
Explanation:
First, we reduce the series into a simpler form.
We know this series converges because
By the Geometric Series Theorem, the sum of this series is given by
### Example Question #1 : Geometric Series
Calculate the sum of a geometric series with the following values:,,. Round the answer to the nearest integer.
Explanation:
This is a geometric series.
The sum of a geometric series can be calculated with the following formula,
, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.
For this question, we are given all of the information we need.
Solution:
Rounding,
### Example Question #6 : Series Of Constants
Calculate the sum, rounded to the nearest integer, of the first 16 terms of the following geometric series:
Explanation:
This is a geometric series.
The sum of a geometric series can be calculated with the following formula,
, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.
We have and n and we just need to find r before calculating the sum.
Solution:
### Example Question #7 : Series Of Constants
Calculate the sum of a geometric series with the following values:
,, ,
rounded to the nearest integer.
Explanation:
This is a geometric series.
The sum of a geometric series can be calculated with the following formula,
, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.
For this question, we are given all of the information we need.
Solution:
Rounding, | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9722512364387512, "perplexity": 558.8381196630252}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948612570.86/warc/CC-MAIN-20171218083356-20171218105356-00078.warc.gz"} |
https://asmedigitalcollection.asme.org/IDETC-CIE/proceedings-abstract/IDETC-CIE2004/46954/407/306455 | This paper shows how the instantaneous invariants for time-independent motions can be obtained from time-dependent motions. Relationships are derived between those parameters that define a time-dependent motion and the parameters that define its geometrically equivalent time-independent motion. The time-independent formulations have the advantage of being simpler than the time dependent ones, and thereby lead to more elegant and parsimonious descriptions of motions properties. The paper starts with a review of the choice of canonical coordinate systems and instantaneous invariants for time-based spherical and spatial motions. It then shows how to convert these descriptions to time-independent motions with the same geometric trajectories. New equations are given that allow the computation of the geometric invariants from time-based invariants. The paper concludes with a detailed example of the third-order motion analysis of the trajectories of an open, spatial R-R chain.
This content is only available via PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9467973709106445, "perplexity": 593.9561907621196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987838289.72/warc/CC-MAIN-20191024012613-20191024040113-00536.warc.gz"} |
http://tex.stackexchange.com/questions/44693/polyline-shading-in-tikz?answertab=oldest | Is it possible to apply a shading effect on one side of a polyline in TikZ? I have the following simple code
\begin{tikzpicture}
\draw (1.06,9.56) -- (1.44,9.78) -- (1.78,9.56) -- (2.14,9.79) -- (2.48,9.29)
-- (2.83,9.79) -- (3.17,9.54) -- (3.54,9.76) -- (3.87,9.56);
\draw (1.06,8.94) -- (1.44,8.72) -- (1.78,8.93) -- (2.14,8.71) -- (2.48,9.21)
-- (2.83,8.72) -- (3.17,8.96) -- (3.54,8.74) -- (3.87,8.94);
\end{tikzpicture}
I want some shading above the upper line and below the lower line, as the lines represent air-solid boundaries and I need to convey this graphically.
-
I believe that a fill or shade requires a closed path up to some \MaxY and down to some \MinY so you can just create one. Below I used a macro to define the \TopPath and \BottomPath so that we can use it twice: Once to draw the line, then to specify the closed area.
\documentclass[border=2pt]{standalone}
\usepackage{tikz}
\begin{document}
\newcommand*{\MaxX}{3.87}%
\newcommand*{\MinX}{1.06}%
\newcommand*{\MaxY}{10.0}%
\newcommand*{\MinY}{8.50}%
\newcommand*{\TopPath}{(\MinX,9.56) -- (1.44,9.78) -- (1.78,9.56) -- (2.14,9.79) -- (2.48,9.29) -- (2.83,9.79) -- (3.17,9.54) -- (3.54,9.76) -- (\MaxX,9.56)}%
\newcommand*{\BottomPath}{(\MinX,8.94) -- (1.44,8.72) -- (1.78,8.93) -- (2.14,8.71) -- (2.48,9.21) -- (2.83,8.72) -- (3.17,8.96) -- (3.54,8.74) -- (\MaxX,8.94)}%
\begin{tikzpicture}
\shade [top color= white, bottom color=brown] \TopPath
-- (\MaxX,\MaxY) -- (\MinX,\MaxY) -- cycle;
\draw [thick] \TopPath;
\shade [top color= blue, bottom color=white] \BottomPath
-- (\MaxX,\MinY) -- (\MinX,\MinY) -- cycle;
\draw [thick] \BottomPath;
\end{tikzpicture}
\end{document}
-
I see what you did here, that's what I need! – MindV0rtex Feb 15 '12 at 23:44
@MindV0rtex: Have updated it so that you can reuse the coordinates without having to specify them twice. – Peter Grill Feb 16 '12 at 0:17
I know that an answer has already been accepted, but I will still give mine.
