url
stringlengths 17
172
| text
stringlengths 44
1.14M
| metadata
stringlengths 820
832
|
|---|---|---|
http://mathoverflow.net/revisions/103843/list | ## Return to Question
2 added 321 characters in body
Given a compact oriented aspherical $3$--manifold $M$ with torus boundary $\partial M\simeq T^2$ (e.g. a knot complement), the condition that the images in $\pi_1 M$ of basis $x,y\in \pi_1 T^2$ under the inclusion generate the boundary of a compact oriented $3$-manifold with fundamental group $\pi_1 M$ (the $3$-manifold $M$ itself, to be precise) is algebraically encoded by the vanishing of their Pontryagin product $\langle x,y\rangle\in H_2 (\pi_1 M)$, which you can think of as the fundamental class of the torus under the inclusion. Recall that the Pontryagin product sends two commuting elements of a group to an element of $H_2$ of the group.
Question: Given a compact oriented aspherical $3$--manifold $M$ with boundary a closed surface $\partial M \simeq \Sigma$ of genus $g\geq 2$, how can I express the statement that $\Sigma$ bounds a $3$--manifold (the $3$-manifold $M$) on the level of fundamental groups? What condition must images under the inclusion of a basis for $\pi_1 \Sigma$ satisfy? How can I express the fundamental class of $\Sigma$ in terms of elements of $\pi_1$, if I know what $\Sigma$ is topologically?
I'm trying to identify some homomorphs of the fundamental group of a knotted theta, with peripheral data, and the answer to this question would, I presume, be something all such homomorphs would have to satisfy.
1
# What is a higher genus analogue of the Pontryagin product?
Given a compact oriented aspherical $3$--manifold $M$ with torus boundary $\partial M\simeq T^2$ (e.g. a knot complement), the condition that the images in $\pi_1 M$ of basis $x,y\in \pi_1 T^2$ under the inclusion generate the boundary of a compact oriented $3$-manifold with fundamental group $\pi_1 M$ (the $3$-manifold $M$ itself, to be precise) is algebraically encoded by the vanishing of their Pontryagin product $\langle x,y\rangle\in H_2 (\pi_1 M)$.
Question: Given a compact oriented aspherical $3$--manifold $M$ with boundary a closed surface $\partial M \simeq \Sigma$ of genus $g\geq 2$, how can I express the statement that $\Sigma$ bounds a $3$--manifold (the $3$-manifold $M$) on the level of fundamental groups? What condition must images under the inclusion of a basis for $\pi_1 \Sigma$ satisfy?
I'm trying to identify some homomorphs of the fundamental group of a knotted theta, with peripheral data, and the answer to this question would, I presume, be something all such homomorphs would have to satisfy. | {"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": 42, "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": 0.9109060764312744, "perplexity_flag": "head"} |
http://www.physicsforums.com/showthread.php?t=119037 | Physics Forums
## Distance between planes in crystals
I'm working on a lab report on powder X-ray diffraction off of some relatively straight-forward crystals (Si, NaCl, CsCl) for an introductary course on modern physics.
I thought it would be useful to include a partial derivation of the formula relating the distance between parallel planes, d, the length of a cell edge, a, and the miller indices (hkl) for a cubic lattice:
$$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$$
I would be happy (and it would be sufficient for my purposes) to do a basic derivation of the spacing between lines in a hypothetical two dimensional square lattice. I've thought a lot about this problem, however, and what I thought would be a clear geometrical fact is turning out to be not so obvious.
Does anyone have any hints or links to a derivation? I got several texts on X-ray diffraction from my college's library, including a text, "Interpretation of x-ray powder diffraction patterns" but none of them include a clear derivation. What I've found online seems to be generally cursory, as well. I've drawn out a two dimensional square lattice and sample parallel lines going through it and I can see that the equation holds, but I'd like a simple proof, from first principles if possible.
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:
Homework Help
Science Advisor
Quote by Stephan Hoyer I'm working on a lab report on powder X-ray diffraction off of some relatively straight-forward crystals (Si, NaCl, CsCl) for an introductary course on modern physics. I thought it would be useful to include a partial derivation of the formula relating the distance between parallel planes, d, the length of a cell edge, a, and the miller indices (hkl) for a cubic lattice: $$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$$ I would be happy (and it would be sufficient for my purposes) to do a basic derivation of the spacing between lines in a hypothetical two dimensional square lattice. I've thought a lot about this problem, however, and what I thought would be a clear geometrical fact is turning out to be not so obvious. Does anyone have any hints or links to a derivation? I got several texts on X-ray diffraction from my college's library, including a text, "Interpretation of x-ray powder diffraction patterns" but none of them include a clear derivation. What I've found online seems to be generally cursory, as well. I've drawn out a two dimensional square lattice and sample parallel lines going through it and I can see that the equation holds, but I'd like a simple proof, from first principles if possible. Thanks for your help.
I am not sure if this helps you but have a look at this site
AM
Recognitions: Gold Member Science Advisor Staff Emeritus I couldn't find it there with a quick look. So, anyway, it's short enough that I can write it down in a few lines. Consider two adjacent planes, one of which goes through the origin. The second plane makes intercepts a/h, b/k, c/l (by definition of the Miller Indices). Let the point on this plane that's nearest the origin (O) be P. Then OP is the required d-spacing. Let the line OP make angles A, B and C with each of the three axes. From trig, we have cos2(A)+cos2(B)+cos2(C)=1 But cos(A) = OP/OX = d/(a/h) = dh/a Similarly, plug in for cos(B) and cos(C) and you will get the required result.
## Distance between planes in crystals
Thanks for you help. It looks like the general proof isn't actually so tedius after all, so I guess I'll include that instead.
Nice visual explanation here. Slides 23+24
Really Thanks. Thank you so much!
Thread Tools
Similar Threads for: Distance between planes in crystals
Thread Forum Replies
Classical Physics 4
Calculus & Beyond Homework 2
General Physics 4
Chemistry 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": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9478905200958252, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/11384/whats-the-name-of-graphs-with-each-vertex-contained-in-a-cycle/15646 | ## What’s the name of graphs with each vertex contained in a cycle?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A tree is a graph with no vertex contained in a cycle.
A non-tree is a graph with some vertex contained in a cyle.
What's the name of graphs with each vertex contained in a cycle?
-
1
I'm wondering why no one had voted up this question (until I just did), even though people were answering and voting up answers. – Jonas Meyer Jan 11 2010 at 1:34
## 5 Answers
Undirected graphs in which every edge is contained in a cycle are called bridgeless or 2-edge-connected. But I don't know of a word for the analogous concept for vertices.
-
Could this mean that this is an unfruitful concept (since if it were fruitful it would have a name X, e.g. to be able to formulate statements like "all graphs that are X are Y" or vice versa?) The other way round: Do you know an interesting theorem which implies graphs with the above mentioned property? – Hans Stricker Jan 11 2010 at 0:48
If this graph property was fruitless: How could this be explained, compared to the "unreasonable" fruitfulness of trees? – Hans Stricker Jan 11 2010 at 0:52
6
Nice pun! Was it intentional? – Qiaochu Yuan Jan 11 2010 at 1:56
I guess it was not (if you refer to the "fruits" of "trees"). But can you - who has used the term "unreasonable" too (in another context) - appreciate the question? – Hans Stricker Jan 11 2010 at 2:21
@David: If every edge is contained in a cyle, doesn't every vertex have to be contained in a cycle, too? And vice versa? That is, isn't the concept "self-dual"? – Hans Stricker Oct 29 at 18:17
show 4 more comments
### 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 don't know a name, but I'll give you a different characterization. Biconnectivity is sufficient but too strong, while "having minimum degree at least 2" is necessary but too weak. I'm almost certain this is a necessary and sufficient condition:
$G$ has minimum degree at least 2, and if v is a cutvertex of $G$, then there is some new connected component of $G - v$ with at least two vertices adjacent to v.
Here's a proof of sufficiency: If v is not a cutvertex of $G$, then pick any two vertices adjacent to v. There's a path between them not going through v (since $G - v$ is connected), so v is contained in a cycle.
If v is a cutvertex of $G$, then pick the two vertices adjacent to v that are in the same connected component of $G - v$. There's a path between them that extends to a cycle containing v.
Now, a proof of necessity. Suppose that $G$ has a cutvertex $v$ whose removal does create deg(v)-1 new connected components. Then $v$ can't lie in a cycle. (This is easy to check.)
This characterization is equivalent to: Removing any vertex of degree d increases the total number of connected components by at most $d-2$. Some generalization of this property may have a name.
-
Thanks. Please see my comments to David's answer. They apply also to yours. – Hans Stricker Jan 11 2010 at 1:07
How can vertices of G-v be adjacent to v? – Hans Stricker Jan 11 2010 at 1:10
Sorry, I meant adjacent in the original graph. – Harrison Brown Jan 11 2010 at 1:16
Sorry for my part, I should have understood it this way. (You treat the one counter-example of David below - the 8-graph - and we are done?) – Hans Stricker Jan 11 2010 at 1:24
These are the graphs that admit "vertex cycle covers". http://en.wikipedia.org/wiki/Vertex_cycle_cover
-
+1, even if this is not really a name but just another characterization. (At least it's something to google for.) – Hans Stricker Jan 25 2010 at 7:27
Assuming that $G$ is connected, we can get a characterization by looking at the block decomposition $B(G)$ of $G$ into 2-connected pieces. That is, we simply look at which vertices of $B(G)$ are $K_2$'s. The required characterization is:
Every vertex of $G$ is contained in a block of $G$ which is not a $K_2$.
Of course, this is the same answer as Harrison's, but it gives a more global view. Also, necessity and sufficiency are trivial.
-
2-connected or biconnected
-
of course, this works only for connected graphs. – Klingonesque Jan 11 2010 at 0:32
A figure-eight graph has every vertex in a cycle but is not biconnected. – David Eppstein Jan 11 2010 at 0:36
Even if those graphs were biconnected, their biconnectivity would have to be proved, but would not be the defining property, for which I am looking for a name. – Hans Stricker Jan 11 2010 at 0:55 | {"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": 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": 0.9498828649520874, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/39451/on-the-double-slit-experiment-with-4-slits | # On the double slit experiment with 4 slits
I'm not a physicist nor a science savvy person, but I was wondering if this experiment was ever performed in a simultaneous fashion on screens with fixed references(marks) and firing different wavelengths particles.
So, to recap, 4 slits, 2 emitters firing at different times (where the $\Delta T$ is known and certain) on two different screens.
I hope this question is not too stupid.
P.S. I'm sorry, I've been too vague. I ask this because I've recently learned that this interference pattern emerge even firing large atom clusters instead of single particles.
This is astonishing and it's clearly stating that there's a limit in mass aggregation where it stops behaving in duality. Cause, you know, firing bricks will hardly produce interference. And, in the end, I was wondering if it's not possible that it's spacetime itself vibrating at a certain $f$ and if there's a limit when mass stop to be affected (relatively) by this vibration. Like pinning a vibrating rubber sheet.
I'm so confused.
-
## 1 Answer
The main reason we can see quantum effects from small objects but not from large objects is due to interactions with the environment.
If a quantised system interacts with anything else it will become entangled with that other object. When this happens you can no longer treat your system on it's own. You have to treat your original system and whatever it's interacted with as a single combined system.
If you're doing the Young's slits experiment with photons or electrons it's not too hard to keep them isolated from the environment while they travel from the source, through the slits and to the screen. However as the objects you're studying get bigger they interact with the environment more strongly and the experiment gets harder to do. To see the Young's slits interference patterns with large molecules like buckyballs is much hard, was was managed in 1999. If you attempted it with pools balls, or even pollen grains, the interactions with the environment would swamp any interference and you'd see no diffraction pattern.
If you want to know more about this try Googling for decoherence, but be warned that it's a complex subject and I don't know of any good popular science level descriptions (except this answer of course ;-).
-
so, in an ipotetical ideal set, firing Jumbos will produce interference on the jumbo-sensitive surface right? It's only a matter of keeping the environment clear from interference(gravitational, magnetic, electromagnetic, whatever)? – vaitrafra Oct 10 '12 at 8:57
Hmm, presumably at some point the internal degrees of freedom would get so large they'd have a similar effect to interaction with the environment. To be honest I don't know if you could theoretically observer diffraction of something as large and complex as a jumbo jet, even if you could isolate it from interactions with the environment. – John Rennie Oct 10 '12 at 9:34 | {"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": 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": 0.9629831910133362, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/114601/geometric-intuition-behind-perverse-coherent-sheaves/114637 | ## Geometric intuition behind perverse coherent sheaves?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I would like to know an intuition behind perverse coherent sheaves. I am aware that it is induced by a heart of another t-structure on the derived category. Are there any better, probably more geometric, way to understand perverse coherent sheaves?
Just in case, let us recall the definition of perverse coherent sheaves. Let $X$ be a projective threefold with at worst Gorenstein terminal singularities and $f:Y\rightarrow X$ be a crepant resolution. Define a full subcategory $\mathrm{Per}(Y/X) \subset \mathrm{D}(Y)$ consisting of objects $E \in \mathrm{D}(Y)$ satisfying the following three conditions;
1. $H^i(E)=0$ unless $i=0,-1$,
2. $R^1f_*H^0(E)=0$ and $R^0f_∗H^{−1}(E)=0$,
3. $Hom_Y(H^0(E),C)=0$ for any sheaf $C$ on $Y$ satisfying $Rf_∗(C)=0$.
We call the objects of $\mathrm{Per}(Y/X)$ perverse coherent sheaves.
-
1
Your definition is far too specific. I don't think there's a really good geometric intuition. They're the natural objects to look at on singular spaces which imitate good "cohomological properties" of smooth spaces (e.g. duality theorems, purity of etale cohomology, etc.). – David Hansen Nov 26 at 23:36
Dear David, The OP is asking about perverse coherent sheaves, not usual perverse sheaves. Incdidentally, I think there is good geometric intuition for usual perverse sheaves (though I am not the right one to convey it). Regards, Matthew – Emerton Nov 27 at 16:55
Dear Matthew: Oops, I didn't realize these were distinct concepts! Thanks for the correction. Best, Dave --- Pooya: Sorry for the tone of my comment. :) – David Hansen Nov 28 at 20:50
## 1 Answer
This is the definition appear in Bridgeland's paper which shows that flops of smooth 3-folds induces equivalence of derived category of coherent sheaves. From your question I think you know the word "perverse" is kind of related to t-structures.
The main theorem of that paper, indicates that for a flop $Y\to X\leftarrow W$, $Per(Y/X)$ will be send to $Coh(W)$ under that isomorphism. In other words, these objects are sheaves on another scheme which you can construct from the data $Y\to X$! In my opinion that's pretty cool, not "perverse" at all. But as for the name, so be it.
BTW, you don't need $X$ and $Y$ to be three fold in the definition. If you check Bridgeland's paper, most of the time he work with birational morphism such that $Rf_*\mathcal{O}_Y=\mathcal{O}_X$ and fibers have dimension at most 1. (For 3-folds that's just a small resolution.)
-
Dear 36min, As you probably know, the adjective "perverse" is inherited from its usage in the usual theory of perverse sheaves, which served as a partial motivation for Bridgeland's work. Regards, Matthew – Emerton Nov 27 at 16:56
I said in the first paragraph the word "perverse" is related to t-structures. – 36min Nov 28 at 2:00 | {"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": 20, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9250008463859558, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/249292/explain-how-to-solve-the-following-anti-derivative/249314 | # Explain how to solve the following anti-derivative
I am trying to figure how to solve the following antiderivative.
$$\int (5x+3)^7 dx \\$$ I've seen the step-by-step solution by WolframAlpha however what they are doing in this part:
For the integrand $(5x+3)^7$, substitute $$u = 5x + 3 \\ du = 5 dx \\$$
Why are they derivating $u$ ?
Thanks.
-
2
– anonymous Dec 2 '12 at 16:55
Yes, I am, I understand now, I had no simple example – Francis Dec 2 '12 at 17:02
## 2 Answers
The general theorem is like this:
$$\int (f\circ g)(x)g'(x)dx=\int f(u)du \Leftrightarrow u=g(x).$$
It may help to note that in calculations, we have $u=\text{something}$ and $du=(\text{something})'dx$. This is because of the definition of the differential. That is, $dy=f'(x)dx$ if and only if $y=f(x)$. (The definition of the differential just comes from the "multiplying" of $dx$ to both sides in $\frac{dy}{dx}=f'(x)$.)
-
Directly:
$$\int (5x+3)^7dx=\frac{1}{5}\int 5(5x+3)^7dx=\frac{1}{5}\int(5x+3)^7d(5x+3)=$$
$$=\frac{1}{5}\frac{(5x+3)^8}{8}+C=\frac{(5x+3)^8}{40}+C$$
- | {"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": 8, "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": 0.3, "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": 0.9419252276420593, "perplexity_flag": "head"} |
http://mathhelpforum.com/algebra/27359-dividing-polynomial-polynomial.html | # Thread:
1. ## dividing polynomial by a polynomial
6x^2+7x-18[3x-4]
thanks!
2. Originally Posted by lenworthmaxwell
i am not sure what you mean. did you mean to type: $\frac {6x^2 + 7x - 18}{3x - 4}$ ??
in any case, make sure you typed what you meant to type. because unless you're doing long division of polynomials or synthetic division in class, this does not have a nice answer. the answer would be: $2x + 5 + \frac 2{3x - 4}$ | {"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": 0.945624828338623, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/85610?sort=votes | Klein Bottle exception to the Heawood Conjecture [closed]
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Possible Duplicate:
The Klein bottle and the Heawood Conjecture
It is well known that the Heawood Conjecture states that the bound for the number of colours which are sufficient to colour a map on a surface of genus $g>0$ is, $$\gamma(g)=\left \lfloor \frac{7 + \sqrt{1+48g}}{2} \right \rfloor$$ as proved by the works of mostly Ringel, Young, and Gustin in the 1950s and 1960s. While two pathological cases were ruled out by the exhaustive proof of the Four Colour Theorem by Appel and Haken in the 1970s, namely the plane and the sphere, the Klein Bottle remains to defy that this condition is also necessary as shown by the Franklin Graph.
Does anyone know of any papers that comment on the status of the Klein Bottle in this regard, in that there is some sort of explanation for why the Heawood Conjecture is true for absolutely every surface of genus $g$, except the Klein Bottle? I know that the Klein Bottle is spectacular in its non-orientable no-boundary way, but are there any known properties about the Klein Bottle that suggest a reason for why it defies the Heawood Conjecture? May it somehow relate to fundamental polygons, cross caps, or the independence polynomial of the Franklin Graph?
Any ideas, references to papers, expository articles, or original explanations would be welcome.
-
Duplicate? mathoverflow.net/questions/51830/… – Gjergji Zaimi Jan 13 2012 at 20:44
I agree with Gjergji. Voting to close. – HW Jan 14 2012 at 8:10
1 Answer
The derrière cri seems to be this paper (in case the link goes away: five coloring graphs on the Klein Bottle, by (wait for it): Chenette, Postle, Streib, Thomas, Yerger.
-
3
I believe you are thinking of Premier Cru. jezebel.com/derriere-cri – Will Jagy Jan 13 2012 at 21:05
I am thinking of derniere cri, but apple insists on correcting me... Still derriere cri has a certain je ne sais quioi. – Igor Rivin Jan 14 2012 at 4:04
The French does not know me. – Will Jagy Jan 14 2012 at 5:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 0.18, "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": 0.9105244874954224, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/167722/kolmogorovs-unit-interval-probability-space | # Kolmogorov's unit interval probability space
Somewhere I've heard that Kolmogorov proved that for all practical purposes, the probability space $$(\Omega,\mathcal F,\mathbb P)$$ that he invented could be taken without loss of generality to be the unit interval endowed with the Lebesgue measure, $$([0,1],\mathcal L,\mu),$$ although the mappings necessary to define random variables on such a space are in general highly contrived and hence not very constructive or intuitive. Does anyone have a reference or a pointer where to find this proof or a translation of it?
Somewhat related: The role of the "hidden" probability space on which random variables are defined
-
– Alex Jul 7 '12 at 9:18
That's right! Now I remember! Thanks for reminding me. That was ages ago and I must have forgotten. It's been 14 years since I took probability. – JL344 Jul 7 '12 at 22:37 | {"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": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.951676070690155, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/293997/interpolation-polynomial | # Interpolation polynomial
Consider the following table of values for a function $j_0(x)$:
$\begin{array}{c|ccccc} x & \delta_0(x) \\ \hline 0.0&1.00000\\0.1&.99833\\0.2&.99335\\0.3&.98507\\0.4&.97355\\0.5&.95885\\0.6&.94107\\0.7&.92031\\0.8&.89670\\0.9&.87036\\1.0&.84147\\1.1&.81019\\1.2&.77670\\1.3&.74120 \end{array}$
What should be the maximum degree of polynomial interpolation used with the table?
I know that I must use the forward difference table so that I can detect the influence of the rounding errors. From that, do I find the polynomial by using $$f[x_0,...,x_n]=\frac{\triangle^nf(x_0)}{n!h^n}$$ where $\triangle f(x)=f(x+h)-f(x)$ and $h$ is the step length? Or do I use newton divided difference? Confused on how I can solve this.
-
If there are no errors, interpolation is the way to go. The forward differences tell you the right degree, but in practice you'd use Newton interpolation (as you have equally spaced data). If there are errors, you'd get the form right and use least squares to get parameters. – vonbrand Feb 4 at 2:37
@vonbrand Can you elaborate a little more please? – user60514 Feb 4 at 3:34
– vonbrand Feb 4 at 3:39
@vonbrand Thanks but that link you gave me says "Not Found" – user60514 Feb 4 at 3:46
## 1 Answer
The data is only given to 5 decimal places. So, we can assume that the data points might be in error by as much as 0.000005. So any curve that deviates from the data points by less than this would be fine. If you plot the data (in Excel, for example) you will see that the curve gently slopes downwards. I would guess that a curve of degree 5 or 6 would fit the points with an error of less than 0.000005. As far as I know, there is no good way to predict the error, other than constructing the curve and measuring. To construct curves, use least-squares fitting: http://mathworld.wolfram.com/LeastSquaresFittingPolynomial.html
To use the Wolfram formulae: In your case, $n=14$, $\{x_1, x_2, \ldots, x_n\} = \{0.0, 0.1, \ldots, 1.3\}$, and $\{y_1, y_2, \ldots, y_n\} = \{1.00000, 0.99833, \ldots, 0.74120\}$. But you can probably find some software to do the fitting for you. For example, you can use the "Fit" function in Mathematica. http://reference.wolfram.com/mathematica/ref/Fit.html
You can do it in Excel, too. Look up the "Add Trendline" function. The function $y = -0.0008x^5 + 0.0096x^4 - 0.0009x^3 - 0.1664x^2 - 0.00002x + 1$ gives a maximum error of $0.00003$.
-
Thank you Bubba, I am confused on how I can plot my table into that wolfram site you provided? What will be in my matrix? – user60514 Feb 4 at 5:50
I added an explanation. – bubba Feb 4 at 6:19
So, on the mathematica site you gave where it says "$data = \{\{0, 1\}, \{1, 0\}, \{3, 2\}, \{5, 4\}\};$ the x side is the left and y on the right. So I plug my points there? – user60514 Feb 4 at 7:06
And is $y = -0.0008x^5 + 0.0096x^4 - 0.0009x^3 - 0.1664x^2 - 0.00002x + 1$ from my data? – user60514 Feb 4 at 8:13
Yes, the Excel answer is for your data. The function allows you to adjust the degree, too. – bubba Feb 4 at 11:14
show 2 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": 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": 0.9071630239486694, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/4243?sort=votes | ## Godel’s 1st incompleteness theorem - clarification.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This should be a trivial question for people who know Gödel's 1st incompleteness theorem. I quote the statement the theorem from wikipedia: "Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true, but not provable in the theory."
My question is: what is the meaning of 'true' in the last sentence? Let me elaborate: the only (introductory) proof of the theorem that I know starts with a specific model of the theory and constructs a sentence which is true in that model, but not provable from the theory.
So does `true arithmetical statement' in the statement of the theorem mean `true in the (implicitly) given model', or true in EVERY model?
-
## 2 Answers
It means true in the usual model. For 1st order logic we have Godel's Completeness theorem, which guarantees that if something is true in every model, then it is actually provable in the theory.
-
Thanks, I guess was confusing true' with valid'. – auniket Nov 6 2009 at 21:59
### 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.
As already pointed out here by Tom Leinster, "true" doesn't make sense without some sort of model. It is a technical word.
-
That assumes a contentious position in the philosophy of mathematics. Some people think that there are no such things as numbers - all there are are various models of arithmetic, and in one context or another we can choose one of these models to be what we're talking about when we use the word "number". (This view gets into a sort of regress problem though when we ask what a model is rather than a number.) – Kenny Easwaran Nov 6 2009 at 6:19
The other view is that the word "number" has an ordinary meaning, and what it means to say that a statement like "every number has property X" is true is just to say that every number actually has property X. On this view, it seems that every arithmetical statement (that is grammatically well-formed) must be either true or false. There may be some model in the set-theoretic sense such that truth-in-that-model coincides with plain old truth, but that's not needed for the notion of truth to make sense. – Kenny Easwaran Nov 6 2009 at 6:19
You raise interesting points. I wasn't meaning to get too deep into the philosophical tangent, but at least in the sense of the question above there's no reason to think of Gödel's "true" as anything other than the strict definition. – Jason Dyer Nov 6 2009 at 16:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9482204914093018, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/48/is-it-feasible-to-build-a-stream-cipher-from-a-cryptographic-hash-function?answertab=oldest | # Is it feasible to build a stream cipher from a cryptographic hash function?
A few years ago I devised a symmetric-key system that worked like so:
````H() = some hashing function
h = the number of bits in the result of H()
key = bits to be used as a key
^ = the XOR operation
+ = concatenation
len() = the length function (in bits)
max = ceiling(len(cipertext) / h)
pad = pad_0 + ... + pad_max
pad_0 = H(key) ^ H(plaintext)
pad_n = H(pad_(n-1))
cipherpart_n = plaintext_n ^ pad_n
ciphertext = H(plaintext) + cipherpart_0 + ... cipherpart_max
````
Basically, it builds up a "pad" based on the hash of the message and hash of the key and XORs the message against it.
I am not actually an expert, so I was wondering why an approach like this isn't very common.
-
## 3 Answers
With the corrected system (which actually uses the key), I see these weaknesses:
• If the attacker can guess some `plaintext_n`, he can derive `pad_n` from `ciphertext_n` and from this all the following `pad_i` - which means that he can read the rest of the message.
• The ciphertext starts with a `H(plaintext)`, which means that an attacker which can guess the plaintext has a means to verify this guess. To avoid this, the plaintext should at least be randomly padded. (But then you could simply use these random bytes as the initialization vector instead of the hash of the plaintext).
• The entropy of your actual key stream (the `pad`) is reduced to `h` bits, even if the key itself had bigger entropy. This might be better if you input the key again to the hash function for creating the higher-order `pad_n`.
• As already said by Jack, the need to hash the plaintext before you can start to encrypt it means that you need two passes over the plaintext, which means that the encrypting algorithm is not an online-algorithm (the decryption is).
Other than this, your algorithm is similar to the Output feedback mode of operation for Block ciphers, with the initialization vector derived from the message, and without new key input for the following steps.
Here is a variant which would not have these weaknesses (and hopefully no others):
````initvec = random
pad_0 = H(key + initvec)
pad_n = H(key + pad_(n-1))
ciphertext = initvec + cipherpart_0 + ... + cipherpart_max
````
(The rest as in your example.)
This does not do integrity-protection ... for this you would append some keyed MAC at the end (so it can be an online algorithm).
Then we have actually the OFB mode of a strange "encryption only" block cipher derived from the hash function. The only problem would be that it would be quite slower than a normal block cipher, as Jack already mentioned.
Disclaimer:
Note that while this construction is secure in the random-oracle model (see this blog entry for a proof of a similar construction, using a counter instead of previous hash), the proof uses properties which are not implied by the collision-resistance and preimage-resistance properties of a hash function (which is what hash functions are constructed for).
So while this might work, it does not have to work with real-world hash functions.
-
This approach, at a high level, is actually fairly common; many stream ciphers operate on this very principle. For instance, Salsa20 uses what is effectively a hash function (a PRF) to convert a secret input (that includes a counter) into the keystream which is XORed with the plaintext. However, this kind of function can be much faster than a secure cryptographic hash, because it does not need to prevent preimages or collisions as a cryptographic hash does. For instance, OpenSSL's SHA-1 runs at about 50 MiB/s for small inputs (on my desktop machine), whereas its AES-128 implementation runs at around 90 MiB/s.
I'm rather confused by your specific scheme, though. You set the first pad to the hash of the plaintext, rather than the key. So given the first piece of the ciphertext (H(plaintext)) one could rederive the entire stream. If this is really what you meant to write, it's also quite slow, since it requires you to read the entire message before you can process any piece of it.
Update: I should mention that if there is any full block of known or guessable plaintext in your message, your scheme is easily broken because given pad_{i} anyone can compute pad_{i+1}. So guess one block of plaintext, XOR it against the ciphertext to recover a possible pad value, then compute the next pad value and see if the recovered ciphertext in that block seems valid.
-
Also, I'm putting the H(plaintext) at the beginning so that it can all stream, and so that we can confirm the decrypted message. – John Gietzen Jul 13 '11 at 16:50
You may be interested in BEAR and LION.
http://www.cl.cam.ac.uk/~rja14/Papers/bear-lion.pdf
The authors basically build block ciphers (with variable block sizes $m$) from a keyed hash function (a function family $H_K : {0,1}^* \to {0,1}^k$) or a normal hash (a function $H': {0,1}^* \to {0,1}^k$) and a stream cipher (a function $S : \{0,1\}^k \to \{0,1\}^n$, for arbitrary $n$ from context; here $n = m-k$).
These work by fist splitting the input block of size $m$ in two parts, one (L) of size $k$ and one (R) of size $m-k$, doing the encryption/decryption, and putting the two parts again together. We also have two keys $K_1$ and $K_2$, each of length $k$.
For BEAR, the encryption is done by
$$L := L ⊕ H_{K_1}(R)$$ $$R := R ⊕ S(L)$$ $$L := L ⊕ H_{K_2}(R)$$
and decryption the other way around:
$$L := L ⊕ H_{K_2}(R)$$ $$R := R ⊕ S(L)$$ $$L := L ⊕ H_{K_1}(R)$$
For LION, we use a un-keyed hash function $H'$, and we have two calls of the stream cipher instead of the hash function instead of two keyed hash calls. Encryption:
$$R := R ⊕ S(L ⊕ K_1)$$ $$L := L ⊕ H'(R)$$ $$R := R ⊕ S(L ⊕ K_2)$$
Decryption:
$$R := R ⊕ S(L ⊕ K_2)$$ $$L := L ⊕ H'(R)$$ $$R := R ⊕ S(L ⊕ K_1)$$
E.g. decryption is in both cases just encryption with swapped key halves.
