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http://math.stackexchange.com/questions/40861/mathematical-notation-for-the-maximum-of-a-set-of-function-values | # Mathematical notation for the maximum of a set of function values
I have a question about the proper notation of the following (simplified) example:
I want to express that I have a value alpha, which is the maximum of a set of n values. Each value in the set is the result of a function $f(x)$, and the range of $x$ is between $1$ and $n$.
So something like
$$\alpha = \max(\{f(x) : x = 1,\ldots,n\}).$$
Is this a proper notation? If not, how would I properly express this? It's too long ago for me studying this sort of thing to convince myself I'm writing it down right.
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## 1 Answer
Your notation looks fine. You could also use the more informal $\alpha = \max(\{f(x_1),\ldots,f(x_n)\})$ or even $\alpha = \max(f(x_1),\ldots,f(x_n))$.
Finally, you could say that $\alpha$ is the maximum (or maximal) value among $f(x_1),\ldots,f(x_n)$, or that $\alpha$ is the maximum (or maximal) value attained by $f$ on the points $x_1,\ldots,x_n$.
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Thank you ever so much. It's one of those things where you want to be sure before sending the document out the door. – Bart May 23 '11 at 16:59 | {"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": 11, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9463759660720825, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/98910/complex-analysis-correspodence-between-h-omega-1-and-h-omega-2?answertab=oldest | # Complex Analysis: Correspodence between $H(\Omega_1)$ and $H(\Omega_2)$.
If we let
$$\Omega_1=\left\{z \in \mathbb{C} : 0<\operatorname{Im}(z)< \pi\}, \quad \Omega_2=\{z \in \mathbb{C}: 0<\operatorname{Im}(z) \right\},$$
can we establish a one-to-one correspondence between $H(\Omega_1)$ and $H(\Omega_2)$ where $H(\Omega)$ represents the set of real-valued harmonic functions on $\Omega \subset \mathbb{C}$?
Thoughts:
Can we appeal to simple conformal mapping and say that, if we can establish a conformal map between $\Omega_1$ and $\Omega_2$ then the correspondence exists as the harmonic nature of the functions is preserved under such a map i.e. the composition of a holomorphic and harmonic map is harmonic? If so, what could this conformal map be?
If we apply $z \mapsto e^z$ initially, this transforms the 'strip'to a 'wedge', but how do we advance from here?
If such a method is incorrect, how else can we demonstrate the existence of the correspondence?
Any help would be greatly appreciated. Best, MM.
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1
If $f(z) = e^z$, then $f(\Omega_1) = \{e^{x+iy} \mid 0<y<\pi\} = \{z\in \mathbb C \mid |z|>0, \; 0< arg(z)<\pi \} = \Omega_2$. That's already the conformal map you are looking for. – Sam Jan 14 '12 at 2:02
@Sam: Thanks! Great help. – Mathmo Jan 14 '12 at 7:31
## 1 Answer
The Riemann Mapping theorem says there is a differentiable map between these two sets that is compositionally invertible. This just might do the job.
- | {"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": 11, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9031987190246582, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/9651/list | ## Return to Answer
3 Texified, since problem was on front-page anyway
space X.
• X $X$.
• $X$ is compact.
• Xκ
• $X^\kappa$ is Lindelöf for any cardinal
• Xω1$\kappa$.
• $X^{\omega_1}$ is Lindelöf.
• for 1 implies 2, since if X $X$ is compact, thenXκ$X^\kappa$ is compact and hence Lindelöf.
So suppose that we have a space X $X$ that is not compact, butXω1$X^{\omega_1}$ is Lindelöf. Itfollows that X $X$ is Lindelöf. Thus, there is a countableU0$U_0 \subset U1U_1 \subset ... Un... \dots U_n \dots$with the union U{ Un$\bigcup\lbrace U_n \; | \; n \in ω} \omega \rbrace = XX$.
For each $J \subset ω1\omega_1$ of size n, $n$, let UJ$U_J$ bethe set {$\lbrace s \in Xω1X^{\omega_1} \; | s(α)\; s(\alpha) \in UnU_n$ for each α $\alpha \in J}. J \rbrace$. As the size of J$J$ increases, the set UJ$U_J$ allows more freedom on thecoordinates in J, $J$, but restricts more coordinates. If J $J$ hassize n, $n$, let us call UJ$U_J$ an open n-box, $n$-box, since itrestricts the sequences on n $n$ coordinates. Let F $F$ be thefamily of all such UJ$U_J$ for all finite $J \subset ω1.\omega_1$
This F $F$ is a cover of Xω1. $X^{\omega_1}$. Tosee this, consider any point $s \in Xω1X^{\omega_1}$. For each α $\alpha \inω1, \omega_1$, there is some n $n$ with s(α) $s(\alpha) \inUnU_n$. Since ω1$\omega_1$ is uncountable,there must be some value of n $n$ that is repeated unboundedlyoften, in particular, some n $n$ occurs at least n $n$ times. Let Jbe the coordinates where this n $n$ appears. Thus, s $s$ is inUJ. $U_J$. So F $F$ is a cover.
Since Xω1$X^{\omega_1}$ is Lindelöf,there must be a countable subcover F0. $F_0$. Let J* $J^*$ bethe union of all the finite J $J$ that appear in theUJ$U_J$ in this subcover. So J* $J^*$ is a countable subsetof ω1. $\omega_1$. Note that J* $J^*$ cannot be finite,since then the sizes of the J $J$ appearing in F0$F_0$Xω1. $X^{\omega_1}$. We may rearrange indicesand assume without loss of generality that J*=ω $J^*=\omega$ isthe first ω $\omega$ many coordinates. So F0$F_0$ isreally a cover of Xω, $X^\omega$, by ignoring the
But this is impossible. Define a sequence $s \inXω1X^{\omega_1}$ by choosing s(n) $s(n)$ to beoutside Un+1, $U_{n+1}$, and otherwise arbitrary. Note thats $s$ is in Un$U_n$ in fewer than n $n$ coordinates belowω, $\omega$, and so s $s$ is not in any n-box $n$-box with $J \subset ω, \omega$, since any such box has n $n$ values in Un.Thus, s $s$ is not in any set in F0, $F_0$, so it is not a
In particular, to answer the question at the end, it suffices to take any uncountable kappa. $\kappa$.
2 fixed typo.
The answer is Yes.
Theorem. The following are equivalent for any Hausdorff space X.
1. X is compact.
2. Xκ is Lindelöf for any cardinal κ.
3. Xω1 is Lindelöf.
Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if X is compact, then Xκ is compact and hence Lindelöf.
So suppose that we have a space X that is not compact, but Xω1 is Lindelöf. It follows that X is Lindelöf. Thus, there is a countable cover having no finite subcover. From this, we may construct a strictly increasing sequence of open sets U0 subset U1 subset ... Un ... with the union U{ Un | n in ω} = X.
For each J subset ω1 of size n, let UJ be the set {s in Xω1 | s(α) in Un for each α in J}. As the size of J increases, the set UJ allows more freedom on the coordinates in J, but restricts more coordinates. If J has size n, let us call UJ an open n-box, since it restricts the sequences on n coordinates. Let F be the family of all such UJ for all finite J subset ω1.
This F is a cover of Xω1. To see this, consider any point s in Xω1. For each α in ω1, there is some n with s(α) in Un. Since ω1 is uncountable, there must be some value of n that is repeated unboundedly often, in particular, some n occurs at least n times. Let J be the coordinates where this n appears. Thus, s is in UJ. So F is a cover.
Since Xω1 is Lindelöf, there must be a countable subcover F0. Let J* be the union of all the finite J that appear in the UJ in this subcover. So J* is a countable subset of ω1. Note that J* cannot be finite, since then the sizes of the J appearing in F0 would be bounded and it could not cover Xω1. We may rearrange indices and assume without loss of generality that J*=ω is the first ω many coordinates. So F0 is really a cover of Xω, by ignoring the other coordinates.
But this is impossible. Define a sequence s in Xω1 by choosing s(n) to be outside Un+1, and otherwise arbitrary. Note that s is in Un in fewer than n coordinates below ω, and so s is not in any n-box with J subset ω, since any such box has n values in Un. Thus, s is not in any set in F0, so it is not a cover. QED
In particular, to answer the question at the end, it suffices to take any uncountable kappa.
1
The answer is Yes.
Theorem. The following are equivalent for any Hausdorff space X.
1. X is compact.
2. Xκ is Lindelöf for any cardinal κ.
3. Xω1 is Lindelöf.
Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if X is compact, then Xκ is compact and hence Lindelöf.
So suppose that we have a space X that is not compact, but Xω1 is Lindelöf. It follows that X is Lindelöf. Thus, there is a countable cover having no finite subcover. From this, we may construct a strictly increasing sequence of open sets U0 subset U1 subset ... Un ... with the union U{ Un | n in ω} = X.
For each J subset ω of size n, let UJ be the set {s in Xω1 | s(α) in Un for each α in J}. As the size of J increases, the set UJ allows more freedom on the coordinates in J, but restricts more coordinates. If J has size n, let us call UJ an open n-box, since it restricts the sequences on n coordinates. Let F be the family of all such UJ for all finite J subset ω1.
This F is a cover of Xω1. To see this, consider any point s in Xω1. For each α in ω1, there is some n with s(α) in Un. Since ω1 is uncountable, there must be some value of n that is repeated unboundedly often, in particular, some n occurs at least n times. Let J be the coordinates where this n appears. Thus, s is in UJ. So F is a cover.
Since Xω1 is Lindelöf, there must be a countable subcover F0. Let J* be the union of all the finite J that appear in the UJ in this subcover. So J* is a countable subset of ω1. Note that J* cannot be finite, since then the sizes of the J appearing in F0 would be bounded and it could not cover Xω1. We may rearrange indices and assume without loss of generality that J*=ω is the first ω many coordinates. So F0 is really a cover of Xω, by ignoring the other coordinates.
But this is impossible. Define a sequence s in Xω1 by choosing s(n) to be outside Un+1, and otherwise arbitrary. Note that s is in Un in fewer than n coordinates below ω, and so s is not in any n-box with J subset ω, since any such box has n values in Un. Thus, s is not in any set in F0, so it is not a cover. QED
In particular, to answer the question at the end, it suffices to take any uncountable kappa. | {"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": 71, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9151735305786133, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/21386?sort=newest | ## What is the nonrecursive formula for the following implicit function?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Given $f(0) = 1-c$, what is the non-recursive $f(n)$ that satify the following equation $f(n)-\frac{1}{2}f(n-1)f(n)+f(n-1) = 1$
for n = 1,2,3,...?
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Did you try just working out what happens for n=1,2,3,... until you find a pattern? – Reid Barton Apr 14 2010 at 20:47
2
f(n) is a fractional linear transformation of f(n-1). Composition of fractional linear transformations is essentially the same thing as matrix multiplication, so you can figure out closed forms by computing eigenvectors and eigenvalues. – Qiaochu Yuan Apr 14 2010 at 21:20
2
@QY: And f(n-1) is the same fractional linear transformation of f(n) :) – Reid Barton Apr 14 2010 at 21:28
1
Which is essentially what Maple did to solve that. It's nice when mathematical knowledge is so well understood it can be fully automated! It frees humans to do the real creative thinking. – Jacques Carette Apr 14 2010 at 21:34
Random sidenote regarding the Maple comment: Mathematica can find vertical asymptotes logically. You can basically say "solve for x: exists M > 0, exists delta > 0, for all y with |x-y|<delta, f(x) > M" and it will output the values of x where f(x) has positive vertical asymptote. I thought this was pretty cool. – Peter Samuelson Apr 14 2010 at 22:22
## 2 Answers
$$f(n) = 2{\frac {1-c+ \left( -1 \right) ^{n}-c \left( -1 \right) ^{n}-\sqrt {2}c+ \left( -1 \right) ^{n}\sqrt {2}c}{2-\sqrt {2}-\sqrt {2}c+2 \left( -1 \right) ^{n}+ \left( -1 \right) ^{n}\sqrt {2}+ \left( -1 \right) ^{n}\sqrt {2}c}}$$
which is a non-trivial pattern to spot! [I used a CAS] The thing to notice is that one can transform this first-order recurrence to the constant coefficient 2nd order linear recurrence $$a(n+2)-2a(n) = 0, a(0) = 1, a(1) = -1-c$$ and then transform back. This can be done as the equation is of Riccati type.
I guess I 'cheated' in that I traced through the CAS's process of solving to extract the above information from it. Though if you don't know what you're looking for, it can be rather difficult information to extract...
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Wow! Not tested yet but it looks like this is it. Thank you so much. – silvanmx Apr 15 2010 at 21:11
What is CAS by the way? – silvanmx Apr 15 2010 at 21:12
CAS = Computer Algebra System. – Jacques Carette Apr 15 2010 at 23:57
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
We can rewrite the given recurrence as $(f(n) - 2)(f(n-1) - 2) = 2$. That makes it clear that the sequence is 2-periodic.
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This answer seems to make a mockery of all the other comments! (other than Reid, who seems to be the only other person who spotted this). +1. – Kevin Buzzard Apr 15 2010 at 7:46 | {"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": 4, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9145044684410095, "perplexity_flag": "middle"} |
http://all-science-fair-projects.com/science_fair_projects_encyclopedia/Combinatorics | # All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
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# Science Fair Project Encyclopedia
For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
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# Combinatorics
Combinatorics is a branch of mathematics that studies finite collections of objects that satisfy specified criteria, and is in particular concerned with "counting" the objects in those collections (enumerative combinatorics) and with deciding whether certain "optimal" objects exist (extremal combinatorics). One of the most prominent combinatorialists of recent times was Gian-Carlo Rota, who helped formalize the subject beginning in the 1960s. The prolific problem-solver Paul Erdős worked mainly on extremal questions. The study of how to count objects is sometimes thought of separately as the field of enumeration.
An example of a combinatorial question is the following: What is the number of possible orderings of a deck of 52 playing cards? That number equals 52! (i.e., "fifty-two factorial"). It is the product of all the natural numbers from one to fifty-two. It may seem surprising that this number, about 8.065817517094 × 1067, is so large. That is a little bit more than 8 followed by 67 zeros. Comparing that number to some other large numbers, it is greater than the square of Avogadro's number, 6.022 × 1023, "the number of atoms, molecules, etc., in a gram mole".
Contents
## Counting functions
Calculating the number of ways that certain patterns can be formed is the beginning of combinatorics. Let S be a set with n objects. Combinations of k objects from this set S are subsets of S having k elements each (where the order of listing the elements does not distinguish two subsets). Permutations of k objects from this set S refer to sequences of k different elements of S (where two sequences are considered different if they contain the same elements but in a different order). Formulas for the number of permutations and combinations are readily available and important throughout combinatorics.
More generally, given an infinite collection of finite sets {Si} typically indexed by the natural numbers, enumerative combinatorics seeks a variety of ways of describing a counting function, f(n), which counts the number of objects in Sn for any n. Although the activity of counting the number of elements in a set is a rather broad mathematical problem, in a combinatorial problem the elements Si will usually have a relatively simple combinatorial description, and little additional structure.
The simplest such functions are closed formulas, which can be expressed as a composition of elementary functions such as factorials, powers, and so on. As noted above, the number of possible different orderings of a deck of n cards is f(n) = n!.
This approach may not always be entirely satisfactory (or practical) for every combinatorial problem. For example, let f(n) be the number of distinct subsets of the integers in the interval [1,n] that do not contain two consecutive integers; thus for example, with n = 4, we have {}, {1}, {2}, {3}, {4}, {1,3}, {1,4}, {2,4}, so f(4) = 8. It turns out that f(n) is the n+2 Fibonacci number, which can be expressed in closed form as:
$f(n) = \frac{\phi^{n+2}}{\sqrt{5}} - \frac{(1-\phi)^{n+2}}{\sqrt{5}}$
where φ = (1 + √5) / 2, the Golden mean. However, given that we are looking at sets of integers, the presence of the √5 in the result may be considered as "unaesthetic" from a combinatoric viewpoint. Alternatively, f(n) may be expressed as the recurrence
f(n) = f(n - 1) + f(n - 2)
which may be more satisfactory (from a purely combinatorial view), since it more clearly shows why the result is as shown.
Another approach is to find an asymptotic formula
f(n) ~ g(n)
where g(n) is a "familiar" function, and where f(n) approaches g(n) as n approaches infinity. In some cases, a simple asymptotic function may be preferable to a horribly complicated closed formula that yields no insight to the behaviour of the counted objects. In the above example, an asymptotic formula would be
$f(n) \sim \frac{\phi^{n+2}}{\sqrt{5}}$
as n becomes large.
Finally, and most usefully, f(n) may be expressed by a formal power series, called its generating function, which is most commonly either the ordinary generating function
$\sum f(n) x^n$
or the exponential generating function
$\sum f(n) \frac{x^n}{n!}$
where the sums are taken for n ≥ 0. Once determined, the generating function may allow one to extract all the information given by the previous approaches. In addition, the various natural operations on generating functions such as addition, multiplication, differentiation, etc., have a combinatorial significance; and this allows one to extend results from one combinatorial problem in order to solve others.
## Results
Some very subtle patterns can be developed and some surprising theorems proved. One example of a surprising theorem is of Frank P. Ramsey:
Suppose 6 people meet each other at a party. Each pair of people either know each other or don't know each other. It is always the case that one can find 3 people out of the 6 such that they either all know each other or that they are all strangers to each other.
The proof is a short proof by contradiction: suppose that there aren't 3 people who either all know each other or all don't know each other. Then consider any one person at the party, hereafter called person A: among the remaining 5 people, there must be at least three who either all know or all do not know A. Without loss of generality, assume three such people all know A. But then among those three people, at least two of them must know each other (otherwise we would have 3 people who all don't know each other). But then those two also know A, so we have 3 people who all know each other. (This is a special case of Ramsey's theorem)
An alternate proof works by double counting: count the number of ordered triples of people (A,B,C) where person B knows person A but does not know person C. Suppose person K knows k of the 5 others. Then he is the B of exactly k(5-k) such triples - A must be one of the k people he knows, and C must be one of the (5-k) people he doesn't. Therefore, he is the B of either 0*5=0, 1*4=4 or 2*3=6 such triples. Since there are 6 people, and each one is the B of at most 6 triples, there are at most 36 triples.
Now consider a triple of people where exactly 1 pair know each other. It is clear that we can turn them into such an (A,B,C) in exactly two ways: let C be the one who is a stranger, and then call one of the others A and the other B. Similarly if exactly 2 pairs know each other, they can be turned into such at triple in exactly two ways: let A be the person who knows both of the others, whilst B and C (in some order) are the two who do not know each other. Therefore, there are at most 36/2=18 triples where either exactly 1 pair or exactly 2 pairs know each other. Since there are 20 triples, there must be at most 2 triples who either all know each other, or are all strangers to each other.
The idea of finding order in random configurations gives rise to Ramsey theory. Essentially this theory says that any sufficiently large configuration will contain at least one instance of some other type of configuration.
## References
• Handbook of Combinatorics, Volumes 1 and 2, R.L. Graham, M. Groetschel and L. Lovász (Eds.), MIT Press, 1996. ISBN 026207169X
• Enumerative Combinatorics, Volumes 1 and 2, Richard P. Stanley, Cambridge University Press, 1997 and 1999, ISBN 0-521-55309-1N
03-10-2013 05:06:04 | {"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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9252883195877075, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/127354/the-real-part-of-a-matrix-under-similarity-transformation | # The real part of a matrix under similarity transformation
I have a question regarding the real part of some matrix A, defined as $$Re\{A\} = \frac{1}{2}\left(A + A^\dagger \right).$$ Where $A^\dagger$ denotes the Hermitian conjugate. One can also assume that the real part is positive semi-definite, i.e. $Re\{A\} \geq 0$ .
Suppose I were to apply a similarity transformation with $S > 0$, and $S$ Hermitian to A as $S^{-1}AS$, what can be said about the real part of this similar matrix? Are there any bounds known of the form $$Re\{S A S^{-1}\} \leq c(S) Re\{A\}$$ for some constant $c(S)$ that can depend on the matrix S? I would guess the constant $c(S)$ is somehow related to the largest eigenvalue of $S$.
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When is a matrix $A$ smaller than a matrix $B$? Do you mean a norm of $\text{Re}\{M\}$? – draks ... Apr 3 '12 at 9:04
Hi draks, Sorry I should have been more clear. No I meant in terms of the partial order for matrices, i.e. matrix $A \geq B$ if the difference between $A - B \geq 0$ is positive semi-definite. So to state the problem differently what is the smallest constant $c(S)$ so that for all vectors $\psi$ we have that $(\psi ,(c(S)Re\{A\} - Re\{SAS^{-1}\})\psi) \geq 0$. – DrMabuse Apr 3 '12 at 12:04
It doesn't really have to be the smallest, any "reasonable" upper bound to $c(S)$ would do. – DrMabuse Apr 3 '12 at 12:11
## 1 Answer
There is little hope here, unless I misunderstood your purpose, even for positive Hermitian matrices.
Assume that $A=\begin{pmatrix}a_1 & 0\\ 0&a_2\end{pmatrix}$ for some positive real numbers $a_1$ and $a_2$ and that $S=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$. Then $SAS^{-1}=\begin{pmatrix}a_2 & 0\\ 0&a_1\end{pmatrix}$ hence $\text{Re}(A)=A$ and $\text{Re}(SAS^{-1})=SAS^{-1}$ but the smallest $c$ such that $SAS^{-1}\leqslant c\cdot A$ in the sense of the Hermitian matrices is $c=\max\{a_1/a_2,a_2/a_1\}$ hence there can exist no finite $c=c(S)$ independent on $A$ such that the upper bound you are interested in holds for every $A$.
If non invertible matrices are allowed things are even simpler: consider the example above with $a_1=1$ and $a_2=0$.
- | {"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": 32, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9516651630401611, "perplexity_flag": "head"} |
http://nrich.maths.org/570/index?nomenu=1 | The areas of the faces are $3$, $12$ and $25$ square centimetres. | {"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": 3, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9126760959625244, "perplexity_flag": "head"} |
http://mathhelpforum.com/statistics/98906-proving-question.html | # Thread:
1. ## Proving question
Hi,
given that: $S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\overline X)^2$.
I want to show that $E[S^2]=\sigma^2$
However, when i reached: $\frac{n}{n-1}(\frac{1}{n}\sum_{i=1}^{n}E[X_i^2]-E[\overline X^2])$, i was puzzled on finding $E[\overline X^2]$?
Since $E[\overline X^2]=Var(\overline X) + [E(\overline X)]^2$ which variance should I use, s.r.s or i.i.d?
or maybe is there any better way to prove it?
2. $\frac{(n-1)S^{2}}{\sigma^{2}}$ is $\chi^{2}_{n-1}$ (chi-square distriubtion with (n-1) degrees of freedom)
If you want to try and prove that statement, use the fact that $S^{2}$ and $\bar{X}$ are independent.
So $E(S^{2}) = E\Big(\frac{\sigma^{2}}{n-1} \ \frac{(n-1)S^{2}}{\sigma^{2}} \Big) = \frac{\sigma^{2}}{n-1} E\Big(\frac{(n-1)S^{2}}{\sigma^{2}} \Big) = \frac{\sigma^{2}}{n-1} (n-1) = \sigma^{2}$
since the expected value of a chi-square distribution with k degrees of freedom is k
3. Hi tks for the reply.
In addition to this, I've read book stating that the best estimator is $(1-\frac{1}{N})s^2$ as compared to this answer showing that $E[s^2]=\sigma ^2$.
Any explaination for this?
4. Here's a much better proof that doesn't involve knowing anything about the chi-square distribution:
$\sum^{n}_{i=1} (X_{i} -\overline{X})^{2} = \sum^{n}_{i=1} X_{i}^{2} -n\overline{X}^{2}$
so $E\Big(\sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) = E\Big(\sum^{n}_{i=1} X_{i}^{2}\Big) - E(n\overline{X}^{2}) = \sum^{n}_{i=1} E(X_{i}^{2}) - nE(\overline{X}^{2})$
$= \sum^{n}_{i=1}(\sigma^{2}+ \mu^{2})-n \Big(\frac{\sigma^{2}}{n}+ \mu^{2}\Big) = n(\sigma^{2}+ \mu^{2})- \sigma^{2}- n\mu^{2} = (n-1)\sigma^{2}$
then $E(S^{2}) = E\Big(\frac{1}{n-1} \sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) = \frac{1}{n-1} E\Big(\sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) =$ $\frac{1}{n-1}(n-1)\sigma^{2} = \sigma^{2}$ | {"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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9597795009613037, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/11117/h-space-structure-on-infinite-projective-spaces | H-space structure on infinite projective spaces
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Any Eilenberg-MacLane space $K(A,n)$ for abelian $A$ can be given the structure of an $H$-space by lifting the addition on $A$ to a continuous map $K(A\times A,n)=K(A,n)\times K(A,n)\to K(A,n)$.
Does somebody know an explicit way to describe this structure in the cases $K({\mathbb Z}/2{\mathbb Z},1)={\mathbb R}P^\infty$ and $K({\mathbb Z},2)={\mathbb C}P^{\infty}$?
-
3 Answers
Look at $\mathbb R^\infty\setminus 0$ as the space of non-zero polynomials, which you can multiply. Pass to the quotient to construct the projective space and, from the multiplication, its $H$-space product.
The complex case is quite the same.
NB: Jason asks in a comment below if this is the same $H$-space structure that Hanno had in mind. To check, we can use the fact that Hanno's is characterised by the fact that if $\mu:K(\mathbb Z_2,1)\times K(\mathbb Z_2,1)\to K(\mathbb Z_2,1)$ is his product and $\alpha\in H^1(K(\mathbb Z_2,1), \mathbb Z_2)$ is the class represented by the identify map $K(\mathbb Z_2,1)\to K(\mathbb Z_2,1)$, then $\mu^*(\alpha)=\alpha\times 1+1\times\alpha$. One should be able to check that this holds for the map given by multiplication of polynomials in a very small skeleton.
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8
As a footnote, the construction does not extend to the quaternionic case since commutativity of multiplication of coefficients is needed in order for the multiplication of polynomials to be well defined modulo scalar multiplication. In the complex case, if you only factor out by scalar multiplication by numbers that are a real number times a p-th root of unity, you get an H-space structure on an infinite-dimensional lens space, a $K({\mathbb Z}_p,1)$. – Allen Hatcher Jan 8 2010 at 7:48
1
Thank you, Mariano & Allen! This is really beautiful, and the structure you described is even strictly associative and unital. What about the (homotopy) inversion of this H-space structure - is there a nice way to describe it, too? – Hanno Becker Jan 8 2010 at 10:14
Is it somehow obvious that this H-space structure and the one Hanno was talking about are the same? – Jason DeVito Jan 8 2010 at 16:41
2
In the complex case, you can also use the fundamental theorem of algebra to replace $\mathbb{CP}^\infty$ to the infinite symmetric power of $\mathbb{CP}^1$. – Reid Barton Jan 8 2010 at 17:17
6
The H-space structure in a $K(A,n)$ is unique up to homotopy since homotopy classes of maps $K(A,n)\times K(A,n) \to K(A,n)$ correspond bijectively with homomorphisms $A\times A \to A$, and the H-space condition says the homomorphism restricts to the identity on each factor so it is just the addition operation in the abelian group $A$. – Allen Hatcher Jan 8 2010 at 17:57
show 1 more comment
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
There's also a different way of writing down the $H$-space structure, that I like for its algebro-geometric flavor. (I'll talk about $\mathbb{C}P^\infty$ here, and $\mathbb{R}P^\infty$ should be analogous.)
Regarding $\mathbb{C}P^\infty$ as a classifying space for complex line bundles, we know that this $H$-space structure is supposed to implement "tensor product of line bundles". In a (not very explicit) sense this tells us the homotopy class of $\mathbb{C}P^\infty \times \mathbb{C}P^\infty \to \mathbb{C}P^\infty$: It represents the line bundle $\mathcal{O}(1,1) = p_1^* \mathcal{O}(1) \otimes p_2^* \mathcal{O}(1)$. We can use this description to write down a much more explicit (and classical) explicit representative.
First, let's recall what the analogous picture looks like for finite projective spaces. The line bundle $\mathcal{O}(1,1)$ determines (upon picking generating sections) the Segre map $\mathbb{C}P^n \times \mathbb{C}P^m \to \mathbb{C}P^{nm+n+m}$ which takes (in homogeneous coordinates)
$([X_0:\ldots:X_n] , [Y_0:\ldots:Y_m]) \mapsto[X_0 Y_0: \ldots : X_i Y_j: \ldots: X_n Y_m]$
where I'm choosing to be vague on the precise ordering of the coordinates. (In the end this won't matter up to homotopy, as the maps will become homotopic upon composing with $\mathbb{C}P^{nm+n+m} \hookrightarrow \mathbb{C}P^\infty$.)
The analogous formula with infinitely many homogeneous coordinate makes just as much sense, one just has to a good ordering of pairs of non-negative integers. Such an infinite Segre map gives another realization of the $H$-space structure.
-
Thank you, Anatoly! I also thought of the Segre map, but didn't notice that the ordering of the factors doesn't matter up to homotopy, therefore struggled when trying to check associativity. – Hanno Becker Jan 8 2010 at 10:27
John Baez has a nice expository page about this called Classifying spaces made easy. About two thirds down the page he talks about multiplicative structure on $\mathbb{C}P^\infty$
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5
Baez is talking about a different version of ${\mathbb C}P^\infty$ from the one topologists usually consider. Namely he takes nonzero rational functions with coefficients in ${\mathbb C}$ modulo scalar multiplication. The rational functions form an infinite dimensional vector space over ${\mathbb C}$, but of uncountable dimension since all the functions $1/(z+a)$ are linearly independent as $a$ ranges over ${\mathbb C}$. This gives a fatter version of ${\mathbb C}P^\infty$ that's actually an abelian group. – Allen Hatcher Jan 9 2010 at 16:59 | {"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": 41, "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}, "config": {"markdown_headings": false, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9235665798187256, "perplexity_flag": "head"} |
http://logiciansdoitwithmodels.com/category/tarskis-theorem/ | # Logicians do it with Models
A self-guided jaunt in philosophical logic
## Archive for the ‘Tarski’s Theorem’ Category
### Hellman’s First Definability Example
September 15, 2010
What Tarski’s Theorem shows is that interpreted formal languages that are interesting (i.e., with enough expressive machinery to represent arithmetic or fragments thereof) cannot contain a predicate whose extension is the set of code numbers (e.g., $\mathsf{Th}(\Omega)^\#$) of sentences true in the interpretation. The extension of any proposed truth predicate in such a system escapes the definitional machinery of the system. Of course, the truth predicate for first-order arithmetic can be defined with appeal to a stronger system, like second-order arithmetic, in the case of the Peano Axioms, etc.
Hellman’s first example is the following. It is a corollary of Tarski’s theorem that a theory in the language, $\textup{L}$, of arithmetic (e.g., an axiom system $\textup{T}$ containing Robinson Arithmetic ($\mathsf{Q}$)) with symbols for zero, successor, addition, and multiplication, when extended with a one place predicate, $\mathsf{Tr}(x)$ (read “true in arithmetic” such that for each closed sentence $\textup{S}$ in $\textup{L}$ a new axiom of the form $\ulcorner \mathsf{Tr}(n) \leftrightarrow \textup{S}\urcorner$ (where $n$ is the numeral for a code number for the sentence $\textup{S}$), the resulting theory $\textup{T}^{*}$ contains no explicit definition of $\mathsf{Tr}(x)$ in $\textup{L}$.
Connecting this to our ongoing discussion of determination of truth and reference in special collections of models, suppose that $\alpha$ is the class of standard $\omega$-models of $\textup{T}^{*}$. Then we have:
• In $\alpha$-structures $\textup{L}$-truth determines $\mathsf{Tr}$-truth.
• In $\alpha$-structures $\textup{L}$-reference determines $\mathsf{Tr}$-reference
Which means that once you have the arithmetical truths in the class $\alpha$, then so are the ‘true-in-arithmetic’ truths and the same goes for the reference of the vocabularies. To avoid collapsing to reductionism via Beth’s theorem, note that there is no first-order theory (like those under discussion) in a language with finitely many non-logical symbols has as it’s models just the models in in $\alpha$.
If you extend $\alpha$ to $\alpha^{*}$ containing all models of $\textup{T}^{*}$, then you do get reductionism, since determination of reference in $\alpha^{*}$ amounts to implicit definability in $\textup{T}^{*}$ -thus showing that there exist non-standard models of arithmetic.
This is a good example because it is clear, based on popular, well established results and firmly shows how determination of truth and reference in one core theory carry over to it’s extension, without thereby reducing the extension to the core.
In the next update I’ll discuss Hellman’s second definability example.
Posted in Robinson Arithmetic, Supervenience and Determination, Tarski's Theorem, Undefinability of Truth | Leave a Comment »
### Tarski’s Theorem
September 14, 2010
Today we’re going to follow Hinman’s proof of Tarski’s theorem. The proof is by contradiction, and employs diagonalization. First, assuming clauses (i)-(iii) on definability, a universal $\textup{U}$-section for the class of $\Omega$-definable sets is established and it’s diagonal is shown to not be definable over $\Omega$.
A $\textup{U}$-section is a set defined for any set $\textup{A}$ and any relation $\textup{U} \subseteq \textup{A} \times \textup{A}$ such that for each $a \in \textup{A}$, $\textup{A}_{a} := \{b : \textup{U}(a, b)\}$. $\textup{U}$ is universal for a class $\mathcal{C} \subseteq \wp (\textup{A})$ if, and only if, every member of the class $\mathcal{C}$ is a $\textup{U}$-section. The diagonal set of $\textup{U}$ is $\textup{D}_{\textup{U}} := \{a : \textup{U}(a, a)\}$.
Second, assuming the definability of $\mathsf{Th}(\Omega)^{\#}$, there is a formula $\phi (y)$ such that for all $p$, $p \in \mathsf{Th}(\Omega)^{\#} \Longleftrightarrow \phi (\dot{p}) \in \mathsf{Th}(\Omega)$. But since the function $\mathsf{Sb}$ is effectively computable, it is definable and so, $\mathsf{Sb}(m) = p \Longleftrightarrow \psi (\dot{m}, \dot{p}) \in \mathsf{Th}(\Omega)$, for some $\psi (x, y)$. But this means that,
$m \in \textup{D}_{\textup{U}} \Longleftrightarrow \exists p [\phi(\dot{p}) \in \mathsf{Th}(\Omega)$ and $\psi(\dot{m}, \dot{p}) \in \mathsf{Th}(\Omega)]$
$m \in \textup{D}_{\textup{U}} \Longleftrightarrow \exists y [\phi(y) \wedge \psi(\dot{m}, y)] \in \mathsf{Th}(\Omega)$,
which means that $\textup{D}_{\textup{U}}$ is definable. This is a contradiction. So, $\mathsf{Th}(\Omega)^{\#}$ is not definable and not effectively countable. Nor is $\mathsf{Th}(\Omega)$ effectively countable.
In the next update (probably on a plane!) I’ll discuss this theorem and get into Hellman’s example.
The theorem showing that the theory determined by the standard model of arithmetic, $\mathsf{Th} (\Omega)$, is undecidable and consequently not decidably axiomatizable (this means that $\mathsf{Th} (\Omega) \not= \mathsf{Th} (\Gamma)$, for some $\Gamma$ consisting of axioms of arithmetic, like those of Peano Arithmetic or one of it’s extensions) can be made stronger by showing that $\mathsf{Th}(\Omega)^{\#} := \{p: \chi_{p} \in \mathsf{Th}(\Omega)\}$, the set of indices of sentences that are true in $\mathsf{Th} (\Omega)$, is not definable over $\Omega$, which means that $\mathsf{Th}(\Omega)$ is not effectively enumerable. This is known as Tarski’s Theorem on the undefinability of truth.
Now set up the (provably effectively computable) function $\mathsf{Sb}(m):= \#(\chi_{m}(\dot{m}))$, which returns the least number $m$ such that $\chi_{m}$ is a formula of $\textup{L}_{\Omega}$. | {"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": 67, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8961449861526489, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/134044-solved-equation-point-line-polar-coords.html | # Thread:
1. ## [SOLVED] Equation for a point on a line in polar coords
[FONT="]Suppose we are given the Polar points P1=(r1, θ1) and P2=(r2, θ2) and we are given a distance D1. How would I find the coordinates of the point P3=(r3, θ3) that lies along the line segment between P1 and P2, and is a distance D1 from point P1?
It would look something like this:
P1--------------P3----------------------P2
|------D1-----|
So we know P1, P2 and distance D1.
I need a formula to find the point P3.
I would like to be able to solve this without converting to Cartesian coordinates because it is in a program that needs to loop as fast as possible - I've done it with the conversion and it is just taking too long. Not to mention the minor precision loss.
Any thoughts?
2. I found the answer.
For anyone else who might be looking for this solution:
The coordinates of the point dividing the line segment P1P2 in the ratio a/b are:
range = $(sqrt[b*b*r_1*r_1+a*a*r_2*r_2+2*a*b*r_1*r_2*cos(theta_2-theta_1)]/[a+b]$
theta = $arctan([b*r_1*sin(theta_1)+a*r_2*sin(theta_2)]/[b*r_1*cos(theta_1)+a*r_2*cos(theta_2)])$
It looks intimidating, but it isn't. The hardest part is making sure you have the correct sign after calculating arctan(). If you are implementing it in Java, you can use:
Code:
```tempTheta = Math.atan2(tempRange, tempAz);
if(tempTheta < 0) tempTheta += (2 * Math.PI);```
Math.atan2(y, x) computes the phase theta by computing an arc tangent of y/x in the range of -pi to pi. In the case of being negative, just add 2*PI.
For more on Points and Lines in Polar Coordinates, look here: Math Forum: Ask Dr. Math FAQ: Polar Coordinates
Thanks anyway! | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9201816916465759, "perplexity_flag": "middle"} |
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### Correct way to compare arrays and do conditional evaluations
I would like to compare two arrays a=: and b= and get: using: (b/.(b_?Positive->a+b))//MatrixForm but this doesn't seem to work? Can ... | {"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": 22, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9126918911933899, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/110274/how-can-we-show-that-i-a-is-invertible | How can we show that $(I-A)$ is invertible?
$A$ is an $n\times n$ matrix with $\|A\|≤a<1$ . I need to prove that the matrix $(I-A)$ is invertible with $\|(I-A)^{-1}\|\le\frac1{(1-a)}$.
It doesn't say anything more. The norm makes me confuse. How can we start to solve this. Could you please help?
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1
Do you know of the geometric series? – Mariano Suárez-Alvarez♦ Feb 17 '12 at 6:39
yes I know geometric series – rose Feb 17 '12 at 6:50
3 Answers
Hint: Show that a certain series converges in the norm $\|\cdot \|$ and that this is an inverse for $I-A$.
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$\bf Hint$: Consider the series $\sum_{n=0}^\infty A^n$
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You can also argue without using the geometric series. The matrix $I-A$ is invertible if and only if $\lambda = 1$ is not an eigenvalue of $A$.
For a contradiction, assume $\lambda = 1$ is an eigenvalue. Then $Ax = x$ for some $x$ with $\|x\| = 1$, so $\|A\| \ge 1$.
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http://math.stackexchange.com/questions/197278/confusion-related-to-context-sensitive-grammar-creation/197282 | # Confusion related to context sensitive grammar creation
I was reading this wikipedia article for generating grammar. They have mentioned that for generating null string in the grammar we can have a rule
$$S \rightarrow \lambda$$
only if $S$ doesn't appear on the right of any production rule.
However in some examples like the following
$L(G) = \{ a^nb^n : n \ge 0 \}$. It has rules like
$$S \rightarrow aSb$$ $$S \rightarrow ab$$ $$S \rightarrow \lambda$$
Here $S$ is appearing on the right. So how come they are using $S \rightarrow \lambda$ rule?
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## 1 Answer
It’s minor sloppiness. Change the productions to
$$\begin{align*} &S\to aTb\\ &S\to ab\\ &T\to aTb\\ &T\to ab\\ &S\to\lambda\;, \end{align*}$$
and you have essentially the same grammar written in a form that meets the requirement.
Added: There are really two different notions involved here. Let $\Sigma$ be the set of terminal symbols and $\mathscr{N}$ the set of non-terminal symbols, and let $S$ be the initial symbol. A context-sensitive production is one of the form $\alpha A\beta\to\alpha\gamma\beta$, where $\alpha,\beta\in(\Sigma\cup\mathscr{N})^*$, $\gamma\in(\Sigma\cup\mathscr{N})^+$, and $A\in\mathscr{N}$; it gets its name because the replacement of $A$ by $\gamma$ occurs only in the context $\alpha\_\beta$. A monotonic production is one of the form $\alpha\to\beta$, where $\alpha,\beta\in(\Sigma\cup\mathscr{N})^*$ and $|\alpha|\le|\beta|$; it gets its name from the fact that the length of the string in a derivation is monotonically non-decreasing. It’s a theorem that a language has a grammar consisting entirely of context-sensitive productions if and only if it has a grammar consisting entirely of monotonic productions. Call such languages purely context-sensitive for the moment.
Unfortunately, these productions don’t allow us to generate any language that contains the empty word. We’d like to make the context-sensitive languages a superset of the context-free languages, some of which do contain the empty word. To do this, we allow the production $S\to\lambda$ provided that $S$ does not appear on the righthand side of any production. It’s a sort of one-time exception to allow us to generate the empty word; it’s not intended to let us generate anything else that wasn’t already able to be generated. In other words, we want to generate only purely context-sensitive languages and languages that would be purely context-sensitive if they didn’t include the empty word.
The reason for not allowing $S$ to appear on the righthand side of any production is that if we don’t make that restriction, we can write grammars that generate languages that would not be purely context-sensitive even if we threw away the empty word. In fact, we could generate every language that can be generated by any formal grammar whatsoever. In terms of the Chomsky hierarchy, we could generate not only all of the type $1$ languages, but also all of the type $0$ languages. The restriction ensures that we really do generate only the type $1$ languages, i.e., those that are purely context-sensitive and those that would be purely context-sensitive if they didn’t include the empty word.
Some grammars that violate the restriction are easily seen to be equivalent to grammars that do not violate it; that’s the case with the one that you asked about, as I showed by replacing it with the grammar above. When that can obviously done, the grammar is sometimes sloppily referred to as a context-sensitive grammar, even though technically it isn’t.
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Yeah thats fine I guess. But I didn't get how $S \rightarrow \lambda$ spoils it. I mean why is it that if S appears on the right we cannot use $S \rightarrow \lambda$ – rajan sthapit Sep 16 '12 at 0:10
@rajan: If you allow $S\to\lambda$ with $S$ on the right, a production like $A\to S$ effectively allows $A\to\lambda$. With that capability you can get Type $0$ formal languages as well as context-sensitive one. – Brian M. Scott Sep 16 '12 at 8:16
But we already have $S \rightarrow \lambda$ which makes it type 0 isn't it. I mean taking your example grammar in the answer. – rajan sthapit Sep 16 '12 at 9:05
@rajan: My example as I rewrote it is context-sensitive. Of course every context-sensitive grammar is also type $0$, but that isn’t what I meant in my previous comment: I meant that you could get all type $0$ langages, as well as the context-sensitive ones. – Brian M. Scott Sep 16 '12 at 9:11
I didn't get it. Can you elaborate more with some examples pls. That will help me understand it. Thanks for your help – rajan sthapit Sep 16 '12 at 13:13
show 3 more comments | {"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": 33, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9415258765220642, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/11647/how-does-human-brain-compare-to-a-modern-cpu-in-energy-per-bit/11654 | how does human brain compare to a modern CPU in energy per bit?
Can someone compare the energy efficiency of human brain as a computer ? What is the energy in joules / flop ? may be some reasonable assumptions on the computational load of common tasks such as pattern recognition or speech synthesis can be used.
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3
– David Zaslavsky♦ Jun 28 '11 at 1:39
The human brain doesn't use bits – JoeHobbit Jun 28 '11 at 5:48
@David that was a very useful link ! – New Horizon Jun 28 '11 at 23:15
– blah Jul 28 '11 at 7:34
2 Answers
Human power consumption can be guesstimated as 100W, similar to the power consumption of an ordinary computer, plus or minus a few orders of magnitude depending on one's idea of "ordinary". A computer can do billions of flops per second, and it would take me many seconds or minutes to perform one with pen and paper, and furthermore I will make many more errors. If we assume that there is some other task which is stacked the opposite way, i.e. a human can perform it a billion times faster than a computer, and that both of these are in some sense extreme cases, then given some more "fair" test we can say that ratio of the efficiency is probably somewhere between $10^{-9}$ and $10^9$.
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6
"somewhere between 10^−9 and 10^9" - a theoretical physicst at work! – Martin Beckett Jun 28 '11 at 15:43
is the 100W a maximum? because it would mean already 2000kcal power consumption a day just by the brain, and I guess this would make every girl happy (just kidding) but seriously speaking what is the power consumption for lazy activities like watching TV compared to doing math or theoretical physics. – Marcel Jul 28 '11 at 9:22
– Ben Crowell Jul 28 '11 at 14:57
@Ben thank you. But also 20W is still a lot... nice to have some idea of the magnitude – Marcel Jul 29 '11 at 12:35
100W is a figure commonly used in A/C design. That's the total heat output of a human awake and at rest, rounded to a nice figure for engineering purposes. 20W sounds about right for the proportion used by the brain. – MSalters Jul 29 '11 at 13:54
The brain is massively parallel, so it tends to come out looking very good. The OP suggested using joules/flop as the measure of (in)efficiency. This leaves considerable ambiguity. I believe the way neurons typically work is that they form something like a weighted average of their binary inputs, and generate a binary output that is based on a threshold value for that average. I would consider this to be the moral equivalent of a floating-point operation. Of course if a floating-point operation means working out a long-division problem using paper and pencil, then the result is going to be horrible -- a kilojoule per flop for me, or infinitely many joules per flop for a kindergartener who hasn't yet learned the long division algorithm. To me it seems perverse to say that a kindergartener's brain has zero efficiency compared to an x86, so I'm going to equate one neuron's weighted-average operation to one flop.
I'm not a big fan of Ray Kurzweil, but he does have a good summary of some relevant data in his 2005 book The Singularity is Near. There's a lot of ambiguity in trying to estimate the number of fundamental operations involved in a certain neurological process. Kurzweil refers to "synaptic transactions," and equates one of those to be something like $10^3$ "arithmetic" operations, where I imagine that he means something roughly similar to my definition of an arithmetic operation above. Anyway, subject to all these ambiguities, the studies he cites estimates of $10^{14}$-$10^{19}$ Hz for the rate of operations per second. If the brain draws ~10 W (Dan's link says 20), then this is an energy consumption of $10^{-13}$-$10^{-18}$ joules per operation. Since a desktop computer currently does $\sim10^9$ arithmetic operations per second, this makes the brain more efficient, as measured by joules per operation, by about a factor of $10^6$-$10^{10}$.
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http://crypto.stackexchange.com/questions/5675/polynomials-and-efficient-computability?answertab=active | # Polynomials and efficient computability
In public key crypto, the popular definitions of security (CPA, CCA1,2) depend on PPT adversaries. I'm trying to understand why adversaries should be PPT.
It's clear that adversaries should be at most probabilistic sub-exponential, because otherwise they could exhaust the message space. It's also clear that adversaries should be at least PPT, because any computer can run low-order polynomial algorithms (barring unusual cases) and there's no good way to distinguish low-order polynomials from very high-order (and practically inefficient) polynomials.
So why not allow the adversaries to be probabilistic quasi-polynomials? Where quasi-polynomial means its time complexity is larger than all polynomials and smaller than all exponentials, like $n^{log(n)}$. Is there some example showing why it would not make sense for an adversary to be probabilitic quasi-polynomial?
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## 2 Answers
The reason I was taught why poly-time is the conventional way to define "efficient" algorithms is that it's the smallest notion closed under various forms of composition.
If we have an algorithm for deciding whether some input is in some language and we let N be the length of the input, then we'd like to call all the following operations efficient:
• Returning "yes" or "no" all the time costs one (or constant) time.
• Reading the input costs N time.
• Calling one efficient algorithm taking time A(N), then passing the result through an efficient algorithm taking time B(N) gives a total time of A(N) + B(N), so our class of "efficient running times" should be closed under addition.
• If an algorithm taking A(N) time on its own is efficient, and you have an algorithm with A(N) steps where each step is a call to another efficient algorithm taking B(N) time, then you end up with A(N)*B(N) time, so "efficiency" should be closed under multiplication.
Plug all that together and you get "polynomial in N" as the smallest possible "efficiency notion" - which does not of course exclude other, larger notions such as the ones mentioned above.
This is not so much a cryptography as a general complexity theory/theoretical CS point, which cryptographers have borrowed for their notions of security.
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There is nothing stopping you from doing this. I don't know of any cryptosystem that is in use today that would be considered unsecure against a quasi-polynomial time adversary. RSA is probably the closest. It is super-polynomial (thanks to GNFS), so it would still be secure in a quasi-poly time adversarial model.
The only argument I see against quasi-polynomial time adversary models is that it obviously does not reflect reality as well as the PT adversary (otherwise we would see much larger RSA keys in use).
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1
"otherwise we would see much larger RSA keys in use" I wouldn't conclude that. Those definitions are asymptotic and don't say anything about concrete key sizes. – CodesInChaos Dec 13 '12 at 8:04
Thank you, but it's not obvious to me why quasi-poly doesn't reflect reality as well as PT? If there's nothing stopping us from taking this definition, then someone would have taken it as it's more general, so I'd like to know why quasi-poly doesn't work in our security definitions. – Stuart Dec 13 '12 at 9:35 | {"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}, "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}, "english_confidence": 0.9622342586517334, "perplexity_flag": "middle"} |
http://mathhelpforum.com/geometry/43983-resolution-triangles.html | # Thread:
1. ## Resolution of triangles?
If in a triangle ABC, length of the sides a,b & c are in Arithmetic Progression, then it is necessary that:
A)2/3 < b/c < 2
B)1/3 < b/c < 2/3
C)2/3 < b/a < 2
D)1/3 < b/a < 2/3
More than one options are correct. How to solve this problem?
2. Hello,
What about something using the law of sines ?
And is there more information about a,b and c ? Because propositions A) and C) are equivalent, and so are B) and D)
3. Hello, fardeen_gen!
I have a start on this problem, but I haven't finished it yet.
. . Maybe this will inspire you . . .
If in a triangle $ABC$, length of the sides $a,b,c$ are in Arithmetic Progression,
then it is necessary that:
. . $\begin{array}{cccccc}(A)& \dfrac{2}{3} & < & \dfrac{b}{c} & < & 2 \\ \\[-2mm]<br /> (B)& \dfrac{1}{3} &<& \dfrac{b}{c} &<& \dfrac{2}{3} \\ \\[-2mm]<br /> (C)& \dfrac{2}{3} &<& \dfrac{b}{a} &<& 2 \\ \\[-2mm]<br /> (D)& \dfrac{1}{3} &<& \dfrac{b}{a} &<& \dfrac{2}{3} \end{array}$
More than one option is correct. How to solve this problem?
Assume that $a,b,c$ are ordered from smallest to largest: . $a \,<\, b \,<\, c$
Since the form an arithmetic progression, there is a common difference $d.$
Then the three sides are: . $b-d,\;b,\;b+d$
The Triangle Inequality says: . $a + b \:>\:c$
. . which gives us: . $(b-d) + b \:>\:b+d\quad\Rightarrow\quad b \:>\:2d \quad\rightarrow\quad d \:<\:\frac{b}{2}$
Add $b$ to both sides: . $b + d \:< \:b + \frac{b}{2} \quad\Rightarrow\quad b+d \:<\:\frac{3b}{2}$
Take reciprocals: . $\frac{1}{b+d} \:>\:\frac{2}{3b}$
$\text{Multiply by }b\!:\;\;\frac{b}{\underbrace{b+d}_{\text{This is }c}} \:<\:\frac{2}{3}$
. . Therefore: . $\frac{b}{c}\:<\:\frac{2}{3}$
We have half of statement (B); I'm still working on the other half.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
We have: . $d \:<\:\frac{b}{2}$
Multiply by -1: . $-d \:>\:-\frac{b}{2}$
Add $b$ to both sides: . $b - d \:>\:b -\frac{b}{2} \quad\Rightarrow\quad b - d \:>\:\frac{b}{2}$
Take reciprocals: . $\frac{1}{b-d} \:<\:\frac{2}{b}$
$\text{Multiply by }b\!:\;\;\frac{b}{\underbrace{b-d}_{\text{This is }a}} \:<\:2$
. . Therefore: . $\frac{b}{a} \:<\:2$
And we have half of statment (C). | {"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": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.8620301485061646, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/105971?sort=oldest | ## How should an analytic number theorist look at Bessel functions?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
(And a related question: Where should an analytic number theorist learn about Bessel functions?)
Bessel functions occur quite frequently in analytic number theory. One example, Corollary 4.7 of Iwaniec and Kowalski, says the following. Let $r(m)$ be the number of representations of $m$ as two squares, and suppose that $g$ is smooth and compactly supported in $(0, \infty)$. Then,
$$\sum_{m = 1}^{\infty} r(m) g(m) = \pi \int_0^{\infty} g(x) dx + \sum_{m = 1}^{\infty} r(m) h(m),$$
where $$h(y) = \pi \int_0^{\infty} g(x) J_0(2 \pi \sqrt{xy}) dx.$$
$J_0(x)$ is a Bessel function, and I+K follow with four equivalent integral expressions -- the equivalence of which is not at all obvious by looking at them. Looking at Lemma 4.17, the relevance appears to be that Bessel functions arise when you take Fourier transforms of radially symmetric functions.
Another example comes from (3.8) of this paper of Miller and Schmid, and the relevance comes from the identity
$$\int_0^{\infty} J_0(\sqrt{x}) x^{s - 1} dx = 4^s \frac{\Gamma(s)}{\Gamma(1 - s)},$$
where the gamma factors come from functional equations of $L$-functions. Okay, if this is true, then I understand why we care, but it seemed a bit deus ex machina to me.
There are many other examples too, for example the Petersson formula in the theory of modular forms, etc. There are $I$-Bessel functions, $K$-Bessel functions, $Y$-Bessel functions, etc., all of which seem to satisfy a dizzying number of highly nontrivial identities, and reading Iwaniec and Kowalski one gets the sense that an expert should have the ability to recognize and manipulate them on sight. They also provide references to, e.g., (23.451.1) of a book by Gradhsteyn and Rizhik, and although I confess I have not looked at it, I can infer from the formula number that it is not the sort of thing I might read on an airport layover.