This answer uses a clip, to remove parts of the boundaries that are drawn (the outside parts). With this approach a path only has to be entered and used once.
Here is the code
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\begin{scope} %only to make sure the clip doesn't act on anything else
\clip ($(0,0)+2*(\pgflinewidth,\pgflinewidth)$) rectangle ($(5,3)-2*(\pgflinewidth,\pgflinewidth)$);
\draw[ultra thick,bottom color=gray,top color = white] (0,2.5) -- (1,2.7) -- (2,2.4) -- (3,2.6) -- (4,2.5) -- (5,2.6) -- (5,3) -- (0,3) -- cycle;
\draw[ultra thick,bottom color=white,top color = gray] (0,0.5) -- (1,0.6) -- (2,0.4) -- (3,0.6) -- (4,0.5) -- (5,0.4) -- (5,0) -- (0,0) -- cycle;
\end{scope}
\end{tikzpicture}
\end{document}
The result is
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9874799847602844, "perplexity": 571.8682234086426}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860116929.30/warc/CC-MAIN-20160428161516-00060-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/centrifical-force-of-earth.96153/ | # Centrifical force of earth
1. Oct 22, 2005
### Petrikovski
How long would a day be if the Earth were rotating so fast that objects at the equator had no weight?
I do Fcnet = m(v^2/r) --> mg = m(v^2/r) --> g = (v^2/r)
g = (v^2/r)
but g = 0 right? so then how do i do this problem? that would make v = 0 and ya...
2. Oct 22, 2005
### Janus
Staff Emeritus
g is the acceleration due to gravity. Its value is independent of any spin of the Earth.
3. Oct 22, 2005
### Petrikovski
ok thanks. unfortunately the answer isnt given so idk if im right until monday. i did:
g = (v^2)/r
9.8 = (v^2)/6.38e6
v =7907
40086722 (distance of Earth in m)/7907 = 5069 seconds
= 1.4 hours per day.
i have one more question and then im done.
A train traveling at a constant speed rounds a curve of radius 275m. A pendulum suspende dfrom the ceiling swings out to an anglr of 17.5 throughot the turn. What is the speed of the train?
Fcnet = m(v^2/r)
Fk = m (v^2/r)
coeff of friction * mg = m(v^2/r)
i dont know the coefficient of friction or v. im supposed to solve for v. am i not supposed to use friction as the force? and how do i factor the 17.5 degree swing fo the pendulum into this?
thanks a lot for the help
4. Oct 23, 2005
### Staff: Mentor
train problem
First figure out the acceleration of the train by analyzing the forces on the pendulum.
5. Oct 23, 2005
### Petrikovski
would this be correct then?
Fnet=ma
mg(cos17.5)=ma
9.8(cos17.5)=a
9.34=a
(v^2)/r = a
(v^2)/275 = 9.34
v^2 = 2568
v = 50.7 m/s
6. Oct 23, 2005
### Staff: Mentor
No. Start by identifying the forces acting on the pendulum mass. (There are two forces.) Then apply Newton's 2nd law to the vertical and horizontal components.
Similar Discussions: Centrifical force of earth | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9010895490646362, "perplexity": 2715.631495636168}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812405.3/warc/CC-MAIN-20180219052241-20180219072241-00260.warc.gz"} |
http://math.stackexchange.com/questions/221128/what-is-left-delta-ab-right-1?answertab=votes | # What is $\left(\delta_{ab}\right)^{-1}$?