The paper proves that these are secure (against certain attacks) as long as even one of the component functions is secure (e.g. breaking the composed primitive allows breaking both of the components). For added resistance against other attacks there is also a version (LIONESS) which uses both keyed hash and stream cipher twice.
I don't really understand why anyone would want a block cipher if they already have a stream cipher, but there you go. See also what Jack Lloyd said, above, about Salsa20.
Here is a rough sketch of a proposal to do the Salsa20 style of counter-mode with a modern secure hash function: http://tahoe-lafs.org/pipermail/tahoe-dev/2010-June/004487.html
-
It looks like the stream cipher term used by the paper is a bit more narrow than the type of stream ciphers one can get from block ciphers using a mode of operation - it is simply a function producing an (arbitrary long) key stream from a key, without any relation to ciphertext and plaintext. (This corresponds to the OFB and CTR modes of block ciphers.) – Paŭlo Ebermann♦ Jul 17 '11 at 12:18
– Paŭlo Ebermann♦ Jul 17 '11 at 12:25
... and of course, an undated paper :-( (From the file date, it is from July 2006 or earlier.) – Paŭlo Ebermann♦ Jul 17 '11 at 12:31
I added a sketch of the algorithms shown in the paper. – Paŭlo Ebermann♦ Jul 17 '11 at 20:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 12, "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": 0.9138273000717163, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/67182/list | ## Return to Question
3 deleted 15 characters in body
I am looking for a generalized version of the Gauss-Green theorem also known as the divergence theorem:
http://en.wikipedia.org/wiki/Divergence_theorem
A quick search on MathSciNet suggests that there are generalizations for bad domains and nonsmooth functions. However, they seem to rely on heavy machinery and not to be suited for the special case I am interested in.
For example, I found this formula on PlanetMath:```$$ \int_E \mathrm{div} f(x)\, dx
= \int_{\partial^* E} \langle \nu_E(x),f(x)\rangle \,d\mathcal H^{n-1}(x)$$```
See http://planetmath.org/?method=l2h&from=objects&name=GaussGreenTheorem&op=getobj for the details.
Let $\Omega \subset \mathbb{R}^n$ be open and bounded and $f\in C^1(\Omega, \mathbb{R}^n) \cap C^0(\overline\Omega, \mathbb{R}^n)$.
Question: What conditions do we have to impose on $\Omega$ (or $f$) to ensure that the divergence theorem holds true?
To clarify my question: I know that requiring the boundary of $\Omega$ to be piecewise regular is is a sufficient condition for the Gauss-Green theorem to be true. I wondered if this condition is also necessary. If so: is the an other "version" of Gauss-Green (e.g. the one cited above) which holds true under weaker conditions and is especially suited for the case of an open and bounded domain
2 added 381 characters in body; added 25 characters in body
I am looking for a generalized version of the Gauss-Green theorem also known as the divergence theorem:
http://en.wikipedia.org/wiki/Divergence_theorem
A quick search on MathSciNet suggests that there are generalizations for bad domains and nonsmooth functions. However, they seem to rely on heavy machinery and not to be suited for the special case I am interested in.
For example, I found this formula on PlanetMath:```$$ \int_E \mathrm{div} f(x)\, dx
= \int_{\partial^* E} \langle \nu_E(x),f(x)\rangle \,d\mathcal H^{n-1}(x)$$```
See http://planetmath.org/?method=l2h&from=objects&name=GaussGreenTheorem&op=getobj for the details.
Let $\Omega \subset \mathbb{R}^n$ be open and bounded and $f\in C^1(\Omega, \mathbb{R}^n) \cap C^0(\overline\Omega, \mathbb{R}^n)$.
Question: What conditions do we have to impose on $\Omega$ (or $f$) to ensure that the divergence theorem holds true?
To clarify my question: I know that requiring the boundary of $\Omega$ to be piecewise regular is is a sufficient condition for the Gauss-Green theorem to be true. I wondered if this condition is also necessary. If so: is the an other "version" of Gauss-Green (e.g. the one cited above) which holds true under weaker conditions and is especially suited for the case of an open and bounded domain
1
# Generalized Gauss-Green theorem
I am looking for a generalized version of the Gauss-Green theorem also known as the divergence theorem:
http://en.wikipedia.org/wiki/Divergence_theorem
A quick search on MathSciNet suggests that there are generalizations for bad domains and nonsmooth functions. However, they seem to rely on heavy machinery and not to be suited for the special case I am interested in.
For example, I found this formula on PlanetMath:```$$ \int_E \mathrm{div} f(x)\, dx
= \int_{\partial^* E} \langle \nu_E(x),f(x)\rangle \,d\mathcal H^{n-1}(x)$$```
See http://planetmath.org/?method=l2h&from=objects&name=GaussGreenTheorem&op=getobj for the details.
Let $\Omega \subset \mathbb{R}^n$ be open and bounded and $f\in C^1(\Omega, \mathbb{R}^n) \cap C^0(\overline\Omega, \mathbb{R}^n)$.
Question: What conditions do we have to impose on $\Omega$ (or $f$) to ensure that the divergence theorem holds true? | {"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": 14, "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": 0.907755970954895, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/247500/positive-cone-definition-for-ordered-rings | # Positive cone definition for ordered rings
There are two definitions of an ordered field: you can define it as a field with a total ordering and certain axioms relating the ordering to the field operations, or as a field with a "positive cone": a subset of the field satisfying certain axioms (see Orderered Field on Wikipedia). These definitions are equivalent: the total orders and the positive cones are in one-to-one correspondence.
I am wondering if the same equivalence holds for rings. I'm taking the definition of a totally ordered ring from Wikipedia, and I am taking the same definition for a positive cone as is given for fields: a positive cone on a ring $R$ is a subset $P$ of $R$ satisfying
1. $P$ is closed under $+$ and $\cdot$.
2. $x^2 \in P$ for all $x \in R$.
3. $-1 \notin P$.
4. either $x \in P$ or $-x \in P$ for all $x \in R$.
I have checked that the function $F$ taking a ring total order $\leq$ to the set $\{x \in R | x \geq 0\}$ and the function $G$ taking a positive cone $P$ to the relation $x \leq y \equiv y - x \in P$ are inverses.
I have also shown that if $\leq$ is a ring total order then $F(\leq)$ is a positive cone. However, I don't see how to show that if $P$ is a positive cone then $G(P)$ is a total order. In particular, I don't see how to show that $G(P)$ is anti-symmetric without using cancellation.
So I am wondering whether this equivalence holds or not. If not, is there a simple tweak to the positive cone style definition that makes it work? I'm interested because I think it is more natural to implement this style of definition in the algebra software I'm designing.
-
## 2 Answers
The equivalence does not hold in $\:\mathbb{Z}\left[\sqrt0\hspace{.01 in}\right]\:$ with $\;\;\;\; P \;\; = \;\; \left\{ \:a+\left(b\cdot \sqrt0\right) \: : \: 0\leq a \: \right\} \;\;\;\;$.
Replacing 3 with $\;\;$ "For all $\:x\in R\:$,$\:$ if $\:x\in P\:$ and $\: -x \in P \:$ then $\:x=0\:$." $\;\;$ will make it work.
(Since that is exactly what you need for anti-symmetry.)
-
Thanks! And that would be a suitable definition for fields as well since in a field (or an integral domain even) the two are equivalent. – mdgeorge Nov 29 '12 at 20:59
The definition has some problems still. First, it's not necessary to suppose that $x^2$ is in $P$ (in the strict ordering variant I describe in my answer, this doesn't quite hold). In any case, if $x\in P$, then we have $x^2\in P$ by (1). Otherwise, $-x\in P$, so $x^2 = (-x)^2 \in P$. In the weak ordering case, $0\in P$ so $0^2\in P$. In the strict ordering case, this doesn't apply. The second problem is that $-1$ only exists in a ring with unity. But in a (non-null) ring with unity, $1≠0$, so $1=1^2 \in P$. Thus $-1 \notin P$ by the antisymmetry condition. – dfeuer Feb 2 at 19:33
I've been messing around with this notion for groups. Suppose you have a group $(G,\circ)$ with identity $e$ and a subset $C$ of $G$ such that for each $x,y\in G$:
$x,y \in C \implies x \circ y \in C$
$x\circ y \in C \implies y \circ x \in C$.
Then if you define a relation $\mathcal R$ by letting $x \mathrel{\mathcal R} y \iff y\circ x^{-1}\in C$, then that relation will be "compatible" with $\circ$ in the sense that if $x \mathrel{\mathcal R} y$ then $x \circ z \mathrel{\mathcal R} y \circ z$ and $z \circ x \mathrel{\mathcal R} z \circ y$. The relation will also be transitive.
If you impose the condition that $C \cap C^{-1} \subseteq \{e\}$, then the relation will be antisymmetric.
If you impose the condition that the group identity $e\in C \cap C^{-1}$, then the relation will be reflexive. Combined with the condition for antisymmetry, this gives a weak partial order. If instead you impose the condition that $e\notin C \cap C^{-1}$, then you get an irreflexive relation which, combined with the antisymmetry condition, gives a strict partial order. These combined conditions can be written as $C\cap C^{-1}=\{e\}$ and $C \cap C^{-1} = \varnothing$, respectively.
If you add to the weak partial order condition the condition that $C\cup C^{-1} = G$, you get a weak total order. If to the strict partial order condition you add the condition that $C\cup C^{-1}\cup\{e\}=G$, you get a strict total order.
All that is good for general groups. In a ring $(R,+,\cdot)$ you need to impose one of the above-described conditions for $(R,+)$, and add the condition that $x,y \in C \implies x\cdot y \in C$.
There are various definitions for a partially ordered field, but the definition that makes the most sense to me requires that $x \in C \implies x^{-1} \in C$. This added condition is automatically satisfied for a totally ordered field.
- | {"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": 68, "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": 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": 0.9331418871879578, "perplexity_flag": "head"} |
http://www.chegg.com/homework-help/questions-and-answers/a-coaxial-cable-resistor-of-length-l-consists-of-two-concentric-cylinders-the-inner-cylind-q3248414 | home / homework help / questions and answers / engineering / electrical engineering / a coaxial cable resistor of...
Close
## Resistance of a coaxial cable
A coaxial cable resistor of length $$l$$ consists of two concentric cylinders. The inner cylinder has radius a and is made of a material with conductivity $$\sigma_1$$ , and the outer cylinder, extending between r=a and r=b, is made of a material with conductivity $$\sigma_2$$ . If the two ends of the resistor are capped with conducting plates, show that the resistance between the two ends is given by
$$R= l/(\pi(\sigma_1a^2+\sigma_1(b^2-a^2)))$$
# Answers (0)
There are no answers to this question yet.
#### Company
Chegg Plants Trees | {"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": 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": 0.9353753328323364, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/206862-finding-real-parts-equation.html | # Thread:
1. ## Finding the real parts of this equation
Finding the real parts to both sides of this equation
I am really stuck trying to find the real parts to both sides of this equation.
1 + exp(theta(i)) + exp(2theta(i)) + .... + exp((n-1)theta(i)) = (exp(ntheta(i)) - 1) / (exp(theta(i)) -1)
I think that the real part of the left hand side is
1 + cos(theta) + cos(2theta) + ..... + cos((n-1)theta)
i have no idea what the real part is for the right.
2. ## Re: Finding the real parts of this equation
Originally Posted by Matt1993
Finding the real parts to both sides of this equation
I am really stuck trying to find the real parts to both sides of this equation.
1 + exp(theta(i)) + exp(2theta(i)) + .... + exp((n-1)theta(i)) = (exp(ntheta(i)) - 1) / (exp(theta(i)) -1)
I think that the real part of the left hand side is
1 + cos(theta) + cos(2theta) + ..... + cos((n-1)theta)
i have no idea what the real part is for the right.
$\displaystyle \begin{align*} \frac{e^{n\theta i} - 1}{e^{\theta i} - 1} &= \frac{\cos{(n\theta)} + i\sin{(n\theta)} - 1}{\cos{(\theta)} + i\sin{(\theta)} - 1} \\ &= \frac{\left[ \cos{(n\theta)} - 1 + i\sin{(n\theta )} \right] \left[\cos{(\theta)} - 1 - i\sin{(\theta)}\right]}{\left[ \cos{(\theta )} - 1 + i\sin{(\theta )} \right] \left[ \cos{(\theta )} - 1 - i\sin{(\theta )} \right]} \end{align*}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.8664016127586365, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions?sort=featured | All Questions
0answers
123 views
+150
Checking whether an element is in all inclusion-wise maximal common independent sets of two matroids
Given two matroids $M$ and $M'$ over the same universe $E$, and some element $x \in E$, I am interested in the importance of $x$ for the intersection (the common independent sets) …
5answers
638 views
+100
Another colored balls puzzle (part II)
The same colleague as in http://mathoverflow.net/questions/130489/another-colored-balls-puzzle asked me the following variant which she called "part II". Imagine you have $n$ ball …
2answers
418 views
+150
Minimum off-diagonal elements of a matrix with fixed eigenvalues
Hello, I am en engineer working in radar research. I came accross a problem I cannot seem to find math literature on it. I can ask it in two different ways. Perhaps depending on … | {"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": false, "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": 0.9061713814735413, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2010/08/06/the-metric-space-of-a-measure-ring/?like=1&source=post_flair&_wpnonce=5b12459335 | # The Unapologetic Mathematician
## The Metric Space of a Measure Ring
Let $(\mathcal{S},\mu)$ be a measure ring. We’ve seen how we can get a metric space from a measure, and the same is true here. In fact, since we’ve required that $\mu$ be positive — that $\mu(E)=0$ only if $E=\emptyset$ — we don’t need to worry about negligible elements.
And so we write $\mathfrak{S}=\mathfrak{S}(\mu)$ for the metric space consisting of the elements $E\in\mathcal{S}$ with $\mu(E)<\infty$. This has a distance function $\rho$ defined by $\rho(E,F)=\mu(E\Delta F)$. We also write $\mathfrak{S}=\mathfrak{S}(\mu)$ for the metric associated with the measure algebra associated with the measure space $(X,\mathcal{S},\mu)$. We say that a measure space or a measure algebra is “separable” if the associated metric space is separable.
Now, if we set
$\displaystyle\begin{aligned}f(E,F)&=E\cup F\\g(E,F)&=E\cap F\end{aligned}$
then $f:\mathfrak{S}\times\mathfrak{S}\to\mathfrak{S}$, $g:\mathfrak{S}\times\mathfrak{S}\to\mathfrak{S}$, and $\mu:\mathfrak{S}\to\mathbb{R}$ itself are all uniformly continuous.
Indeed, if we take two pairs of sets $E_1$, $E_2$, $F_1$, and $F_2$, we calculate
$\displaystyle\begin{aligned}\mu\left((E_1\cup F_1)\setminus(E_2\cup F_2)\right)&=\mu\left((E_1\cup F_1)\cap(E_2^c\cap F_2^c)\right)\\&=\mu\left((E_1\cup F_1)\cap(E_2^c\cap F_2^c)\right)\\&=\mu\left((E_1\cap E_2^c\cap F_2^c)\cup(F_1\cap E_2^c\cap F_2^c)\right)\\&\leq\mu(E_1\cap E_2^c\cap F_2^c)+\mu(F_1\cap E_2^c\cap F_2^c)\\&\leq\mu(E_1\setminus E_2)+\mu(F_1\setminus F_2)\end{aligned}$
Similarly, we find that $\mu\left((E_2\cup F_2)\setminus(E_1\cup F_1)\right)\leq\mu(E_2\setminus E_1)+\mu(F_2\setminus F_1)$. And thus
$\displaystyle\begin{aligned}\rho\left((E_1\cup F_1),(E_2\cup F_2)\right)&=\mu\left((E_1\cup F_1)\Delta(E_2\cup F_2)\right)\\&=\mu\left((E_1\cup F_1)\setminus(E_2\cup F_2)\right)+\mu\left((E_2\cup F_2)\setminus(E_1\cup F_1)\right)\\&\leq\mu(E_1\setminus E_2)+\mu(F_1\setminus F_2)+\mu(E_2\setminus E_1)+\mu(F_2\setminus F_1)\\&=\mu(E_1\Delta E_2)+\mu(F_1\Delta F_2)\\&=\rho(E_1,E_2)+\rho(F_1,F_2)\end{aligned}$
And so if we have control over the distance between $E_1$ and $E_2$, and the distance between $F_1$ and $F_2$, then we have control over the distance between $E_1\cup F_1$ and $E_2\cup F_2$. The bounds we need on the inputs uniform, and so $f$ is uniformly continuous. The proof for $g$ proceeds similarly.
To see that $\mu$ is uniformly continuous, we calculate
$\displaystyle\begin{aligned}\left\lvert\mu(E)-\mu(F)\right\rvert&=\left\lvert\left(\mu(E)-\mu(E\cap F)\right)-\left(\mu(F)-\mu(F\cap E)\right)\right\rvert\\&=\left\lvert\mu\left(E\setminus(E\cap F)\right)-\mu\left(F\setminus(F\cap E)\right)\right\rvert\\&=\left\lvert\mu(E\setminus F)-\mu(F\setminus E)\right\rvert\\&\leq\mu(E\setminus F)+\mu(F\setminus E)\\&=\mu(E\Delta F)\\&=\rho(E,F)\end{aligned}$
Now if $(X,\mathcal{S},\mu)$ is a $\sigma$-finite measure space so that the $\sigma$-ring $\mathcal{S}$ has a countable set of generators, then $\mathfrak{S}(\mu)$ is separable. Indeed, if $\{E_n\}$ is a countable sequence of sets that generate $\mathcal{S}$, then we may assume (by $\sigma$-finiteness) that $\mu(E_n)<\infty$ for all $n$. The ring generated by the $\{E_n\}$ is itself countable, and so we may assume that $\{E_n\}$ is itself a ring. But then we know that for every $E\in\mathfrak{S}(\mu)$ and for every positive $\epsilon$ we can find some ring element $E_n$ so that $\rho(E,E_n)=\mu(E\Delta E_n)<\epsilon$. Thus $\{E_n\}$ is a countable dense set in $\mathfrak{S}(\mu)$, which is thus separable.
### Like this:
Posted by John Armstrong | Analysis, Measure Theory
## 5 Comments »
1. [...] of the Metric Space of a Measure Space Our first result today is that the metric space associated to the measure ring of a measure space is [...]
Pingback by | August 9, 2010 | Reply
2. [...] of Metric Spaces of Measure Rings Today we collect a few facts about the metric space associated to a measure ring [...]
Pingback by | August 12, 2010 | Reply
3. [...] Metric Spaces and Absolutely Continuous Measures I If is the metric space associated to a measure space , and if is a finite signed measure that is absolutely continuous with respect to [...]
Pingback by | August 16, 2010 | Reply
4. [...] continuous finite signed measure on a measure space defines a continuous function on the associated metric space , and that a sequence of such finite signed measures that converges pointwise is actually uniformly [...]
Pingback by | August 17, 2010 | Reply
5. [...] the metric space has a diameter of — no two sets can differ by more than , and — we can find a dense [...]
Pingback by | August 25, 2010 | Reply
« Previous | Next »
## 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": 50, "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": 0.9334895014762878, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/27193/consideration-of-static-atomic-displacements-in-electronic-structure-calculation | # Consideration of static atomic displacements in electronic structure calculations
I am hoping to discuss some details of electronic structure calculations. I am not an expert on this topic, so please forgive any abuse of terminology. It is my understanding that first principles electronic structure calculations are based on perfect translational symmetry of a crystalline lattice. Further, that the first round of calculations traditionally treat the atomic nuclei as fixed points due to the large disparity in nuclei and electron velocities, and after the initial electronic structure is solved, one can then turn on the nuclear motions and understand the interaction between the nuclear and electronic motions. The heart of my question is: is it possible to investigate local distortions of structures using existing techniques? By a local distortion, I mean a distortion which does not disrupt the long range crystalline periodicity, but results in a very different local atomic environment?
-
Since I was looking at this I took the opportunity to put some tags on it, but feel free to fix the tags if there are better choices. – David Zaslavsky♦ May 16 '12 at 18:19
## 1 Answer
A technique is to perform a supercell calculation where in the calculation is performed over several cells instead of a single unit cell. The desired change is made to a single cell within this larger structure. This does make your change periodic, but if the supercell is large enough, then the change can be made essentially local. This method has been used to generate electronic structures for substitutional series like FeSi$_{(1-x)}$Ge$_x$ and $\text{X}_x\text{W}_{(1-x)}\text{O}_3$ (X=Nb,V,Re) (link). So, a small structural change is not out of the question. I know wien2k and quantum espresso have a scripts for generating supercell, but I don't know what other major packages do. VASP does not seem to.
As to how big a supercell to make, I don't know. I have not actively looked for one, but I have not found a resource giving such guidance. I'd suggest starting as small as possible, possibly $2\times2\times1$ or $2\times2\times2$, and increasing the size until the effect seems localized to the interior of the supercell. Obviously, this will put a strain on your computing power, so the smaller the better.
-
1
the preview of the second formula looks correct, but the "published" version does not. Hmmm. – rcollyer Dec 24 '11 at 12:16 | {"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": 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": 0.9479842185974121, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/182950/what-is-x-and-1n | # What is $|X|$ and $1n$?
I'm doing a course in calculus but I'm quite young so haven't learnt all the mathematical terms one should perhaps no before learning calculus, and these 2 terms keep coming up and I have no idea in the world of what they are:
"$|X|$" and "1n".
I am really sorry that this probably seems ridiculously elementary but I really want to know what they are.
What are they?
-
For the meaning of absolute value, and information about it, please follow this link. I leave the search for natural logarithm to you. – André Nicolas Aug 15 '12 at 19:48
## 2 Answers
$|x|$ is the absolute value of $x$. $\ln(x)$ (that's ell, not one) is the natural logarithm of $x$.
-
That depends on the context. $\lvert X\rvert$ is usually used to denote the cardinality of the set $X$ (esp. since you used capital letter, which are usually used for sets). On the other hand, if $X$ is a real (or complex) number, it is almost surely the absolute value of $X$.
If the symbol you refer to is $\ln$ as Robert guessed, it is the natural (base $e$) logarithm. If it is $1_n$, then I would guess it is the characteristic function of $\{n\}$, the function which takes value $1$ at $n$ and $0$ elsewhere...
- | {"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": 0.18, "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": 0.956048309803009, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/112892/club-in-a-singular-cardinal/112911 | # Club in a singular cardinal.
Let $\kappa$ be a singular cardinal with $\operatorname{cf}\kappa = \lambda > \omega$.
Let $C = \{ \alpha_{\zeta} \mid \zeta < \lambda \}$ be a strictly increasing continuous sequence of cardinals with limit $\kappa$. I want to show this is a closed unbounded set in $\kappa$.
Obviously it is unbounded, because $\sup C = \kappa$, but how do I show it is closed? Is it to do with continuity? Also I am used to dealing with clubs in regular cardinals, are there any differences for clubs in singular cardinals?
Thanks very much.
-
## 2 Answers
Yes, you are correct. The fact that $C$ is closed has everything to do with the sequence $\{ \alpha_\zeta \}_{\zeta < \lambda}$ being continuous.
The main idea is that since the sequence is strictly increasing, then for any limit ordinal $\gamma < \lambda$ we have that $\lim_{\zeta < \gamma} \alpha_\zeta = \sup ( \{ \alpha_\zeta : \zeta < \gamma \} )$. Therefore, if $\alpha < \kappa$ is a limit ordinal such that $\sup ( C \cap \alpha ) = \alpha$, let $\gamma = \min \{ \zeta < \lambda : \alpha_\zeta \geq \alpha \}$. We can show that $\gamma$ must be a limit ordinal, and that $C \cap \alpha = \{ \alpha_\zeta : \zeta < \gamma \}$. Using continuity we have that $$\alpha_\gamma = \lim_{\zeta < \gamma} \alpha_\zeta = \sup ( \{ \alpha_\zeta : \zeta < \gamma \} ) = \sup ( C \cap \alpha ) = \alpha,$$ and so $\alpha \in C$.
-
Thanks, but why is $\gamma$ a limit ordinal? I can't find a contradiction if I suppose it is a successor. Even it if it's not a limit ordinal, surely $\alpha_{\gamma} = \lim_{\zeta \to \gamma} \alpha_{\zeta}$ still holds? – Paul Slevin Feb 27 '12 at 15:12
No sorry, that is wrong. If $\gamma = \beta + 1$, then $\alpha_\beta = \lim_{\zeta \to \gamma}\alpha_{\zeta}$. I think the answer to my question follows after showing that $C \cap \alpha = \{ \alpha_{\zeta} \mid \zeta < \gamma \}$, and then we get a contradiction if we assume $\gamma$ is a successor. – Paul Slevin Feb 27 '12 at 15:22
@Paul: You're correct that the contradiction will follow once you show that $C \cap \alpha = \{ \alpha_\zeta : \zeta < \gamma \}$. This little observation stems from the definition of $\gamma$ and the fact that the original sequence is strictly increasing. – Arthur Fischer Feb 27 '12 at 16:24
Recall the definition of a closed and unbounded set in $\kappa$:
1. For every $\alpha<\kappa$ there is some $\alpha_\zeta\in C$ such that $\alpha<\alpha_\zeta$ (unboundedness);
2. For every limit ordinal $\delta<\lambda$ we have $\sup\{\alpha_\zeta\mid\zeta<\delta\}=\alpha_\delta$ (continuity).
Now if $\sup C=\kappa$ then it is clearly unbounded in $\kappa$, and continuity is exactly the demand that the sequence is closed. As you may recall, a continuous sequence is the range of a continuous function with respect to the order topology.
As for clubs of singular cardinals, those are indeed a bit different. In regular cardinals being a club means being a very large set in every possible way, while clubs in singular cardinals can be quite small in terms of cardinality.
-
I forget, Paul, did we discuss the continuity in the previous answer or on the chat? – Asaf Karagila Feb 24 '12 at 16:35
you defined a new sequence in your previous answer, which was continuous – Paul Slevin Feb 27 '12 at 14:12
Or if you mean in terms of continuity in continuous functions, we talked about it in chat a little bit – Paul Slevin Feb 27 '12 at 14:28 | {"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": 32, "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": 0.9585100412368774, "perplexity_flag": "head"} |
http://nrich.maths.org/7474 | nrich enriching mathematicsSkip over navigation
Weekly Challenge 43: A Close Match
Can you massage the parameters of these curves to make them match as closely as possible?
Weekly Challenge 44: Prime Counter
A weekly challenge concerning prime numbers.
Weekly Challenge 28: the Right Volume
Can you rotate a curve to make a volume of 1?
Weekly Challenge 22: Combinations of Two
Stage: 4 and 5 Short Challenge Level:
Can you use this diagram to prove that the number of different pairs of objects which can be chosen from six objects, $^6C_2$, is $$1 + 2 + 3 + 4 + 5?$$
Generalise this to show that the number of ways of choosing pairs from $n$ objects is
$$^nC_2 = 1 + 2 + ...+ (n-1) = \frac{1}{2}n (n - 1).$$
Did you know ... ?
The sum of the first $n$ whole numbers is called a triangle number because this sum can be represented geometrically by a triangular array of dots. The sum is easily found by working out the number of dots in the parallelogram formed by putting two triangular arrays side by side.
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": 3, "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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9100754857063293, "perplexity_flag": "middle"} |
http://www.reference.com/browse/game+theory | Definitions
Nearby Words
game theory
Branch of applied mathematics devised to analyze certain situations in which there is an interplay between parties that may have similar, opposed, or mixed interests. Game theory was originally developed by John von Neumann and Oscar Morgenstern in their book The Theory of Games and Economic Behavior (1944). In a typical game, or competition with fixed rules, “players” try to outsmart one another by anticipating the others' decisions, or moves. A solution to a game prescribes the optimal strategy or strategies for each player and predicts the average, or expected, outcome. Until a highly contrived counterexample was devised in 1967, it was thought that every contest had at least one solution. Seealso decision theory; prisoner's dilemma.
Encyclopedia Britannica, 2008. Encyclopedia Britannica Online.
Game theory is a branch of applied mathematics that is used in the social sciences (most notably economics), biology, engineering, political science, computer science (mainly for artificial intelligence), and philosophy. Game theory attempts to mathematically capture behavior in strategic situations, in which an individual's success in making choices depends on the choices of others. While initially developed to analyze competitions in which one individual does better at another's expense (zero sum games), it has been expanded to treat a wide class of interactions, which are classified according to several criteria. Today, "game theory is a sort of umbrella or 'unified field' theory for the rational side of social science, where 'social' is interpreted broadly, to include human as well as non-human players (computers, animals, plants)" .
Traditional applications of game theory attempt to find equilibria in these games—sets of strategies in which individuals are unlikely to change their behavior. Many equilibrium concepts have been developed (most famously the Nash equilibrium) in an attempt to capture this idea. These equilibrium concepts are motivated differently depending on the field of application, although they often overlap or coincide. This methodology is not without criticism, and debates continue over the appropriateness of particular equilibrium concepts, the appropriateness of equilibria altogether, and the usefulness of mathematical models more generally.
Although some developments occurred before it, the field of game theory came into being with the 1944 book by John von Neumann and Oskar Morgenstern. This theory was developed extensively in the 1950s by many scholars. Game theory was later explicitly applied to biology in the 1970s, although similar developments go back at least as far as the 1930s. Game theory has been widely recognized as an important tool in many fields. Eight game theorists have won Nobel prizes in economics, and John Maynard Smith was awarded the Crafoord Prize for his application of game theory to biology.
Representation of games
The games studied in game theory are well-defined mathematical objects. A game consists of a set of players, a set of moves (or strategies) available to those players, and a specification of payoffs for each combination of strategies. Most cooperative games are presented in the characteristic function form, while the extensive and the normal forms are used to define noncooperative games.
Extensive form
The extensive form can be used to formalize games with some important order. Games here are often presented as trees (as pictured to the left). Here each vertex (or node) represents a point of choice for a player. The player is specified by a number listed by the vertex. The lines out of the vertex represent a possible action for that player. The payoffs are specified at the bottom of the tree.
In the game pictured here, there are two players. Player 1 moves first and chooses either F or U. Player 2 sees Player 1's move and then chooses A or R. Suppose that Player 1 chooses U and then Player 2 chooses A, then Player 1 gets 8 and Player 2 gets 2.
The extensive form can also capture simultaneous-move games and games with incomplete information. To represent it, either a dotted line connects different vertices to represent them as being part of the same information set (i.e., the players do not know at which point they are), or a closed line is drawn around them.