Meanwhile, Wikipedia tells me that they naturally arise as solutions of certain partial differential equations. Looks extremely interesting, although I'm afraid I am not an expert in PDE.
As an analytic number theorist, how might I make friends with these objects? How should I look at them, and what conceptual frameworks do they fit in? Thank you!
(ed. Thanks to everyone for informative answers! I could only accept one answer but +1 all around)
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You might want to look at Siegel's work on transcendental numbers. – Franz Lemmermeyer Aug 30 at 17:16
My impression is that many families of special functions fit naturally into the setting of representation theory; they are matrix coefficients of suitable representations of suitable groups. A simple example is the trigonometric functions, which are matrix coefficients of $S^1$. Bessel functions are apparently related in this way to $\text{Isom}(\mathbb{R}^2)$. (But I don't know how much analytic number theorists like representation theory.) – Qiaochu Yuan Aug 30 at 17:21
Zach Kent once told me that he had to figure this out for himself as part of his thesis. Maybe get in contact with him? – stankewicz Aug 30 at 17:54
Actually, analytic number theorists like representation theory a lot! For example, modular forms can be described in terms of representations of $GL_2$ on certain function spaces. – Frank Thorne Aug 30 at 18:32
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Bessel functions appear as coefficients in series expansions of automorphic forms on $GL_2(K)$, where $K$ is an imaginary quadratic field, in much the same way that exponential functions appear in the $q$-expansions of modular forms. This is explained rather nicely in Shai Haran's 1987 Compositio paper on p-adic L-functions. (This comment is more or less a special case of Qiaochu's comment.) – David Loeffler Aug 30 at 18:50
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## 6 Answers
Radial Fourier transforms provide a good, consistent perspective on most of the theory. The Fourier transform $\widehat{f}(t)$ of a function $f \colon \mathbb{R}^n \to \mathbb{R}$ is given by the integral of $f(x) e^{2\pi i \langle x,t \rangle} \, dx$ over $x \in \mathbb{R}^n$. If $f$ is a radial function (i.e., $f(x)$ depends only on $|x|$), then we can radial symmetrize everything and the exponential function averages out to a radial function. Specifically, we get $$\widehat{f}(t) = 2\pi |t|^{-(n/2-1)} \int_0^\infty f(r) J_{n/2-1} (2 \pi r |t|) r^{n/2} \, dr.$$ The precise factors are a little annoying, but basically this just means $J_{n/2-1}$ is what you get when you radially symmetrize an exponential function in $n$ dimensions. It's easy to see that if you symmetrize $e^{2\pi i \langle x,t \rangle}$ by averaging over all $x$ on a sphere, then you get a radial function of $t$, and furthermore as you vary the radius of the sphere you just rescale the function. So the one function $J_{n/2-1}$ captures all of this, modulo scaling.
One consequence is that Bessel functions inherit the orthogonality of the exponential functions (i.e., the different scalings are orthogonal), so they also inherit all the consequences of orthogonality. For example, this is really where the differential equation comes from. There's a strong analogy between Bessel functions and orthogonal polynomials, where rescaling the Bessel function corresponds to varying the degree of the polynomial.
You also get certain qualitative results for free: for example, the product of two Bessel functions should be an integral of Bessel functions with positive coefficients, since this corresponds to saying the product of two radial, positive-definite functions remains positive definite. You can write down the coefficients explicitly, but sometimes all you need is nonnegativity, and in any case this point of view makes it easy to believe that there should be an explicit formula.
This is basically a low-brow version of the representation theory approach. Basically, ordinary Fourier analysis studies $L^2(\mathbb{R}^n)$ under the action of the translation group $\mathbb{R}^n$. If you look at the full group of isometries of $\mathbb{R}^n$ (including the orthogonal group), then it's just a little more elaborate, and the Bessel functions arise as zonal spherical functions. It's worthwhile working through this perspective, but in practice just thinking about radial Fourier analysis gives you most of the benefits with less machinery.
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Isn't there some $f$ missing in the formula for $\hat f$? – Dirk Aug 30 at 20:42
Oops, thanks! I just added it. – Henry Cohn Aug 30 at 20:56
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I like this, of course, though it still does not account for Bessel functions whose order is not in $\frac12\bf Z$. – Noam D. Elkies Aug 31 at 0:09
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From the point of view of analytic number theory, the point usually would be asymptotic behaviour. This is typically well understood, and is in the massive book of Watson. Apart from that, yes, numerous identities (which are not that deep, I feel), and an analogy with Kloosterman sums, though some would read that the other way round. A part of special function theory that appears quite well explored.
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Expanding on Qiaochu's and David's comments, from the point of view of automorphic forms and automorphic representations, a good reference is "Special Functions and the Theory of Group Representations" by N. Ja. Vilenkin and published by the AMS.
"Special Functions" by Andrews, Askey, and Roy published by CUP is another good reference with an anayltic number theory slant.
EDIT: The point here is that for a finite group $G$ and a representation $\pi$ into a finite dimensional vector space $V$ with orthonormal basis $\{\vec{e_i}\}$, the function $g\to \langle \vec{e_i},\pi(g)\vec{e_j}\rangle$ gives the $i,j$ coefficient of the matrix associated to $\pi(g)$, or 'matrix coefficient function'. For a Lie group $G$ with an infinite dimensional representation $\pi$ in a Hilbert space $H$, there is a natural analog, and the functions which arise this way play an obviously important role in harmonic analysis in $G$. Bessel functions can be interpreted this way.
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One way is to think of Bessel functions as some sort of non-abelian analogues of the exponential function. In fact, you can form a complete orthogonal system of $L^2 (\mathbb{R^+}, x^{-1}d x)$ using the Bessel functions. See Chapter 16 of Iwaniec-Kowalski (page 411).
For manipulations, the formulae given in Appendix B of Iwaniec's Spectral Methods book usually suffice.
Yes, Dan Brown is more appropriate than Gradhsteyn and Rizhik for airport reading.
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There are many, many things that can be said here! I think there are two slightly different mechanisms that conjure up Bessel functions of various types, namely, Euclidean Laplacians and separation of variables, and $SL_2(\mathbb R)$ (and orthogonal groups $O(n,1)$) Casimir or Laplace-Beltrami operator and separation of variables. The fact that there cannot be a greater variety of second-order ordinary differential equations with certain control and types of singular points goes back to Riemann.
A very tangible connection with down-to-earth examples from automorphic forms: just as the function $z\rightarrow y^s$ is the spherical vector in an unramified principal series, the integral $\int_{-\infty}^\infty e^{-ix} {y^s\over |cz+d|^{2s}}\;dx$ that is the starting point for integral expressions for $K$-type Bessel functions is the image of the spherical vector $y^s$ under the obvious (from the viewpoint of "Mackey theory") intertwining from that principal series to the Whittaker space.
Not only in the Euclidean case, but also generally, very many "formulas" are manifestations either of a provably unique intertwining operator (given by an integral), or of a Plancherel theorem for the situation at hand.
(The fact that Mellin transforms of Bessel functions (for the Fourier expansions of waveforms) are expressible in terms of Gamma is not typical of integral transform methods on larger groups, unfortunately. The classical computations of archimedean integrals, for Rankin-Selberg, etc., for $SL_2(\mathbb R)$ are done in a self-contained fashion in http://www.math.umn.edu/~garrett/m/v/standard_integrals.pdf, for example.)
The asymptotics are more systematically understandable from the more general results on asymptotics of solutions of second-order ordinary differential equations. The regular singular point theory is very old, and even the good-irregular singular point case has been essentially understood since Poincare. It turns out that fairly simple heuristics are provably correct, and therefore function as excellent mnemonics. (I wrote up some examples and proofs in a more contemporary style in some course notes: http://www.math.umn.edu/~garrett/m/mfms/notes_c/reg_sing_pt.pdf, http://www.math.umn.edu/~garrett/m/mfms/notes_c/irreg_sing_pt.pdf, and http://www.math.umn.edu/~garrett/m/mfms/notes_c/frobenius_ode.pdf ... In particular, this is not about PDEs, but about the ODEs obtained after various separations of variables.)
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For an analytic number theorist the most useful way to look at Bessel functions and related special functions (e.g. Whittaker functions or products of two such functions) is through their Mellin transforms. The Mellin transform explains:
(1) the shape of gamma factors in various automorphic $L$-functions
(2) the shape of Voronoi summation (as in your example)
(3) the behavior of the Bessel function at zero and at infinity
(4) the Taylor expansion of the Bessel function
The differential equation is indispensable when one needs to estimate the Hankel-type transforms arising in (2), or the Bessel transforms that arise in the Petersson and Kuznetsov formulae. Techniques involve integration by parts or passing to the Mellin side and deforming the contour.
Studying the formulae in Gradshteyn-Ryzhik is very useful for an analytic number theorist, and here I mean the whole book, not just the parts with Bessel functions!
For a conceptual understanding I recommend Part II of Representation theory and noncommutative harmonic analysis 2 (Enc.Math.Sci.59, Springer): Representations of Lie groups and special functions by Klimyk and Vilenkin.
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http://www.physicsforums.com/showthread.php?t=16742 | Physics Forums
## batteries
What does it mean when it's said that the positive terminal of a battery is at a higher potential than the negative end? How does electric potential (voltage) relate to electric current/electric fields?[o)]
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Recognitions: Science Advisor Batteries (voltaic cells if you want to google it) work by having 2 chemical half reactions happen at the same time. Have a look at this table: reduction table The reactions can go either left or right, the voltage is reversed if the reaction goes right to left. Reactions that go right to left happen at the negative terminal. Reactions that go left to right happen at the positive terminal. The voltage of the cell is the sum of the voltages for each of the half reactions. Now as to why the positive can have higher voltage, suppose you had a cell made of perchloric acid (fifth from the top on the left side) and iron (15 from the bottom on the right side). The perchloric acid reaction looks like this: $$ClO_4^- + 8H^+ + 8e^- ---> Cl^- + 4H_2O$$ (1.39 volts) The iron reaction looks like this $$Fe ---> Fe^{2+} + 2e^-$$ (0.45 volts) You can see that the positive terminal (the acid) has a higher potential than the negative terminal. Voltage is sort of like the force trying to make electrons flow. Higher voltage means higher current.
You sound like a chemist…god bless you!!! What do you mean when you say “2 chemical half reactions”? Do you mean that there is a chemical reaction in each terminal? How do the electrons know which is the positive and negative terminals? Also, how can the reactions be reversible? I thought that was only the case for rechargeable batteries? Can you tell I’m not a chemist? Keep it simple for the aspiring physicist over here... [;)]
Recognitions:
Science Advisor
## batteries
What do you mean when you say “2 chemical half reactions”? Do you mean that there is a chemical reaction in each terminal?
That's exactly what it means. The cell is separate into 2 sides. For the perchloric acid and iron cell I mentioned, the perchloric acid is only on the positive side, and the iron is only on the negative side.
The reaction that happens at each terminal is called a half-reaction because 2 half-reactions are needed before anything happens. The perchloric acid and iron cannot just break down on their own. The acid needs the iron before it can react and the iron needs the acid before it can react.
How do the electrons know which is the positive and negative terminals?
It's based on how spontaneous the reaction is. Look at that table again, Fe 2+ is listed twice.
$$Fe^{2+} + 2e^- ---> Fe$$ Is lower on the table
$$Fe^{3+} + e^- ---> Fe^{2+}$$ is higher on the table.
In one half-reaction, it's being oxidized, in the other reaction, it's being reduced. The terminals on the cell are positive or negative because one of the half-reactions is a strong oxidation reaction and one is a strong reduction half-reaction. In the perchloric acid and iron cell, the acid is able to pull the electrons off the iron. If iron was able to pull electrons away from the acid, the iron would be the positive terminal and the acid would be the negative terminal. That doesn't happen because the acid has a stronger affinity for electrons than the iron.
Also, how can the reactions be reversible? I thought that was only the case for rechargeable batteries?
The reactions can be reversed if you supply a voltage that is higher than the voltage the cell outputs. If a cell supplies 1 volt, you can reverse the reaction by supplying more than 1 volt to the cell.
To the reverse the reaction, positive on the cell should be connect to the positive on the power supply. Negative on the cell should be connected to the negative on the power supply. If you hook it up the other way, the cell will be drained of all energy very quickly.
thank you so much for clearing that up!!! ps i love undergrads =p
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http://en.wikipedia.org/wiki/L1-norm | # Taxicab geometry
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Taxicab geometry versus Euclidean distance: In taxicab geometry all three pictured lines (red, yellow, and blue) have the same length (12) for the same route. In Euclidean geometry, the green line has length 12/√2 ≈ 8.48, and is the unique shortest path.
Taxicab geometry, considered by Hermann Minkowski in the 19th century, is a form of geometry in which the usual distance function or metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the absolute differences of their coordinates. The taxicab metric is also known as rectilinear distance, L1 distance or $\ell_1$ norm (see Lp space), city block distance, Manhattan distance, or Manhattan length, with corresponding variations in the name of the geometry.[1] The latter names allude to the grid layout of most streets on the island of Manhattan, which causes the shortest path a car could take between two intersections in the borough to have length equal to the intersections' distance in taxicab geometry.
## Formal definition
The taxicab distance, $d_1$, between two vectors $\mathbf{p}, \mathbf{q}$ in an n-dimensional real vector space with fixed Cartesian coordinate system, is the sum of the lengths of the projections of the line segment between the points onto the coordinate axes. More formally,
$d_1(\mathbf{p}, \mathbf{q}) = \|\mathbf{p} - \mathbf{q}\|_1 = \sum_{i=1}^n |p_i-q_i|,$
where
$\mathbf{p}=(p_1,p_2,\dots,p_n)\text{ and }\mathbf{q}=(q_1,q_2,\dots,q_n)\,$
are vectors.
For example, in the plane, the taxicab distance between $(p_1,p_2)$ and $(q_1,q_2)$ is $| p_1 - q_1 | + | p_2 - q_2 |.$
## Properties
Taxicab distance depends on the rotation of the coordinate system, but does not depend on its reflection about a coordinate axis or its translation. Taxicab geometry satisfies all of Hilbert's axioms (a formalization of Euclidean geometry) except for the side-angle-side axiom, as one can generate two triangles each with two sides and the angle between them the same, and have them not be congruent.
### Circles in Taxicab geometry
Circles in discrete and continuous taxicab geometry
A circle is a set of points with a fixed distance, called the radius, from a point called the center. In taxicab geometry, distance is determined by a different metric than in Euclidean geometry, and the shape of circles changes as well. Taxicab circles are squares with sides oriented at a 45° angle to the coordinate axes. The image to the right shows why this is true, by showing in red the set of all points with a fixed distance from a center, shown in blue. As the size of the city blocks diminishes, the points become more numerous and become a rotated square in a continuous taxicab geometry. While each side would have length √2r using a Euclidean metric, where r is the circle's radius, its length in taxicab geometry is 2r. Thus, a circle's circumference is 8r. Thus, the value of a geometric analog to $\pi$ is 4 in this geometry. The formula for the unit circle in taxicab geometry is $|x| + |y| = 1$ in Cartesian coordinates and
$r = \frac{1}{| \sin \theta| + |\cos\theta|}$
in polar coordinates.
A circle of radius r for the Chebyshev distance (L∞ metric) on a plane is also a square with side length 2r parallel to the coordinate axes, so planar Chebyshev distance can be viewed as equivalent by rotation and scaling to planar taxicab distance. However, this equivalence between L1 and L∞ metrics does not generalize to higher dimensions.
Whenever each pair in a collection of these circles has a nonempty intersection, there exists an intersection point for the whole collection; therefore, the Manhattan distance forms an injective metric space.
A circle of radius 1 (using this distance) is the von Neumann neighborhood of its center.
## Applications
### Measures of distances in chess
In chess, the distance between squares on the chessboard for rooks is measured in Manhattan distance; kings and queens use Chebyshev distance, and bishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop's case).
## References
• Eugene F. Krause (1987). Taxicab Geometry. Dover. ISBN 0-486-25202-7. | {"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": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9236451387405396, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/28803?sort=votes | ## What is the proper initiation to the theory of motives for a new student of algebraic geometry?
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A preliminary apology is in order: I realize that most of my contributions to this site are in the form of reference requests. I understand that this makes it seem as though I do nothing more than sit around most of the time, soaking in as much advanced mathematics as possible, despite my position as a lowly undergraduate. In all actuality, this couldn't be more accurate; I really do just sit around reading maths most of the time.
Alright, now that I put that out there, I am curious as to where I might find a coherent treatment of the theory of motives; one which is below the level of a professional mathematician and roughly suited for readers of Hartshorne or Eisenbud/Harris's wonderful scheme theory text. That is, I want to understand the discipline which I hear extolled as beautiful and complex by researchers in the field, but which is notoriously abstruse and difficult to learn/understand. I wonder if expositions of the theory of motives are necessarily highly technical, or if it is approachable to the ambitious advanced undergraduate.
Thank you again, MO community, for imparting your wisdom regarding good references. It is very much appreciated =)
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Your question is like asking for a treatment of Deligne's work on mixedness of $\ell$-adic sheaves at the level of Hartshorne, or the Langlands Program at the level of Serre's book on representations of finite groups. Wait until you've mastered several different cohomology theories in algebraic geometry (e.g., coherent cohomology, etale cohomology, topological cohomology) and the relations among them. Then you'll be in position to appreciate the point (and why it doesn't necessarily matter to even have a definition of "motive" for the idea to be useful). – Boyarsky Jun 20 2010 at 4:59
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You can find references in the answers to Ilya's question: mathoverflow.net/questions/2146/… Unfortunately, none of them are aimed at new students of algebraic geometry. – S. Carnahan♦ Jun 20 2010 at 5:34
## 8 Answers
Asking for the moon, in my view. Here are 10 "heuristics" that try to place the theory. NB that many people stop at #1, as if this were enough. None of these points is particularly easy to track in the literature, AFAIK.
1. An irreducible variety V is going to be treated as a "molecule" in this theory, not an "atom". Motives are in a sense parts of varieties. If you count points over finite fields this can look combinatorial (e.g. Euler's formula for the triangulations of a sphere) but has to go a lot deeper.
2. In cohomology of a variety of dimension n, the top relevant dimension has to be 2n, for reasons that are easy to see over the complex numbers (real dimension), but in general have to do with topological intuitions, such as ramification taking place in codimension 2.
3. Grothendieck's big-scale pattern of thought involves defining a whole category at once, and understanding it by means of category-level structures and concepts. The "category of motives" is to be understood, in particular its Hom-sets. These are to be modelled on the idea of algebraic correspondence. So it's morally a category of relations.
4. Algebraic cycles (i): Generally in homology theory, the modern approach is to start with a very abstract definition and worry later about how to represent a class concretely. Here the opposite approach is useful - algebraic cycles are traced on varieties by combinations of subvarieties.
5. Algebraic cycles (ii): Algebraic cycles need to be subject to equivalence relations, such as linear equivalence for divisors. There are significant technical issues here (vaguely replacing homotopies).
6. Algebraic cycles (iii): There is (or may be) a paucity of algebraic cycles. Cf. the Hodge conjecture. In other words we lack existence proofs in general.
7. Problem-solving (i): Assume enough about a good category of motives and you get a conditional proof of the Weil conjectures.
8. Problem-solving (ii): Motives can conjecturally account (coarsely, Lie algebra level) for the images of Galois representations on l-adic cohomology.
9. Top-down view: Motives solve the problem of what would be the "universal Weil cohomology", at least in the best of all possible worlds.
10. Grothendieck's period conjecture: a concrete out-turn in transcendence theory is the conjectural upper bound for the transcendence degree of the periods of abelian varieties. Motives can "catch" enough algebraic cycles to do this.
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You're probably right--carriage before the horse, if you'll excuse the cliche. However, I am sure that it is understandable for one to want to look at sophisticated constructions and methods used by modern research in a particular subject area to provide motivation. I want to experience the wonderment that I felt when passing from calculus to analysis, or basic number theory/algebra to more advanced material. With a subject so rich as AG, it is more difficult to 'see your endgame' as it were. I am sure that you will understand my enthusiasm here on some level. We were all naive once :) – lambdafunctor Jun 22 2010 at 4:39
Should really add that the original theory is really not known to work (after 45 years). There are ways round this (absolute Hodge cycles, motivic cohomology), but these are less accessible. – Charles Matthews Jun 22 2010 at 7:30
Right, I look at things which aren't accessible to me (yet) as a proverbial gauntlet having been thrown down. I have a love for abstruse math, for some reason. But again, I partially just want to see what all the 'fuss' is about with motives (and the work of Voevodsky, Beilinson, et al.)--that is, I realize that it is brilliant from what I can gather, and I'm just looking for the 'why'. – lambdafunctor Jun 23 2010 at 5:20
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Some say motive as in motivation, some say motif as in recurring theme in music. In some concrete cases it is actually not so hard to see what is going on: the category is defined from algebraic correspondences by splitting idempotents (Karoubi construction), which is not so terrible. The explicit examples such as Fermat varieties tend to fall in the motif class, while the broad categorical properties hoped for in the motive class. The flamboyant Nick Katz talk ihes.fr/jsp/site/… shows why these ideas live on. – Charles Matthews Jun 23 2010 at 7:33
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There is a very friendly introduction to motivic homotopy theory starting even below Hartshorne level: The lecture notes from the Nordfjordeid Summer school on Motivic Homotopy Theory. They consist of three chapters:
1.Topological Prerequisites (Dundas)
2.Algebro-Geometric Prerequisites (Levine)
3.Motivic Homotopy Theory (Voevodsky/Röndigs/Østvær) which is not legally available online but has almost the same content as Voevodsky's ICM talk
Note that it is on the level you asked for and from this source you learn in some detail about one modern approach, but not about classical motives and all the "yoga" and motivation and intuition connected to number theory. The closest to being a coherent source going more into that direction is probably Yves André's book (the link gives you the table of contents), otherwise I just know of scattered notes and articles
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Yes, André's book review: smf4.emath.fr/Publications/Gazette/2007/111/… is IMO a very beautifull introduction including mixed motives, even if some people react allergic to it. – Thomas Riepe Jun 20 2010 at 13:31
Manin's old article (the first publication on motives, acc. to the author "an exercise" by Grothendieck) is very readable and beautifully written. Very readable and good too are Kleimann's "Motives" in "Algebraic geometry", Oslo 1970 (Proc. Fifth Nordic Summer-School in Math., Oslo, 1970), pp. 53--82 , and Demazure's article of that time.
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By far the best introduction I've seen is Milne's introduction Motives-Grothendieck's dream from his webpage. Beware though that there are a lot of things that you ought to know before you can fully appreciate everything in the paper. But even a superficial reading is highly rewarding and motivating!
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Motivating... No pun intended, right? – lambdafunctor Jun 20 2010 at 17:10
^my shamelessly poor sense of humor aside, I now absolutely love this article! I never knew that this gorgeous theory plays such a fundamental role in things like Birch & Swinnerton-Dyer or the Langlands Program, or so many other unsolved problems. Of course, you are right; I do not understand everything (a forteriori, a great deal) of what is contained therein, but it is certainly highly rewarding and... well, motivating. – lambdafunctor Jun 20 2010 at 17:15
Actually, the pun was uninteded (but oh how much cooler would I appear if it wasn't!? ;) ) Glad to be of assistance. – Daniel Larsson Jun 20 2010 at 19:17
Where's the beef? Looks like just definitions and formalism. The philosophy of "thinking motivically" is influential, but until there is a proof of the standard conjectures I don't see what to get excited about (like Goldfeld's strategy before Gross-Zagier). The definition looks too easy to tell us something deep; in what sense is it even a "theory"? Does work of Voevodsky et al. rest on a more sophisticated concept of "motive"? If so, maybe that would be more compelling (on par with etale cohomology, which I know very well). Why is lambdafunctor so excited? Sorry for being a party pooper... – Boyarsky Jun 21 2010 at 2:15
Well I don't think that I'm any more excited than many in the community get when approaching the methods of Voevodsky et al, or the conjectures of Beilinson, etc. I mean, if there weren't something to get excited about, as it were, then I don't think that so many brilliant mathematicians would spend their time thinking about it. I don't understand all of the mechanisms at play here, mind you. I am just partial because my undergraduate adviser is an A^1 homotopy theorist who has also done a lot of work in motivic cohomology. I just wanted to get to a level where I could understand it better. – lambdafunctor Jun 22 2010 at 4:35
show 3 more comments
The word "motive" has a lot of different (although highly related) meanings. I suggest you go ahead to learn about "pure Chow motives" first, before looking at the more complicated theory of mixed motives.
For motivation, it is necessary to have seen at least one Weil-cohomology theory, so you might want to have a look at the Weil conjectures, too.
For technical stuff, you should know what an abelian category is (and then learn the rest along the way).
Related to the theory of motives are also: K-Theory, (stable) homotopy theory of schemes, intersection theory (Chow groups). If you have an interest in any of these topics, it might be good to look at a treatment that covers the relationsship between this and motives, to give a little bit more motivation.
Since there is no abelian category of mixed motives yet, but instead what "feels like" it's derived category, you might want to learn a little bit about derived categories and triangulated categories before walking to (Voevodsky's theory of) mixed motives.
Of course, there is also the AMS Notices article What is ... a motive? by Barry Mazur.
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Well, this is not specifically motives but I liked reading these notes http://www.math.northwestern.edu/~eric/lectures/zurich/ by Eric Friedlander. These lectures introduce a lot of things you want to know if you are interested in motives.
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I would suggest you: A Pamphlet on Motivic Cohomology, Luca Barbieri-Viale, http://arxiv.org/abs/math/0508147 (or the published version) and/or Bruno Kahn's http://people.math.jussieu.fr/~kahn/preprints/kcag.pdf
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Here is a very understandable introductory article by R. Sujatha. For a beginning student this is good.
In my case, after that article, my next encounter with motives was with the more precise definition of a motive from the initial parts of Deligne's monograph “Le Groupe Fondamental de la Droite Projective Moins Trois Points”. It even sort of defines a mixed motive; in fact it is the only definition of mixed motive that I know.
Read the Mathscinet review and also, Jordan Ellenberg's opinion on this remarkable paper of Deligne. I myself was astonished when I first looked into it and saw how much stuff was contained in it.
Deligne's paper "Formes modulaires et représentations $l$-adiques" proving that the Weil conjectures imply the Ramanujan conjecture, is almost close to the theory of motives even though it does not explicitly mention motives. Here the representations of the absolute Galois group on the étale, or rather on the $\ell$-adic, cohomology is considered. This might give some starting insight into the Galois representations approach to motives.
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http://physics.stackexchange.com/questions/36249/speed-of-sound-and-the-potential-energy-of-an-ideal-gas-goldstein-derivation | speed of sound and the potential energy of an ideal gas; Goldstein derivation
I am looking the derivation of the speed of sound in Goldstein's Classical Mechanics (sec. 11-3, pp. 356-358, 1st ed). In order to write down the Lagrangian, he needs the kinetic and potential energies.
He gets the kinetic energy very easily as the sum of the kinetic energies of the individual particles (the sum going over to an integral in the limit). Let $\eta_i, i=1,2,3$ be the components of the displacement vector (each $\eta_i = \eta_i(x,y,z)$ being a function of position). So the kinetic energy density is ${\cal T}=(\mu_0/2) (\dot{\eta}_1^2+\dot{\eta}_2^2+\dot{\eta}_3^2)$, where $\mu_0$ is the equilibrium mass density.
For the potential energy, he uses a thermodynamic argument, relying on the work done in a PV diagram, and using the equation $PV^\gamma = C$. His ultimate result, after several steps, is
${\cal V} = -P_0 \nabla\cdot\vec{\eta}+\frac{\gamma P_0}{2}(\nabla\cdot\vec{\eta})^2$
Here, $P_0$ is the equilibrium pressure, and $\gamma$ is the ratio of specific heats.
He later shows that the term $P_0 \nabla\cdot\vec{\eta}$ has no effect on the equations of motion, and so he drops it. So his final formula for the Lagrangian density is:
${\cal L} = (1/2)(\mu_0\dot{\vec{\eta}}^2 - \gamma P_0(\nabla\cdot\vec{\eta})^2$
and the Lagrangian of course is the integral of this over all space.
Now in the case of an ideal gas (or better yet, a perfect gas), my understanding is that the internal energy is entirely kinetic. Naively, the statistical model is a bunch of non-interacting point particles racing around, bouncing off the walls of the container. (For simplicity, ignore gravity.)
This seems contradictory. Shouldn't we get the same results from a microscopic and a macroscopic viewpoint?
To put it another way, this suggests that in a gas made up of non-interacting point particles, with no external forces except for the hard-wall forces, sound waves could not propagate (since the Lagrangian density would reduce to the kinetic part). That doesn't seem right.
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1
where is in the book the thing you are talking about? – Yrogirg Sep 12 '12 at 15:25
1
and why didn't you put his Lagrangian here? – Yrogirg Sep 12 '12 at 15:28
Sorry! I've expanded the entry to include this info. – Michael Weiss Sep 12 '12 at 21:43
1 Answer
He gets the kinetic energy very easily as the sum of the kinetic energies of the individual particles (the sum going over to an integral in the limit). Let $\eta_i, i=1,2,3$ be the components of the displacement vector (each \eta_i = $\eta_i(x,y,z)$ being a function of position). So the kinetic energy density is ${\cal T}=(\mu_0/2) (\dot{\eta}_1^2+\dot{\eta}_2^2+\dot{\eta}_3^2)$, where is the equilibrium mass density.
Naively, the statistical model is a bunch of non-interacting point particles racing around, bouncing off the walls of the container. (For simplicity, ignore gravity.)
Just to clarify, it seems you understand it, the "particles" in these two cases are different. In the second they are just molecules, in the first they are macroscopic pieces of gas, small volumes.
As for your question getting from microscopic energy to the macroscopic. I'll show how do they related.
$E_\text{mic} = \sum_{i=1}^N \frac{1}{2} m \boldsymbol c_i^2$
In that approach we'll have to deal with $N$ molecules, that's like finding individual trajectories of all the molecules, which we don't want. There should be also the term to reflect either the interaction of molecules (collisions) either with each other or with the walls, but it doesn't matter right now. To reduce the number of degrees of freedom we introduce one-particle distribution function:
$$f(\boldsymbol r, \boldsymbol c, t)$$
The quantaty $f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol r d\boldsymbol c$ tells us how many molecules are there in the phase space element $d\boldsymbol r d\boldsymbol c$
$$E_\text{mac}(\boldsymbol r, t) = \int \frac{1}{2} m \boldsymbol c^2 f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c$$
As you remember we want to express macroscopic energy in terms of macroscopic kinetic energy, i.e. we need to introduce macroscopic speed $\boldsymbol v$:
$$\rho(\boldsymbol r, t) = \int m f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c$$
$$\rho \boldsymbol v(\boldsymbol r, t) = \int m \boldsymbol c f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c$$
Then
$$\begin{multline} E_\text{mac}(\boldsymbol r, t) = \underbrace{\int \frac{1}{2} m (\boldsymbol c - \boldsymbol v + \boldsymbol v)^2 f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c}_{\text{Averaging microscopic kinetic energy}} = \\ \underbrace{\frac{1}{2} \rho \boldsymbol v^2}_{\text{Macroscopic kinetic energy}} + \underbrace{\int \frac{1}{2} m (\boldsymbol c - \boldsymbol v)^2 f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c}_{\text{Macroscopic potential energy}} \end{multline}$$
The part with $\boldsymbol v - \boldsymbol c$ was omitted for good reasons, but going into details would be too much, I've already essentially telling the kinetic theory. For the details you can consult appropriate books.
So, we've extracted the macroscopic kinetic energy, what is left is macroscopic potential energy, pressure if you want.
To put it another way, this suggests that in a gas made up of non-interacting point particles, with no external forces except for the hard-wall forces, sound waves could not propagate (since the Lagrangian density would reduce to the kinetic part). That doesn't seem right.
Ideal gas model does not assume particles to be non-colliding, it says that the potential energy of the molecular interaction can be neglected. Abrupt hard wall potential with delta functions have zero overall energy. You can view gas made of non-interacting particles only for the equilibrium case. And even then you have to consider collisions with the walls. For the non-equilibrium case you have to take into account collisions. If they were not interacting, indeed, there would be no sound.
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Thank you, that helps somewhat, but I want to clarify about the potential energy and collisions. I'm new to this site; should I edit my original question, or your answer, or post a new follow-up question? – Michael Weiss Sep 13 '12 at 13:02
@MichaelWeiss Just ask here (in comments) what exactly do you want to clarify? – Yrogirg Sep 13 '12 at 16:41
I have a few questions, I'll see if they all fit in one comment. – Michael Weiss Sep 13 '12 at 17:51
They didn't, and it timed out when I tried to edit. I'll use several comments. (1) With the model of point particles, no interactions except for collisions, at the microscopic level it seems as though all the energy is kinetic. (It even says so in Wikipedia.) So then where does the potential energy appear at the macroscopic level? My guess: collisions imply a delta potential between each pair of particles, so with coarse-graining we do a time and space average of these delta potentials, getting an average potential energy at the macro (fluid description) level. Is this right? – Michael Weiss Sep 13 '12 at 18:01
(2) With point particles, the probability of collision is zero for any finite time. I guess we just ignore this. If there were no collisions, then sound would not propagate, so the speed of sound depends somehow (inversely) on the mean-free-path -- is that all correct? – Michael Weiss Sep 13 '12 at 18:05
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http://mathhelpforum.com/calculus/209818-forecast-cubic-spline-print.html | # Forecast with cubic spline
Printable View
• December 14th 2012, 06:25 AM
igorbaldacci
Forecast with cubic spline
Hi all,
I have a simple question: is it possible to make forecast with cubic spline? And, if yes, how?
Frecast, for me, meens that I can understend the value of my interpolation function OUTSIDE the data field that I have.
Thank you, every help will be appreciated. (Itwasntme)
Igor
• December 14th 2012, 09:06 AM
hollywood
Re: Forecast with cubic spline
I think the word you're looking for is extrapolate. It means (in one dimension) that you are given f(x) for various x in [a,b] and want to estimate f(x) for some x>b or some x<a.
Of course, you can use any method, and the accuracy you get depends on the situation. If I understand correctly, a cubic spline would end up being just a cubic function approximation, and so would assume the third derivative of the function is approximately constant. Your accuracy will depend on how well this actually fits the situation.
As an example, here
Taylor series - Wikipedia, the free encyclopedia
there is a graph of various polynomial approximations to the function $\sin{x}$. You can see that the cubic approximation is pretty good for $-\frac{\pi}{2}<x<\frac{\pi}{2}$ but really gets bad as you go further from 0.
- Hollywood
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http://mathoverflow.net/questions/30113/geometric-model-for-classifying-spaces-of-alternating-groups/30115 | ## Geometric model for classifying spaces of alternating groups
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The classifying space of the nth symmetric group $S_n$ is well-known to be modeled by the space of subsets of $R^\infty$ of cardinality $n$. Various subgroups of $S_n$ have related models. For example, $B(S_i \times S_j)$ is modeled by subsets of $R^\infty$ of cardinality $i + j$ with $i$ points colored red and $j$ points colored blue. More fun: the wreath product $S_i \int S_j \subset S_{ij}$ has classifying space modeled by $ij$ points partitioned into $i$ sets of cardinality $j$ (but these sets are not "colored").
My question: is there a geometric model, preferably related to these, for classifying spaces of alternating groups? [Note: since any finite group is a subgroup of a symmetric group one wouldn't expect to find geometric models of arbitrary subgroups, but alternating groups seem special enough...]
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2
For any subgroup $G\subset S_n$ you can proceed by: Take $E$ to be the universal cover of any classifying space of $S_n$. Then $E/G$ is a classifying space for $G$. This is implicit in the answers and in your examples. – Bruce Westbury Jul 1 2010 at 3:10
Yes, but to get to more "natural" models requires at least a bit more than such a general construction (and presumably won't be possible for arbitrary subgroups). – Dev Sinha Jul 1 2010 at 5:13
I don't understand the comment about "natural" models: the classifying space $BG$ is only defined up to homotopy and $BH=EG/H$ for $H<G$ is rather natural! – Victor Protsak Jul 30 2010 at 7:31
Natural is perhaps the wrong word (because of the connotations of functoriality). What I mean is more "found in nature." For example you wouldn't want to understand the classifying space of Z/p \times Z/p as some kind of "configurations with ordering up to equivalence" - one model found in nature is the product of Lens spaces. – Dev Sinha Jul 30 2010 at 8:15
## 2 Answers
$n$ linearly independent points in $R^\infty$ together with an orientation of the $n$-plane which they span.
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Nice - I had started thinking along the lines Greg suggests but this is what I was looking for (and should have thought of myself). – Dev Sinha Jun 30 2010 at 23:17
The two solutions are hardly different. Kevin is exactly describing a sign-ordering of the points, the only difference being that he removes linearly dependent sets of points. This deletion doesn't hurt anything because it has infinite codimension, but it also isn't necessary for the answer. – Greg Kuperberg Jun 30 2010 at 23:50
My initial idea was the same as Greg's, but I didn't want to have to explain what a sign-ordering was. I also thought a sign-ordering might be exotic enough to disqualify the answer as simple. Also, note that the linear independence requirement arises naturally if we think of `A_n` as a subgroup SO(n). BSO(n) is oriented n-planes in `R^\infty`, and an `(SO(n)/A_n)`-bundle over this grassmanian gives the same answer as above. – Kevin Walker Jul 1 2010 at 0:20
That's fair. The common theme in all of these constructions is to obtain one classifying space from another one by passing to a subgroup. (But strictly speaking you are using $A_n \subset SL(n)$. $A_n \subset SO(n)$ would give you orthogonal vectors rather than l.i. vectors.) – Greg Kuperberg Jul 1 2010 at 0:43
Yes, SL(n) not SO(n) -- thanks. – Kevin Walker Jul 1 2010 at 0:55
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Probably the right thing to do is to express the classifying space of $A_n$ as the non-trivial double cover of the classifying space of $S_n$. A point in the classifying space is then a set of $n$ points in $\mathbb{R}^\infty$ with a "sign ordering". A sign ordering is an equivalence class of orderings of the points, i.e., ways to number them from 1 to $n$, up to even permutations. I coined the term "sign ordering" by analogy with a cyclic ordering. But that name aside, the idea comes up all the time in various guises. For instance an orientation of a simplex is by definition a sign ordering of its vertices.
This is in the same vein as your other examples and you can of course do something similar with any subgroup $G \subseteq S_n$. You can always choose an ordering of the points up to relabeling by an element of $G$.
A bit more whimsically, you could call the configuration space of $n$ sign-ordered points in a manifold "the configuration space of $n$ fermions". Although a stricter model of the $n$ fermions is the local system or flat line bundle on $n$ unordered points, in which the holonomy negates the fiber when it induces an odd permutation of the points. This local system is similar to the sign-ordered space in the sense that the sign-ordered space is the associated principal bundle with structure group $C_2$.
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http://mathhelpforum.com/advanced-math-topics/137523-big-o-proof.html | Thread:
1. BIG O proof
How do I prove the following:
O(n^2 logn) is also O(n^3)?
2. Originally Posted by taurus
How do I prove the following:
O(n^2 logn) is also O(n^3)?
$f(n)\in O(n^2 \log(n))$
means there exists a constant $k>0$ and an $N>0$ such that for all $n>N$:
$|f(n)|<k n^2 \log(n)$
Now for $n>0$ we have $n>\log(n)$ so for $n>N>0$:
$|f(n)|<k n^3$
etc.
CB
3. You will have to expand on that, I have no clue how to go on...
4. Originally Posted by CaptainBlack
$f(n)\in O(n^2 \log(n))$
means there exists a constant $k>0$ and an $N>0$ such that for all $n>N$:
$|f(n)|<k n^2 \log(n)$
Now for $n>0$ we have $n>\log(n)$ so for $n>N>0$:
$|f(n)|<k n^3$
etc.
CB
Originally Posted by taurus
You will have to expand on that, I have no clue how to go on...
So there exists a constant $k>0$ and an $N>0$ such that for all $n>N$:
$|f(n)|<k n^3$
hence $f(n)\in O(n^3)$
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http://physics.stackexchange.com/questions/19970/does-the-scientific-community-consider-the-loschmidt-paradox-resolved-if-so-wha/19976 | # Does the scientific community consider the Loschmidt paradox resolved? If so what is the resolution?
Does the scientific community consider the Loschmidt paradox resolved? If so what is the resolution?
I have never seen dissipation explained, although what I have seen a lot is descriptions of dissipation (i.e. more detailed pathways/mechanisms for specific systems). Typically one introduces axioms of dissipation for example:
entropy $S(t_1) \geq S(t_0) \Leftrightarrow t_1 \geq t_0$ (most often in words)
These axioms (based on overwhelming evidence/observations) are sadly often considered proofs. I have no problem with usefull axioms (and I most certainly believe they are true), but I wonder if it can be proven in terms of other (deeper and already present) axioms. I.e. is the axiom really independent? or is it a corrollary from deeper axioms from say logic (but not necessarily that deep).
(my opinion is that a proof would need as axioms some suitable definition of time (based on connection between microscopic and macroscopic degrees of freedom))
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5
It would be useful to clarify your question by stating what you consider to be Loschmidt's paradox. I assume you mean the fact that thermodynamics is time asymmetric when the underlying laws of physics are symmetric. – Philip Gibbs Jan 25 '12 at 8:16
Note that if CPT symmetry holds (and no violations are known) the existence of CP violation implies that the fundamental laws of physics are not fully time symmetric. – dmckee♦ Jan 25 '12 at 18:02
2
@dmckee but note that in that case Loschmidt's paradox is just as much (or as little) of a problem as before, because every trajectory still has an "antitrajectory" that's effectively the same but with time reversed. The only difference is that the antitrajectory has the charge and parity reversed, as well as velocities. – Nathaniel Aug 4 '12 at 20:47
1
– Emilio Pisanty Aug 5 '12 at 1:53
## 6 Answers
Loschmidt's paradox is that the laws of thermodynamics are time asymmetric because entropy always increases, but the underlying laws of physics are symmetric under time reversal. It should not therefore be possible to derive the second law of thermodynamics from first principles. Opinions in the scientific community differ as to whether this has been resolved (which implies that it has not been resolved) One commonly held opinion is that entropy increases only because it was low at the big bang, but that we don't know why it had to be low at the beginning. There are other possible explanations some of which also have significant support.
One point to make is that physics is not about axioms and proofs. These belong to mathematics which can be used to understand physical models and theories, but it makes no sense to declare axioms for physics. Any model must be tested against experiment and nothing is as absolute in science as an axiom. Thermodynamics in particular is a statistical science so its laws may only apply in closed systems of many degrees of freedom moving between states of equilibrium.
Some people still think that Boltzmann's H theorem explains why entropy always increases, but as Loschmidt's paradox implies, it must have a hidden time asymmetric assumption to work. You cannot get asymmetric solutions from symmetric equations unless there is a mechanism of spontaneous symmetry breaking (which the H theorem does not have) Boltzmann assumed that the initial state has low entropy and that there is nothing to constrain the future states to have low entropy. This leaves open the question as to why the initial state of the universe had low entropy. Since we do not yet have a complete theory of the initial state of the universe we cannot expect to be able to answer this question yet.
There are other ways that the paradox might be resolved with varying degrees of support from physicists. Here are three of them:
• CPT is most likely an exact symmetry of nature but CP and T are not. It could be that this small asymmetry drives the second law of dynamics by leaving the universe dominated by matter rather than anti-matter.
• It could be that the time asymmetry of the universe is driven by the laws of quantum mechanics through the measurement process which is time-asymmetric.
• In the theory of eternal inflation spacetime is always expanding. This is itself time-asymmetric and could be considered as a mechanism of spontaneous symmetry breaking that drives the arrow of time.
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"It could be that the time asymmetry of the universe is driven by the laws of quantum mechanics through the measurement process which is time-asymmetric." it seems like time and knowledge of an entity embedded in the system (which performs and learns information) is getting towards an answer – propaganda Jan 25 '12 at 9:00
has the statistical mechanics relationship between microscopic variables (unknown to an entity within the system) and macroscopic variables (some of which form the "knowledge" of an entity in the system) been combined with Kripke semantics to explain the arrow of time? – propaganda Jan 25 '12 at 9:04
Feel free to vote me down if you think the scientific community all agrees with your solution to this paradox. – Philip Gibbs Jan 25 '12 at 9:48
actually I had already voted you up because: the second bullet looks a lot like a different interpretation of what I have in mind, and because it is a summary of what different directions the scientific community explores on the topic, which was part of the question – propaganda Jan 25 '12 at 10:51
First of all, it's strange how the OP jumps from the Loschmidt "paradox" to dissipation. It makes it very unclear what he or she is actually asking because dissipation has no direct relationship to the Loschmidt "paradox" except that both of them are issues concerned with irreversibility in statistical physics or thermodynamics. The existence of dissipation is indisputable and demonstrable and all axioms or non-axioms in physics have to agree with this existence.
Irreversibility "paradox"
The Loschmidt "paradox" was an objection that Johann Loschmidt raised against (his younger colleague) Ludwig Boltzmann's claims about the statistical origin of entropy. In particular, Loschmidt claimed that Boltzmann shouldn't be able to prove the H-theorem – the increasing nature of entropy, a mathematical incarnation of the second law of thermodynamics (which implies a future-past asymmetry, the so-called thermodynamic arrow of time) – from microscopic laws that are invariant under the time reversal.
However, as Boltzmann understood, the objection is really invalid because all probabilistic reasoning in physics inevitably depends on the so-called logical arrow of time – which really says that the future is (fully or statistically but predictably) determined by the past but not in the other way around. For example, it follows from pure logic applied to events in time that if there are $N_0$ initial microstates and $N_1$ final microstates, the probability to get from the initial ensemble to the final ensemble must be averaged over the initial microstates but summed over the final microstates.
This really follows from pure logic; no other physical assumption is needed. We sum the probabilities over final states because we don't care which of them will occur and $P(A{\rm\,\,or\,\,} B)=P(A)+P(B)$ for mutually exclusive outcomes. We average the probabilities over the initial states because we don't know which of them was the right one and their prior probabilities have to satisfy $P(A)+P(B)+\dots = 1$. The asymmetry between the initial (past) state and the final (future) state doesn't depend on any details of the dynamics; it's pure logic. The logical arrow of time. It boils down to the asymmetry that the assumptions about the past and the claims about the future play in the Bayes formula. Implications in logic, $A\Rightarrow B$, aren't symmetric in $A,B$.
Note that the transition probability is therefore $${\rm Prob} = \sum_{i=1}^{N_0} \sum_{f=1}^{N_1} \frac{1}{N_0} {\rm Prob} (i\to f)$$ The factors $N_0$ and $N_1$ enter asymmetrically. The very fact that only $1/N_0$ is added is the reason why the evolution prefers a higher number of final states relatively to the initial states. One may compute the probability of the time-reverted process (or CPT-reverted process, to be more precise in QFT), and the factor will be $1/N_1$ instead. The ratio of these probabilities is therefore $N_1/N_0$ which is $\exp[(S_1-S_0)/k]$: and this ratio of probabilities which is extremely large for any macroscopic system guarantees that only the evolution in the direction where the entropy is increasing may occur with a detectably nonzero probability; the reverted process is impossible. Even though some people don't understand it, the rules for retrodiction are completely different from the rules for prediction: retrodiction is a form of (Bayesian) inference that, unlike predictions, always depends on (to some extent) arbitrary and subjective priors. Some people are making retrodictions according to the rules that only hold for predictions – and then they are surprised that they end up with absurd conclusions.
Ludwig Boltzmann organized the proof differently but he understood very well that his proof was actually a proof that the thermodynamic arrow of time is inevitably correlated with the logical arrow of time. People discovered quantum mechanics and lots of new reformulations of these arguments and proofs were written down but the essence hasn't changed. All physicists who understand and take statistical physics seriously understand that the Loschmidt "paradox" was already resolved by Boltzmann and there is no paradox. But much like 100 years ago, there exist people who don't understand the logic behind similar proofs in statistical physics and who keep on repeating misconceptions that there exists a Loschmidt "paradox". This is a purely social phenomenon that will probably not go away; 100 years ago, physics has simply become so advanced and abstract that most people, even those who manage to get "some" physics education, are already unable to get to the cutting edge (and even "not so cutting edge"). The situation is even more striking in the case of quantum mechanics.
At any rate, the relevant answer is that the competent part of the scientific community (especially most of the people who are statistical physics experts) agrees that the Loschmidt "paradox" was already addressed and resolved more than 100 years ago while a broader "community" is split about this issue.
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I think Motl may be the only person I know who thinks that the H-theorem resolves this issue directly. – Philip Gibbs Jan 25 '12 at 9:34
"the probability to get from the initial ensemble (at t1) to the final ensemble (at t2) must be averaged over the initial microstates but summed over the final microstates" you are using english "initial" and "final" to introduce a (hidden?) axiom which assumes t1 =< t2, i.e. it should really read "the probability to get from the ensemble A to the ensemble B must be averaged over A's microstates but summed over the B's microstates if and only if timeA <= time B" I believe this is true, as I said the evidence is overwhelming. But my question is: is this extra axiom necessary? – propaganda Jan 25 '12 at 9:36
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Could it not be that this axiom is not independent from more fundamental axioms (from say logic), i.e. it does not harm to add it as an axiom as it wont conflict, but it may be a corrolary of widely used deeper axioms, of which we dont know the proof yet? – propaganda Jan 25 '12 at 9:38
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Otherwise your games with the "definition" of initial and final states and with the sign of $t$ are completely immaterial. "Initial" and "final" states are, according to logic, qualitatively different things, and the usual convention for the sign of $t$ is that $t_{\rm initial}<t_{\rm final}$. But I have never even used this convention. Even if I had, it wouldn't matter. One can easily rewrite all proofs to the opposite convention by replacing $t$ with $-t$; all those things are physically vacuous. The non-vacuous claim is that the future and past don't play symmetric roles in logic. – Luboš Motl Jan 25 '12 at 9:49
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initial and final are no symbols from logic! although you may certainly call them symbols from common sense: we are macroscopic subsystems in a larger system – propaganda Jan 25 '12 at 9:50
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The Loschimidt paradox does not state that reversible laws of motion can not imply irreversible processes which sounds like a philosophical objection. It rather observes that Boltzmann H-Theorem leads to the following physical contradiction: Take a system that starts at H_1 and evolves to H_2 and finally to H_3. The theorem states that H_3 < H_2 < H_1. Now take the microstate which correspond to H_2 and reverse the direction of all the velocities. We should all agree on the fact that at that point we would observe the system going back to H_1. Unfortunately the H-theorem states that the system will go to H_3 regardless of our intervention on the velocities. This is does not make sense at all, and this is why Loschmidt paradox is a real paradox, and not a solved paradox. A solved paradox is not a paradox. The reaction of Boltzmann indeed was not to try convince anyone that this paradox can be solved. His reaction was to leave the H-theorem in favor of a new prospective based on the combinatorial argument. Consider the classic Gibbs book for instance; you don't find anything similar to the H-theorem in his theory. What you find instead is the observation that in order to describe irreversible processes, you need to ignore the nature of mechanics expressed by the Liouville Theorem, and you need to introduce some different approach based on the coarse grain.. which is the same idea that Boltzmann had after Loschmidt objection.
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I think most people would say the paradox is resolved - but, as the answers to this question make clear, they wouldn't necessarily agree about who resolved it or what precisely the resolution is. For my money the paradox was elegantly resolved by Edwin Jaynes in this 1965 paper. In Jaynes' argument, the symmetry is broken by the fact that we, as experimenters, have the ability to directly intervene in the initial conditions of an (isolated) system, but we can only affect the final conditions indirectly, by changing the initial conditions.
Of course, this then leaves open the question of why our ability to interact with physical systems is time-asymmetric in this way. This is not a paradox but rather a physical fact in need of explanation. So while the mystery is not entirely solved by Jaynes' argument, at least the aparrent paradox can be laid to rest.
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Time asymetry appears in the solution of the Boltzmann's equation because its solution depends exponientally on the initials conditions. After a few caracteristic relaxation times, the initial conditions becomes exponentially small. So, although the microscopic particles obey Hamiltonian dynamics (with trajectories depending on initial conditions), as a whole this Hamiltonian caracteristic disappears and a new dynamics appears which for neutral gas is well modelized by the Boltzmann's equation. It is fundamental to understand that in statistical physics one can not think in terms of a single test particle. A single particle is a set with zero mesure which is irrelevant. There is a more problematic theorem: the Poincare's theorem which roughly states that any mechanical system goes back to its initial state. However, the time it takes to do it is for large system far greater than the age of the universe.
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Could you please illustrate or point to the specific derivation/formulae which show that "its solution depends exponientally on the initials conditions."? – propaganda Aug 13 '12 at 4:54
It is in one of those books. The demonstration is pretty long. Look first into: Radu Balescu: EQUILIBRIUM AND NONEQUILIBRIUM STATISTICAL MECHANICS : John Wiley & Sons, New York, 1975 Or into the vol 1 of those: Radu Balescu: TRANSPORT PROCESSES IN PLASMAS Vol. 1: CLASSICAL TRANSPORT Vol. 2: NEOCLASSICAL TRANSPORT – Shaktyai Aug 13 '12 at 6:22
Although summarized as an objection of macroscopic irreversibility when microscopic laws are reversible, Loschmidt's objection originally points that there has to be something breaking the time reversal symmetry in Boltzmann's derivation of the $H$-theorem.
I think that Boltzmann's answer was to say that high $H$ states (in absence of external driving) are more the exception than the rule. This is betrayed by the fact that inverting time in the $H$-theorem still leads to a decrease in $H$.
I think it is important to stress that Boltzmann's equation (from which derives the $H$-theorem) only looks at a very coarse grained quantity, namely the one-particle density and most rationals for the asymmetry are put at this coarse grained level.
Yet, mathematicians are still working on the problem (see here and there ).
But as a physicist, and for a picture beyond physics of gases, I think that this article on relevant entropies gives a lot of insights about these things in general.
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http://math.stackexchange.com/questions/157445/the-limit-of-integral | # The limit of integral
Let $1 \le p < \infty$ and assume $f \in L^p(\mathbb{R})$. I'm trying to prove the limit of integral $$\lim_{x \to \infty} \int^{x+1}_x f(t)dt =0.$$
Can I use Riesz Theorem for Banach spaces?