I have an expression that involves the Wigner 3j coefficient:
$$\left(\matrix{a&b&0\\0&0&0}\right)^{-1}$$
This simplifies to:
$$\left[\frac{\left(-1\right)^{a}\delta_{ab}}{\sqrt{2a+1}}\right]^{-1}$$
Which, in turn would be:
$$\frac{\sqrt{2a+1}}{\left(-1\right)^{a}\delta_{ab}}$$
What I'm unsure about is how the Kronecker Delta behaves when it's in the denominator. This would seem to me to indicate that the expression is finite when $a=b$ and infinite otherwise, is that correct?
-
Whether it makes sense to regard the value of expression as infinite will depend on the context; in some contexts "undefined" might be a better description. It usually makes sense to regard something as infinite when it approaches infinity, e.g. $1/r$ as $r$ approaches $0$ from above; but in this case there's no approach, and in particular there's no reason to prefer calling this $+\infty$ over calling it $-\infty$. This has nothing specific to do with the Kronecker symbol, which is just a convenient notation that captures the fact that this Wigner $3$-$j$ symbol vanishes unless $a=b$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9806928634643555, "perplexity": 103.2174159074796}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065375.30/warc/CC-MAIN-20150827025425-00060-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.gradesaver.com/textbooks/science/physics/physics-principles-with-applications-7th-edition/chapter-3-kinematics-in-two-dimensions-vectors-problems-page-70/36 | ## Physics: Principles with Applications (7th Edition)
Choose x = 0 and y = 0 to be the point of launch with speed v and initial angle $\theta$, and upward to be the positive y direction. Assume the boy in the tree starts at x = d, y = h. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. Write the equations of motion for the balloon and for the boy. $$x_{balloon} = (v cos \theta)(t)$$ $$y_{balloon} = 0 + (v sin \theta)(t) – 0.5 g t^{2}$$ $$y_{boy} = h – 0.5 g t^{2}$$ Use the balloon's horizontal motion to find the time for the balloon to travel a distance d. $$t = \frac{d}{(v cos \theta)}$$ Use that time and find the balloon’s vertical position. Use the same time to find the boy’s vertical position. They are exactly the same, showing that the boy is splashed by the balloon. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9239467978477478, "perplexity": 347.0853438798307}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125937193.1/warc/CC-MAIN-20180420081400-20180420101400-00102.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/221480-semi-direct-product-proof-print.html | # Semi-Direct product proof
• August 28th 2013, 09:43 PM
akarimi
Semi-Direct product proof
Suppose N is a normal subgroup of G and H is a subgroup of G, and G is the semi-direct product of N and H.
Then is there a nice way to prove that N(and)H = {1}
Thanks.
• August 28th 2013, 09:48 PM
Drexel28
Re: Semi-Direct product proof
I'm confused by your question. I'm going to assume by your wording that you're considering groups of the form $G=N\rtimes_\varphi H$? Then, we identify $N$ with $N\times\{1\}$ inside $N\times H$ and similarly for $H$. In this context, your question should be easily solved.
If this is not what you mean, could you please elaborate?
• August 28th 2013, 10:07 PM
akarimi
Re: Semi-Direct product proof
Hmm I'm not sure if it's the same as what you're saying but it could just be my lack of knowledge, this is my first time reading about semi-direct products but I have yet to see their connection to cross products.
The question I asked was part of a bigger Lemma: Let N <| G and H < G. Then G = N ><| H if and only if G = NH and N(and)H = {1}.
The part I'm having trouble with is using the fact that G = N ><| H to implicate that N(and)H = {1}.
I Hope I've made it a bit clearer.
• September 4th 2013, 05:46 PM
Deveno
Re: Semi-Direct product proof
Hint: Show the mapping G → Nx|H given by nh → (n,h) is an isomorphism iff G = NH and N∩H = {e}. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8019659519195557, "perplexity": 500.291936013739}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257822172.7/warc/CC-MAIN-20160723071022-00053-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=2877517 | ## Solving for a Surjective Matrix
I saw this in a book as a Proposition but I think it's an error:
Assume that the (n-by-k) matrix, $$A$$, is surjective as a mapping,
$$A:\mathbb{R}^{k}\rightarrow \mathbb{R}^{n}$$.