Normal form
The normal (or strategic form) game is usually represented by a matrix which shows the players, strategies, and payoffs (see the example to the right). More generally it can be represented by any function that associates a payoff for each player with every possible combination of actions. In the accompanying example there are two players; one chooses the row and the other chooses the column. Each player has two strategies, which are specified by the number of rows and the number of columns. The payoffs are provided in the interior. The first number is the payoff received by the row player (Player 1 in our example); the second is the payoff for the column player (Player 2 in our example). Suppose that Player 1 plays Up and that Player 2 plays Left. Then Player 1 gets a payoff of 4, and Player 2 gets 3.
When a game is presented in normal form, it is presumed that each player acts simultaneously or, at least, without knowing the actions of the other. If players have some information about the choices of other players, the game is usually presented in extensive form.
Characteristic function form
In cooperative games with transferable utility no individual payoffs are given. Instead, the characteristic function determines the payoff of each coalition. The standard assumption is that the empty coalition obtains a payoff of 0.
The origin of this form is to be found in the seminal book of von Neumann and Morgenstern who, when studying coalitional normal form games, assumed that when a coalition $C$ forms, it plays against the complementary coalition ($Nsetminus C$) as if they were playing a 2-player game. The equilibrium payoff of $C$ is characteristic. Now there are different models to derive coalitional values from normal form games, but not all games in characteristic function form can be derived from normal form games.
Formally, a characteristic function form game (also known as a TU-game) is given as a pair $\left(N,v\right)$, where $N$ denotes a set of players and $v:2^Nlongrightarrowmathbb\left\{R\right\}$ is a characteristic function.
The characteristic function form has been generalised to games without the assumption of transferable utility.
Partition function form
The characteristic function form ignores the possible externalities of coalition formation. In the partition function form the payoff of a coalition depends not only on its members, but also on the way the rest of the players are partitioned .
Application and challenges
Game theory has been used to study a wide variety of human and animal behaviors. It was initially developed in economics to understand a large collection of economic behaviors, including behaviors of firms, markets, and consumers. The use of game theory in the social sciences has expanded, and game theory has been applied to political, sociological, and psychological behaviors as well.
Game theoretic analysis was initially used to study animal behavior by Ronald Fisher in the 1930s (although even Charles Darwin makes a few informal game theoretic statements). This work predates the name "game theory", but it shares many important features with this field. The developments in economics were later applied to biology largely by John Maynard Smith in his book .
In addition to being used to predict and explain behavior, game theory has also been used to attempt to develop theories of ethical or normative behavior. In economics and philosophy, scholars have applied game theory to help in the understanding of good or proper behavior. Game theoretic arguments of this type can be found as far back as Plato.
Political science
The application of game theory to political science is focused in the overlapping areas of fair division, political economy, public choice, positive political theory, and social choice theory. In each of these areas, researchers have developed game theoretic models in which the players are often voters, states, special interest groups, and politicians.
For early examples of game theory applied to political science, see the work of Anthony Downs. In his book An Economic Theory of Democracy , he applies a hotelling firm location model to the political process. In the Downsian model, political candidates commit to ideologies on a one-dimensional policy space. The theorist shows how the political candidates will converge to the ideology preferred by the median voter. For more recent examples, see the books by Steven Brams, George Tsebelis, Gene M. Grossman and Elhanan Helpman, or David Austen-Smith and Jeffrey S. Banks.
A game-theoretic explanation for democratic peace is that public and open debate in democracies send clear and reliable information regarding their intentions to other states. In contrast, it is difficult to know the intentions of nondemocratic leaders, what effect concessions will have, and if promises will be kept. Thus there will be mistrust and unwillingness to make concessions if at least one of the parties in a dispute is a nondemocracy .
Economics and business
Economists have long used game theory to analyze a wide array of economic phenomena, including auctions, bargaining, duopolies, fair division, oligopolies, social network formation, and voting systems. This research usually focuses on particular sets of strategies known as equilibria in games. These "solution concepts" are usually based on what is required by norms of rationality. In non-cooperative games, the most famous of these is the Nash equilibrium. A set of strategies is a Nash equilibrium if each represents a best response to the other strategies. So, if all the players are playing the strategies in a Nash equilibrium, they have no unilateral incentive to deviate, since their strategy is the best they can do given what others are doing.
The payoffs of the game are generally taken to represent the utility of individual players. Often in modeling situations the payoffs represent money, which presumably corresponds to an individual's utility. This assumption, however, can be faulty.
A prototypical paper on game theory in economics begins by presenting a game that is an abstraction of some particular economic situation. One or more solution concepts are chosen, and the author demonstrates which strategy sets in the presented game are equilibria of the appropriate type. Naturally one might wonder to what use should this information be put. Economists and business professors suggest two primary uses.
Descriptive
The first known use is to inform us about how actual human populations behave. Some scholars believe that by finding the equilibria of games they can predict how actual human populations will behave when confronted with situations analogous to the game being studied. This particular view of game theory has come under recent criticism. First, it is criticized because the assumptions made by game theorists are often violated. Game theorists may assume players always act in a way to directly maximize their wins (the Homo economicus model), but in practice, humans behaviour is often contrary to this model. Explanations of this phenomenon are many; irrationality, new models of deliberation, or even different motives (like that of altruism). Game theorists respond by comparing their assumptions to those used in physics. Thus while their assumptions do not always hold, they can treat game theory as a reasonable scientific ideal akin to the models used by physicists. However, additional criticism of this use of game theory has been levied because some experiments have demonstrated that individuals do not play equilibrium strategies. For instance, in the centipede game, guess 2/3 of the average game, and the dictator game, people regularly do not play Nash equilibria. There is an ongoing debate regarding the importance of these experiments.
Alternatively, some authors claim that Nash equilibria do not provide predictions for human populations, but rather provide an explanation for why populations that play Nash equilibria remain in that state. However, the question of how populations reach those points remains open.
Some game theorists have turned to evolutionary game theory in order to resolve these worries. These models presume either no rationality or bounded rationality on the part of players. Despite the name, evolutionary game theory does not necessarily presume natural selection in the biological sense. Evolutionary game theory includes both biological as well as cultural evolution and also models of individual learning (for example, fictitious play dynamics).
Prescriptive or normative analysis
On the other hand, some scholars see game theory not as a predictive tool for the behavior of human beings, but as a suggestion for how people ought to behave. Since a Nash equilibrium of a game constitutes one's best response to the actions of the other players, playing a strategy that is part of a Nash equilibrium seems appropriate. However, this use for game theory has also come under criticism. First, in some cases it is appropriate to play a non-equilibrium strategy if one expects others to play non-equilibrium strategies as well. For an example, see Guess 2/3 of the average.
Second, the Prisoner's dilemma presents another potential counterexample. In the Prisoner's Dilemma, each player pursuing his own self-interest leads both players to be worse off than had they not pursued their own self-interests.
Biology
Unlike economics, the payoffs for games in biology are often interpreted as corresponding to fitness. In addition, the focus has been less on equilibria that correspond to a notion of rationality, but rather on ones that would be maintained by evolutionary forces. The best known equilibrium in biology is known as the Evolutionarily stable strategy or (ESS), and was first introduced in . Although its initial motivation did not involve any of the mental requirements of the Nash equilibrium, every ESS is a Nash equilibrium.
In biology, game theory has been used to understand many different phenomena. It was first used to explain the evolution (and stability) of the approximate 1:1 sex ratios. suggested that the 1:1 sex ratios are a result of evolutionary forces acting on individuals who could be seen as trying to maximize their number of grandchildren.
Additionally, biologists have used evolutionary game theory and the ESS to explain the emergence of animal communication . The analysis of signaling games and other communication games has provided some insight into the evolution of communication among animals. For example, the Mobbing behavior of many species, in which a large number of prey animals attack a larger predator, seems to be an example of spontaneous emergent organization.
Biologists have used the hawk-dove game (also known as chicken) to analyze fighting behavior and territoriality.
Maynard Smith, in the preface to Evolution of the Theory of Games writes, "[p]aradoxically, it has turned out that game theory is more readily applied to biology than to the field of economic behaviour for which it was originally designed." Evolutionary game theory has been used to explain many seemingly incongruous phenomena in nature.
One such phenomena is known as biological altruism. This is a situation in which an organism appears to act in a way that benefits other organisms and is detrimental to itself. This is distinct from traditional notions of altruism because such actions are not conscious, but appear to be evolutionary adaptations to increase overall fitness. Examples can be found in species ranging from vampire bats that regurgitate blood they have obtained from a night’s hunting and give it to group members who have failed to feed, to worker bees that care for the queen bee for their entire lives and never mate, to Vervet monkeys that warn group members of a predator’s approach, even when it endangers that individual’s chance of survival. All of these actions increase the overall fitness of a group, but occur at a cost to the individual.
Evolutionary game theory explains this altruism with the idea of kin selection. Altruists discriminate between the individuals they help and favor relatives. Hamilton’s rule explains the evolutionary reasoning behind this selection with the equation cRecent applications of biological game theory to humans has garnered some criticism because evolutionary analysis cannot provide a value-neutral evaluation of a given cultural situation. The valuations of whether an action is good or bad constitute a normative judgment of whether an action is altruistic. Altruism also has a different socially constructed meaning in the context of human society because altruistic actions within culture are not all instinctually driven and do not always result in increased fitness for a group.
Computer science and logic
Game theory has come to play an increasingly important role in logic and in computer science. Several logical theories have a basis in game semantics. In addition, computer scientists have used games to model interactive computations. Also, game theory provides a theoretical basis to the field of multi-agent systems.
Separately, game theory has played a role in online algorithms. In particular, the k-server problem, which has in the past been referred to as games with moving costs and request-answer games . Yao's principle is a game-theoretic technique for proving lower bounds on the computational complexity of randomized algorithms, and especially of online algorithms.
Radio networks
Game theory has recently become a useful tool for modeling and studying interactions between cognitive radios envisioned to operate in future communications systems. Such terminals will have the capability to adapt to the context they operate in, through possibly power and rate control as well as channel selection. Software agents embedded in these terminals will potentially be selfish, meaning they will only try to maximize the throughput/connectivity of the terminal they function for, as opposed to maximizing the welfare (total capacity) of the system they operate in. Thus, the potential interactions among them can be modeled through non-cooperative games. The researchers in this field often strive to determine the stable operating points of systems composed of such selfish terminals, and try to come up with a minimum set of rules (etiquette) so as to make sure that the optimality loss compared to a cooperative - centrally controlled setting- is kept at a minimum.
Philosophy
Game theory has been put to several uses in philosophy. Responding to two papers by , used game theory to develop a philosophical account of convention. In so doing, he provided the first analysis of common knowledge and employed it in analyzing play in coordination games. In addition, he first suggested that one can understand meaning in terms of signaling games. This later suggestion has been pursued by several philosophers since Lewis (). In ethics, some authors have attempted to pursue the project, begun by Thomas Hobbes, of deriving morality from self-interest. Since games like the Prisoner's dilemma present an apparent conflict between morality and self-interest, explaining why cooperation is required by self-interest is an important component of this project. This general strategy is a component of the general social contract view in political philosophy (for examples, see and .
Other authors have attempted to use evolutionary game theory in order to explain the emergence of human attitudes about morality and corresponding animal behaviors. These authors look at several games including the Prisoner's dilemma, Stag hunt, and the Nash bargaining game as providing an explanation for the emergence of attitudes about morality (see, e.g., and ).
Some assumptions used in some parts of game theory have been challenged in philosophy; psychological egoism states that rationality reduces to self-interest—a claim debated among philosophers. ()
Types of games
Cooperative or non-cooperative
A game is cooperative if the players are able to form binding commitments. For instance the legal system requires them to adhere to their promises. In noncooperative games this is not possible.
Often it is assumed that communication among players is allowed in cooperative games, but not in noncooperative ones. This classification on two binary criteria has been rejected .
Of the two types of games, noncooperative games are able to model situations to the finest details, producing accurate results. Cooperative games focus on the game at large. Considerable efforts have been made to link the two approaches. The so-called Nash-programme has already established many of the cooperative solutions as noncooperative equilibria.
Hybrid games contain cooperative and non-cooperative elements. For instance, coalitions of players are formed in a cooperative game, but these play in a non-cooperative fashion.
Symmetric and asymmetric
A symmetric game is a game where the payoffs for playing a particular strategy depend only on the other strategies employed, not on who is playing them. If the identities of the players can be changed without changing the payoff to the strategies, then a game is symmetric. Many of the commonly studied 2×2 games are symmetric. The standard representations of chicken, the prisoner's dilemma, and the stag hunt are all symmetric games. Some scholars would consider certain asymmetric games as examples of these games as well. However, the most common payoffs for each of these games are symmetric.
Most commonly studied asymmetric games are games where there are not identical strategy sets for both players. For instance, the ultimatum game and similarly the dictator game have different strategies for each player. It is possible, however, for a game to have identical strategies for both players, yet be asymmetric. For example, the game pictured to the right is asymmetric despite having identical strategy sets for both players.
Zero sum and non-zero sum
Zero sum games are a special case of constant sum games, in which choices by players can neither increase nor decrease the available resources. In zero-sum games the total benefit to all players in the game, for every combination of strategies, always adds to zero (more informally, a player benefits only at the equal expense of others). Poker exemplifies a zero-sum game (ignoring the possibility of the house's cut), because one wins exactly the amount one's opponents lose. Other zero sum games include matching pennies and most classical board games including Go and chess.
Many games studied by game theorists (including the famous prisoner's dilemma) are non-zero-sum games, because some outcomes have net results greater or less than zero. Informally, in non-zero-sum games, a gain by one player does not necessarily correspond with a loss by another.
Constant sum games correspond to activities like theft and gambling, but not to the fundamental economic situation in which there are potential gains from trade. It is possible to transform any game into a (possibly asymmetric) zero-sum game by adding an additional dummy player (often called "the board"), whose losses compensate the players' net winnings.
Simultaneous and sequential
Simultaneous games are games where both players move simultaneously, or if they do not move simultaneously, the later players are unaware of the earlier players' actions (making them effectively simultaneous). Sequential games (or dynamic games) are games where later players have some knowledge about earlier actions. This need not be perfect information about every action of earlier players; it might be very little knowledge. For instance, a player may know that an earlier player did not perform one particular action, while he does not know which of the other available actions the first player actually performed.
The difference between simultaneous and sequential games is captured in the different representations discussed above. Often, normal form is used to represent simultaneous games, and extensive form is used to represent sequential ones; although this isn't a strict rule in a technical sense.
Perfect information and imperfect information
An important subset of sequential games consists of games of perfect information. A game is one of perfect information if all players know the moves previously made by all other players. Thus, only sequential games can be games of perfect information, since in simultaneous games not every player knows the actions of the others. Most games studied in game theory are imperfect information games, although there are some interesting examples of perfect information games, including the ultimatum game and centipede game. Perfect information games include also chess, go, mancala, and arimaa.
Perfect information is often confused with complete information, which is a similar concept. Complete information requires that every player know the strategies and payoffs of the other players but not necessarily the actions.
Infinitely long games
Games, as studied by economists and real-world game players, are generally finished in a finite number of moves. Pure mathematicians are not so constrained, and set theorists in particular study games that last for an infinite number of moves, with the winner (or other payoff) not known until after all those moves are completed.
The focus of attention is usually not so much on what is the best way to play such a game, but simply on whether one or the other player has a winning strategy. (It can be proven, using the axiom of choice, that there are games—even with perfect information, and where the only outcomes are "win" or "lose"—for which neither player has a winning strategy.) The existence of such strategies, for cleverly designed games, has important consequences in descriptive set theory.
Discrete and continuous games
Much of game theory is concerned with finite, discrete games, that have a finite number of players, moves, events, outcomes, etc. Many concepts can be extended, however. Continuous games allow players to choose a strategy from a continuous strategy set. For instance, Cournot competition is typically modeled with players' strategies being any non-negative quantities, including fractional quantities.
Differential games such as the continuous pursuit and evasion game are continuous games.
Metagames
These are games the play of which is the development of the rules for another game, the target or subject game. Metagames seek to maximize the utility value of the rule set developed. The theory of metagames is related to mechanism design theory.
History
The first known discussion of game theory occurred in a letter written by James Waldegrave in 1713. In this letter, Waldegrave provides a minimax mixed strategy solution to a two-person version of the card game le Her. It was not until the publication of Antoine Augustin Cournot's Recherches sur les principes mathématiques de la théorie des richesses (Researches into the Mathematical Principles of the Theory of Wealth) in 1838 that a general game theoretic analysis was pursued. In this work Cournot considers a duopoly and presents a solution that is a restricted version of the Nash equilibrium.
Although Cournot's analysis is more general than Waldegrave's, game theory did not really exist as a unique field until John von Neumann published a series of papers in 1928. While the French mathematician Émile Borel did some earlier work on games, Von Neumann can rightfully be credited as the inventor of game theory. Von Neumann was a brilliant mathematician whose work was far-reaching from set theory to his calculations that were key to development of both the Atom and Hydrogen bombs and finally to his work developing computers. Von Neumann's work in game theory culminated in the 1944 book by von Neumann and Oskar Morgenstern. This profound work contains the method for finding mutually consistent solutions for two-person zero-sum games. During this time period, work on game theory was primarily focused on cooperative game theory, which analyzes optimal strategies for groups of individuals, presuming that they can enforce agreements between them about proper strategies.
In 1950, the first discussion of the prisoner's dilemma appeared, and an experiment was undertaken on this game at the RAND corporation. Around this same time, John Nash developed a criterion for mutual consistency of players' strategies, known as Nash equilibrium, applicable to a wider variety of games than the criterion proposed by von Neumann and Morgenstern. This equilibrium is sufficiently general, allowing for the analysis of non-cooperative games in addition to cooperative ones.
Game theory experienced a flurry of activity in the 1950s, during which time the concepts of the core, the extensive form game, fictitious play, repeated games, and the Shapley value were developed. In addition, the first applications of Game theory to philosophy and political science occurred during this time.
In 1965, Reinhard Selten introduced his solution concept of subgame perfect equilibria, which further refined the Nash equilibrium (later he would introduce trembling hand perfection as well). In 1967, John Harsanyi developed the concepts of complete information and Bayesian games. Nash, Selten and Harsanyi became Economics Nobel Laureates in 1994 for their contributions to economic game theory.
In the 1970s, game theory was extensively applied in biology, largely as a result of the work of John Maynard Smith and his evolutionarily stable strategy. In addition, the concepts of correlated equilibrium, trembling hand perfection, and common knowledge were introduced and analysed.
In 2005, game theorists Thomas Schelling and Robert Aumann followed Nash, Selten and Harsanyi as Nobel Laureates. Schelling worked on dynamic models, early examples of evolutionary game theory. Aumann contributed more to the equilibrium school, introducing an equilibrium coarsening, correlated equilibrium, and developing an extensive formal analysis of the assumption of common knowledge and of its consequences.
In 2007, Roger Myerson, together with Leonid Hurwicz and Eric Maskin, was awarded of the Nobel Prize in Economics "for having laid the foundations of mechanism design theory." Among his contributions, is also the notion of proper equilibrium, and an important graduate text: Game Theory, Analysis of Conflict .
See also
• Analytic narrative
• Game
• Glossary of game theory
• List of games in game theory
• , in which Adam Curtis examines the rise of game theory during the Cold War
References
Textbooks and general references
• .
• . Suitable for undergraduate and business students.
• . Suitable for upper-level undergraduates.
• . Acclaimed reference text, public description
• . Suitable for advanced undergraduates.
*Published in Europe as .
• . Presents game theory in formal way suitable for graduate level.
• . Snippets from interviews
• . Suitable for a general audience.
• . Undergraduate textbook.
• . A general history of game theory and game theoreticians.
• . A modern introduction at the graduate level.
• Praised primer and popular introduction for everybody, never out of print.
Historically important texts
• Aumann, R.J. and Shapley, L.S. (1974), Values of Non-Atomic Games, Princeton University Press
*reprinted edition:
*reprinted edition:
• Shapley, L.S. (1953), A Value for n-person Games, In: Contributions to the Theory of Games volume II, H.W. Kuhn and A.W. Tucker (eds.)
• Shapley, L.S. (1953), Stochastic Games, Proceedings of National Academy of Science Vol. 39, pp. 1095-1100.
Other print references
• , ISBN 978-0-631-23257-5 (2002 edition)
• . A layman's introduction.
Websites
• Paul Walker: History of Game Theory Page
• David Levine: Game Theory. Papers, Lecture Notes and much more stuff.
• Alvin Roth: Game Theory and Experimental Economics page - Comprehensive list of links to game theory information on the Web
• Adam Kalai: Game Theory and Computer Science - Lecture notes on Game Theory and Computer Science
• Mike Shor: Game Theory .net - Lecture notes, interactive illustrations and other information.
• Jim Ratliff's Graduate Course in Game Theory (lecture notes).
• Valentin Robu's software tool for simulation of bilateral negotiation (bargaining)
• Don Ross: Review Of Game Theory in the Stanford Encyclopedia of Philosophy.
• Bruno Verbeek and Christopher Morris: Game Theory and Ethics
• Chris Yiu's Game Theory Lounge
• Elmer G. Wiens: Game Theory - Introduction, worked examples, play online two-person zero-sum games.
• Marek M. Kaminski: Game Theory and Politics - syllabuses and lecture notes for game theory and political science.
• Web sites on game theory and social interactions
• Kesten Green's Conflict Forecasting - See Papers for evidence on the accuracy of forecasts from game theory and other methods.
• McKelvey, Richard D., McLennan, Andrew M., and Turocy, Theodore L. (2007) . | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9505350589752197, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/tagged/algebra-precalculus?page=2&sort=faq&pagesize=15 | # Tagged Questions
linear, exponential, logarithmic, polynomial, rational, and trigonometric functions, conic sections, binomial, surds, graphs and transformations of graphs, equation- and system-solving, and other symbolic-manipulation topics
3answers
479 views
### Why everytime the final number comes the same?
I have come across an interesting puzzle. Write $20$ numbers. Erase any two number say $x$ and $y$ and and replace with $\text{Number}_{new} = xy/(x + y)$ OR $\text{Number}_{new}= x + y + xy$ ...
5answers
1k views
### No radical in the denominator — why?
Why do all school algebra texts define simplest form for expressions with radicals to not allow a radical in the denominator. For the classic example, $1/\sqrt{3}$ needs to be "simplified" to ...
2answers
574 views
### Is this a known algebraic identity?
In the course of analyzing a certain Markov chain, I once had to prove the following algebraic identity. Is there a slick or known proof? For $n$-tuples $(x_1,x_2,\dots, x_n)$ of positive real ...
11answers
5k views
### Derivation of the formula for the vertex of a Parabola
I'm taking a course on Basic Conic Sections, and one of the ones we are discussing is of a parabola of the form $y = a x^2 + b x + c$ My teacher gave me the formula: $x = -\frac{b}{2a}$ as the $x$ ...
2answers
304 views
### How can I compute $\sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k}$?
This sum is difficult. How can I compute it, without using calculus? $$\sum_{k = 1}^n \frac1{k + 1}\binom{n}{k}$$ If someone can explain some technique to do it, I'd appreciate it. Or advice using ...
6answers
332 views
### The existence of partial fraction decompositions
I'm sure you are all familiar with partial fraction decomposition, but I seem to be having trouble understanding the way it works. If we have a fraction f(x)/[g(x)h(x)], it seems only logical that it ...
5answers
283 views
### How to “Re-write completing the square”: $x^2+x+1$
The exercise asks to "Re-write completing the square": $$x^2+x+1$$ The answer is: $$(x+\frac{1}{2})^2+\frac{3}{4}$$ I don't even understand what it means with "Re-write completing the square".. ...
3answers
645 views
### How to find the sum of the following series
How can I find the sum of the following series? $$\sum_{n=0}^{+\infty}\frac{n^2}{2^n}$$ I know that it converges, and Wolfram Alpha tells me that its sum is 6 . Which technique should I use to ...
4answers
402 views
### Steps to solve this system of equations
I want to solve this system of equations, I have been out of Maths for a long !! $$\sqrt{x} + y = 7$$ $$\sqrt{y} + x =11.$$ Just wondering easiest step to find values for $x$ and $y$ from the above ...
1answer
76 views
### What is the minimum number of blocks to build this?
A rectangular solid is built using $N$ cubes of a side length of 1cm. When viewed from such an angle such that only 3 of the sides of the rectangular solid are visible, there are 231 $cm^2$ of area ...
4answers
274 views
### Algorithms for “solving” $\sqrt{2}$
The very first words out of my mouth need to be this... "Solving" is the wrong term since I am speaking about irrational numbers. I just don't know which word is the correct word... So that can be ...
4answers
818 views
### Geometric mean never exceeds arithmetic mean
This was a mathematical induction question proposed in a textbook, and I've exhausted multiple approaches (proving RHS - LHS > 0, splitting the fraction, fractional exponents, etc.) The geometric ...
5answers
1k views
### How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$?
I'm doing exercise on discrete mathematics and I'm stuck with question: If $f:Y\to Z$ is an invertible function, and $g:X\to Y$ is an invertible function, then the inverse of the composition \$(f ...
2answers
217 views
### If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$?
If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$ ? I would be very grateful if somebody show me how to factor this polynomial by hand, as of now I have used to ...
3answers
276 views
### Question about solving absolute values
I solved the following problem from by book, but the answer of this problem at the end of book is x < = 3 Please tell me how can i get this answer.
3answers
209 views
### Finding the error in a proof
I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof: If x = y, then \begin{eqnarray} x^2 &=& xy \nonumber \\ x^2 - y^2 &=& xy - ...
4answers
445 views
### Solving a quadratic inequality
I am solving the following inequality, please look at it and tell me am i write or not because this is an example in Howard Anton's book, i solved it on my own as given below but in the book it is ...
2answers
94 views
### How do I find the delta analytically for $f(x)$ with a degree other than $1$
So I know how to find the $\delta$ in $f(x)$ that can be factored into a degree of $1$, or I can solve it by solving $L + \epsilon = f(x)$ then finding the distance from $a$. But how do I find the ...
4answers
269 views
### Power summation of $n^3$ or higher [duplicate]
Possible Duplicate: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ If I want to find the formula for $$\sum_{k=1}^n k^2$$ I would do the next: (n+1)^2 = a(n+1)^3 ...
1answer
163 views
### Maximum inequality
Can anyone show that ...
4answers
177 views
### the sum of powers of $2$ between $2^0$ and $2^n$
Lately, I was wondering if there exists a closed expression for $2^0+2^1+\cdots+2^n$ for any $n$?
2answers
82 views
### Problem with Abel summation
Let $$S_n=\sum_{i=1}^{n}\sin k,\quad S_0=0.$$ Then $$\sum_{k=1}^{n}\frac{\sin k}{k}=\sum_{k=1}^{n}\frac{S_k-S_{k-1}}{k}=\frac{S_n}{n}+\sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}.$$ Could someone please ...
2answers
230 views
### How to find all real solution to satisfy this equation without casework or bruteforce?
How to find all real solution to satisfy this equation without casework or bruteforce? a+b+c=abc Thanks in advance!
16answers
2k views
### Interesting calculus problems of medium difficulty?
I would like this to be a community wiki. But, I don't know how to do that here. I would like to know sources, and examples of good "challenge" problems for students who have studied pre-calculus ...
5answers
403 views
### Solving a peculiar system of equations
I have the following system of equations where the $m$'s are known but $a, b, c, x, y, z$ are unknown. How does one go about solving this system? All the usual linear algebra tricks I know don't apply ...
4answers
392 views
### How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$?
Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$ Apart from induction,I tried with Wolfram Alpha to check the validity,but ...
2answers
406 views
### Given $n! = c$, how to find $n$?
I'm dealing with a time-complexity problem in which I know the running time of an algorithm: $$t = 1000 \mathrm{ms} .$$ I also know that the algorithm is upper bounded by $O(n!)$. I want to know ...
7answers
1k views
### Algebraic Identity $a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$
Prove the following: $\displaystyle a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$. So one could use induction on $n$? Could one also use trichotomy or some type of combinatorial ...
4answers
201 views
### Should the domain of a function be inferred?
It is a common practice to have students of elementary algebra infer the domain of a function as an exercise. I believe this is contrary to the spirit of the definition of a function as a collection ...
2answers
166 views
### Determining the number $N$
Let $1 = d_1 < d_2 <\cdots< d_k = N$ be all the divisors of $N$ arranged in increasing order. Given that $N=d_1^2+d_2^2+d_3^2+d_4^2$, determine $N$. The divisors include $N$. It seems that ...
4answers
465 views
### Solve $\cos^{n}x-\sin^{n}x=1$ with $n\in \mathbb{N}$.
Solve $\cos^{n}x-\sin^{n}x=1$ with $n\in \mathbb{N}$ I have no idea how to deal with this crazy question. One idea came into my mine is factorization, but I can't go on... Can anyone help me please? ...
2answers
256 views
### Algebraic equation problem - finding $x$
$$(x^2 +100)^2 =(x^3 -100)^3$$ How to solve it?
2answers
263 views
### Prove the following property of $f(x)$?
Let $$f(x)=|a_1\sin(x)+a_2\sin(2x)+a_3\sin(3x)+...+a_n\sin(nx)|.$$ Given that $f(x)$ is less than or equal to $|\sin(x)|$ for all $x$, prove that $|a_1+a_2+a_3+....|$ is less than or equal to ...
6answers
423 views
### What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$?
Find the limit: $$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$$ I am not able to find it because I don't know how to prove or disprove $0$ is the answer.
5answers
475 views
### Prove this number fact
Prove that $x \neq 0,y \neq 0 \Rightarrow xy \neq 0$. Suppose $xy = 0$. Then $\frac{xy}{xy} = 1$. Can we say that $\frac{xy}{xy} = 0$ and hence $1 = 0$ which is a contradiction? I thought ...
4answers
109 views
### Working with proofs help?
I'm trying to study for my midterm and doing some random practise questions to work with proofs. However I'm stuck on, as the only way I know how to prove it is through plugging in numbers, however as ...
1answer
243 views
### One sum of squares and two Diophantine equations
This question comes from trying to see why 24 is the only non-trivial value of $n$ for which $$1^2+2^2+3^2+\cdots+n^2$$ is a perfect square. To this end, let $m,n \in \mathbb N$ be such that ...
3answers
492 views
### Gaussian proof for the sum of squares?
There is a famous proof of the Sum of integers, supposedly put forward by Gauss. $$S=\sum\limits_{i=1}^{n}i=1+2+3+\cdots+(n-2)+(n-1)+n$$ $$2S=(1+n)+(2+(n-2))+\cdots+(n+1)$$ $$S=\frac{n(1+n)}{2}$$ ...
3answers
1k views
### The logic behind the rule of three on this calculation
First, to understand my question, checkout this one: Calculating percentages for taxes. Second, consider that I'm a layman in math. So, after trying to understand the logic used to get the final ...
3answers
93 views
### How to expand $(a_0+a_1x+a_2x^2+…a_nx^n)^2$?
I know you can easily expand $(x+y)^n$ using the binomial expansion. However, is there a simple summation formula for the following expansion? $$(a_0+a_1x+a_2x^2+...+a_nx^n)^2$$ I found something ...