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Are you using the Lebesgue integral? What is your definition of it? – Alex Becker Jun 12 '12 at 16:54
I learned $\int f d\mu = sup\{ \int s d\mu :0 \le s \le f, s \in S\}$. Is it the key for this problem? – Collins Jun 12 '12 at 17:00
$s$ is a simple function? – Alex Becker Jun 12 '12 at 17:04
Yes that was what I mean. – Collins Jun 12 '12 at 17:11
## 3 Answers
Note that we need only prove this for real-valued $f\geq 0$, since $$\left|\int_x^{x+1}f(x)dx\right|\leq \int_x^{x+1}|f(x)|dx.$$ Let $\epsilon>0$, and choose some simple function $s\leq f$ such that $\int fd\mu<\int sd\mu+\epsilon/2$. Since $s$ is simple and integrable, we have that $$s(x)=\sum\limits_{n=0}^k \alpha_n\chi_{A_n}(x)$$ for some collection $\{A_n\}$ of measurable sets, and each $A_n$ has finite measure except $A_0$ (for which $\alpha_0=0$). Thus $A=\bigcup\limits_{n=1}^k A_n$ has finite measure, so we have some interval $I$ such that $\mu(A\setminus I)<\epsilon/2\max\{|\alpha_1|,\ldots,|\alpha_n|\}$. Thus for sufficiently large $x$ we have $(x,x+1)\cap I=\emptyset$ so $$\int_x^{x+1} f(x)dx<\int_x^{x+1}s(x)dx+\epsilon/2< \max\{|\alpha_1|,\ldots,|\alpha_n|\}\cdot \epsilon/2\max\{|\alpha_1|,\ldots,|\alpha_n|\}+\epsilon/2=\epsilon$$ i.e. we have $\lim\limits_{x\to \infty} \left|\int_x^{x+1}f(x)dx\right|<\epsilon$. Since this is true for all $\epsilon>0$, the limit must be $0$.
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We want to have limit of x→+∞ but is that ϵ→0 come to limit of x? – Collins Jun 12 '12 at 17:20
@Collins I edited to clarify. Hope that helps. – Alex Becker Jun 12 '12 at 17:26
I got it. Thanks for your very clear and nice answer. – Collins Jun 12 '12 at 17:29
Not a full solution I just provide some hints:
• Using Hölder's inequality (or Jensen), we get $$\left|\int_{[x,x+1]}f(t)dt\right|^p\leq \int_{[x,x+1]}|f(x)|^pdx,$$ hence we just have to deal with the case $f\in L^1$.
• Write $\int_{[x,x+1]}f(t)dt=\int_{(-\infty,x+1]}f(t)dt-\int_{(-\infty,x)}f(t)dt$. What is the limit, when $x\to +\infty$, of each term?
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Oh, you already give me almost full solution not just hints. – Collins Jun 12 '12 at 17:22
I would argue that $f(t) \chi_{[x,x+1]}(t)$ is a sequence of functions that converge pointwise to $0$ for $x\rightarrow\infty$. Then use theorem of dominated convergence with upper bound $f$ (assume $f$ is non-negative real-valued function first, then extend to all functions).
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But this is false, there are even functions in $L^p(\mathbb{R})$ that are unbounded at $\pm\infty$. – AD. Jun 12 '12 at 19:52
Example: For integers $|n|>0$ put $f(x) = |n|$ when $x\in (n,n+1/|n|^3)$, and $f(x)=0$ elsewhere; then $$\int |f(x)|dx=\sum_{|n|>0}\frac{1}{n^2}<\infty$$ so $f\in L^1$ but there is no point wise limit at $\pm\infty$. – AD. Jun 12 '12 at 19:59
Another example I saw somewhere on math.SE is $$f(x)=xe^{-x^2|\sin x|}$$ what happens at $x =2\pi n$? – AD. Jun 13 '12 at 4:52
Dear A.D.: I don't think that my argument is wrong. I think you are mixing up the variables. I call $t$ what you call $x$. Let $f$ be either one of your functions. Fix $t\in \mathbb R$. Then still $f(t)\chi_{[x,x+1]}(t)$ converges to zero for $x\rightarrow\infty$, even though your functions have large values for large $t$. But $t$ doesn't get large, $t$ stays fixed. That is what pointwise convergence means. – h.h.543 Jun 30 '12 at 12:35
You are right! I mixed up the variables, what you mean is of course $g_x = f \cdot \chi_{[x,x+1]}$, and then I agree. Thanks! (+1) – AD. Jul 1 '12 at 19:26 | {"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": 48, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9348288774490356, "perplexity_flag": "head"} |
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## Shapiro time delay?
Can someone clear up what is observed in Shapiro time delay experiments.
As I understand it a delay is seen in the round trip signal time of a radar signal sent to a distant planet on the opposite side of the sun due to gravitational effects. The signal path would be longer than the straight line path to the calculated position of the planet in its orbit due to deflection by the gravitational field of the sun. This longer path would introduce a geometric delay. Would it be right to assume the Shapiro time delay is a measured delay over and above the the simple geometric delay?
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Recognitions: Science Advisor Staff Emeritus As I understand it, the Shapiro time delay will include both the longer geometric path, and the effects of gravitational time dilation.
Quote by kev Can someone clear up what is observed in Shapiro time delay experiments. As I understand it a delay is seen in the round trip signal time of a radar signal sent to a distant planet on the opposite side of the sun due to gravitational effects. The signal path would be longer than the straight line path to the calculated position of the planet in its orbit due to deflection by the gravitational field of the sun. This longer path would introduce a geometric delay. Would it be right to assume the Shapiro time delay is a measured delay over and above the the simple geometric delay?
Yes, there is some additional delay due to the non-straight path of light. But it is much smaller than the main effect. In my opinion, the main origin of the Shapiro delay is the fact that light propagates slower in the vicinity of massive objects.
Eugene.
## Shapiro time delay?
From the vantage point of the earth observer, the speed of light passing near the sun's surface of radius r is c[(1-2GM/(c^2)r] but the local speed for an infinitely small physical distance and time is still c.
Quote by yogi From the vantage point of the earth observer, the speed of light passing near the sun's surface of radius r is c[(1-2GM/(c^2)r] but the local speed for an infinitely small physical distance and time is still c.
This might be true, however I don't think the speed of light in different gravitational potentials was ever measured with sufficient accuracy. We know for sure that the rate of all physical processes (e.g., the frequency of atomic clocks) slows down in the gravitational potential according to f' = f[1-GM/(c^2)r]. This fact can be reconciled with the variation of the speed of light c' = c[1-2GM/(c^2)r] if we assume that all distances decrease according to d' = d[1-GM/(c^2)r].
Eugene.
Recognitions:
Science Advisor
Staff Emeritus
Quote by yogi From the vantage point of the earth observer, the speed of light passing near the sun's surface of radius r is c[(1-2GM/(c^2)r] but the local speed for an infinitely small physical distance and time is still c.
Quote by meopemuk This might be true, however I don't think the speed of light in different gravitational potentials was ever measured with sufficient accuracy. We know for sure that the rate of all physical processes (e.g., the frequency of atomic clocks) slows down in the gravitational potential according to f' = f[1-GM/(c^2)r]. This fact can be reconciled with the variation of the speed of light c' = c[1-2GM/(c^2)r] if we assume that all distances decrease according to d' = d[1-GM/(c^2)r]. Eugene.
While direct tests of the speed of light may or may not have been done to any particular degree of accuracy, it is a fundamental theoretical prediction of General Relativity that the local speed of light is constant.
If the local speed of light were not constant, General Relativity would be falsified.
General Relativity to date has a fairly impressive amount of experimental support, certainly it does a good job of modeling transit times and light paths in the solar system, including the shapiro effect.
Note that "the speed of light" near the sun as seen from Earth is not even measurable unless one devises some synchronization scheme. What Yogi computes above is the "coordinate speed" of light in the Schwarzschild coordinate system, i.e. dr/dt, where r is the Schwarzschild r coordinate, and t is the Schwarzschild t coordinate. This can be and usually is interpreted as the coordinate speed of light viewed from an "observer at infinity", (so Yogi is talking about the Earth as if it were infnitley far away from the sun here), but there are conceptual ambiguities about talking about speeds at distant events.
Measuring dr/dt is not really a very fundamental way to measure speed, because (among other reasons) r is not really a distance measure, it is just a coordinate. The situation is rather similar to someone who claims that the speed of naval ships increases near the poles of the earth, because d(longigutude)/dt is greater, i.e. the ship can cover one degree of longitude in a shorter time near the poles than it can at the equator. If you look at the actual local speed of the naval ship relative to the ocean, it is constant - the apparent increase in d(longitude)/dt near the poles is an artifact of the curvature of the Earth.
Philosphically, lattitude (usually denoted by $\phi$) and longitude (usually denoted by $\lambda$) on the Earth are coordinates, not distances. The metric of the curved Earth's surface converts small changes in these coordinates into actuall distances. Because the Earth's surface is curved, the metric coefficients are not constant, assuming the Earth's surface is constant one can write this metric as
$$ds^2 = r^2 d\lambda^2 + r^2 \cos^2 \, \phi d \phi^2$$
(I've taken the liberty of assuming one converts from degrees to radians rather than write out the conversion explicitly)
here ds is the distance on the Earth's surface, and $\phi$ and $\lambda$ are the lattitude and longitude as previously mentioned.
Compare this to the Schwarzschild metric: it's very similar as far as the angular terms go. (Note that equator of the earth is $\phi=0$ while the equator in Schwarzschild coordinates is $\theta=\pi/2$.)
Similarly, the local speed of light is a constant everywhere - what could be called the "change" of dr/dt with r is an artifact of the curvature of space-time, not an physical change in speed, just as the change in $d\lambda/dt$ for our naval ship as it got closer to the poles (i.e. as $\phi$ increased) was.
The confusion comes about in large part by not distinguishing between coordinates: on the Earth, lattitude and longitude, in space: Schwarzschild r, theta, phi, and t coordinates, and distances.
Distances are computed from small changes in the coordinates via the metric. Note that in relativity, one actually computes the value of the space-time interval.
Hi pervect, you are correct that my post about the varying speed of light is not consistent with the ideas of general relativity. I simply wanted to note that the Shapiro time delay can be explained without GR's assumption of the geometry change in gravitational fields. An alternative explanation - that the speed of light slows down according to $$c' = c(1-\frac{2GM}{c^2 r})$$ also works. Eugene.
Recognitions: Science Advisor Staff Emeritus http://xxx.lanl.gov/abs/astro-ph/0006423 takes a somewhat similar point of view, I think. We like to talk about peer-reviewed theories here. This is a peer reviewed theory that I think (hope) is simlar to what you are talking about. It's also an example of a non-geometrical interpretation of something that is locally the same as GR. Unfortunately, this is not quite the same as being equivalent to GR. It is a different thoery because it doesn't necessarily have the same global topology as GR does. The above paper unfortunately doesn't go into or even mention the topology issues. There used to be some criticism of this theory, informally called "funky fields in a Minkowski space-time" that I would refer people to on the WWW, but it disappeared :-(. You can probably find some past PF discussions if you google. I think the idea has promise for getting rid of pesky time machines in GR, but that's a personal opinion, and there are still some unanswered questions as to how one deals with the topological issues, and it is also not quite clear how the theory deals with black holes. (Do said "funky fields" become infinite on the event horizon? Or what?).
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Thanks for all the thoughtful replies and sorry for the delay replying (family issues).
We all seem to be agreed there would be a measurable time delay compared to a constant speed of light over the the coordinate distance. Yogi and Eugene interpret the cause as due to the speed of light slowing down in a gravitational field while Pervect has the more formal point of view that the speed of light is constant in a gravitational field when measured in 4 dimensional curved spacetime rather than the 3 dimensional space coordinates we are used to.
Quote by yogi From the vantage point of the earth observer, the speed of light passing near the sun's surface of radius r is c[(1-2GM/(c^2)r] but the local speed for an infinitely small physical distance and time is still c.
Quote by meopemuk Hi pervect, you are correct that my post about the varying speed of light is not consistent with the ideas of general relativity. I simply wanted to note that the Shapiro time delay can be explained without GR's assumption of the geometry change in gravitational fields. An alternative explanation - that the speed of light slows down according to $$c' = c(1-\frac{2GM}{c^2 r})$$ also works. Eugene.
Shouldn't that be $$c = c'\sqrt(1-\frac{2GM}{c^2 r})$$ ?
Quote by meopemuk This might be true, however I don't think the speed of light in different gravitational potentials was ever measured with sufficient accuracy. We know for sure that the rate of all physical processes (e.g., the frequency of atomic clocks) slows down in the gravitational potential according to f' = f[1-GM/(c^2)r]. Eugene.
Did you intend f' = f[1-GM/(c^2)r] (the excellent approximation of f' = f sqrt[1-2GM/(c^2)r] in a week field that is the first term of a binomial expansion) or f' = f[1-GM/(c^2)r] ?
This aproximation has the interesting property that the solutions for r < GM/(c^2) do not involve imaginary numbers. It also implies negative gravity and negative time for r < GM/(c^2). I wonder if any real experiments have been carried out that can distinguish f' = f[1-GM/(c^2)r] from f' = f sqrt[1-2GM/(c^2)r] ?
Quote by meopemuk This fact can be reconciled with the variation of the speed of light c' = c[1-2GM/(c^2)r] if we assume that all distances decrease according to d' = d[1-GM/(c^2)r]. Eugene.
Assuming we all agreed that the speed of light as measured by a local observer is always c then a slowing down of light as observed by a distant observer would be more likely to be reconciled with an increase of horizontal distances? For example, say we had a fibre optic that took one second for a photon to traverse in a very weak gravitational field. If we lowered that optical fibre ruler into a strong gravity well where the gravitational time dilation factor was 2 then a photon would take 0.5 seconds to traverse that ruler as measured by a local observer with slow clocks (assuming no length change). If the ruler had "length dilated" to twice its proper length, then the local observer would still measure the time taken to traverse the ruler as one second and the speed of light would still appear to be c to the local observer.
I would like to suggest a thought experiment with 2 scenarios to explore the possibilities and further clarify what is thought to be happening in a strong gravitational field.
Thought experiment:
Suppose we had optical fibre ring that had a radius of $$\frac{1}{2\pi}$$ light seconds when the ring is far from any gravitational fields. A device is spliced into the ring that can inject a photon into the optical fibre ring and times the interval for the photon to complete one loop and this should presumably be one second. R is calculated by a distant observer from the angle subtended by the ring when viewed through a telescope and the distance of the observer from the ring. At this point both local and distant observers agree on the radius of the ring, the apparent speed of light and that the circumference is given by $${2\pi r}$$ .
Now suppose we had a convenient (non rotating) gravitational body which just happens to be a snug fit for our optical ring around the equator of the body. Assume the mass of the body is such that the gravitational time dilation factor is 1.25. Also assume there is atmosphere that would give rise to optical effects. Please ignore practical considerations such as this body is probably a neutron star and all local instruments and observers would be crushed ;) What would the local and distant observers measure?
Scenario A: (Assuming horizontal gravitational length dilation).
Here we assume the circumference of the ring has expanded by a factor of 1.25
The local observer measures the time for a photon to travel within the optical fibre ring as one second due to his slow clocks relative to the distant observer.
The distant observer measures the photon as taking 1.25 seconds to travel a distance of 1.25 light seconds. Both agree on the same number for the speed of light in metres per second so neither sees an apparent slowing down of the speed of light in a gravitational field.
Scenario B: (No length dilation)
The distant observer measures the circumnavigation time of the photon within the optical fibre ring on the equator of the gravitational body as 1.25 seconds. From the distant observers point of view the circumference is 1 light second so the speed of light seems to have slowed down to 0.8c
If the distant observer defines distance by light travel times then he could say that that the circumeferance of the body is not given by $${2\pi r}$$
The local observers on the surface of the body (with the slower clock) measures the photon travel time as 1 second and so the local speed of light appears to be c to them.
Note: Self gravitational lensing
I have not taken self gravitational lensing into account in either scenario. It should be apparent that the strong gravitational field of the body will magnify the apparent size of the body to a distant observer. The body and ring will appear to subtend a larger angle in the telescope the distant observer. I can find formulas for the amount light is bent by a gravitational field as it passes a massive body. However I can not find a formula for how much a body would be appear to be magnified by its own gravity. It may be possible that when the distance of the observer from the body, the mass of the body and the gravitational dilation factor on the surface of the body and at the location of the distant observer (assuming he is not at infinity) are all taken into account then the optical magnification would make the speed of light be c to both local and distant observers. For example, in scenario B the distant observer would see the radius and circumference of the body optically magnified by a factor of 1.25 so that the circumnavigation time of 1.25 seconds would be consistent with an unchanged speed of light even to the distant observer.
So, is scenario A or B (or neither) more consistent with the accepted wisdom?
Does anyone have a formula for self gravitational lensing of a body?
(My interpretation is that it would be something like $$s = s'\frac{\sqrt(1-\frac{2GM}{c^2 d}}{ \sqrt(1-\frac{2GM}{c^2 r}}$$ where s is the self gravitational optical magnification of the body and d is the distance of the observer from the body)
In measuring r and d I have assumed there is no dilation or contraction of length in the vertical direction in a gravitational field. Is that a reasonable assumption?
Thanks
Quote by kev Did you intend f' = f[1-GM/(c^2)r] (the excellent approximation of f' = f sqrt[1-2GM/(c^2)r] in a week field that is the first term of a binomial expansion) or f' = f[1-GM/(c^2)r] ? This aproximation has the interesting property that the solutions for r < GM/(c^2) do not involve imaginary numbers. It also implies negative gravity and negative time for r < GM/(c^2). I wonder if any real experiments have been carried out that can distinguish f' = f[1-GM/(c^2)r] from f' = f sqrt[1-2GM/(c^2)r] ?
Hi kev,
I used f' = f[1-GM/(c^2)r] instead of f' = f sqrt[1-2GM/(c^2)r] simply because (as far as I know) all existing experimental tests have precision of not better than $c^{-2}$. So, keeping only the first order in $c^{-2}$ seems to be a prudent approach.
Your ideas about experimental tests for gravity effects on length and speed of light look interesting, however (you would probably agree) completely unrealistic. Is it possible to modify these tests, so that they become realizable with modern technology?
Eugene.
Blog Entries: 6
Quote by meopemuk Hi kev, I used f' = f[1-GM/(c^2)r] instead of f' = f sqrt[1-2GM/(c^2)r] simply because (as far as I know) all existing experimental tests have precision of not better than $c^{-2}$. So, keeping only the first order in $c^{-2}$ seems to be a prudent approach. Eugene.
Hi Eugene,
In 2005 I posted a document on the web that conjectured "What if f' = f[1-GM/(c^2)r] if the true relationship rather than the conventional f' = f sqrt[1-2GM/(c^2)r] ?" As you point out the accuracy of existing tests can not rule out f' = f[1-GM/(c^2)r] (as far as I know). In strong gravity fields such as near a black hole the differences become significant and give rise to a different type of black hole. Using f' = f[1-GM/(c^2)r] the event horizon is half the conventional Schwarzschild radius. Below the event horizon gravity is negative giving rise to a solid shell with radius GM/(c^2) rather than the singularity of the conventional interpretation. Quantum fluctuations at the surface of the shell would allow radiation from the surface of black hole but that radiation would be at very long wavelengths and very faint due to extreme time dilation at the surface. (The alternative black hole would still be a very dark object) Gravitational lensing due to a black hole would probably be less than the conventional interpretation. One interesting thought is that if all the mass of the known universe was contained within a Planck radius there would extreme negative repulsive gravity giving rise to a simple model of the inflation/expansion of the big bang. The negative time dilation at such extremes in this model is not a problem as it is the cause of negative gravity. Objects fall away from each other instead of towards each other in negative time. Anyway, I have not posted a link to my document as I admit it is all idle speculation and conjecture. (Just an interesting thought)
Quote by meopemuk Your ideas about experimental tests for gravity effects on length and speed of light look interesting, however (you would probably agree) completely unrealistic. Is it possible to modify these tests, so that they become realizable with modern technology? Eugene.
I agree the thought experiments are unrealistic (as are most thought experiments) and being non local there is no possibility of testing them in a lab. However observations of X ray emissions from a black hole accretion disc and possibly measurements of red shifted emissions in the radio spectrum from nearer the event horizon of the black hole may give an indication of the actual "radius" of the black hole. Comparing these measurements to the black hole mass calculated from the orbit of a companion binary star may give some indications of physics at work in these extreme gravitational fields. Quite possibly, this has already been done.
In my last post I asked a lot of questions (perhaps too many).
To keep things simple the main question I am asking at his point is:
Do objects length contract or expand in a strong gravitational field as measured by a distant observer ...or not?
Any expert in GR should be able to answer that simple question :P
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Quote by kev Can someone clear up what is observed in Shapiro time delay experiments.
What textbooks have you studied so far? I'll hopefully assume that you are familiar with the contents of a standard textbook such as O'Hanian and Ruffini, Gravitation and Spacetime, which has a nice exposition of the Shapiro time delay effect. For the reader's convenience I'll review the essential ideas here.
Quote by kev As I understand it a delay is seen in the round trip signal time of a radar signal sent to a distant planet on the opposite side of the sun due to gravitational effects.
Correct. Consider the Earth, Venus, and the Sun. The experiment is best performed when all three are almost aligned, but Venus is on the other side of the Sun relative to the Earth. Then a radar pip is sent from Earth toward Venus, passing very near the Sun; it is reflected from Venus and returns to Earth, again passing very near the Sun.
Idealize Earth and Venus to have neglible mass, i.e. imagine that the ambient gravitational field is due to the Sun. This suggests treating the problem in the (exterior of the) Schwarzschild vacuum solution by considering two world lines representing the motion of the Earth and Venus as test particles. Since the round trip takes on the order of ten minutes we can even idealize them as static test particles. The problem is now reduces to studying null geodesics connecting the two world lines. By symmetry, we can consider the two legs of the journey to be almost identical.
The line element written in the standard Schwarzschild chart is
[tex]
ds^2 = -(1-2m/r) \, dt^2 \, + \, \frac{dr^2}{1-2m/r} \, + \, r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),
[/tex]
[tex]
-\infty < t < \infty, \; 2 \, m < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi
[/tex]
The world lines of static observers in this chart have the form $r=r_0, \, \theta=\theta_0, \, \phi=\phi_0$. The spatial hyperslices orthogonal to such observers have the form $t=t_0$ and they are all identical (by the time translation symmetry of the Schwarzshild vacuum). The null geodesics are planar, so without loss of generality we can suppress the coordinate $\theta$ by setting $\theta=\pi/2$ (the locus of the equatorial plane).
If we project the null geodesic arc representing outward leg of the journey of the radar pip into $t=0$, we obtain a curve which I'll call "the path". It is the projection of a null geodesic in spacetime, but it is not a geodesic in the metric on t=0 which is induced by restricting the Schwarzschild metric tensor to this hyperslice (with $\theta$ suppressed):
[tex]
d\sigma^2 = \frac{dr^2}{1-2m/r} \, + \, r^2 \, d\phi^2, \; 2 \, m < r < \infty, \; -\pi < \phi < \pi
[/tex]
This induced metric can be considered a perturbation of the usual polar coordinate chart for the euclidean plane, and also a good approximation to it, in the region we are considering (outside the surface of the Sun).
The path appears to be "bent" slightly as it passes near the origin (the location of the Sun). As everyone knows, extending the path to infinity in both directions we get a angle between two asymptotes, which Einstein computed ("light bending"). The two asymptotes in fact are rather close to the path itself, so to simplify the computation we can (subject to justification upon demand!) treat the path as two line segments.
We can write any straight line segment in a polar chart in the form $R = r\, \cos(\phi)$, where R is the distance of closest approach to the origin. Differentiating gives $r \, d\phi = \cot(\phi) \, dr$ and plugging this relation into the line element of the Schwarzschild vacuum (still working in the equatorial plane) and setting $ds^2 = 0$ (since we are working with a null geodesic arc) gives
[tex]
dt = \frac{r}{r^2-R^2} \left( 1 + 2m/r - 2mR/r^3 \right) \, dr \; + \; O(m^2)
[/tex]
Now suppose that the Earth is at $r=R_1$ (angle irrelevant) and integrate from $R_1$ to $R$. Proceding similarly for $R$ to $R_2$, where Venus is at $r=R_2$ (angle irrelevant), and adding, we obtain an expression consisting of the flat spacetime "height" (travel time) of our broken null geodesic path, namely $\sqrt{R_1^2-R^2}+\sqrt{R_2^2-R^2}$, where of course $2 \, m \ll R \ll R_1, \, R_2$, plus some correction terms, of which the largest is a logarithmic term. (Closer examination would show that it overwhelms the small errors introduced by the broken line approximation.)
By symmetry, the return leg should give the same result, so twice the logarithmic correction term gives the desired time delay. This is the increased time it takes for the round trip journey, as measured by an ideal clock on Earth, in case the radar pip travels near the limb of the Sun, as compared to when it does not. It's a very concrete effect!
Here's a pointer for your original question: suppose we model bending a paper clip using some one parameter family of curves. The tip moves quite a bit as we vary the parameter, but the change of length due to the bending is comparatively modest.
Quote by kev The signal path would be longer than the straight line path to the calculated position of the planet in its orbit due to deflection by the gravitational field of the sun. This longer path would introduce a geometric delay. Would it be right to assume the Shapiro time delay is a measured delay over and above the the simple geometric delay?
To find out, and also to justify the broken line approximation, compute the length of
• a straight line path "from Earth to Venus" in $t=0$, using the euclidean metric or the induced metric,
• the broken line path, likewise,
• the actual path, likewise
(I'll let you attempt this before I say more.)
Quote by meopemuk This might be true, however I don't think the speed of light in different gravitational potentials was ever measured with sufficient accuracy. We know for sure that the rate of all physical processes (e.g., the frequency of atomic clocks) slows down in the gravitational potential according to f' = f[1-GM/(c^2)r]. This fact can be reconciled with the variation of the speed of light c' = c[1-2GM/(c^2)r] if we assume that all distances decrease according to d' = d[1-GM/(c^2)r].
This is the same misconception which I just characterized as a FAQ in another thread! Time does not slow down and distances do not shorten! (Neither of these claims even makes sense.) Rather--- well, see this post. Well, ditto Pervect generally
kev, I urge you to simply ignore what meopemuk said; IMO he's adding confusion, not clearing it up.
Quote by kev For example, say we had a fibre optic that took one second for a photon to traverse in a very weak gravitational field. If we lowered that optical fibre ruler into a strong gravity well where the gravitational time dilation factor ... Now suppose we had a convenient (non rotating) gravitational body which just happens to be a snug fit for our optical ring around the equator of the body. Assume the mass of the body is such that the gravitational time dilation factor is 1.25...Here we assume the circumference of the ring has expanded by a factor of 1.25 The local observer measures the time for a photon to travel within the optical fibre ring as one second due to his slow clocks relative to the distant observer.
Careful now! I've seen even professional physicists fall into fallacy with this kind of thinking The basic problem is that it is rarely straightforward to compare two objects in two different spacetime manifolds and to declare these objects to be "equivalent". IMO you need to think much harder about your thought experiments. For example, your ring cannot remain rigid when you "lower" it, so you need to think about how it responds to the changing stresses.
Due to the difficulty of nonlinear mathematics, approximations are often neccessary to make even limited progress. Unfortunately, approximation is a tricky art form and this is where fallacies most often seem to creep into "mathematical physics discourse". Unfortunately for the present context, approximations which are somewhat tricky even in flat spacetime often become much trickier in curved spacetimes.
There are in fact a number of "fiber optic" thought experiments offered in arXiv eprints, some of which IMO are misleading and have led to incorrect conclusions. Some authors are much more careful than others; preprints posted to the gr-qc section seem to exhibit a particularly large range of quality
Quote by kev In my last post I asked a lot of questions (perhaps too many).
Yes, IMO meopemuk is confusing you and leading you away from your original question, which at least makes sense to me.
Quote by kev the main question I am asking at his point is: Do objects length contract or expand in a strong gravitational field as measured by a distant observer ...or not?
All that stuff meomepuk told you about length shortening and time slowing is just wrong. Gtr says no such thing. (See any good textbook.) The best answer to this question is that it doesn't make sense. It's like asking: "In the game of baseball, when the batter turns up an ace of spades, does that result in an automatic advancement of the player holding the office of second base to the office of first base?" It sounds like a reasonable question, but only if you know nothing about baseball!
Quote by meopemuk This fact can be reconciled with the variation of the speed of light c' = c[1-2GM/(c^2)r] if we assume that all distances decrease according to d' = d[1-GM/(c^2)r].
Quote by kev Assuming we all agreed that the speed of light as measured by a local observer is always c then a slowing down of light as observed by a distant observer would be more likely to be reconciled with an increase of horizontal distances? For example, say we had a fibre optic that took one second for a photon to traverse in a very weak gravitational field. If we lowered that optical fibre ruler into a strong gravity well where the gravitational time dilation factor was 2 then a photon would take 0.5 seconds to traverse that ruler as measured by a local observer with slow clocks (assuming no length change). If the ruler had "length dilated" to twice its proper length, then the local observer would still measure the time taken to traverse the ruler as one second and the speed of light would still appear to be c to the local observer.
Let me see if you would agree with the following logic:
The claim is that from the point of view of distant observer D objects in the gravity field (F) have the following properties:
1. All clocks go [1+GM/(c^2)r] slower, i.e., the duration of one F second is equal to [1+GM/(c^2)r] D seconds
2. All objects become shorted, i.e., what F observer perceives as 1-meter rod, the D observer sees as a [1-GM/(c^2)r] meter rod
3. Light is slower in the gravitational field, i.e., observer D measures the speed of light near observer F as c[1-2GM/(c^2)r], where c is the speed of light in the vicinity of D.
If these conditions are satisfied, then measurements of observer F would also yield the speed of light as c. If F takes a rod of 300.000.000m long (which, according to D is shorter than necessary) and sends a light pulse (which, according to D is slower than usual) along it, then signal would arrive to the other end exactly in one F-second (which, according to D is longer than normal). So, both D and F would measure the speed of light as 300.000.000 m/s in their local frames.
Eugene.
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Quote by meopemuk Let me see if you would agree with the following logic: ... All clocks go [1+GM/(c^2)r] slower ... All objects become shorted... Light is slower in the gravitational field
Wrong, wrong (even by your own view since you forgot about radial versus nonradial orientations), wrong. GTR says nothing like this. Meopemuk, IMO you are not helping here.
kev, to state the obvious, it makes good sense to master the mainstream view before exploring/debunking fringe viewpoints (should you desire to do that).
Quote by Chris Hillman Time does not slow down and distances do not shorten! ...GTR says nothing like this.
I never claimed that what I said is a part of GTR. I simply looked at experimental observations and tried to understand them without prejudice in a simplest possible model.
I wouldn't say that "time slows down" in the gravitational potential (because I don't know what's the meaning of the expression "speed of time"). However, it is well established that identical clocks run slower in the gravitational potential than far away from it (e.g., the Pound-Rebka experiment, GPS,...). This difference in the clock rates is described well by formula
$$T' = T(1+\frac{GM}{c^2r})$$
where T' is the period of one clock tick in the potential.
It is also known that light takes longer time to propagate between two points (e.g., Earth and Venus) if there is a massive body (Sun) on its way. This Shapiro time delay can be explained within GTR, as you demonstrated. But one can offer a simpler explanation as well. One can simply assume that the speed of light depends on the gravitational potential as
$$c' = c(1-\frac{2GM}{c^2r})$$
(independent on the light direction) It can be shown that the numerical value of the Earth-Venus-Earth time delay comes out exactly as in GTR.
Another fact is that the speed of light appears the same to observers in different gravitational potentials. All these facts can be reconciled by the assumption that the length of any rod decreases in the potential as
$$d' = d(1-\frac{GM}{c^2r})$$
The slowing-down of clocks and the reduction of the speed of light can be explained within a simple Newton-like theory of gravity, where gravitational interactions are distance- and velocity-dependent. Of course, this theory has nothing to do with GTR, and doesn't belong to the mainstream. However, in contrast to GTR, it is perfectly compatible with quantum mechanics, so, in my opinion, it can't be dismissed lightly.
I have no idea how to explain the shortening of lengths in the field. I think it would be great to design an experiment to show whether this length shortening exists or not. This question makes sense no matter which theory of gravity is correct. And it would be a great independent check on the general theory of relativity. Here I agree with kev.
Eugene.
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Quote by meopemuk I never claimed that what I said is a part of GTR. I simply looked at experimental observations and tried to understand them without prejudice in a simplest possible model.
It would be very helpful if you found peer-reviewed papers and/or respectable popularizations with a viewpoint similar to your own, and pointed them out.
http://xxx.lanl.gov/abs/astro-ph/0006423 might be a place to start. However, it's at a level that I wouldn't want to throw it at the average lay person. It also is arguably flawed, to boot, though I think it's interesting nonetheless for the advanced reader. It would unfortunately probably only confuse someone who didn't already know quite a bit about GR :-(.
It's problematic with respect to PF guidelines to promote ideas unless they've been published in a peer-reviewed journal. Because GR is very technical, I think that respectable popularizations are also acceptable.
It's also very handy to be able to point interested readers at respectable published sources. They can usually explain things in a lot more depth than a post can, and they are also something that interested readers can point to when *they* want to talk about the topic with someone else.
Quote by meopemuk Yes, there is some additional delay due to the non-straight path of light. But it is much smaller than the main effect. In my opinion, the main origin of the Shapiro delay is the fact that light propagates slower in the vicinity of massive objects. Eugene.
I agree that the non-straight path of light is much smaller than the main effect.
But the space can be curved (stretched or compressed) even if it is straight
for example, rubber band can be stretched without bending
You can say that space is flat but ruler shrinks in gravitational field
Anyway, if you measure the length of the path with rulers, you definitely need more rulers
In my opinion, if you ignore the curvature of space, you will get only half of the Shapiro delay
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http://mathoverflow.net/questions/34442/sum-of-sets-modulo-a-square/34471 | Sum of sets modulo a square
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I would be glad to see a reference to the following easy lemma in additive combinatorics: if $A_1$ and $A_2$ are two sets of remainders modulo $n^2$, each has cardinality $n > 1$ and all elements of $A_i$ are different modulo $n$ (for $i=1,2$), then $A_1+A_2$ is not equal to the set of all remainders modulo $n^2$.
Maybe, it is a partial case of more general and deep:) result.
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For $n$ even it is easy to get a contradiction by summing up, but for $n$ odd? – darij grinberg Aug 3 2010 at 22:24
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Well, the easy proof is as follows: assuming contrary, consider all $n$ remainders with remainder $r$ modulo $n$, sum up. We get that sum of all elements of $A_1$ and $A_2$ equals $r+(r+n)+\dots+r+(n^2-n)$. But this does depend on $r$. Contradiction! – Fedor Petrov Aug 4 2010 at 6:39
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@Fedor, I don't follow your proof. – Gerry Myerson Aug 4 2010 at 7:09
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It is cute! Line up the elements of A_1 by increasing residue mod n. underneath them put the elements of A_2 in decreasing order (mod n). The sums (reduced mod n^2) are 0, n,2n,...,(n-1)*n in some order for a grand total of n*n*(n-1)/2 which is 0 mod n^2. Now cyclically shift the second row. The n sums (reduced mod n^2) would be 1 n+1, ...,(n-1)n+1 for a grand total of n*n*(n-1)/2+n*1 which is n mod n^2. But the grand total should remain unchanged since it is the sum sum of 2n integers. – Aaron Meyerowitz Aug 4 2010 at 7:40
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$n^2(n-1)/2$ is not always $0$ modulo $n^2$, but it always differs from $n^2(n-1)/2+n$ – Fedor Petrov Aug 4 2010 at 9:12
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2 Answers
There must be an easier proof but here is a nice approach which can indeed lead to deeper results (feel free to edit for math display, I tried): Techniques with characteristic polynomials and roots of unity can be very powerful. I like the way that the appropriate lemmas are explained in my paper with Ethan Coven "Tiling the Integers with Translates of One Finite Set" http://arxiv.org/abs/math/9802122 or Journal of Algebra v 212 (1988) p 161-174. One does not need their full generality for this problem but perhaps for deeper results.
I'll sketch this result which implies what was asked for: Suppose that A and B are sets of size #A and #B so that A+B is a complete set of residues mod N=#A#B. Let p be a prime dividing N. Then exactly one of the sets has its members equally distributed mod p.
digression: Lemma 3.2 from the paper above (not needed here) shows that at least one of the following is true:
1) No member of A-A is relatively prime to #B
2) No member of B-B is relatively prime to #A end of digression
Consider the corresponding polynomials $A(x)=\sum_{a \in A}x^a$ and $B(x)=\sum_{b \in B}x^b$. Then
i) A(1)=#A and B(1)=#B
ii) A(x)B(x) is a sum of N distinct powers of x, one from each residue class.
iii) `$A(x)B(x)=(x^N-1)Q(x)+\frac{x^N-1}{x-1}$` for some polynomial Q(x).
iv) Every irreducible polynomial dividing $\frac{x^N-1}{x-1}$ divides at least one of $A(x)$ and $B(x)$
As an example consider A={0,9,13,16,29,32} B={0,10,12,22,24,34} with A+B a complete set of residues mod N=36.
`$$\frac{x^{36}-1}{x-1}=(x+1)(x^2+x+1)(x^2+1)(x^2-x+1)(x^4+x^2+1)(x^4-x^2+1)(x^{18}-x^9+1)$$`
evaluated at $x=1$ this becomes 36=2 * 3 * 2 * 1 * 3 * 1
In general the irreducible polynomial divisors of $\frac{x^N-1}{x-1}$ are the cyclotomic polynomials corresponding to the divisors of N. Evaluated at x=1 each is either 1 (composite divisor) or a prime p (prime power divisor) and the primes have product N. Since A(1)B(1)=N and A(x)B(x) is divisible by all the prime power cyclotomic divisors of $\frac{x^N-1}{x-1}$ and these evaluated at 1 also have product N, each divides just one of A(x) or B(x) and all other polynomial divisors evaluate to 1 at 1. In particular: for each prime divisor of N, only one of A(X), B(x) divides by $\frac{x^p-1}{x-1}$ and only that one has corresponding set equidistributed mod p.
In our example A is a complete set of residues mod 6 so A(x) divides by (1+x) and by (1+x+x^2). Since A(1)=6 , A(x) can't have either of (1+x^2) and (1+x^2+x^4) as factors. But they do divide A(x)B(x) and hence they divide B(x). This means that neither (1+x) nor (1+x+x^2) can divide B(x), again since B(1)=6. Hence, B is not equidistributed mod 2 (or mod 3) and certainly not mod 6.
By the way, $B(x)=(x^{10}+1)(x^{24}+x^{12}+1)$ and $A(x)=(x^{13}+1)(x^{32}+x^{16}+1)$ (mod $x^{36}-1$)
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I don't see how any of the facts you wrote help with this problem. Here we need to show that $x^{n^2}-1$ does not divide $A(x)B(x)$ in your notation... – Gjergji Zaimi Aug 4 2010 at 7:03
Oh, Aaaron, thanks! That's what I actually needed. Gjergji, I think Aaaron is correct: we assume that $A+B$ is full set of residues modulo $N$ ($N=n^2$ in my case), then we have such equality and that's it. – Fedor Petrov Aug 4 2010 at 7:20
I show that if $x^{n^2}-1$ does divide A(x)B(x) and A(x) is equally distributed mod n then B(x) is not equally distributed mod n. – Aaron Meyerowitz Aug 4 2010 at 7:25
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Replace each set by a sum of powers of x. Let p be a prime like 5 dividing n. Under your condition 1+ x + x^2 + x^3 + x^4 would divide both polynomials. Show it only divides the product once. I'd be less coy but I am typing this on a phone in a power outage! I've used those ideas to great effect. If n is prime then one set not only is not distinct mod n but actually has all elements equal mod n.
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Ok, this is messed up. {0,1,4,5,8,9}+{0,2,12,14,24,26} is a counter-example. But note that mod 6 the second set is {0,2,0,2,0,2}. I'll post a clearer answer and take this one down. – Aaron Meyerowitz Aug 4 2010 at 4:25
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@Aaron, Fedor calls it an "easy lemma" so I infer he already has a proof and just wants a citation. – Gerry Myerson Aug 4 2010 at 5:50
I suppose you are right. The approach I give shows (inter alia) that if A+B is a complete set of residues mod N=#A#B and both A and B contain 0 (no loss of generality in assuming that) then at least one of the two contains no elements relatively prime to N. – Aaron Meyerowitz Aug 4 2010 at 6:48 | {"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": 40, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9246813654899597, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/72986/alternate-definition-of-ordinals | ## Alternate definition of ordinals ?
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Hello, recently I came upon some personal notes I'd made several years ago while reviewing some basic set theory (ordinals, transfinite recursion, inaccessible cardinals etc.), and I stumbled upon a loose thread which I obviously had not resolved at the time, and which I would like to lay to rest: Assuming some standard set theory (say ZF, even though I prefer NBG), without the Axiom of Foundation (preferably), one may define an ordinal $\alpha$ (von Neumann's definition) as a transitive set whose elements are well-ordered with respect to the membership relation $\in$. This is seen to be equivalent to the statement that $\alpha$ is transitive, all its $\beta\in\alpha$ are transitive too, and (as we cannot rely on foundation) for each non-empty $x\subseteq\alpha$ there exists some $\beta\in x$ such that $x\cap\beta=\emptyset$ (except for the last condition, this is as in Schofield's book on Mathematical Logic). One then goes on to prove that the class of all ordinals is well-ordered with respect to membership etc.; along the way a useful intermediate step is to prove that any ordinal $\alpha$ is (ad hoc definition) $\textbf{strange}$ in the sense that one has $x\in\alpha$ for any transitive $x\subsetneq\alpha$. My question finally (as this would provide an alternate definition of ordinal sets): are elements of strange sets themselves strange, or at least transitive ? Thanks in advance for any useful comments ! Kind regards, Stephan F. Kroneck.
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## 3 Answers
Strange sets are the same thing as ordinals. Given a strange set $\alpha$, let $\beta$ be the smallest ordinal such that $\beta\notin\alpha$. Then $\beta\subseteq\alpha$. If $\beta\subsetneq\alpha$, then $\beta\in\alpha$ as $\alpha$ is strange, which contradicts the definition of $\beta$. Thus $\beta=\alpha$.
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Emil, it looks like we hit on the same observation. – Joel David Hamkins Aug 16 2011 at 14:53
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Indeed. And I find the observation somewhat interesting in that it provides a definition of ordinals in ZF which works in the absense of the foundation axiom, despite that it does not refer to well-foundedness (or well-order). (It does, however, quantify over subsets of $\alpha$, which is to be expected, as ordinals are not definable by a bounded formula in ZF minus foundation.) – Emil Jeřábek Aug 16 2011 at 15:04
Well, the way that well-foundedness works its way in is in the definition of ordinal. I have seen alternative definitions of ordinal in the anti-foundation context simply as "hereditarily transitive" sets, dropping the well-founded part of the definition. I'm not sure, but I think that some intuitionists use something like this also, and perhaps they (or you?) can tell us. – Joel David Hamkins Aug 16 2011 at 15:12
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@Joel: I think we are talking past each other. I’m not interested in a definition of alternative ordinals (such as hereditarily transitive sets, which do not have any useful properties without the foundation axiom), but alternative definitions of the standard class of ordinals. Its usual definitions explicitly involve the condition of well-foundedness in one way or another (such as by demanding that the set be well-ordered by ∈), which is rather inelegant. In contrast the definition of a strange set does not invoke well-foundedness in any way, it only refers to its transitive subsets. ... – Emil Jeřábek Aug 16 2011 at 15:37
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... As for intuitionistic set theories: the usual ones (IZF, CZF) include the foundation axiom (in the form of $\in$-induction). – Emil Jeřábek Aug 16 2011 at 15:38
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Theorem. Every strange set is an ordinal.
Proof. Suppose that $\alpha$ is strange. Let $\beta$ be the smallest ordinal such that $\beta\notin\alpha$. Such a $\beta$ exists, because no set can contain all the ordinals, and this does not require the foundation axiom to prove. It follows that $\beta\subset\alpha$ and $\beta$ is transitive. Thus, if $\beta\neq\alpha$, we would have $\beta\in\alpha$, contradicting the choice of $\beta$. Hence $\beta=\alpha$ and $\alpha$ is an ordinal. QED
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Update: I haven't given the whole matter terribly much thought recently, apart from the fact that I'm able to prove that strange sets are necessearily hereditarily transitive (what a mouthful !) and that no element of a strange set has itself as a member (directly using the definition, avoiding the whole ordinal machinery), and (what is then quite easy) that strange sets are ordered linearly w.r.t. subset inclusion. What remains is the original loose end: avoiding ordinals, just using the definition, why are elements of strnage sets again strange ? (ctd. below) – Stephan F. Kroneck Sep 8 2011 at 13:21
(ctd.) Thus I wish to leave the question open (and prefer not to hand out an "answered" just yet). Kind regards to all who have contributed their time and insights so far ! S.F. Kroneck. – Stephan F. Kroneck Sep 8 2011 at 13:22
Would like to close this question, as I've answered its parallel on Stackexchange: "Alternate definition of ordinal sets". Kind regards - Stephan F. Kroneck.
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Why would you close it? Other people may benefit from it or have something to say later, even if you are satisfied with the answers obtained so far. I would suggest that instead you click the check mark next to one of the answers, so the questions does not appear as "unsolved" any longer. – Andres Caicedo Sep 11 2011 at 21:09
@ Andreas Caicedo: Thank you for your comment ! It just seems a bit weird ticking my own answer ... however, it is the only one up till now that satisfies my admittedly "strange" standards. Kind regards - Stephan F. Kroneck. – Stephan F. Kroneck Sep 12 2011 at 7:56 | {"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": 33, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9424322247505188, "perplexity_flag": "middle"} |
http://en.m.wikibooks.org/wiki/Strength_of_Materials/Loading_of_Beams | Strength of Materials/Loading of Beams
Strength of Materials
Strength of Materials in Engineering Mechanics
Introduction | Introductory Concepts | Loading of Beams | Torsion | General State of Stress | Weakest link determination by use of three parameter Weibull statistics
One of the areas where solid mechanics as discussed in this book is most effective is in the case beam loading. The loads on a beam can be point loads, distributed loads, or varying loads. There can also be point moments on the beam. The beam itself is supported at one or more points. The conditions at the support depend on the kind of support used. If the support is a roller, it can only have a reaction perpendicular to the motion of the roller. If the support is a pin, it cannot carry a moment. If the support is fixed, then it can have a reaction in any direction and support a moment as well.
In the above image, a simple beam is loaded at the center by a load P. It has a pinned contact at one end, and a rolling contact at the other end.
The above image shows an ideal moment acting at the center of a beam. An ideal moment is one which is not associated with a force.
The general method is analyzing beam problems is to find the loads, reactions, and moments, and come up with the values for the loads and moments at each section. This, in general will be a piecewise function of the distance along the beam.
For loads, we set up axes and this decides the sign of the force. For moments, we set the convention that the clockwise moment is positive.
Examples
1. Consider the case of the beam discussed earlier shown above.
The beam has reactions R1 and R2 acting on each of the supports.
R1 + R2 = P
Applying symmetry, we have,
R1 = R2
So that,
R1 = R2 = P/2
The above image shows the shear diagram for this problem. Note that the positive and negative directions are conventions, but it is important to choose one direction for positive shear and stick to it. As can be seen, the shear value changes at the point of application of the load.
The above image shows the moment diagram for the beam. The moment varies linearly from the supports to the middle. (The value at the middle should be PL/4). Note that the moment is maximum at the middle and zero at the ends, so that the effect of the moment is maximum at the center. In later chapters we will see that the stresses and strains due to moments are the most important ones for beams.
Exercises
1. In the above beam, find the reactions in the supports and the shear force at a position x. Also, find the moment at that point.
2. The above beam shows loading by two separate point loads. Find the shear and moment at location x from one end of the beam.
3. Find the shear and moment at each point along the beam with a point load and moment acting at two different points. How can a point moment occur in practice?
Calculus for Solving Beam Problems
Methods of calculus can be used to deal with continuous load functions. However these methods can be extended to point loads and moments by the use of the Dirac delta function.
The above shows a beam with uniform load per unit length w. Such loads are used to model the self weight of the beam where it acts uniformly throughout its length. This can also be used for the loading of, say, a bridge due to all the vehicles on it. The individual point loads acting through the tires can be modeled as a continuous load if the number of vehicles is large.
Consider a beam with load per unit length q(x) for any point x along its length. Consider an elemental length dx of the beam. Applying force equilibrium for this segment, we have,
V + dV = q dx + V
where V is the shear at the point x. Thus, we have the relation
dV/dx = q
Applying moment equilibrium to the same segment, we have,
M + dM − V dx − M − q dx dx/2 = 0
Neglecting the second order terms in dx, we have,
dM/dx = V
Or,
d2M/dx2 = q
The above differential equations can be integrated with appropriate boundary conditions to get the shear and moment at each point.
Examples
Consider a beam of length L supported at its ends by two pins, with a uniform load per unit length of w.
The total load on the beam is thus wL. Thus, the reaction at each support is wL/2.
We have, the general relation for shear
$V = \int_0^x q dx + C_1$
The shear at origin is just the reaction at that point (=wL/2). If we take the vertical direction as the positive direction, we have, the shear at the origin is wL/2. Also, q = −w. Thus, we have, V = −wx + wL/2 for any point x.
The Moment at the centre should be wL^2/8
Similarly, we have the moment M
$M = \int_0^x V dx + C_2$
It is easy to see that the moment at the origin is zero. Thus, we have M = w x (L − x)/2.
Calculus for Point Loads
The above works well for well for continuous, smooth load functions. But in real world situations, we have to deal with point loads and moments. Thus we use the Dirac functions for a point load P at location a as
$P\left\langle x - a \right\rangle^{-1}$
and for a point moment M at location a as
$M \left\langle x - a \right\rangle^{-2}$
Now the previously stated equations can be used under the rules for the Dirac function. The function itself is defined as
$\left\langle x - a \right\rangle^n = \left\{ \begin{matrix} 0 & x < a\\ (x - a)^n & x > a \end{matrix} \right.$
Consider a beam of length L with supports at both ends and a point load P at the center. From the Dirac definition, we have,
$q = -P\left\langle x - L/2 \right\rangle^{-1}$
The shear at one end is just the reaction P/2. The shear at any point is given by
$V = \int_0^x q dx = \int_0^x -P\left\langle x - L/2 \right\rangle^{-1} dx = -P \left\langle x - L/2 \right\rangle^{0} + P/2$
The moment at one end is zero. The moment at any point is given by
$M = \int_0^x V dx = -P \left\langle x - L/2 \right\rangle + \frac{Px}{2}$
Applying the definition of the Dirac function, we have, the shear
$V = \left\{ \begin{matrix} P/2 & x < L/2\\ -P/2 & x > L/2 \end{matrix} \right.$
$M = \left\{ \begin{matrix} \frac{Px}{2} & x < L/2\\ \frac{P(L - x)}{2} & x > L/2 \end{matrix} \right.$
Note that the above is a special case of the general method called Laplace Transforms.
Exercises
1. Find the shear and moment in the above beam using Dirac methods.
2. The beam shown above has two loads which can be modeled as shown. Find the shear and moment at any point along the beam.
3. Consider the beam shown above with an overhang. Find the shear and moment at points along the axis.
Stresses and Strains in Beams
Given the loads and moments at each cross section, we can calculate the stress and strain at each location.
Consider the bending of a slender beam (one for which the cross section is much smaller than the length). A moment acting on the beam causes a deformation called flexure. Let the radius of the osculating circle of the beam be ρ. Consider an elemental length ds in the neutral plane (for which the deformation is zero). This element subtends an angle θ at the center of curvature, so that
ds/dθ = ρ
If we move a distance y along the radius, we have the length of the arc subtended would be (ρ − y) dθ. Thus the elemental extension would be y dθ. For a small enough curvature, we have the distance along the neutral plane would be the same as the initial undeformed length, dx. This gives us the axial strain at any point x as
εx = y/ρ
Also, using the Hooke's law, we have,
σx = E εx = Ey/ρ
Now, we know that a pure moment M acts on the beam, so that there is no axial force. Or,
∑ Fx = 0
∫ σx dA = 0
∫ y dA = 0
The above equation gives us the location of the neutral plane.
Further, applying the moment conservation part of the equilibrium relations, we have,
M − ∫ σx dA y = 0
Now, we know,
σx = Ey/ρ
and
I = ∫ y2 dA
where I is the moment of inertia, so that
M = E/ρ I
Or
E/ρ = M/I
Thus, we have the expression for the stress due to a pure moment as
σx = My/I
Suppose the maximum value of y is c (the distance from the neutral plane). Then, we have,
σmax = Mc/I
The quantity I/c is called the section moment of a beam and it denoted by S.
σmax = M/S
Beams with Arbitrary Moments
So far we have considered beams which were acted upon by moments in the Mz. Suppose there is another moment in the My direction. Note that a moment in the Mx direction would simply be a torsional effect as the axis is in the x direction. Now, it is easy to see that the combination of moments is, in fact, equivalent to a moment acting on a beam of arbitrary cross section. It is fairly straightforward to show that, in this case, the stress at any point (y, z) on the face of the section for the beam is a function of Iz, Iy, and Iyz.
Shear in Beams
Earlier, we saw the methods to calculate the shear forces on a beam. Now we can analyze the stresses due shear forces, like we did for stresses due to bending moments. We showed that for the shear force V is given by V = dM/dx, where M is the moment acting at the point x.
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http://math.stackexchange.com/questions/28286/general-introduction-to-functional-and-other-mathematic-notations/29030 | # General Introduction to Functional and other Mathematic Notations
I've been a programmer for a good while now. Fairly experienced at a bit of math as far as coming up with algorithms and such but I am far far behind on understanding quite a deal of notation.
Here and there I run into an issue where someone will notify me that I've reinvented some piece of calculus, trig or some other fields. Occasionally this makes for some interesting code and all, but I've begun to think that I could very often avoid this by being able to read and write standard notation more fluently.
When it comes to this area, I'm honestly a complete newb. Are there any good introductions or resources that can help get me on a clear path to understanding?
I have some concept on simple functions, but not much. Tendency in study has been that I'll find myself too deep in something too complicated too quickly and forget everything.