For any $$y \in \mathbb{R}^{n}$$, consider the optimization problem
$$min_{x \in \mathbb{R}^{k}}\left{||x||^2\right}$$
such that $$Ax = y$$.
Then, the following hold:
(i) The transpose of $$A$$, call it $$A^{T}$$ is injective.
(ii) The matrix $$A^{T}A$$ is invertible.
(iii) etc etc etc....
I have a problem with point (ii), take as an example the (2-by-3) surjective matrix
$$A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}$$
$$A^{T}A$$ in this case is not invertible.
Can anyone confirm that part (ii) of this Proposition is indeed incorrect ?
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I would agree with you that part (ii) of the proposition is incorrect (unless matrices are not acting on vectors from the left). If you look at bijection part http://en.wikipedia.org/wiki/Bijecti...and_surjection it reads: "If g o f is a bijection, then it can only be concluded that f is injective and g is surjective." Working right to left with matrices and composition of functions says if A^{T}A was invertible (i.e. a bijection) then A would be injective and A^{T} would be surjective. Thus something is wrong! P.S. I didn't see the bit where it clearly said the matrices were acting from the left so I would say that it is definitely wrong.
Recognitions:
Quote by lauratyso11n Can anyone confirm that part (ii) of this Proposition is indeed incorrect ?
Yes, you are right. (i) is true but (ii) is false. But (ii) is true if A^TA is replaced by AA^T, so maybe it's a typo? I don't understand what "the optimization problem" has to do with this, or are the other parts of the proposition about this?
## Solving for a Surjective Matrix
Quote by Landau Yes, you are right. (i) is true but (ii) is false. But (ii) is true if A^TA is replaced by AA^T, so maybe it's a typo? I don't understand what "the optimization problem" has to do with this, or are the other parts of the proposition about this?
The full Proposition is as follows:
Assume that the (n-by-k) matrix, $$A$$, is surjective as a mapping,
$$A:\mathbb{R}^{k}\rightarrow \mathbb{R}^{n}$$.
For any $$y \in \mathbb{R}^{n}$$, consider the optimization problem
$$min_{x \in \mathbb{R}^{k}}\left{\left||x|\right|^2\right}$$
such that $$Ax = y$$.
Then, the following hold:
(i) The transpose of $$A$$, call it $$A^{T}$$ is injective.
(ii) The matrix $$A^{T}A$$ is invertible.
(iii) The unique optimal solution of the minimum norm problem is given by
$$(A^TA)^{-1}A^Ty$$
I have a problem with point (ii), take as an example the (2-by-3) surjective matrix
$$A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}$$
$$A^{T}A$$ in this case is not invertible.
Quote by Landau Yes, you are right. (i) is true but (ii) is false. But (ii) is true if A^TA is replaced by AA^T, so maybe it's a typo?
I don't think it's a typo as he uses this result to carry some other analysis further. The crux of it is:
$$\sigma\lambda = \alpha - \bf{r}$$
where $$\sigma \in \mathbb{R}^{n-by-k}$$ is surjective, and $$\lambda \in \mathbb{R}^{k}$$, $$\alpha , \bf{r} \in \mathbb{R}^{n}$$.
How would you solve for $$\lambda$$ ? Isn't is critical that the 'typo' has to be correct to be able to solve for this ?
The author's solution as you might expect is $$\lambda = \left(\sigma^{T}\sigma\right)^{-1}\sigma^{T}\left[\alpha - \bf{r}\right]$$.
So it's either a HUGE mistake on his part or I'm missing something. The author is actually quite insightful, and this error would be quite out of character for him.
BTW, thanks for making the effort to look at the problem. Much appreciated.
Blog Entries: 1 Recognitions: Homework Help Looking at dimension counting with your example, AtA is a 3x3 matrix, which means (AtA)-1 A wouldn't make sense even if the inverse was defined because the sizes of the matrices don't match up. On the other hand A At and A are compatible matrices which suggests that he just put the transpose on the wrong one and carried the error through.