2answers
73 views
### Show that equation has no solution in $(0,2\pi)$
Hi I want to show that the equation $2=2 \cos(x)+x \sin(x)$ has no solution in $(0,2 \pi)$. Since it is algebraically impossible to solve this equation for $x$ I wanted to ask you whether one of you ...
4answers
391 views
### Problems with Inequalities
It seems like I am facing some confusion while handling with inequalities,I was doing some tasks where it is asked to find the interval of the variable,after some steps I deduced the the necessary ...
4answers
632 views
### Proving that $\sum\limits_{i=1}^k i! \ne n^2$ for any $n$ [duplicate]
Possible Duplicate: How to prove that the number 1!+2!+3!+…+n! is never square? Show that $\displaystyle\sum\limits_{i=1}^k i!$ is never a perfect square for $k\ge4$ I could prove ...
5answers
366 views
### How to prove that $\sum\limits_{i=0}^p (-1)^{p-i} {p \choose i} i^j$ is $0$ for $j < p$ and $p!$ for $j = p$
Let $p \in \mathbf{N}$. I don't know how to prove that $$\sum_{i=0}^p (-1)^{p-i} {p \choose i} i^j=0 \textrm{ for } j \in \{0,\ldots,p-1\},$$ and $$\sum_{i=0}^p (-1)^{p-i} {p \choose i} i^p=p!$$ ...
2answers
891 views
### What is the best way to solve an equation involving multiple absolute values?
An absolute value expression such as $|ax-b|$ can be rewritten in two cases as $|ax-b|=\begin{cases} ax-b & \text{ if } x\ge \frac{b}{a} \\ b-ax & \text{ if } x< \frac{b}{a} \end{cases}$, ...
8answers
443 views
### How to factor quadratic $ax^2+bx+c$?
How do I shorten this? How do I have to think? $$x^2 + x - 2$$ The answer is $$(x+2)(x-1)$$ I don't know how to get to the answer systematically. Could someone explain? Does anyone have a link to ...
2answers
372 views
### Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ [duplicate]
Possible Duplicate: Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction How do I prove the following? \sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + ...
2answers
157 views
### Maximum of the difference
What is the maximum value of $f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{2012})$ where $x_{1}, x_{2}, … , x_{2012}$ are distinct integers in the set ${1, 2, 3, …, 2012}$ and $f$ is the absolute ...
1answer
220 views
### $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$
How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$? The polynomial is allowed to have repeated roots.
3answers
264 views
### There isn't a product operation that is commmutative on $\mathbb{R}^{n}$ that satisfies all the field axioms for $n \geq 3$.
This proof is broken down into simple easy algebra and vector questions. I would like to discuss different answers and approaches. Please see pg 162-163 on books.google.ca/books?isbn=0387290524 ... | {"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": 112, "mathjax_display_tex": 23, "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": 0.944786548614502, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/9705/why-is-fracdxdt-0-in-this-average-momentum-calculation | # Why is $\frac{dx}{dt}=0$ in this average momentum calculation?
In the following excerpt from S. Gasiorowicz's Quantum Physics, he derives an expression for the average momentum of a free particle. $\psi(x,t)$ is the wave function of a free particle, $\psi^*$ denotes its complex conjugate.
We try the following: since classically,
$$p = mv = m\frac{dx}{dt}$$
we shall write
$$<p> = m\frac{d}{dt}<x> = m\frac{d}{dt}\int{dx \psi^*(x,t) x \psi(x,t)}$$
This yields
$$<p> = m\int_{-\infty}^\infty{dx\left( \frac{\partial\psi^*}{\partial t} x \psi + \psi^* x \frac{\partial\psi}{\partial t} \right)}$$
Note that there is no $dx/dt$ under the integral sign. The only quantity that varies with time is $\psi(x,t)$, and it is this variation that gives rise to a change in $x$ with time.
I seem to have trouble understanding the difference between the position $x$ and the average position $<x>$. Why can it be assumed that $\frac{dx}{dt}=0$? What is x?
-
## 2 Answers
The confusion seems to stem from a) not understanding what kind of objects you are dealing with and b) usual custom of not writing (all) arguments of functions when they are understood.
To clarify a) note that the position operator $\hat x$ does not depend on time, and so also its kernel $\left< x \right | \hat{x} \left | x' \right> = x\delta(x-x')$ with respect to "position vectors" $\left | x \right >$ also doesn't depend on time. This $x$ is the one that is present in your integral. So in particular ${{\rm d} x \over {\rm d} t} = 0$.
On the other hand, the average of the operator $\hat A$ in the state $\psi$ (which depends on time) obviously depends on the state $\psi$: $\left< \hat {A} \right> := \left< \psi \right | \hat {A} \left | \psi \right>$ and so if you perform averages on a family of vectors $\psi(t)$ so also the average will depend on time. In your case, this should be written $\left< \hat{x} \right > (t)$ to make it obvious that one is dealing with a function of time. But this dependence is usually understood and omitted.
-
+1, Marek. Interestingly, the question contains the first formula for $\langle p \rangle$ which shows perfectly what the objects depend upon. It says that $\psi$ and $\psi^*$ depend on $x,t$ while $x$ doesn't depend on anything, especially not on $t$. So despite this detailed notation, the dependence was ignored by the OP. Moreover, to claim that $x$ depends on $t$ in those formulae would be totally preposterous because $x$ is being integrated over in the formula - it takes all real values simultaneously and "all real values" (the real axis) clearly can't depend on $t$. – Luboš Motl May 11 '11 at 9:08
This $x$ is a position in the reference frame's coordinate system, which is just static by design. You can imagine it as a ruler, with a probability cloud in a foreground; the ruler stays on its place while the cloud moves and deforms changing its mean position.
- | {"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": 31, "mathjax_display_tex": 3, "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": 0.9549856185913086, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/104140/deriving-the-approximation-formula | # Deriving the approximation formula
$f'(x) \thickapprox$ $\frac{1}{2h} [ 4f(x+h) - 3 f(x) + f(x + 2h)]$
I need to derive the approximation formula for the function above. And I need to show that it's error term is of the form $\frac{1}{3}h^2 f'''(\xi)$
How do I go around doing this? I've been suggested to use the Central Difference Formula or Forward or Backward Approximation
-
2
You'll have a tough time, since the formula is wrong, or not well written. Let $f(x)=x$. We get nowhere near $1$. But for example if you interchange $4$ and $3$, you get something more plausible. – André Nicolas Jan 31 '12 at 3:52
I double checked, I have the correct formula – user906153 Jan 31 '12 at 3:54
2
Do you mean (on top) $4f(x+h)-(3f(x)+f(x+2h))$? Please note the parentheses. – André Nicolas Jan 31 '12 at 3:59
The way I wrote it in the question is how I it is written in the book that I have. It is apparently poorly written then – user906153 Jan 31 '12 at 4:17
## 2 Answers
Let $f(x)=x$. Then $$4f(x+h)-3f(x)+f(x+2h)=4(x+h)-3x+(x+2h)=2x-2h.$$ When you divide by $2h$, you get $\dfrac{x}{h}-1$. This is nowhere near the derivative. Actually, there is also terrible behaviour for the function $f(x)=1$!
How to modify so that things look roughly like the given expression? There are lots of possibilities. For sure the coefficients need to have sum $0$. My two guesses as to what might have been meant have error behaviour worse than the one asked for.
Probably the most popular approximation formula that in general behaves better than the ordinary $(f(x+h)-f(x))/h$ is $(f(x+h)-f(x-h))/2h$. It has error behaviour of the kind you want, with I think $1/6$ instead of $1/3$.
There are many others, involving more complicated combinations of differences.
Remark: Suppose that a numerical differentiation procedure has error behaviour of the type that you are looking for. Then since the third derivative of $1$, $x$, and $x^2$ is identically $0$, the procedure must be exact for $f(x)=1$, $f(x)=x$, and $f(x)=x^2$.
-
In general, symmetric differences have better numerical properties than unsymmetric differences. Also, they are much more suited to further processing like Richardson extrapolation than their unsymmetric counterparts. – J. M. Jan 31 '12 at 4:31
Following André Nicolas' remark one may search for all formulas of the type $$f'(x)\doteq {1\over 2h}\bigl( a f(x)+ b f(x+h)+ c f(x+2h)\bigr)\qquad (|h|\ll 1)$$ that are exact for $f(x)=x^k$ $\ (k=0,1,2)$. There is just one such formula, namely $$f'(x)={1\over2h}\bigl(4 f(x+h)-f(x+2h)-3 f(x)\bigr)\ .$$ To obtain an error estimate fix $x$ and consider the auxiliary function $$g(h):=4f(x+h)-f(x+2h)-3f(x)-2h f'(x)\ .$$ Then $$g'(h)=4f'(x+h)-2f'(x+2h)-2f'(x), \quad g''(h)=4f''(x+h)-4f''(x+2h)\ .$$ It follows that $g(0)=g'(0)=g''(0)=0$; furthermore $|g''(h)|\leq 4 M |h|$ where $M:=\sup|f'''(t)|$ in the $t$-interval with endpoints $x$ and $x+2h$. From this we conclude that $$|g'(h)|\leq 2 M h^2$$ and then $$|g(h|\leq {2\over3} M |h|^3\ .$$ It follows that $$\left|{1\over 2h}\bigl(4f(x+h)-f(x+2h)-3f(x)\bigr)-f'(x)\right|={|g(h)|\over 2|h|}\leq {1\over 3} h^2\,M\ ,$$ as stated by the OP (albeit without the $\xi \thinspace$).
- | {"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": 34, "mathjax_display_tex": 8, "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": 0.9429510831832886, "perplexity_flag": "head"} |
http://lukepalmer.wordpress.com/2008/07/13/ch-isomorphism-duality/ | # Luke Palmer
Functional programming and mathematical philosophy with musical interludes
# The Curry-Howard isomorphism and the duality of → and ×
By on July 13, 2008
To people who are familiar with the Curry-Howard isomorphism, this post will probably be trivial and obvious. However, it just clicked for me, so I’m going to write about it. I had passively heard on #haskell that → and × (functions and tuples) were dual, so I played with types a bit, finding dual types, and it never really came together.
I’m not actually sure what I’m talking about here is exactly the CH isomorphism. I’ve been thinking about dependent types, and converting propositional formulae into that calculus. But if it’s not, it’s close.
Let’s use some sort of type bounded quantification in our logic. I was thinking about the following formula:
$\forall n \in \mathbb{N} \, \exists p \in \mathbb{N} \, (p > n \wedge \text{Prime}(p) \wedge \text{Prime}(p+2))$
The corresponding type is:
$(n : \mathbb{N}) \rightarrow [ (p : \mathbb{N}) \times (p > n) \times \text{Prime}\, p \times \text{Prime} (p+2) ]$
In other words, the proposition “for every natural, there is a twin prime greater than it” corresponds to the type “a function which takes a natural and returns a twin prime greater than it”. × corresponds to existentials, because it is the proof: it tells you which thing exists. The first element of the tuple is the thing, the second element is the proof that it has the desired property.
It’s interesting that ∀ corresponds to →, since logical implication also corresponds to →. The difference is that the former is a dependent arrow, the latter is an independent one. Beautifully in the same way ∃ corresponds to ×, and so does ∧.
Oh right, the duality. Knowing this, let’s take the above type and negate it to find its dual.
$(n : \mathbb{N}) \times [ (p : \mathbb{N}) \rightarrow (\neg (p > n) + \neg\text{Prime}\,p + \neg\text{Prime}(p + 2)) ]$
i didn’t bother expanding the encoding of not on the inner levels, because it doesn’t really make anything clearer.
The dual is a pair: a number n and a function which takes any number and returns one of three things: a proof that it is no greater than n, a proof that it is not prime, or a proof that the number two greater than it is not prime. Intuitively, n is a number above which there are no twin primes. If this pair exists, the twin prime conjecture is indeed false.
So yeah, that’s how → and × are dual. It’s pretty obvious now, actually, but it took a while to make sense.
### Share this:
Posted in: Uncategorized | Tagged: code, math
# Post navigation
### 3 Responses
1. Evgeny Makarov July 13, 2008 at 8:51 pm
Hello,
You are not quite right. What you are saying basically is that
not (A -> B /\ C) = A /\ (B -> not C)
(B is p : N and C is (p > n) /\ Prime p /\ Prime (p + 2))
This is correct. First, however, you could have written the negation of a symmetric formula (B /\ C) as (C -> not B), or as (not B \/ not C), in addition to (B -> not C). The duality between /\ and -> should probably be based on something more unambiguous. Second, negation is not the only way to understand “dual”. For example, according to Wikipedia (Boolean algebra), the dual of (A /\ B) is (A \/ B). Also, for a formula (A -> B), its contrapositive (not B -> not A), which is equivalent to the original formula, is sometimes called “dual”.
More importantly is that the relationship between negation and the transformations shown above, on the one hand, and Haskell and Curry-Howard isomorphism, on the other hand, is not nearly as direct. Speaking mathematically, the formula ((not (A -> B)) -> A /\ not B), as well as its corollary ((not (A -> B)) -> A), is true only in classical, not constructive logic. “Classical” means using the law of excluded middle (A \/ not A). If we have this law we can prove the second formula. Indeed, if A is true, then we are done. If A is false, then A -> B is true and we get a contradiction with the assumption.
In constructive logic, on the other hand, proving a formula is the same as writing a program that has this formula as the type (this is Curry-Howard isomorphism). Now, (not A) is understood as (A -> _|_), where _|_ is falsehood. Can we construct a program with the type ((A -> B) -> _|_) -> A? No: if we have an operator that converts a function A -> B into some new type _|_ (or even empty type), then there is no way to construct a value of type A.
Where were we? The duality of /\ and -> is a constructive thing; it does not require classical logic. Further, Curry-Howard isomorphism is normally considered with constructive logic, because, as shown above, there is no way to write a purely functional program with some types which are valid classical formulas.
The duality of /\ and -> means simply this: A /\ B -> C is the same thing as A -> (B -> C) (viewed as types, this is of course currying). Why is this important, I don’t know very well, except that this is a special case of what is called “adjunction” in category theory, which has many examples throughout mathematics.
Finally, I said that there is no *purely* functional program of type ((A -> B) -> _|_) -> A. Let’s consider a similar type ((A -> B) -> A) -> A, which is also derived only in classical logic. The type of the argument (A -> B) -> A is the type of a term t that accepts a *continuation*. Namely, if a place in the big program (whose final type is B)requires an subprogram of type A, t asks for the rest of the computation (A -> B) and returns A. Now, (callcc t) has type A, so it can be this subprogram. When we reach (callcc t), we take the rest of the computation (of type A -> B), plug it in t, and get A, which the current place in the program expects! Therefore, callcc has type ((A -> B) -> A) -> A.
So, the types of purely functional programs are (via Curry-Howard isomorphism) formulas provable in constructive logic. And the types of programs with callcc are formulas provable in classical logic! (More precisely, classical minimal logic.) That is, callcc extends CHI from constructive to classical logic—isn’t this amazing?
2. Michael Ben-Yosef July 14, 2008 at 3:04 am
Terminology police: a category theorist would say they are “adjoint”, not dual. “Dual” typically means “what you get when you reverse all the arrows”, so to speak. The dual of products in categories are coproducts. As Evgeny said, the dual of (A /\ B) is (A \/ B), those two being respectively the product and the coproduct of A and B in a Boolean algebra regarded as a category.
On the other hand, in any cartesian closed category, hom(A, _) is the right adjoint functor of _ × A. In the specific cases you’re concerned with, this gives you what you want, e.g.
B × A → C is isomorphic to B → (A → C) in categories of types (called currying and uncurrying in functional programming)
and
B /\ A → C is equivalent to B → (A → C) in Boolean algebras (which I’ve heard called the “import-export law”).
Of course, ∃ and ∀ can also be seen as left and right adjoints respectively in an appropriate categorical framework, and the mathematical reason that “∃ corresponds to ×” as you say is that they are both left adjoints and so interact well with each other.
So I suggest the concept you’re looking for is “Adjoint functors”. Take a look at the Wikipedia article of that name.
3. Michael: Thanks. I’ve been dipping my toes into category theory for a while, but have had great difficulty learning. This is another concrete example, which is very helpful. As useful as category theory seems to be, I notice a surprising lack of concrete examples in articles meant to teach it…
### Recent Comments
• H2s on Polyamory and Respect
• Luke on Polyamory and Respect
• akos on Polyamory and Respect
• wren ng thornton on Polyamory and Respect
• Luke on Polyamory and Respect
### Twitter
• RT @pigworker: Statutory "Ult" tweet. A tweet prefixed with "Ult" is a comment on the immediately preceding retweet. We all need something … 2 days ago | {"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": 3, "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": 0.9373643398284912, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/1570/calculating-de-broglie-wavelength/1571 | # Calculating de Broglie wavelength
Hey, trying to finish an assignment but having some trouble with it. I will show all my work. The topic is on wave/particle dualty, uncertainty principle (second year modern physics course).
So the question is:
Calculate the de Broglie wavelength of a 5.7 MeV α particle emitted from an atomic nucleus whose diameter is approximately $1.6 \times 10^{-14} m$.
So I know the de broglie wavelength is $\lambda = h / p$. But we dont know p in this case, however we do know the energy. So to relate energy and momentum I use the following formula:
$$E^2 = p^2c^2+m^2c^4$$
using $m = 6.644 \times 10^{-27}$ and using $E = 9.13\times 10^{-13}$ gives me a complex solution. (Check WolframAlpha)
Now I've tried leaving the energy in terms of eV and MeV and they give me real solutions however the answer is not correct.
The answer is 6.02 fentometers.
Anyone have any advice?
-
I am having trouble understanding why the diameter of the the nucleus is given. Maybe it is to be compared later with $\lambda$. – ja72 Dec 3 '10 at 0:13
## 2 Answers
When the kinetic energy is given (5.7 MeV) and the mass is known, calculating momentum is very straightforward. I would like to point to the part of the assignment that you missed which is the nucleus diameter. It tells you something about the position of the particle and thus prevents you from knowing the momentum (and hence the de Broglie wavelength) all too well.
-
The assignment is asking to calculate the debroglie. every formula I've tried dosnt give me the right results. What am I missing? – masfenix Dec 4 '10 at 1:19
You are missing a formula to convert kinetic energy to momentum which is very simple to figure out if you remember that E=mv^2/2 and p=mv. Taking h/p gives me 6.015 fm. – gigacyan Dec 7 '10 at 13:46
Normally I'd knock on you for simply asking a homework question, but in this case you're dealing with a poorly-worded problem. The rest mass of an alpha particle is 3.7 GeV; you can't have an alpha particle with less energy than its rest mass.
The 5.7 MeV is supposed to be interpreted as the kinetic energy of the alpha particle, not its total energy. I guess you can take it from there.
-
The rest mass is technically 3.7 GeV/c^2, but yeah... – Noldorin Dec 3 '10 at 2:03
@Noldorin Depends on what units you use. Energy is a completely legitimate unit for mass. – Lagerbaer Dec 3 '10 at 3:04
1
@Lagerbaer: Incorrect. The units are already specified. The quantity is dimensionally wrong unless you specify `/c^2` too. – Noldorin Dec 3 '10 at 3:14
1
@masfenix G = Giga = 10^9; M = mega = 10^6. The rest mass of the particle is about a thousand times the kinetic energy in this case. You'll have to think about the implications there for what equations you should use. – Mark Eichenlaub Dec 3 '10 at 7:08
2
I think you can define a consistent unit system in which energy and mass are measured in the same unit (and trust me, I'm the biggest stickler for units I know so I don't say this lightly) - these would be the natural units with $c=1$. Of course, in a certain sense it's not as useful a system as SI or CGS, but it's so darn convenient that everyone caves eventually ;-) – David Zaslavsky♦ Dec 3 '10 at 8:44
show 7 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": 6, "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": 0.9335362911224365, "perplexity_flag": "middle"} |
http://stats.stackexchange.com/questions/24382/binomial-distribution-confidence-interval-for-log-plot | Binomial distribution confidence interval for log plot
In simulating iterative decoding of low-density parity-check codes there may be (for a certain signal-to-noise-ratio of a noisy channel) for example 10 decoding failures out of $10^6$ trials. The log of probability of decoding failure is plotted against signal-to-noise-ratio.
I quote from the Appendix of MacKay's paper Good Error-Correcting Codes based on Very Sparse Matrices to show how he deals with creating error bars on such plots:
The experiments result in a certain number of block decoding failures $r$ out of a number of trials $n$. We report the maximum likelihood estimate of the block error probability, $\hat{p}=r/n$, and a confidence interval $[p_-,p_+]$, defined thusly: if $r \geq 1$ then $p_{\pm}=\hat{p}\exp(\pm 2\sigma_{\log p})$ where $\sigma_{\log p}=\sqrt{(n-r)/(rn)}$; else if $r=0$ then $p_+=1-\exp(-2/n)$ and $p_-=0$.
I've read the wikipedia page on Binomial proportion confidence interval and a number of helpful answers on this website. Mackay's formula can be derived by approximating the normal approximation to a 2 sigma confidence interval: $\hat{p}\pm z_{1-\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n} = r/n \pm 2\sqrt{r(n-r)/(n^3)}=(r/n)(1\pm2\sqrt{(n-r)/(rn)})$ which is approximately equal to $(r/n)\exp(\pm2\sqrt{(n-r)/(rn)})$ when the argument of the exponential is small.
His notation also suggests he is really doing statistics with $\log p$. His formulas give symmetric error bars on log plots. I don't know whether this is a reasonable goal.
So I have a couple of questions:
1. Is my derivation the best way to understand what's going on with this formula?
2. Is it the most appropriate confidence interval?
- | {"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": 14, "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": 0.8825671076774597, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2009/12/30/integrals-are-additive-over-regions/?like=1&_wpnonce=84af3d0b4c | # The Unapologetic Mathematician
## Integrals are Additive Over Regions
Before I state today’s proposition, I need to define what I mean by saying that two sets in $\mathbb{R}^n$ are “nonoverlapping”. Intuitively, we might think that this means they have no intersection, but that’s not quite it. We’ll allow some intersection, but only at boundary points. Since the regions we’re interested in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries in an integral will make no difference.
Let $\left\{S_i\right\}_{i=1}^n$ be a collection of bounded regions in $\mathbb{R}^n$, so that any two of these regions are nonoverlapping. We define their union
$\displaystyle S=\bigcup\limits_{i=1}^nS_i$
and let $f:S\rightarrow\mathbb{R}$ be a bounded function defined on this union. Then $f$ is integrable on $S$ if and only if it’s integrable on each $S_i$, and we find
$\displaystyle\int\limits_Sf\,dx=\sum\limits_{i=1}^n\int\limits_{S_i}f\,dx$
Indeed, if $R$ is some $n$-dimensional interval containing $S$, then it will also contain each of the smaller regions $S_i$, and we can define
$\displaystyle\int\limits_Sf\,dx=\int\limits_Rf(x)\chi_S(x)\,dx$
and
$\displaystyle\int\limits_{S_i}f\,dx=\int\limits_Rf(x)\chi_{S_i}(x)\,dx$
We can use Lebesgue’s condition to establish our assertion about integrability. On the one hand, the discontinuities of $f$ in $S_i$ must be contained within those of $S$, so if the latter set has measure zero then so must the former. On the other hand, the discontinuities of $f$ consist of those within each of the $S_i$, and maybe some along the boundaries. Since the boundaries have measure zero, and we assume that the discontinuities in each $S_i$ are of measure zero, their (countable) union will also have measure zero. And then so must the set of discontinuities of $f$ in $S$ have measure zero as well.
The inclusion-exclusion principle tells us that we can rewrite the characteristic function of $S$:
$\displaystyle\begin{aligned}\chi_S&=\chi_{S_1\cup\dots\cup S_n}\\&=\sum\limits_{1\leq i\leq n}\chi_{S_i}-\sum\limits_{1\leq i<j\leq n}\chi_{S_i\cap S_j}+\sum\limits_{1\leq i<j<k\leq n}\chi_{S_i\cap S_j\cap S_k}-\dots+(-1)^n\chi_{S_1\cap\dots\cap S_n}\end{aligned}$
We can put this into our integral and use the fact that integrals are additive with respect to finite sums in the integrand
$\displaystyle\begin{aligned}\int\limits_Sf\,dx&=\int\limits_Rf(x)\chi_S(x)\,dx\\&=\sum\limits_{1\leq i\leq n}\int\limits_Rf(x)\chi_{S_i}(x)\,dx-\sum\limits_{1\leq i<j\leq n}\int\limits_Rf(x)\chi_{S_i\cap S_j}(x)\,dx+\dots+(-1)^n\int\limits_Rf(x)\chi_{S_1\cap\dots\cap S_n}(x)\,dx\\&=\sum\limits_{1\leq i\leq n}\int\limits_{S_i}f\,dx-\sum\limits_{1\leq i<j\leq n}\int\limits_{S_i\cap S_j}f\,dx+\dots+(-1)^n\int\limits_{S_1\cap\dots\cap S_n}f\,dx\end{aligned}$
But we assumed that all the $S_i$ are nonoverlapping, so any intersection of two or more of them must lie only along their boundaries. And since these boundaries all have Jordan content zero the integrals over them must come out to zero. We’re left with only the sum over each subregion, as we wanted.
### Like this:
Posted by John Armstrong | Analysis, Calculus
No comments yet.
« Previous | Next »
## 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": 27, "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": 0.9398483037948608, "perplexity_flag": "head"} |
http://quant.stackexchange.com/questions/7077/copula-models-and-the-distribution-of-the-sum-of-random-variables-without-monte/7099 | # Copula models and the distribution of the sum of random variables without Monte Carlo
There is a vast literature on copula modelling. Using copulas I can describe the joint law of two (and more) random variables $X$ and $Y$, i.e. $F_{X,Y}(x,y)$. Very often in risk management (credit risk, operational risk, insurance) the task is to model a sum $Z=X+Y$ and find its distribution $F_Z(z) = F_{X+Y}(z)$.
I know several approaches that do not directly use copulas (e.g. commons shock models and mixed compound Poisson models) but how can I elegantly combine a copula model and a model for the sum (without Monte Carlo of course - otherwise it would be easy).
Is there some useful Fourier-transform approach? I had the feeling that in the case of Archemedian copulas there could be a chance (looking at this mixture representation as in e.g. in Embrechts, Frey, McNeil). Who has an idea? Are there any papers on this?
-
1
Check the article "Fast computation of the distribution of the sum of two dependent random variables" by Embrechts and Puccetti (searcheable via google) – Alexey Kalmykov Jan 24 at 18:39
@AlexeyKalmykov Thanks for the link. This looks very interesting. I will read it soon. Do you know anything more applied too? If you make your comment an answer then I will accept if nothing else comes in. Thanks! – Richard Jan 25 at 8:32
Does anybody have an idea which is more copula based? Do copulas and sums don't go together well? I mean - the answer by Julian Wergieluk shows that this is not true in general - but using copulas we do not gain any insight to the sum? No (Fourier/moment/ ...) transform trick? – Richard Jan 29 at 13:16
– Richard Jan 29 at 13:25
Let's see whether a bounty attracts new ideas. – Richard Feb 7 at 13:02
show 2 more comments
## 2 Answers
In general setting this is quite a tough problem and it looks like just switching from regular multivariate probability to copulas doesn't make it easier. In general case you need to rely on numerical methods for integration.
There is a nice overview of the problem in Copula Theory and Its Applications: Proceedings of the Workshop Held in Warsaw, 25-26 September 2009, Part I, Section 5.3, "The Calculation of the Distribution of the Sum of Risks".
If you want to avoid Monte Carlo methods, you can look at a new deterministic method AEP, which was specially designed to tackle this problem. Recursive and FFT methods are compared here.
I haven't seen any work that deals with this problem specifically for Archimedean copulas.
-
– Richard Feb 13 at 21:47
@Richard Yes, there is no way to avoid numerical integration (MC or any other method). – Alexey Kalmykov Feb 13 at 22:00
So copulas do not facilitate the calculation of the law of sums. This is somewhat disappointing. In many contexts I am interested in the sum ... thanks for your links. – Richard Feb 13 at 22:05
If the density of $(X,Y)$ is known, then you may obtain the density of the sum $X+Y$ simply by applying the Jacobi's transformation formula, which describes the density of the transformed random variable $g(X,Y)$ for $g(x,y) = (x+y, x)$. Integrating out the $x$-component yields the density of $X+Y$. See Jacod/Protter Probability Essentials ch. 12 for details.
I am not sure, if this is an answer you are expecting. Perhaps you could provide us with some more specific description of your modelling task.
-
thanks for your comment. This looks nice. With my question I don't have a specific application in mind. I was just wondering whether there is no useful alternative to MC in this case. – Richard Jan 25 at 11:27
By the way: do you have a link to the internet to the procedure that you propose? I don't have the book that you mention - thanks! – Richard Jan 25 at 11:28
– Julian Wergieluk Jan 25 at 20:05
1
Assume the densoty of $(X,Y)$ is given by $f_{X,Y}(x,y)$. Then you propose $u=x+y$ and $v=x$ which gives $x=v$ and $y = u-v$. The applying Jacobi we get $$f_{U,V}(u,v) = f_{X,Y}(v,u-v) \cdot |-1|$$ ... now how can I proceed? I can't integrate over $u-v$ straight forward. Can you help me? Futhermore: Julian aren't you working in credit risk. Isn't there anything nice in the case of Archemidean copulas? – Richard Jan 27 at 21:17
Well thinking about it again the solution depends on the specific form. Then we must integrate over $v$ and note $u-v$ then we are done. I think I got it :) Further more the following helped me: math.uiuc.edu/~r-ash/Stat/StatLec1-5.pdf and www2.econ.iastate.edu/classes/econ671/hallam/documents/… – Richard Jan 27 at 21:25 | {"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": 20, "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": 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": 0.909641444683075, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/2106/what-is-the-purpose-of-using-different-hash-functions-for-the-leaves-and-interna/2107 | # What is the purpose of using different hash functions for the leaves and internals of a hash tree?
I just learned that the THEX hash tree specification which is widely used in P2P requires that two different hash functions be used: one for the leaf nodes (hashes of input data) and one for the internal hashes (hashes of hashes).
In order to protect against collisions between leaf hashes and internal hashes, different hash constructs are used to hash the leaf nodes and the internal nodes. The same hash algorithm is used as the basis of each construct, but a single '1' byte in network byte order, or 0x01 is prepended to the input of the internal node hashes, and a single '0' byte, or 0x00 is prepended to the input of the leaf node hashes.
By contrast, the proposed (though not yet widely adopted) Simple Merkle Hashes extension for BitTorrent just uses unmodified SHA-1 for all hashes. It's conceivable that this was a trade-off of security for simplicity, but that wasn't mentioned in the proposal.
What is the benefit of using two different hashes in this scheme?