• For instance, to borrow from another open bounty at this time, I cannot read the following:
$$\sum_{n=-\infty}^\infty J_n(x) J_{n+m}(x) = \delta(m)$$
My mind is stuck in code, help me out of my cave! :)
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I think you should try picking up first year calculus textbook, it usually introduces concepts like sets, sequences, limits, series, derivatives, integrals etc. I'm afraid I can't suggest anything in particular except Diodonne's book that I haven't myself read :) – Alexei Averchenko Mar 25 '11 at 2:12
@Alexei: Will look into this one way or another! My math is so rough around the edges >_<; – Garet Claborn Mar 25 '11 at 4:29
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@Alexei Averchenko You mean Dieudonné ;) – Klaus Mar 25 '11 at 13:15
## 5 Answers
As a programmer myself, I feel your pain in this.
I think it's safe to say, any well-seasoned programmer could pick up a reference book on just about any programming language in existence and write competently in that language in short order. Would you agree?
Math is like that, but it doesn't have the same flow and nuances as programming. Spoken languages have a certain rhythm to them; learning one Asian language makes it easier to pick up other Asian languages but not necessarily European languages.
Picking up a 'quick' introduction book to mathematical notation won't help you. You need experience, as you did with programming.
There are some good resources out there, but I can't think of a better resource than your own hard work and study. Find the best classwork-style books you can on the topics I list below, and you'll do just fine:
• Linear Algebra (2010, MIT OCW)
• single variable calculus (2010, MIT OCW)
• multi-variable calculus (2010, MIT OCW)
• differential equations (2010, MIT OCW)
• introduction to algorithms (2005, MIT OCW)
• This is a good one for CS types. They cover asymptotics as it relates to algorithm analysis & design: searching & sorting; binary trees, red/black trees, skip layer structures, bubble sort, etc. They spend a significant amount of time showing you how to analyze the run-time of an algorithm, its memory requirements, number of arithmetic operations or conditionals, etc.
• The MIT OCW site has some other math classes for scientific computing, but the format is hard to follow; lots of chalk dust, hard to read what's written, etc.
• boundary value problems and/or partial differential equations
• This, or a book on signal processing will help you with the equation you quoted. Unless I'm mistaken, that is an eigenvector/eigenvalue problem.
• graph theory
• A must-have for any CS person
• I recommend Applied Combinatorics by Alan Tucker. It's a pretty compact book, very readable, and most of it is easily approachable with little more than a decent understanding of Algebra.
• probability & statistics (not the easy freshman/sophomore books)
• I don't have a good book or lecture recommendation. The book I used in school was painful. I won't inflict that on you.
• numerical analysis
• This is a big one for CS people wanting math exposure
• Two books I'll recommend (with a caveat):
• Numerical Analysis by Burden & Faires
• This book is incredibly readable until the algorithms start to involve many parameters/variables. The notation gets irritating near the end of the book.
• [Most of?] the first 4 chapters should be easy to follow without a strong calculus background. Without exposure to ODEs, I recommend going through chapters 1-4, and 6-8; possibly chapter 9 (those come from the 7th edition. I don't know the layout of the latest edition)
• Numerical Optimization by Nocedal & Wright
• This is a good one to get later on when you're starting to feel more comfortable with this stuff. It's usually used as a graduate textbook.
• Both of these books have pretty readable pseudo-code for all their algorithms
-Brian
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@Brian "Math doesn't have the same flow and nuances as programming."? What does that mean? – Myself Mar 26 '11 at 1:27
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My point is it isn't just some set of axioms and rules. It IS a language with a logical flow or rhythm. If sentence words threw I order any did, I together a in then the sentence wouldn't be readable to you. The same thing is true in math. Just because you know the words and symbols doesn't mean you know how to string them together in a manner that is readable. – Brian Vandenberg Mar 26 '11 at 7:25
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or string the ideas together into a coherent and correct proof. – Brian Vandenberg Mar 26 '11 at 7:30
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@Brian Vandenberg: I think you make a good point, I hadn't realised the OP's point of view about math may be oversimplified in the way that you mean. As a programmer, apparently you understand the OP's concerns and desires better than I do. – Myself Mar 26 '11 at 11:40
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@Brian - yes, I definitely relate to picking up languages. I'm basically out hunting for the 'spec' for the 'math framework'. Doing all manner of crazy things with numeric and representative values is all fine and good in code but sometimes very difficult for me to convey human to human. And thanks for the OpenCourseWare links, don't know why I didn't think to check there earlier. – Garet Claborn Mar 27 '11 at 8:40
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Concrete Mathematics: A Foundation for Computer Science by Graham, Knuth, and Patashnik
Here is a paragraph from the preface which talks about the purpose of the book:
"One of the present authors had embarked on a series of books called The Art of Computer Programming, and in writing the first volume he [Knuth] had found that there were mathematical tools missing from his repertoire; the mathematics he needed for a thorough, well-grounded understanding of computer programs was quite different from what he'd learned as a mathematics major in college. So he introduced a new course, teaching what he wished somebody had taught him."
The book goes over recurrence relations, sums (including a section on finite differences mentioned by @Alex), number theory, binomial coefficients, special numbers like Stirling numbers, Bernoulli numbers, and Fibonacci numbers, generating functions, probability, and asymptotics. I have read most of the book and it's fun reading and it has helped me a lot (though I am a math grad student trying to study number theory and a lot of this stuff is helpful there). The level is not too high (I know it's going to be easier for me as a math grad student but I still don't think it's at an incredibly high level). There are a lot of exercises, ranging from very easy to research problems. Solutions are in the back of the book for most of them. There are almost no errors in this book, which is good for someone learning on their own. Knuth pays people \$2.56 for any error they find and it's sort of a prestige thing to get a check from Knuth. He also does the same sort of thing for "The Art of Computer Programming" and also for the computer write-up language he invented, TeX.
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This is a fantastic book. I highly recommend it. I respectfully disagree with Numth about how easy it is. I would describe it as not very technical but very deep. Reading it carefully will take a while. – Sam Lisi Mar 25 '11 at 18:08
This sounds like a great read, I will probably look into it. Only, it sounds more like learning math concepts than math notation, is that true? I realize they will be fairly hand in hand in most resources but the concepts aren't my weak point. – Garet Claborn Mar 28 '11 at 6:38
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@Garet Yes, it is a math textbook so the point is learning the math, but it has a lot of notation as well. – Graphth Mar 29 '11 at 19:38
http://dlmf.nist.gov/ is a good site for special functions, etc. Thus, for the given example, searching for J(x) in the search box quickly leads to the sections about Bessel functions.
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– Garet Claborn Mar 30 '11 at 21:40
Personally, I've found "Modern Algebra: An Introduction" by John Durbin a very interesting and accessible introduction to abstract math, including a lot of information about how to think of sets, functions and the like. I'm not sure if that's exactly what you're after, but I'd say it's at least worth a look.
If all you're interested in is learning about how mathematicians define functions and the notation that they use, a quick google search revealed two promising tutorials: This one which explains functions of a single variable, how to think about them, and different notations used to define them, and this one, a slideshow explaining functions of multiple variables.
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Thanks. To clarify I'm talking generally about most of the things you end up needing MathML to express :) I can read very basic f(x)=... functions but not much past that (and not always even that >< ) Once you get deep into the sigma, lamda, nested functions and potatoes..I'm out of scope. I have most of the concepts if I could read them :) – Garet Claborn Mar 24 '11 at 23:10
I'm not entirely sure if functions is all I'm looking for because I really don't know what all there is. I'm trying to 'formalize' my math skills. Thanks for the extra links, will check them out. – Garet Claborn Mar 24 '11 at 23:22
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Ah. That is a long process, and you will probably need to read more than one book on the subject, because if you haven't been introduced to the notation then you are probably unfamiliar with the concepts being represented. – Alex Becker Mar 24 '11 at 23:25
I've had a somewhat odd flow of education. I'm sure there are some concepts I don't know, but many I have used quite a bit writing 3D and physics, but I think my team mates are getting tired of translating to and from C++ for me. More than one book is fine though, any suggestions of the areas I need to cover? – Garet Claborn Mar 24 '11 at 23:29
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Linear algebra is a must for a programmer, calculus would also be helpful (especially finite differences). – Alex Becker Mar 24 '11 at 23:35
If you're also interested in the history of mathematical symbols (you probably don't, because you're in trouble now, but maybe you're curious...), you can read the classical book by F. Cajori, A History of Mathematical Notations.
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Oh I'm not in too much trouble lol, thanks for the reference. In my last place we all spoke the same language and it just seems it would make life easier for everyone. – Garet Claborn Mar 25 '11 at 4:27
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@Garet: Also the book by Howson, A Handbook of Terms used in Algebra and Analysis, Cambridge University Press, is a nice one. – Pacciu Mar 25 '11 at 12:31 | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9541071653366089, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/9240/what-are-fields/9245 | # What are fields?
I'm following my first course in field theory and the professor began, like many books do, by introducing the scalar field. However, I am a bit hesitant about the physical idea of fields. My question is: what is the physical meaning of the fields? Why they are introduced? I read the introduction of the books of Peskin and Weinberg but I'm not satisfied.
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I was tempted to Answer, at a relatively philosophical level, just that "Fields are introduced because of a preference for local causality as an explanation for events", but I decided that your reference to Peskin&Schroeder and Weinberg leaves me wondering what kind of justification you want. Perhaps you could clarify your Question? You could try Tian Yu Cao, "Conceptual Developments of 20th Century Field Theories", Cambridge UP, 1997, or I could give you other references into the Philosophical literature on field theory. – Peter Morgan Apr 28 '11 at 17:40
I'm just a student but from what I've learned so far it seems to me that the fields are introduced as a trick to make things work and I'm missing the "essence" of this mechanism. This is why I posted the question. – Andrea Amoretti Apr 28 '11 at 17:47
You know what a harmonic oscillator is, right? You know how two SHOs can be coupled - in series or in parallel. I'm assuming you've solved these problems in classical mechanics courses. Now take $n$ of oscillators and lay them out end to end in series. Take the limit $n \rightarrow \infty$. The resulting system has a Lagrangian of a scalar field in one dimension (1+1 to be precise). This is shown quite explicitly in Goldstein's mechanics book, though perhaps not in the newer versions. You can construct field theories in any dimension in this manner. – user346 Apr 28 '11 at 18:07
– pcr Apr 29 '11 at 7:38
## 3 Answers
Probably the most fundamental and simple idea of the field arises from heat equation. You have a heat source and heat diffusion through media. It is described by field of temperature. It is the simplest scalar field I can imagine. But it has nontrivial equation of motion - has it? From that simplest cases more complicated arises: force fields introduced without any complicated mathematics by well known genius Michael Faraday, who just draw lines of equal potential from one charge to another in order to understand how electrostatic and magnetic forces works. Of course You may imagine even more complicated fields - general tensor or spinor ones. Strict definition of fields is of course a function which for given point on the manifold assign tensor or spinor to it. Physical fields often has strong continuity and differential conditions on them and obeys complicated differential or integral equations, but the most fundamental idea is still the same: lines which was drawn by Faraday, temperature in continuous media, velocity field of the fluid passing through water canal. Try to understand that simple ideas and return to "quantum-compacted-26-dimensional-theories of everything".
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Fields are important in both classical and quantum physics.
At first it was thought they are just some imaginary mapping of the forces of gravitation or electric and magnetic forces. Gradually it became clear one can not keep it away from physics. They are more than imagination.
Why? It is because electromagnetic effects take some finite nonzero time to reach from one point to other. A moving charge takes a nonzero time to influence another charge or a magnetic needle. If we want to preserve the laws of conservation of momentum and energy during that nonzero time we must have to take the concept of field as real and it should be thought of as the seat of energy and momentum.
At first people tried to understand field in terms of classical mechanics. It was believed at first that there must be some mechanical medium whose stress and strain must be the manifestation of field. The medium was named as aether. But all such efforts failed.
It became clear that field is a fundamental entity of nature just as matter is. Or maybe matter is itself a kind of concentrated field!
Quantum theory has given rise to quantum field theories and the classical fields have been quantized in QFT. It is now understood that fields can be thought of as consist of field quantas or particles with integer spin (bosons) and the so called matters are particles of half interger spin (fermions).
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I am a layperson who knows little about science but am eager to learn. Perhaps a field can be described as an invisible medium that enables action between objects and the effects of which can be measured.
Brian Silver, The Ascent of Science, gives basic and more advanced descriptions of this concept.
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FYI Please see user1355's answer: A field is not an aether. – John Feb 15 '12 at 14:21 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9574593901634216, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/101253/surjectivity-of-operators-on-linfty/101612 | ## surjectivity of operators on l^infty
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Can anyone give me an example of an bounded and linear operator `$T:\ell^\infty\to \ell^\infty$` (the space of bounded sequences with the usual sup-norm), such that T has dense range, but is not surjective?
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Despite the two quick votes to close, I don't find this a trivial question. Am I missing something? – Bill Johnson Jul 3 at 21:41
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I guess you mean also that $A$ should be closed and map its domain back into itself. I would have to review semigroup theory (or think more than I care to right now) to see if that is correct. Anyway, how do you get such an $A$? – Bill Johnson Jul 3 at 22:48
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Surely, Yemon; for the weak$^*$ topology the problem is trivial. – Bill Johnson Jul 3 at 23:25
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There is a theorem of Lotz stating that there are no strongly continuous semigroups on $l^{\infty}$, meaning that if semigroup is strongly continuous, then the generator is bounded. – András Bátkai Jul 4 at 17:27
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Here is a much easier question: Does there exist a non surjective bounded linear operator from some Banach space into $\ell_\infty$ that has dense range? I see how to do this but the argument uses something that is not elementary. Is there a simple reason such an operator exists? – Bill Johnson Jul 5 at 16:21
show 5 more comments
## 2 Answers
Here is an answer to an easier but related question.
Proposition. There is a one to one operator $T$ from $\ell_1(2^{\aleph_0})$ into $\ell_\infty$ that has dense range.
Of course, such an operator cannot be surjective because $\ell_1(2^{\aleph_0})$ is not isomorphic to $\ell_\infty$.
My proof of the Proposition uses an old result of Bill Davis and mine (Remark 4 in
Davis, W. J.; Johnson, W. B. On the existence of fundamental and total bounded biorthogonal systems in Banach spaces. Studia Math. 45 (1973), 173–179):
$\ell_\infty$ has a biorthogonal system $(x_\alpha,x_\alpha^*)_{\alpha<2^{\aleph_0}}$ with $\|x_\alpha\|=1$ and $\sup_\alpha \|x_\alpha^*\|<\infty$ such that the linear span of $(x_\alpha)$ is dense in $\ell_\infty$.
To prove the Proposition, define $T$ to be the norm one linear extension of the map $e_\alpha \mapsto x_\alpha$, where $(e_\alpha)$ is the unit vector basis for $\ell_1(2^{\aleph_0})$. This mapping obviously has dense range and is one to one because every biorthogonal system is countably linearly independent.
Here is a variation on the OP's question:
Is there a one to one bounded linear operator from $\ell_\infty$ into itself that has dense range but is not surjective?
The interest in the variation is that this question is easily seen to be equivalent to:
Are there quasi-complementary copies of $\ell_\infty$ in $\ell_\infty$ that are not complementary?
(Recall that two closed subspaces of a Banach space are said to be quasi-complementary if their sum is dense and their intersection is ${0}$.)
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@Bill: The existence of "quasi-complementary but not complementary subspaces" in $l_{\infty}$ is equivalent to $l_{\infty}$ has infinite dimension separable quotient. \\ But the existence of "quasi-complementary but not complementary subspaces" in $l_{\infty}$ is not equivalent to the existence of "one to one bounded linear operator form $l_{\infty}$ into itself that has dense range but is not surjective". – Qingping Zeng Aug 27 at 13:58
The former condition is in fact equivalent to the existence of "one to one bounded linear operator form some Banach space $X$ into $l_{\infty}$ that has dense range but is not surjective". \\ And, it is well known that $l_{\infty}$ has infinite dimension separable quotient. So, the existence of "one to one bounded linear operator form some Banach space $X$ into $l_{\infty}$ that has dense range but is not surjective" can also be deduced from the above facts. – Qingping Zeng Aug 27 at 14:03
@Qingping: What I said is that the existence of two quasi-complementary subspaces of `$\ell_\infty$` that are isomorphic to `$\ell_\infty$` and are not complementary is equivalent to the existence of a one to one bounded linear operator from `$\ell_\infty$` into itself that has dense range but is not surjective. – Bill Johnson Dec 7 at 18:34
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I could prove that if $T$ has dense then $T$ is surjective, in the cases where $T=S^{*}+W$, $W$ is weakly compact and $S:l^1\rightarrow l^1$ or when $T$ has totally disconnected spectrum.
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If you merge your existing accounts then you will be able to add this information to the existing question. Please try not to use the "answer" boxes for comments, as MO is not set up like a forum or blog thread. – Yemon Choi Jul 8 at 0:36
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I merged the two users, but this will not be very useful since Amir does not have a registered user. @Amir: Please consider registering a user; it will help make sure that you can edit all of your posts in the future. – Anton Geraschenko♦ Jul 8 at 18:23 | {"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": 39, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9539454579353333, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-algebra/170970-more-algebraic-geometry-posted.html | # Thread:
1. ## More Algebraic Geometry
Again, I know this one looks like it should be easy (easier, in fact, even than what I posted before). My brain is just completely fried on this subject right now for some reason...
A hyperplane in $\mathbb{P}^N$ is a hypersurface [which means its dimension (as a topological space in the Zariski topology on $\mathbb{P}^N)$ is $\dim\mathbb{P}^N-1=N-1$], which is defined by a single linear polynomial. That is, our hyperplane is $H=\mathcal{Z}(f)\subseteq\mathbb{P}^N$, where $f$ is a linear, homogeneous (since we are in projective space) polynomial in $S=k[x_0,x_1,\ldots,x_N]$ ( $k$ an algebraically closed field) . The problem says to determine that $\mathbb{P}^N\setminus H$ is an affine variety.
2. Okay... I'm really not trying to beat this thing to death, but I have a preliminary exam coming up soon, and I'm very nervous because I still feeling like I'm struggling with what should be the most basic of questions.
I DO know that, if the hyperplane is defined simply by one of the coordinates (for sake of example, say $H=\mathcal{Z}(x_0)$), then the statement is true. On the complement of $H$, we know that the $x_0$ coordinate is nonzero. Since we are in projective space, we may assume that $x_0=1$ (because in $\mathbb{P}^n,(x_0,\ldots,x_n)\equiv (\lambda x_0,\ldots,\lambda x_n)\, \forall \, \lambda\in k\setminus\{0\}$; if we are restricting to points where $x_0\neq 0$, we may as well "normalize" our points by dividing through by $x_0$). In this case, after this normalization, we may define a map
$\varphi:\mathbb{P}^n\setminus H\rightarrow\mathbb{A}^n$
$(1,x_1,\ldots,x_n)\mapsto(x_1\ldots,x_n)\in\mathbb {A}^n$
This turns out to be an isomorphism $\mathbb{P}^n\setminus H\cong\mathbb{A}^n$.
Is there some sort of way I can generalize this to the complement of an arbitrary hyperplane (that is, to a set defined by the equation $a_0x_0+\cdots+a_nx_n\neq 0$,each $a_i\in k$)? Maybe some sort of (projective..) linear change of coordinates (which I don't know how to do...)? Would the mapping, for example, $x_0\mapsto f,x_1\mapsto x_1,\ldots,x_n\mapsto x_n$ be a legal isomorphism on the points of $\mathbb{P}^n$, in which case I could apply the idea above? Or is there something completely different I should be looking at?
I'm sorry for posting this stuff over and over when it seems no one really is able to help anyway... but I'm.. well, don't wanna say desperate, but... yea. I really feel like I am understanding the concepts, but I'm not able to apply them at all to solve anything.
EDIT: Am I asking this in the appropriate forum, or would another section be more appropriate? I just figure it would go here since it uses a good deal of abstract algebra... | {"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": 22, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9582042098045349, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/67728/correspondence-between-normal-subgroups-of-a-group-and-normal-subgroups-of-a-quo | # Correspondence between normal subgroups of a group and normal subgroups of a quotient groups
Suppose $G$ be a group and $N$ be a proper normal subgroup such that it is not contained in any other proper normal subgroup. Does that mean $G/N$ is simple?
Is there one-one correspondence between the normal subgroups of $G/N$ and normal subgroups of $G$ containing $N$?
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Yes. The correspondence between the subgroups of $G/N$ and those of $G$ which contain $N$ maps normal subgroups to normal subgroups. – Mariano Suárez-Alvarez♦ Sep 26 '11 at 17:58
## 1 Answer
The lattice isomorphism between subgroups of $G$ that contain $N$ and subgroups of $G/N$ is not only inclusion-preserving, but also normality-preserving.
If $N\lt H\lt G$ and $H\triangleleft G$, then for all $h\in H$ and $g\in G$ we have $ghg^{-1}\in H$. Then for all $gN\in G/N$ and all $hN\in H/N$ we have $(gN)(hN)(gN)^{-1} = gNhNg^{-1}N = ghg^{-1}N\in HN$, so $gNhN(gN)^{-1}\in H/N$. Hence, the image of a normal subgroup under the correspondence is normal in the quotient.
Conversely, if $H/N$ is normal in $G/N$, and $g\in G$, $h\in H$, then we know that $ghg^{-1}N\in HN$, so there exists $h'\in H$ and $n\in N$ such that $ghg^{-1}=h'n$. But $N\subseteq H$, so $h'n\in H$. Hence, $ghg^{-1}\in H$ for all $h\in G$, $g\in G$, so $H\triangleleft G$; showing that if the image of the subgruop is normal, then the original subgroup is normal as well.
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http://mathoverflow.net/revisions/103809/list | ## Return to Answer
2 added strengthening
Let $\vec{a} \in A^{<\omega}$. There are uncountably many automorphisms $f$ and only countably many possible values for $f(\vec{a})$, so there must be two different automorphisms $f_1$ and $f_2$ with $f_1(\vec{a}) = f_2(\vec{a})$. Then $f_2^{-1} \circ f_1$ is a nontrivial automorphism fixing $\vec{a}$.
In fact, there are uncountably many automorphisms moving $\vec{a}$ the same way, so fixing one and composing the others with its inverse gives uncountably many nontrivial automorphisms fixing $\vec{a}$.
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Let $\vec{a} \in A^{<\omega}$. There are uncountably many automorphisms $f$ and only countably many possible values for $f(\vec{a})$, so there must be two different automorphisms $f_1$ and $f_2$ with $f_1(\vec{a}) = f_2(\vec{a})$. Then $f_2^{-1} \circ f_1$ is a nontrivial automorphism fixing $\vec{a}$. | {"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": 18, "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}, "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}, "english_confidence": 0.9202134013175964, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/79898/list | 2 switched tags
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# zeros of a holomorphic function in several variable
Let $f$ be a holomorphic function in $n$ complex variables on a domain $D\subset \mathbb C^n$. Let $S$ be a subset of $D$ such that for a polynomial $P$ in $n$ variable, $P(S)=(a,b)$ for an interval $(a,b)$. Can $f$ be zero on $S$?
What happens if $P(S)=(a,b)\cap \mathbb Q$? | {"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": 12, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7617427110671997, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/133025-solved-just-need-help-setting-up-equation.html | # Thread:
1. ## [SOLVED] Just need help setting up the equation
In the section of integration by parts
Find the area of the region enclosed by y=lnx, the line x=e and the x-axis
2. Originally Posted by vinson24
In the section of integration by parts
Find the area of the region enclosed by y=lnx, the line x=e and the x-axis
I expect that this is the region enclosed by the function, the line $x = e$ and the $x$ intercept of the function (which is $x = 1$).
So you are evaluating
$\int_1^e{\ln{x}\,dx} = \int_1^e{1\cdot \ln{x}\,dx}$.
Now set $u = \ln{x}$ and $dv = 1$.
3. thanks i had the integral set from 0 to e and was getting the wrong answer | {"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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9378781318664551, "perplexity_flag": "middle"} |
http://mathhelpforum.com/number-theory/197765-generator-z-p.html | 3Thanks
• 1 Post By Sylvia104
• 2 Post By Deveno
# Thread:
1. ## generator of Z*p
if p,q are primes with p=2q+1, and we pick a random number g in Z*p, what is the probability that g is a generator of Z*p?
2. ## Re: Generator of Z*p
$\mathbb Z_p^\times$ is cyclic of order $p-1=2q$ and so has $\varphi(2q)$ generators. The probability of picking a generator is thus
$\frac{\varphi(2q)}{2q}\ =\ \left\{ \begin{array} {cl} \dfrac12 & \text{if}\ q=2 \\\\ \dfrac{q-1}{2q} & \text{if}\ q>2 \end{array} \right.$
3. ## Re: generator of Z*p
note that if q is very large this is "nearly" 1/2 (even for the relatively small q = 3 it's already 1/3 (approx. 33%) and for q = 5 it's jumped to 2/5 (40%)). these odds are good enough that it often suffices to check 2,3,5 and 7 as possible generators for small values of p (i think 109 is the first prime where you have to use something else, and 6 works there, 191 is the smallest prime that has a relatively "large" smallest primitive element (19)...note that both of these are not of the form considered here since neither 54 nor 95 is prime).
this is good news, since it's not always easy to tell if a given number *is* a generator. as far as i know, trial-and-error is still the best method. | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9518063068389893, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/34839/how-do-nasas-curiosity-determine-the-elemental-composition-of-mars-using-spectr?answertab=votes | # How do NASA's Curiosity determine the elemental composition of Mars using spectrometer?
The resultant flash of glowing plasma is viewed by the system’s 4.3-inch aperture telescope, which sends the light down an optical fiber to a spectrometer located in the body of the rover. There, the colours of light from the flash are recorded and then sent to Earth, enabling scientists to determine the elemental composition of the vaporized material.
How do NASA's Curiosity determine the elemental composition of Mars using spectrometer? Do just examining the colors of light enabled scientist to determine the elemental composition of any object?
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– Chris White Aug 24 '12 at 22:24
## 3 Answers
Yes, with a few tricks of course.
Put very simply: when you let white light pass through a gas and examine the light afterwards, you'll find that all sorts of colours are "missing" from the light. Precisely which colours are missing depends for the most part on the exact constituents of the gas.
It also works the other way around: if you heat the gas up to the point where it starts emitting light, this light will not be perfectly white, but consist mostly of the colours that were missing before.
Have a read about Fraunhofer lines, and spectroscopy for a general overview. Then, for more background about spectroscopic instruments, read up on prisms for a nice introduction, and diffraction gratings for a few examples of more advanced techniques.
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Yes, the response can be analysed to detect the tiny differences between wavelengths - which is used to identify atoms:
From the Encyclopedia Britannica,
Spectroscopic techniques are extremely sensitive. Single atoms and even different isotopes of the same atom can be detected among 1020 or more atoms of a different species. (Isotopes are all atoms of an element that have unequal mass but the same atomic number. Isotopes of the same element are virtually identical chemically.) Trace amounts of pollutants or contaminants are often detected most effectively by spectroscopic techniques. Certain types of microwave, optical, and gamma-ray spectroscopy are capable of measuring infinitesimal frequency shifts in narrow spectroscopic lines. Frequency shifts as small as one part in 1015 of the frequency being measured can be observed with ultrahigh resolution laser techniques. Because of this sensitivity, the most accurate physical measurements have been frequency measurements.
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This is a sample image provided by NASA...
The Chemistry & Camera instrument (ChemCam) in Curiosity uses a laser to generate a flash of ionized plasma on the target sample (like Rock) & observes it with spectrometers receiving light through a telescope. Peaks in the intensity of light emitted at specific wavelengths are indicators of specific chemical elements in the target. The horizontal scale is wavelengths, in nanometers and the vertical scale is the Intensity.
The main advantage of spectrometers is that we could find different elements present on a sample in more efficient way. Every element has its own emission and absorption spectra. In case of hot bodies, the spectra would be continuous. In case of atoms in gaseous state (such as $Na$ or $Hg$ vapor), it would be line spectrum. In case of molecules, it would be band spectrum. Even Wikipedia has a good article on Emission and Absorption spectra. To go even in brief, refer Fraunhofer Lines. It would explain the absorption spectra obtained from sun - which was helpful to discover various elements present in sun's atmosphere by comparing the absorption spectra of elements here.
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http://en.wikipedia.org/wiki/Speed_of_sound | # Speed of sound
For other uses, see Speed of sound (disambiguation).
Sound measurements
Sound pressure p, SPL
Particle velocity v, SVL
Particle displacement ξ
Sound intensity I, SIL
Sound power Pac
Sound power level SWL
Sound energy
Sound energy density E
Sound energy flux q
Acoustic impedance Z
Speed of sound c
Audio frequency AF
The speed of sound is the distance travelled during a unit of time by a sound wave propagating through an elastic medium. In dry air at 20 °C (68 °F), the speed of sound is 343.2 metre per second (1,126 ft/s). This is 1,236 kilometres per hour (768 mph), or about one kilometer in three seconds or approximately one mile in five seconds.
In fluid dynamics, the speed of sound in a fluid medium (gas or liquid) is used as a relative measure of speed itself. The speed of an object divided by the speed of sound in the fluid is called the Mach number. Objects moving at speeds greater than Mach1 are traveling at supersonic speeds.
The speed of sound in an ideal gas is independent of frequency, but does vary slightly with frequency in a real gas. It is proportional to the square root of the absolute temperature, but is independent of pressure or density for a given ideal gas. Sound speed in air varies slightly with pressure only because air is not quite an ideal gas. In addition, for different gases, the speed of sound is inversely proportional to the mean molecular weight of the gas, and is affected to a lesser extent by the number of ways in which the molecules of the gas can store heat from compression, since sound in gases is a compression wave. Although (in the case of gases only) the speed of sound is expressed in terms of a ratio of both density and pressure, these quantities cancel in ideal gases at any given temperature, composition, and heat capacity. This leads to a velocity formula for ideal gases which includes only the latter independent variables.
In common everyday speech, speed of sound refers to the speed of sound waves in air. However, the speed of sound varies from substance to substance. Sound travels faster in liquids and non-porous solids than it does in air. It travels about 4.3 times as fast in water (1,484 m/s), and nearly 15 times as fast in iron (5,120 m/s), than in air at 20 degrees Celsius. Sound waves in solids are composed of compression waves (just as in gases and liquids), but there is also a different type of sound wave called a shear wave, which occurs only in solids. These different types of waves in solids usually travel at different speeds, as exhibited in seismology. The speed of a compression sound wave in solids is determined by the medium's compressibility, shear modulus and density. The speed of shear waves is determined only by the solid material's shear modulus and density.
## Basic concept
The transmission of sound can be illustrated by using a toy model consisting of an array of balls interconnected by springs. For real material the balls represent molecules and the springs represent the bonds between them. Sound passes through the model by compressing and expanding the springs, transmitting energy to neighboring balls, which transmit energy to their springs, and so on. The speed of sound through the model depends on the stiffness of the springs (stiffer springs transmit energy more quickly). Effects like dispersion and reflection can also be understood using this model.
In a real material, the stiffness of the springs is called the elastic modulus, and the mass corresponds to the density. All other things being equal (ceteris_paribus), sound will travel more slowly in spongy materials, and faster in stiffer ones. For instance, sound will travel 1.59 times faster in nickel than in bronze, due to the greater stiffness of nickel at about the same density. Similarly, sound travels about 1.41 times faster in light hydrogen (protium) gas than in heavy hydrogen (deuterium) gas, since deuterium has similar properties but twice the density. At the same time, "compression-type" sound will travel faster in solids than in liquids, and faster in liquids than in gases, because the solids are more difficult to compress than liquids, while liquids in turn are more difficult to compress than gases.
Some textbooks mistakenly state that the speed of sound increases with increasing density. This is usually illustrated by presenting data for three materials, such as air, water and steel, which also have vastly different compressibilities which more than make up for the density differences. An illustrative example of the two effects is that sound travels only 4.3 times faster in water than air, despite enormous differences in compressibility of the two media. The reason is that the larger density of water, which works to slow sound in water relative to air, nearly makes up for the compressibility differences in the two media.
### Compression and shear waves
Pressure-pulse or compression-type wave (longitudinal wave) confined to a plane. This is the only type of sound wave that travels in fluids (gases and liquids)
Transverse wave affecting atoms initially confined to a plane. This additional type of sound wave (additional type of elastic wave) travels only in solids, and the sideways shearing motion may take place in any direction at right angles to the direction of wave-travel (only one shear direction is shown here, at right angles to the plane). Furthermore, the right-angle shear direction may change over time and distance, resulting in different types of polarization of shear-waves
In a gas or liquid, sound consists of compression waves. In solids, waves propagate as two different types. A longitudinal wave is associated with compression and decompression in the direction of travel, which is the same process as all sound waves in gases and liquids. A transverse wave, called a shear wave in solids, is due to elastic deformation of the medium perpendicular to the direction of wave travel; the direction of shear-deformation is called the "polarization" of this type of wave. In general, transverse waves occur as a pair of orthogonal polarizations. These different waves (compression waves and the different polarizations of shear waves) may have different speeds at the same frequency. Therefore, they arrive at an observer at different times, an extreme example being an earthquake, where sharp compression waves arrive first, and rocking transverse waves seconds later.
The speed of a compression wave in fluid is determined by the medium's compressibility and density. In solids, the compression waves are analogous to those in fluids, depending on compressibility, density, and the additional factor of shear modulus. The speed of shear waves, which can occur only in solids, is determined simply by the solid material's shear modulus and density.
## Basic
In general, the speed of sound c is given by the Newton-Laplace equation:
$c = \sqrt{\frac{K}{\rho}}\,$
where
K is a coefficient of stiffness, the bulk modulus (or the modulus of bulk elasticity for gases),
$\rho$ is the density
Thus the speed of sound increases with the stiffness (the resistance of an elastic body to deformation by an applied force) of the material, and decreases with the density. For ideal gases the bulk modulus P is simply the gas pressure multiplied by the adiabatic index.
For general equations of state, if classical mechanics is used, the speed of sound $c$ is given by
$c^2=\left(\frac{\partial p}{\partial\rho}\right)_s$
where $p$ is the pressure and $\rho$ is the density and the derivative is taken adiabatically, that is, at constant entropy per particle (s).
If relativistic effects are important, the speed of sound is calculated from the relativistic Euler equations.
In a non-dispersive medium sound speed is independent of sound frequency, so the speeds of energy transport and sound propagation are the same for all sound frequencies. For audible sounds, the mixture of oxygen and nitrogen constitutes a non-dispersive medium. But air does contain a small amount of CO2 which is a dispersive medium, and it introduces dispersion to air at ultrasonic frequencies (> 28 kHz).[1]
In a dispersive medium sound speed is a function of sound frequency, through the dispersion relation. The spatial and temporal distribution of a propagating disturbance will continually change. Each frequency component propagates at its own phase velocity, while the energy of the disturbance propagates at the group velocity. The same phenomenon occurs with light waves; see optical dispersion for a description.
## Dependence on the properties of the medium
The speed of sound is variable and depends on the properties of the substance through which the wave is travelling. In solids, the speed of transverse (or shear) waves depend on the shear deformation under shear stress (called the shear modulus), and the density of the medium. Longitudinal (or compression) waves in solids depend on the same two factors with the addition of a dependence on compressibility.
In fluids, only the medium's compressibility and density are the important factors, since fluids do not tolerate shear stresses. In heterogeneous fluids, such as a liquid filled with gas bubbles, the density of the liquid and the compressibility of the gas affect the speed of sound in an additive manner, as demonstrated in the hot chocolate effect.
In gases, adiabatic compressibility is directly related to pressure through the heat capacity ratio (adiabatic index), and pressure and density are inversely related at a given temperature and composition, thus making only the latter independent properties (temperature, molecular composition, and heat capacity ratio) important. In low molecular weight gases, such as helium, sound propagates faster compared to heavier gases, such as xenon (for monatomic gases the speed of sound is about 75% of the mean speed that molecules move in the gas). For a given ideal gas the sound speed depends only on its temperature. At a constant temperature, the ideal gas pressure has no effect on the speed of sound, because pressure and density (also proportional to pressure) have equal but opposite effects on the speed of sound, and the two contributions cancel out exactly. In a similar way, compression waves in solids depend both on compressibility and density—just as in liquids—but in gases the density contributes to the compressibility in such a way that some part of each attribute factors out, leaving only a dependence on temperature, molecular weight, and heat capacity ratio (see derivations below). Thus, for a single given gas (where molecular weight does not change) and over a small temperature range (where heat capacity is relatively constant), the speed of sound becomes dependent on only the temperature of the gas.
In non-ideal gases, such as a van der Waals gas, the proportionality is not exact, and there is a slight dependence of sound velocity on the gas pressure.
Humidity has a small but measurable effect on sound speed (causing it to increase by about 0.1%-0.6%), because oxygen and nitrogen molecules of the air are replaced by lighter molecules of water. This is a simple mixing effect.
## Altitude variation and implications for atmospheric acoustics
Density and pressure decrease smoothly with altitude, but temperature (red) does not. The speed of sound (blue) depends only on the complicated temperature variation at altitude and can be calculated from it, since isolated density and pressure effects on sound speed cancel each other. Speed of sound increases with height in two regions of the stratosphere and thermosphere, due to heating effects in these regions.
In the Earth's atmosphere, the chief factor affecting the speed of sound is the temperature. For a given ideal gas with constant heat capacity and composition, sound speed is dependent solely upon temperature; see Details below. In such an ideal case, the effects of decreased density and decreased pressure of altitude cancel each other out, save for the residual effect of temperature.
Since temperature (and thus the speed of sound) decreases with increasing altitude up to 11 km, sound is refracted upward, away from listeners on the ground, creating an acoustic shadow at some distance from the source.[2] The decrease of the sound speed with height is referred to as a negative sound speed gradient.
However, there are variations in this trend above 11 km. In particular, in the stratosphere above about 20 km, the speed of sound increases with height, due to an increase in temperature from heating within the ozone layer. This produces a positive sound speed gradient in this region. Still another region of positive gradient occurs at very high altitudes, in the aptly-named thermosphere above 90 km.
## Practical formula for dry air
Approximation of the speed of sound in dry air based on the heat capacity ratio (in green) against the truncated Taylor expansion (in red).
The approximate speed of sound in dry (0% humidity) air, in meters per second (m·s−1), at temperatures near 0 °C, can be calculated from:
$c_{\mathrm{air}} = (331{.}3 + 0{.}606 \cdot \vartheta) \ \mathrm{m \cdot s^{-1}}\,$
where $\vartheta$ is the temperature in degrees Celsius (°C).
This equation is derived from the first two terms of the Taylor expansion of the following more accurate equation:
$c_{\mathrm{air}} = 331.3\,\mathrm{m \cdot s^{-1}} \sqrt{1+\frac{\vartheta}{273.15}}$
Dividing the first part, and multiplying the second part, on the right hand side, by $\sqrt{273.15}$ gives the exactly equivalent form:
$c_{\mathrm{air}} = 20.0457\,\mathrm{m \cdot s^{-1}} \sqrt{{\vartheta}+ {273.15\;}}$
The value of 331.3 m/s, which represents the speed at 0 °C (or 273.15 K), is based on theoretical (and some measured) values of the heat capacity ratio, $\gamma$, as well as on the fact that at 1 atm real air is very well described by the ideal gas approximation. Commonly found values for the speed of sound at 0 °C may vary from 331.2 to 331.6 due to the assumptions made when it is calculated. If ideal gas $\gamma$ is assumed to be 7/5 = 1.4 exactly, the 0 °C speed is calculated (see section below) to be 331.3 m/s, the coefficient used above.
This equation is correct to a much wider temperature range, but still depends on the approximation of heat capacity ratio being independent of temperature, and for this reason will fail, particularly at higher temperatures. It gives good predictions in relatively dry, cold, low pressure conditions, such as the Earth's stratosphere. The equation fails at extremely low pressures and short wavelengths, due to dependence on the assumption that the wavelength of the sound in the gas is much longer than the average mean free path between gas molecule collisions. A derivation of these equations will be given in the following section.
A graph comparing results of the two equations is at right, using the slightly different value of 331.5 m/s for the speed of sound at 0°C.
## Details
### Speed in ideal gases and in air
For a gas, K (the bulk modulus in equations above, equivalent to C, the coefficient of stiffness in solids) is approximately given by
$K = \gamma \cdot p\,$
thus
$c = \sqrt{\gamma \cdot {p \over \rho}}\,$
Where:
$\gamma$ is the adiabatic index also known as the isentropic expansion factor. It is the ratio of specific heats of a gas at a constant-pressure to a gas at a constant-volume($C_p/C_v$), and arises because a classical sound wave induces an adiabatic compression, in which the heat of the compression does not have enough time to escape the pressure pulse, and thus contributes to the pressure induced by the compression.
p is the pressure.
$\rho$ is the density
Using the ideal gas law to replace $p$ with nRT/V, and replacing ρ with nM/V, the equation for an ideal gas becomes:
$c_{\mathrm{ideal}} = \sqrt{\gamma \cdot {p \over \rho}} = \sqrt{\gamma \cdot R \cdot T \over M}= \sqrt{\gamma \cdot k \cdot T \over m}\,$
where
• $c_{\mathrm{ideal}}$ is the speed of sound in an ideal gas.
• $R$ (approximately 8.3145 J·mol−1·K−1) is the molar gas constant.[3]
• $k$ is the Boltzmann constant
• $\gamma$ (gamma) is the adiabatic index (sometimes assumed 7/5 = 1.400 for diatomic molecules from kinetic theory, assuming from quantum theory a temperature range at which thermal energy is fully partitioned into rotation (rotations are fully excited), but none into vibrational modes. Gamma is actually experimentally measured over a range from 1.3991 to 1.403 at 0 degrees Celsius, for air. Gamma is assumed from kinetic theory to be exactly 5/3 = 1.6667 for monoatomic molecules such as noble gases).
• $T$ is the absolute temperature in kelvin.
• $M$ is the molar mass in kilograms per mole. The mean molar mass for dry air is about 0.0289645 kg/mol.
• $m$ is the mass of a single molecule in kilograms.
This equation applies only when the sound wave is a small perturbation on the ambient condition, and the certain other noted conditions are fulfilled, as noted below. Calculated values for $c_{\mathrm{air}}$ have been found to vary slightly from experimentally determined values.[4]
Newton famously considered the speed of sound before most of the development of thermodynamics and so incorrectly used isothermal calculations instead of adiabatic. His result was missing the factor of $\gamma$ but was otherwise correct.
Numerical substitution of the above values gives the ideal gas approximation of sound velocity for gases, which is accurate at relatively low gas pressures and densities (for air, this includes standard Earth sea-level conditions). Also, for diatomic gases the use of $\ \gamma\, = 1.4000$ requires that the gas exist in a temperature range high enough that rotational heat capacity is fully excited (i.e., molecular rotation is fully used as a heat energy "partition" or reservoir); but at the same time the temperature must be low enough that molecular vibrational modes contribute no heat capacity (i.e., insignificant heat goes into vibration, as all vibrational quantum modes above the minimum-energy-mode, have energies too high to be populated by a significant number of molecules at this temperature). For air, these conditions are fulfilled at room temperature, and also temperatures considerably below room temperature (see tables below). See the section on gases in specific heat capacity for a more complete discussion of this phenomenon.
For air, we use a simplified symbol $\ R_* = R/M_{\mathrm{air}}$.
Additionally, if temperatures in degrees Celsius(°C) are to be used to calculate air speed in the region near 273 kelvin, then Celsius temperature $\vartheta = T - 273.15$ may be used. Then:
$c_{\mathrm{ideal}} = \sqrt{\gamma \cdot R_* \cdot T} = \sqrt{\gamma \cdot R_* \cdot (\vartheta + 273.15)}\,$
$c_{\mathrm{ideal}} = \sqrt{\gamma \cdot R_* \cdot 273.15} \cdot \sqrt{1+\frac{\vartheta}{273.15}}\,$
For dry air, where $\vartheta\,$ (theta) is the temperature in degrees Celsius(°C).
Making the following numerical substitutions:
$\ R = 8.314510 \cdot \mathrm{J \cdot mol^{-1}} \cdot K^{-1}\,$
is the molar gas constant in J/mole/Kelvin;
$\ M_{\mathrm{air}} = 0.0289645 \cdot \mathrm{kg \cdot mol^{-1}}\,$
is the mean molar mass of air, in kg; and using the ideal diatomic gas value of $\ \gamma\, = 1.4000\,$
Then:
$c_{\mathrm{air}} = 331.3 \ \mathrm{\frac{m}{s}} \sqrt{1+\frac{\vartheta^{\circ}\mathrm{C}}{273.15\;^{\circ}\mathrm{C}}}\,$
Using the first two terms of the Taylor expansion:
$c_{\mathrm{air}} = 331.3 \ \mathrm{\frac{m}{s}} (1 + \frac{\vartheta^{\circ}\mathrm{C}}{2 \cdot 273.15\;^{\circ}\mathrm{C}})\,$
$c_{\mathrm{air}} = ( 331{.}3 + 0{.}606\;^{\circ}\mathrm{C}^{-1} \cdot \vartheta)\ \mathrm{ \frac{m}{s}}\,$
The derivation includes the first two equations given in the Practical formula for dry air section above.
### Effects due to wind shear
The speed of sound varies with temperature. Since temperature and sound velocity normally decrease with increasing altitude, sound is refracted upward, away from listeners on the ground, creating an acoustic shadow at some distance from the source.[2] Wind shear of 4 m·s−1·km−1 can produce refraction equal to a typical temperature lapse rate of 7.5 °C/km.[5] Higher values of wind gradient will refract sound downward toward the surface in the downwind direction,[6] eliminating the acoustic shadow on the downwind side. This will increase the audibility of sounds downwind. This downwind refraction effect occurs because there is a wind gradient; the sound is not being carried along by the wind.[7]
For sound propagation, the exponential variation of wind speed with height can be defined as follows:[8]
$\ U(h) = U(0) h ^ \zeta\,$
$\ \frac {dU} {dH} = \zeta \frac {U(h)} {h}\,$
where:
$\ U(h)$ = speed of the wind at height $\ h$, and $\ U(0)$ is a constant
$\ \zeta$ = exponential coefficient based on ground surface roughness, typically between 0.08 and 0.52
$\ \frac {dU} {dH}$ = expected wind gradient at height $h$
In the 1862 American Civil War Battle of Iuka, an acoustic shadow, believed to have been enhanced by a northeast wind, kept two divisions of Union soldiers out of the battle,[9] because they could not hear the sounds of battle only 10 km (six miles) downwind.[10]
### Tables
In the standard atmosphere:
• T0 is 273.15 K (= 0 °C = 32 °F), giving a theoretical value of 331.3 m·s−1 (= 1086.9 ft/s = 1193 km·h−1 = 741.1 mph = 644.0 knots). Values ranging from 331.3-331.6 may be found in reference literature, however.
• T20 is 293.15 K (= 20 °C = 68 °F), giving a value of 343.2 m·s−1 (= 1126.0 ft/s = 1236 km·h−1 = 767.8 mph = 667.2 knots).
• T25 is 298.15 K (= 25 °C = 77 °F), giving a value of 346.1 m·s−1 (= 1135.6 ft/s = 1246 km·h−1 = 774.3 mph = 672.8 knots).
In fact, assuming an ideal gas, the speed of sound c depends on temperature only, not on the pressure or density (since these change in lockstep for a given temperature and cancel out). Air is almost an ideal gas. The temperature of the air varies with altitude, giving the following variations in the speed of sound using the standard atmosphere - actual conditions may vary.
Effect of temperature on properties of air
Temperature T in °C Speed of sound c in m·s−1 Density of air ρ in kg·m−3 Acoustic impedance Z in N·s·m−3
+35 351.88 1.1455 403.2
+30 349.02 1.1644 406.5
+25 346.13 1.1839 409.4
+20 343.21 1.2041 413.3
+15 340.27 1.2250 416.9
+10 337.31 1.2466 420.5
+5 334.32 1.2690 424.3
0 331.30 1.2922 428.0
−5 328.25 1.3163 432.1
−10 325.18 1.3413 436.1
−15 322.07 1.3673 440.3
−20 318.94 1.3943 444.6
−25 315.77 1.4224 449.1
Given normal atmospheric conditions, the temperature, and thus speed of sound, varies with altitude:
| | | | | | |
|---------------------------------------------------------------------------------------|-----------------|-------|--------|-----|-------|
| Altitude | Temperature | m·s−1 | km·h−1 | mph | knots |
| Sea level | 15 °C (59 °F) | 340 | 1225 | 761 | 661 |
| 11,000 m−20,000 m (Cruising altitude of commercial jets, and first supersonic flight) | −57 °C (−70 °F) | 295 | 1062 | 660 | 573 |
| 29,000 m (Flight of X-43A) | −48 °C (−53 °F) | 301 | 1083 | 673 | 585 |
## Effect of frequency and gas composition
The medium in which a sound wave is travelling does not always respond adiabatically, and as a result the speed of sound can vary with frequency.[11]
The limitations of the concept of speed of sound due to extreme attenuation are also of concern. The attenuation which exists at sea level for high frequencies applies to successively lower frequencies as atmospheric pressure decreases, or as the mean free path increases. For this reason, the concept of speed of sound (except for frequencies approaching zero) progressively loses its range of applicability at high altitudes.:[4] The standard equations for the speed of sound apply with reasonable accuracy only to situations in which the wavelength of the soundwave is considerably longer than the mean free path of molecules in a gas.
The molecular composition of the gas contributes both as the mass (M) of the molecules, and their heat capacities, and so both have an influence on speed of sound. In general, at the same molecular mass, monatomic gases have slightly higher sound speeds (over 9% higher) because they have a higher $\gamma$ (5/3 = 1.66...) than diatomics do (7/5 = 1.4). Thus, at the same molecular mass, the sound speed of a monatomic gas goes up by a factor of
${ c_{\mathrm{gas: monatomic}} \over c_{\mathrm{gas: diatomic}} } = \sqrt{{{{5 / 3} \over {7 / 5}}}} = \sqrt{25 \over 21}$ = 1.091...
This gives the 9% difference, and would be a typical ratio for sound speeds at room temperature in helium vs. deuterium, each with a molecular weight of 4. Sound travels faster in helium than deuterium because adiabatic compression heats helium more, since the helium molecules can store heat energy from compression only in translation, but not rotation. Thus helium molecules (monatomic molecules) travel faster in a sound wave and transmit sound faster. (Sound generally travels at about 70% of the mean molecular speed in gases).
Note that in this example we have assumed that temperature is low enough that heat capacities are not influenced by molecular vibration (see heat capacity). However, vibrational modes simply cause gammas which decrease toward 1, since vibration modes in a polyatomic gas gives the gas additional ways to store heat which do not affect temperature, and thus do not affect molecular velocity and sound velocity. Thus, the effect of higher temperatures and vibrational heat capacity acts to increase the difference between sound speed in monatomic vs. polyatomic molecules, with the speed remaining greater in monatomics.
## Mach number
Main article: Mach number
U.S. Navy F/A-18 traveling near the speed of sound. The white halo consists of condensed water droplets formed by the sudden drop in air pressure behind the shock cone around the aircraft (see Prandtl-Glauert singularity).[12]
Mach number, a useful quantity in aerodynamics, is the ratio of air speed to the local speed of sound. At altitude, for reasons explained, Mach number is a function of temperature. Aircraft flight instruments, however, operate using pressure differential to compute Mach number, not temperature. The assumption is that a particular pressure represents a particular altitude and, therefore, a standard temperature. Aircraft flight instruments need to operate this way because the stagnation pressure sensed by a Pitot tube is dependent on altitude as well as speed.
## Experimental methods
A range of different methods exist for the measurement of sound in air.
The earliest reasonably accurate estimate of the speed of sound in air was made by William Derham, and acknowledged by Isaac Newton. Derham had a telescope at the top of the tower of the Church of St Laurence in Upminster, England. On a calm day, a synchronized pocket watch would be given to an assistant who would fire a shotgun at a pre-determined time from a conspicuous point some miles away, across the countryside. This could be confirmed by telescope. He then measured the interval between seeing gunsmoke and arrival of the noise using a half-second pendulum. The distance from where the gun was fired was found by triangulation, and simple division (time / distance) provided velocity. Lastly, by making many observations, using a range of different distances, the inaccuracy of the half-second pendulum could be averaged out, giving his final estimate of the speed of sound. Modern stopwatches enable this method to be used today over distances as short as 200–400 meters, and not needing something as loud as a shotgun.
### Single-shot timing methods
The simplest concept is the measurement made using two microphones and a fast recording device such as a digital storage scope. This method uses the following idea.
If a sound source and two microphones are arranged in a straight line, with the sound source at one end, then the following can be measured:
1. The distance between the microphones (x), called microphone basis. 2. The time of arrival between the signals (delay) reaching the different microphones (t)
Then v = x / t
### Other methods
In these methods the time measurement has been replaced by a measurement of the inverse of time (frequency).
Kundt's tube is an example of an experiment which can be used to measure the speed of sound in a small volume. It has the advantage of being able to measure the speed of sound in any gas. This method uses a powder to make the nodes and antinodes visible to the human eye. This is an example of a compact experimental setup.
A tuning fork can be held near the mouth of a long pipe which is dipping into a barrel of water. In this system it is the case that the pipe can be brought to resonance if the length of the air column in the pipe is equal to ({1+2n}λ/4) where n is an integer. As the antinodal point for the pipe at the open end is slightly outside the mouth of the pipe it is best to find two or more points of resonance and then measure half a wavelength between these.
Here it is the case that v = fλ
## Non-gaseous media
### Speed of sound in solids
#### Three-dimensional solids
In a solid, there is a non-zero stiffness both for volumetric and shear deformations. Hence, it is possible to generate sound waves with different velocities dependent on the deformation mode. Sound waves generating volumetric deformations (compressions) and shear deformations are called longitudinal waves and shear waves, respectively. In earthquakes, the corresponding seismic waves are called P-waves and S-waves, respectively. The sound velocities of these two type waves propagating in a homogeneous 3-dimensional solid are respectively given by:[13]
$c_{\mathrm{l}} = \sqrt {\frac{K+\frac{4}{3}G}{\rho}} = \sqrt {\frac{Y (1-\nu)}{\rho (1+\nu)(1 - 2 \nu)}}$
$c_{\mathrm{s}} = \sqrt {\frac{G}{\rho}}$
where K and G are the bulk modulus and shear modulus of the elastic materials, respectively, Y is the Young's modulus, and $\nu$ is Poisson's ratio. The last quantity is not an independent one, as $Y = 3K(1-2\nu)$. Note that the speed of longitudinal/compression waves depends both on the compression and shear resistance properties of the material, while the speed of shear waves depends on the shear properties only.