Recognitions:
Quote by lauratyso11n The author's solution as you might expect is $$\lambda = \left(\sigma^{T}\sigma\right)^{-1}\sigma^{T}[\alpha - \bf{r}]$$.
This is correct provided that $\sigma^{T}\sigma$ is invertible, i.e. provided that $\left(\sigma^{T}\sigma\right)^{-1}$ makes sense.
The least square solution of Ax=y satisfies the normal equation:
$$A^TAx=A^Ty.$$
This solution is unique if and only if $A^TA$ is invertible. In this case, it is given by
$$x=(A^TA)^{-1}A^Tb$$, just like the author asserts.
But $A^TA$ is not necessarily invertible, contrary to the proposition. Note that $A^TA:\mathbb{R}^k\to\mathbb{R}^k$ is bijective if and only if its rank equals k. But since $A^TA$ and A always have equal rank, this happens if and only if A has rank k. Since $A:\mathbb{R}^k\to\mathbb{R}^n$ is assumed to be surjective, it has rank n and we must have $k\geq n$. So in fact, $A^TA$ is invertible if and only if k=n! Indeed, in your counter-example k and n are not equal.
Quote by Office_Shredder which means (AtA)-1 A wouldn't make sense even if the inverse was defined because the sizes of the matrices don't match up.
This is not the expression the author uses; it is (AtA)-1 At.
Recognitions: Science Advisor Perhaps you could post a link to the book?
Blog Entries: 1
Recognitions:
Homework Help
Quote by Landau This is not the expression the author uses; it is (AtA)-1 At.
I assumed the * was just referring to matrix multiplication
Recognitions: Science Advisor I assumed * means adjoint, so (in this real case) the transpose. This is also what lauratyso11n writes in post #5 (where A is called sigma).
Quote by Office_Shredder I assumed the * was just referring to matrix multiplication
Quote by Landau I assumed * means adjoint, so (in this real case) the transpose. This is also what lauratyso11n writes in post #5 (where A is called sigma).
SORRYYYYYY, the $$A^*$$ is actually an $$A^T$$. I've corrected it.
The correct statement is that if $$A$$ is surjective then $$A^T$$ is injective and $$AA^T$$ is invertible. The formula for the optimal $$x$$ is $$\hat{x}=A^T(AA^T)^{-1}y$$
Recognitions:
Quote by lauratyso11n The correct statement is that if $$A$$ is surjective then $$A^T$$ is injective and $$AA^T$$ is invertible. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9348123669624329, "perplexity": 484.45140216917474}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368696382261/warc/CC-MAIN-20130516092622-00065-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://economics.stackexchange.com/questions/33298/can-you-take-the-moving-average-of-quarterly-data-of-an-explanatory-variable-in | # Can you take the moving average of quarterly data of an explanatory variable in a regression to smoothen noise and get more accurate coefficients?
I'm trying to use acceleration of quarterly data on household debt (the difference in the difference in debt) in a regression on unemployment (only concerned with correlation) but quarterly data is much noisier than annual data, which may weaken the correlation. I was wondering if transforming the already differenced (twice) data as a moving average would change my coefficients at all and if they would be more accurate. Is this a common technique in econometrics?
## 2 Answers
Let's say $$D_t$$ is the stock of debt at time $$t$$.
The first difference is $$D_t - D_{t-1}$$.
The second difference is $$(D_t - D_{t-1}) - (D_{t-1}-D_{t-2}) = D_t - 2D_{t-1} + D_{t-2}$$.
The [4-period] moving average is then given by
\begin{align*} &= \frac 1 4 (D_t-2D_{t-1}+D_{t-2}+D_{t-1}-2D_{t-2}+D_{t-3}\\ &\qquad+D_{t-2}-2D_{t-3}+D_{t-4}+D_{t-3}-2D_{t-4}+D_{t-5})\\ &= \frac 1 4 (D_t - D_{t-1} - D_{t-4} + D_{t-5})\\ &= \frac 1 4 ((D_t - D_{t-4}) - (D_{t-1} - D_{t-5})) \end{align*}
Which would be the yearly increase in the stock of debt this period compared to the same thing a quarter ago. Only you could tell if this is what you want in your regression. Something to consider is only to use the first difference: $$D_t - D_{t-4}$$.