-
## 2 Answers
If the same standard hash function was used for both leaves and branch nodes, it would be easy to generate collisions and even second preimages.
For example, let $M$ be a message which is longer than the segment size of the hash tree, but (for simplicity) no more than two segments long. Then the hash value of $M$ is calculated as $$H(M) = H_I(H_L(M_0) \,\|\, H_L(M_1)),$$ where $H_I$ is the internal hash function, $H_L$ is the leaf hash function, and $M_0$ and $M_1$ are the first and second segments of $M$. We'll also assume that the segment size of the hash tree is at least twice as long as the length of the output of $H_L$.
Now let $M' = H_L(M_0) \,\|\, H_L(M_1)$. Since $M'$ is, by assumption, at most one segment long, its hash will be calculated as $$H(M') = H_L(M') = H_L(H_L(M_0) \,\|\, H_L(M_1)).$$ If $H_L = H_I$, then $H(M') = H(M)$, and we've just found a second preimage for $H(M)$.
-
I realized that the reason BitTorrent's Simple Merkel Trees aren't vulnerable to this is that they're padded full by adding `0`-leaves (until the number of them reaches a power of two). On the other hand THEX allows unpaired leaf hashes to "float up" the tree into a spot where an internal hash would otherwise be expected. – Jeremy Banks Mar 17 '12 at 15:20
1
Just padding the number of segments to a power of 2 doesn't help: note that $M$ and $M'$ in my example have 2 and 1 segments respectively. However, padding the last segment itself to full length would indeed suffice to differentiate leaf hashes from internal hashes, as long as the segment length is strictly greater than the internal hash input length. (But do note that such padding itself, if not done carefully enough, may allow the construction of collisions.) – Ilmari Karonen Mar 17 '12 at 15:32
1
Oops! Thank you for the clarification. New interpretation: The trees are secure in the context of BitTorrent because it already knows the total length of the input, so it's impossible for an attacker to add or remove leaves as would be necessary. – Jeremy Banks Mar 17 '12 at 15:51
## Did you find this question interesting? Try our newsletter
email address
The torrent tree hash is vulnerable to second pre-image attacks by itself, even with `00` padding. I won't repeat Ilmari Karonen's answer, who already explained that part very well.
But it isn't used to identify the data by itself:
The original publisher of the content-file set creates a so-called Merkle torrent which is a torrent file that contains a root hash key in its info part instead of a pieces key
This means the infohash, which serves as a unique ID for a torrent isn't just based on the root of the tree, but also includes the filenames and filesizes which are kept in the info dictionary.
Knowing the total size of the torrent prevents such attacks. I still don't like their design decision, since it can easily lead to bugs in torrent clients which aren't aware of this issue. For example if a client forgot to validate the size of the piece, and only checked the hash, it's still be vulnerable.
IMO this design is also flawed in non security related ways.
In particular, the hash tree still crosses file boundaries. Having one root per file would have been nicer. But I guess they tried to stay as close as possible to the original format.
And they decided to use the leaf-size as the piece-size. I would have chosen a small constant leaf-size and leave the piece-size independent from the leaf-size. This would allow changes in piece-size without changing the hash.
- | {"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": 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": 0.9506747722625732, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/243793/geometrico-harmonic-progression | # Geometrico-Harmonic Progression
Like Arithmetico-geometric series, is there anyway to calculate in closed form of Geomtrico-harmonic series like $$\sum_{1\le r\le n}\frac{y^r}r$$ where $n$ is finite.
We know if $n\to \infty,$ the series converges to $-\log(1+y)$ for $-1\le y<1$
The way I have tried to address it is as follows:
we know, $$\sum_{0\le s\le n-1}y^s=\frac{y^n-1}{y-1}$$
Integrating either sides wrt $y$, we get $$\sum_{1\le r\le n}\frac{y^r}r=\int \left(\frac{y^n-1}{y-1}\right) dy$$
but how to calculate this integral in the closed form i.e., without replacement like $z=(y-1)$
-
1
I don't think that there exists a closed form. I don't know if you would allow the solution obtained from wolfram. It uses hypergeometric functions – Amr Nov 24 '12 at 16:33
@Amr, what is the solution from wolfram? – lab bhattacharjee Nov 24 '12 at 16:35
2
Even when $y = 1$, where your sum reduces to the famous harmonic number, there is no known simple closed form. – sos440 Nov 24 '12 at 16:56
## 1 Answer
It doesn't appear that a simple closed form for this sum exists. As @sos440 suggests in the comments, we shouldn't expect to find one. As noted by @Amr, the sum is related to the hypergeometric function.
We have $$\begin{eqnarray*} \sum_{1\le r\le n}\frac{y^r}{r} &=& \sum_{r=1}^\infty\frac{y^r}{r} - \sum_{r=n+1}^\infty\frac{y^r}{r} \\ &=& -\log(1-y) - y^{n+1}\sum_{k=0}^\infty \frac{y^k}{n+k+1}. \end{eqnarray*}$$ The ratio of successive terms in the hypergeometric series $${}_2 F_1(a,b;c;y) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)}{\Gamma(c+k)k!} y^k$$ is $$\frac{(a+k)(b+k)y}{(c+k)(k+1)}.$$ Thus, the sum above is a hypergeometric series with $a = n+1$, $b=1$, and $c = n+2$. There is an overall factor of $1/(n+1)$, so $$\begin{eqnarray*} \sum_{1\le r\le n}\frac{y^r}{r} &=& -\log(1-y) - \frac{y^{n+1}}{n+1} \, {}_2 F_1(n+1,1;n+2;y). \end{eqnarray*}$$
- | {"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": 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": 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": 0.9583497047424316, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/99873/list | ## Return to Question
3 changed title and changed "raise" to "bet"
# HowcanIminimizethisfunction-whichisanintegralofFinding the maximumNashEquilibrium of 4expressions-bychoosingavalueandafunction$0-1$pokerwithonebettinground
While working on a hobby project I encountered a difficult math problem. Or at least, difficult for me. Here is the problem:
Given an $a > 0$, find all pairs of a value $λ \in [0,1]$ and a function $f \colon [0,1] \to [0,1]$ such that $z$ is minimal.
```$
b(v) = \int_0^v f(x)\ \mathrm{d}x \\
z=\int_0^1{
\max \left(
\begin{array}{l}
(1 + 2a)λ, \\
(2 + 2a)x - λ, \\
2 a b(x) + 1, \\
(2 + 2a)x + b(1) - 2b(x) \\
\end{array}
\right)
}
\mathrm{d}x
$```
That is the complete problem, but a partial solution would already help me a lot, for example:
• A solution only for certain values of $a$
• Finding values of $z$ without knowing $λ$ and $f$
• Only one pair $(λ, f)$ for an $a$ instead of all the pairs
• Are there even multiple pairs for one $a$? (considering different $f$'s that yield the same $b$ as equal)
• Is it possible that all values of $a$ have at least one pair in which the $f$ has `$\{0, 1\}$` as its range?
I'll give you some info on what this formula is about. The goal of the project I'm working on is to have a better understanding of poker by finding Nash equilibria of very simplified versions of poker. The game I'm now trying to find equilibria for is a two player game that goes as follows: In the beginning both players are required to bet a certain amount, the ante, variable $a$ in the formula. Both players get a 'card', which is a uniformly distributed number between 0 and 1. Then the one and only betting round follows. Both players have only one coin with a value of 1 that they can use. Player 1 starts. There are five different ways the betting round can go:
• Check > Check
• Check > Raise Bet > Fold
• Check > Raise Bet > Call
• Raise
• Bet > Fold
• Raise
• Bet > Call
After the betting round the payout is done. If the betting round ends with check or call, there will be checked who of the players has the highest card. The winner will get the ante, or the ante plus one if a raise bet followed by a call occurred. The goal for each player is to have as much expected profit as possible.
When player 1 uses the best response to the strategy of player 2, the expected value of the game will be $z - 1 - a$. A positive value indicates profit for player 1, a negative value profit for player 2. If player 1 checks, then player 2 will check with probability $f(c)$ and raise bet otherwise, where $c$ is the card player 2 has. If player 1 raisesbets, then player 2 will call if his card is at least $λ$.
If I have the solution to this problem and didn't make any mistakes in making the formula, then I have the optimal strategy for player 2 and the expected value in the Nash equilibrium. And then I only have to do something similar for player 1.
Any ideas on minimizing $z$?
To
I'll give you any idea some info on why you are thinking about what this problem right now I will tell you something formula is aboutthe project I'm working on. The goal of the project I'm working on is to have a better understanding of poker by finding Nash equilibria of very simplified versions of poker. The game I'm now trying to find equilibria for is a two player game that goes as follows:In the beginning both players are required to bet a certain amount, the ante, variable $a$ in the formula. Both players get a 'card', which is a uniformly distributed number between 0 and 1. Then the one and only betting round follows. Both players have only one coin with a value of 1 that they can use. Player 1 starts. There are five different ways the betting round can go:
After the betting round the payout is done. If the betting round ends with check or call, there will be checked who of the players has the highest card. The winner will get the ante, or the ante plus one if a raise followed by a call occurred. The goal for each player is to have an expected value that is as high much expected profit as possible.
I won't give you
When player 1 uses the relation best response to the strategy of player 2, the poker game to expected value of the above formulagame will be $z - 1 - a$. A positive value indicates profit for player 1, since then I would have to explain a lot morenegative value profit for player 2.But If player 1 checks, then player 2 will check with probability $f(c)$ and raise otherwise, where $c$ is the card player 2 has. If player 1 raises, then player 2 will call if his card is at least $λ$.
If I have the solution to the this problem I'm 3/4 on and didn't make any mistakes in making the way to formula, then I have a the optimal strategy for player 2 and the expected value in the Nash equilibrium, if . And then I didn't make any mistakesonly have to do something similar for player 1.
1
# How can I minimize this function - which is an integral of the maximum of 4 expressions - by choosing a value and a function
While working on a hobby project I encountered a difficult math problem. Or at least, difficult for me. Here is the problem:
Given an $a > 0$, find all pairs of a value $λ \in [0,1]$ and a function $f \colon [0,1] \to [0,1]$ such that $z$ is minimal.
```$
b(v) = \int_0^v f(x)\ \mathrm{d}x \\
z=\int_0^1{
\max \left(
\begin{array}{l}
(1 + 2a)λ, \\
(2 + 2a)x - λ, \\
2 a b(x) + 1, \\
(2 + 2a)x + b(1) - 2b(x) \\
\end{array}
\right)
}
\mathrm{d}x
$```
That is the complete problem, but a partial solution would already help me a lot, for example:
• A solution only for certain values of $a$
• Finding values of $z$ without knowing $λ$ and $f$
• Only one pair $(λ, f)$ for an $a$ instead of all the pairs
• Are there even multiple pairs for one $a$? (considering different $f$'s that yield the same $b$ as equal)
• Is it possible that all values of $a$ have at least one pair in which the $f$ has `$\{0, 1\}$` as its range?
To give you any idea on why you are thinking about this problem right now I will tell you something about the project I'm working on. The goal of the project is to have a better understanding of poker by finding Nash equilibria of very simplified versions of poker. The game I'm now trying to find equilibria for is a two player game that goes as follows: In the beginning both players are required to bet a certain amount, the ante. Both players get a 'card', which is a uniformly distributed number between 0 and 1. Then the one and only betting round follows. Both players have only one coin with a value of 1 that they can use. Player 1 starts. There are five different ways the betting round can go:
• Check > Check
• Check > Raise > Fold
• Check > Raise > Call
• Raise > Fold
• Raise > Call
After the betting round the payout is done. If the betting round ends with check or call, there will be checked who of the players has the highest card. The winner will get the ante, or the ante plus one if a raise followed by a call occurred. The goal for each player is to have an expected value that is as high as possible.
I won't give you the relation of the poker game to the above formula, since then I would have to explain a lot more. But if I have the solution to the problem I'm 3/4 on the way to have a Nash equilibrium, if I didn't make any mistakes.
Any ideas on minimizing $z$? | {"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": 43, "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": 0.9542278051376343, "perplexity_flag": "head"} |
http://mathhelpforum.com/differential-geometry/114753-analytic-functions.html | # Thread:
1. ## Analytic Functions
How would I prove that the sum and product of analytic functions are analytic?
With an analytic function being defined as...
A function $f$ is (real) analytic on an open set D in the real line if for any $x_0$ in D one can write...
$f(x) = \sum_{n=0}^\infty a_n \left( x-x_0 \right)^n = a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 + a_3 (x-x_0)^3 + \cdots$
in which the coefficients $a_0$, $a_1$, ... are real numbers and the series is convergent to f for x in a neighborhood of $x_0$.
Thinking something along the lines of grouping the x terms together, then can find epsilon so that it converges after the nth term..? This along the right lines?
2. You can use that $f$ is analytic on $D\subseteq\mathbb{R}$ if and only if there exists an open set $G\subseteq C$ containing $D$ such that
there exists $g$ holomorphic on $G$ extending $f$. Then the assertion follows from the fact that sums and products of holomorphic functions are holomorphic.
Also you can use that $f$ is analytic on $D\subseteq\mathbb{R}$ if and only if for each compact subset
$K$ of $D$ there exists $M>0$ such that $\left|\frac{f^{(n)}(x)}{k!}\right|\leq M^k$ for each $k\in\mathbb{N}, x\in K$. Here the sum is almost immediate but for the product you will need Leibnitz rule.
If you want a direct proof for the product, puff, let say using Cauchy Series could be done, but anyway you will need the fact that every analytic expansion converges absolutely, which follows from the characterization above. | {"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": 20, "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": 0.9252558350563049, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/40593?sort=oldest | ## Hausdorff dimension of the boundary of an open set in the Euclidean space - lower bound
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I consider a bounded open set $A$ in ${\mathbb R}^d$. Is the Hausdorff dimension of the boundary of $A$ at least $d-1$ ? I thought I would have found a result on this problem in any textbook about Hausdorff dimension but I failed. As you may guess I have never work with Hausdorff dimension.
-
## 2 Answers
If a compact set $C$ separates a connected space $X$ of dimension $d$ (with some homogeneity properties, for example, an euclidean space) in two connected components, then it must have topological dimension at least $d-1$ (see for example the book by Hurewicz and Wallman on topological dimension, Theorem IV.4). In general, Hausdorff dimension exceeds (or is equal to) topological dimension (see chapter VII.2 of the book by Hurewicz-Wallman mentioned above).
-
Sorry, english is not my native language. I shall correct it. – rpotrie Sep 30 2010 at 11:21
Thank you very much for your answer ! – Hugh J Sep 30 2010 at 11:52
By definition (in Hurewicz and Wallman, if I remember correctly), the topological dimension of separable metric space $A$ is the dimension of the boundary (frontier) of $A$ plus one. Since an open set in $\mathbb{R}^d$ has dimension $d$, you are done. But if you start with the definition of topological dimension as Lebesgue's covering dimension, then you might use the theorem above. – Robert Bell Sep 30 2010 at 12:09
rbell: Actually that is usually called the "inductive dimension". – Pietro Majer Sep 30 2010 at 15:42
### 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.
Here is a simple proof for the Hausdorff dimension. Consider the orthogonal projection to $\mathbb R^{d-1}$. Since $A$ is bounded, the projection of the boundary contains the projection of $A$. The latter is open in $\mathbb R^{d-1}$ and hence has Hausdorff dimension $d-1$. On the other hand, the projection map does not increase Hausdorff dimension since it does not increase distances.
-
That's great ! This is really simple. Thank you ! – Hugh J Sep 30 2010 at 19:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9052267670631409, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/197375/example-of-an-infinite-group | # example of an infinite group
find an infinite group that has exactly two elements with order 4?
let G be an infinite group for all R_5 (multiplication mod 5) within an interval [1,7). so |2|=|3|=4. any other suggestions please.
-
## 2 Answers
The one we bump into most often is the multiplicative group of non-zero complex numbers. There are all sorts of minor variants of the idea, such as the complex numbers of norm $1$ under multiplication. One can disguise these groups as matrix groups, or geometric transformation groups.
-
what about group of R/Z which is the isomorphic of U. does this group has only 2 elements with order 4? – david Sep 16 '12 at 5:41
Yes david, because $\mathbb R / \mathbb Z$ is in fact isomorphic to the multiplicative group of complex numbers of norm $1$. – Niccolò Sep 16 '12 at 5:43
Yes, that's isomorphic to the circle group I mentioned in the answer. – André Nicolas Sep 16 '12 at 5:44
can you give a specific example of a cyclic group. – david Sep 17 '12 at 16:57
@david: The comment said circle group, which is a standard name for the group of complex numbers of norm $1$, And, as pointed out, it is isomorphic to your $\mathbb{R}/\mathbb{Z}$. There is no infinite cyclic group with the desired property. There are finite cyclic groups, like $\mathbb{Z}_4$, $\mathbb{Z}_8$, $\mathbb{Z}_{12}$, but you were asked for infinite. – André Nicolas Sep 17 '12 at 17:06
## Did you find this question interesting? Try our newsletter
Sign up for our newsletter and get our top new questions delivered to your inbox (see an example).
email address
Pick $G$ an infinite group with no torsion (e.g. $\mathbb Z$ or $\mathbb Q$), then $\mathbb Z / 4 \mathbb Z \times G$ works.
As a non-abelian example, you can also take the free product $\mathbb Z / 4 \mathbb Z * G$.
-
The free product has infinitely many elements of order 4. They form two conjugacy classes. – Jack Schmidt Sep 16 '12 at 6:33
Right, I didn't think about conjugation. Could you give an example of a non-abelian group with exactly two elements of order 4? – Niccolò Sep 16 '12 at 21:13
A silly example is $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z} \times (3\ltimes 7)$. It would be nice if the group was sort of "generated" by the elements of order 4, but with only 2 that is impossible (they generate a cyclic group of order 4). So in some sense all examples are silly, but probably some are less silly than mine. – Jack Schmidt Sep 17 '12 at 20:41 | {"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": 14, "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": 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": 0.9251609444618225, "perplexity_flag": "middle"} |
http://matthewkahle.wordpress.com/2010/08/12/gods-number-is-20/?like=1&source=post_flair&_wpnonce=18b5244a97 | # God’s number is 20
After thirty years of people thinking about it, a famous open problem about the Rubik’s Cube has finally been resolved: God’s number is 20. In other words, every Rubik’s Cube, no matter how mixed up, can be solved in 20 moves or less. A nice summary of the work can be found at: http://www.cube20.org/
This enormous calculation was completed by a team of Morley Davidson, John Dethridge, Herbert Kociemba, and Tomas Rokicki, and oh yes, 35 CPU-years of idle computer time donated by Google. The preprint should be available by early September sometime.
It should be noted that for quite a while “God’s algorithm” has been known — in other words computers could quickly solve any given position with a minimal number of moves. But even with the algorithm in hand, to get God’s number by brute force would take too long, since there are $4.3 \times 10^{19}$ possible positions and one would have to check that they can all be solved in 20 moves or less. To get around checking this humongous number of positions required a lot of ingenuity and not only computing muscle, and the team should be warmly congratulated.
For the rest of this note, I’d like to briefly discuss a tiny bit of the mathematics involved, and an open problem or two.
First of all we should be clear about what we mean by “20 moves” — this is in the half-turn metric (HTM), where a move consists of turning any side, by either a quarter turn or a half turn. So to mathematicians, “God’s number” is the diameter of the Cayley graph for the generating set $\langle L, R, U, D, F, B, L^2, R^2, U^2, D^2, F^2, B^2 \rangle$. This is the metric of choice for cubers, and probably has been for thirty years. To cubers turning a side is turning a side.
Of course a half turn is just two quarter turns, and this leads us to consider the quarter-turn metric (QTM), where we only allow turning any side a quarter turn. Of course you can still give it a half turn, but now this counts as two moves. What is the diameter of the Cayley graph for the generating set $\langle L, R, U, D, F, B \rangle$? This is probably a more natural generating set to mathematicians, who might resist including both $x$ and $x^2$ in the list of generators for a group.
It is now known (due to a calculation by Rokicki) that 29 moves are always sufficient, and he suspects that the methods used to show that 20 suffice for HTM could probably easily bring this down to 28.
On the other hand, there is essentially only one position known that requires 26 moves in the QTM, and only two in that require 25 moves. This is a very different situation than HTM, where there are roughly 300 million positions distance 20 from solved. Extensive searches by Rokicki have failed to find any more positions requiring 25 or 26 moves. So the conjecture is that the diameter of the Cayley graph with the QTM generators is 26, but it might be a while before we know that for sure. In particular the methods applied to solve HTM apparently don’t work quite as cleanly as they do for QTM.
So this is one problem: show that God’s number for quarter-turn metric is 26.
What might be nice to see, for either metric, is a description of the geometry or topology of the Rubik’s cube. One could imagine, for example, adding higher dimensional cells to extend the Cayley graph to a polyhedral complex, and studying the topology of this complex.
Finally, the following also seems to be wide open: if one makes random moves on the cube (i.e. takes a random walk on one of the Cayley graphs), what can be said about the Markov chain mixing time? A famous result of Bayer and Diaconis is that it takes “seven shuffles” to mix up a deck of 52 cards, and it would be mathematically nice to have such an answer for the Cube. (The diameter of the Cayley graph being 20 puts an upper bound on the mixing time, but the bound obtained in this naive way is much too large to be close to the truth.)
But unlike the card shuffling, which actually has some practical application, at least for (say) casinos who want to make sure their decks are randomized, this would be purely for knowledge and enjoyment. Apparently this kind of mixing time question was once an issue for speed-solving tournaments, as they would make a sequence of random moves to mix up the cube, and one might want to know how many random moves to make. But this is no longer an issue — instead of choosing a sequence of random moves, now they have the computer choose a random position uniformly from all $4.3 \times 10^{19}$ positions, and then use God’s algorithm to put the cube into that position.
(Special thanks to Tom Rokicki for very helpful conversations in writing this note.)
### Like this:
This entry was posted in puzzles and tagged God's number, Rubik's cube.
# Post navigation
## 5 thoughts on “God’s number is 20”
1. Is God’s number 20 for HTM or QTM? It seems a bit confused in the text above…
2. matthewkahle says:
You’re right Mikael — I got jumbled up in writing. Hopefully it is fixed now — God’s number is 20 in HTM.
3. [...] Kähle writes God’s number is 20 at mathematical musings. “Every Rubik’s Cube, no matter how mixed up, can be solved in 20 [...]
4. Anonymous says:
Isn’t the “God’s number” the diameter of the Cayley graph for the generating set
{L, R, U, D, F, B, L^2, R^2, U^2, D^2, F^2, B^2, L^{-1}, R^{-1}, U^{-1}, D^{-1}, F^{-1}, B^{-1} } having the cardinality of 18?
• Anonymous says:
Yes, what you say is true. There ar 18 generators
# Mathematical art
Blog at WordPress.com. | Theme: Dusk To Dawn by Automattic.
%d bloggers like this: | {"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": 6, "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": 0.9425672292709351, "perplexity_flag": "middle"} |
http://mathramble.wordpress.com/category/algebraic-topology/ | # Random Math
Home » Algebraic Topology
# The Mayer-Vietoris sequence in sheaf cohomology
Posted on February 1, 2013 by |
In this post, I will prove the Mayer-Vietoris Sequence for sheaf cohomology.
In what follows, ${X}$ is a topological space and ${\mathcal F, \mathcal G, \mathcal H}$ are sheaves of abelian groups on ${X}$. Let ${Z}$ be a closed subset of ${X}$. We let ${\Gamma_Z(X,\mathcal F)}$ denote the global sections of ${\mathcal F}$ with support in ${Z}$. The functor ${\Gamma_Z(X, -)}$ is a left-exact additive functor from sheaves on ${X}$ to abelian groups, and its right derived functors, denoted ${H^i_Z(X, -)}$, is the ${i}$-th cohomology of ${X}$ with support in ${Z}$. If ${\mathcal F}$ is a sheaf, the presheaf ${U \mapsto \Gamma_{Y \cap U}(U, \mathcal F)}$ is also a sheaf on ${X}$, denoted ${\mathcal H^0_Y(\mathcal F)}$ and called the “subsheaf of ${\mathcal F}$ with support in ${Y}$“.
The Mayer-Vietoris sequence, for a sheaf ${\mathcal F}$ and for a pair of closed subsets ${Y,Z \subseteq X}$, is the long exact sequence of cohomology with supports
$\displaystyle \dots \rightarrow H^i_{Y \cap Z}(X, \mathcal F) \rightarrow H^i_Y(X, \mathcal F) \oplus H^i_Z(X, \mathcal F) \rightarrow H^i_{Y \cup Z}(X, \mathcal F) \rightarrow H^{i+1}_{Y \cap Z}(X, \mathcal F) \rightarrow \dots$
We will prove the existence of this sequence in several steps.
Lemma 1 Let ${\mathcal E}$ be a flasque sheaf, ${Y}$ a closed subset of ${X}$, and ${U=X-Y}$. Then the sequence
$\displaystyle 0 \rightarrow \Gamma_Y(X, \mathcal E) \rightarrow \Gamma(X, \mathcal E) \rightarrow \Gamma(U, \mathcal E) \rightarrow 0$
is exact.
Proof: Trivial. $\Box$
Lemma 2 Let ${\mathcal E}$ be a flasque sheaf, and let ${Y, Z}$ be closed subsets of ${X}$. Then the sequence
$\displaystyle 0 \rightarrow \Gamma_{Y \cap Z}(X, \mathcal E) \rightarrow \Gamma_Y(X, \mathcal E) \oplus \Gamma_Z(X, \mathcal E) \rightarrow \Gamma_{Y \cup Z}(X, \mathcal E) \rightarrow 0$
is exact (where the first map is the diagonal embedding, and the second map is ${(s, t) \mapsto s-t}$).
Proof: Exactness is clear except possibly on the right. Let $U=X-Y, V=X-Z$, and Let ${D, E}$ be the short exact sequences
$\displaystyle 0 \rightarrow \Gamma(X, \mathcal E) \rightarrow \Gamma(X, \mathcal E) \oplus \Gamma(X, \mathcal E) \rightarrow \Gamma(X, \mathcal E) \rightarrow 0$
and
$\displaystyle 0 \rightarrow \Gamma(U \cup V, \mathcal E) \rightarrow \Gamma(U, \mathcal E) \oplus \Gamma(V, \mathcal E) \rightarrow \Gamma(U \cap V, \mathcal E) \rightarrow 0$
where the maps are defined similarily as in the statement of the Lemma. There is an obvious morphism of short exact sequences ${D \rightarrow E}$. Since ${\mathcal E}$ is flasque, this morphism is surjective onto each term of ${E}$. By the snake lemma, and using Lemma 1, we get the desired short exact sequence. $\Box$
Now we are ready to prove the existence of the Mayer-Vietoris sequence for ${\mathcal F}$. Let
$\displaystyle 0 \rightarrow \mathcal F \rightarrow \mathcal E^0 \rightarrow \mathcal E^1 \rightarrow \dots$
be a flasque resolution of ${\mathcal F}$. By the lemma, we have a short exact sequence of complexes
$\displaystyle 0 \rightarrow \Gamma_{Y \cap Z}(X, \mathcal E^\bullet) \rightarrow \Gamma_Y(X, \mathcal E^\bullet) \oplus \Gamma_Z(X, \mathcal E^\bullet) \rightarrow \Gamma_{Y \cup Z}(X, \mathcal E^\bullet) \rightarrow 0.$
The long exact sequence of cohomology associated to this short exact sequences of complexes is precisely the Mayer-Vietoris sequence.
Posted in Algebraic Geometry, Algebraic Topology
# Pages I suggest
Blog at WordPress.com. | Theme: My Life by Justin Tadlock. | {"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": 46, "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": 0.9330979585647583, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/76596/what-is-a-basic-textbook-to-studying-symmetric-spaces/79730 | ## What is a basic textbook to studying symmetric spaces?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I want to study basic properties of symmetric spaces.
What is a basic textbook?
-
2
It would help people to give better answers if you said a bit about your background or existing knowledge, e.g. Lie group theory, Riemannian geometry, PDEs, or whatever – Yemon Choi Sep 28 2011 at 3:21
4
Someone is going to mention Helgason's... – Mariano Suárez-Alvarez Sep 28 2011 at 3:28
4
...and to feel guilty immediately afterwards – Piero D'Ancona Sep 28 2011 at 6:27
3
What's wrong with Helgason? :) – José Figueroa-O'Farrill Sep 28 2011 at 8:05
1
Helgason is awesome, but not really "basic"... – S. Sra Sep 28 2011 at 8:32
show 3 more comments
## 5 Answers
I do not now really elementary text, but for the non-positively curved ones you can have a look at Eberlein's Geometry of nonpositively curved manifolds (Chicago Lectures in Mathematics, University of Chicago Press, 1996).
By the way, I do not see why similar answers where stated as comments. Maybe you should repost them as answers so that one can be accepted?
-
### 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.
My favorite text is chapter 8 in Joseph Wolf's book "Spaces of constant curvature". I have a copy of the 5th edition, published by Publish of Perish, but a 6th edition by AMS Chelsea has recently come out. I don't claim it is an easy read, you need to work a lot on the details, but it gets to the point very efficiently. Perhaps it is fair to say that one can use it as a guide and complement the arguments as needed using the books of Helgason and Loos (2nd volume).
In particular, the classification of symmetric spaces is done in a rather elementary way, up to the case of involutions of $E_6$ which requires a bit of theory of roots (this part is best looked up in Loos' book).
-
You can start with the chapter in the second volume of Kobayashi-Nomizu; both volumes are written in a reader friendly manner, and normally proofs are well detailed with precise references to the previous used results. Helgason is quite tough I think. Neither of these two are geometric enough in my opinion. The best for that are Loos's books (Symmetric space 1 and 2); these are my favourite !
-
As a very basic reference, you might enjoy the easy-to-read Arvanitoyeorgos' book (see Chapter 6 for symmetric spaces): An Introduction to Lie Groups and the Geometry of Homogeneous Spaces (Student Mathematical Library, V. 22). Another good introductory text is the Chapter XI of Kobayashi & Nomizu, Foundations of Differential Geometry, vol II. Of course more advanced references are the ones mentioned before: Helgason, Wolf, Loos etc.
-
"Metric rigidity theorems on Hermitian locally symmetric manifolds" by Ngaiming Mok is good, but may not be basic.
- | {"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": 0.3, "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": 0.9352188110351562, "perplexity_flag": "middle"} |
http://nrich.maths.org/2129 | ### Your Number Is...
Think of a number... follow the machine's instructions. I know what your number is! Can you explain how I know?
### First Forward Into Logo 6: Variables and Procedures
Learn to write procedures and build them into Logo programs. Learn to use variables.
### First Forward Into Logo 8: More about Variables
Write a Logo program, putting in variables, and see the effect when you change the variables.
# Special Numbers
##### Stage: 3 Challenge Level:
One day our teacher asked us a puzzling question:
Can you find a special number?
Can you find more than one?
Can you find them all?