Typically, compression or P-waves travel faster in materials than do shear waves, and in earthquakes this is the reason that onset of an earthquake is often preceded by a quick upward-downward shock, before arrival of waves that produce a side-to-side motion. For example, for a typical steel alloy, K = 170 GPa, G = 80 GPa and $\rho$ = 7700 kg/m3, yielding a longitudinal velocity cl of 6000 m/s.[13] This is in reasonable agreement with cl=5930 m/s measured experimentally for a (possibly different) type of steel.[14]
The shear velocity cs is estimated at 3200 m/s using the same numbers.
#### Long rods
The speed of sound for longitudinal waves in stiff materials such as metals is sometimes given for "long, thin rods" of the material in question, in which the speed is easier to measure. In rods where their diameter is shorter than a wavelength, the speed of pure longitudinal waves may be simplified and is given by:
$c_{\mathrm{l}} = \sqrt {\frac{Y}{\rho}}$
This is similar to the expression for shear waves, save that Young's modulus replaces the shear modulus. This speed of sound for longitudinal waves in long, thin rods will always be slightly less than the 3-D, longitudinal wave speed in an isotropic materials, and the ratio of the speeds in the two different types of objects depends on Poisson's ratio for the material.
### Speed of sound in liquids
In a fluid the only non-zero stiffness is to volumetric deformation (a fluid does not sustain shear forces).
Hence the speed of sound in a fluid is given by
$c_{\mathrm{fluid}} = \sqrt {\frac{K}{\rho}}$
where K is the bulk modulus of the fluid. This value typically decreases with temperature for non-polar fluids: the speed of sound in ultra-wave frequency range is inverse proportional to the cube of the volume of a fixed amount of the fluid.[15]
#### Water
The speed of sound in water is of interest to anyone using underwater sound as a tool, whether in a laboratory, a lake or the ocean. Examples are sonar, acoustic communication and acoustical oceanography. See Discovery of Sound in the Sea for other examples of the uses of sound in the ocean (by both man and other animals). In fresh water, sound travels at about 1497 m/s at 25 °C. See Technical Guides - Speed of Sound in Pure Water for an online calculator.
#### Seawater
Sound speed as a function of depth at a position north of Hawaii in the Pacific Ocean derived from the 2005 World Ocean Atlas. The SOFAR channel is centered on the minimum in sound speed at ca. 750-m depth.
In salt water that is free of air bubbles or suspended sediment, sound travels at about 1560 m/s. The speed of sound in seawater depends on pressure (hence depth), temperature (a change of 1 °C ~ 4 m/s), and salinity (a change of 1‰ ~ 1 m/s), and empirical equations have been derived to accurately calculate sound speed from these variables.[16] Other factors affecting sound speed are minor. Since temperature decreases with depth while pressure and generally salinity increase, the profile of sound speed with depth generally shows a characteristic curve which decreases to a minimum at a depth of several hundred meters, then increases again with increasing depth (right).[17] For more information see Dushaw et al.[18]
A simple empirical equation for the speed of sound in sea water with reasonable accuracy for the world's oceans is due to Mackenzie:[19]
c(T, S, z) = a1 + a2T + a3T2 + a4T3 + a5(S - 35) + a6z + a7z2 + a8T(S - 35) + a9Tz3
where T, S, and z are temperature in degrees Celsius, salinity in parts per thousand and depth in meters, respectively. The constants a1, a2, ..., a9 are:
a1 = 1448.96, a2 = 4.591, a3 = -5.304×10-2, a4 = 2.374×10-4, a5 = 1.340,
a6 = 1.630×10-2, a7 = 1.675×10-7, a8 = -1.025×10-2, a9 = -7.139×10-13
with check value 1550.744 m/s for T=25 °C, S=35 parts per thousand, z=1000 m. This equation has a standard error of 0.070 m/s for salinity between 25 and 40 ppt. See Technical Guides - Speed of Sound in Sea-Water for an online calculator.
Other equations for sound speed in sea water are accurate over a wide range of conditions, but are far more complicated, e.g., that by V. A. Del Grosso[20] and the Chen-Millero-Li Equation.[18][21]
### Speed in plasma
The speed of sound in a plasma for the common case that the electrons are hotter than the ions (but not too much hotter) is given by the formula (see here)
$c_s = (\gamma ZkT_e/m_i)^{1/2} = 9.79\times10^3\,(\gamma ZT_e/\mu)^{1/2}\,\mbox{m/s}\,$
where $m_i$ is the ion mass, $\mu$ is the ratio of ion mass to proton mass $\mu = m_i/m_p$; $T_e$ is the electron temperature; Z is the charge state; k is Boltzmann's constant; K is wavelength; and $\gamma$ is the adiabatic index.
In contrast to a gas, the pressure and the density are provided by separate species, the pressure by the electrons and the density by the ions. The two are coupled through a fluctuating electric field.
## Gradients
Main article: sound speed gradient
When sound spreads out evenly in all directions in three dimensions, the intensity drops in proportion to the inverse square of the distance. However, in the ocean there is a layer called the 'deep sound channel' or SOFAR channel which can confine sound waves at a particular depth.
In the SOFAR channel, the speed of sound is lower than that in the layers above and below. Just as light waves will refract towards a region of higher index, sound waves will refract towards a region where their speed is reduced. The result is that sound gets confined in the layer, much the way light can be confined in a sheet of glass or optical fiber. Thus, the sound is confined in essentially two dimensions. In two dimensions the intensity drops in proportion to only the inverse of the distance. This allows waves to travel much further before being undetectably faint.
A similar effect occurs in the atmosphere. Project Mogul successfully used this effect to detect a nuclear explosion at a considerable distance.
## References
1. ^ a b Everest, F. (2001). The Master Handbook of Acoustics. New York: McGraw-Hill. pp. 262–263. ISBN 0-07-136097-2.
2. "CODATA Value: molar gas constant". Physics.nist.gov. Retrieved 2010-10-24.
3. ^ a b U.S. Standard Atmosphere, 1976, U.S. Government Printing Office, Washington, D.C., 1976.
4. Uman, Martin (1984). Lightning. New York: Dover Publications. ISBN 0-486-64575-4.
5. Volland, Hans (1995). Handbook of Atmospheric Electrodynamics. Boca Raton: CRC Press. p. 22. ISBN 0-8493-8647-0.
6. Singal, S. (2005). Noise Pollution and Control Strategy. Alpha Science International, Ltd. p. 7. ISBN 1-84265-237-0. "It may be seen that refraction effects occur only because there is a wind gradient and it is not due to the result of sound being convected along by the wind."
7. Bies, David (2004). Engineering Noise Control; Theory and Practice. London: Spon Press. p. 235. ISBN 0-415-26713-7. "As wind speed generally increases with altitude, wind blowing towards the listener from the source will refract sound waves downwards, resulting in increased noise levels."
8. Cornwall, Sir (1996). Grant as Military Commander. Barnes & Noble Inc. ISBN 1-56619-913-1 pages = p. 92 Check `|isbn=` value (help).
9. Cozzens, Peter (2006). The Darkest Days of the War: the Battles of Iuka and Corinth. Chapel Hill: The University of North Carolina Press. ISBN 0-8078-5783-1.
10. "APOD: 19 August 2007- A Sonic Boom". Antwrp.gsfc.nasa.gov. Retrieved 2010-10-24.
11. ^ a b L. E. Kinsler et al. (2000), Fundamentals of acoustics, 4th Ed., John Wiley and sons Inc., New York, USA
12. J. Krautkrämer and H. Krautkrämer (1990), Ultrasonic testing of materials, 4th fully revised edition, Springer-Verlag, Berlin, Germany, p. 497
13. Padmini, P. R. K. L.; Ramachandra Rao (12). "Molar Sound Velocity in Molten Hydrated Salts". Nature 191 (4789): 694–695. Bibcode:1961Natur.191..694P. doi:10.1038/191694a0. Retrieved 10 February 2012.
14. "How fast does sound travel?". Discovery of Sound in the Sea. University of Rhode Island. Retrieved 2010-11-30.
15. ^ a b Dushaw, Brian D.; Worcester, P.F.; Cornuelle, B.D.; and Howe, B.M. (1993). "On equations for the speed of sound in seawater". Journal of the Acoustical Society of America 93 (1): 255–275. Bibcode:1993ASAJ...93..255D. doi:10.1121/1.405660.
16. Mackenzie, Kenneth V. (1981). "Discussion of sea-water sound-speed determinations". Journal of the Acoustical Society of America 70 (3): 801–806. Bibcode:1981ASAJ...70..801M. doi:10.1121/1.386919.
17. Del Grosso, V. A. (1974). "New equation for speed of sound in natural waters (with comparisons to other equations)". Journal of the Acoustical Society of America 56 (4): 1084–1091. Bibcode:1974ASAJ...56.1084D. doi:10.1121/1.1903388.
18. Meinen, Christopher S.; Watts, D. Randolph (1997). "Further evidence that the sound-speed algorithm of Del Grosso is more accurate than that of Chen and Millero". Journal of the Acoustical Society of America 102 (4): 2058–2062. Bibcode:1997ASAJ..102.2058M. doi:10.1121/1.419655. | {"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": 64, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8923704028129578, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?t=194650 | Physics Forums
## difference between a function being continuos at at point and havinga limit there...
I'm confused about a point that my book on real analysis is making about the difference between the definition of a function being continuos at at point and the definition of a function having a limit L at a point.
My rough understanding of the matter is that in order for a function to be continuous at a point the function must have a limit L at the point (from both sides) and the values of the function at that point must be L. However it is possible to have a limit at a point, even if the value of that function at the point is not the same as the limit.
From the book:
DEFINITION: Suppose E is a subset of R and $$f: E \rightarrow R$$. If $$x_0 \in E$$, then f is continuous at $$x_0$$ iff for each $$\epsilon > 0$$, there is a $$\delta > 0$$ such that if
$$|x-x_0|< \delta$$, with $$x \in E$$,
then
$$|f(x)-f(x0)|< \epsilon$$.
Compare this with the definition of the limit of a function at a point $$x_0$$ . First of all, for continuity at $$x_0$$, the number must belong to E but it need not be an accumulation point of E. ( Is this saying that f(x) = L? If not what is it saying? ) Indeed, if $$f: E \rightarrow R$$ with $$x_0 \in E$$ and $$x_0$$ not an accumulation point of E, then there is a $$\delta > 0$$ such that if $$|x - x_0 | < \delta$$ and $$x \in E$$, then $$x=x_0$$ .
That last bit has me lost. Why would $$x = x_0$$ if $$x_0$$ is not an accumulation point of E?
I think that $$x_0$$ , not being an accumulation point means that there is at least one neighborhood of $$x_0$$ that contains a finite number, possibly zero vales of f(x) when x is in a delta neighborhood of $$x_0$$. But how does this force $$x = x_0$$?
The book defines accumulation points interms of series:
A real number A is an accumulation point of S iff every neighborhood of A contains infinitely many points of S.
So, not an accumulation point would mean that every neighborhood of A contains finite points of S, or zero ponits of S... ?
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Quote by futurebird I think that $$x_0$$ , not being an accumulation point means that there is at least one neighborhood of $$x_0$$ that contains a finite number, possibly zero vales of f(x) when x is in a delta neighborhood of $$x_0$$. But how does this force $$x = x_0$$?
Let's say that $x_0 \in E$ is not an accumulation point of $E$. Then there is some $\delta$ so that:
[tex]E \cap (x_0 - \delta,x_0+\delta) = x_0[/itex]
Then the hypothesis:
$$|x-x_0|<\delta$$
and
$$x \in E$$
is equivalent to
$$x_0 \in E \cap (x_0 - \delta,x_0+\delta) = x_0$$
which implies
$$x_0=x$$
(and obviously)
$$f(x_0)-f(x)=0$$
Quote by NateTG ..... is equivalent to $$x_0 \in E \cap (x_0 - \delta,x_0+\delta) = x_0$$ ....
Shouldn't it be $$x \in E \cap (x_0 - \delta,x_0+\delta) = x_0$$?
## difference between a function being continuos at at point and havinga limit there...
A function is vacuously continuous at isolated points.
Quote by zhentil A function is vacuously continuous at isolated points.
That helps a lot. But let me see if I know what you mean. If I have the set of points N, the natural numbers then it's vacuously continuous a say 7?
That's odd.
Is that what this is all about???
The point is trivial. x_0 an accumulation pt in E iff every nbd of x_0 contains a point in E iff every neighbourhood of x_0 contains infinitely many points in E. If x_0 is not an accumulation pt in E then there exists a nbd of x_0 such that it only contains finitely many points in E. Trivial point: Well, if x_0 is not an accumulation point in E then there are finitely many x's in E within some delta>0 distance from x_0. Take the closest point to x_0, this exists as we're taking the minimum of some finite set. Call the distance between x_0 and this closest point delta_min. So the only points in E with the property that |x-x_0|<delta_min is the one point x_0. (Draw a picture: this immediately clear). Some authors absolutely refuse to talk about limits of functions at non accumulation points. The author of whatever you're quoting is obviously not one of them. The problem arises when you look at the difference between continuity and limits. Limits talk only about the elements getting closer to some point (but not the point itself), continuity talks about elements getting closer to some point AND the point itself. (0<|x-x_0|<delta => |f(x)-f(x_0)|<epsilon for limits and |x-x_0|<delta => |f(x)-f(x_0)|<epsilon for continuity) Saying that functions are vacuously continuous at non accumulation is awkward as the limits NEVER exist, yet they will always satisfy the latter criterion for continuity above. Take your example. The integers with the identity function. What's the limit of id(x) as x->7? If it did exist then for every epsilon there would be a delta such that 0<|x-7|<delta (x an integer) =>|x-7|<epsilon. Take epsilon=1/2. It's clear that |x-7|<1/2 and x integral iff x=7. But x cannot be 7, as then |x-7|=|7-7|=0. Hence id(x) does not have a limit as x->7. But it does satisfy the delta-epsilon definition of continuity at x=7: since 7 is not an accumulation point (only one point in any ball centered at 7 with radius strictly less than 1) we can always pick delta<1 and so |x-7|<delta<1=>x=7=>|id(7)-id(7)|=0<epsilon for any epsilon>0. Do you see how continuity at non accumulation points is weird? Heuristically, we think of continuity as getting closer and closer to some point. But if we're talking about an non accumulation point you can't get closer and closer! There's just one lone point. Non-accumulation points are called isolated points -- this should help you remember the definition. Summary: f(x)->f(x_0) as x->x_0 is equivalent to continuous at x_0 (epsilon/delta defn) if x_0 is an accumulation point
Quote by futurebird That helps a lot. But let me see if I know what you mean. If I have the set of points N, the natural numbers then it's vacuously continuous a say 7? That's odd. Is that what this is all about???
Yes. $$|x-7| < 1/2 \Rightarrow |f(x)-f(7)| = 0.$$
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http://mathhelpforum.com/calculus/33786-complex-variables.html | # Thread:
1. ## Complex Variables
Can anyone help me with this pls?
How can you prove that the integral of f(z) around the contour z= 1 is 0
where f(z) is Log(z+5)
Thx
2. I know Log(z) is ln r + i (theta). But i dont know how that applies to this situation.
Also, do I solve it as a normal integral or use the Cauchy Goursat theorem to prove that its integral is zero?
3. Originally Posted by stillsoulsearching
Can anyone help me with this pls?
How can you prove that the integral of f(z) around the contour z= 1 is 0
where f(z) is Log(z+5)
Thx
Because $\log (z+5)$ is analytic on $\mathbb{C} - (\infty,-5]$. This is a simply connected domain and Cauchy's theorem tells us that the integral over a (rectifiable) curve of an analytic function on a simply connected domain is zero. | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9193804264068604, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/81887/proving-sequence-equality-using-the-binomial-theorem | # Proving sequence equality using the binomial theorem
The problem: Prove that for $n \in \mathbb N$:
$$\left(1 + \frac{1}{n} \right)^n = 1 + \sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right).$$
The hint is to use the binomial theorem. So the left side can become:
$$\sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m$$
I don't really know where to go from here, I've tried manipulating the expressions to make them look similar but I'm not really getting anywhere.
-
Here's a hint: $$\sum_{m=0}^n \frac{n!}{m!(n-m)!}\left(\frac{1}{n}\right)^m = \sum_{m=0}^n \frac{1}{m!} n(n-1)\dots(n-(m-1))\left(\frac{1}{n}\right)^m$$ – Dimitrije Kostic Nov 14 '11 at 4:34
## 2 Answers
Take your second sum $$\sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m$$ and write it as $$1+\sum_{m=1}^n \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m$$ to get the indices to match.
In your first sum $$\sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right)$$ ignoring the $\frac{1}{m!}$ for now, notice $$\left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right)=\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{n-m+1}{n}\right).$$ Multiplying by $1=\frac{n}{n}$ gives $$\frac{n}{n}\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{n-m+1}{n}\right)=\dots$$ Can you take it from there?
-
Hint. $\displaystyle 1 - \frac{k}{n} = \frac{n-k}{n}$. So $$\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right) = \frac{(n-1)(n-2)\cdots(n-(m-1))}{(n)(n)\cdots(n)}.$$
- | {"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": 4, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9289632439613342, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/16817/penetration-of-armor-plate/16821 | # Penetration of armor plate
Is there a simple mathematical expression for the stopping power of a given thickness of armor, given the thickness of armor plate, the radius of a cannon ball, the density of the cannonball and the armor, the tensile strength and/or toughness of the armor, and the speed of the cannonball? For simplicity assume the cannonball is a solid metal sphere and that the armor plate is homogeneous. I realize that in modern warfare the projectiles are pointed and armor plate isn't a homogeneous slab, but I want to understand the simple case. (My question is inspired in part from reading about Civil War ironclads, but I also saw the question about chain mail and thought that if that question was legitimate this one should be more so.)
In case my question isn't clear, what I'm asking is something like the following. Suppose it took 10 cm of iron armor to stop a 20 cm diameter cannonball moving at 300 meters/sec. How thick would the armor have to be to stop a 40 cm cannonball moving at the same speed? Or what if you doubled the speed? Or what if you doubled the tensile strength of the armor? Etc...
-
a simple equation can be put like, $$x = E_0.e^{-kh}$$, but then all the pain would be in finding the proper value of $k$. Which would again be dependent on the initial energy, $E_0$. – Vineet Menon Nov 11 '11 at 5:55
## 1 Answer
People spend their whole lives answering this question. The best I can do is give you a method for making a very rough estimate of the stopping power of a material. I'm going to assume a uniform slab of armor and a cubic bullet that will not deform at all. The relevant material property is the toughness. Toughness is defined as the amount of energy per unit volume a material can absorb before fracturing. It depends heavily on the rate at which the material is deformed and is typically determined empirically.
The reason I've chosen a cubic projectile is to have a uniform, time independent stress being applied to the armor. Let's also assume that the projectile applies force only to the region of the armor directly in its path. Under these many simplifications, the energy required to penetrate the armor can be estimated as $T\times t_{\text{armor}} \times A_{\text{bullet}}$, in which $T$ is the toughness of the armor, $t$ is the thickness of the armor, and $A$ is the area of the bullet. This is essentially the energy required to rip all the armor blocking the projectile out of the way.
This paper gives the high-strain rate (i.e., impact) toughness of one particular steel as roughly 2000 MPa, which translates to 2000 J/cm$^3$. So for 1 cm thick armor, a 50 gram bullet with a 1 cm cross section would need to have 2000 J of kinetic energy to make it through. Solving
$$2000 = \frac{1}{2}mv^2$$
gives $v = 282\text{m/s}$ to just penetrate.
I'm not a ballistics expert, but there are bullets that can make it through a cm of steel and bullets generally have velocities measured in the 100s of m/s so this all seems pretty reasonable.
-
You might also assume that C17 steel or C19 iron (depends whose civil war you are talking about!) is 1/2 or 1/4 as strong as modern pipe steel and so 1/4 to 1/16 the speed of projectile – Martin Beckett Nov 11 '11 at 5:28
1
Usual army rifle bullets have velocities of 800 to 900 m/s , but never will penetrate one cm of steel. If those bullets were made from tungsten, they might penetrate such armour. In practice such questions are business for engieers and are solved experimentally. Maybe very sophisticated fined element calculations can be used. An important thing in such processes is that the velocity of the impactor is not far from sound velocity in the target material. – Georg Nov 11 '11 at 10:38 | {"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": 8, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9432708024978638, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/110320/list | ## Return to Answer
3 added 2 characters in body
Not sure if this question really qualifies for MO.
Anyway, the answer very much depends on the group $G$. In most cases $n_p(G)\ne n_q(G)$ for distinct prime divisors of the group order. However, there are infinitely many examples very where equality occurs: If $r$ is an odd prime, then $n_p(\text{PSL}(2,r))=r(r+1)/2$ for each odd prime divisor $p$ of $r-1$.
But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizer normalizers of the $3$-Sylows and $5$-Sylows have order $60$.
2 added 5 characters in body
Not sure if this question really qualifies for MO.
Anyway, the answer very much depends on the group $G$. In most cases $n_p(G)\ne n_q(G)$ for distinct prime divisors of the group order. However, there are infinitely many examples very equality occurs: If $r$ is an odd prime, then $n_p(\text{PSL}(2,r))=r-1$ n_p(\text{PSL}(2,r))=r(r+1)/2$for each odd prime divisor$p$of$r-1\$.
But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizer of the $3$-Sylows and $5$-Sylows have order $60$.
1
Not sure if this question really qualifies for MO.
Anyway, the answer very much depends on the group $G$. In most cases $n_p(G)\ne n_q(G)$ for distinct prime divisors of the group order. However, there are infinitely many examples very equality occurs: If $r$ is an odd prime, then $n_p(\text{PSL}(2,r))=r-1$ for each odd prime divisor $p$ of $r-1$.
But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizer of the $3$-Sylows and $5$-Sylows have order $60$. | {"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": 30, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9298058748245239, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/19592/how-to-read-the-temperature-of-an-abstract-system/19662 | # How to “read” the temperature of an abstract system?
How can I interpret the parameter temperature $T$, if I'm not given the description of the system in terms of the equation of state, $E(S,V\ )$ or $S(E,V\ )$ and so on.
In many systems it makes sense to think of it as an "energy" itself, e.g. when the entropy is such, that $T$ basically represents the mean kinetic energy of particles. The general definition goes like $$\frac 1 T=\small{\left(\frac{\partial S(E,V\ )}{\partial E}\right)_V}.$$
However, very often, especially when I'm reading about phase transitions, examples for systems with critical exponents and so on, they usually talk about the parameter $T$ and an associated critical $T_c$ without taking any specific system into accound. There is usually some abstract free energy $F$, which pops out abstract quantities. Or in the Boltzman distribution and derived quantities, often something gets activated when the value of $kT$ catches up with some system specific energy value $E_0$. And it gets more complicated when it takes QFT like form.
What is the temperature in a general setting.
How do I read this kind of things and what should I have in mind when reading these kinds of texts?
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1
You wrote the answer. This is the definition of $T$. In the Ising model. for example, there is no sense in talking about "average kinetic energy of the spins" or anything of that sort. I'd say that a good way to think about the temperature in this case is "how far above the ground state can I go", which is roughly equivalent to "how much energy can I pay in order to buy some entropy"? – yohBS Jan 16 '12 at 21:03
Could you elaborate on the sentences in parentheses? Why "can I go"? Can I go with what? I have a probability for all energy levels to be to taken. And why "pay"? Why "buy" entropy? Do I want entropy? – Nick Kidman Jan 17 '12 at 8:36
1
Systems flow towards probable states, that is - states that have a large number of micro-states. The (log of the) number of micro states is the entropy. So systems "want" to be in (="go to") states with high entropy, but that usually means states with high energy. The trade-off between entropy and energy is exactly the temperature. In evaporating liquid, for example, the vapor state has a higher entropy, but also a higher energy. Therefore, at low $T$ the system is liquid, but when $T$ is high enough the system prefers to be at a higher energetic state (="pay") because its entropy is higher. – yohBS Jan 17 '12 at 12:00
@yohBS: Your explanation is formulated in terms of the system, which is not in equilibrium. Would you say one should view and describe the meaning of the temperature $T$ functionally, i.e. in that its the value, which equals out for coupled subsystems, if the whole system reaches equilibrium? – Nick Kidman Jan 17 '12 at 17:10
I posted a real, elaborate answer. Hope that gets the job done. – yohBS Jan 17 '12 at 21:20
show 3 more comments
## 3 Answers
The most fundamental definition of temperature is derived from the zeroth law of thermodynamics.
The zeroth law declares thermal equilibrium an equivalence relationship, and thus we can tag each equivalence class with a number that we call temperature. Or in less mathematical term, temperature is a physical quantity tagged to each thermodynamic system such that any two systems with the same temperature would stay in thermal equilibrium when they contact.
The exact way of assigning temperature to a system is called a temperature scale. There were multiple scales before, most based on thermal properties of a particular substance. Then Kelvin devised a scale based solely on thermodynamic principles, which we call "absolute scale".
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Interesting points! I wonder if there zeroth law can't be deduced within thermodynamics, do you (if no) know why not? Regarding your last paragraph, I think the Kelvin definition and the absolute isn't necessary if (via statistical mechanics) you can compute the absolute entropy. – Nick Kidman Jan 19 '12 at 11:18
@Nick Kidman: The zeroth law is "zeroth" because it was formulated after the first and second law when people realized it is implicit in their understanding, yet an independent and fundamental law. So no, you cannot deduce it within thermodynamics. As for "absolute scale", or thermodynamical scale, there are multiple definitions, all equivalent. Kelvin's version is the first one, though. – C.R. Jan 20 '12 at 2:13
Stat Mech is all about taking averages with the proper measure.
The simplest measure is the micro-canonical measure (ensemble), in which you assume that the system has a given energy, and that all the states in this energy shell are equally probable. The assumption that the energy is known basically mean that the system does not exchange energy with a bath.
A much more useful measure is the canonical measure, in which you assume that the average energy is known, but that the system may exchange energy with a heat bath. It can be shown (and it is shown here) that if you assume that in each energy shell the states are equi-probable, this probability is proportional to $e^{-E/k_bT}$.
This shows you that $T$ measures the weight that is given to states, according to their energy. When $T$ is high, the weight of a state with a given energy becomes higher, and when $T$ is low it becomes lower. It is easily seen that for $T\to\infty$ all the states become equi-probable and for $T\to 0$ only the ground state(s) is (are) counted.
You might ask then, "so why isn't the system almost always at the ground state? It is the most probable one!". Here entropy goes into play. By definition of the entropy $S(E)$, the number of states with an energy $E$ is $e^{S(E)/k_b}$, and it is a rapidly increasing function of $E$. Therefore, when you average over energies you should use the measure $$e^{-\beta E}e^{S(E)/k_b}=e^{-\beta(E-TS)}$$ because higher energies have a much larger number of states on the shell. This is what I meant when I said that $T$ is a measure of "how much energy can I pay in order to buy some entropy" - high-entropy states are more probable from state-counting considerations, but are less probable from energetic considerations. $T$ is a measure of the relative balance of these two.
Also, your suggestion of functionally defining $T$ as "the thing that is equal for two systems in thermal contact" is great, and is actually used it in this textbook which gives offers a thorough, insightful introduction to "what is temperature". I highly recommend it for beginners.
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So $\int_{\text{states}} e^{-\beta E}=\int_{\text{energies}} e^{-\beta E}\Omega(E)=\int_{\text{energies}} e^{-\beta E}e^{S}=\int_{\text{energies}} e^{-\beta F}$, where $\Omega(E)$ is the phase space volume with energy $E$? I see that $E-TS$ must be minimized, but since I don't see the lower bound of this expression without knowing $E$ or $S$ I don't know how to minimize it. Also, if you're taking the canonical ensamble with a real bath, then you can't just choose $T$. In that case there is no battle between energy and entropy, since $T$ will just take the value it has to from outside. – Nick Kidman Jan 17 '12 at 21:43
The point is showing that coupling your system to an external bath results in a probability that goes like $e^{-\beta E}$, for some $\beta$. This is te definition of $T$, and is crux of the matter. Until you've shown this you don't know that "T will just take its value" from the bath, because $T$ is not even defined. – yohBS Jan 17 '12 at 21:51
forget what I wrote about f. It is confusing and wrong. In this formulation the free energy is given by $f=-k_bT\log Z=k_bT\log \sum_i e^{-\beta E_i}$, and the mean energy is calculated by averaging $\langle E\rangle=Z^{-1}\sum E_i e^{-\beta E_i}=-\partial_\beta \log(Z)$ – yohBS Jan 17 '12 at 23:00
I don't understand how "this is the definition of $T$". How do you determine $T$ from this formalism, if it's not defined as the entropy derivative with respect to $E$? And eventually do you agree that with the bath, $T$ will just take the value of the temperature of the outer bath? And what is "confusing and wrong", I don't know what you're referring to. – Nick Kidman Jan 18 '12 at 8:03
When you couple your system to a bath, you actually allow it to exchange heat with it. The total system (your system and the bath) will be in the most probable state - the one with the highest entropy. Since the total entropy is the sum of the bath's and the system's entropy, the total entropy is maximal when $$\frac{\partial S_{bath}}{\partial E}=\frac{\partial S_{system}}{\partial E}\ .$$ This quantity is defined to be $1/T$, and the probability to find your system in an energy $E$ goes like $e^{-E/kT}$, as can be seen by taking a first order Taylor series around the maximum. – yohBS Jan 18 '12 at 9:38
Your definition is the thermodyanmics definition of temperature. Most discussions of phase transitions are coming from a statisical mechanics point of view. In this paradigm, the definition of temperature follows from the definition of entropy $S = k_b \log \Omega(E)$ $$\frac{1}{kT} = \frac{d \log \Omega(E)}{dE}$$ Where $\Omega(E)$ is the number of microstates with energy $E$.
Wikipedia has a nice derivation of the connection between these two approaches.
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Thanks for the comment, I'm familiar with the definition. I don't know if "the multiplicative inverse of the relative phase space volume change with energy" is all too helpful though. – Nick Kidman Jan 17 '12 at 8:37 | {"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": 50, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9482046365737915, "perplexity_flag": "head"} |
http://mathhelpforum.com/algebra/201315-help-homework-quadratics.html | Thread:
1. Help on homework!(Quadratics)
Hello...
I have this problem on my homework, I have to solve the quadratics by factorising;
x(x-7)=0
If you could explain how to do it (simple terms!) I would be grateful
2. Re: Help on homework!(Quadratics)
Originally Posted by RonnieSaha
Hello...
I have this problem on my homework, I have to solve the quadratics by factorising;
x(x-7)=0
If you could explain how to do it (simple terms!) I would be grateful
That's already "factored". Now you need the "zero product rule": if ab= 0 then either a= 0 or b= 0 (or both).
3. Re: Help on homework!(Quadratics)
I really don't understand the process you have explained above... could you please go through the stages of solving x(x-7) with the zero product rule?
4. Re: Help on homework!(Quadratics)
$x(x-7) = 0$
Zero times any (finite) number is zero. This means that at least one of the factors on the LHS must be zero. Hence either $x = 0$ or $x-7 = 0 \Rightarrow x = 7$.
5. Re: Help on homework!(Quadratics)
Originally Posted by richard1234
$x(x-7) = 0$
Zero times any (finite) number is zero. This means that at least one of the factors on the LHS must be zero. Hence either $x = 0$ or $x-7 = 0 \Rightarrow x = 7$.
Thank-you for that. Could you just do this one as I think I am on the right lines;
3a(a-1)=0
6. Re: Help on homework!(Quadratics)
Originally Posted by richard1234
Zero times any (finite) number is zero. This means that at least one of the factors on the LHS must be zero.
The second claim is certainly true for real numbers, but it does not follow from the first one. In every ring we have 0 * x = 0, but in some rings x * y = 0 does not imply that x = 0 or y = 0.
7. Re: Help on homework!(Quadratics)
So how would I do 3a(a-1)=0 if I had to solve it. The same with x^2-6x=0
8. Re: Help on homework!(Quadratics)
Originally Posted by RonnieSaha
So how would I do 3a(a-1)=0 if I had to solve it.
As has been said in post #4,
Originally Posted by richard1234
at least one of the factors on the LHS must be zero.
9. Re: Help on homework!(Quadratics)
I understand what the process is but I don't understand how to apply it to my equations
10. Re: Help on homework!(Quadratics)
I would be willing to talk about this if in exchange you explain in a very detailed way what exactly you don't understand about this problem. I find it hard to believe that you don't know how to solve 3a(a - 1) = 0 after you've been shown how to solve x(x - 7) = 0 and have been told that a product is zero if and only if one of the factors is zero. It may be that you don't know what a factor is or you don't know how to solve linear equations. I'd like to know what your reasoning is before further explanations.
11. Re: Help on homework!(Quadratics)
Okay... I can see that the left hand side must equal to the right hand side which is 0. I don't know how you get from the quadratic to the point of finding out what x or a could be. I don't know what the factors are and so don't understand where to put the 0 in either a or b. In fact, i don't know where a or b are in x(x - 7) or 3a(a - 1).
12. Re: Help on homework!(Quadratics)
Originally Posted by RonnieSaha
I don't know what the factors are
Ah, now we are talking. The left-hand side, 3a(a - 1), is a product of three numbers: 3, a and a - 1. Here a is some number that we don't know yet and have to find. Each of those three numbers, i.e., 3, a and a - 1, are called factors. Now, multiplication has this property that if the product equals zero, then one or more of the factors also equal zero.
13. Re: Help on homework!(Quadratics)
xa^2 - ya = 0
a(xa - y) = 0 : this is the "factoring" step
now, a = 0 and/or xa - y = 0 ; a = y/x
14. Re: Help on homework!(Quadratics)
I understand that the 3 parts you suggested are a, b and c. However I don't know how to solve the quadratic anymore?
15. Re: Help on homework!(Quadratics)
Originally Posted by RonnieSaha
I understand that the 3 parts you suggested are a, b and c. However I don't know how to solve the quadratic anymore?
If you are talking about 3a(a - 1), I did not mentioned b and c. They are not part of the problem. The three factors in question are 3, a and a - 1. To continue, you have to say what follows from the fact that the product of these three numbers equals zero. | {"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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9693079590797424, "perplexity_flag": "head"} |
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## Double slit experiment (one slit closed)
Hi,
In young experiment, say one slit is completely closed, what observed is looks like a single band light. But why is not a diffraction pattern? I wonder whether the reason is, the slit width(don't mean distance between slits) more narrow than single slit diffraction in Young experiment?
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Having one slit completely closed off is equivalent to the single-slit diffraction experiment. Yes, there should be diffraction if the slit width is of the appropriate size relative to wavelength. Many introductory physics discuss Young's interference experiment before diffraction effects, and they are usually in separate chapters. So if you are somewhere between these chapters then the presentation may only consider interference effects and this would explain the single-band intensity pattern. Both effects are present and there should be a section in your text that discusses it.
Hmm, you say there is diffraction too. Can you suggest me a web site which explain diffraction in Young experiment? I couldn't find a good source. BTW, my books aren't explaining diffraction effect in Young. Thanks.
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## Double slit experiment (one slit closed)
Hyperphysics has a good section on slit diffraction and interference: Fraunhofer Diffraction
In particular, look at the links marked single slit and double slit.
Great. Good illustrations. So we can not see the diffraction "directly" in multiple slit interference but the envelop is the evidence of diffraction. I will study this, good start point for me. But an early question(guest); if one slit closed in Young experiment, we will see a diffraction but the center of diffraction will shift to the side of open slit and has less intensity(approx. 1/4 of double slit), is it?
And one more question; What will we observe if one of the slit narrow? 1. Intensity will lessen 2. The center of peak will not change(not sure this) 3. more diffrence... mmm... :P sorry
Quote by Volcano And one more question; What will we observe if one of the slit narrow? 1. Intensity will lessen 2. The center of peak will not change(not sure this) 3. more diffrence... mmm... :P sorry
I'll assume your initial scenario (Young's experiment with two slits open and then we close one off) when giving the answers.
1. Two things are going here: (1) A narrower slit implies greater diffraction, so you will see the light-bands have a larger spread. (2) You can imagine the slit to be a point source emitter of light. So at larger radial distances from the point source, the intensity falls off. Your physics text may make the assumption that the light intensity on viewing screen is constant for all points (ignoring 2 then), and therefore, the resulting intensity pattern is due solely to diffraction effects.
2. If you compare the central peak from Young's interference experiment (two slits open) to the scenario where you close one slit of the previous setup, you will see a shift in the central peak. In the former scenario, the central point of the peak will occur exactly between the two slits, and in the latter, the peak will occur directly across the open slit. If you only perform the single-slit experiment and gradually narrower the slit, the location of the central peak will not change.
A little explanation for my first question:
http://tr.wikipedia.org/wiki/%C3%87i...%C4%B1k_deneyi
(left side: Sun light ; right side: monocromatic light)
above link has a picture. And if one of slit closed(for monocromatic), observed pattern is not look like a diffraction. So asked the first question. Now I belive that, there is diffraction while a slit closed in Young experiment. The picture incapable. Don't know where is the original place of that picture.
1.
If you compare the central peak from Young's interference experiment (two slits open) to the scenario where you close one slit of the previous setup, you will see a shift in the central peak.
Agree. This is ok.
2.
If you only perform the single-slit experiment and gradually narrower the slit, the location of the central peak will not change
No, I compare for double slit. One of them is narrower.
For the second question(one of them narrower): I think the central peak will stay in place. Central peak(and rest) intencity will lessen. Finally light bands will have a larger spread but not sure about the symmetry. I guess, will symmetrical too.
Summarize: I need help about the symmetry and band spread.
Quote by Volcano above link has a picture. And if one of slit closed(for monocromatic), observed pattern is not look like a diffraction. So asked the first question. Now I belive that, there is diffraction while a slit closed in Young experiment. The picture incapable. Don't know where is the original place of that picture.
I see in the right-most pictures (monochromatic light) that there is only one band. There can be other bands, but the intensity of these bands will be some small fraction of the main, central band. Are you familiar with the equation that gives the intensity as function of angle from the central axis for single-slit diffraction:
$$I\left(\theta\right) = I_{max}\left(\frac{sin\left(\alpha\left(\theta\right)\right)}{\alpha\le ft(\theta\right)\right)}\right)^{2}$$
where
$$\alpha\left(\theta\right)\right) = \frac{\pi a}{\lambda}sin\left(\theta\right)$$
"a" is the slit width and everything else should be obvious. With these equations, you can determine the fractional intensity of the other bands relative to the main band. The picture may not be accurate, or the modeled parameters may produce those results. If you have Mathcad, I can send you some simulations that model single-slit diffraction and double-slit interference.
Quote by Volcano 2. No, I compare for double slit. One of them is narrower. For the second question(one of them narrower): I think the central peak will stay in place. Central peak(and rest) intencity will lessen. Finally light bands will have a larger spread but not sure about the symmetry. I guess, will symmetrical too.
I never fully analyzed this situation, but it's an interesting question nonetheless. My thoughts are that the central peak would remain in the same location, and also, the intensity pattern would be skewed left or right depending on which slit width is narrowed. To make this last point clearer, the distances from the maximum point of the central band to the points associated with the first minimum on each side will not be the same distance. If your familiar with the derivations, you may be able to create a simulation that models this.
EDIT: See link for derivation of double-slit diffraction: http://cnx.org/content/m12916/latest/
Thank you buffordboy23, I don't have Mathcad but if you suggest, I can try. Anyway, I know those formulas and don't think have a problem about the intensity as function of angle from the central axis for diffraction. I can not imagine the effect of different width slits.
My thoughts are that the central peak would remain in the same location, and also, the intensity pattern would be skewed left or right depending on which slit width is narrowed.
Agree this.
I looked the link which you show. But It is for calculation mostly and much imported is for identical widths. After your words, thought again and reach such an approach: let's narrow one slit continuously. What will we observe? Initially, will observe a known Young interference+diffraction then finally only the diffraction of single slit(other slit almost closed). In time, we will observe a pattern between two of them. Thus, if I'm not wrong, only difference will the distribution symmetry of brightness. It will be favour of widest slit. Is it?
Quote by Volcano I looked the link which you show. But It is for calculation mostly and much imported is for identical widths.
It appears from a simple glance that all you have to do is make some new substitutions for the two different slit widths. The intensity is then proportional to the square of the electric field at that point. It is an interesting question to ask (although I can't think of any useful applications for the result), so when I have the time I will develop a simulation to model the behavior and upload the results (it probably won't happen for a day or so).
Thank you again for to share your time and knowledge. Greets
I put together a quick and simple simulation and here are the results for slit widths of two different sizes (remember that intensity is proportional to the square of the electric field). As one slit is made larger while the other is held constant, the intensity of the central peak increases (because one slit has less spread in diffraction), and there appears to be no effect regarding the symmetry of the intensity plot with respect to the central axis or maxima. I can post the full-derivation used in constructing the simulation. Just let me know and I will post it later. Attached Thumbnails
Thank you very much. Impressed two times. What a good software. But how did you use two different slit widths above formula? I would like to see the background of this graph, this is your skill, great. Your graphs opposite of what we talk about. One of slit width is increasing opposite of decreasing. Thus, this help to comment on the opposite too. I suppose, we may say: Intencity will decrease and diffraction envelop will spread(I think). On of impressive thing is, certain, hard symmetry of diffraction envelop. Honestly, this not exactly what I expected. Glad too see the graphs. This helped too much. buffordboy23, don't know what will I say; how can I thank to you. My mind relaxed :) Can you show me the calculations and the thing what you used with that software? I liked it.
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Quote by buffordboy23 I put together a quick and simple simulation and here are the results for slit widths of two different sizes (remember that intensity is proportional to the square of the electric field). As one slit is made larger while the other is held constant, the intensity of the central peak increases (because one slit has less spread in diffraction), and there appears to be no effect regarding the symmetry of the intensity plot with respect to the central axis or maxima.
Are you saying that the location of the central maxima remains fixed regardless of the relative width of the slits? That seems a bit counter-intuitive. What if you kept reducing the size of one slit--and thus the intensity of the light through it--until it was essentially gone?
Quote by Volcano Your graphs opposite of what we talk about. One of slit width is increasing opposite of decreasing. Thus, this help to comment on the opposite too. I suppose, we may say: Intencity will decrease and diffraction envelop will spread(I think).
Yes, the graphs are opposite to what we talked about (I should of paid more attention) but qualitatively the results are the same. I verified this again.
Quote by Volcano On of impressive thing is, certain, hard symmetry of diffraction envelop. Honestly, this not exactly what I expected. Glad too see the graphs. This helped too much.
Me neither. I used the Fraunhofer Diffraction model as in the link that I submitted earlier.
Quote by Volcano Can you show me the calculations and the thing what you used with that software? I liked it.
Yes, when I get the time. This will be a somewhat lengthy endeavor, but it should be done so that the posted data can be criticized accordingly.
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http://mathoverflow.net/questions/80284/what-are-good-examples-of-string-manifolds | ## What are “good” examples of string manifolds?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Based on this mathoverlow question, I would like to have a similar list for the case of string manifolds. An $n$-dim. Riemannian manifold $M$ is said to be string, if the classifying map of its bundle of orthonormal frames $M \to BO(n)$ lifts to a map $M \to BO(n)<8> = BString(n)$, which is the case if and only if the class $\frac{p_1}{2} \in H^4(M, \mathbb{Z})$ vanishes. There are a lot of models that yield geometric realizations of $String(n)$ either as a topological group (see Stolz-Teichner), infinite-dimensional Lie group (see Nikolaus-Sachse-Wockel) or a 2-group (see Schommer-Pries).
What are enlightening examples of string manifolds? What are non-examples? When do you have a geometric interpretation of the obstruction class?
So far, I am aware of the list given at the end of Douglas-Henriques-Hill. What else is out there?
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3
There is always a geometric interpretation of the obstruction class, namely as the Chern-Simons 2-gerbe associated to the frame bundle of M and the basic gerbe over Spin. – Konrad Waldorf Nov 7 2011 at 13:10
2
Here are the string manifolds from the cited paper by Douglas, Henriques, and Hill: RP^{16n+7}, G_2/T, F_4/Spin(9), F_4/G_2. – Dmitri Pavlov Nov 7 2011 at 16:21
## 1 Answer
Qingtao Chen and Fei Han have constructed [arxiv:0612055] tons of examples of string manifolds and non-string manifolds as complete intersections in products of complex projective spaces.
For integers $s$ and $n_q$ with $1\leq q \leq s$ consider the product $$Z:=\mathbb{C}P^{n_1} \times ... \times \mathbb{C}P^{n_s}$$ and the projection $pr_q:Z \to \mathbb{C}P^{n_q}$ to the $q$th factor. For further integers $t$ and $d_{pq}$ with $1 \leq p \leq t$ and $1 \leq q \leq s$, consider for $1 \leq p \leq t$ the line bundle $$E_p := \bigotimes_{q=1}^s \;\; pr_q^{*}\mathcal{O}^{d_{pq}}_{q}$$ over $Z$, where $\mathcal{O}_q$ is the canonical line bundle over $\mathbb{C}P^{n_q}$. Now let $V_{d_{pq}}$ be the intersection of the zero loci of smooth global sections of $E_1,...,E_t$.
$V_{d_{pq}}$ is always a smooth manifold, and the statement of Proposition 3.1 of the above paper is:
Theorem. Let $m_q$ be the number of non-zero elements in the $q$th column of the matrix $D := (d_{pq})$. Suppose that $m_q +2 \leq n_q$. Then, $V_{d_{pq}}$ is string if and only if $$D^t D = diag(n_1 + 1,...,n_s+1).$$
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http://mathhelpforum.com/trigonometry/191665-question-about-periodicity.html | # Thread:
1. ## Question about periodicity?
Ok as we know, if $f(x+P) = f(x)$ then the function is periodic.
So if the function is not periodic, $f(x+p) = f(x)$, what it the value of p?
2. ## Re: Question about periodicity?
It's just a positive real number, for example the $f(x)=\sin(x)$ is periodic with periodicity $2\pi$ which is a positive real number.
3. ## Re: Question about periodicity?
But in that: case f(x)is periodic.
for a function not periodic , the period is a positive real number too?
4. ## Re: Question about periodicity?
Originally Posted by Fabio010
Ok as we know, if $f(x+P) = f(x)$ then the function is periodic.
So if the function is not periodic, $f(x+p) = f(x)$, what it the value of p?
If this always true $f(x+0)=f(x)~?$
5. ## Re: Question about periodicity?
So you are telling me that in not periodic functions P = 0?
6. ## Re: Question about periodicity?
If you know that $f(x)$ is not periodic but for all $x$ you have $f(x+P)=f(x)$ then $P=~?$
7. ## Re: Question about periodicity?
P = 0
so it is not periodic | {"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": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7851361632347107, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/45904/if-f-is-differentiable-in-1-infty-and-lim-x-to-infty-fx-l-infty/45908 | # If $f$ is differentiable in $(1,\infty )$ and $\lim_{x\to\infty }f'(x)=L<\infty$ then $\lim_{x\to\infty }f(x)=l\le\infty$?
I need to prove or disprove this:
If $f$ is differentiable on $(1,\infty)$ and $\lim\limits_{x\to\infty}f'(x)=L\lt \infty$, then $\lim\limits_{x\to\infty}f(x)=\ell\leq\infty$.
After I didn't find any function to disprove with,
I started to think that if $\lim\limits_{x\to\infty }f'(x)=L<\infty$ so $f'$ is bounded and therfore $f$ is uniformly continuous but it doesn't mean that $\lim\limits_{x\to\infty }f(x)=l\leq\infty$, for example : $\sin x$ which is uniformly continuous, but it's limit when $x\to\infty$ does not exist.
What do you think?
Thank you.
-
## 1 Answer
Look at $f(x) = x^{1 \over 2} \sin(x^{1 \over 3})$. Then $f'(x) = {1 \over 2} x^{-{1 \over 2}}\sin(x^{1 \over 3}) + {1 \over 3}x^{-{1 \over 6}}\cos(x^{1 \over 3})$. The limit as $x$ goes to infinity of $f(x)$ doesn't exist; the $x^{1 \over 2}$ factor increases to infinity while the sine factor modulates it. On the other hand, the limit of $f'(x)$ is zero.
-
what you defined doesn't have a limit at infinity, both $f$ and $f'$. – timhortons Jun 17 '11 at 15:53
ok corrected it, this should work – Zarrax Jun 17 '11 at 15:55
4
You don't need the $\sqrt{x}$. Just take $f(x) = \sin(x^p)$ where $0 < p < 1$. – Robert Israel Jun 17 '11 at 19:33 | {"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": 25, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9509176015853882, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/55333/is-the-quantization-of-gravity-necessary-for-a-quantum-theory-of-gravity-part-i | # Is the quantization of gravity necessary for a quantum theory of gravity? Part II
(At the suggestion of the user markovchain, I have decided to take a very large edit/addition to the original question, and ask it as a separate question altogether.)
Here it is:
I have since thought about this more, and I have come up with an extension to the original question. The answers already given have convinced me that we can't just leave the metric as it is in GR untouched, but at the same time, I'm not convinced we have to quantize the metric in the way that the other forces have been quantized. In some sense, gravity isn't a force like the other three are, and so to treat them all on the same footing seems a bit strange to me. For example, how do we know something like non-commutative geometry cannot be used to construct a quantum theory of gravity. Quantum field theory on curved non-commutative space-time? Is this also a dead end?
-
1
– twistor59 Feb 27 at 17:57
@twistor59: you could plausibly argue that LQG treats gravity on an equal footing with the other forces, though one of the main issues with LQG is figuring out how to couple the other interactions to the theory. – Jerry Schirmer Feb 27 at 18:35
1
– user1504 Feb 27 at 18:38
@user1504: or that there's a flaw with Euclidean gravity analysis. LQG people would agree that LQG has a $\propto \hbar^{2}$ correction to the Hawking formula, which seems to be Sen's result from my fast reading of the abstract and the discussion beginning on page 27 of the preprint. – Jerry Schirmer Feb 27 at 18:50
2
– twistor59 Feb 27 at 20:22
show 3 more comments
## 1 Answer
Quantum field theory on curved non-communative spacetime is exactly what is used in the "semiclassical gravity" approach, an early example of which is Hawking's original derivation of the Hawking radiation effect. The limitation of this approach should be obvious--when the Hawking radiation ends up having a mass comparable to the original black hole, how can you trust the result? The Hawking radiation has mass and energy, too, and obviously, this should be factored into the result, but the semiclassical problem explicitly ignores it. (and if you try to do an iterative approach, factoring in the Hawking radiation as a source, and calculating the result, you quickly run into a LOT of complexity)
Non-communative geometry is an active area of research and a potential solution, albeit one chosen by a minority of researchers
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http://www.physicsforums.com/showthread.php?t=262942 | Physics Forums
## Log in a river
1. The problem statement, all variables and given/known data
Two rivers of unequal width (64m and 125m) meet at a right angle, forming an L-shaped channel. What is the longest possible log you can float on it?
2. Relevant equations
3. The attempt at a solution
I tried for equal widths and built a recursive algorithm for turning the log. But it is getting stuck if the log is $$2\sqrt{2}(Width)$$
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the log will turn around the inner corner. just find a formula for finding the length of a line entirely in the water which passes through that point. then find the minimum
Recognitions: Gold Member Science Advisor Staff Emeritus Start by drawing a picture. Draw two channels of width 64 and 125 and draw a straight line from the outside edges just touching the inside corner. You should be able to find a formula for the length of that line in terms of x, the distance from the outside corner to one of the ends of the line. Then find x which minimizes length. You MIGHT be able to get that by completing the square but I seem to remember a problem like this requiring a derivative. Are you sure this is "precalculus"?
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http://www.physicsforums.com/showthread.php?p=3786878 | Physics Forums
## questions on inductive definitions in a proof
Hi
I was trying to solve the following problem from Kenneth Ross's Elementary Analysis book.
here is the problem.
Let S be a bounded nonempty subset of $\mathbb{R}$ and suppose that
$\mbox{sup }S\notin S$. Prove that there is a non decreasing sequence
$(s_n)$ of points in S such that $\lim s_n =\mbox{sup }S$.
Now the author has provided the solution at back of the book. I have attached the snapshot of the proof. I am trying to understand it. He is using induction here in the proof. Now in induction, we usually have a statement P(n) , which depends upon the natutal number n. And then we use either weak or strong induction. So what would be P(n) in his proof. I am trying to understand the logical structure of the proof. Thats why I decided to post in this part of PF.
thanks
Attached Thumbnails
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It's induction in the sense that given the n-1 term he can construct the nth term. He starts with the 1st term and shows you how to construct the 2nd term. He could then, just as well, have said "proceeding in this manner". Notice that below he just says "therefore the construction continues".
Sorry, maybe I didn't answer your question. He shows that the first term exists. Then he shows that given that the n-1 term exists then the nth term exists by his construction. Therefore all terms exist.
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## necessary and sufficient conditions: Fourier Transform
I don’t know if the question belongs to engineering or math but here it goes.
I was taught that a sufficient (not necessary) condition for existence of Fourier transform of f(t) is f(t) is absolutely integratble. I was wondering what are the “necessary and sufficient conditions” for FT of f(t) to exist. Some textbook said “it can be very involved” but what is that?
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Recognitions: Homework Help Science Advisor This depends on what you mean by "Fourier transform." Usually (but there are many exceptions), the Fourier transform F is defined on the space L^1 of absolutely integrable functions. Are you trying to find the 'largest' domain for F?
Quote by morphism This depends on what you mean by "Fourier transform."
The Fourier Transform which I was taught is:
[tex]
F(\omega)=\int _{-\infty}^{+\infty}f(t)e^{-j\omega t}dt
[/tex]
the one that I use to convert signals from timedomain to analyze and process in freq domain. All physically realizable signals that I have to deal with are "absolutely integrable".
Other forms of FT that I'm familiar with are those 2D or 3D FT that are used to solve linear multi-dimensional PDEs.
Quote by morphism Usually (but there are many exceptions), the Fourier transform F is defined on the space L^1 of absolutely integrable functions.
Can you please show an example of exceptions or point me to some URLs for further reading?
Quote by morphism Are you trying to find the 'largest' domain for F?
No. I was trying to find a signal (function of time) that doesn't satisfy "absolutely integrable" condition, but some sort of FT exists for such signal, for the sake of curiosity. As an analytical electrical engineer, I love engineering math in general but certainly can't afford much time digging into hardcore math.
thanks again.