The word "noise" implies that there is error in the measurement of household debt, but there is no good reason to believe this is so. In fact, I argue that a finer time period would give you more detail, and a more accurate measurement of the correlation between two variables. You don't want to find a moving average because that throws away valuable variation in the data.
What you should be more concerned about is which lag is most correlated with the current unemployment rate, and you can find those coefficients by including multiple degrees of lag in the multivariate regression.
Overall, no, it is not a common technique in econometrics. Don't do it. Rather use the standard errors to determine how accurate your estimates are. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9226527810096741, "perplexity": 681.1929583986922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323586043.75/warc/CC-MAIN-20211024142824-20211024172824-00068.warc.gz"} |
http://wiki.zcubes.com/Manuals/calci/MATRIXSCALARDIVIDE | # Manuals/calci/MATRIXSCALARDIVIDE
MATRIXSCALARDIVIDE (a,afactor)
• is any matrix.
• is any constant value.
## Description
• This function is dividing the scalar value with the given matrix.
• In , is any matrix.
• Suppose when we are dividing two matrices,the order of the matrices to be considered.
• But while dividing with a scalar order of matrix is not considered.
• Normally we could not do the matrix division directly.
• To divide a matrix by a scalar which is a single number,that simply divide each element in the matrix by this number.
• So all the entries in the given matrix are divide by the scalar value.
## Examples
1. MATRIXSCALARDIVIDE([[6,7,8],[10,12,-22],[7,17,23]],2)
3 3.5 4 5 6 -11 3.5 8.5 11.5
2. MATRIXSCALARDIVIDE([[1,2,3,4],[7,8,-9,10],[11,12,-13.56,14],[10,12,17,18],[21,22,23,24]],6.4)
0.15625 0.3125 0.46875 0.625 1.09375 1.25 -1.40625 1.5625 1.71875 1.875 -2.11875 2.1875 1.5625 1.875 2.65625 2.8125 3.28125 3.4375 3.59375 3.75
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http://annals.math.princeton.edu/articles/10911 | # A structure theorem for actions of semisimple Lie groups
### Abstract
We consider a connected semisimple Lie group $G$ with finite center, an admissible probability measure $\mu$ on $G$, and an ergodic $(G,\mu)$-space $(X,\nu)$. We first note (Lemma 0.1) that $(X,\nu)$ has a unique maximal projective factor of the form $(G/Q,\nu_0)$, where $Q$ is a parabolic subgroup of $G$, and then prove:
1. Theorem 1. If every noncompact simple factor of $G$ has real rank at least two, then the maximal projective factor is nontrivial, unless $\nu$ is a $G$-invariant measure.
2. Theorem 2. For any $G$ of real rank at least two, if the action has positive entropy and fails to have nontrivial projective factor, then $(X,\nu)$ has an equivariant factor space with the same properties, on which $G$ acts via a real-rank-one factor group.
3. Theorem 3. Write $\nu = \nu_0\ast \lambda$, where $\lambda$ is a $P$-invariant measure, $P = MSV$ a minimal parabolic subgroup [F2], [NZ1]. If the entropy $h_\mu(G/P,\nu_0)$ is finite, and every nontrivial element of $S$ is ergodic on $(X,\lambda)$ (or just a well chosen finite set, Theorem 9.1), then ($X,\nu)$ is a measure-preserving extension of its maximal projective factor.
4. The foregoing results are best possible (see Section 11, in particular Theorem 11.4).
We also give some corollaries and applications of the main results. These include an entropy characterization of amenable actions, an explicit entropy criterion for the invariance of $\nu$, and construction of a projective factor for an action of a lattice in $G$ on a compact metric space.
Amos Nevo
Robert J. Zimmer | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9786674380302429, "perplexity": 258.72057516974917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646375.29/warc/CC-MAIN-20180319042634-20180319062634-00597.warc.gz"} |
https://brilliant.org/problems/fractions-sum-to-1/ | # Fractions sum to 1?
Find number of solutions in positive integers of the following equation,
$$\large \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$$ with $$a\le b\le c \le d$$.
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