One way to convince yourself that you've found all the special numbers is to use a little algebra...
Any two-digit number can be represented algebraically as $10a + b$.
$a$ is the tens digit and $b$ is the units digit.
Write down the algebraic expressions for the sum and the product of the digits.
Now write an equation that all special numbers will satisfy. See the hint if you're not sure how to do this.
Solve the equation, and use it to convince yourself that you have found all the special numbers.
There are other sorts of special two-digit numbers...
• I add twice the tens digit to the units digit, then add this to the product of the digits. I get back to my original number.
• I add three times the tens digit to the units digit, then add this to the product of the digits. I get back to my original number.
• I add four times... or five times... or...
Can you use algebra to help you to find these special numbers? | {"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": 0.8454151749610901, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/126093/order-of-a-normal-subgroup?answertab=votes | # Order of a normal subgroup
I need help showing this result:
"Let $G$ be a group such that $|G|=nm$ where $m$ and $n$ are relatively prime. Suppose that there exists a normal subgroup H of G such that $|H|=n$. Show that $H$ is the only subgroup with order $n$."
Can someone give me a light?
-
## 1 Answer
Consider the order of the image of any alleged subgroup of order $n$ under the canonical map from $G$ to $G/H$.
-
Can I say that $|\pi (H')|=n$ ($\pi$ is the canonical map)? Because I think I should think about $ker (\pi)$ which, when restricted to $H'$ (where $H'$ is the group other than $H$ with order $n$) as being the intersection of $H$ and $H'$ which is not necessarily $\{1\}$. – Gustavo Marra Mar 30 '12 at 2:13
If your notation means what I think it means, then $\pi(H')$ is a subgroup of $G/H$, and that has implications for its order. – Gerry Myerson Mar 30 '12 at 2:59
Yes, then it must divide the order of $G/H$, but that does not mean that it must divide m or n because they are not prime numbers themselves. – Gustavo Marra Mar 30 '12 at 9:45
And what is the order of $G/H$? – Gerry Myerson Mar 30 '12 at 11:25
It must be $m$. I think I got this, but my mind is clouded with some (personal life) problems right now. – Gustavo Marra Mar 30 '12 at 14:05
show 2 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": 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": 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": 0.9577310681343079, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/74961-graph-function.html | # Thread:
1. ## Graph of a function
List all the asymptotes of the graph of the function and the approximate coordinates of each local extremum.
f(x)=x(2^x)
I've had a hard time trying to solve this graph , I used my graphing calculator, but it didn't help me.
Acording to my book, the negative x axis is an asymptote. Can someone explain me how to interpret and graph the function properly? Thanks in advance!
2. Hi
$f(x) = x\:2^x = x\:e^{2\ln x}$
$f'(x) = e^{2\ln x} (1 + x \ln 2)$
Therefore f is decreasing from $-\infty$ to $-\frac{1}{\ln 2}$ and increasing from $-\frac{1}{\ln 2}$ to $+\infty$
The minimum has coordinates $(-\frac{1}{\ln 2} ; -\frac{1}{e\:\ln 2})$
The limit in $-\infty$ is 0 which shows that x-axis is an asymptot
The limit in $+\infty$ is $+\infty$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9343103766441345, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/169998/figure-out-if-a-fourth-point-resides-within-an-angle-created-by-three-other-poin?answertab=oldest | # Figure out if a fourth point resides within an angle created by three other points
If I have a point that is considered the origin and two lines that extend outwards infinitely to two other points, what is the best way to determine whether or not a fourth point resides on or within the angle created by those points?
The process I'm currently using is to get the angle of all three lines that extend out from the origin and then check to see whether the third angle is within the range of the first two.
Grid space is defined as Screen Space, that is, 2D Cartesian with the Y-Axis flipped so "up" is negative y and the origin is the upper left corner.
-
1
Your "fourth point" would be contained on either the acute angle or the obtuse angle determined by your lines... – J. M. Jul 12 '12 at 16:34
Exactly. What is the fastest way mathematically to determine whether this is the case? I'm currently working out three angles and determining whether the test angle is within the other two but hoping that there's a way to do this in less steps than I'm currently using. As in, determining three angles. – Steve Jul 12 '12 at 17:09
## 4 Answers
The two lines divide the plane into four wedges. The fourth point will be in one of these four wedges. To find which one, you can use the Atan2 function to find the angle of each line and the angle of the fourth point from the origin. You can look to see which ones it is between. You must remember that one of the wedges will cross $\theta=0$ and need to add $2\pi$ as required when checking.
-
I forgot to mention that I'm using this for a computer program. This is the process I'm currently using to determine whether the fourth point is in the wedge that the previous two points create. I'm trying to figure out if there's a faster way to get to the result than to get the angles of three lines using two points a piece and also compensating for the addition of 2pi. – Steve Jul 12 '12 at 16:52
@Steve: Then I don't have any better. The point is that you suggest there will be some points not in the angle, but there are four different angles and the point will be in one of them (unless it is on one of the lines). – Ross Millikan Jul 12 '12 at 16:54
If I understand you correctly you want to know whether the fourth point lies in the cone generated by the three other points with the origin as its apex. In that case, you can do this: Let $O,A,B$ denote the given vectors where $O$ is the origin and $C$ the fourth one. Unless $A$ and $B$ are linearly dependent you can uniquely write $C$ as a linear combination $C=\lambda_1 A + \lambda_2 B$. Now, if both $\lambda_1$ and $\lambda_2$ are positive $C$ lies within the cone spanned by $A$ and $B$. The other cases for values of $\lambda_1$ and $\lambda_2$ will give similar results such that $C$ lies on one of the rays spanned by $A$ or $B$ or will lie in the interior of the complement of the cone.
-
You understand my question perfectly, but I'm having difficulty understanding your solution. Do I need to reduce the vectors A, B and C to one unit in your equation? If not, this would run much better in my program than other solutions. Even if they do need to be reduced, I'd like to understand your method to compare it to the other ones I'm currently using for processing speed. Having A, B and C, I'm unsure how I am to figure out what λ1 and λ2 are. Also, can your method be used to determine whether C resides on the ray spanned by A or B as well as determining if it resides within the cone?ty – Steve Jul 13 '12 at 21:41
Finding λ1 and λ2 simply reduces to solving the 2x2 system of linear equation C=λ1A+λ2B. In this simple case, the solution can be written out explicitly: $\lambda_1=\frac{c_1b_2−b_1c_2}{a_1b_2−b_1a_2},\lambda_2=\frac{a_1c_2−c_1a_2}{a_1b_2−b_1a_2}$. This always works when A and B are linearly independent. The signs of λ1 and λ2 then uniquely define the location of C: both positive means in the interior of the cone, if one is zero it lays on one of the lines. There are $9$ different cases corresponding to the different combinations of signs. A sketch really helps to clarify this. – erlking Jul 13 '12 at 22:50
Shift your coordinates so that the point of the angle is the origin. Normalize the other points (so we're replacing all other points by ones that are a unit distance from the "origin". Say the cone is made by $O,A,B$ (in the new coordinates) and the point to find is $C$. If $A\cdot C$ and $B\cdot C$ are greater than $A\cdot B$, then it's between them.
EDIT: formula, with $O,A,B,C$ as the original, non-shifted points: $$D_{AB}=\frac{(A_x-O_x)(B_x-O_x)+(A_y-O_y)(B_y-O_y)}{\sqrt{(A_x-O_x)^2+(A_y-O_y)^2}\sqrt{(B_x-O_x)^2+(B_y-O_y)^2}}$$
If $D_{AB}<D_{AC}$ and $D_{AB}<D_{BC}$, then $C$ is in the angle. Otherwise, it's outside. The matrix method given on the stackoverflow page is also good, and may be easier. In this, $A,B,C$ have already been shifted.
$$C_x=\alpha A_x+\beta B_x$$ $$C_y=\alpha A_y+\beta B_y$$
if $\alpha$ and $\beta$ are both positive, then it's in the interior.
$$\alpha=\frac{C_yB_x-C_xB_y}{A_yB_x-A_xB_y}$$ $$\beta=\frac{C_yA_x-C_xA_y}{B_yA_x-B_xA_y}$$
My only suggestion is that the algorithm hasn't been implemented properly, possibly because of some subtle language/code issue. I'd advise you to post your written code on the stackoverflow question, it's harder to help without it.
-
I follow up until the multiplication and greater than. I'll try to explain what I get and hopefully you'll be able to clarify what I don't. I subtract points A, B and C individually by point O so that they are all shifted to be in the same position they would be in if the angle's point were at (0,0). After that is where I'm confused. How do I determine whether one point is greater than another? And in what way should I be multiplying the existing points? I think you understand what I'm asking and this definitely seems like it'll process faster than what I'm currently using =) – Steve Jul 12 '12 at 21:44
– Robert Mastragostino Jul 12 '12 at 22:21
I implemented this and for the most part, it works nicely but I keep running into a strange problem where this also proves to be true for a range that is outside of the cone formed by points O, A and B. For A=(0.9913462,-0.1312741), B=(0.4740998,0.8804711) and C=(-0.5978239,-0.8016275). C should be outside of the cone formed by O, A and B but when plugged in, it is less than both AdotC and BdotC. Is there a reason for this and is it something that I can fix? – Steve Jul 13 '12 at 21:35
Do you know roughly what the range is where it fails? That seems kind of bizarre. – Robert Mastragostino Jul 13 '12 at 22:00
youtu.be/e9h16q-n8Mk I went ahead and put together a video that should illustrate where it seems to go wrong. I've also tried two other formulas that I received here stackoverflow.com/questions/11456671/… If you could take a look and maybe help me figure out what's going wrong, that'd be immensely. . . well, helpful =) Please let me know if you have any questions about what's going on -_-;; – Steve Jul 14 '12 at 8:50
show 1 more comment
No need to solve systems of linear equations, call trigonometric functions, or even normalize the vectors. Let $a$, $b$, and $c$ be the vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\overrightarrow{OC}$, and denote $$u \wedge v = \begin{vmatrix}u_x & v_x \\ u_y & v_y\end{vmatrix} = u_x v_y - u_y v_x$$ for any two vectors $u$ and $v$. The sign of $u \wedge v$ tells you which side of the line parallel to $u$ the vector $v$ lies on, and vice versa. Then $C$ lies in the wedge between the rays $OA$ and $OB$ if and only if $a \wedge c$ and $c \wedge b$ have the same sign as $a \wedge b$.
-
Beautifully simple! And easy to program too. – TonyK Nov 22 '12 at 13:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 6, "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": 0.9556941986083984, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/118618/list | ## Return to Answer
4 misprint corrected
Regarding the question of pairs, here is the answer (hope, I'm not mistaken).
Let me first change the notation a little bit. Suppose, we are looking for irreps $V_1=\Sigma^{\lambda}V$ and $V_2=\Sigma^{\mu}V$ where $\lambda = (\lambda_1,\ldots,\lambda_n)$ and $\mu = (\mu_1,\ldots,\mu_n)$ resp., such that there exists an embedding $S^kV\to V_1^*\otimes V_2$.
It's immediate that such an embedding exists iff $V_2=\Sigma^{\mu}V$ is an irreducible factor in $V_1\otimes S^kV=\Sigma^{\lambda}V\otimes S^kV$. The Littlewood-Richardson rule gives us the answer. Firstly, $k=|\mu|-|\lambda|$, where $|\lambda| = \sum_i\lambda_i$. Secondly, one has the following inequalities:
• $\mu_1\geq \lambda_1$,
• $\lambda_1\geq \mu_2 \geq \lambda_2$,
• $\ldots$
• $\lambda_{n-1}\geq \mu_n \geq \lambda_n$.
Finally, if such an embedding exists, it's unique by the very same L-R rule.
UPD: Forgot to mention that the main question still seems to be very hard as even checking that $\dim V_1=\dim V_2$ is a huge problem.
UPD 2: Let me answer your main question (despite UPD). Let's forget about representation theory and do some linear algebra. What we have is a surjective map $V_1\otimes U\to V_2$, where $\dim V_1 = \dim V_2$. Let the corresponding bilinear map be $h:V_1\times U\to V_2$. We want to find such $u\in U$ that $h_u = h(-, u)$ is invertible, equivalently, of maximal rank.
Well, this seems to be quite obvious: let $r$ be the maximal rank of $h_u$ among all $u\in U$. Suppose $r<\dim V_2$. Let $u_0\in U$ be such that $\mathrm{rk}\ h_{u_0}=r$. Take any $v\in V_2\setminus\mathrm{Im}\ h_{u_0}$. Then there exists such $u_1\in U$ that $v\in \mathrm{Im}\ h_{u_1}$. Now it's easy to see that the rank of a general linear combination $\alpha h_{u_0} + \beta h_{u_1} = h_{\alpha u_0+\beta u_1}$ will be greater then than $r$.
3 added 832 characters in body
Regarding the question of pairs, here is the answer (hope, I'm not mistaken).
Let me first change the notation a little bit. Suppose, we are looking for irreps $V_1=\Sigma^{\lambda}V$ and $V_2=\Sigma^{\mu}V$ where $\lambda = (\lambda_1,\ldots,\lambda_n)$ and $\mu = (\mu_1,\ldots,\mu_n)$ resp., such that there exists an embedding $S^kV\to V_1^*\otimes V_2$.
It's immediate that such an embedding exists iff $V_2=\Sigma^{\mu}V$ is an irreducible factor in $V_1\otimes S^kV=\Sigma^{\lambda}V\otimes S^kV$. The Littlewood-Richardson rule gives us the answer. Firstly, $k=|\mu|-|\lambda|$, where $|\lambda| = \sum_i\lambda_i$. Secondly, one has the following inequalities:
• $\mu_1\geq \lambda_1$,
• $\lambda_1\geq \mu_2 \geq \lambda_2$,
• $\ldots$
• $\lambda_{n-1}\geq \mu_n \geq \lambda_n$.
Finally, if such an embedding exists, it's unique by the very same L-R rule.
UPD: Forgot to mention that the main question still seems to be very hard as even checking that $\dim V_1=\dim V_2$ is a huge problem.
UPD 2: Let me answer your main question (despite UPD). Let's forget about representation theory and do some linear algebra. What we have is a surjective map $V_1\otimes U\to V_2$, where $\dim V_1 = \dim V_2$. Let the corresponding bilinear map be $h:V_1\times U\to V_2$. We want to find such $u\in U$ that $h_u = h(-, u)$ is invertible, equivalently, of maximal rank.
Well, this seems to be quite obvious: let $r$ be the maximal rank of $h_u$ among all $u\in U$. Suppose $r<\dim V_2$. Let $u_0\in U$ be such that $\mathrm{rk}\ h_{u_0}=r$. Take any $v\in V_2\setminus\mathrm{Im}\ h_{u_0}$. Then there exists such $u_1\in U$ that $v\in \mathrm{Im}\ h_{u_1}$. Now it's easy to see that the rank of a general linear combination $\alpha h_{u_0} + \beta h_{u_1} = h_{\alpha u_0+\beta u_1}$ will be greater then $r$.
2 added 139 characters in body
Regarding the question of pairs, here is the answer (hope, I'm not mistaken).
Let me first change the notation a little bit. Suppose, we are looking for irreps $V_1=\Sigma^{\lambda}V$ and $V_2=\Sigma^{\mu}V$ where $\lambda = (\lambda_1,\ldots,\lambda_n)$ and $\mu = (\mu_1,\ldots,\mu_n)$ resp., such that there exists an embedding $S^kV\to V_1^*\otimes V_2$.
It's immediate that such an embedding exists iff $V_2=\Sigma^{\mu}V$ is an irreducible factor in $V_1\otimes S^kV=\Sigma^{\lambda}V\otimes S^kV$. The Littlewood-Richardson rule gives us the answer. Firstly, $k=|\mu|-|\lambda|$, where $|\lambda| = \sum_i\lambda_i$. Secondly, one has the following inequalities:
• $\mu_1\geq \lambda_1$,
• $\lambda_1\geq \mu_2 \geq \lambda_2$,
• $\ldots$
• $\lambda_{n-1}\geq \mu_n \geq \lambda_n$.
Finally, if such an embedding exists, it's unique by the very same L-R rule.
UPD: Forgot to mention that the main question still seems to be very hard as even checking that $\dim V_1=\dim V_2$ is a huge problem.
1
Regarding the question of pairs, here is the answer (hope, I'm not mistaken).
Let me first change the notation a little bit. Suppose, we are looking for irreps $V_1=\Sigma^{\lambda}V$ and $V_2=\Sigma^{\mu}V$ where $\lambda = (\lambda_1,\ldots,\lambda_n)$ and $\mu = (\mu_1,\ldots,\mu_n)$ resp., such that there exists an embedding $S^kV\to V_1^*\otimes V_2$.
It's immediate that such an embedding exists iff $V_2=\Sigma^{\mu}V$ is an irreducible factor in $V_1\otimes S^kV=\Sigma^{\lambda}V\otimes S^kV$. The Littlewood-Richardson rule gives us the answer. Firstly, $k=|\mu|-|\lambda|$, where $|\lambda| = \sum_i\lambda_i$. Secondly, one has the following inequalities:
• $\mu_1\geq \lambda_1$,
• $\lambda_1\geq \mu_2 \geq \lambda_2$,
• $\ldots$
• $\lambda_{n-1}\geq \mu_n \geq \lambda_n$.
Finally, if such an embedding exists, it's unique by the very same L-R rule. | {"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": 87, "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": 0.943615198135376, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2010/05/27/indefinite-integrals/?like=1&source=post_flair&_wpnonce=c1ce40f6e1 | # The Unapologetic Mathematician
## Indefinite Integrals
We can use integrals to make a set function $\nu$ out of any integrable function $f$.
$\displaystyle\nu(E)=\int\limits_E f\,d\mu$
We call this set function the “indefinite integral” of $f$, and it is defined for all measurable subsets $E\subseteq X$. This isn’t quite the same indefinite integral that we’ve worked with before. In that case we only considered functions $f:\mathbb{R}\to\mathbb{R}$, picked a base-point $a$, and defined a new function $F$ on the domain. In our new language, we’d write $F(x)=\nu\left([a,x]\right)$, so the two concepts are related but they’re not quite the same.
Anyhow, the indefinite integral $\nu$ is “absolutely continuous”. That is, for every $\epsilon>0$ there is a $\delta$ so that $\lvert\nu(E)\rvert<\epsilon$ for all measurable $E$ with $\mu(E)<\delta$. Indeed, if $c$ is an upper bound for $\lvert f\rvert$, then we can show that
$\displaystyle\lvert\nu(E)\rvert=\left\lvert\int\limits_Ef\,d\mu\right\rvert\leq\int\limits_E\lvert f\rvert\,d\mu\leq c\mu(E)$
And so if we make $\mu(E)$ small enough we can keep $\lvert\nu(E)\rvert$ small.
Further, an indefinite integral $\nu$ is countably additive. Indeed, if $f=\chi_S$ is a characteristic function then countable additivity of $\nu$ follows immediately from countable additivity of $\mu$. And countable additivity for general simple functions $f$ is straightforward by writing each such function as a finite linear combination of characteristic functions.
With the exception of this last step, nothing we’ve said today depends on the function $f$ being simple, and so once we generalize our basic linearity and order properties we will immediately have absolutely continuous indefinite integrals.
### Like this:
Posted by John Armstrong | Analysis, Measure Theory
## 19 Comments »
1. I been following your series of posts on measure and integrals. This is getting really interesting now, thanks for taking the time to put these posts up.
Comment by ip | May 30, 2010 | Reply
2. I’m not sure why this called an indefinite integral though, considering E is a subset of X
Comment by ip | May 30, 2010 | Reply
3. Glad to do it, ip, and glad you’re enjoying them.
As for “indefinite integral”, the idea is just as you said yesterday: “integrating” a function $f$ gives us a set function. Once you specify which set to integrate over, you get a number. In the same way, when we consider Riemann integrals we have to specify what interval to integrate over to actually get a number.
The old “indefinite integral” is a special, restricted form of the new version, where the only variation in the region is the choice of one endpoint.
Comment by | May 30, 2010 | Reply
4. [...] Integrals and Convergence I Let’s see how the notion of an indefinite integral plays with sequences of simple functions in the [...]
Pingback by | May 31, 2010 | Reply
5. I think that you want to write “measurable subsets E \subset X” in the second sentence. Great series of posts.
Comment by | May 31, 2010 | Reply
6. Thanks emile. Fixed.
Comment by | May 31, 2010 | Reply
7. [...] their indefinite integrals converge to the same limiting set function. That is, if and are the indefinite integrals of and [...]
Pingback by | June 1, 2010 | Reply
8. [...] is the indefinite integral of , and is the indefinite integral of . Then if we use to define the integral of we [...]
Pingback by | June 2, 2010 | Reply
9. [...] if an integrable function is a.e. nonnegative, then its indefinite integral is [...]
Pingback by | June 3, 2010 | Reply
10. [...] measure, and we’ve shown that convergence in mean implies uniform absolute continuity of the indefinite integrals. All we need to show in the first direction is that if converges in mean to , then the indefinite [...]
Pingback by | June 9, 2010 | Reply
11. [...] this is a statement about finite measure spaces; the function is here so that the indefinite integral of will give us a finite measure on the measurable space to replace the (possibly non-finite) [...]
Pingback by | June 14, 2010 | Reply
12. [...] the Integral Given an integrable function , we’ve defined the indefinite integral to be the set [...]
Pingback by | June 21, 2010 | Reply
13. [...] The Jordan Decomposition of an Indefinite Integral So, after all our setup it shouldn’t be surprising that we take an integrable function and define its indefinite integral: [...]
Pingback by | June 29, 2010 | Reply
14. [...] Continuity I We’ve shown that indefinite integrals are absolutely continuous, but today we’re going to revise and extend this notion. But first, to review: we’ve [...]
Pingback by | July 1, 2010 | Reply
15. [...] Continuity II Now that we’ve redefined absolute continuity, we should tie it back to the original one. That definition makes precise the idea of “smallness” as being bounded in size below [...]
Pingback by | July 2, 2010 | Reply
16. [...] let be a -finite signed measure on which is absolutely continuous with respect to . Then is an indefinite integral. That is, there is a finite-valued measurable function so [...]
Pingback by | July 6, 2010 | Reply
17. [...] define a signed measure as the indefinite integral of . Immediately we know that is totally -finite and that . And, obviously, is the Radon-Nikodym [...]
Pingback by | July 13, 2010 | Reply
18. Nice Site; I’ve learned a few things in just this read!. Re the proof of absolute continuity, what do you do if f is not bounded in E;
f may even take the value oo there, right (albeit in a subset of E of measure zero)? How would the argument c*m(E)->0 ; where c is
an upper-bund proceed then?.
Comment by carl | May 11, 2013 | Reply
19. To be honest I’m not sure offhand. I’m not really an analyst by training.
Comment by | May 11, 2013 | Reply
« Previous | Next »
## 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": 27, "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": 0.9004607796669006, "perplexity_flag": "middle"} |
http://mathhelpforum.com/advanced-statistics/209425-binomial-probability.html | # Thread:
1. ## Binomial Probability
Hi there,
I am having a few problems with this topic. I am actually not sure if it is on binomial.
I have independent trials - 50. I know the probability of pass and fail.
What I need to find is the probability of AT LEAST a certain number passing.
I cannot get my head around it - I have finally decided it is best to use the nCx formula, but it is impractical to calculate over 30 values individually and sum them.
What is the faster way?
Thanks!
2. ## Re: Binomial Probability
Originally Posted by SamanthaJane
Hi there,
I am having a few problems with this topic. I am actually not sure if it is on binomial. I have independent trials - 50. I know the probability of pass and fail.
What I need to find is the probability of AT LEAST a certain number passing.
If we have $N$ independent trials each of which has probability $p$ and $0<k\le N$
the probability of AT LEAST $k$successes is:
$\sum\limits_{k = 1}^N {\binom{N}{k}p^k \left( {1 - p} \right)^{N - k} }$
3. ## Re: Binomial Probability
Thanks for the response
This is what I have already done, excluding the sum.
I have gotten the answer to the nCx - but how do I go about the sum? I feel like I need to look back at my work on series' - ? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9521673917770386, "perplexity_flag": "head"} |
http://cms.math.ca/Events/summer11/abs/ps | 2011 CMS Summer Meeting
University of Alberta, Edmonton, June 3 - 5, 2011 www.cms.math.ca//Events/summer11
Plenary Speakers
[PDF]
LEAH EDELSTEIN-KESHET, UBC
Mathematics of Cell Polarization and Motility [PDF]
In my talk, I will introduce the phenomena of cell motility and explain what are some of the open problems and challenges in this interdisciplinary field. I will then survey some interesting mathematical and computational issues that such research motivates. I will describe work in my group on the partial differential equations that we used to describe how a cell polarizes (decides where is its front and back). Such PDEs describe a phenomenon of wave-pinning stemming from bistable kinetics, substrate depletion, and large disparity in rates of diffusion. I will also briefly describe our computations of 2D cell motility models and what we learned about the interaction between reaction-diffusion and geometry in the evolving cell.
This work is joint with Yoichiro Mori, Alexandra Jilkine, AFM (Stan) Maree, Ben Vanderlei, and William Holmes.
OLGA HOLTZ, UC Berkeley; TU Berlin
Zonotopal algebra: approximation theory meets algebra and combinatorics [PDF]
What do 1) integer points in polyhedra 2) hyperplane arrangements 3) parking functions on graphs 4) multivariate polynomial interpolation 5) kernels of differential operators have in common?
I will discuss recent mathematical developments motivated by the multivariate spline theory that demonstrate surprising connections between these (and some other) seemingly unrelated subjects in algebra, analysis and combinatorics.
FRANÇOIS LALONDE, University of Montreal, CRM
The New Real Algebraic Geometry [PDF]
Recently, Jean-Yves Welschinger introduced new Gromov-Witten types of invariants for real algebraic geometry. This led to a revolution in our understanding of real algebraic geometry. I will describe these invariants and other invariants introduced by Shengda Hu and myself that relate the locus of Lagrangian submanifolds to the geometry of the ambient spaces.
BJORN POONEN, MIT
$x^2 + y^3 = z^7$ [PDF]
There are 16 solutions to $x^2 + y^3 = z^7$ in relatively prime integers, one of which is $(21063928,-76271,17)$ (joint work with Ed Schaefer and Michael Stoll). I will explain why the existence of such solutions is not surprising, and I will sketch how one proves statements like this.
ROMAN VERSHYNIN, University of Michigan
Random matrices: invertibility, structure, and applications [PDF]
At the heart of random matrix theory lies the realization that the spectrum of a random matrix $H$ tends to stabilize as the dimensions of $H$ grow to infinity. This phenomenon is captured by the limit laws of random matrix theory, in particular by Wigner's semicircle law, Girko's circular law, and Marchenko-Pastur law. These limit laws offer us a clear global and asymptotic picture of the spectrum of $H$.
In the last few years, a considerable progress was made on the more difficult local and non-asymptotic regimes. In the non-asymptotic regime, the dimensions of $H$ are fixed rather than grow to infinity. In the local regime, one zooms in on a small part of the spectrum of $H$ until one sees individual eigenvalues. The location of the eigenvalue nearest zero determines the invertibility properties of $H$. This essentially determines whether the matrix $H$ is well conditioned, which is a matter of importance in numerical analysis.
Examples of recent developments include the proofs that a random matrix $H$ with independent entries (whether symmetric or not) is singular with an exponentially small probability, that the condition number of $H$ is linear in the dimension, and that the eigenstructure of $H$ is delocalized and unstructured -- the eigenvectors are spread out and their coefficients are highly incommensurate.
We will see some examples of heuristics, results, and problems of the non-asymptotic random matrix theory in the local regime. Applications and problems in related areas will be discussed, in particular for covariance estimation in statistics. | {"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": 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": 0.8845545649528503, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?p=4161670 | Physics Forums
Page 1 of 2 1 2 >
## Order of learning Mathematics
Hey there , Im a 14 year old teaching myself mathematics and I was wondering what comes next.. I have completed everything uptil Precalculus(That includes Algebra 1,2 Geometry and Triginometry) .. Id appreciate it if one you could list the order of learning mathematics , I know that math is interconnected (eg: You need Abstract alg to understand Linear Alg and you need multivariable Calc and whatnot) , say I was on a road to getting a degree in Maths . What maths will I be taught and in what order. Thanks
and It'd be great if some you seniors could suggest good books on self teaching some of the more advanced Math(that which comes after Pre-cal)
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
Blog Entries: 2 After someone has completed Algebra I&II, Geometry, and Trig, the natural progression seems to be Calculus. I'm sure a lot of people would recommend a lot of different books, but http://ocw.mit.edu/resources/res-18-...2005/textbook/ this is free and online. So enjoy. After calculus, the normal progression seems like this: Calc I, II, III, Linear Algebra, Differential Equations, and then a bunch of electives with an abstract algebra and real analysis course thrown into the mix.
Thanks , Could you also tell me what the prerequisites for number theory are? and how id go about to start learning it?
Blog Entries: 8
Recognitions:
Gold Member
Science Advisor
Staff Emeritus
## Order of learning Mathematics
Quote by MrApex Thanks , Could you also tell me what the prerequisites for number theory are? and how id go about to start learning it?
Elementary number theory doesn't really have any prereqs. You have to be well-versed in proofs though. You shouldn't be mystified with proofs by contradiction or induction and you should be able to form basic mathematical arguments. Furthermore, number theory can be abstract, so you might need to some mathematical maturity.
If you go deeper into number theory, then you are going to need abstract algebra and real analysis. But for a first course, I don't think those things are strictly needed (although a knowledge of abstract algebra makes a lot of things easier to grasp).
@Micromass The only proofing Ive done was when I was learning Geometry , Could you perhaps suggest a few books on how to write proofs ? and a few books to get me started with number theory? Thanks.
I don't really know that you need a dedicated proof book. If you pick some books that require you to work a lot of proofs, I feel this is sufficient... especially if you have someone who is willing to critique your answers (for instance, post your work up on here). I've read a couple of proof books but, other than giving a basic familiarity with proof techniques, I really feel that having someone rip apart your work and then you go back and correct/learn from the mistakes is the fastest way to learn. This is my perspective/experience, so take it as you wish.
If you feel like you want a book, you can find other recommendations by searching the forums. There are also some free books available online, for example: http://www.people.vcu.edu/~rhammack/BookOfProof/ I have no idea if it is any good, but the price is right for trying it out! jason
Merci Beaucoup !
Most people would advise you to study Calculus. It builds intuition for analysis ("calculus done right"). However, I would recommend studying a little bit of topology and set theory as well. The earlier you learn these subjects, the better of you will be in the future. I find I learn the most when I am most confused. And at first, these subjects can be very confusing.
Blog Entries: 8
Recognitions:
Gold Member
Science Advisor
Staff Emeritus
Quote by lucid_dream Most people would advise you to study Calculus. It builds intuition for analysis ("calculus done right"). However, I would recommend studying a little bit of topology and set theory as well. The earlier you learn these subjects, the better of you will be in the future. I find I learn the most when I am most confused. And at first, these subjects can be very confusing.