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## necessary and sufficient conditions: Fourier Transform
Hmm. I don't know if this is what you're looking for, but try to look into Fourier transforms of distributions.
Hmm, let's see. My problem is to find a function that is NOT "absolutely integrable" but its fourier transform exists. If I can find such function, that means "absolutely integrable" is not necessary condition for FT to exist. Given the problem, let's say $$F(\omega)=\int_{-\infty}^{+\infty}f(t)e^{-j\omega t}dt$$ $$f(t)=\dfrac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)e^{j\omega t}d\omega$$ let $$f(t)=\cos{\Omega t}$$ And clearly, f(t) is not absolutely integrable, but its FT does exist: $$F(\omega)=\pi \delta (\omega -\Omega)+ \pi \delta (\omega+\Omega)$$ Make sense?
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Quote by klondike And clearly, f(t) is not absolutely integrable, but its FT does exist: $$F(\omega)=\pi \delta (\omega -\Omega)+ \pi \delta (\omega+\Omega)$$ Make sense?
That depends very much on what you mean by 'exist'! At the most elementary level (i.e. the math you learned in your basic calculus sequence in college), that Fourier transform doesn't exist.
To be able to use 'generalized functions' (such as the Dirac delta), you have to set up some additional mathematical scaffolding. And then to use the Fourier transform on them (or even things like derivatives), that's a bit more scaffolding.
If Wikipedia is to be trusted, if the Fourier transform is of primary interest, then the right scaffolding is the space of tempered distributions. The natural definition for the Fourier transform of a tempered distribution is implicitly through the following integral equation:
$$\int_{-\infty}^{+\infty} \mathcal{F} \{ f \} \phi = \int_{-\infty}^{+\infty} f \mathcal{F} \{ \phi \}$$
where f denotes any tempered distribution and $\phi$ denotes any 'well-behaved' function. It turns out that the Fourier transform of a tempered distribution is a tempered distribution -- so in this context, Fourier transforms always exist.
Every 'somewhat well-behaved' function can be viewed as a tempered distribution -- the basic requirement is that it doesn't grow 'too fast' at infinity. And every tempered distribution can be written as a limit of distributions that came from somewhat well-behaved functions. (e.g. you do this when you write the Dirac delta as a 'limit' of ordinary functions)
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http://mathoverflow.net/questions/21168?sort=newest | ## How to classify the algebras C^∞(M)?
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This continues my question about smooth Gelfand-duality. In the book
Juan A. Navarro González & Juan B. Sancho de Salas, C∞-Differentiable Spaces, LNM 1824
it is shown that $M \mapsto C^\infty(M)$ is a fully faithfull contravariant functor from the category of manifolds (smooth, separable and without boundary) to the category of $\mathbb{R}$-algebras. Isn't this nice? It would be even more nice if there is an algebraic description of the essential image of this functor, so that we have an antiequivalence of categories between manifolds and certain $\mathbb{R}$-algebras. Thus my question is:
• Which $\mathbb{R}$-algebras $A$ are isomorphic to $C^{\infty}(M)$ for some manifold $M$?
Of course, you could just formulate that $Spec_r(A)=Hom(A,\mathbb{R})$ with the obvious structure sheaf is a manifold and that the canonical map $A \to C^{\infty}(M)$ is an isomorphism in terms of the ring structure of $A$. But this does not seem to be handy at all. I want some nontrivial purely algebraic formulation. If possible avoiding structure sheaves at all.
Here are some necessary conditions:
• If $f \neq g$ in $A$, then there is some $\mathbb{R}$-homomorphism $\phi : A \to \mathbb{R}$ such that $\phi(f) \neq \phi(g)$. In particular, $A$ is reduced.
• For every $p \in Spec_r(A)$ with corresponding maximal ideal $m_p$, then the maximal ideal $\overline{m_p}$ of $A_{\mathfrak{m}_p}$ is finitely generated, say by elements $f_1,...,f_n$, and the canonical map $\mathbb{R}[t_1,...,t_n] / (t_1,...,t_n)^{r+1} \to A_{\mathfrak{m}_p} / \overline{m_p}^{r+1}, t_i \mapsto f_i$ is an isomorphism for all $r \geq 0$.
• With the notation above, the canonical map $A/m_p^{r+1} \to A_{\mathfrak{m}_p} / \overline{m_p}^{r+1}$ is an isomorphism.
• The function $Spec_r(A) \to \mathbb{N}, p \to \dim_\mathbb{R} \mathfrak{m}_p/{\mathfrak{m}_p}^2$ is locally constant.
Are they sufficient [no, see Michael's answer]? Finally [solved by Dmitri's answer]:
• How can we characterize the algebras (at least within all the $C^{\infty}(M)$'s), that come from compact manifolds?
You might admit that "$Spec_r(A)$ is compact with the Gelfand topolgy" is not a satisfactory answer ;-).
Addendum: At first glance, it appears too optimistic to find an algebraic characterization. But many famous problems started like that and involved unexpected methods. I don't claim that this applies to my problem. But at least I invite you to think about it. The properties of the algebras above are just an approximation. Even if we add some of the conditions in the answers (such as $\cap_{r} \overline{m}_p^{r+1} \neq 0$), it would be a great surprise that the conditions are sufficient. But I'm not convinced of the contrary as soon someone provides a counterexample. It is fun trying to deduce some of the differential geometric theorems such as IFT from the properties above (if $A \to B$ is an isomorphism in one tangent space, then it is a local isomorphism). Perhaps a first step is to characterize the local rings $C^{\infty}_p(\mathbb{R}^n)$.
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Shouldn't the right condition be something like "is locally isomorphic to C^{\infty}(U) for some open subset of R^n"? I don't know if one can avoid mentioning R^n somewhere in the answer... – Qiaochu Yuan Apr 13 2010 at 0:38
3
Does this help? arxiv.org/abs/math.GT/9404228 – Jonas Meyer Apr 13 2010 at 0:39
2
I asked something similar earlier - you may find the references there helpful. mathoverflow.net/questions/5344/… – Jason DeVito Apr 13 2010 at 0:44
2
@Jason: ok you already posted the same question (but already assuming $M$ to be compact) but the answer basically consists of abstraction typical for MO ... I want something concrete, you know. – Martin Brandenburg Apr 13 2010 at 0:47
2
"At first glance, it appears too optimistic to find an algebraic characterization. But many famous problems started like that and involved unexpected methods." I don't know how to convince you that this is the problem Connes has solved! He has characterised function algebras of compact smooth manifolds by saying that these algebras have a canonical projective module whose properties one can recognise. The cohomology of a compact manifold satisfies Poincare duality, and $C^\infty(M)$ knows this via the structure of its Hochschild/cyclic homology - hence Connes' invocation of Hochschild chains. – Tim Perutz Apr 15 2010 at 18:15
show 8 more comments
## 4 Answers
There is a strongly geometric characterization of those algebras which arise as $C^\infty(M)$ for $M$ compact and orientable, recently proved by Connes, see here. This has come up on MO before, e.g. in Joel Fine's answer to this question:
http://mathoverflow.net/questions/5344/algebraic-description-of-compact-smooth-manifolds
Like the proofs of most major theorems in differential topology, Connes's approach invokes (a) Riemannian metrics, and (b) hard analysis. Spectral geometry is not my area, so this will be an amateurish explanation...
If $M$ is a compact smooth manifold, $C^\infty(M)$ is represented faithfully on the Hilbert space $H$ of $L^2$ sections of any hermitian vector bundle $S$. If $D$ is a first-order differential operator acting (unboundedly) on $H$ then we can recover the projective $C^\infty(M)$-module of sections $C^\infty(M;S)$ within $H$ as $\bigcap_{k>0}{dom(D^k)}$ (these domains will actually by Sobolev spaces, I believe). The algebraic counterpart of being first-order is that $[[D,f],g]=0$ for any $f,g \in C^\infty(M)$.
$D$ has particularly nice properties when it's elliptic. There's no canonical elliptic operator over a smooth manifold until one chooses a Riemannian metric; there's then the signature operator $D=d+d^\ast$ acting on the complexified differential forms. This is an example of a Dirac operator (its square is a Laplacian - this is a condition on the symbol of the operator). As such, it's formally self-adjoint, Fredholm, and its (real) spectrum has known growth rate depending on $\dim(M)$.
Connes (see Theorem 11.4) shows that a commutative $\mathbb{R}$-algebra $A$ arises as $C^\infty(M)$ for a smooth manifold structure on the space $M$ Gelfand-dual to $A$ provided that it's part of a "spectral triple" $(A,H,D)$ of the right kind. This means that $A$ should act on a Hilbert space $H$ carrying an unbounded symmetric operator $D$ satisfying various properties. I've hinted at some of these; the most sophisticated property is an "orientation" condition invoking a Hochschild cycle $c\in Z_{\dim M}(A,A)$. This cycle is something like a volume form, and from its components Connes rebuilds local charts.
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Tim, I'm no geometer, but doesn't Connes' construction assume more and (hence) obtain more? The original question asks about $M$ with only a smooth structure and not a choice of Riemannian metric. (For what it's worth, I'm not claiming this approach is better or worse; just slightly different.) – Yemon Choi Apr 13 2010 at 21:02
Yemon - this issue was raised earlier on MO, but I don't quite see the objection. Slightly different to what? Is there a comparable theorem that invokes only the bare algebra? I don't see one mentioned in Connes' paper. The roles of the metric here - in defining a Hilbert-space completion of a projective module of smooth sections of a vector bundle, and in defining a differential operator from which we can find our original module inside that Hilbert space - seem very natural to me, both geometrically and analytically. – Tim Perutz Apr 13 2010 at 21:31
Tim - I actually agree that the spectral triple formulation is very natural, and I have no problem with it. It just seemed to me that the original post's wording, "I want some nontrivial purely algebraic formulation, if possible avoiding structure sheaves at all", was ambitiously looking for something only using the bare algebra. In other words, a bunch of conditions that can be formulated solely in terms of ${\mathbb C}$-algebra notions. But perhaps I misunderstood. – Yemon Choi Apr 14 2010 at 20:04
Yemon: I agree with your reading of Martin's question. All the same, and notwithstanding my previous comment, Connes' condition ARE conditions on the algebra! No sheaves are invoked. They aren't conditions on the structure of its ideals, but instead on its representation theory. – Tim Perutz Apr 14 2010 at 20:41
very impressive theorem, though I won't understand it. thanks Tim for the reference! as yemon pointed out, I'm hoping for a purely algebraic formulation. and yes, this seems to be only a dream. – Martin Brandenburg Apr 16 2010 at 14:44
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This is to expand further on something I wrote in the comments. Martin wrote:
Here are some necessary conditions: [...] Are they sufficient?
I think no since a polynomial algebra $\mathbb{R}[x_1,\ldots,x_n]$ satisfies all these conditions but is not isomorphic to the algebra of smooth functions on a manifold. A short explanation for this is that smooth functions satisfy Borel's lemma while polynomial algebras don't. Here's a more detailed explanation:
Let $m_p$ the maximal ideal corresponding to a point $p\in Spec_{\mathbb{R}}A$. Then for polynomial algebras as well as algebras of the form $C^\infty(M)$, the $m_p$-completion $\hat{A}=\lim_{r \geq 1} A/m_p^r$ is (after fixing local coordinates) a formal power series algebra $R[[x_1,\ldots,x_n]]$. The natural map $A\to \hat A$ may be interpreted as associating to a function its Taylor expansion at $p$. One version of the lemma of Borel says that for smooth function algebras this map is surjective. In other words: for every power series I give you (even non convergent) you can find a smooth function which has it as its Taylor series. Obviously this does not hold for polynomial algebras. So this gives you another necessary condition.
In the comments I said that polynomial algebras are finitely generated while algebras $C^\infty(M)$ are not. You asked me how to see this. I don't know if there is a simpler proof, but I would apply the same lemma of Borel: finitely generated algebras have a countable basis as vector spaces. But formal power series have no countable basis, and since the map $A\to \hat A$ is surjective also $A$ cannot have a countable basis. (Obviously if you just wanted to know that polynomial algebras are not isomorphic to smooth function algebras you didn't need this anymore).
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Note also that, if $M$ is positive dimensional, then $C^{\infty}(M)$ is not Noetherian, since if $\mathfrak m$ is a maximal ideal corresponding to a point $x \in M$, then $\bigcap_{n \geq 0} \mathfrak m^n$ is non-zero (since there are non-zero smooth functions whose Taylor series is identically zero). This also rules out polynomial or power-series rings. – Emerton Apr 14 2010 at 17:12
thanks michael and emerton. I've added some remarks in my original question. – Martin Brandenburg Apr 14 2010 at 23:46
I think the appropriate category here is not that of algebras but algebras with derivations (linear maps satisfying Leibniz's product rule). If you don't look at the derivations you're forgetting the differentiable structure of the manifold and all the manifolds homeomorphic (possibly not diffeomorphic) to your manifold support your algebra of functions. I posted an answer along the same lines in your previous questions.
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2
the fact that $C^{\infty}$ is full faithful implies that the manifold can be reconstructed from the algebra and every additional structure needed on this algebra can be recovered from the algebra structure. your claim that homeomorphic manifolds yield the same algebra is wrong. $C^{\infty}(M)$ and $C^{\infty}(N)$ are isomorphic as algebras iff $M,N$ are diffeomorphic. – Martin Brandenburg Apr 13 2010 at 11:32
How can we characterize the algebras (at least within all the C^∞(M)'s), that come from compact manifolds?
An algebra of the form C^∞(M) corresponds to a compact manifold if and only if all of its maximal ideals have codimension 1.
If the manifold is non-compact, consider the ideal of all functions with compact support and extend it to some maximal ideal using Zorn's Lemma. Clearly, this ideal cannot correspond to any point of the manifold, hence its codimension must be greater than 1.
To prove the converse, choose a maximal ideal I in C^∞(M) for some compact M. Denote by A the set of all points x∈M such that all elements of I vanish at x. A must be nonempty, because otherwise we can cover M by preimages of R \ {0} of functions in I, choose a finite subcover of this open cover, and then observe that the corresponding finite set of functions generates C^∞(M) as an ideal. A cannot consist of more than one point, because otherwise I is not maximal. For the same reason I consists of all functions that vanish at A, hence it has codimension 1.
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what is the codimension of an ideal? perhaps the codimension of the corresponding zero set? – Martin Brandenburg Apr 13 2010 at 2:14
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Martin: I assume Dmitri means codimension as a linear subspace – Yemon Choi Apr 13 2010 at 2:49
1
@Martin: Yemon is right. – Dmitri Pavlov Apr 13 2010 at 3:27
alright. thus the maximal ideals of codim. 1 are exactly the points of the real spectrum (aka the rational points). – Martin Brandenburg Apr 13 2010 at 3:32
@dmitri: ok this is just as in usual Gelfand-duality. fine :) – Martin Brandenburg Apr 13 2010 at 3:33 | {"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": 83, "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}, "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}, "english_confidence": 0.9337260127067566, "perplexity_flag": "head"} |
http://cms.math.ca/10.4153/CMB-1997-013-7 | Canadian Mathematical Society
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# Continuous Self-maps of the Circle
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http://dx.doi.org/10.4153/CMB-1997-013-7
Canad. Math. Bull. 40(1997), 108-116
Published:1997-03-01
Printed: Mar 1997
• J. Schaer
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## Abstract
Given a continuous map $\delta$ from the circle $S$ to itself we want to find all self-maps $\sigma\colon S\to S$ for which $\delta\circ\sigma = \delta$. If the degree $r$ of $\delta$ is not zero, the transformations $\sigma$ form a subgroup of the cyclic group $C_r$. If $r=0$, all such invertible transformations form a group isomorphic either to a cyclic group $C_n$ or to a dihedral group $D_n$ depending on whether all such transformations are orientation preserving or not. Applied to the tangent image of planar closed curves, this generalizes a result of Bisztriczky and Rival [1]. The proof rests on the theorem: {\it Let $\Delta\colon\bbd R\to\bbd R$ be continuous, nowhere constant, and $\lim_{x\to -\infty}\Delta(x)=-\infty$, $\lim_{x\to+\infty}\Delta (x)=+\infty$; then the only continuous map $\Sigma\colon\bbd R\to\bbd R$ such that $\Delta\circ\Sigma=\Delta$ is the identity $\Sigma=\id_{\bbd R}$.
MSC Classifications: 53A04 - Curves in Euclidean space 55M25 - Degree, winding number 55M35 - Finite groups of transformations (including Smith theory) [See also 57S17] | {"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": 17, "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}, "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}, "english_confidence": 0.7386491894721985, "perplexity_flag": "middle"} |
http://alanrendall.wordpress.com/2009/03/30/the-fitzhugh-nagumo-model/ | # Hydrobates
A mathematician thinks aloud
## The FitzHugh-Nagumo model
As mentioned in a previous post the FitzHugh-Nagumo model is a simplified version of the Hodgkin-Huxley model describing the propagation of signals in nerve cells. In this post only the spatially homogeneous case is discussed. It is possible to consider a corresponding system including a diffusion term for one of the unknowns ($x$ in the system below), a subject touched on in a previous post. A different ODE reduction can be obtained by looking at traveling wave solutions. The equations in the homogeneous case form a system of two ODE. It is supposed to capture the essential qualitative behaviour of the system of four ODE defining the Hodgkin-Huxley model while otherwise being as simple as possible. Here the notation of the original paper of FitzHugh (Biophys. J. 1, 445) will be used.The system is
$\dot x=c(y+x-x^3/3+z)$
$\dot y=-(x-a+by)/c$
with constant parameters $a$, $b$ and $c$. The quantity $z$ is in general a prescribed function of $t$. The unknown $x$ corresponds to the voltage in the Hodgkin-Huxley system while $y$ corresponds to the other variables. The parameter $z$ plays the role of the external current in the Hodgkin-Huxley system. The relation to the van der Pol oscillator, as written in the previous post, is easily seen. The parameters $a$ and $b$ and the function $z$ are new. If they are set to zero we get the van der Pol equation in slightly different variables. FitzHugh gives an intuitive description of the dynamics with phase plane pictures. Depending on the values of the parameters there may be a stable stationary solution which is a global attractor, or an unstable stationary solution together with a stable periodic solution. There is more discussion of the qualitative behaviour in Murray’s book ‘Mathematical Biology’.
Theorems about the van der Pol oscillator can be found in many textbooks. Corresponding material on the FitzHugh-Nagumo model seems to be much rarer. There is, nevertheless, a big literature out there. There is a thesis of Matthias Ringqvist online which collects a lot of interesting material on the subject and can serve as a point of entry to the literature for the uninitiated, like myself. He considers the properties of a dynamical system which contains the FitzHugh-Nagumo model as a special case. He motivates this generalization by noting that the more general system includes a number of different systems of interest in different problems in applied mathematics. He discusses the presence of periodic solutions in various parameter regimes and the bifurcations they are involved in. Hopf bifurcations play an important role. Another type of bifurcation which occurs in this context is the Bautin bifurcation. This differs from Hopf case in that one of the non-degeneracy conditions (the one which does not only depend on the linearization at the point of interest) fails. There can be coexistence of more than one periodic solution.
To what extent does the FitzHugh-Nagumo model capture the dynamics present in the Hodgkin-Huxley model? Ringqvist mentions numerical work of Guckenheimer and Oliva from 2002 which suggests that the HH model exhibits chaotic dynamics.The authors do not claim to have proved rigorously that chaos is present but this is a warning that it would be foolish to think that the dynamics of the HH system might be ‘essentially understood’. Chaos is observed in the forced van der Pol oscillator, i.e. the system obtained from the van der Pol oscillator by adding a prescribed function of time. It is more impressive for me to see chaos coming up in a system which is autonomous and which occurs naturally in an application. Of course the Lorenz system, one of the icons of chaos, satisfies the first condition and, at least in a weak sense, the second.
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### One Response to “The FitzHugh-Nagumo model”
1. Hopf bifurcations and Lyapunov numbers « Hydrobates Says:
January 10, 2010 at 1:14 pm | Reply
[...] there the Field-Noyes model also exhibits Hopf bifurcations. Hopf bifurcations occur in the FitzHugh-Nagumo and Hogdkin-Huxley systems. Thus they are potentially relevant for electrical signalling by [...]
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http://mathhelpforum.com/differential-geometry/81560-poisson-summation-formula-proof.html | # Thread:
1. ## Poisson Summation Formula Proof
I'm having trouble understanding a couple of parts of the proof presented here in Knapp's Basic Real Analysis : Book link from Google (pgs 389-390).
He defines $c_n=\frac{1}{2L}\int_{-L}^{L}f(t)e^{-2\pi int/L}dt$. He then takes the Fourier series for L=1 on $F(x)=\sum_{n=-\infty}^{\infty}f(x+n)$ and gets $F(x)=\sum_{n=-\infty}^{\infty}e^{2\pi i nx}\int_{0}^{1}F(t)e^{-2\pi i nt}dt$ which seemingly identifies $c_n=\int_{0}^{1}F(t)e^{-2\pi i nt}dt$. This means to me that $F(t)e^{-2\pi int}$ is an even function of t, but this is not mentioned; am I missing something here?
My second question is when he seemingly substitutes $t+k\rightarrow t$ in the integral $\int_0^1 f(t+k)e^{-2\pi int}dt=\int_{k}^{k+1}f(t)e^{-2\pi int}dt.$ However shouldn't the exponential in the integrand be $e^{-2\pi in(t-k)}$?
Thank you for any help! | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9146446585655212, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/102829/is-the-mostowski-collapse-natural | ## Is the Mostowski collapse natural?
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The Mostowski collapse lemma (see here for a quick ref) is one of the key basic tools in the set-theory arsenal. I wonder if the collapse is natural, in the functorial sense.
More precisely, is this a reflection from the large category of well-founded models of ZF to the subcategory of transitive models?
My taste would say yes, but I have not thought it through ((apologies if the answer turns out to be trivial).
MOTIVATION: still thinking a bit about the MULTIVERSE Category. If the answer is affirmative, then it makes good sense to simply work with the subcategory of transitive models of ZF, which is certainly more manageable, and simpler to ponder.
ADDENDUM TO THE MOTIVATION: on a quick after-thought, I partially retract what I just said: there could still be some interest in considering the larger category of not necessarily well-founded models. In this case, perhaps someone could provide some speculations as to this larger cat and what can be found there (exotic models)
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I've asked almost the same question here: mathoverflow.net/questions/37639/… – Martin Brandenburg Jul 23 at 9:27
yes, almost. Thanks for the link Martin. Take a look at Andreas' answer and subsequent comments, I am sure you will find it to your taste. That the mostowksy collpase is a reflection (actually an equivalence) is no surprise, but I confess I never thought of the other answer Andreas gave me, namely that taking the wekk-founded part is (likely) a coreflection into the well-founded universe (and therefore into the mostowsky cat). That, I think, needs further investigation, something cool lurks there. – Mirco Mannucci Jul 23 at 10:00
## 2 Answers
You didn't say what the morphisms in your categories are supposed to be; Trevor assumed you meant elementary embeddings, but you could also have meant mere embeddings, or something else. Nevertheless, unless you make a very strange choice of morphisms, the answer to your question is yes. Not only is the Mostowski collapse a reflection, it's an equivalence of categories. The transitive models constitute a skeleton of the category of all well-founded models; that is, every well-founded model $M$ is isomorphic to exactly one transitive model. Better yet, the isomorphism is uniquely determined by $M$. (All this information is part of the full statement of Mostowski's collapsing theorem.) So, from a category-theoretic point of view, it makes no difference whether you work with arbitrary well-founded models or with only the transitive models. Note, though, that in some situations, non-transitive well-founded models arise naturally, for example as elementary substructures of transitive ones, and in such cases your desire to work only with transitive models would require you to immediately apply the Mostowski collapse as soon as such a model enters your considerations.
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Yes, Andreas, I was admittedly sloppy (see my comment to Trevor). Accepted without reservation! Now, can you tell me what is the "categorical status" of the not-mostowkian models? How does the category of well founded models (and its equivalent version, the transitive universe) sits in the category of models at large? I suspect there is no simple characterization, and if this is the case, looks like that in a theory of the multiverse(s) one cannot reduce oneself to well founded ones. Thoughts? – Mirco Mannucci Jul 21 at 18:58
In your comment to Trevor, you described the intended morphisms as "all maps between models, not necessarily elementary." I assume you intend "all maps" preserving at least the membership relation. (Otherwise, we might as well forget that the objects are models of ZF and regard them as unstructured sets.) Is that right? If so, do you also intend that the maps preserve non-membership? that they be one-to-one? – Andreas Blass Jul 21 at 19:14
Note that non-well-founded models of ZF can look very different from well-founded ones. For example, they can fail to satisfy Con(ZF). On the other hand, every model M has a well-founded part, its largest well-founded submodel containing all members (in the sense of M) of its members. With a suitable notion of morphism, this might provide a coreflection from arbitrary models to well-founded ones. – Andreas Blass Jul 21 at 19:17
1) Yes, I meant all maps AS models of ZF, so they do respect membership. In other words, my cat is simply the cat of models of the theory ZF, as, say, the category of groups is MOD(Theory of groups). No, I do not demand anything else but being a map of models. 2) As for the second part, again great!!! I see, the "take my well founded part" provides a (candidate, but likely) co-reflection from the large multiverse to its well founded part. Very very interesting... then there is probably more to dig here (coalgebras). But for now I have my answers. Thanks again: your answers are gold – Mirco Mannucci Jul 21 at 19:33
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
If I understand the definition correctly at https://en.wikipedia.org/wiki/Reflective_subcategory, the question boils down to showing that every elementary embedding $f: B \to A$ between wellfounded models uniquely factors through the transitive collapse of $B$. This is true: It factors through the transitive collapse of $B$ because the transitive collapse map is an isomorphism. Uniqueness follows from the fact that isomorphic wellfounded models (in particular, $B$, its transitive collapse, and the range of $f$) are uniquely isomorphic.
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Trevor, as usual my question was a bit sloppy. I mentioned the category of models of ZF without saying what the maps are. This category is simply the category of $L_ZF$ structures which happens to be models of ZF, so the maps are all maps between models, not necessarily elementary. However, I think your point remains true, so you have my vote (which will become an "accepted" if there are no objections). Now, I am curious to know about the not well founded models (ie the remainder of the large cat of all models). The candidate reflection is not there, but is there something weaker in place? – Mirco Mannucci Jul 21 at 18:34 | {"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": 8, "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}, "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}, "english_confidence": 0.9438402056694031, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/97969/convex-functions-from-surfaces-of-revolution | ## convex functions from surfaces of revolution
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Li-Tam (Ann of math 1987) proved some estimates on nonnegative harmonic functions on manifolds with sec $\geq 0$ outside a compact set. I just want to get some feeling on how nontivial those estimates are. so I tried surface of revolution with positive curvature. In this simple case, those estimates are reduced the following question on convex functions from the graph of the surface.
Assume that $f$ is a smooth convex function on $[0,+\infty]$ with $f(0)=f^{\prime}(0)=0$ and $\lim_{x \rightarrow +\infty} f(x)=\lim_{x \rightarrow +\infty} f^{\prime}(x)=+\infty$ ($f^{\prime}$ is stricty increasing),
Then can we show:
$$\lim_{x \rightarrow +\infty} \frac{\log[\int_{C}^{x} \frac{f^{\prime}(\tau)}{\tau} d\tau \int_{C}^{x} \tau f^{\prime} (\tau) d\tau]} {\log[f(x)]}=2$$ for any fixed number $C>0$?
At first I thought it was just a simple calculus and $\liminf \geq 2$ by Cauchy-Schwarz, but I tried and failed to get the $\limsup$ part. I can only show that $\limsup \leq 2$ when $\lim_{x \rightarrow +\infty} \frac{f(x)}{xf^{\prime}(x)}$ exists. However, this assumption seems not true in general (although I don't have a counterexample yet). I will greatly appreciate if anyone suggests a solution.
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Maybe the computation is made simpler if we change variable, that makes the limit in terms of the inverse $g$ of $f$: $$\lim_{x\to\infty}\frac{\log\left(\int_c^xg(s)ds\int_c^x\frac{1}{g(s)}ds\right)}{\log x}=2\\ .$$ – Pietro Majer May 25 at 22:06 | {"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": 13, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9354435801506042, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2009/01/27/ | # The Unapologetic Mathematician
## Determining Eigenvalues
Yesterday, we defined eigenvalues, eigenvectors, and eigenspaces. But we didn’t talk about how to actually find them (though one commenter decided to jump the gun a bit). It turns out that determining the eigenspace for any given eigenvalue is the same sort of problem as determining the kernel.
Let’s say we’ve got a linear endomorphism $T:V\rightarrow V$ and a scalar value $\lambda$. We want to determine the subspace of $V$ consisting of those eigenvectors $v$ satisfying the equation
$T(v)=\lambda v$
First, let’s adjust the right hand side. Instead of thinking of the scalar product of $\lambda$ and $v$, we can write it as the action of the transformation $\lambda1_V$, where $1_V$ is the identity transformation on $V$. That is, we have the equation
$T(v)=\left[\lambda1_V\right](v)$
Now we can do some juggling to combine these two linear transformations being evaluated at the same vector $v$:
$\left[T-\lambda1_V\right](v)=0$
And we find that the $\lambda$-eigenspace of $T$ is the kernel $\mathrm{Ker}(T-\lambda1_V)$.
Now, as I stated yesterday most of these eigenspaces will be trivial, just as the kernel may be trivial. The interesting stuff happens when $\mathrm{Ker}(T-\lambda1_V)$ is nontrivial. In this case, we’ll call $\lambda$ an eigenvalue of the transformation $T$ (thus the eigenvalues of a transformation are those which correspond to nonzero eigenvectors). So how can we tell whether or not a kernel is trivial? Well, we know that the kernel of an endomorphism is trivial if and only if the endomorphism is invertible. And the determinant provides a test for invertibility!
So we can take the determinant $\det(T-\lambda1_V)$ and consider it as a function of $\lambda$. If we get the value ${0}$, then the $\lambda$-eigenspace of $T$ is nontrivial, and $\lambda$ is an eigenvalue of $T$. Then we can use other tools to actually determine the eigenspace if we need to.
Posted by John Armstrong | Algebra, Linear Algebra | 4 Comments
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
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http://mathematica.stackexchange.com/questions/17181/numbered-symbols?answertab=active | # Numbered symbols
I work with an exterior algebra over $R^n$. I have the basis $\{1,\omega_i\}_{i=1}^n$ in this algebra, and my differential operator is defined as
$$d\omega_k=\sum_{i>j>0,i+j=k} (i-j)w_i\wedge w_j.$$
As you see, my sum depends directly on the indexes of the elements of the basis. I surely can define the wedge and the differential operator in Mathematica, and they will calculate the derivative for any polynome I want. My problem is to find a good way to describe the elements of the basis, which will permit me use these formulas easily. It can not be just the integers $1, .. n$ for every element $\omega_1, .. \omega_n$, because Mathematica will multiply them as normal numbers, not treating them as elements of the exterior algebra; it cannot be just symbols such as `w1`, `w2`, ... because I cannot easily extract the numbers from them to calculate the derivatives (or I just don't know how).
I tried to represent them with `{1}`, `{2}`, and so on. I also used noncommutative multiplication with additional relations as the exterior multiplication. Example of what my notebook can calculate:
````In[30]:= d[{5} ** {4}]
Out[30]= {4} ** {3} ** {2} + 2 {5} ** {3} ** {1}
````
That's nice. I defined multiplication and differentiation without problems, except for one small thing -- multiplying by a scalar. For example, consider the expression $2\omega_4+\omega_5$. In this notation it appears as `2 {4} + {5}`. Mathematica proceeds to `{8} + {5}` and then to `{13}`, which is completely wrong.
````In[31]:= 2 {4} + {5}
Out[31]= {13}
````
This only happens when working with dimension 1. For higher grades everything's fine because Mathematica's multiplication doesn't misbehave when working with non-commutative elements:
````In[32]:= 2 {4} ** {3} + 6 {5} ** {2}
Out[32]= 2 {4} ** {3} + 6 {5} ** {2}
In[33]:= (2 {4} ** {3} + 6 {5} ** {2}) ** {3}
Out[33]= -6 {5} ** {3} ** {2}
````
But this is still not what I want -- lower dimensions are important to me, so my question is: how do I represent the basis elements without these problems? I can try to partially clear `Times`, but I don't think this is best way, as the multiplication of lists may be vital later in the code.
-
3
Why not w@1 ... w@n ? – belisarius Jan 2 at 18:36
The answer by helen to Differential geometry add-ons for Mathematica describes how Atlas 2 can handle this - but that's a commercial product and I can understand if you want to build your own version. However, maybe you can find some useful ideas in that link. – Jens Jan 2 at 18:37
@belisarius Thanks a lot, this works fine. – kriokamera Jan 2 at 18:46
1
Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!. And please,change your userid to something easier to remember! – belisarius Jan 2 at 18:56
## 1 Answer
Just do
````lst = w /@ Range[10]
(*{w[1], w[2], w[3], w[4], w[5], w[6], w[7], w[8], w[9], w[10]}*)
````
You can extract indices like this
````lst /. w[n_] :> n
(*{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}*)
````
(this works on each element, so eg `w[3] /. w[n_] :> n` evaluates to `3` and so on).
-
lang-mma | {"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": 5, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.903971791267395, "perplexity_flag": "middle"} |
http://mathhelpforum.com/pre-calculus/22656-coordinate-geometry-question.html | # Thread:
1. ## Coordinate geometry question
Just a really quick one, suppose I have a line of equation: y=1/2x+5/2
The next part of the question asks for this in the form of ax+by+c=0
How do I convert from y=mx+c to ax+by+c=0?
Thank youuu!
2. Originally Posted by hymnseeker
Just a really quick one, suppose I have a line of equation: y=1/2x+5/2
The next part of the question asks for this in the form of ax+by+c=0
How do I convert from y=mx+c to ax+by+c=0?
Thank youuu!
bring everything to one side and leave 0 on the other side and (in this case) multiply through by 2 when done
3. Thanks for the reply, so just let y=0?
1/2x+5/2=0
Does this satisfy what the question wants in asking for ax+by+c=0?
4. Originally Posted by hymnseeker
Thanks for the reply, so just let y=0?
1/2x+5/2=0
Does this satisfy what the question wants in asking for ax+by+c=0?
did you read what i said? i said bring everything over to one side, how does that translate to equating y to zero?
$y = \frac 12x + \frac 52$ ............subtract y from both sides
$\Rightarrow 0 = \frac 12x - y + \frac 52$ ...............multiply through by 2
$\Rightarrow 0 = x - 2y + 5$
which is the same thing as
$x - 2y + 5 = 0$
here a = 1, b = -2 and c = 5
5. Thank you for your help. Needed to see it visually - and in answer to your question (although I suspect it was rhetorical), yes, I did read what you said first off.
6. Originally Posted by hymnseeker
Thank you for your help. Needed to see it visually - and in answer to your question (although I suspect it was rhetorical), yes, I did read what you said first off.
yes, it was rhetorical. you're welcome, and good luck with your class | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9477589726448059, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/108430/free-resolution-of-this-determinantal-variety | ## Free Resolution of this determinantal variety.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am looking for a free resolution of the ideal generated by $2\times 2$-minors of a $3\times 3$ -matrix. More precisely let $M$ be a matrix (sorry but I cannot write a matrix for some TeX technical reason) $$M=\begin{bmatrix} x_{1}& x_{2}& x_{3} \\ x_{4}& x_{5}& x_{6} \\ x_{7}& x_{8}& x_{9} \end{bmatrix}$$ whose entries are indeterminates. I would like to find a free resolution of the ideal generated by $2\times 2$-minors of $M$ in the ring $\mathbb{C}[x_{1},\dots,x_{9}]$.
Does anyone know the reference for this resolution? Or make one for me.
-
The issue isn't TeX, it's Markdown, which the software processes before processing TeX and which interprets backslashes a certain way. This can be fixed by using three backslashes instead of two. – Qiaochu Yuan Sep 30 at 1:24
1
For this specific case, see the answer below. For the general case, in characteristic 0, Lascoux wrote a paper in the late 70s which gave an very explicit answer up to unknown scalar constants. (He does not even prove these constants exist). Weyman and Pragacz wrote a 100 page paper in the late 80s which is explicit enough for someone who really understands it to write down matrices. Weyman also wrote a book which explains what is going on conceptually with these free resolutions but is a bit less explicit in terms of getting matrices. Much is still unknown in positive characteristic. – Alexander Woo Sep 30 at 5:51
@Qiaochu Thank you for fixing the question and clarifying what's going on. Do you know where I can read about Markdown for MO? I would like to know how to use TeX on MO for next time (or it may be written somewhere in the instruction on MO). – Ronagh Sep 30 at 6:46
@Alexander Thanks you for the information. I will take a look at it. – Ronagh Sep 30 at 6:47
## 2 Answers
````Macaulay2, version 1.3.1
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra, SchurRings,
TangentCone
i1 : S = QQ[x_(1,1)..x_(3,3)]
o1 = S
o1 : PolynomialRing
i2 : M = transpose genericMatrix(S,x_(1,1),3,3)
o2 = {-1} | x_(1,1) x_(1,2) x_(1,3) |
{-1} | x_(2,1) x_(2,2) x_(2,3) |
{-1} | x_(3,1) x_(3,2) x_(3,3) |
3 3
o2 : Matrix S <--- S
i3 : I = minors(2,M)
o3 = ideal (- x x + x x , - x x + x x , - x x + x x , - x x + x x , - x x + x x , -
1,2 2,1 1,1 2,2 1,2 3,1 1,1 3,2 2,2 3,1 2,1 3,2 1,3 2,1 1,1 2,3 1,3 3,1 1,1 3,3
----------------------------------------------------------------------------------------------------------------------------
x x + x x , - x x + x x , - x x + x x , - x x + x x )
2,3 3,1 2,1 3,3 1,3 2,2 1,2 2,3 1,3 3,2 1,2 3,3 2,3 3,2 2,2 3,3
o3 : Ideal of S
i4 : (res I).dd
1 9
o4 = 0 : S <------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- S : 1
| x_(1,2)x_(2,1)-x_(1,1)x_(2,2) x_(1,3)x_(2,1)-x_(1,1)x_(2,3) x_(1,3)x_(2,2)-x_(1,2)x_(2,3) x_(1,2)x_(3,1)-x_(1,1)x_(3,2) x_(1,3)x_(3,1)-x_(1,1)x_(3,3) x_(2,2)x_(3,1)-x_(2,1)x_(3,2) x_(2,3)x_(3,1)-x_(2,1)x_(3,3) x_(1,3)x_(3,2)-x_(1,2)x_(3,3) x_(2,3)x_(3,2)-x_(2,2)x_(3,3) |
9 16
1 : S <----------------------------------------------------------------------------------------------------------------------------------------------------------- S : 2
{2} | -x_(1,3) x_(2,3) 0 -x_(3,1) 0 x_(3,2) 0 0 x_(3,3) 0 0 x_(3,3) 0 0 0 0 |
{2} | x_(1,2) -x_(2,2) 0 0 -x_(3,1) 0 0 x_(3,2) 0 x_(3,3) 0 -x_(3,2) 0 0 0 0 |
{2} | -x_(1,1) x_(2,1) 0 0 0 0 -x_(3,1) 0 0 0 0 0 -x_(3,2) x_(3,3) 0 0 |
{2} | 0 0 -x_(1,3) x_(2,1) 0 -x_(2,2) -x_(2,3) 0 -x_(2,3) 0 0 0 0 0 x_(3,3) 0 |
{2} | 0 0 x_(1,2) 0 x_(2,1) 0 x_(2,2) -x_(2,2) 0 -x_(2,3) 0 0 0 0 -x_(3,2) 0 |
{2} | 0 0 0 -x_(1,1) 0 x_(1,2) 0 x_(1,3) 0 0 -x_(2,3) 0 0 0 0 x_(3,3) |
{2} | 0 0 0 0 -x_(1,1) 0 0 0 x_(1,2) x_(1,3) x_(2,2) 0 0 0 0 -x_(3,2) |
{2} | 0 0 -x_(1,1) 0 0 0 0 0 0 0 0 x_(2,1) x_(2,2) -x_(2,3) x_(3,1) 0 |
{2} | 0 0 0 0 0 0 -x_(1,1) x_(1,1) -x_(1,1) 0 -x_(2,1) -x_(1,1) -x_(1,2) x_(1,3) 0 x_(3,1) |
16 9
2 : S <-------------------------------------------------------------------------------------------- S : 3
{3} | x_(3,1) -x_(3,2) 0 -x_(3,3) 0 0 0 0 0 |
{3} | 0 0 x_(3,1) 0 -x_(3,3) x_(3,2) 0 0 0 |
{3} | -x_(2,1) x_(2,2) 0 x_(2,3) 0 0 0 0 0 |
{3} | -x_(1,3) 0 x_(2,3) 0 0 0 -x_(3,3) 0 0 |
{3} | x_(1,2) 0 -x_(2,2) 0 0 0 x_(3,2) 0 0 |
{3} | 0 -x_(1,3) 0 0 0 -x_(2,3) 0 x_(3,3) 0 |
{3} | -x_(1,1) 0 x_(2,1) 0 0 0 0 x_(3,2) -x_(3,3) |
{3} | -x_(1,1) x_(1,2) 0 0 x_(2,3) 0 0 0 -x_(3,3) |
{3} | x_(1,1) 0 0 -x_(1,3) 0 x_(2,2) 0 -x_(3,2) 0 |
{3} | 0 0 0 x_(1,2) -x_(2,2) 0 0 0 x_(3,2) |
{3} | 0 0 -x_(1,1) 0 x_(1,3) -x_(1,2) 0 0 0 |
{3} | -x_(1,1) 0 0 0 x_(2,3) -x_(2,2) -x_(3,1) 0 0 |
{3} | 0 x_(1,1) 0 0 0 x_(2,1) 0 -x_(3,1) 0 |
{3} | 0 0 0 -x_(1,1) x_(2,1) 0 0 0 -x_(3,1) |
{3} | 0 0 0 0 0 0 x_(2,1) x_(2,2) -x_(2,3) |
{3} | 0 0 0 0 0 0 -x_(1,1) -x_(1,2) x_(1,3) |
9 1
3 : S <------------------------------------------ S : 4
{4} | -x_(2,3)x_(3,2)+x_(2,2)x_(3,3) |
{4} | -x_(2,3)x_(3,1)+x_(2,1)x_(3,3) |
{4} | -x_(1,3)x_(3,2)+x_(1,2)x_(3,3) |
{4} | x_(2,2)x_(3,1)-x_(2,1)x_(3,2) |
{4} | x_(1,2)x_(3,1)-x_(1,1)x_(3,2) |
{4} | x_(1,3)x_(3,1)-x_(1,1)x_(3,3) |
{4} | -x_(1,3)x_(2,2)+x_(1,2)x_(2,3) |
{4} | x_(1,3)x_(2,1)-x_(1,1)x_(2,3) |
{4} | x_(1,2)x_(2,1)-x_(1,1)x_(2,2) |
1
4 : S <----- 0 : 5
0
o4 : ChainComplexMap
````
-
1
In other words, Macaulay is your friend! – Mariano Suárez-Alvarez Sep 30 at 1:40
1
IIRC Miller and Sturmfels construct resolutions for these ideals at the end of their Combinatorial Commutative Algebra. – Mariano Suárez-Alvarez Sep 30 at 1:49
2
@Mariano - that is incorrect - the current known answer is much too complicated to put in a book. – Alexander Woo Sep 30 at 5:46
@Graham Thank you for the code. I totally forgot about Macaulay! – Ronagh Sep 30 at 6:48
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The standard way to construct a resolution is the following. Let $X = A^9$ and let $E$ be a 3-dim (trivial) vector bundle on $X$ such that $M \in End(E)$. Consider the relative Grassmannian $p:Gr(2,E) \to X$ and let $U$ be its tautological rank 2 bundle. Let $Z$ be the zero locus of the morphism $U \to p^*E \stackrel{\ M\ }\to p^*E$ on $Gr(2,E)$. Then it is clear that you ask for a resolution of $p(Z)$. What is good is that it is easy to write a resolution for $Z$ on $Gr(2,E)$, and then you can ush it forward to $X$.
Indeed, $Z$ is the zero locus of a regular section of the vector bundle $U^*\otimes p^*E$. Consequently, $O_Z$ has a resolution by the Koszul complex ```$$
\dots \to \Lambda^2(U\otimes p^*E^*) \to U\otimes p^*E^* \to O_{Gr(2,E)} \to O_Z\to 0.
$$``` It remains to note that `$\Lambda^k(U\otimes p^*E) = \oplus \Sigma^\alpha U\otimes p^*\Sigma^{\alpha^T}E^*$`, the sum is over all Young diagrams in a $2\times 3$ rectangle, $\alpha^T$ is the transposed diagram, and $\Sigma^\alpha$ is the Schur functor. To push forward to $X$ one can use the projection formula and Borel--Bott--Weil to compute the cohomology on $Gr(2,E)$. As a result you will get ```$$
0 \to O_X \to
E\otimes E^* \to
sl(E) \oplus sl(E) \to
E^*\otimes E \to
O_X \to O_{p(Z)} \to 0.
$$```
- | {"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": 27, "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}, "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}, "english_confidence": 0.7521757483482361, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/119416-standard-confusion-picturing-work-empty-tank.html | # Thread:
1. ## Standard confusion on picturing work to empty tank
Yes, I have searched the forums and google but nothing quite yet what I was looking for.
Problem:
A water tank in the shape of a cone is underground with the distance above ground being called D, the base of the cone is horizontal with the ground ie pointing straight up, The height in feet of the tank is called H, which has a base radius r in feet, the tank is filled up to some point h. Set up the integral to calculate the total work done. No calculations.
My assessments
The equation to find the volume of my slice obviously would be: $\pi r^2 dh$
The distance to move the slice is to go from the current level to the top of the tank to the distance above ground thus giving $(D+H-h)$
D being distance above ground, H height of the tank, and h being current distance from top of the tank at the water level
Next I thought of my vertical being of h and the change in height was of dh, and thought of my horizontal axis r, radius. where 2r(diameter) is the width of the cone, this is perhaps where I confused my self not placing it on a x-y coordinate plan to make some sort of distinction. Even If i had placed it on a x-y plane the, apex, or peak would be at some value of y, I always noticed my radius could be at some point x or y, which usually results in the equation of the slope of the line r/h * x, but this when people place one of the cone's half horizontal to the x-axis, so the slope can be positive.
In general in need some help sorting out this problem, it's been bugging me such a simple concept can get so confusing.
2. See attachment
Attached Thumbnails
3. Except you see, the cone is inverted, as the base is on the "x-axis" so as you lift each piece out of the cone the radius will increase as to where, in your picture as you lift each slice out the radius gets smaller and smaller, that's what threw me off. Perhaps I am over complicating things, and if even the cone is inverted the the function of the radius/height remains the same? even if radius increasing when I think about it
4. Let me seeif I have this right--the apex is at ground level?
If so see attachment
Attached Thumbnails
5. Yes, thank you, I even hard the terms written out but I wasn't complete sure. I knew it wasn't that complicated | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9610593914985657, "perplexity_flag": "middle"} |
http://2brains.tumblr.com/tagged/mathematics | Belief
A true scientist must always be reminded that beliefs are a necessary evil. But in all situations he would do well to remember that his strongest and most fervent beliefs are always in those matters in which his ignorance is greatest.
Stupid Questions
It is often said that there are no stupid questions. Is this true? Or is it just a way to avoid discouraging seekers?
I can think of at least one circumstance where a question might be considered stupid. And that is to first misquote a statement, or misunderstand it, and then with conviction, question it’s validity.
Is this a stupid question? I’ll let you decide.
And is this important? It is important since the foundation of all scientific endeavors has never been to find the right answers so much as it has been to ask the right questions. And in looking back at the history of science, every single great discovery, every theory, every law, has really been a question in disguise.
Consider one of the most well known. Sir Isaac Newton looked at the moon. Made a few observations, and by sheer genius realized that the Earth was falling. But that is really the question isn’t it? Why is the Earth Falling? He could only ask that question if he had, by an even greater force of genius, realized that perhaps the Earth should NOT be falling. Had no one ever asked that question, (which by the way still awaits a definitive answer) we would never have flown from the earth, let alone reached the moon. And yes, by falling, as much, if not more than flying.
Mistakes
It is a kind of arrogance to willfully repeat mistakes.
For a function that is the sum of cosine theta and i sine theta
the derivative is the difference of i cosine theta and sine theta
factoring out the square root of negative one from the difference
the derivative of the function is the product of the function and i
solving the differential…
Interesting patter, African village built using fractal geometry.
Theories
All scientific theories fail or eventually will as they are refined or revised. Only in Mathematics are theories indelible. Which for me makes physics all the more mysterious for even what has been well established can and must be revisited.
To eliminate clutter
It is the wonderful task of mathematics, as well as it’s immense power, to simplify all matters by the elimination of all that is unnecessary.
Faith
Many people who encounter me as a scientist like to say that I have no faith. This is a tremendously erroneous accusation. In response I will outline a great many beliefs that I have based entirely on faith for I have no proof to offer in support of these postulates.
1. I believe in Evolution. I believe that, given enough time, new species can be manifest from sufficient mutation. And I believe that sufficient time has been given.
2. I believe that there is intelligent life in this universe, perhaps even in our galaxy. The fact that we almost are gives me hope that beings that actually are intelligent may exist.
3. I believe that the secret to happiness is hope which can only be secured by hard work and discipline. That those who would attempt to raise my self esteem by encouraging me to lower my standards are really trying to help themselves at the expense of my proxy, and as such end up becoming the problem by trying to fix what’s not really broken.
4. I believe that there is a relationship between electromagnetism and gravity.
5. I believe that space consists of discrete quanta. And that we will discover that this will give dimension to the fine structure constant
$\alpha = \frac{e^2}{(4 \pi \varepsilon_0)\hbar c} = \frac{e^2 c \mu_0}{2 h} = \frac{k_\mathrm{e} e^2}{\hbar c},$
and essentially replace the speed of light as the bridge between reference frames. Or perhaps even eliminate reference frames.
6. I believe that there can be an entirely separate form of mathematics that could simplify what is considered difficult but will severely complicate the trivial. I further believe that transforms can be developed that will allow a change in mathematics in the same way that Lorentz transforms allow us to exchange reference frames.
7. I believe that a new mathematics may change the role of mathematics in science, which up until now has only been used to describe models and quantify observations. Thus in the same way observation can affect physical phenomena, mathematics may play a role in affecting the actual relationships of physical phenomena which we have so far only used numbers to quantify.
8. I believe that our planet has never been visited by alien life forms. There are far too many mysteries on earth to be leaping to the conclusion that anything we don’t understand must be from outer space. I have a far greater faith in our ignorance of terrestrial matters than I do in our knowledge of flying saucers.
9. I believe that there is an upper limit to the intelligence of a species. That individuals that approach or go beyond this limit are not sufficiently successful to pass on those genetic traits.
10. I believe that the egg came first. (Laid by something that was not quite a chicken yet)
Conservation
It is interesting to note that those who are complacent in their belief that scientists will find a way to fix everything generally never listen to scientists when they say “Stop what you’re doing now!!!”
A critical point in the polution of our environment will at some point be reached. But those hellbent on getting there are convinced that the scientists whom they hope will save us from ourselves are lying about the need for conservation.
Hot Coffee
I get up in the morning and make hot coffee and I am ready to pour it into a thermos to take to work. The thermos is not perfect and so the coffee will lose some of it’s heat while sitting in the thermos. I also like to add creamer. The creamer is cold. If I add the creamer it will also cool down the coffee to some lower temperature.
I could add the creamer right away before pouring the coffee into the thermos and then take it to work.
I could put the coffee into the thermos without the creamer and wait to add the creamer when I am at work.
The question is, which scenario will yield a warmer cup of coffee when I am at work? Put the creamer in right away? or wait to add the creamer later? Or does it not matter, the coffee will lose the same amount of heat either way. | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9644651412963867, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/5415/is-this-how-padding-can-work/5427 | # Is this how padding can work?
So for block ciphers you need a fixed size block.
If the plaintext length is not a multiple of the block length then you need to pad it.
One way you could do this is that for the last block you just pad the blocks with 0s or some random characters except for the last character of the last block which represents the number of characters padded.
Of course the problem is what happens when the plaintext is a multiple of the block length (i.e. fits exactly). In this case you just add a dummy block of 0s or 1s. Then when you decrypt you know that if you encounter such a dummy block then the previous block has no padding.
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So, what is the problem? Last byte will contain the block length, so you will know that entire last block is a padding. – Pavel Ognev Nov 19 '12 at 7:16
But it might have that value anyway. How do you distinguish between the padding and the actual ciphertext? – winterfell Nov 19 '12 at 9:34
Padding is applied to plaintext, not ciphertext. – Pavel Ognev Nov 19 '12 at 17:30
## 2 Answers
You normally want to use a fully invertible padding scheme, i.e. a padding scheme with an associated unpadding scheme such that $unpad(pad(X)) = X$ for every $X$.
Assuming you only append data at the end, and don't do different things depending on the content of the message, this means that you always have to append something, even if the message is already of the right length.
In your scheme, in the case of a message which has already the length of a multiple of the block length, you'll have to append a full block ending with a 16 (assuming a 16-byte block cipher like AES). Your unpadding function then reads the last byte, and knows it has to strip off 16 bytes.
You shouldn't do any special-casing with a block of a certain structure only in the case of a full block appended.
But note that you should preferably use standardized schemes to help interoperability - for example PKCS#7-padding works just like your scheme, but with a bit different filling of the padded data. (This, just like your scheme, requires that the number of appended bytes is encodable in a byte, i.e. a block size of less than 256 bytes = 2048 bits. I'm not aware of any common block cipher with a larger block size, but you would need a different padding scheme in this case.)
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PKCS#7 requires the block size to be no more than 2048 bits, though. Not a problem yet but it might become one in the future. – Thomas Nov 19 '12 at 14:46
@Thomas Yes, of course, it depends on the block size being representable in one byte. The scheme proposed in the question has the same problem. One could modify this by using two bytes for the length, but block ciphers with blocks larger than 2048 are not that often used. (For public key encryption you'll want to use other padding schemes anyways.) – Paŭlo Ebermann♦ Nov 19 '12 at 14:56
Agreed, I think the widest block cipher I've seen yet is Threefish-1024 and nobody really uses that outside Skein. Just thought I'd mention it :) – Thomas Nov 19 '12 at 15:41
It looks fine though you are better off sticking to established standards which already have parsers written. That way you don't have any problems.