You're suggesting him to study topology before studying calculus. Am I understanding you right?
Topology should be very straightforward (that's not the same as saying it is trivial). If you find topology confusing, then you don't have the right prereqs or maturity.
Quote by micromass You're suggesting him to study topology before studying calculus. Am I understanding you right? Topology should be very straightforward (that's not the same as saying it is trivial). If you find topology confusing, then you don't have the right prereqs or maturity.
In my opinion, topology is confusing. In particular, I feel that many of the proofs to theorems require a high degree of creativity, and they are not always intuitive. In your opinion, if I find topology confusing, I don't have the right prereqs or maturity. But I do have the right prereqs, as I have taken graduate level classes in analysis, number theory, algebra and set theory before topology. Perhaps in your opinion I don't have the right maturity, but many of my professors (who have received PHds from Ivy League universities and the University of Chicago) have stated they found topology (in particular, algebraic topology) confusing when they first learnt it as well. This would indicate to me you may be of exceptional mathematical maturity, and not have the same perception of an average, or even above average student.
I am indeed encouraging him to study topology while he is studying calculus. I do think it may be difficult, but if he sticks with it, he will quickly build his mathematical maturity. In my opinion, mathematicians should not even bother with calculus.
Blog Entries: 8
Recognitions:
Gold Member
Science Advisor
Staff Emeritus
Quote by lucid_dream In my opinion, topology is confusing. In particular, I feel that many of the proofs to theorems require a high degree of creativity, and they are not always intuitive. In your opinion, if I find topology confusing, I don't have the right prereqs or maturity. But I do have the right prereqs, as I have taken graduate level classes in analysis, number theory, algebra and set theory before topology. Perhaps in your opinion I don't have the right maturity, but many of my professors (who have received PHds from Ivy League universities and the University of Chicago) have stated they found topology (in particular, algebraic topology) confusing when they first learnt it as well. This would indicate to me you may be of exceptional mathematical maturity, and not have the same perception of an average, or even above average student. I am indeed encouraging him to study topology while he is studying calculus. I do think it may be difficult, but if he sticks with it, he will quickly build his mathematical maturity. In my opinion, mathematicians should not even bother with calculus.
I don't really understand what you would find so hard about topology. Most of the terms and definitions are straightforward generalizations of things that you already know about. If you have the right prerequisites, then you should know about the things topology is trying to generalize. And if you do, then I don't see why topology is so hard.
I don't have an exceptional maturity at all. But I don't think you need one anyway.
I do agree that topology tends to have many definitions which make things confusing. I never remember what "accumulation point", "close point", "limit point", "complete accumulation point" are for example. But I can just look up those definitions. But yes, it might be common for new people to get tangled in a web of definitions they need to know. But that's about the only problem I really see.
The book "Mathematical Proofs: A Transition to Advanced Mathematics" by Chartrand et al. is a pretty good introduction to thinking like a mathematician, and the content in that book furnishes the minimum required background for studying subjects like number theory and topology. Once you've written a decent number of proper mathematical proofs, you ought to be able to work through a more abstract subject.
Quote by lucid_dream In my opinion, topology is confusing. In particular, I feel that many of the proofs to theorems require a high degree of creativity, and they are not always intuitive. In your opinion, if I find topology confusing, I don't have the right prereqs or maturity. But I do have the right prereqs, as I have taken graduate level classes in analysis, number theory, algebra and set theory before topology. Perhaps in your opinion I don't have the right maturity, but many of my professors (who have received PHds from Ivy League universities and the University of Chicago) have stated they found topology (in particular, algebraic topology) confusing when they first learnt it as well. This would indicate to me you may be of exceptional mathematical maturity, and not have the same perception of an average, or even above average student. I am indeed encouraging him to study topology while he is studying calculus. I do think it may be difficult, but if he sticks with it, he will quickly build his mathematical maturity. In my opinion, mathematicians should not even bother with calculus.
IN that case could you suggest a few books on topology which you think would be a good place to start? and what are the prereqs for topology?
and thank you all for the feedback (@Micromass and @Lucid Dream)
Everything in topology has a bunch of different definitions that are all equivalent, but still can be confusing, and sometimes using one definition and not another is actually helping in understanding a problem more clearly. But it's also interesting, and is a little bit like walking through an exotic zoo -- lots of weird counterexamples. I would suggest that you learn how to do proofs as soon as possible. Reading and writing proofs is really where the fun is in Mathematics, and it will be your gateway to all advanced mathematics. Then there are several paths you can take: Linear Algebra, Abstract Algebra, Analysis, or Topology. I took Analysis before I took topology and that helped. I also think that most topology books assume that you've taken analysis. At the very least it helps to know what limits are rigorously before taking topology. Abstract Algebra has very few prerequisites, mostly elementary Number Theory, which is covered in the beginning of most Algebra books. Linear Algebra has almost no prerequisites (except for what you've done so far), and is a nice and gentle mix between computing things and proving things. So you might want to look into that first. Analysis is very wide -- it has the computational aspect, which is very important and broad, but it also has the more theoretical part where you just" prove theorems. It's possible to learn one and then the other, but most people go with the computational part (i.e. Calculus) and then the theoretical part which is what people usually call Analysis. I think that I would have benefited from a mix of the two using something like Spivak's Calculus book. To put it in one line: First learn how to read and write proofs, then pick whichever subject you're interested in, but be reasonable -- don't try to learn Algebraic Geometry first.
Quote by micromass I don't really understand what you would find so hard about topology. Most of the terms and definitions are straightforward generalizations of things that you already know about. If you have the right prerequisites, then you should know about the things topology is trying to generalize. And if you do, then I don't see why topology is so hard. I don't have an exceptional maturity at all. But I don't think you need one anyway. I do agree that topology tends to have many definitions which make things confusing. I never remember what "accumulation point", "close point", "limit point", "complete accumulation point" are for example. But I can just look up those definitions. But yes, it might be common for new people to get tangled in a web of definitions they need to know. But that's about the only problem I really see.
What I find most confusing about topology is that there seem to be many similar (or related) definitions and concepts and not enough examples for them. It helps a lot when I know an example that comes with definition because even when I forget parts of the definition I can always fill the blanks by considering an example case. Another thing is that there is little to be done with the knowledge you gain for some time. When you learn calculus for instance you always do something with the definitions you learn which helps to clarify things quite a bit.
Quote by MrApex IN that case could you suggest a few books on topology which you think would be a good place to start? and what are the prereqs for topology?
There is an interesting book with very minimal prerequisites (mostly just traditional high school maths) called The Knot Book. I believe the author's last name is Adams. Really fun, VERY interesting read. I would also recommend Principles of Mathematics by Allendoerfer. The first time I attempted to go through it, I didn't really like it... but I have since come to like it a lot and have worked through the whole thing. It will cover a great deal of material assuming no more prerequisite knowledge than you already possess.
Page 1 of 2 1 2 >
Tags
math, order
Thread Tools
| | | |
|----------------------------------------------------|----------------------|---------|
| Similar Threads for: Order of learning Mathematics | | |
| Thread | Forum | Replies |
| | Educators & Teaching | 43 |
| | General Math | 2 |
| | Academic Guidance | 1 |
| | Academic Guidance | 12 |
| | General Math | 1 | | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "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": 0.966624915599823, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/136526/int-c-fz-dz-leq-frac-pi-rr2-a2-ra | # $|\int_{C} f(z)\, dz|\leq \frac{\pi R}{R^{2}-a^{2}}$, $R>a$
Could someone help me through this problem?
Let $C$ be an open (upper) semicircle of radius R withits center at the origin, and consider $$\displaystyle\int_{C} f(z)\, dz.$$ Let $\displaystyle f(z)=\frac{1}{z^{2}+a^{2}}$ for real $a>0$. Show that $$|f(z)|\leq \frac{1}{R^{2}-a^{2}}~\text{for }R>a,$$ and $$\left|\displaystyle\int_{C} f(z)\, dz \right|\leq \frac{\pi R}{R^{2}-a^{2}}~\text{for }R>a$$
-
1
Consider the reverse triangle inequality – Alex Apr 25 '12 at 0:10
## 2 Answers
For $z\in C$, we have $|z|= R>a$. Hence, we have $$|f(z)|=\frac{1}{|z^2+a|^2}\leq\frac{1}{|z|^2-a^2}=\frac{1}{R^2-a^2}.$$ Now, using this inequality, we have $$\left|\int_{C} f(z)\, dz\right|\leq \left|\int_{C} |f(z)|\, dz\right|\leq\frac{1}{R^2-a^2}\left|\int_{C}\, dz\right|=\frac{\pi R}{R^{2}-a^{2}}$$ since $C$ is an open (upper) semicircle of radius R with its center at the origin.
-
We have $|z^2 + a^2| \geq |z|^2 - |a|^2 = R^2 - a^2$ for $z\in C$. Taking reciporicals, we get the first inequality
For the second, we have by definition of a path integral that
$$|\int_C f(z) dz| = |\int_0 ^\pi f(Re^{i\theta}) Re^{i\theta} d\theta| \leq \int_0 ^\pi R|f(Re^{i\theta})|d\theta \leq \frac{\pi R}{R^2 - a^2}$$
using the first inequality.
- | {"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": 10, "mathjax_display_tex": 6, "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": 0.9387935996055603, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/130881/taylor-expansion-for-matrices | # Taylor expansion for matrices
Is it possible to define a Taylor expansion for matrices ? Can I use functional derivative ?
More precisely I have to calculate something like : $\ln(A+B)$ using a Taylor expansion, where $A$ and $B$ are hermitian matrices which depend also on $x\in \mathbb{R}^3$. My idea is to use functional derivative but I don't know if the result is correct!
-
## 1 Answer
In one particular case there is very good approach. Denote $C=A+B-I$, if $\operatorname{Sp}(C)\subset\mathbb{D}$ then we can use Tailor expansion for the $$\ln(I+C)=\sum\limits_{n=1}^\infty\frac{(-1)^{n+1}}{n}C^n=\sum\limits_{n=1}^\infty\frac{(-1)^{n+1}}{n}(A+B-I)^n$$ If this conditions are not satisfied you should apply general theory - holomorphic functional calculus. In order to $\ln(A+B)$ make sense it is necessary that $\operatorname{Sp}(A+B)\subset\mathbb{C}^\times$, because $\ln\in\mathcal{O}(\mathbb{C}^\times)$. In this case we take arbitrary contour $\gamma$ containing $\operatorname{Sp}(A+B)$ and "simply" calculate $$\ln(A+B)=\frac{1}{2i\pi}\int\limits_\gamma\ln(\lambda)(\lambda I-A-B)^{-1}d\lambda$$
-
More explicitly: the series for the matrix logarithm is valid if and only if the eigenvalues of $C$ are within the disk of convergence of the usual scalar series for the logarithm... – J. M. May 13 '12 at 8:53
– J. M. May 13 '12 at 8:56 | {"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": 12, "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": 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": 0.8757988214492798, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/94365/isoperimetric-inequality/94558 | ## Isoperimetric inequality
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $M$ be a complete, non-compact, simply connected Riemannian manifold of dimension $n$ whose sectional curvatures are bounded above by $\kappa<0$. I want to prove that for any open subset $\Omega\subset M$ whose closure in $M$ is compact, the following inequality holds: $$\frac{Vol(\Omega)}{Vol(\partial \Omega)}\leq \frac{1}{(n-1)\sqrt{-\kappa}}$$
The constant on the right gives a lower bound for the first Dirichlet eigenvalue of the Laplace operator. If the metric on $M$ is given by $ds^2=g_{ij}dx^idx^j$, then $$\Delta=\frac{1}{\sqrt{\det g}}\sum_{i,j} \frac{\partial}{\partial x^i}\left(\sqrt{\det g} g_{ij} \frac{\partial}{\partial x^j}\right)$$ If $0<\lambda_1<\lambda2<\cdots$ are the Dirichlet eigenvalues of $-\Delta$, by a theorem of Mckean we have an inequality $$\lambda_1(M)\geq \frac{1}{4}(n-1)^2k$$ for a Riemannian manifold satisfying the conditions above.
Is there a way to relate the first eigenvalue to the ratio of volumes so as to prove the isoperimetric inequality above or is all this the wrong strategy?
Thanks in advance for any insight.
-
1
I don't think using eigenvalue estimates is the right way to go. What about just using the Bishop-Gromov inequality? – Deane Yang Apr 18 2012 at 8:42
3
Bishop-Gromov is in the wrong direction; the right inequality is Günther's one (sometimes called Bishop-Günther). – Benoît Kloeckner Apr 18 2012 at 12:58
Thanks for the correction! – Deane Yang Apr 18 2012 at 13:40
## 2 Answers
I do not know if there is a way to get the isoperimetric inequality from the spectral gap, but both can be proven in almost the same way. The classical references for the linear isoperimetric inequality are S.-T. Yau, "Isoperimetric constants and the first eigenvalue of a compact Riemannian manifold", Ann. Sci. École Norm. Sup. (4) 8 (1975), no. 4, 487–507 and Yurii D. Burago and Victor A. Zalgaller, "Geometric inequalities".
I like this proof so let me give it here (this is Burago-Zalgaller presentation). For any unit tangent vector $u$ and positive real $r$, let $s(u,r)$ be the "candle function" defined by `$$dy = s(u,r) \,du \,dr$$` when $y=\exp_x(ru)$ and $u\in UT_xM$. Up to a normalization, this is simply the jacobian of the exponential map. The curvature hypothesis implies $(\log s(u,r))'\geqslant \sqrt{-\kappa}(n-1)$ where the prime denotes derivative with respect to $r$ (this is a consequence of Günther's inequality).
$\Omega$ is contained in the union of all geodesic rays from any fixed point $x_0$ to $\partial \Omega$. Let $U\subset UT_{x_0}M$ be the set of unit vectors generating geodesics that intersect $\Omega$, and for $u\in U$ let $r_u$ be the last intersection time of the geodesic generated by $u$ with $\Omega$. Then
$$\mathrm{Vol}(\partial \Omega) \geqslant \int_U s(u,r_u) \,du$$ and $$\mathrm{Vol}(\Omega) \leqslant \int_U \int_0^{r_u} s(u,t) \,dt\,du.$$
Now, writing $s(u,r_u)=\int_0^{r_u} s'(u,t) \,dt$ and using Günther's inequality, the desired result comes.
I cannot help self-advertising: in fact, the same conclusions (Günther inequality, hence both the linear isoperimetric inequality and the spectral gap of MCKean) hold under a weaker curvature bound (some higher but non-positive sectional curvature can be compensated by enough more negative sectional curvature in other directions). This is explained in an arXiv paper with Greg Kuperberg, "A refinement of Günther's candle inequality" [arXiv:1204.3943].
-
Thanks for the advertising. I will definitely have to take a look. – Deane Yang Apr 18 2012 at 14:49
1
That would be Yuri Burago and Victor Zalgaller. – alvarezpaiva Apr 18 2012 at 19:29
@alvarezpaiva: you are right, thanks a lot. I do not know where we got the wrong reference. – Benoît Kloeckner Apr 19 2012 at 8:03
C'est mon erreur. I wanted full names in the bibliography, and I did not realize that there is both a Yuri D. Burago and a Dmitri Yu. Burago working in metric geometry. (Who might well be related.) – Greg Kuperberg Apr 19 2012 at 14:31
2
Dmitri (who is at Penn State) is Yuri's son. – Deane Yang Apr 19 2012 at 14:35
show 2 more comments
### 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.
Benoit gave a good answer in the sense that he revised the question so that it can have a good answer. Instead of trying to derive an isoperimetric inequality from McKean's inequality, the natural thing to do is to prove them both from a logarithmic candle inequality, the property that is discussed in our new paper. But suppose that you still wanted to prove an isoperimetric inequality just from McKean's inequality. Then I don't think that it's possible, because I think that there are manifolds with a large spectral gap, but which have domains with a lot of volume and very little boundary.
Suppose that you have a hyperbolic $n$-disk $D$ with radius $\epsilon$ and extreme negative curvature $\kappa \ll 0$. Then you can glue together two copies of $D$ along their boundary, i.e., the doubling construction. This is already a Riemannian manifold with at least a continuous metric; you can also smooth the metric in a tiny neighborhood of the common circle to make it $C^\infty$. If you send $|\kappa|$ to infinity first, before sending $\epsilon$ to 0, then it looks like this surface has a big eigenvalue gap. Since it is a topological sphere, it has a big disk inside with small boundary, namely the complement of a small disk. You can also take a connected sum with this manifold and another manifold, connecting at a small disk.
Okay, this type of example is not non-positively curved. So one remaining question is, if $M$ is non-positively curved, then does McKean's inequality imply a linear isoperimetric inequality as stated above? I don't know the answer to that; it could be a good question. But if the intent of the original question was how to prove the existing isoperimetric inequality of $K < \kappa$ manifolds, rather than whether one can prove a new isoperimetric inequality, then I don't see much possibility for McKean's inequality to lead to simplifications.
Here is a possibly related fact: There exist amenable groups with exponential growth. Non-amenability and exponential growth are two other properties that both look something like the properties discussed here, and it turns out that they are not equivalent.
-
There is a way to relate spectral gap to isoperimetric inequalities: given a domain, construct a function $f$ that approaches its characteristic function, and apply a Poincaré inequality (obtained from the spectral gap by considering a Rayleigh quotient). However, if one considers the usual action of $\Delta$ on $L^2$, one gets a $2,2$ Poincaré inequality and the $\int |\nabla f|^2$ blows up. It may be possible to get a different Poincaré inequality that relates $\int |\nabla f|$ to a $L^p$ norm of $f$ and gives the desired conclusion, and this must be classical. – Benoît Kloeckner Apr 20 2012 at 18:57 | {"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": 44, "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": 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": 0.9263814687728882, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/72029-box-square-base-has-surface-area-including-top-3-m-2-a.html | # Thread:
1. ## A box with a square base has a surface area including the top of 3 m^2.
1. A box with a square base has a surface area including the top of 3 m^2. Express the volume V of the box as a function of the width w of the base.
2. A stone is thrown into a lake and t seconds later after the splash the diameter of the circle of the ripples is t meters.
a. Express the circumference C of this circle as a function of t.
b. Express the area A of this circle as a function of t.
2. Originally Posted by Jasmina8
1. A box with a square base has a surface area including the top of 3 m^2. Express the volume V of the box as a function of the width w of the base.
surface area ... $2w^2 + 4wh = 3$
use the surface are equation to solve for h in terms of w, then substitute for h in the volume equation ...
$V = w^2h$
2. A stone is thrown into a lake and t seconds later after the splash the diameter of the circle of the ripples is t meters.
a. Express the circumference C of this circle as a function of t.
b. Express the area A of this circle as a function of t.
you are given that $d = t$ ...
a. hint ... $C = \pi d$
b. two hints ... $A = \pi r^2$ and $r = \frac{d}{2}$ | {"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": 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": 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": 0.9181196689605713, "perplexity_flag": "middle"} |
http://quant.stackexchange.com/questions/tagged/collective-risk-model?sort=votes&pagesize=15 | Tagged Questions
The collective-risk-model tag has no wiki summary.
There is a vast literature on copula modelling. Using copulas I can describe the joint law of two (and more) random variables $X$ and $Y$, i.e. $F_{X,Y}(x,y)$. Very often in risk management (credit ... | {"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": 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": 0.8622814416885376, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/246939/what-is-the-function-field-of-a-hyperplane | # What is the function field of a hyperplane?
If $X\subset\mathbb{P}^n$ is a hyperplane, what is the function field $K(X)$ ? Is $K(X)$ just k?
-
6
A hyperplane is isomorphic to $\mathbb{P}^{n-1}$. Do you know what the function field of $\mathbb{P}^{n-1}$ is? – Qiaochu Yuan Nov 29 '12 at 0:55 | {"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": 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": 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": 0.8639553785324097, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/171285-find-trigonometric-values-theta.html | # Thread:
1. ## Find the trigonometric values of theta.
My teacher wasn't very thorough with the lesson and I'm pretty lost on how to go about doing this.
Basically I have to find all of the trigonometric values of the following:
$\tan \theta = -4 , \sin \theta > 0$
What steps do I need to take in order to find sin, cos, tan, csc, sec, and cot using the following information?
2. Hello, dagbayani481!
$\tan \theta = \text{-}4 ,\;\;\sin \theta > 0$
$\text}Find all of the trignometric values of }\theta.$
We know that tangent is negative in quadrants 2 and 4.
We know that sine is positive in quadrants 1 and 2.
. . Hence, $\,\theta$ is in quadrant 2.
Code:
``` *
\
\
\ @
- - - - * - - -```
We know that: . $\tan\theta \:=\:-\dfrac{4}{1} \:=\:\dfrac{opp}{adj}$
So the diagram looks like this:
Code:
``` *
|\
4 | \
| \ @
- - * - * - - -
-1```
We have: . $opp = 4,\;adj = \text{-}1$
Pythagorus says: . $hyp \:=\:\sqrt{4^2 + (\text{-}1)^2} \:=\:\sqrt{17}$
Now you can write all six trigonometric values.
3. Thanks that really helped!
I tried working it out myself before I saw the answer. I got something similar except instead of having the opp. side as 4 and the adj. side as -1, I had the opp. side as -4 and the adj. side as 1. Does it make a difference where you chose to put the negative?
4. Originally Posted by dagbayani481
Does it make a difference where you chose to put the negative?
It does as that would put this triangle in either the 3rd or 4th quadrant.
But in this case you could get away with it as it would not change the result.
5. Yes, it does change the result. If you take "opposite side" as -4 and "near side" as +1, you get the sine, "opposite over hypotenuse" negative when the probelem specifically says it is positive. You would have the wrong signs for sine, cosine, secant, and cosecant. | {"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": 7, "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": 0.9545930624008179, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/13320/cool-problems-to-impress-students-with-group-theory/13434 | ## Cool problems to impress students with group theory [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Since this forum is densely populated with algebraists, I think I'll ask it here.
I'm teaching intermediate level algebra this semester and I'd like to entertain my students with some clever applications of group theory. So, I'm looking for problems satisfying the following 4 conditions
1) It should be stated in the language having nothing whatsoever to do with groups/rings/other algebraic notions.
2) It should have a slick easy to explain (but not necessarily easy to guess) solution using finite (preferrably non-abelian) groups.
3) It shouldn't have an obvious alternative elementary solution (non-obvious alternative elementary solutions are OK).
4) It should look "cute" to an average student (or, at least, to a person who is curious about mathematics but has no formal education).
An example I know that, in my opinion, satisfies all 4 conditions is the problem of tiling a given region with given polyomino (with the solution that the boundary word should be the identity element for the tiling to be possible and various examples when it is not but the trivial area considerations and standard colorings do not show it immediately)
I'm making it community wiki but, of course, you are more than welcome to submit more than one problem per post.
Thanks in advance!
-
1
Are group actions allowed? – Harry Gindi Jan 29 2010 at 2:41
Certainly. But not in the statement of the problem itself. – fedja Jan 29 2010 at 2:46
1
The tiling problem is a great example; I can't help but think there's more to it than meets the eye, but I don't know anywhere that it's written down and developed in detail. Do you have a good reference? – Qiaochu Yuan Jan 29 2010 at 3:00
1
Aren't the tiling groups usually infinite? jstor.org/pss/2324578 – Douglas Zare Jan 29 2010 at 3:10
The point is one can take quotients and hope that they are 1) finite, and 2) still prove that tiling is impossible. The example I know is dominos of size 1x2 and 2x1, where a certain quotient of the tiling group is S_3. – Qiaochu Yuan Jan 29 2010 at 3:19
show 3 more comments
## 28 Answers
In the TV show "Futurama", there's an episode named "The Prisoner of Benda" in which two of the characters swap bodies using a machine with a fundamental flaw: no two people can use it to swap bodies twice. This means they can't simply use the machine again to swap back.
They spend the rest of the episode trying to return to their original bodies. Eventually, a scientist (who also happens to be a Harlem globetrotter) figures out how to help them using group theory.
With a little reflection, one can see that this problem can be recast into one about the symmetric group.
-
3
Said problem is apparently based on a theorem in the episode writer's PhD thesis. – Jonathan Kiehlmann Nov 30 2010 at 8:48
theinfosphere.org/Futurama_theorem – Martin Brandenburg Aug 9 2011 at 12:49
oz.plymouth.edu/~dcernst/Talks/DeckJS/GordonTalk/… – Andres Caicedo Jan 1 2012 at 16:03
Andres, that like doesn't work. – Dan Ramras Jun 7 at 21:05
This should be more stable: Some talks by Dana Ernst on this subject: danaernst.com/tag/futurama – Andres Caicedo Apr 13 at 6:27
### 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.
Here is a striking application of a particular finite non-abelian group.
Explain to your students the issue of check digits as an error-detecting device on credit cards, automobile identification numbers, etc. Two common errors in communicating strings of numbers is a single-digit error (...372... --> ...382...) or an adjacent transposition error (...32... ---> ...23...). We want to design a check digit protocol in such a way that these two common errors are both detected (though not necessarily corrected: an error sign may flash in practice and the person is just prompted to enter the numbers all over again). The simplest check digit protocol uses modular arithmetic, as follows.
If we have an alphabet of m symbols and we agree that our strings of symbols to be used all have n terms, say they are written as a_1a_2...a_n, introduce a set of weights w_1,...,w_n and a valid string is one where
w_1a_1 + ... + w_na_n = 0 mod m.
In practice we take w_n = 1 or -1 and the unique choice of a_n that fits the congruence given all the other data is the check digit.
Theorem 1: All single digit errors are caught iff (w_i,m) = 1 for all i.
That means if a valid string -- one satisfying the above congruence -- has a single term changed then the result will not satisfy the congruence and thus the error is detected.
Theorem 2: All adjacent transposition errors are caught iff (w_{i+1}-w_i,m) = 1 for all i. (Wrap around when i = n.)
For example, say m = 10 (using the symbols 0,1,2,...,9). If all single digit errors are caught then each w_i has to be taken from {1,3,5,7}, but the difference of any two of these is even, so Theorem 2 won't apply.
The conclusion is that "no check digit protocol exists on Z/10 (for strings of length greater than 1) which detects all single digit errors and all adjacent transposition errors.
Maybe you think we are just not being clever enough in our check digit protocol mod 10. For example, instead of those scaling operations a |---> w_ia which are put together by addition, we could just define in some other way a set of permutations s_i of Z/m and declare a string a_1....a_n to be valid when
s_1(a_1) + ... + s_n(a_n) = 0 mod m.
We can use this congruence to solve for a_n given everything else, so we can make check digits this way too.
Theorem 3. When m = 10, or more generally when m is even, there is an adjacent transposition error -- and in fact a transposition error in any two predetermined positions for some string -- that won't be caught.
The proof is a clever argument by contradiction, but I won't type up the details here.
Since in practice we'd like to use 10 digits (or 26 letters -- still even) for codes, Theorem 3 is annoying. The book community with their ISBN code got around this by using m = 11 with a special check digit of X (a few years ago they switched to m = 13). It is natural to ask: is there some check digit protocol on 10 symbols?
Answer: Yes, using the group D_5 (non-abelian of order 10) in place of Z/10.
This was found by Verhoeff in 1969. It has hardly been adopted anywhere, due to inertia perhaps, even though the mechanism of it would in practice always be hidden in computer code so the user wouldn't really need to know such brain-busting group theory like D_5.
S. J. Winters, Error Detecting Schemes Using Dihedral Groups, The UMAP Journal 11 (1990), 299--308.
The only bad thing about this article by Winters is the funny use of the word scheme, e.g., the third section of the article is called (I am not making this up) "Dihedral Group Schemes". I recommend using the word "protocol" in place of "scheme" for this check digit business since it it more mathematically neutral.
By the way, Theorem 3 should not be construed as suggesting there is no method of using Z/10 to develop a check digit protocol which detects both of the two errors I'm discussing here. See, for example,
K. A. S. Abdel-Ghaffar, Detecting Substitutions and Transpositions of Characters, The Computer Journal 41 (1998) 270--277.
Section 3.4 is the part which applies to modulus 10. I have not read the paper in detail (since I'm personally not interested enough in it), but the end of the introduction is amusing. After describing what he will be able to do he says his method "is easier to understand compared to the construction based on dihedral groups". What the heck is so hard about dihedral groups? Sheesh.
-
2
You are right that the 10 digit long ISBN uses a modulo 11 check digit. However, the new 13 digit long ISBN uses a modulo 10 check digit which does not catch all adjacent transpositions (but catches all single-digit modifications). The weights used are alternating 1 and 3. – Zsbán Ambrus Jun 24 2011 at 12:38
8
The old German bills (in D-Mark, before the Euro) used to have checksums defined over the dihedral group $D_5$. – jp Jun 24 2011 at 15:10
An obvious choice is the enumeration of orbits of finite group actions, which show up everywhere in middle- and high-school competitions in disguise. The "cute" example here is coloring a cube or a regular polygon up to rotation. I have written a few blog posts on the subject and there are several examples given in the third post. One of my favorite applications here is Lucas' theorem, and once you bring that up you can casually mention the Sierpinski triangle as well.
Edit: And again I feel the need to tell you things most AoPSers already know, but another nice class of examples is provided by the application of linear algebra over $\mathbb{F}_2$ to problems in combinatorics, see e.g. this post by Tim Gowers or Jacob Steinhardt's excellent solution to USAMO 2008 #6.
Edit #2: Although it is possible to state all this in the simpler language of equivalence relations, one of my favorite ways to show that one integer divides another is to show that there exist a pair of groups $H, G$ with the corresponding cardinalities, one of which is a subgroup of the other. I know a few examples related to the symmetric groups in particular:
• $S_{m+n}$ contains the subgroup $S_m \times S_n$, hence $m! n! | (m+n)!$.
• $S_{mn}$ contains the subgroup $S_m \wr S_n$, hence $m!^n n! | (mn)!$. (See also the Young tableaux example I wrote about above.)
• The Sylow $p$-subgroups of $S_n$ are the iterated wreath product of cyclic groups of order $p$, as follows: divide $n$ up into $\lfloor \frac{n}{p} \rfloor$ blocks of $p$ elements and consider the permutations which preserve these blocks, then divide the blocks up into $\lfloor \frac{n}{p^2} \rfloor$ blocks and toss in permutations of these blocks, etc. Consequently $v_p(n!) = \sum_{k \ge 1} \lfloor \frac{n}{p^k} \rfloor$.
-
1
Very interesting answer! – Grétar Amazeen Jan 29 2010 at 9:48
A colleague of mine gave a talk on using Burnside's lemma to count sudoku puzzles. – Victor Protsak May 23 2010 at 10:26
Mattress Turning! Don't say it's not 'cute'!