To remove the padding, read the last byte, then move that many values back and truncate. In python it would look something like this:
```
p = decrypt(key, ciphertext)
pad_len = p[len(p)-1]
return p[0:len(p)-pad_len]
```
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Your code is unfortunately not complete. You need to validate the padding bytes are correct and within a valid range for your blocksize. PKCS#7 works fairly well as an integrity test for decryption - if the padding characters are ill-formed, either the message itself was corrupted, or a wrong key was used to decrypt. It has to be handled properly, of course, or this could open up some kind of padding oracle attack similar to those seen in CRIME/BEAST. – John Deters Nov 21 '12 at 15:21 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9182125329971313, "perplexity_flag": "middle"} |
http://cms.math.ca/Reunions/hiver12/abs/de | Réunion d'hiver SMC 2012
Fairmont Le Reine Elizabeth (Montréal), 7 - 10 décembre 2012
Équations différentielles fonctionnelles non linéaires
Org: Hermann Brunner (Memorial)
[PDF]
URI ASCHER, UBC
Stochastic algorithms for inverse problems involving PDEs and many measurements [PDF]
Inverse problems involving systems of partial differential equations (PDEs) can be very expensive to solve numerically. This is so especially when many experiments, involving different combinations of sources and receivers, are employed in order to obtain reconstructions of acceptable quality. The mere evaluation of a misfit function (the distance between predicted and observed data) often requires hundreds and thousands of PDE solves. We develop and assess dimensionality reduction methods, both stochastic and deterministic, to reduce this computational burden.
We present in detail our methods for solving such inverse problems for the famous DC resistivity and EIT problems. These methods involve incorporation of a priori information such as piecewise smoothness, bounds on the sought conductivity surface, or even a piecewise constant solution. A more general random subset method is proposed first. We then assume that all experiments share the same set of receivers and concentrate on methods for reducing the number of combinations of experiments, called simultaneous sources, that are used at each stabilized Gauss-Newton iteration. Algorithms for controlling the number of such combined sources are proposed and justified. Evaluating the misfit approximately, except for the final verification for terminating the process, always involves random sampling. Methods for selecting the combined simultaneous sources, involving either random sampling or truncated SVD, are proposed and compared. Highly efficient variants of the resulting algorithms are identified.
JACQUES BÉLAIR, Université de Montréal
DISTRIBUTION OF DELAYS IN A PHARMACODYNAMIC MODEL [PDF]
Time delays occur naturally in pharmacokinetic and pharmacodynamic (PK/PD) processes, but the form in which they are introduced in the models is not always entirely obvious, the distribution of delays being typically ignored or represented empirically. We present a model of chemotherapy-induced myelosuppression using differential equations with distributed delays, to take into account the delay between administration of the drug and the observed effect. The transit compartment yields a single differential equation with a bimodal distribution of delays. We discuss the stability of this system, obtaining a stability chart in a two parameter space, and possible oscillatory solutions.
Joint work with Andreea Rimbu Pruncut.
SUE ANN CAMPBELL, University of Waterloo
Plankton Models with Time Delay [PDF]
We consider a three compartment (nutrient-phytoplankton-zooplankton) model with nutrient recycling. When there is no time delay the model has a conservation law and may be reduced to an equivalent two dimensional model. We consider how the conservation law is affected by the presence of time delay (both discrete and distributed) in the model. We study the stability and bifurcations of equilibria when the total nurient in the system is used as the bifurcation parameter. This is joint work with Matt Kloosterman and Francis Poulin.
TONY HUMPHRIES, McGill University
Stability and numerical stability of a model state-dependent DDE [PDF]
We consider the stability properties of the model state-dependent delay differential equation (DDE), $$\dot{u}(t) = \mu u(t) + \sigma u(t - a - cu(t)),$$ and its numerical discretization by the backward Euler method. The stability region for the DDE itself is well-known in the constant delay case ($c=0$), and is the basis for several numerical stability definitions. Recent results in state-dependent DDEs show that (with the possible exception of points on the boundary) the state-dependent DDE ($c\ne0$) has the same stability region as the constant delay DDE. We thus propose this as a model problem in the numerical analysis of state-dependent DDEs, and extend the numerical stability definitions accordingly.
To study the stability properties of the backward Euler method, we first study the DDE itself and use a Lyapunov-Razumikhin approach to directly prove stability of the state-dependent DDE in parts of its stability region including the entire delay-independent portion and parts of the delay-dependent portion. The Lyapunov-Razumikhin approach is further generalised to study the stability of the backward Euler method solution, and we establish stability for all step-sizes in the part of the stability region for which the direct proof showed stability of the DDE, and stability for sufficiently large step-size in the entire stability region.
Joint work with Felicia Magpantay (York) \& Nicola Guglielmi (L'Aquila)
MICHAEL MACKEY, McGill University
Modeling chemotherapy effects on hematopoiesis [PDF]
The delivery of chemotherapy has, almost without exception, profound side effects on many physiological systems and the regulation of hematopoiesis is not an exception. In this talk I will outline the work that is ongoing in our attempts to minimize the hematopoietic side effects of chemotherapy. This modeling work is naturally framed within the context of functional differential equations with state dependent delays, and has revealed some of the many potential dynamical effects of forcing systems so described with external perturbations like those due to chemotherapy. The dynamic effects that the numerical solutions of these equations have revealed offer important clues about how clinicians may be able to avoid some of the side effects of chemotherapy. Specifically we find that the mathematical model for the regulation of hematopoiesis shows significant resonance effects at certain periods of chemotherapy administration that are probably associated with especially adverse reactions in patients.
FELICIA MAGPANTAY, York University
An age-structured population model with state-dependent delay: derivation and numerical integration [PDF]
We present an age-structured population model that accounts for the following aspects of complex life cycles: (1) There are juvenile and adult stages, (2) only the adult stage is capable of reproducing, (3) cohorts of juveniles can transition to the adult stage when they have consumed enough nutrition and (4) the juvenile and adult populations consume different limited food sources. Taking all of these into account leads to a new mathematical model that cannot be directly analyzed using the established framework of functional differential equations. The model consists of a partial differential equation with a nonlinear boundary condition and state-dependent delay due to a threshold condition. In this talk we present the derivation of the model, its properties and a numerical scheme to integrate the equations.
This is joint work with Nemanja Kosovalic and Jianhong Wu.
CHUNHUA OU, Memorial University
Delayed reaction diffusion models and traveling waves for spatial spread of rabies in Europe: a re-visit [PDF]
In this talk, spatial spread of rabies in Europe is re-visited with the consideration of the impacts of the incubation and its interaction with the spatial movement of the susceptible and the incubative. First, a delayed reaction diffusion model is constructed with the incorporation of the incubation only. The minimal spreading speed is derived by the classical stability analysis and it is shown to be a decreasing function of the incubation time. A new method based on the integration of wave pulses is proposed which yields the existence of wave patterns for all values of the incubation time. In addition, by incorporating the spatial movement of the incubative foxes, a non-local reaction diffusion system is constructed. Rigorous proof of the existence of the wave patterns is shown via the integration of the wave pulses coupled with the Fredholm alternative theorem.
GAIL WOLKOWICZ, McMaster University
Dynamics of the chemostat and classical predator-prey models with time delay [PDF]
The dynamics of the classical predator-prey model and the predator-prey model based in the chemostat are studied and compared to see whether a discrete time delay in the conversion process can lead to sustained oscillatory behaviour, when no such behaviour is possible when delay is ignored. A surprising similarity between the possible attractors of the classical predator-prey model and the attractors of the Mackey-Glass equation are demonstrated. The analogous integro-differential equations models are also considered.
JIANHONG WU, York University
Global dynamics of the Nicholson's blowflies equation revisited: onset and termination of nonlinear oscillations [PDF]
We revisit the Nicholson's blowflies model with natural death rate incorporated into the delay feedback. We consider the delay as a bifurcation parameter and examine the onset and termination of Hopf bifurcations of periodic solutions from a positive equilibrium. We show that the model has only a finite number of Hopf bifurcation values and we describe how branches of Hope bifurcations are paired so the existence of periodic solutions with specific oscillation frequencies occurs only in bounded delay intervals. The bifurcation analysis and the Matlab package DDE-BIFTOOL developed by Engelborghs et al guide some numerical simulations to identify ranges of parameters for coexisting multiple attractive periodic solutions. This is a joint work with Hongying Shu and Lin Wang.
YUAN YUAN, Memorial University of newfoundland
Stability and Bifurcation in FDE with Distributed Delay [PDF]
A set of sufficient conditions for the global and local stability is established for a large class of functional differential equation with distributed delay. With the loss of the stability at the boundary of the stability regions, we discuss the Hopf bifurcation, there the computation of the coefficients are given in the form of the corresponding characteristic equation explicitly. Then these analytic results are applied to the mathematical models of white blood cell production. Numerical simulations are presented to illustrate the stability regions of parameters and to address the effect of the distributed time delay in the physiological oscillations.
This talk is based on my joint work with Drs. Jacques Belair and Xiaoqiang Zhao.
XIAOQIANG ZHAO, Memorial University of Newfoundland
Global Dynamics of A Time-Delayed Reaction and Diffusion Malaria Model [PDF]
In this talk, I will report our recent research on a vector-bias malaria model with incubation period and diffusion. We first prove the global stability of the disease-free or endemic equilibrium for the spatially homogeneous time-delayed system. Then we establish the threshold dynamics for the spatially heterogeneous system in terms of the basic reproduction ratio. We also obtain a set of sufficient conditions for the global attractivity of the positive steady state. This talk is based on my joint work with Zhiting Xu.
XINGFU ZOU, University of Western Ontario
On a DDE model describing malaria transmission dynamics in a patch environment [PDF]
I will present some results on a DDE model that describes the transmission dynamics of malaria over a patchy environment. The model incorporates two important factors into the classic Ross-McDonand model: disease latencies in both humans and mosquitoes, and dispersal of humans between patches. The basic reproduction number $\mathcal{R}_0$ of model is identified by the theory of the next generation operator for structured disease models and the dynamics of the model is investigated in terms of $\mathcal{R}_0$. It is shown that the disease free equilibrium is asymptotically stable if $\mathcal{R}_0<1$, and it is unstable if $\mathcal{R}_0>1$; in the latter case, the disease is endemic in the sense that the variables for the infected compartments are uniformly persistent. For the case of two patches, more explicit formulas for $\mathcal{R}_0$ are derived by which, impacts of the dispersal rates as well as the latency delays on disease dynamics are explored. Some numerical computations for $\mathcal{R}_0$ in terms of dispersal rates are carried out, which visually show that the impacts could be very complicated: in certain range of the parameters, $\mathcal{R}_0$ is increasing with respect to a dispersal rate while in some other range, it can be decreasing with respect to the same dispersal rate.
This is a joint work with Dr. Yanyu Xiao. | {"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": 9, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9020251035690308, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/38624/determining-the-center-of-mass-of-a-cone/38630 | # Determining the center of mass of a cone
I'm having some trouble with a simple classical mechanics problem, where I need to calculate the center of mass of a cone whose base radius is $a$ and height $h$..!
I know the required equation. But, I think that I may be making a mistake either with my integral bounds or that $r$ at the last..! $$z_{cm} = \frac{1}{M}\int_0^h \int_0^{2\pi} \int_0^{a(1-z/h)} r \cdot r \:dr d\phi dz$$
'Cause, once I work this out, I obtain $a \over 2$ instead of $h \over 4$...!
Could someone help me?
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1
– Ϛѓăʑɏ βµԂԃϔ Sep 29 '12 at 10:16
They are not homework, and I did do the sum just making a mistake somewhere. And asking for help with that.. – Coolcrab Sep 29 '12 at 10:55
1
What!? $a/2 := h/4$!? I don't think so... Increasing $a$ has no effect on $h$, this assertation is plane false. – Killercam Sep 29 '12 at 11:14
Ofc its not, and thats my problem. It's either my bounds or the r should be a z. I'm not sure. – Coolcrab Sep 29 '12 at 11:16
@Killercam: Hi guys, I understood it. But, I just assumed that the center of mass would be somewhere along the axis of the cone...! Isn't that right? – Ϛѓăʑɏ βµԂԃϔ Sep 29 '12 at 11:36
show 1 more comment
## 1 Answer
I am not sure about this formula. Lets start by taking the vertex of the solid cone to be $O(0, 0, 0)$ in cylindrical coordinates ($r$, $\theta$, $z$). Then take the height of the cone to be $h$ and the base of the cone to have radius $a$. In this case the we know that
$r = \frac{a}{h} z$.
The formula for the center of mass of this cone can be written as
$Mz_{M} = \int^{h}_{0} z \mathrm{d}m$,
where $M$ is the total mass of the (solid) cone and $z_{M}$ is the location of the center of mass. We can write $\mathrm{d}m$ as
$\mathrm{d}m = \pi \rho \frac{a^{2}}{h^{2}}z^{2}\mathrm{d}z$,
where we have considered $\mathrm{d}m$ to be the mass of a thin disk at height $z$ and of radius $r$, with thickenss $\mathrm{d}z$. Now we can write the full equation for the center of mass as
$Mz_{M} = \pi\rho\int^{h}_{0}\frac{a^{2}}{h^{2}}z^{3}\mathrm{d}z$,
this becomes
$Mz_{M} = \rho Vz_{M} = \frac{1}{4}\pi\rho a^{2}h^{2}$.
We know that the volume of a cone $V = \frac{1}{3}\pi a^{2} h$, so we find
$z_{M} \rho \frac{1}{3}\pi a^{2} h = \frac{1}{4}\pi\rho a^{2}h^{2}$,
so
$z_{M} = \frac{3}{4} h$.
Which is the distance from the vertex of the cone.
I 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": 29, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9226498007774353, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/53846/when-does-the-distributive-propertive-hold-for-inner-products | # When does the distributive propertive hold for inner products?
Going though some papers in clustering in machine learning, I often find the following claim:
$(a-b)(a-b) = \|a\|^2 + \|b\|^2 - 2a \cdot b$
My question is: When does this hold? (i.e. what type of norms or inner products satisfy this)? Is this only true for the $L_2$ inner product and its induced norm?
Also, would you call the above an example of distributivity in inner products? Does distributivity hold for inner products?
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2
A minor LaTeX thing: it's a bit better to use \| or \Vert instead of || to get double norm-strokes: $\|$. – t.b. Jul 26 '11 at 15:09
Thanks @Theo. That makes sense. – user815423426 Jul 26 '11 at 15:09
## 1 Answer
All inner products (over $\mathbb{R}$) satisfy this, if by the LHS you mean $(a - b) \cdot (a - b)$. It follows from bilinearity (which you can call distributivity if you want, but it would be better to call it bilinearity because the result of an inner product is a scalar, not a vector) and symmetry. You can call it the law of cosines if you want.
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Thanks, I fixed the dot product in the question. So I guess all inner products over $\mathbb{R}$ induce norms, and the above always satisfies for them? – user815423426 Jul 26 '11 at 15:08
1
@AmV: yes. For complex inner products one instead gets $||a||^2 + ||b||^2 - 2 \text{Re}(a \cdot b)$ (exercise). – Qiaochu Yuan Jul 26 '11 at 15:12
1
To add to what Qiaochu said, let $\odot$ be any bilinear operation on a vector space (that is, if $\vec{v},\vec{w},\vec{u}$ are vectors and $a,b,c$ are scalars, then we require $(a\vec{v}+b\vec{w})\odot (c\vec{u}) = ac (\vec{v}\odot \vec{u}) + bc (\vec{w}\odot \vec{u})$ and similarly $(a\vec{v})\odot(b\vec{w} + c\vec{u}) = ab (\vec{v}\odot \vec{w}) + ac (\vec{v}\odot \vec{u})$), then necessarily $(\vec{v}+\vec{w})\odot(\vec{v}+\vec{w}) = \vec{v}\odot \vec{v} + \vec{w}\odot \vec{w} + \vec{v}\odot\vec{w} + \vec{w}\odot\vec{v}$. – Willie Wong♦ Jul 26 '11 at 15:14 | {"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": 13, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8899670243263245, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/271141/number-of-roots-of-a-complex-equation-rouches-theorem | # Number of roots of a complex equation/ Rouche's theorem
For $n\geq2$ consider the equation $z^n+z+n=0$ for $z\in \mathbb C$. Show that if $k$ is an integer with $1\leq k \leq n$ then inside the sector
$$S_k=\{z\in \mathbb C: 0< Arg(z) < \dfrac{2\pi k}{n} \}$$
There are exactly $k$ roots of the above equation. $Arg (z)$ is the principal argument of $z$. (Hint: Prove that $x^n+n>x$ for real $x$)
The only thing I can think of is Rouche's theorem but then the region needs to be bounded to be able to use that. Can anybody give some pointers as to how I should proceed here. Thanks.
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1
If you make the substitution $z = n^{1/n} \zeta$, you get the equation $\zeta^n + \frac{n^{1/n}}{n}\zeta + 1 = 0$. Does this give you an idea of where to look for your roots? – Antonio Vargas Jan 5 at 21:41
@AntonioVargas: I still don't see it. when $n$ is large the cofficient of $\zeta$ goes to zero. But that is not relevant, I think. We need to show this for all $n$ – user54755 Jan 5 at 23:28
Exactly, when $n$ is large the equation is very similar to $\zeta^n+1=0$. So you should be looking for the roots of $\zeta^n + \frac{n^{1/n}}{n}\zeta + 1 = 0$ near the roots of $\zeta^n+1=0$. This should give you an idea of what region to use in Rouché's theorem. – Antonio Vargas Jan 5 at 23:48 | {"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": 19, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9503758549690247, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/246761/find-a-basis-for-the-subspace-and-prove-it-is-a-basis-state-the-dimension | # find a basis for the subspace and prove it is a basis, state the dimension
Each of the following sets spans a subspace of the space of all functions from $\bf R$ to $\bf R$.
$\exp x, \sin(x), \cos(x)$
$\exp x, \cosh(x), \sinh(x)$
this is just an extra credit problem, but I have no clue how to go about it. If someone please could help me:)
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Doesn't any set of vectors span a subspace of the space containing it...? – tomasz Nov 28 '12 at 20:27
## 2 Answers
I will provide some hints.
For the first family, show that it forms a linearly independent set. So it's a basis of the generating subspace of functions from $\Bbb R$ to itself.
For the second one, recall that $\cosh x+\sinh x=e^x$, so the family is not linearly independent. But it is one you remove $\exp x$ for example.
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an additional hint: To prove that $f_1 , f_2, f_3$ are linearly independent, assume $\lambda_1 f_1 + \lambda_2 f_2 + \lambda_3 f_3 = 0$ for some $\lambda_1, \lambda_2, \lambda_3 \in \mathbb{R}$ , that is for ALL $x \in \mathbb{R}$: $$\lambda_1 f_1(x) + \lambda_2 f_2(x) + \lambda_3 f_3(x) = 0$$. Then prove that $\lambda_1, \lambda_2, \lambda_3$ must all be $0$. To do this, it is usually enough to plug in some different values for $x$ and solve the system of linear equations you get from that. – Brusko651 Nov 28 '12 at 20:14
Thank you!!! i will try to do that:)) – Andriy Lysak Nov 28 '12 at 20:20
for the second one i dont understand the part about removing exp x, could you please explain a bit more? – Andriy Lysak Nov 28 '12 at 22:26
The generated family will be the same. – Davide Giraudo Nov 29 '12 at 9:40
The set of vectors $\{\sinh(x),\cosh(x),e^{x}\}$ is a linearly dependent set because the last vector (function) can be written as a linear combination of the two previous ones, more specifically as the sum of the previous two.
Hence if we remove it and consider only the functions spanned by $\{\sinh(x),\cosh(x)\}$ we get the same subspace!
This is true in general. If you have a linear dependent set of vectors, you can always remove a vector that is in the span of the others, and the new (smaller) set of vectors will generate the same subspace.
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Oh I got it now, Thank You!!! – Andriy Lysak Nov 28 '12 at 22:51 | {"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": 16, "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}, "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}, "english_confidence": 0.9142177700996399, "perplexity_flag": "head"} |
http://cstheory.blogoverflow.com/2011/07/an-original-proof-of-a-folk-theorem/ | # An original proof of a folk theorem
2011-07-25 by Aaron Sterling. 4 comments
A folk theorem is a mathematical fact that is generally known (at least by experts), even though its proof may never have appeared in print, or may be difficult to locate. Recently, cstheory user Eli asked for a citation to a proof of the following folk theorem.
If $$G$$ is a graph with maximum degree 3 and is a minor of $$H$$, then $$G$$ is a topological minor of $$H$$.
Wikipedia mentions this to be true, and refers to Reinhard Diestel’s excellent book on graph theory. Diestel’s book asserts its truth as well, but does not give a proof.
Graph $$G$$ is a minor of graph $$H$$, if $$G$$ can be obtained from $$H$$ by removing edges in $$H$$ and identifying the vertices that were previously the endpoints of the now-deleted edge. So, in a sense, we can say that $$H$$ “has local structures that looks like” $$G$$ if $$G$$ is a minor of $$H$$. It is sometimes possible to prove strong properties about the global structure of $$H$$ if we are guaranteed that $$H$$ has no minors of a certain type, because then no local area of $$H$$ can look a certain way, so it is possible to say something about the entire graph. The most famous example of this is Wagner’s Theorem, which states that a graph is planar iff neither $$K_5$$ nor $$K_{3,3}$$ is a minor of it.
A topological minor is a related notion. Informally, $$G$$ is a topological minor of $$H$$ if we can start with $$G$$, and transform it into a subgraph of $$H$$ by only adding vertices into edges that already exist (i.e., subdividing edges, so edge $$(u,v)$$ becomes two edges, $$(u,x)$$ and $$(x,v)$$). Figure 1 on this web page shows examples of a minor and a topological minor. (Note: that figure may be behind a paywall. If anyone knows a good diagram explaining minor vs. topological minor that is freely available, please put it in the comments.)
Fortunately for Eli, Robin Kothari and co-authors had needed this folk theorem for a paper, and they had been unable to find a proof of it either, so they reproved it themselves. Kothari’s proof (and answer to Eli’s question) appears here. It is an example of a growing number of original proofs that appear on cstheory. That proof by Kothari is definitely my “favorite answer of the week.” As Eli put it, “This is stellar.”
|
## 4 Comments
• Alexey says:
“G is a topological minor of H if we can start with G, and transform it into H”
would it be
“G is a topological minor of H if we can start with G, and transform it into minor of H”
• Alexey says:
oh, sorry… i mean “G is a topological minor of H if we can start with G, and transform it into subgraph of H”
• says:
Yes, quite so! I will fix the post now.
• [...] posts about graph minors, we reviewed the definitions of graph minors and topological minors, then looked at a relationship between the two; and explored whether there was a relationship between the computational hardness of a set of [...] | {"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": 25, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9523643255233765, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/98708-limit-problem.html | # Thread:
1. ## limit problem
lim [f(x) - 8] / [x - 1] = 10
x ~> 1
What is the limit as x approaches 1 of f(x)?
(Hint: let g(x) = [f(x) - 8] / [x - 1])
i'm not sure how the hint helps me in solving this problem. all i did was multiply both sides of the equation [f(x) - 8] / [x - 1] = 10 by (x - 1) and then added 8 to solve for f(x). then i got f(x) = 10x - 2 and i found that the limit of f(x) as x approaches 1 = 8.
is this the correct method of doing the problem?
in case it looked confusing up there, it should read: "the limit as x approaches 1 of (f(x) - 8) / (x - 1) equals 10."
2. Originally Posted by oblixps
lim [f(x) - 8] / [x - 1] = 10
x ~> 1
What is the limit as x approaches 1 of f(x)?
(Hint: let g(x) = [f(x) - 8] / [x - 1])
i'm not sure how the hint helps me in solving this problem. all i did was multiply both sides of the equation [f(x) - 8] / [x - 1] = 10 by (x - 1) and then added 8 to solve for f(x). then i got f(x) = 10x - 2 and i found that the limit of f(x) as x approaches 1 = 8.
is this the correct method of doing the problem?
in case it looked confusing up there, it should read: "the limit as x approaches 1 of (f(x) - 8) / (x - 1) equals 10."
If you let $g(x)=\frac{f(x)-8}{x-1}$, we have $f(x)=g(x)(x-1)+8$.
Therefore, $\lim_{x\to1}f(x)=\lim_{x\to1}g(x)(x-1)+8=0+8=8$.
Does this make sense? | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9503807425498962, "perplexity_flag": "middle"} |
http://nrich.maths.org/5749/clue | ### Plane to See
P is the midpoint of an edge of a cube and Q divides another edge in the ratio 1 to 4. Find the ratio of the volumes of the two pieces of the cube cut by a plane through PQ and a vertex.
### Four Points on a Cube
What is the surface area of the tetrahedron with one vertex at O the vertex of a unit cube and the other vertices at the centres of the faces of the cube not containing O?
### Instant Insanity
We have a set of four very innocent-looking cubes - each face coloured red, blue, green or white - and they have to be arranged in a row so that all of the four colours appear on each of the four long sides of the resulting cuboid.
# Cheese Cutting
##### Stage: 5 Challenge Level:
Simplify the problem by making the initial cuts all at right angles so that the large cube is sliced into$8$ smaller cubes.
Align the cube so that the $8$ corners lie at the points $(\pm 1, \pm 1, \pm 1).$
Now consider the plane $x+y+z=0$.
How many of the $8$ smaller cubes does this plane pass through?
What can you say about any cubes that this plane does not pass through? How can you alter this plane to try to get more intersections?
For the second part of the problem don't attempt to work out the theoretical maximum number of pieces as this is exceedingly difficult even for $5$ cuts! You can, however, consider how many pieces each small piece of cheese can be cut into per slice. You can also experiment with various pre-determined 'cutting schemes' to try to find a guaranteed minimum number of possible pieces.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found 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": 6, "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}, "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}, "english_confidence": 0.9377076029777527, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/14064/can-a-black-hole-be-formed-by-radiation | Can a black hole be formed by radiation?
I'm trying to find out if black holes could be created by focusing enough light into a small enough volume.
So far I have found (any or all may be incorrect):
• Maxwell's equations are linear, dictating no interaction of radiation.
• The Kerr effect and self-focusing has been observed in mediums, but not vacuums.
• Masses bending light have been observed per general relativity.
• Photons are said to have no rest mass, just energy and momentum (???).
• General relativity seems to provide for energy to energy interaction.
This leads to a more specific question:
Does radiation or energy curve space like mass curves space?
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– droud Aug 28 '11 at 16:42
The answer is yes, and it is obvious. Mathematicians have recently made a small industry of rigorously proving it by complicated methods, but the methods are not insightful. – Ron Maimon Aug 28 '11 at 20:45
2 Answers
The answer to your first question is yes.
Building on Demetrios Christodoulou's seminal work showing that black holes can form "generically" from focusing of gravitational waves starting from an initial space-time that is arbitrarily close to flat, Pin Yu has recently shown that one can also dynamically (and generically, in the sense that the formation is stable under small perturbations) form a black hole starting with only electromagnetic waves.
Of course, the interaction between electromagnetism and gravity means that as soon as you set the thing in motion, you will pick up gravitational radiation. And also that since a precise covariant notion of local gravitational energy is not available, the idea that the space-time starts out with only electromagnetic waves is a specific, frame dependent mathematical definition; one should keep that in mind before trying to draw too much physical significance out of the casual statement of the theorem.
For your specific second question, the answer is also yes. Einstein's equation specifies that $$G_{\mu\nu} = T_{\mu\nu}$$ the left hand side, the Einstein tensor, is purely geometrical, and reflects the curvature of space-time. The right hand side comes from the energy-momentum contributions from the matter fields. The standard way of coupling electromagnetic waves to general relativity (Einstein-Maxwell theory) gives that the right hand side is zero only when the electromagnetic field vanishes. So the content of Einstein-Maxwell theory is based on that electromagnetic radiation can curve space-time.
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-1: (I hate Christodoulou's methods) This is not great as a physics answer, because both results are ridiculously obvious as physics. They are marginally less obvious as mathematics, but there was no doubt, before Christodoulou, that radiation produces black holes. – Ron Maimon Aug 28 '11 at 20:59
In general relativity, the quantities that can act as sources of spacetime curvature are the ingredients of the energy momentum tensor. There is a nice diagram of these ingredients in the Wikipedia article.
Since radiation has such attributes (energy, momentum etc), it can act as a source for the gravitational field.
Edit: I seem to have omitted the very first part of your question. See the discussion here about whether or not there is a "critical energy density" which determines black hole formation.
- | {"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}, "config": {"markdown_headings": false, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9360403418540955, "perplexity_flag": "middle"} |
http://www.reference.com/browse/Matched_filter | Related Searches
Definitions
# Matched filter
In telecommunications, a matched filter is obtained by correlating a known signal, or template, with an unknown signal to detect the presence of the template in the unknown signal. This is equivalent to convolving the unknown signal with a time-reversed version of the template (cross-correlation). The matched filter is the optimal linear filter for maximizing the signal to noise ratio (SNR) in the presence of additive stochastic noise. Matched filters are commonly used in radar, in which a known signal is sent out, and the reflected signal is examined for common elements of the out-going signal. Pulse compression is an example of matched filtering. Two-dimensional matched filters are commonly used in image processing, e.g., to improve SNR for X-ray pictures.
## Derivation of the matched filter
The matched filter is the linear filter, $h$, that maximizes the output signal-to-noise ratio.
$y\left[n\right] = sum_\left\{k=-infty\right\}^\left\{infty\right\} h\left[n-k\right] x\left[k\right].$
Though we most often express filters as the impulse response of convolution systems, as above (see LTI system theory), it is easiest to think of the matched filter in the context of the inner product, which we will see shortly.
We can derive the linear filter that maximizes output signal-to-noise ratio by invoking a geometric argument. The intuition behind the matched filter relies on correlating the received signal (a vector) with a filter (another vector) that is parallel with the signal, maximizing the inner product. This enhances the signal. When we consider the additive stochastic noise, we have the additional challenge of minimizing the output due to noise by choosing a filter that is orthogonal to the noise.
Let us formally define the problem. We seek a filter, $h$, such that we maximize the output signal-to-noise ratio, where the output is the inner product of the filter and the observed signal $x$.
Our observed signal consists of the desirable signal $s$ and additive noise $v$:
$x=s+v.,$
Let us define the covariance matrix of the noise, reminding ourselves that this matrix has Hermitian symmetry, a property that will become useful in the derivation:
$R_v=E\left\{ vv^H \right\},$
where $.^H$ denotes Hermitian (conjugate) transpose, and $E$ denotes expectation. Let us call our output, $y$, the inner product of our filter and the observed signal such that
$y = sum_\left\{k=-infty\right\}^\left\{infty\right\} h^*\left[k\right] x\left[k\right] = h^Hx = h^Hs + h^Hv = y_s + y_v.$
We now define the signal-to-noise ratio, which is our objective function, to be the ratio of the power of the output due to the desired signal to the power of the output due to the noise:
$SNR = frac\left\{|y_s|^2\right\}\left\{E\left\{|y_v|^2\right\}\right\}.$
We rewrite the above:
$SNR = frac\left\{|h^Hs|^2\right\}\left\{E\left\{|h^Hv|^2\right\}\right\}.$
We wish to maximize this quantity by choosing $h$. Expanding the denominator of our objective function, we have
$E\left\{ |h^Hv|^2 \right\} = E\left\{ \left(h^Hv\right)\left\{\left(h^Hv\right)\right\}^H \right\} = h^H E\left\{vv^H\right\} h = h^HR_vh.,$
Now, our $SNR$ becomes
$SNR = frac\left\{ |h^Hs|^2 \right\}\left\{ h^HR_vh \right\}.$
We will rewrite this expression with some matrix manipulation. The reason for this seemingly counterproductive measure will become evident shortly. Exploiting the Hermitian symmetry of the covariance matrix $R_v$, we can write
$SNR = frac\left\{ | \left\{\left(R_v^\left\{1/2\right\}h\right)\right\}^H \left(R_v^\left\{-1/2\right\}s\right) |^2 \right\}$
{ {(R_v^{1/2}h)}^H (R_v^{1/2}h) },
We would like to find an upper bound on this expression. To do so, we first recognize a form of the Cauchy-Schwarz inequality:
$|a^Hb|^2 leq \left(a^Ha\right)\left(b^Hb\right),,$
which is to say that the square of the inner product of two vectors can only be as large as the product of the individual inner products of the vectors. This concept returns to the intuition behind the matched filter: this upper bound is achieved when the two vectors $a$ and $b$ are parallel. We resume our derivation by expressing the upper bound on our $SNR$ in light of the geometric inequality above:
$SNR = frac\left\{ | \left\{\left(R_v^\left\{1/2\right\}h\right)\right\}^H \left(R_v^\left\{-1/2\right\}s\right) |^2 \right\}$
{ {(R_v^{1/2}h)}^H (R_v^{1/2}h) }
` leq`
` frac{ left[`
{(R_v^{1/2}h)}^H (R_v^{1/2}h)
` right]`
` left[`
{(R_v^{-1/2}s)}^H (R_v^{-1/2}s) right] } { {(R_v^{1/2}h)}^H (R_v^{1/2}h) }.
` `
Our valiant matrix manipulation has now paid off. We see that the expression for our upper bound can be greatly simplified:
$SNR = frac\left\{ | \left\{\left(R_v^\left\{1/2\right\}h\right)\right\}^H \left(R_v^\left\{-1/2\right\}s\right) |^2 \right\}$
{ {(R_v^{1/2}h)}^H (R_v^{1/2}h) } leq s^H R_v^{-1} s.
` `
We can achieve this upper bound if we choose,
$R_v^\left\{1/2\right\}h = alpha R_v^\left\{-1/2\right\}s$
where $alpha$ is an arbitrary real number. To verify this, we plug into our expression for the output $SNR$:
$SNR = frac\left\{ | \left\{\left(R_v^\left\{1/2\right\}h\right)\right\}^H \left(R_v^\left\{-1/2\right\}s\right) |^2 \right\}$
{ {(R_v^{1/2}h)}^H (R_v^{1/2}h) } = frac{ alpha^2 | {(R_v^{-1/2}s)}^H (R_v^{-1/2}s) |^2 } { alpha^2 {(R_v^{-1/2}s)}^H (R_v^{-1/2}s) } = frac{ | s^H R_v^{-1} s |^2 } { s^H R_v^{-1} s } = s^H R_v^{-1} s.
` `
Thus, our optimal matched filter is
$h = alpha R_v^\left\{-1\right\}s.$
We often choose to normalize the expected value of the power of the filter output due to the noise to unity. That is, we constrain
$E\left\{ |y_v|^2 \right\} = 1.,$
This constraint implies a value of $alpha$, for which we can solve:
$E\left\{ |y_v|^2 \right\} = alpha^2 s^H R_v^\left\{-1\right\} s = 1,$
yielding
$alpha = frac\left\{1\right\}\left\{sqrt\left\{s^H R_v^\left\{-1\right\} s\right\}\right\},$
giving us our normalized filter,
$h = frac\left\{1\right\}\left\{sqrt\left\{s^H R_v^\left\{-1\right\} s\right\}\right\} R_v^\left\{-1\right\}s.$
If we care to write the impulse response of the filter for the convolution system, it is simply the complex conjugate time reversal of $h$.
Though we have derived the matched filter in discrete time, we can extend the concept to continuous-time systems if we replace $R_v$ with the continuous-time autocorrelation function of the noise, assuming a continuous signal $s\left(t\right)$, continuous noise $v\left(t\right)$, and a continuous filter $h\left(t\right)$.
## Alternative derivation of the matched filter
Alternatively, we may solve for the matched filter by solving our maximization problem with a Lagrangian. Again, the matched filter endeavors to maximize the output signal-to-noise ratio ($SNR$) of a filtered deterministic signal in stochastic additive noise. The observed sequence, again, is
$x = s + v,,$
with the noise covariance matrix,
$R_v = E\left\{vv^H\right\}.,$
The signal-to-noise ratio is
$SNR = frac\left\{|y_s|^2\right\}\left\{ E\left\{|y_v|^2\right\} \right\}.$
Evaluating the expression in the numerator, we have
$|y_s|^2 = \left\{y_s\right\}^H y_s = h^H s s^H h.,$
and in the denominator,
$E\left\{|y_v|^2\right\} = E\left\{ \left\{y_v\right\}^H y_v \right\} = E\left\{ h^H v v^H h \right\} = h^H R_v h.,$
The signal-to-noise ratio becomes
$SNR = frac\left\{h^H s s^H h\right\}\left\{ h^H R_v h \right\}.$
If we now constrain the denominator to be 1, the problem of maximizing $SNR$ is reduced to maximizing the numerator. We can then formulate the problem using a Lagrange multiplier:
$h^H R_v h = 1$
$mathcal\left\{L\right\} = h^H s s^H h + lambda \left(1 - h^H R_v h \right)$
$nabla_\left\{h^*\right\} mathcal\left\{L\right\} = s s^H h - lambda R_v h = 0$
$\left(s s^H\right) h = lambda R_v h$
which we recognize as an eigenvalue problem
$h^H \left(s s^H\right) h = lambda h^H R_v h = lambda.$
Since $s s^H$ is of unit rank, it has only one nonzero eigenvalue. It can be shown that this eigenvalue equals
$lambda_\left\{max\right\} = s^H R_v^\left\{-1\right\} s,$
yielding the following optimal matched filter
$h = frac\left\{1\right\}\left\{sqrt\left\{s^H R_v^\left\{-1\right\} s\right\}\right\} R_v^\left\{-1\right\} s.$
This is the same result found in the previous section.
## Example of matched filter in radar and sonar
Matched filters are often used in signal detection (see detection theory). As an example, suppose that we wish to judge the distance of an object by reflecting a signal off it. We may choose to transmit a pure-tone sinusoid at 1 Hz. We assume that our received signal is an attenuated and phase-shifted form of the transmitted signal with added noise.
To judge the distance of the object, we correlate the received signal with a matched filter, which, in the case of white (uncorrelated) noise, is another pure-tone 1-Hz sinusoid. When the output of the matched filter system exceeds a certain threshold, we conclude with high probability that the received signal has been reflected off the object. Using the speed of propagation and the time that we first observe the reflected signal, we can estimate the distance of the object. If we change the shape of the pulse in a specially-designed way, the signal-to-noise ratio and the distance resolution can be even improved after matched filtering: this is a technique known as pulse compression.
Additionally, matched filters can be used in parameter estimation problems (see estimation theory). To return to our previous example, we may desire to estimate the speed of the object, in addition to its position. To exploit the Doppler effect, we would like to estimate the frequency of the received signal. To do so, we may correlate the received signal with several matched filters of sinusoids at varying frequencies. The matched filter with the highest output will reveal, with high probability, the frequency of the reflected signal and help us determine the speed of the object. This method is, in fact, a simple version of the discrete Fourier transform (DFT). The DFT takes an $N$-valued complex input and correlates it with $N$ matched filters, corresponding to complex exponentials at $N$ different frequencies, to yield $N$ complex-valued numbers corresponding to the relative amplitudes and phases of the sinusoidal components.
## Example of matched filter in digital communications
The matched filter is also used in communications. In the context of a communication system that sends binary messages from the transmitter to the receiver across a noisy channel, a matched filter can be used to detect the transmitted pulses in the noisy received signal.
Imagine we want to send the sequence "0101100100" coded in non polar Non-return-to-zero (NRZ) through a certain channel.
Mathematically, a sequence in NRZ code can be described as a sequence of unit pulses or shifted rect functions, each pulse being weighted by +1 if the bit is "1" and by 0 if the bit is "0". Formally, the scaling factor for the $k^mathrm\left\{th\right\}$ bit is,
$a_k =$
begin{cases} 1, & mbox{if bit } k mbox{ is 1}, 0, & mbox{if bit } k mbox{ is 0}. end{cases}
We can represent our message, $M\left(t\right)$, as the sum of shifted unit pulses:
$M\left(t\right) = sum_\left\{k=-infty\right\}^infty a_k times$
Pi left( frac{t-kT}{T} right).
where $T$ is the time length of one bit.
Thus, the signal to be sent by the transmitter is
If we model our noisy channel as an AWGN channel, white Gaussian noise is added to the signal. At the receiver end, for a Signal-to-noise ratio of 3dB, this may look like:
A first glance will not reveal the original transmitted sequence. There is a high power of noise relative to the power of the desired signal (i.e., there is a low signal-to-noise ratio). If the receiver were to sample this signal at the correct moments, the resulting binary message would possibly belie the original transmitted one.
To increase our signal-to-noise ratio, we pass the received signal through a matched filter. In this case, the filter should be matched to an NRZ pulse (equivalent to a "1" coded in NRZ code). Precisely, the impulse response of the ideal matched filter, assuming white (uncorrelated) noise should be a time-reversed complex-conjugated scaled version of the signal that we are seeking. We choose
$h\left(t\right) = Pileft\left(frac\left\{t\right\}\left\{T\right\} right\right).$
In this case, due to symmetry, the time-reversed complex conjugate of $h\left(t\right)$ is in fact $h\left(t\right)$, allowing us to call $h\left(t\right)$ the impulse response of our matched filter convolution system.
After convolving with the correct matched filter, the resulting signal, $M_mathrm\left\{filtered\right\}\left(t\right)$ is,
$M_mathrm\left\{filtered\right\}\left(t\right) = M\left(t\right) * h\left(t\right)$
where $*$ denotes convolution.
Which can now be safely sampled by the receiver at the correct sampling instants, and compared to an appropriate threshold, resulting in a correct interpretation of the binary message.
Since the matched filter is the filter that maximizes the signal-to-noise ratio it can be shown that it also minimizes the Bit error ratio (BER), which is the ratio of the number of bits that the receiver interprets incorrectly as a fraction of the total number of bits sent.
## References
• Melvin, Willian L. "A STAP Overview." IEEE Aerospace and Electronic Systems Magazine 19 (1) (January 2004): 19-35.
• Turin, George L. "An introduction to matched filters." IRE Transactions on Information Theory 6 (3) (June 1960): 311- 329. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 73, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9147199988365173, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/244743/singular-value-proofs | # Singular value proofs
1.) Let $A$ be a nonsingular square matrix.
• a.) Prove that the product of the singular values of $A$ equals the absolute value of its determinant: $\sigma_1\sigma_2...\sigma_n=|detA|$.
• b.) Does the sum equal the absolute value of the trace? And a matrix whose determinant is very small is ill-conditioned?
• c.) Show that if $|detA|< 10^{-k}$, then its minimal singular value satisfies $\sigma_n <10^{-\frac{k}{n}}$. Can you construct an ill-conditioned matrix with $detA = 1$?
2.) Let A be a square matrix. Prove that its maximum eigenvalue is smaller than its maximal singular value.
My attempt:
a.) I know that the equation for singular values is $A = U \dot\ E \dot\ V$ so $|det(A)| = |det(U) \dot\ det(E) \dot\ det(V)| = |\pm1 \dot\$ (product of singular values) | = product of singular values. Is that correct?
b.) True for the first part. For the second part, do they mean a matrix can have ill-conditioned determinant if its determinant is small? Not sure what they are asking.
c.) I do not know how to do.
2.) Not sure how to do because I thought that the singular value should be smaller than the maximum eigenvalue?
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Should there be an absolute value around product of singular values in (a)? – Stuart Nov 26 '12 at 4:28
@Stuart: singular values are non-negative. – Martin Argerami Nov 26 '12 at 4:30
## 2 Answers
a) is correct
b) No, it's not true. If $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, then the sum of the singular values is $2$, while the absolute value of the trace is zero.
To discuss whether your matrix is ill-conditioned, you need to say which norm you are talking about. Assuming we are talking about the operator norm (=largest singular value), if the determinant is small it means that some singular values are small; then the inverse will have big singular values and the condition number will be large.
c) You have $\sigma_1\sigma_2\cdots\sigma_n<10^{-k}$; if all $\sigma_j\geq 10^{-k/n}$, then $$|\det A|=\sigma_1\cdots\sigma_n\geq(10^{-k/n})^n=10^{-k};$$ so at least one singular value is less than $10^{-k/n}$.
2) This is not well phrased, because they can be equal. The maximum singular value is $\|A^TA\|^{1/2}$. Now let $\lambda$ be an eigenvalue of $A$ with unit eigenvector $v$. Then $$|\lambda|=\|\lambda v\|=\|Av\|=(v^TA^TAv)^{1/2}\leq\|A^TA\|^{1/2}(v^Tv)^{1/2}=\|A^TA\|^{1/2}.$$ So every eigenvalue is smaller in absolute value than the biggest singular value.
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Thank you very much for clearing that up for me, Martin! – diimension Nov 26 '12 at 5:28
1
You are very welcome. – Martin Argerami Nov 26 '12 at 5:31
For $1(c)$, make use of the fact that $\sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_n$. Hence, we get that $$\sigma_n^n \leq \sigma_1 \sigma_2 \cdots \sigma_n = \det(A) < 10^{-k}$$ Hence, we get that $$\sigma_n < 10^{-k/n}$$ The maximal eigen value is given by \begin{align} \left \vert \lambda_{\max} \right \vert & = \max_{\Vert x \Vert_2 = 1} \left \vert x^T A x \right \vert\\ & = \underbrace{\max_{\Vert x \Vert_2 = 1} \left \vert x^T U \Sigma V^T x \right \vert \leq \max_{\Vert x \Vert_2 = 1} \Vert x^T U \Vert_2 \Vert \Sigma \Vert_2 \max_{\Vert x \Vert_2 = 1} \Vert V^T x \Vert_2}_{\text{By sub-multiplicativity of matrix norm}}\\ & = 1 \times \Vert \Sigma \Vert_2 \times 1 = \sigma_1 \end{align}
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Thank you very much, Marvis! – diimension Nov 26 '12 at 5:28 | {"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": 19, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7932907938957214, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/29948/applications-of-topological-and-diferentiable-stacks | ## Applications of topological and diferentiable stacks
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What are some examples of theorems about topology or differential geometry that have been proven using topological/differentiable stacks, or, some examples of proofs made easier by them? I'm well aware of several statements made more beautiful in the LANGUAGE of stacks, but, I'm looking for a concrete application.
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## 2 Answers
I would like to point out that stacks are "just" higher analogues of sheaves - a very basic tool to arrange structure. The same is true for topological or differentiable stacks. So I think everybody who expects amazing applications of stacks should be able to name an equally amazing application of a sheaf. (I am not saying that those don't exist!)
That said, let me mention an application. In view of the fact that you didn't get any answers so far (apart from your own), I hope it's not too inappropriate to take one from my own research. It applies abelian gerbes with connection to lifting problems for principal bundles.
I hope the following specifications qualify the theorem below as application: its statement does not involve any stacks or gerbes, just "basic" differential geometry. Its proof, however, is a simple composition of two gerbe-theoretical theorems.
Theorem. Let $M$ be a connected smooth manifold, let $P$ be a principal $G$-bundle with connection over $M$, let $\hat G$ be a central extension of $G$ by an abelian Lie group $A$, and let $\rho \in \Omega^2(M,\mathfrak{a})$ be a 2-form. Then, there exists a principal $A$-bundle $\mathcal{L}_P$ over $LM$ with a connection and with a fusion product, and a bijection between
1. isomorphism classes of lifts of the structure group of $P$ from $G$ to $\hat G$ with compatible connection of scalar curvature $\rho$, and
2. smooth sections of $\mathcal{L}_P$ that preserve the fusion product and pull back the connection to the transgressed 1-form $L\rho \in \Omega^1(LM,\mathfrak{a})$.
Of course some concepts that appear here would need some more explanation - but that's not the point. Let me better point out how gerbes with connection come into the picture. We employ two results from gerbe theory:
1. Associated to every lifting problem posed by a bundle $P$ is an $A$-gerbe over $M$, called the "lifting gerbe" and denoted $\mathcal{G}_P$. This gerbe represents geometrically the obstruction against lifts. Moreover, the actual lifts are in equivalence with trivializations of $\mathcal{G}_P$. The same works if one wants to include connections into the lifting problem. These are results of Murray and Gomi.
2. The category of $A$-gerbes with connection over $M$ is equivalent to a certain category of principal $A$-bundles with connection over $LM$ which are additionally equipped with "fusion products". The equivalence is established by a transgression functor, which has been introduced by Brylinski and McLaughlin. It takes trivializations of gerbes to sections of bundles.
Now, define $\mathcal{L}_P$ as the transgression of $\mathcal{G}_P$. Since transgression is an equivalence of categories, it is a bijections on Hom-sets, and this bijection is exactly the statement of the theorem.
Ok, in order to complete my claim that this is an application, I should probably mention an example where the theorem is useful. That's the case for $spin$ and $spin^c$ structures on manifolds, and I have learned about it from Stephan Stolz and Peter Teichner. In the case of $spin$ structures, $\mathcal{L}_P$ is a $\mathbb{Z}_2$-bundle over $LM$ and plays the role of the orientation bundle of $LM$. Since $\mathbb{Z}_2$ is discrete, all the connections disappear and forms are identically zero. So, the theorem says that isomorphism classes of $spin$ structures on $M$ are in bijection to "fusion-preserving orientations" of $LM$. In the $spin^c$ case, a similar statement follows that additionally includes the scalar curvature of the $spin^c$-structures.
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Sorry for taking so long to reply to this, I fell off of MO during August and forgot to check this. Do you have a reference for this? I'm in the middle of TeXing some gerbe stuff as well... – David Carchedi Oct 20 2010 at 10:13
The story is in my paper "A loop space formulation for geometric lifting problems" (arxiv.org/abs/1007.5373). There you also find the references to papers of Murray, Gomi, Brylinski-McLaughlin and Stolz-Teicher mentioned above. – Konrad Waldorf Oct 20 2010 at 11:10
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Whilst asking this, I nearly forgot that one application does come to mind:
http://www.math.fsu.edu/~aluffi/archive/paper325.pdf
In this paper Behrang Noohi shows how to use topological stacks to calculate the fundamental group of the quotient of a topological space by a group(oid) action by using fixed-point data.
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http://mathoverflow.net/questions/31767/indexing-schemes-of-binary-sequences/31812 | ## Indexing schemes of binary sequences
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I am looking for "low-complexity" indexing methods to enumerate binary sequences of a given length and a given weight.
Formally, let $T_k^n = \{x_1^n \in \{0,1\}^n: \sum_{i=1}^n x_i = k\}$. How to construct a bijective mapping $f: T_k^n \to \{1, 2, \ldots, \binom{n}{k}\}$ such that computing each $f(x_1^n)$ needs small number of operations?
For example, one could do lexicographical ordering, that is, e.g., $0110 < 1010$. Then this gives the following scheme:
$f(x_1^n) = \sum_{k=1}^n x_k \binom{n-k}{w_k}$
where $w_k=\sum_{i=k}^n x_i$. Computing $n$ binomial coefficients can be quite demanding. Any other ideas? Or is it impossible to avoid?
-
## 2 Answers
You want Volume 4, Fascicle 3 of Knuth's The Art of Computer Programming, chapter 7.2.1.3: "Generating All Combinations" - I won't include links because everyone has a favorite online bookseller, but AFAIK all the major ones have it in stock. Highly, highly recommended!
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thanks. will take a look! i heard before that this enumeration problem needs at least $O(n^2)$ binary operations. The above lexicographic enumeration needs to be computed using $n$ operations on a $n$-bit register, which is $O(n^2)$. Does DEK talk about this? – mr.gondolier Jul 13 2010 at 23:29
I think there was some discussion about the complexity of generation, but I'm not sure if your specific issue was discussed - I'll check when I get home. Also, even if you have $O(n^2)$ worst-case performance, keep in mind that amortized performance may be much better (think of enumerating the numbers from $1$ to $2^n$ in a single register - you may have to change up to $n$ bits on any given operation, but the total number of bits changed is still just $2^{(n+1)}$, not $n2^n$) – Steven Stadnicki Jul 14 2010 at 0:21
If you need frequently compute number of sequence you can pre-calculate all $\binom{i}{j}$. This would make you algorithm run in $O(n)$ time and $O(nk)$ memory. – falagar Jul 14 2010 at 5:30
I had some chance to look at this last night - Knuth's primary focus is on iteration, not on generating the indexing you mention, but he has a number of schemes that can iterate over all combinations with very few bit operations per iteration; if you don't explicitly need indices (and my gut instinct is that you're unlikely to), then his algorithms very definitely bear looking into. – Steven Stadnicki Jul 14 2010 at 18:23
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You'll probably want to use the combinatorial number system or combinadic. La Wik has a useful overview and a number of references including Knuth's The Art of Computer Programming
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http://mathhelpforum.com/calculus/156708-question-involving-collection-closed-intervals.html | # Thread:
1. ## Question involving a collection of closed intervals
Let $F$ be the collection of closed intervals $A_n=[1/n, 1-1/n]$ for $n=3,4,5,...$. What do you notice about $\bigcup F$? Is it closed, open, both, or neither?
I'm afraid the wording and available material has got me stumped on this, and being sick at the moment doesn't help.
2. Notice that $\left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1$.
Can either 1 or 0 be included?
3. Originally Posted by Plato
Notice that $\left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1$.
Can either 1 or 0 be included?
I should note, the original question said that $n=1,2,3,...$, but our Prof. told us to change it to $n=3,4,5,...$ when he realized that when $n=1, A_n=[1,0]$ and when $n=2, A_n=[1/2,1/2]$ (or at least that's what I heard him say). I wrote the question word-for-word (besides the change our Prof. gave us), so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.
4. Originally Posted by Runty
so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.
That is the whole point: neither 0 nor 1 can be included.
So what is the union?
5. Originally Posted by Plato
That is the whole point: neither 0 nor 1 can be included.
So what is the union?
I am presently guessing, but here is my answer thus far.
$\bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]$
My guess so far indicates that the union is both open and closed, though I could easily be wrong.
6. Originally Posted by Runty
I am presently guessing, but here is my answer thus far.
$\bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]$
For any $k$ does $1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?$
If not how can the union be $(0,1]?$
7. Originally Posted by Plato
For any $k$ does $1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?$
If not how can the union be $(0,1]?$
Point taken. I guess it's a completely open interval, then.
EDIT: One more thing, I've looked at some work done on this in another forum, for which I will provide the link.
http://www.mymathforum.com/viewtopic...=15805&start=0
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How might I work with the following functions on a list, if the list is a disk based file? Partition[] DeleteDuplicates[] Split[] SortBy[] The file is too ... | {"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": 7, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9036803841590881, "perplexity_flag": "middle"} |
http://mathhelpforum.com/statistics/124584-variance-problem-i-think.html | # Thread:
1. ## variance problem (i think)
Sorry, I was having difficulty with the equation editor on here. I've attached my problem as a picture file.
I need to show that the two equations are equal, and I'm not sure where to begin.
Thanks!
Attached Thumbnails
2. Assuming that $\overline{z}$ is the conjugate of the complex number $z$, that statement is false.
$\overline{({z-\overline{z}})^2~}=\overline{z~}^2 -2\overline{z}z+z^2$.
Note that $\overline{\overline{z}}=z$ and $\overline{z^2}=\overline{z}^2$
3. Originally Posted by chemekydie
Sorry, I was having difficulty with the equation editor on here. I've attached my problem as a picture file.
I need to show that the two equations are equal, and I'm not sure where to begin.
Thanks!
$\overline{(x-\overline{x})^2}=\overline{x^2-2x\overline{x}+\overline{x}^2}$ $=\overline{x^2}-2\overline{x}\overline{x}+\overline{x}^2$
etc.
CB
4. @Plato - x-bar is not a conjugate
@CaptainBlack -
In your process, what happens to the x double bar? How does that get factored out?
5. Originally Posted by chemekydie
@Plato - x-bar is not a conjugate
@CaptainBlack -
In your process, what happens to the x double bar? How does that get factored out?
OK, then what is $\overline{x}?$
6. Originally Posted by Plato
OK, then what is $\overline{x}?$
Expectation of $x$ (the clue is in the thread title)
CB
7. Originally Posted by chemekydie
@Plato - x-bar is not a conjugate
@CaptainBlack -
In your process, what happens to the x double bar? How does that get factored out?
$\overline{x}$ is just a number its expected value is itself :
$\overline{(\overline{x})}=\overline{x}$
CB
8. x-bar is the mean.
I was under the assumption that x double bar is the average of the means. I'm not aware that x double bar can be transformed back to regular x bar.
Am I missing something?
9. Originally Posted by chemekydie
x-bar is the mean.
I was under the assumption that x double bar is the average of the means. I'm not aware that x double bar can be transformed back to regular x bar.
Am I missing something?
x double bar as you call it is the mean of the mean, but the mean is just a number, its mean is itself.
To see this just look at your definition of a mean.
CB
10. Originally Posted by CaptainBlack
$\overline{(x-\overline{x})^2}=\overline{x^2-2x\overline{x}+\overline{x}^2}$ $=\overline{x^2}-2\overline{x}\overline{x}+\overline{x}^2$
etc.
CB
I'm getting a $+\overline{x}^2$ instead of negative. And I'm not sure how to factor out the 2.
11. Originally Posted by chemekydie
I'm getting a $+\overline{x}^2$ instead of negative. And I'm not sure how to factor out the 2.
$\overline{-2x\overline{x}}=-2\overline{x}~\overline{\overline{x}}=-2\overline{x}~\overline{x}=-2\overline{x}^2$ | {"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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.962039053440094, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/1659/locating-the-quadrant-containing-a-point-on-an-n-sphere | # Locating the quadrant containing a point on an n-sphere
Suppose I have a point $x \in \mathbb{R}^n$ on an n-sphere. Suppose I divide the n-sphere into 4 sections (I think this makes sense in $n$ dimensions), how do I know which section $x$ lies on?
-
6
No, it doesn't make sense in n dimensions unless n=2. What are the 4 quadrants in 3 dimensions? You can talk of the 8 octants instead, and the 2^n orthants in n dimensions (it's an $(n-1)$-sphere, BTW), and if that's what you want, then the answer to your question is simply to look at the signs of the coordinates in x. – ShreevatsaR Aug 5 '10 at 20:13
1
@Shreevatsa does it make sense to say "if I split a sphere into 8 equal hemispheres" ? – Jonathan Fischoff Aug 5 '10 at 20:16
@Shreevatsa: In 3 dimensions, you have 2 axis aligned planes dividing the sphere into 4 quadrants (one plane divides it into two hemispheres and the other divides each hemisphere). – Jacob Aug 5 '10 at 20:17
@Shreevatsa: Also, is (n-1) in LaTeX? If so, how did you format it? I couldn't find any formatting tips regarding LaTeX! – Jacob Aug 5 '10 at 20:19
2
@Jacob: Actually if you wait for 4 seconds the equations will appear. – KennyTM Aug 5 '10 at 20:24
show 4 more comments
## 1 Answer
This is just a rephrasing of ShreevastasR's answer; no credit to me. It does make sense to divide an $n$-sphere into quadrants, as you explain in $\mathbb{R}^3$: partition by two coordinate planes. But then deciding which quadrant is, as ShreevastasR says, simply looking at the signs of the coordinates of $x$. If $x_1$ and $x_2$ are both positive, you are in the first, $++$, quadrant; if $x_1$ is negative and $x_2$ positive, you are in the second, $-+$, quadrant. And so on. If instead you partition the sphere into $2^n$ orthants, then you consider all the signs of the coordinates of $x$.
-
My understanding from you question is that you have $x$ in your hands, presumably in coordinate form. So you look at the first coordinate, the second, the third, etc. I guess I don't understand your question! – Joseph O'Rourke Aug 5 '10 at 21:20 | {"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": 16, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9208903908729553, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/70611?sort=oldest | ## Reference for proof that $C_b^* = rba$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The following theorem seems to have folk status:
The topological dual of the space $C_b(X)$ of bounded continuous functions on a topological space $X$ is isomorphic to the space $rba(X)$ of finite, regular, finitely additive Borel set functions.
This fact is often mentioned (for instance in the answer to http://mathoverflow.net/questions/44183/dual-of-bounded-uniformly-continuous-functions) but I'm having great difficulty actually finding a reference. Often Dunford & Schwartz is mentioned as a reference; D&S defines $rba$, but doesn't prove the connection to the dual of $C_b$. Hildebrandt 1934 proves a characterization in terms of limits of Stieltjes integrals, but that is still some steps away from the characterization above. I haven't been able to find anything coming closer than this.
Does anyone know of a real proof of this statement? Am I maybe overlooking a very simple proof?
-
1
What's wrong with the discussion in IV.6, p.261ff in DS? They prove the duality at least for $X$ normal. – Theo Buehler Jul 18 2011 at 11:51
What hypotheses (if any) are you putting on your space $X$? – Yemon Choi Jul 18 2011 at 12:56
Perhaps confusion: Dunford & Schwartz notation $C(X)$ is the space of bounded continuous functions on $X$. The dual (for $X$ normal, but probably completely regular will also work) is Theorem IV.6.2, as Theo says. – Gerald Edgar Jul 18 2011 at 13:47
Thanks @Theo and @Gerard - I obviously overlooked Theorem IV.6.2, possibly because of the different notation. Thanks for pointing that out! – Mark Peletier Aug 14 2011 at 18:14
## 2 Answers
The topological dual of the space of bounded continuous functions on a topological space X is isomorphic to the space of finite, zero set regular, finitely additive Baire set functions; see: R. F. Wheeler, A survey of Baire measures and strict topologies, Exposition. Math. 2 (1983), 97–190 (a proof is on pp. 115-117).
-
1
The point being: in exotic spaces, if you use Baire sets (not Borel sets) and zero sets (not closed sets) you get what you want. A reference for more discussion: Gillman & Jerison, Rings of Continuous Functions. – Gerald Edgar Aug 11 2011 at 15:36
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
In the answer you mentioned, the space $X$ is metrizable, hence normal, so the proof from Dunford & Schwartz that appeared in the comments is aplicable in that case.
- | {"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": 12, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9244573712348938, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/295728/whats-the-solution-of-the-functional-equation/295759 | # What's the solution of the functional equation?
I need help with this:
"Find functions $f$, $g : \mathbb{Z} \rightarrow \mathbb{Z}$, knowing that $g$ is injective and such that: $$f(g(x)+y) = g(f(x)+x), \mbox{ for all } x, y \in \mathbb{Z}.$$
-
Did you maybe mean $g(f(x)+y)$, or something to that effect? Otherwise set $x=0$ to get $f(g(0)+y)=g(f(0))$, so that $f$ is constant. – Alex Feb 5 at 20:46
## 1 Answer
There must be an error on the question. As the second member doesn't depend on $y$ then $f$ must be a constant $c$.
$f(g(0) + y) = g(f(0)) = c, \mbox{ for all } y \in \mathbb{Z}$
Then the first member of the equation is a constant, $f(g(x) + y)=c$, so $g$ must also be constant, $c = g(c + x) \mbox{ for all } x \in \mathbb{Z}$.
Maybe you mean $f(g(x)+y) = g(f(y)+x), \mbox{ for all } x, y \in \mathbb{Z}$ ?
- | {"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": 15, "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}, "config": {"markdown_headings": true, "markdown_code": false, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9544192552566528, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/107192?sort=oldest | ## Dehn function for undistorted subgroups of a product of free groups
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $G$ be a finitely generated subgroup of a product of two finite rank free groups $F_m \times F_n$. If there is a Lipschitz retraction $F_m \times F_n \to G$ with respect to word metrics, then $G$ is undistorted in $F_m \times F_n$, and the Dehn function of $G$ has a quadratic upper bound.
Suppose now that we require only that $G$ is undistorted in $F_m \times F_n$. Is it still true that the Dehn function of $G$ has a quadratic upper bound?
-
Lee: All subgroups G are either free or direct products of free groups or are not finitely-presented. Among latter, I do not know any examples of undistorted ones, do you? In fact, the only undistorted subgroups of semi simple Lie groups that I know are retracts, non-uniform lattices and Anosov (in the sense of Guichard and Wienhard) hyperbolic groups (the latter could in the end turn out to be retracts, this is unknown). – Misha Sep 14 at 17:32
...I forgot one more class of undistorted subgroups: Some polycyclic groups which are peripheral subgroups of non-uniform lattices. However, one should regard them as lattices in solvable Lie groups. – Misha Sep 14 at 17:41
why you closed my question, please reopen it – Hassan Jolany Dec 21 at 17:47
## 2 Answers
The Bieri-Stallings subgroup of $F_n\times F_n$ is undistorted and finitely generated, but not finitely-presented, so in some sense it has an infinite Dehn function. It's the kernel of the map $F_n\times F_n\to \mathbb{Z}$ which sends each generator to 1, and it's generated by elements of the form $g_ih_j^{-1}$ where $g_i$ and $h_j$ are generators of the two different factors.
-
1
Welcome to MO Robert! – Andy Putman Sep 14 at 18:23
How does one know, or find a reference for, the statement that this subgroup is undistorted? – Lee Mosher Sep 14 at 18:32
2
There's a proof for the theorem that Yves mentions in Olshanskii, Sapir, "Length and area functions on groups and quasi-isometric Higman embeddings" (Theorem 2). For this subgroup, you can construct words explicitly -- if $$w=w_1(g_1,\dots, g_n)w_2(h_1,\dots,h_n)$$ and $w$ lies in the kernel, we can rewrite $w$ as $$w=w_1(g_1 h_1^{-1},\dots, g_n h_1^{-1})(h_1 g_1^{-1})^k w_2(h_1g_1^{-1},\dots,h_n g_1^{-1})$$ where $k$ is the sum of the exponents in $w_1$. This is a product of generators of the subgroup and its length increases by at most a constant. – Robert Young Sep 14 at 19:37
That's pretty. – Lee Mosher Sep 14 at 21:18
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The answer is no: actually by Baumslag-Roseblade (JLMS 1984) either $G$ is commensurable to a product of free groups (hence has linear or quadratic Dehn function), or is not finitely presented (so the Dehn function is infinite, or not defined, as you wish). The latter case occurs if $H$ is an infinite word hyperbolic group and $f:F_m\to H$ is a non-bijective surjection and $H$ is the fibre product $[(g,h)\in F_m\times F_m:f(g)=f(h)]$.
-
Sorry, I can't type braces so I put brackets. – Yves Cornulier Sep 14 at 17:44
@Yves, I have the same question here as for Robert Young's answer: how does one obtain nondistortion in these examples? – Lee Mosher Sep 14 at 18:36
In general the Dehn function of $H$ is equivalent to the distortion of $G$. This is not completely formal (there's a little Van Kampen diagram cuisine) but in the case of $H=\mathbf{Z}$ it's simple to verify by hand. I saw the general result written somewhere but I can't remember right now. – Yves Cornulier Sep 14 at 18:42 | {"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": 26, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.931372344493866, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?s=ea56b5c1a20722da15d3a5427a9a5a88&p=4085070 | Physics Forums
## Coexistence in phase diagram
In a nutshell, I am trying to see how to derive the conditions of coexistence between a solid and a liquid from a phase diagram.
The situation is as follows:
Consider a mixture of hard spheres of diameter σ. The potential energy
for a hard sphere system is given by
[itex]\beta U(r) = 0 (r > \sigma)
[/itex]
[itex]
∞ (r ≤ \sigma)
[/itex]
The packing fraction (η) of the system is the amount of space occupied
by the particles.
The equation of state for the hard sphere fluid is approximately
[itex]
\frac{P_{liq}V}{Nk_{B}T}= \frac{1+ \eta + \eta^2 - \eta^3}{ (1 - \eta)^3 }
[/itex]
Another similar equation is given for $P_{sol}$, the pressure in the solid state.
By integration I managed to calculated the free energy as a function of the packing density, using given boundary conditions. This resulted in the following diagram:
Here the free energy is plotted against the packing density. The red line corresponds to the solid phase and the blue line to the liquid phase.
In general I would think one could calculate the minima of the free energy and then draw a common tangent line between them, but in this case there doesn't seem to be any minima. Also, it appears that the free energy of the solid phase is lower than that of the liquid phase even for low densities. Is this even possible? Or should I conclude that my plots are incorrect?
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> Nanocrystals grow from liquid interface>> New insights into how materials transfer heat could lead to improved electronics
Recognitions: Homework Help Science Advisor Are you sure those equations are meaningful in the full [0,1] range? Especially: How do you get a packing density η > 0.741? (I think it is not an accident that the spike of the solid line is there). There are materials which do not have a liquid phase at a specific temperature, independent of the packing density.
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http://en.wikipedia.org/wiki/Magnetic_domain | # Magnetic domain
Several grains of NdFeB with magnetic domains made visible with a Kerr microscope. The domains are the light and dark stripes visible within each grain.
A magnetic domain is a region within a magnetic material which has uniform magnetization. This means that the individual magnetic moments of the atoms are aligned with one another and they point in the same direction. When cooled below a temperature called the Curie temperature, the magnetization of a piece of ferromagnetic material spontaneously divides into many small regions called magnetic domains. The magnetization within each domain points in a uniform direction, but the magnetization of different domains may point in different directions. Magnetic domain structure is responsible for the magnetic behavior of ferromagnetic materials like iron, nickel, cobalt and their alloys, ferrites etc. such as the formation of permanent magnets. The regions separating magnetic domains are called domain walls, where the magnetization rotates coherently from the direction in one domain to that in the next domain. The study of magnetic domains is called micromagnetics.
## Development of domain theory
Magnetic domain theory was developed by French physicist Pierre-Ernest Weiss[1] who in 1906 suggested existence of magnetic domains in ferromagnets.[2] He suggested that large number of atomic magnetic moments (typically 1012-1018)[citation needed] were aligned parallel. The direction of alignment varies from domain to domain in a more or less random manner although certain crystallographic axis may be preferred by the magnetic moments, called easy axes. Weiss still had to explain the reason for the spontaneous alignment of atomic moments within a ferromagnetic material, and he came up with the so-called Weiss mean field : he assumed that a given magnetic moment in a material experienced a very high effective magnetic field due to the magnetization of its neighbors. In the original Weiss theory the mean field was proportional to the bulk magnetization M, so that
$H_e = \alpha\ M$
where $\alpha\$ is the mean field constant. However this is not applicable to ferromagnets due to the variation of magnetization from domain to domain. In this case, the interaction field is
$H_e = \alpha\ M_s$
Where $M_s$ is the saturation magnetization at 0K.
Later, the quantum theory made it possible to understand the microscopic origin of the Weiss field. The exchange interaction between localized spins favored a parallel (in ferromagnets) or an anti-parallel (in anti-ferromagnets) state of neighboring magnetic moments
## Domain structure
How dividing a ferromagnetic material into magnetic domains reduces the magnetostatic energy
### Why domains form
The reason a piece of magnetic material such as iron spontaneously divides into separate domains, rather than exist in a state with magnetization in the same direction throughout the material, is to minimize its internal energy.[3] A large region of ferromagnetic material with a constant magnetization throughout will create a large magnetic field extending into the space outside itself (diagram a, right). This requires a lot of magnetostatic energy stored in the field. To reduce this energy, the sample can split into two domains, with the magnetization in opposite directions in each domain (diagram b right). The magnetic field lines pass in loops in opposite directions through each domain, reducing the field outside the material. To reduce the field energy further, each of these domains can split also, resulting in smaller parallel domains with magnetization in alternating directions, with smaller amounts of field outside the material.
The domain structure of actual magnetic materials does not usually form by the process of large domains splitting into smaller ones as described here. When a sample is cooled below the Curie temperature, for example, the equilibrium domain configuration simply appears. But the description of domains splitting is often used to reveal the energy tradeoffs in domain formation.
### Size of domains
A domain which is too big is unstable, and will divide into smaller domains. But a small enough domain will be stable and will not split, and this determines the size of the domains created in a material. This size depends on the balance of several energies within the material.[3] Each time a region of magnetization splits into two domains, it creates a "domain wall" between the domains, where magnetic dipoles (molecules) with magnetization pointing in different directions are adjacent. The exchange interaction which creates the magnetization is a force which tends to align nearby dipoles so they point in the same direction. Forcing adjacent dipoles to point in different directions requires energy. Therefore creating a domain wall requires extra energy, called the "exchange energy", which is proportional to the area of the wall.
Thus the net amount that the energy is reduced when a domain splits is equal to the difference between the magnetic field energy saved, and the additional energy of the domain wall created. The field energy saved is proportional to the cube of the domain size, while the domain wall energy is proportional to the square of the domain size. So as the domains get smaller, the net energy saved by splitting decreases. The domains keep dividing into smaller domains until the energy cost of creating an additional domain wall is just equal to the field energy saved. Then the domains of this size are stable. In most materials the domains are microscopic in size, around 10-4 - 10-6 m.
### Magnetic anisotropy
Micrograph of surface of ferromagnetic material, showing the crystal grains, each divided into several domains parallel to its "easy" axis of magnetization, with the magnetization in alternating directions (red and green areas).
Animation showing how magnetostriction works. A changing external magnetic field causes the magnetic dipoles to rotate, changing the dimensions of the crystal lattice.
An additional way for the material to further reduce its magnetostatic energy is to form domains with magnetization at right angles to the other domains (diagram c, right), instead of just in opposing parallel directions.[3] These domains, called flux closure domains, allow the field lines to turn 180° within the material, forming closed loops entirely within the material, reducing the magnetostatic energy to zero. However, forming these domains incurs two additional energy costs. First, the crystal lattice of most magnetic materials has magnetic anisotropy, which means it has an "easy" direction of magnetization, parallel to one of the crystal axes. Changing the magnetization of the material to any other direction takes additional energy, called the "magnetocrystalline anisotropy energy".
### Magnetostriction
The other energy cost to creating domains with magnetization at an angle to the "easy" direction is caused by the phenomenon called magnetostriction.[3] When the magnetization of a piece of magnetic material is changed to a different direction, it causes a slight change in its shape. The change in magnetic field causes the magnetic dipole molecules to change shape slightly, making the crystal lattice longer in one dimension and shorter in other dimensions. However, since the magnetic domain is "squished in" with its boundaries held rigid by the surrounding material, it cannot actually change shape. So instead, changing the direction of the magnetization induces tiny mechanical stresses in the material, requiring more energy to create the domain. This is called "magnetoelastic anisotropy energy".
To form these closure domains with "sideways" magnetization requires additional energy due to the aforementioned two factors. So flux closure domains will only form where the magnetostatic energy saved is greater than the sum of the "exchange energy" to create the domain wall, the magnetocrystalline anisotropy energy, and the magnetoelastic anisotropy energy. Therefore most of the volume of the material is occupied by domains with magnetization either "up" or "down" along the "easy" direction, and the flux closure domains only form in small areas at the edges of the other domains where they are needed to provide a path for magnetic field lines to change direction (diagram c, above).
### Grain structure
The above describes magnetic domain structure in a perfect crystal lattice, such as would be found in a single crystal of iron. However most magnetic materials are polycrystalline, composed of microscopic crystalline grains. These grains are not the same as domains. Each grain is a little crystal, with the crystal lattices of separate grains oriented in random directions. In most materials, each grain is big enough to contain several domains. Each crystal has an "easy" axis of magnetization, and is divided into domains with the axis of magnetization parallel to this axis, in alternate directions.
### "Magnetized" states
It can be seen that, although on a microscopic scale almost all the magnetic dipoles in a piece of ferromagnetic material are lined up parallel to their neighbors in domains, creating strong local magnetic fields, energy minimization results in a domain structure that minimizes the large-scale magnetic field. The domains point in different directions, confining the field lines to microscopic loops between neighboring domains, so the combined fields cancel at a distance. Therefore a bulk piece of ferromagnetic material in its lowest energy state has little or no external magnetic field. The material is said to be "unmagnetized".
However, the domains can also exist in other configurations in which their magnetization mostly points in the same direction, creating an external magnetic field. Although these are not minimum energy configurations, due to a phenomenon where the domain walls become "pinned" to defects in the crystal lattice they can be local minimums of the energy, and therefore can be very stable. Applying an external magnetic field to the material can make the domain walls move, causing the domains aligned with the field to grow, and the opposing domains to shrink. When the external field is removed, the domain walls remain pinned in their new orientation and the aligned domains produce a magnetic field. This is what happens when a piece of ferromagnetic material is "magnetized" and becomes a permanent magnet.
## Landau-Lifshitz energy equation
Electromagnetic dynamic magnetic domain motion of grain oriented electrical silicon steel
Moving domain walls in a grain of silicon steel caused by an increasing external magnetic field in the "downward" direction, observed in a Kerr microscope. White areas are domains with magnetization directed up, dark areas are domains with magnetization directed down.
The contributions of the different internal energy factors described above is expressed by the free energy equation proposed by Lev Landau and Evgeny Lifshitz in 1935 [1], which forms the basis of the modern theory of magnetic domains. The domain structure of a material is the one which minimizes the Gibbs free energy of the material. For a crystal of magnetic material, this is the Landau-Lifshitz free energy, E, which is the sum of these energy terms:[4]
$E = E_{ex} + E_D + E_{\lambda} + E_k + E_H\,$
where
• Eex is exchange energy: This is the energy due to the exchange interaction between magnetic dipole molecules in ferromagnetic, ferrimagnetic and antiferromagnetic materials. It is lowest when the dipoles are all pointed in the same direction, so it is responsible for magnetization of magnetic materials. When two domains with different directions of magnetization are next to each other, at the domain wall between them magnetic dipoles pointed in different directions lie next to each other, increasing this energy. This additional exchange energy is proportional to the total area of the domain walls.
• ED is magnetostatic energy: This is a self-energy, due to the interaction of the magnetic field created by the magnetization in some part of the sample on other parts of the same sample. It is dependent on the volume occupied by the magnetic field extending outside the domain. This energy is reduced by minimizing the length of the loops of magnetic field lines outside the domain. For example, this tends to encourage the magnetization to be parallel to the surfaces of the sample, so the field lines won't pass outside the sample. Reducing this energy is the main reason for the creation of magnetic domains.
• Eλ is magnetoelastic anisotropy energy: This energy is due to the effect of magnetostriction, a slight change in the dimensions of the crystal when magnetized. This causes elastic strains in the lattice, and the direction of magnetization that minimizes these strain energies will be favored. This energy tends to be minimized when the axis of magnetization of the domains in a crystal are all parallel.
• Ek is magnetocrystalline anisotropy energy: Due to its magnetic anisotropy, the crystal lattice is "easy" to magnetize in one direction, and "hard" to magnetize in others. This energy is minimized when the magnetization is along the "easy" crystal axis, so the magnetization of most of the domains in a crystal grain tend to be in either direction along the "easy" axis. Since the crystal lattice in separate grains of the material is usually oriented in different random directions, this causes the dominant domain magnetization in different grains to be pointed in different directions.
• EH is Zeeman energy: This is energy which is added to or subtracted from the magnetostatic energy, due to the interaction between the magnetic material and an externally applied magnetic field. It is proportional to the negative of the cosine of the angle between the field and magnetization vectors. Domains with their magnetic field oriented parallel to the applied field reduce this energy, while domains with their magnetic field oriented opposite to the applied field increase this energy. So applying a magnetic field to a ferromagnetic material generally causes the domain walls to move so as to increase the size of domains lying mostly parallel to the field, at the cost of decreasing the size of domains opposing the field. This is what happens when ferromagnetic materials are "magnetized". With a strong enough external field, the domains opposing the field will be swallowed up and disappear; this is called saturation.
Rotation of orientation and increase in size of magnetic domains to an externally applied field (compare Zeeman energy).
Some sources define a wall energy EW equal to the sum of the exchange energy and the magnetocrystalline anisotropy energy, which replaces Eex and Ek in the above equation.
A stable domain structure is a magnetization function M(X), considered as a continuous vector field, which minimizes the total energy E throughout the material. To find the minimums a variational method is used, resulting in a set of nonlinear differential equations, called Brown's equations after William Fuller Brown Jr. Although in principle these equations can be solved for the stable domain configurations M(X), in practice only the simplest examples can be solved. Analytic solutions do not exist, and numerical solutions calculated by the finite element method are computationally intractable because of the large difference in scale between the domain size and the wall size. Therefore micromagnetics has evolved approximate methods which assume that the magnetization of dipoles in the bulk of the domain, away from the wall, all point in the same direction, and numerical solutions are only used near the domain wall, where the magnetization is changing rapidly.
## Observing domains
There are a number of microscopy methods which can make the magnetization at a surface of a magnetic material visible, revealing the magnetic domains. Each method has a different application because not all domains are the same. In magnetic materials, domains can be circular, square, irregular, elongated, and striped, all of which have varied sizes and dimensions. Large domains, within the range of 25-100 micrometers can be easily seen by Kerr microscopy, which uses the magneto-optic Kerr effect, which is the rotation of the polarization of light reflected from a magnetized surface. Smaller domains, down to the scale of a few nanometers, can be viewed by the use of magnetic force microscopy.
Magnetooptical images of different domain structures
Domain structure of a shape-memory alloy (recorded using CMOS-MagView)
Domain structure of an examplary meander domain (recorded using CMOS-MagView)
Domain structure of an examplary magnetic bubble domain (recorded using CMOS-MagView)
## References
1. P. Weiss (1906) La variation du ferromagnetisme du temperature, Comptes Rendus, 143, p.1136-1149, cited in Cullity, 2008, p.116
2. Cullity; C. D. Graham (2008). Introduction to Magnetic Materials, 2nd ed.. New York: Wiley–IEEE. p. 116. ISBN 0-471-47741-9. .
3. ^ a b c d Feynman, Richard P.; Robert B. Leighton, Matthew Sands (1963). The Feynman Lectures on Physics, Vol. I. US: California Inst. of Technology. pp. 37.5–37.6. ISBN 0-201-02117-XP.
4. Carey R., Isaac E.D., Magnetic domains and techniques for their observation, The English University Press Ltd, London, (1966).
• Jiles, David (1998). Introduction to magnetism and magnetic materials. London: Chapman & Hall. ISBN 0-412-79860-3. | {"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": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9017481803894043, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/255624/borel-sets-including-infty | # Borel sets including $\infty$
Can someone give me a hint how to show the equality $\{ B \cup F \mid B\in \sigma(\{(-\infty,a) \mid a\in \mathbb{R}\}),\ F\subseteq\{-\infty,+\infty\} \}=\sigma(\{ [-\infty,a) \mid a \in \mathbb{R} \})$ ?
The set $\sigma(\{(-\infty,a) \mid a\in \mathbb{R}\})$ represents the usual Borel sets on the real line (and $\{(-\infty,a)\mid a\in \mathbb{R}\}$ is its generator).
What puzzles me is the fact that for example $(-\infty,1)\cup \{+\infty \}$ is in the set on the LHS of the equality, but I don't know how to produce this set on the RHS of the equality (making me wonder, if the above equality is true.)
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Do want $F \subseteq \{ - \infty , + \infty \}$ in the definition on the left-hand-side? (Otherwise $[ - \infty , + \infty ]$ is not in the collection defined on the left-hand-side, but it is in the collection defined on the right-hand-side.) – Arthur Fischer Dec 10 '12 at 18:16
@ArthurFischer Ah, yes. Sorry. – user52145 Dec 10 '12 at 18:18
## 2 Answers
I will denote by $\overline{\mathbb{R}}$ the set $\mathbb{R} \cup \{ - \infty , + \infty \}$.
To produce $( - \infty , 1 ) \cup \{ + \infty \}$ on the right-hand-side, note that
• for all $a \in \mathbb{R}$ we have that $[ a , + \infty ] = \overline{\mathbb{R}} \setminus [ - \infty , a )$, and so these sets belong to the $\sigma$-algebra on the right-hand-side.
Then $$( - \infty , 1 ) \cup \{ + \infty \} = \left( \bigcup_{n=1}^\infty [ - \infty , 1-{\textstyle \frac{1}{n}} ] \cap \bigcup_{n=1}^\infty [ -n , + \infty ] \right) \cup \bigcap_{n=1}^\infty [ n , + \infty ].$$
To show equality, you can proceed as follows:
1. Show that the collection on the LHS is itself a $\sigma$-algebra. (Showing closure under countable unions should be quite easy, and complementation will follow after noting that $\overline{\mathbb{R}} \setminus ( B \cup F ) = ( \mathbb{R} \setminus B ) \cup ( \{ - \infty , + \infty \} \setminus F)$ for $B \subseteq \mathbb{R}$ and $F \subseteq \{ - \infty , + \infty \}$.) (This can be seen as taking the "join" of two $\sigma$-algebras on disjoint sets.)
2. To show RHS $\subseteq$ LHS: Simply note the every set in the generating set for the RHS is an element of the LHS. (Should be very easy.) Since LHS is a $\sigma$-algebra, the set inclusion is done.
3. To show LHS $\subseteq$ RHS: Show, using ideas similar to the example above, that every set of the form $( - \infty , a )$ belongs to the RHS, as well as $\{ - \infty \}$ and $\{ + \infty \}$. (From this you will be able to show that all Borel subsets of $\mathbb{R}$ belong to RHS, as well as all subsets of $\{ - \infty , + \infty \}$.)
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To see that this the RHS contains the LHS:
It suffices to show that this set contains $\infty$, $-\infty$ and the open intervals (since it is by assumption a $\sigma$-algebra)
$-\infty$ :
It contains $[- \infty, -n)$ for all $n$ and is closed under intersection.
$\infty$:
It is closed under compliment so contains $[n, \infty]$ for every $n$ and again is closed under intersection.
$(a,b)$ :
It contains $[-\infty, a]$ by taking the intersection of $[-\infty, a + \frac{1}{n})$ and it contains $[-\infty, b)$. By virtue of being a $\sigma$-algebra it is then closed under the difference of these two sets, so it contains $(a,b)$ .
To see that the LHS contains the RHS:
Since the LHS contains $[-\infty, a)$ for all $a$, you just need to show that the LHS is actually a $\sigma$-algebra.
You should try do this, all you need to do is show that the set is closed under countable unions and compliments. You should do this using the fact that the Borel sets are a $\sigma$-algebra, plus possibly some cases based on if you include $-\infty, +\infty$ or both.
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http://mathoverflow.net/questions/58856/advances-and-difficulties-in-effective-version-of-thue-roth-siegel-theorem/58877 | ## Advances and difficulties in effective version of Thue-Roth-Siegel Theorem
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A fundamental result in Diophantine approximation, which was largely responsible for Klaus Roth being awarded the Fields Medal in 1958, is the following simple-to-state result:
If $\alpha$ is a real algebraic number and $\epsilon > 0$, then there exists only finitely many rational numbers $p/q$ with $q > 0$ and $(p,q) = 1$ such that $$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^{2 + \epsilon}}$$ This result is famous for its vast improvement over previous results by Thue, Siegel, and Dyson and its ingenious proof, but is also notorious for being non-effective. That is, the result nor its (original) proof provides any insight as to how big the solutions (in $q$) can be, if any exists at all, or how many solutions there might be for a given $\alpha$ and $\epsilon$.
I have come to understand that to date no significant improvement over Roth's original proof has been made (according to my supervisor), and that the result is still non-effective. However, I am not so sure why it is so hard to make this result effective. Can anyone point to some serious attempts at making this result effective, or give a pithy explanation as to why it is so difficult?
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What about the converse? That is, for any fixed $r > 2$, consider the set $S$ of polynomials in $\mathbb{Z}[X]$ which have a real root $\alpha$ for which there exists a rational approximation $\vert\alpha -p/q\vert < q^{-r}$. Is it plausible that $S$ might not be computable? Maybe, even, it could be reduced to the Halting problem. Has this ever been considered? – George Lowther Mar 18 2011 at 23:32
Alternatively, similar to the case of the rank of an elliptic curve and the Birch and Swinnerton-Dyer conjecture, are there any conjectures which are generally believed to be true and would lead to an effective Thue-Siegel-Roth theorem? – George Lowther Mar 18 2011 at 23:47
Well, Felipe's response answers my question. An effective Thue-Siegel-Roth theorem is implied by the abc conjecture. – George Lowther Mar 19 2011 at 2:49
## 5 Answers
The non-effectivity, as far as I understand, is already present in Thue's Theorem, thus to understand it, one can look a the proof of the latter. The issue is roughly that, to show that there are not many "close rational approximations" $p/q$, one starts with the assumption that there exists one very close one $p_0/q_0$, and show that this very good approximation "repulses" or excludes other similarly or better ones. This of course doesn't work if the first $p_0/q_0$ doesn't exist... but such an assumption also gives the result! The ineffectivity is that we have no way of knowing which of the two alternatives has led to the conclusion.
There is a well-known analogy with the Siegel (or Landau-Siegel) zero question in the theory of Dirichlet $L$-functions. Siegel -- and it is certainly not coincidental that this is the same Siegel as in Thue--Siegel--Roth, though Landau did also have crucial ideas in that case -- proved an upper bound for real-zeros of quadratic Dirichlet $L$-functions by (1) showing that if there is one such $L$-function with a zero very close to $1$, then this "repulses" the zeros of all other quadratic Dirichet $L$-functions (this phenomenon is fairly well-understood under the name of Deuring-Heilbronn phenomenon), thus obtaining the desired bound; (2) arguing that if the "bad" $L$-function of (1) did not exist, then one is done anyway.
Here the ineffectivity is clear as day: the "bad" character of (1) is almost certainly non-existent, because it would violate rather badly the Generalized Riemann Hypothesis. But as far as we know today, we have to take into account the possibility of the existence of these bad characters... a possibility which however does have positive consequences, like Siegel's Theorem...
(There's much more to this second story; an entertaining account appeared in an article in the Notices of the AMS one or two years ago, written by J. Friedlander and H. Iwaniec.)
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Denis, are you referring to “What is … the parity phenomenon?” (ams.org/notices/200907/rtx090700817p.pdf)? – L Spice Jul 3 2011 at 3:48
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
This is a kind of a late response, but the OP said "That is, the result nor its (original) proof provides any insight as to how big the solutions (in q) can be, if any exists at all, or how many solutions there might be for a given α and ϵ," and no one has addressed his comment about how many solutions. In fact, Roth's proof combined with a simple gap principle (and a lot of bookkeeping) gives a completely explicit upper bound for the number of solutions as a function of ϵ and the height of α. One can find versions of this in various places, including my paper: A quantitative version of Siegel's theorem: Integral points on elliptic curves and Catalan curves J. Reine Angew. Math. 378 (1987), 60-100. I have a vague recollection that Davenport may have been the first to point this out (maybe just for Thue or Siegel's theorem). There are also deep generalizations giving upper bounds for the number of exceptional subspaces in Schmidt's Subspace Theorem, see for example: A quantitative version of the absolute subspace theorem, J.-H. Evertse and H. P. Schlickewei, J. Reine Angew. Math. 548 (2002), 21–127.
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One way of looking at the issue is this: it is quite easy to transform the question of good rational approximations to algebraic numbers into a question about integral points on certain affine curves (Thue equations). Now the same issues come up: are there any integer points at all, are there only a finite number, and if so can you find them effectively? The finiteness here was taken care of by Siegel, when the genus is > 1. That was an ineffective proof also (sure, it came down to Siegel's version of TSR).
When Baker's method came along in the 1960s, some diophantine problems on curves of this general type became solvable, not just in theory (because Baker's methods are effective) but also in practice. Those were, however, examples where something special was working in our favour.
It is important to understand that while the proof method of Roth simply won't allow effectiveness, Baker's big advance was to remedy that defect by transcendence techniques. Baker's method has given some rather slim general effective results on TSR. Considering that Roth and Baker were both students of Davenport, I think it can be imagined that the techniques were looked over thoroughly. Decades have now gone by.
There isn't a slick answer here, I think. It is typical of tough research problems that we can say that a real advance will depend on a new idea, about which we don't have much clue right now.
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Others have mentioned the Baker-Feldman theorem and similar results coming from transcendence, which is the major source of weak effective general bounds in diophantine approximation. There is also the following paper, which deals with some special cases:
E. Bombieri, AJ van der Poorten, and JD Vaaler, Effective measures of irrationality for cubic extensions of number fields, Ann. Scuola Norm. Sup. Pisa Cl. Sci. 23 (1996), 211-248.
I don't think it is possible to adequately survey a big area of research in a MO post. If we knew why it is difficult to make it effective we might be able to prove something.
I should also add that both the ABC conjecture and Vojta's conjecture (which is a generalization of ABC, I guess) imply effective (perhaps up to some constant) versions of Roth's theorem, so we kind of know what to expect.
BTW this question is a duplicate of http://mathoverflow.net/questions/58569/question-related-to-diophantine-approximations-and-roths-theorem/
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The techniques are often some sort of Padé approximation, or can be put in a framework of that. The work of the Chudnovskys had a linear diff equation in the background. This gives some cases that "work" I think, but I am no expert. springerlink.com/content/j02t25105r35g171 – Junkie Mar 19 2011 at 4:44
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@Felipe: Do you mean that an effective ABC conjecture implies an effective version of Roth's theorem? If one only knows, say, that $\max(|a|,|b|,|c|) < K*N(abc)^2$, for some constant $K$, but no bound for $K$ is effectively known, can you deduce an effective version of Roth's theorem? One can certainly conceive of a Diophantine approximation style proof of the ABC conjecture that would lead to an ineffective constant $K$. – Joe Silverman Jul 3 2011 at 2:42
@Joe: Yes, it all depends on what is meant by abc conjecture. See remarks 14.4.18 and 14.4.20 in the book of Bombieri and Gubler. – Felipe Voloch Jul 3 2011 at 11:35
@Joe: And theorem 12.2.9 of op.cit. – Felipe Voloch Jul 3 2011 at 11:40
@Felipe: Thanks for supplying the reference where Bombieri and Gubler say explicitly that one needs effective $ABC$ to get effective Roth. For those who don't have [BG], here's the exact quote from page 497: "Remark 14.4.20. The proof we have given actually shows that an effective version of the $K$-rational $abc$-conjecture implies effective versions of Roth's and Faltings's theorems." – Joe Silverman Jul 4 2011 at 1:12
There has been progress towards effective Roth's theorem. Notably, Fel'dman was first to prove an effective power saving over Liouville's bound.
In the nutshell the source of ineffectivity comes from the following kind of argument. One obtains a sequence of positive real numbers $x_1,x_2,\dotsc,$ with the property that the product of any two distinct $x$'s is at most $1$. It immediately follows that the sequence is bounded, but of course this information does not yield any actual bound. In the Thue's proofs, one argues that no two rational approximation can be very good simultaneously, which is where such a sequence of $x$'s arises.
In my opinion, a good introduction to effective methods in transcendental number theory is in the notes by Waldschmidt.
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http://mathhelpforum.com/advanced-applied-math/187921-permutation-tensor-explanation-needed.html | # Thread:
1. ## Permutation Tensor (Explanation Needed)
Can somebody help me understand the Permutation Tensor?
I know what it means when it says "if i=j or j=k or i=k, then the permutation tensor is 0".
However, I don't understand the concepts of even and odd permutation (ie, those cases when the permutation tensor equals -1 or 1). Can somebody explain?
Thanks!!!
Attached Thumbnails
2. ## Re: Permutation Tensor (Explanation Needed)
A permutation is when you've swapped two indices. So, if I start with the ordering (1,2,3), and I perform one swap (a swap has to be on adjacent numbers), I could get, for example, (2,1,3). Since I've performed one swap, the permutation (2,1,3) is an odd permutation. Now, let's suppose I do another swap from (2,1,3) to get to (2,3,1). I've now done two swaps, so (2,3,1) is an even permutation. The even-ness or odd-ness of a permutation is identified with the even-ness or odd-ness of the number of swaps I must do to get to that permutation. It's a theorem of abstract algebra that, although there are non-unique ways to get to a particular permutation (for example, I might do two swaps and get right back to where I started), the even-ness or odd-ness of a permutation is constant.
Does that help? Incidentally, in practice, the most important thing to know about the Levi-Civita symbol $\varepsilon_{ijk}$ is that
$\sum_{i}\varepsilon_{ijk}\varepsilon_{imn}=\delta_ {jm}\delta_{kn}-\delta_{jn}\delta_{km}.$
And you should know the relationship between the vector cross product and the Levi-Civita symbol as well. | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9200381636619568, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/tagged/perturbation-theory+solar-system | # Tagged Questions
4answers
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### Staying in orbit - but doesn't any perturbation start a positive feedback?
I am not a physicist; I am a software engineer. While trying to fall asleep recently, I started thinking about the following. There are many explanations online of how any object stays in orbit. The ...
2answers
120 views
### Most suitable metric for the Solar system?
If I wanted to solve the Einstein equations for the solar system, which choice of $g_{\mu\nu}$ and $T_{\mu\nu}$ is more suitable? I thought about using a Schwarzschild metric near each planet, but ... | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9334160089492798, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2009/05/06/the-category-of-inner-product-spaces/?like=1&source=post_flair&_wpnonce=5b12459335 | # The Unapologetic Mathematician
## The Category of Inner Product Spaces
So far we’ve been considering the category $\mathbf{Vect}$ of vector spaces (over either $\mathbb{R}$ or $\mathbb{C}$) and adding the structure of an inner product to some selected spaces. But of course there should be a category $\mathbf{Inn}$ of inner product spaces.
Clearly the objects should be inner product spaces, and the morphisms should be linear maps, but what sorts of linear maps? Let’s just follow our noses and say “those that preserve the inner product”. That is, a linear map $T:V\rightarrow W$ is a morphism of inner product spaces if and only if for any two vectors $v_1,v_2\in V$ we have
$\displaystyle\langle T(v_1),T(v_2)\rangle_W=\langle v_1,v_2\rangle_V$
where the subscripts denote which inner product we’re using at each point.
Of course, given any inner product space we can “forget” the inner product and get the underlying vector space. This is a forgetful functor, and the usual abstract nonsense can be used to show that it creates limits. And from there it’s straightforward to check that the category of inner product spaces inherits some nice properties from the category of vector spaces.
Most of the structures we get this way are pretty straightforward — just do the same constructions on the underlying vector spaces. But one in particular that we should take a close look at is the biproduct. What is the direct sum $V\oplus W$ of two inner product spaces? The underlying vector space will be the direct sum of the underlying vector spaces of $V$ and $W$, but what inner product should we use?
Well, if $v_1$ and $v_2$ are vectors in $V$, then they get included into $V\oplus W$. But the inclusions have to preserve the inner product between these two vectors, and so we must have
$\displaystyle\langle\iota_V(v_1),\iota_V(v_2)\rangle_{V\oplus W}=\langle v_1,v_2\rangle_V$
and similarly for any two vectors $w_1$ and $w_2$ in $W$ we must have
$\displaystyle\langle\iota_W(w_1),\iota_W(w_2)\rangle_{V\oplus W}=\langle w_1,w_2\rangle_W$
So the only remaining question is what do we do with one vector from each space? Now we use a projection from the biproduct, which must again preserve the inner product. It lets us calculate
$\displaystyle\langle\iota_V(v),\iota_W(w)\rangle_{V\oplus W}=\langle\pi_V(\iota_V(v)),\pi_V(\iota_W(w))\rangle_V=\langle v,0\rangle_V=0$
Thus the inner product between vectors from different subspaces must be zero. That is, distinct subspaces in a direct sum must be orthogonal. Incidentally, this shows that the direct sum between a subspace $U\subseteq V$ and its orthogonal complement $U^\perp$ is also a direct sum of inner product spaces.
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Posted by John Armstrong | Algebra, Linear Algebra
## 6 Comments »
1. Sorry to have to say this, but I think there’s a bit of a problem here. This category is not abelian; for one thing, there’s no zero object. The trouble is that when we impose
$\langle T v, T w \rangle = \langle v, w \rangle$
it is automatic that $T$ is injective, because we have
$|T v|^2 = |v|^2$
so $T$ cannot take a nonzero vector to a zero vector.
Comment by | May 6, 2009 | Reply
2. damn. On this one you’re right.
Comment by | May 6, 2009 | Reply
3. But another possibility might be to consider the category whose morphisms consist of adjoint pairs of linear maps $T: V \to W$, $T^*: W \to V$. I haven’t thought about this thoroughly, but it’s connected with other interesting categories I’ve thought about, e.g., Chu spaces.
Comment by | May 6, 2009 | Reply
4. Please excuse me if I’m speaking out of turn on this, but I’ve had a moment to reflect on my last comment further, and maybe I could share a bit.
As many readers will be aware, the adjoint of a linear map $T: V \to W$ is the (uniquely determined) linear map $T^*: W \to V$ such that
\$\langle T^* w, v \rangle_V = \langle w, T v \rangle_W\$
Let me call Inner the category of inner product spaces whose morphisms $V \to W$ are such adjoint pairs \$(T: V \to W, T^*: W \to V)\$. (Throughout this discussion I’ll stick to finite-dimensional spaces.) Now one observation is that this category isn’t all that exciting, because the forgetful functor
Inner $\to$ Vect,
taking a morphism $(T, T^*)$ to $T$, is an equivalence. (The functor is full, faithful, and surjective on objects, which is enough.) But at least this assures that Inner is abelian.
However, there is no canonical quasi-inverse to the forgetful functor, so there may be some things which we can say canonically with regard to Inner which cannot be canonically expressed in Vect. This applies in particular to the inner product on biproducts that John was alluding to above.
I’ll spell this out a bit. We have an evident contravariant functor
$(-)^*:$ Inner $\to$ Inner
which takes a morphism $(T, T^*)$ to $(T^*, T)$. This is an involution in the sense that $(-)^{**}$ equals the identity functor (on the nose). Being a contravariant involution, it takes coproducts to products and vice-versa.
It thus takes biproducts to biproducts as well, but there’s a slightly subtle point that unless we choose the inner product structure on the biproduct “correctly”, this transpose functor $(-)^*$ changes the biproduct structure. To be more precise, suppose that
$i_V: V \to V \oplus W \qquad i_W: W \to V \oplus W$
are the coproduct injections of a biproduct, and
$\pi_V: V \oplus W \to V \qquad \pi_W: V \oplus W \to W$
are the product projections, so that
$\pi_V i_V = 1_V \qquad \pi_W i_W = 1_W$
$\pi_V i_W = 0 \qquad \pi_W i_V = 0$
(compatibility between coproduct and product structures). Then, a priori, there is no guarantee that we have
$i_V^* = \pi_V \qquad i_W^* = \pi_W$
i.e., the transpose functor \$(-)^*\$ may take the coproduct data on a biproduct to a different product structure than the one we started with (hence change the biproduct structure).
But it may be checked that we do get these equations by endowing the biproduct $V \oplus W$ with the orthogonal direct sum that John was describing in his post.
Comment by | May 6, 2009 | Reply
5. [...] vector spaces affect forms on those spaces. We’ve seen a hint when we talked about the category of inner product spaces: if we have a bilinear form on a space and a linear map , then we can “pull back” the [...]
Pingback by | July 24, 2009 | Reply
6. [...] They both start the same way: given root systems and in inner-product spaces and , we take the direct sum of the vector spaces, which makes vectors from each vector space orthogonal to vectors from the [...]
Pingback by | January 25, 2010 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://physics.stackexchange.com/questions/39616/what-is-the-physical-interpretation-of-force-times-area | # What is the physical interpretation of force times area?
I know that $\text{Force} \times \text{Distance = Work}$.
But, what would be the physical meaning of $\text Force \times \text Area?$
Is such a quantity used in physics?
-
An equation by itself is meaningless. You need context to be able to assign physical meaning. – ja72 Oct 23 '12 at 13:16
## 2 Answers
No, not really. The reason why force x distance is a useful quantity is because work is typically defined on a moving particle: force x distance is really a special case of $W = \int F \,dx$. In other words, the particle has to have moved for there to be work.
Because movement is only one dimensional (distance) and not 2D (area), there is no clear interpretation of what force*area is. It is obvious what I mean when I say "a particle has moved 3 cm," but it is nonsensical to say that "a particle has moved 4 square miles."
-
You have not address the question of $\int \vec{F}\cdot \vec{n} {\rm d}A$. A summed force through an area may have usefulness somewhere, I just don't know where. – ja72 Oct 23 '12 at 13:19
Well lets start with something similar that would be meaningful to the general physics population: Pressure = Force / area. As you may know this is a common measurement. It is easily calculated, and is a concept that many in the physics world will understand and recognize. The concept of Work is out of date. If you move something in the vertical direction and then let it go in a gravitational field, it will fall until it hits a surface that doesn't allow it to fall any farther, such as the ground or surface of the earth. Then when the "Work" is calculated, the answer becomes zero, because the distance that the object have moved is a net of zero. The "work" function assumes a net result of a change of distance for the object. Energy expenditure is different in my opinion, as it took energy to move the object to a height, K.E., then object had P.E. while at the height, then when released, the P.E. was turned into K.E., via gravity, and then the K.E. was converted into an Impulse when it hit the ground. Work = 0. Energy = depending on when and where the object was at the time.
Now lets look at your question: Force * Area =?
Force = mass * acceleration ; acceleration = distance /(time * time) area = distance * distance. Putting the three concepts together gets you : (mass * distance * distance * distance) / (time * time).
Now we can combine the units into useful concepts that a common physics person might understand. Mass is the concept of a unit-vector with ‘punch’. Mass is a tangible abstract concept. Mass is what gives, matter and energy, reality. Mass is a fundamental property of matter, energy, momentum, inertia, and gravity. Mass is not “inertia”. Mass has been given a scalar physical unit of measure, in SI units, which is the amount of matter in a Platinum / Iridium cylinder with the label of “kilogram”, and is the ‘K’ in MKS. Mass is a concept that has misleading terminology. For the purpose of weights and measures this was needed to standardize measurement so that trade could be conducted without major confrontations as were going on before it was standardized. Once accepted, the world around, it became the standard and a kilogram in Europe was the same in Australia, South America, and in the USA.
The concept of “mass”, as I use it here, is not a scalar quantity, nor a static-past quantity, instead it is considered as the Causative Vector, as it causes the future to be the present, and holds onto the past. Mass is an ‘at Cause’ unit-vector that forms physical, structure producing, relationships between the various primary concepts of dimension by extension and time. Mass is that property that gives solidness to matter. Mass is what gives, and takes, the effort behind the action of change. Mass is the ‘Woof’ Thread of reality. Mass is what causes stuff to be.
Then we have (distance*distance*distance) which can easily be seen as the concept of space or the units of the concept of volume. Given that we also have 1/(time*time) one can use various math transformational properties:
(distance/time) * (distance*time) = velocity^2 and if you remember the speed of light 'c' is a velocity, then you can say 'c^2'. So now we have mc^2 * distance or Energy * distance. This formula of units may have a concept attached to it already with some kind of name for that concept, but I think that it could be a "Practical Work Function" as energy expended in any direction, even in two directions opposite each other, resulting in a net displacement of zero, still has a positive non-zero value.
-
</randomgarbage> – Rody Oldenhuis Oct 23 '12 at 19:09 | {"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": 4, "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}, "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}, "english_confidence": 0.9610813856124878, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2011/08/10/functoriality-of-cubic-singular-homology/?like=1&source=post_flair&_wpnonce=b0679ce6ea | # The Unapologetic Mathematician
## Functoriality of Cubic Singular Homology
We want to show that the cubic singular homology we’ve constructed is actually a functor. That is, given a smooth map $f:M\to N$ we want a chain map $C_k(f):C_k(M)\to C_k(N)$, which then will induce a map on homology: $H_k(f):H_k(M)\to H_k(N)$.
The definition couldn’t be simpler. We really only need to define the image of a singular $k$-cube $c$ in $M$ and extend by linearity. And since $c:I^k\to M$ is a function, we can just compose it with $f$ to get a singular $k$-cube $f\circ c:I^k\to N$. What’s the $(i,j)$ face of this singular $k$-cube? Why it’s
$\displaystyle(f\circ c)_{i,j}=(f\circ c)\circ I^k_{i,j}=f\circ(c\circ I^k_{i,j})=f\circ c_{i,j}$
and so we find that this map commutes with the boundary operation $\partial$, making it a chain map.
We should still check functoriality. The identity map clearly gives us the identity chain map. And if $f:M_1\to M_2$ and $g:M_2\to M_3$ are two smooth maps, then we can check
$\displaystyle\left[C_k(g\circ f)\right](c)=(g\circ f)\circ c=g\circ(f\circ c)=\left[C_k(f)\circ C_k(g)\right](c)$
which makes this construction a covariant functor.
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8775453567504883, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?t=632953 | Physics Forums
## question on regular group actions
Hello,
we know that if $\bullet : G\times A \rightarrow A$ is a regular action of G on the set A, then the action $\bullet$ is isomorphic to the action of G on itself, where the action on G is given by the group operation, that is: $\circ : G\times G \rightarrow G$ is defined as $g\circ g'=gg'$.
My question is: if we have a regular action of G on A, and a isomorphism $f:A\rightarrow G$ of actions, can we find a binary operation * for the set A, such that (A,*) is a group isomorphic to G?
***** EDIT: ******
My attempt to a possible solution was to define: $* : A\times A \rightarrow A$ as follows: $$a*a' = f(f(a) \bullet a')$$ which yields:
$$f(a) \circ f(a') = f(a)f(a') = gg'$$
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