The set of sensible orientations of your mattress on a bed (probably) has 4 elements. If your mattress is prone to sagging due to the weight of its users then you need to periodically turn it into another one of the 4 configurations. It'd be nice if there were a simple operation you could perform on the mattress once a month, say.
Now comes the group theory: the 4 element group of configurations is the Klein viergruppe, not a cyclic group. So you can't find one transformation that you can repeat to get all configurations. You need to have a more complex procedure where the transformation varies from month to month.
It gets harder, of course, if you have a cubical mattress. See Group Theory in the Bedroom for more information.
-
2
That's a great article! – Victor Protsak May 23 2010 at 10:21
Here's the first "group theory" problem I ever saw (before I knew anything about groups). Consider the following tiling of a 4x4 square with 15 numbered square tiles and one empty square:
``` 1 2 3 4
5 6 7 8
9 10 11 12
13 15 14 []
```
The `[]` is the empty square. You can slide any adjacent tile into the empty square, leaving its previous space empty. Is there any way to put the tiles in numerical order with the lower left corner empty (i. e., switch the 15 and 14)?
The solution is simple (but hardly obvious) even without explicitly mentioning group theory, but it's especially simple if you know a little bit about permutation groups.
-
2
That's more of a "grupoid theory" problem :) – Mariano Suárez-Alvarez Jan 29 2010 at 3:17
12
The simple solution can also be explained to people with no group theory, but who know that switching two rows in a determinant introduces a minus sign. – Allen Knutson Jan 29 2010 at 3:37
1
Yakov Perelman demonstrated the solution without explicitly introducing any math above elementary school level at all :) – Kallikanzarid Oct 6 2010 at 14:00
1
This problem - the Fifteen puzzle - is one of my all-time favourites. It has an interesting history too, en.wikipedia.org/wiki/Fifteen_puzzle – AL Jan 31 2011 at 16:27
1
This has not really something to do with groups, but only with the notion of the signum of a permutation. This was known decades before abstract groups came into life. – Martin Brandenburg Aug 9 2011 at 12:49
show 1 more comment
Fermat's little theorem, which is easy to prove with cyclic groups, can be used to prove that an integer is composite without producing a factorization.
-
This summer I developed a minicourse for high school students in group theory. We discussed the game of 15, SET, walpaper patterns, and Rubik's cube. Maybe this could be helpful:
www-math.mit.edu/~etingof/groups.pdf
in particular, we used this diagram for recognition of the 17 plane patterns:
www-math.mit.edu/~etingof/Patternsdiagram.pdf
-
What a great introduction to group theory! – Martin Brandenburg Aug 9 2011 at 12:51
This set of notes is excellent! – Jon Bannon Apr 12 at 17:23
One of my favourite easy group theory problems:
Any prime number $p$ divides $f_{2p(p^2-1)}$, where $f_n$ is the Fibonacci sequence.
Proof: Let
$$G:= \{ A \in M_2 (Z_p) | \det(a)= \pm1 \}$$
and let
$$F=\left( \begin{array}{c c} 1 & 1 \\ 1 & 0 \\ \end{array}\right)$$
Then $F \in G$ and $G$ is a group of order $2p(p^2-1)$.
Thus
$$\left( \begin{array}{c c} f_{n+1+2p(p^2-1)} & f_{n+2p(p^2-1)} \\ f_{n+2p(p^2-1)} & f_{n+2p(p^2-1)-1} \end{array}\right)$$
$$= F^{n+2p(p^2-1)}= F^n = \left( \begin{array}{c c} f_{n+1} & f_{n} \\ f_{n} & f_{n-1} \end{array}\right) \mod p \,.$$
P.S. Better periods can be obtained by solving the linear reccurence in $Z_p$ if $p =\pm1 \mod 5$ or in an algebraic extension if $p =\pm 2 \mod 5$, but that's exactly the same thing as calculating the order of the matrix $F$ by diagonalizing it.
The same idea can be used for any linear recurrence, but one has to replace $G$ by $GL_2(Z_p)$, and discuss the cases when $p$ divides or doesn't divide the free term of the polynomial associated to teh recurrence.
PPS: Can anyone please fix my matrices.
-
This thread has been dormant for some time, and the class which motivated it is probably by now over, but I've just stumbled across it and can't help but mention an interesting and surprising example of elementary group theory making an appearance in anthropology.
This comes from an appendix, "On the Algebraic Study of Certain Types of Marriage," that Andre Weil wrote for a book of Claude Levi-Strass.(!) (In turn, my knowledge of it stems from a mention of it in the book "Algebra" by T.T. Moh.) I can't find a good summary online, so:
Anthropological Part
The following kinship system (the "Murngin system") is found apparently often in primitive societies:
(1) Every member of the society is assigned a type, with marriage only permissible for couples of the same type
(2) A person's type is a determined solely by their gender and the type of their parents; in the opposite direction, the type of a person's parents can be determined by their gender and type.
(3) Blood relations only determine whether or not two people have the same type. (So if there is one instance of a father and son having the same type, all fathers and sons have the same type -- if this is opaque, it will be rephrased more transparently using group theoretic language below.)
(4) The type of brothers and sisters is always different.
(5) It is always possible for some descendants of any two people to get married.
An anthropologist might observe that first cousins whose mothers are sisters never marry, or that when there less than 4 types of people in a tribe a marriage between first cousins whose parents are oppositely gendered siblings is permissible, and wonder whether these are separate empirical observations, or already implied by observations made above. Some elementary group theory gives the answer.
Group Theoretic Part
(1)&(2) rephrased: If we let the types of people be $1, 2, ..., n$, then by (2) we can define the type of the son of a couple of type $i$ to be $s(i)$, and the type of a daughter to be $d(i)$. The second part of (2) is just the statement that $s$ and $d$ are group actions on the set of types.
(3) rephrased: If $H$ is the subgroup of $S_n$ generated by $s$ and $d$, we can rewrite (3) as the statement that for any element $g$ of $H$, if $g(i) = i$ for some $i$, then $g(i) = i$ for all $i$ (in other words $g=e$).
(4) rephrased: $s(i) \neq d(i)$, or equivalently $d^{-1}s(i) \neq i$, for all $i$.
(5) rephrased: The orbit of any number $i$ under $H$ is the entire set of types ${1,2,...,n}$.
We can see right away then that if cousins share mothers who are sisters, they cannot marry, as if the type of the male is $i$, the type of the female is $ddd^{-1}s^{-1}(i) = ds^{-1}(i) \neq i$. This of course is not hard to see without group theory, but we can do more.
Plainly if the tribe is going to perpetuate itself there must be more than 1 type. If there are 2 types, then $s$ must be $(12)$ (it cannot be $e$), and likewise for $d$, a contradiction.
In the case of 3 types, then it is a matter of running through possibilities to show $s = d^{-1} = (123)$ up to permutation of type.
In the case of 4 types, a similar but more lengthy computation yields that up to permutation of type, there are the following four possibilities:
1. $s = (1234)$, $d=e$
2. $s = (1234)$, $d = (13)(24)$
3. $s = (1234)$, $d = (1432)$
4. $s = (12)(34)$, $d = (13)(24)$.
In both the case of 3 different types and 4 different types, it is now easy to verify that the group generated by $s$ and $d$ is commutative, proving that the statement about permissible marriages made earlier will always hold for such a type system.
...Hopefully at least a little interesting. I suppose some of the more geometric examples above might make for a more visceral application of group theory, at least on first exposure.
-
There is also a short informal discussion of this problem in "The artist and the mathematician" by Amir D. Aczel. – Gunnar Magnusson May 24 2010 at 7:30
I'm not seeing why we can't have just two types with s=(1 2) and d=e. – Eric Hall Jan 28 2011 at 22:10
Sorry to take so long to respond. My memory isn't fresh, so I can't remember what I was thinking -- if anything -- in saying "s must be (12) (it cannot be e)," but it seems to me you are right. It is still true in this case that the group generated by s and d is commutative, for all that it's worth. – Brad Rodgers Mar 26 2011 at 21:33
There's a slightly eccentric (but entertaining) book by Budden called "The Fascination of Groups" that has an extensive chapter about the application of groups to the British practice of "change ringing" (this actually is featured prominently in the Dorothy Sayers, Lord Peter Wimsey mystery, "The Nine Tailors"). Change Ringing is the practice of ringing all the permutations of some number of church bells (usually 5 or 6, but can be as many as 9) with the requirement that successive permutations differ by a two-cycle (i.e. just swapping the order of two particular bells). Why this practice arose, I don't really know, but it sets up interesting algorithmic and mathematical problems.
-
3
Quite interesting. Wikipedia tells me that in 1963 the full set of permutations on 8 bells was played, for 18 hours (!) – Mariano Suárez-Alvarez Feb 4 2010 at 17:49
Another book discussing this practice is "Adventures in group theory" by D. Joyner. The book explores group theory through permutation games (Rubik's cube, the 15 puzzle, ...), which I suppose also qualify as "cool" uses of the theory. – Andres Caicedo Jan 6 2012 at 7:30
This is related to many of the answers already here, but a little different. When I was an undergrad, I got fixated on the following problem:
Given a deck of $n$ cards, how many perfect shuffles does it take to get back to the starting position. Does it matter if this preserves the top and bottom card (what I called an "out shuffle") or mixed them in ("an in shuffle")? This leads to the discrete logarithm problem and other stuff with cyclic groups.
-
2
It might be construed as fairly shocking that it takes just 8 (I think) perfect riffle shuffles on a 52-card deck to return it to the original. Persi Diaconis can execute this, I believe. :) – paul garrett Nov 1 at 22:52
Theorem 5.10 of http://www-math.mit.edu/~rstan/algcomb.pdf is proved by elementary group theory and linear algebra. It states the following:
(a) Fix $m\geq 1$. Let $p_i$ be the number of
nonisomorphic simple graphs with $m$ vertices and $i$ edges. Then the
sequence $p_0, p_1,\dots,p_{{m\choose 2}}$ is symmetric and unimodal. (This means that $p_i=p_{{m\choose 2}-i}$ and $p_0\leq p_1 \leq \cdots \leq p_{\lfloor \frac 12{m\choose 2}\rfloor}$. The symmetry propery is trivial, but unimodality is another story.)
(b) Let $T$ be a collection of nonisomorphic simple graphs with $m$ vertices such that no element of $T$ is isomorphic to a spanning subgraph of another element of $T$. Then $|T|$ is maximized by taking $T$ to consist of all nonisomorphic simple graphs with $\lfloor \frac{1}{2}{m\choose 2}\rfloor$ edges.
Many similar results can be proved using the techniques of the above link.
-
The set of rational tangles has a nice identification with $PSL(2,Z)$. Here's a formal treatment: Rational Tangles and the Modular Group. And here's an easy version aimed at teachers: http://www.geometer.org/mathcircles/tangle.pdf.
Rather than repeat what's written in this papers I'll just say this: We can use the isomorphism, and a bit of continued fractions, to give us a procedure for undoing tangles. I think it gives a wonderful example of how the same abstract group structure appears in two entirely different looking places: topology and (elementary) arithmetic.
-
We did this for the kids at the UW math circle and it was very successful! – Dylan Wilson Jan 28 2011 at 21:35
@Dylan Cool. I've often wondered how well it would work out with younger participants. – Dan Piponi Apr 26 2011 at 15:36
One can use finite group theory to understand the Rubik's cube. This might not be as elementary as you're looking for, but it's usually pretty entertaining.
-
1
I can bring the cube to the class (I believe I still have it somewhere in my junk box). The question is then what exactly I should do with it, given that I do not want to spend the whole 75 minute class just talking about Rubik's cube but still want to demonstrate something non-trivial and funny. Any suggestions for, say, a 25 minute presentation? – fedja Jan 29 2010 at 4:02
4
Well, the standard fact is that if you break the cube up into pieces and reassemble, the probability that you can get back to the proper configuration is 1/12. There are two factors of 1/2 from standard even/odd permutation considerations, while the last 1/3 comes from a Z/3Z invariant. – aorq Jan 29 2010 at 8:03
3
I quite like the fact that one can apply a sequence of arbitrary moves (aka an arbitrary group element) a finite number of times to get back to where you started from (because, of course, every element in your group has finite order). This is quite easy to demonstrate (given finite time...) It is also trivial (but interesting!) to note that there is no single sequence of moves (aka group element) which will take any position and return it to the `solved' cube. This is basically asking if your group is cyclic, and it obviously isn't. – AL Jan 31 2011 at 16:33
This may be a bit frivolous, but I was told the following puzzle by Peter Cameron:
A prison has $2n$ prisoners. One day the warden decides to set them a challenge. The prisoners' names are put in boxes, one in each box. Each prisoner is then independently allowed to look in up to $n$ of the boxes - he looks inside one box, then another, and so on, with a free choice of box at each step. After this the warden will ask each of the prisoners to say which box contains their name. The prisoners succeed if all of them give the correct answer, and all fail if any of them gives an incorrect answer. Before the challenge starts, the prisoners can agree a strategy, but once they start looking in boxes, no information passes between them (so they don't know for instance which boxes have been looked in). Random strategies are allowed.
There is a strategy for the prisoners that gives them a probability of success bounded away from $0$ for $n$ arbitrarily large. (I won't give the answer away, but suffice to say it involves permutations.)
-
The calculation of the probability of getting a graph isomorphic to a specific graph $\Gamma$ when picking a uniformly random graph of with the same number of vertices as $\Gamma$.
-
There's a group-action proof of the famous Morley's theorem in Euclidean Geometry: http://www.ems-ph.org/journals/newsletter/pdf/2004-12-54.pdf. It's near the end of Alain Connes's article. This proof very convincingly shows that geometry is about symmetry which is what groups are about.
-
Another nice arithmetic application of (cyclic) group theory is the fact that the multiplicative group of a finite field is cyclic, or (in down-to-earth terms) that one can obtain every non-zero residue class modulo a prime just taking consecutive powers of a single well-chosen one.
I always give this example in my undergraduate Algebra course because it also gives me the occasion to show that in maths there are still questions which are unanswered but can be easily understood by a beginner (obviously, I'm referring to Artin's conjecture in this context).
-
I love the proof of Wilson's theorem through consideration of the multiplicative group of integers modulo-p. The solution is very slick and it wasn't until I saw this early in an undergraduate course that I truly understood the power of group theory and its wide range of applicability. It also has a nice bit of history to it: After neither Wilson nor his mentor Warring was able to give a proof of the assertion, Gauss remarked in Disquisitiones Arithmeticae: "And Waring confessed that the proof seemed the more difficult, since one cannot imagine any notation to express a prime number. – In our opinion, however, such truths should be extracted from notions rather than from notations".
-
For an extension of this proof from $\mathbb{Z}$ to rings of integers in number fields, see arxiv.org/abs/0711.3879 – Chandan Singh Dalawat Jan 31 2010 at 4:47
Peg solitaire can be analyzed by putting elements of the Klein group on the diagonals of the board so that whenever you make a legal move the product of the elements with pegs is invariant. You can prove that European solitaire is unsolvable (when the central peg is missing) or that if solve the English version and the last peg isn't in the center, your last move was very, very stupid.
-
Using permutations group to study solving and/or showing the impossibility of solving given initial configurations of the 15-puzzle (roughly, one can only solve "even" permutations of the puzzle). Cute, non-abelian, and non-obvious -- and most students will have seen the puzzle.
Edit: Ah, sorry for the repeat. Darsh beat me to it mid-writeup.
-
I always liked knot coloring and how it related to knot groups and classifications of knots. The property of if a knot is n-colorable or not is an property of the knot (so it doesn't change under any movement that doesn't cross the strands of the knot). The colors of a presentation of a knot's strands can be assigned labels which are related to each other according to whoever moves over/under a strand, which can lead to a generalized way to label the strands of a knot in a way that assigns (almost uniquely) a group to each knot (usually defined in terms of the fundamental group of $\mathbb{R}^3-K$, but you can get at it through a particular presentation, too.)
An example with pictures, http://en.wikipedia.org/wiki/Fox_n-coloring
There are also associated group-like structures, http://unapologetic.wordpress.com/2007/05/02/more-knot-coloring/
-
Knot colorings are essentially about certain representations of the knot complement's fundamental group into dihedral groups. – Jim Conant Oct 6 2010 at 10:18
Suppose one has a rectangle and a rhombus neither of which are squares. Which is more symmetric?
I think that issues about symmetry naturally lead to ideas that show how powerful thinking about the theory of groups is.
In addition to the above question there is the issue of how many different kinds of patterns one can have on a strip (frieze patterns). Though visually there appears to be great complexity there is a natural sense in which there are only 7 types of patterns. One can generalize to patterns in the plane (17 wallpaper groups) and to patterns where colors are allowed.
Details and lovely pictures can be found in:
Grunbaum and Shephard, Tilings and Patterns Conway, Burgiel, and Goodman-Strauss, The Symmetries of Things
-
Consider the Westminister Chimes. If you consider the 4 notes as your elements and the operation of switch (where you swap any two notes), the westminister chimes have all the characteristics of a group. One note of caution....there appear to be a number of 'modified' westminister chimes that alter one note here or there....you have to find the original one). This is particularly effective with math students who are from the fine arts area. They see that math is really infused in everything that is done.
-
Quintic equations?
-
Let $A$, $B$, $C$ be subsets of a finite set $X$. Let $A \Delta B$ denote the symmetric difference of $A$ and $B$, i.e. the set of elements of $X$ lying in exactly one of $A$ and $B$. Then $A \Delta B$ = $C \Delta D$ if and only if $A \Delta C = B \Delta D$.
This has a one-line proof: take the symmetric difference of both sides of either equation with $B \Delta C$. Underlying this is the fact that the set of all subsets of $X$ is an elementary abelian group of order $2^{|X|}$ under symmetric difference, with the empty set as the identity.
-
This problem might be related to some of the ones suggested above, I did not check carefully.
Assume that you have to prepare the calendar of a basketball league to which $2n$ teams take part, where $n$ is an integer $\geq 1$. The full season consists of $2n-1$ dates where in each date each of the team has to play against another one (different than itself :-)). This has to be done in such a way that, overall, every team meets every other team exactly once (we all know what the calendar of a basketball league look like).
A solution is given in the comment below. The solution I had first suggested was wrong.
-
1
Your group $G$ has $2^n$ elements and not $2n$, so this doesn't work. Usually this is done as follows: Take an abelian group $A$ of order $2n-1$ and identify $A$ with the matchdays. Choose a "jokerteam" and identify the other teams with $A$. Then on day $a\in A$, team $x$ is to play against team $a-x$, and team $2^{-1}a$ is to play against the jokerteam. At least, this system is used in the german Fußballbundesliga. – F. Ladisch May 10 2011 at 22:59
Oh thanks! I did remember a group of ODD order was used in the solution. And yes, what I wrote above is non-sense... – Tommaso Centeleghe May 11 2011 at 15:21
How about Galois theorem that a prime degree equation is solvable if and only if all the roots are rational functions in two of the roots. I have read that Galois stated it this way, so that the statement does not involve the notion of a solvable group, but the proof uses the result that a transitive subgroup of $S_p$ ($p$ prime), is contained in the normalizer of the group generated by a $p$-cycle.
- | {"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": 128, "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": 0.9335719347000122, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/tagged/homework?sort=votes&pagesize=50 | # Tagged Questions
Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright.
learn more… | top users | synonyms
1answer
566 views
### Mechanics + Thermodynamics: Bouncing Ball
In preparation for an exam, I'm revisiting old exam questions. This one seems neat, but also quite complicated: A soccer ball with Radius $R=11cm$ is inflated at a pressure of \$P =9 \times 10^4 ...
4answers
286 views
### Trace and adjoint representation of $SU(N)$
In the adjoint representation of $SU(N)$, the generators $t^a_G$ are chosen as $$(t^a_G)_{bc}=-if^{abc}$$ The following identity can be found in Taizo Muta's book "Foundations of Quantum ...
9answers
5k views
### What's the core difference between the electric and magnetic forces?
I require only a simple answer. One sentence is enough... (It's for high school physics)
1answer
249 views
### One strategy in a snowball fight
Here's a common college physics problem: One strategy in a snowball fight is to throw a first snowball at a high angle over level ground. While your opponent is watching the first one, you ...
2answers
836 views
### Transformation of angles in special relativity
My question is about the problem below Depicted are two space ships (the USS Voyager and the USS Enterprise), each with velocity $v=c/2$ relative to the space station (Babylon 5). At the exact ...
2answers
271 views
### Equivalence principle and radiation from falling particle
I am currently having a hard time solving a problem of GR from Lasenby's book. I can't make it more clear than by quoting the exercise: 7.2 A charged object held stationary in a laboratory on the ...
4answers
119 views
### Integrating radial free fall in Newtonian gravity
I thought this would be a simple question, but I'm having trouble figuring it out. Not a homework assignment btw. I am a physics student and am just genuinely interested in physics problems involving ...
2answers
140 views
### Radioactive Decay
Problem:Nuclei of a radioactive element $\Bbb X$ having decay constant $\lambda$ , ( decays into another stable nuclei $\Bbb Y$ ) is being produced by some external process at a constant rate ...
2answers
144 views
### Why does the quantum Heisenberg model become the classical one when $S\to\infty$?
The Hamiltonian of the spin $S$ quantum Heisenberg model is $$H = J\sum_{<i,j>}\mathbf{S}_{i}\cdot\mathbf{S}_{j}$$ I have read that when the spin quantum number $S\to\infty$, quantum fluctuation ...
3answers
1k views
### Charge density in concentric spheres
Question: If there are two conducting spherical shells and the inner shell is grounded, what will be >the charge density in the inner shell if there is a charge Q placed on the outer shell? ...
2answers
749 views
### How do Leptons arise from Lambda decay?
I have a question for an assignment: Use your understanding of the quark model of hadrons and the boson model of the weak nuclear interaction to explain how leptons can arise from lambda decay, ...
2answers
234 views
### Conservation of energy in a non-linear oscillator
I have a homework question about a "non-linear oscillator". I actually have an answer to this question, but the answer I get is stronger than what is needed according to the question. The question ...
0answers
217 views
### exponential potential $\exp(|x|)$
For $a$ being positive what are the quantization conditions for an exponential potential? $$- \frac{d^{2}}{dx^{2}}y(x)+ ae^{|x|}y(x)=E_{n}y(x)$$ with boundary conditions $$y(0)=0=y(\infty)$$ I ...
0answers
221 views
### Capacitance of this unusual capacitor
This capacitor is composed of two half spherical shelled conductors both with radius $r$. There is a very small space between the two parts seeing to that no charge will exchange between them. ...
3answers
312 views
### Why isn't average speed equal to displacement over time?
I'm in an introductory Physics course and I need help! During a one-hour trip, a small boat travels 80.0km north and then travels 60.0km east. What is the boat's average speed during the one-hour ...
2answers
5k views
### What is the relationship between kinetic energy and momentum?
I can't seem to figure out the relationship between $E_k$ and $p$ or $F$. I understand that the units are pretty different. But for example: A bullet with a mass of 10.0g is moving at the speed of ...
3answers
97 views
### Horizontal $E$-field for a charged conducting disk
For part of a simulation I am writing, I need to know the electric field emitted from a charged conducting disk. If the disk was laid out in the $x$-$y$ plane, I am interested in the field in that ...
4answers
3k views
### Derivation of self-inductance of a long wire
Currently I am stuck, trying to derive the self-inductance of a long wire. According to literature it should be $$L=\frac{\mu_r\mu_0l}{8\pi}$$ and in literature its derived by looking at the energy ...
3answers
207 views
### experimental technique for measuring temperature of an ant
I am taking a course on thermodynamics. I have a question from my text(halliday & resnick,physics-1). They asked me to measure temperature of an ant or an insect or a small body,like a small ...
3answers
220 views
### Force between two charged rods?
Suppose that we have two rods of length $l_1, l_2$ connected at one end but free to rotate. These rods have charge density $\lambda$ uniformly distributed, so the total charge of rod $i$ is \$\lambda ...
1answer
5k views
### How do you find the uncertainty of an weighted average?
The following is taken from a practice GRE question: Two experimental techniques determine the mass of an object to be $11\pm 1\, \mathrm{kg}$ and $10\pm 2\, \mathrm{kg}$. These two ...
2answers
284 views
### relativistic acceleration equation
A Starship is going to accelerate from 0 to some final four-velocity, but it cannot accelerate faster than $g_M$, otherwise it will crush the astronauts. what is the appropiate equation to constraint ...
2answers
175 views
### Can the process $u\overline{u} \rightarrow s\overline{s}$ be mediated by the EM interaction?
I’m working on a homework problem which asks for the dominant contribution (e.g. EM, strong, or weak) to the process $p + \overline{p} \rightarrow \Lambda + \overline{\Lambda}$). I know that the EM ...
3answers
3k views
### Weight on carpet vs. hard floor
Will the scale say I weigh more, less, or the same on a carpet as compared to a hard floor? You can assume the scale works via spring mechanism. Free body diagrams encouraged!
3answers
472 views
### Relationship between height and velocity in conservation of mechnical energy
I'm a high school physics student, and we recently did a lab on the conservation of energy where we measured the speed of a marble at varying heights on a rollercoaster track. We were supposed to ...
2answers
272 views
### Gauge invariant Chern-Simons Lagrangian
I have to prove the (non abelian) gauge invariance of the following lagrangian (for a certain value of $\lambda$): -\frac14 F^{\mu\nu}_aF_{\mu\nu}^a + ...
2answers
152 views
### Boy and box on ice
I have a troubles. Can you help me? I have a task, called "slippery activity". Boy (he has $m=45kg$) Stands at ice and tries to move a big box ($M=90kg$) with a string (rope). Boy's Slip ratio is ...
2answers
464 views
### High school double lens optics question
This is a first year high school homework question (in the Finnish high school), and I'm having serious trouble solving it. I apologize for possible non-standard terms: I'm doing the translation from ...
1answer
180 views
### Thermo homework (HW) problem
I do not believe this question can be solved with the thermodynamics knowledge that I have learned thus far, but someone correct me if I am wrong: A 0.2-m³ tank containing helium at 15 bar and ...
1answer
204 views
### What is the change in flux through a loop that has been rotated?
We have a number of field lines perpendicular to one loop of wire with an area $A = 10\textrm{ cm}^2$. The magnetic field is$B= 7.2\times10^{-5}\textrm{ T}$. You turn the loop and the flux decreases ...
1answer
184 views
### few fermions in a harmonic trap — position density matrix from diagrammatics
I'm trying to calculate the momentum distribution of a 1D system of non-interacting identical fermions in a harmonic trap. Given Feynman's answer (from his Statistical Mechanics book) for the ...
0answers
118 views
### Find $x$ operator given $p$
This is problem $1.2$ of Molecular Quantum Mechanics by Atkins, 4th edition. I'm given the momentum operator $$p=\sqrt{\frac{\hbar}{2m}}(A+B)$$ with $$[A,B]=1$$ and I need to find $x$ in this ...
1answer
262 views
### Lagrangian density for a Piano String
So I'm trying to do this problem where I'm given the Lagrangian density for a piano string which can vibrate both transversely and longitudinally. $\eta(x,t)$ is the transverse displacement and ...
3answers
803 views
### Charge Distribution on a Parallel Plate Capacitor
If a parallel plate capacitor is formed by placing two infinite grounded conducting sheets, one at potential $V_1$ and another at $V_2$, a distance $d$ away from each other, then the charge on either ...
1answer
412 views
### Magnetic field from a half-cylinder [closed]
I am preparing for an exam, on this problem I had the opposite direction of the magnetic field. A conductor-cylinder with radius R has been cut in half ($\phi \in [0,\pi]$) A DC current $I$ runs ...
2answers
1k views
### Help me get out of a speeding ticket
I was driving uphill from a complete stop for a distance of .4 miles estimated to take 1 minute in a navigation app. I was pulled over right after cresting the hill. The cop had me on radar going ...
5answers
355 views
### The approximate uncertainty in $r$
The volume of a cylinder is given by the expression $V=\pi r^2 h$ The uncertainties for $V$ and $h$ are as shown below $\begin{align} V&\pm 7\%\\ h&\pm 3\% \end{align}$ What is the ...
3answers
2k views
### Why can't a single photon produce an electron-positron pair?
In reading through old course material, I found the assignment (my translation): Show that a single photon cannot produce an electron-positron pair, but needs additional matter or light quanta. ...
5answers
3k views
### What would be the effective resistance of the ladder of resistors having n steps
I'm a tutor. This is a high school level problem. In high school, every one have might have solved a problem of effective resistance of a ladder of resistors having infinite steps. Now the problem is ...
2answers
820 views
### Make a semi transparent mirror with copper
The question: How would you make a semi transparent mirror (50% reflection, 50% transmission) with glass with a layer of copper. For light $\lambda$ = 500nm Try to be as realistic as possible What ...
2answers
92 views
### What's the physical difference between the quantities $\langle v_{i}v_{j}\rangle$ and $\langle v_{i}\rangle\langle v_{j}\rangle$?
What's the physical difference between the quantities $\left\langle v_{i}v_{j}\right\rangle$ and $\left\langle v_{i}\right\rangle \left\langle v_{j}\right\rangle$? Where \$\left\langle ...
2answers
202 views
### Is the proper interpretation of temperature missing in this book?
In Randall T. Knight’s textbook “Physics for Scientists and Engineers” in the first chapter on thermodynamics (Ch. 16: A Macroscopic Description of Matter) one of the first conceptual questions is ...
3answers
238 views
### How does capacitance work?
I have a circuit whit a AC source a capacitor and a resistance all in series. I find that the difference of potential between the capacitor leads begin to change after some instants as it should. But ...
4answers
132 views
### How to prove that the symmetrisation Operator is hermitian?
Let $\mathcal{H}_N$ be the $N$ particle Hilbert space. So a quantum state $\left| \Psi \right>$ may be representated by \left| \Psi \right> = \left| k_1 \right>^{(1)}\left| k_2 ...
2answers
388 views
### Uniform chain falls off table Diff EQ
I really need some assistance setting up this problem. any assistance would be a Godsend: a uniform heavy chain of length a initially has length ...
4answers
2k views
### Infinitely charged wire and Differential form of Gauss' Law
I have tried calculating the potential of a charged wire the direct way. If lambda is the charge density of the wire, then I get \phi(r) = \frac{\lambda}{4 \pi \epsilon_0 r} \int_{-\infty}^\infty ...
2answers
88 views
### Proof for commutator relation $[\hat{H},\hat{a}] = - \hbar \omega \hat{a}$
I know how to derive below equations found on wikipedia and have done it myselt too: \begin{align} \hat{H} &= \hbar \omega \left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)\\ \hat{H} &= ...
2answers
163 views
### A wonky gravitational potential and its critical points
I have tough problem I am not sure how to solve: For this question, we are confined to a plane. Consider a gravitational field that is proportional to $\frac{1}{r^3}$ instead of $\frac{1}{r^2}$, and ...
1answer
508 views
### Hooke's Law question
I have the following question to answer: a force of160 N stretches a spring 0.050m from its equilibrium position. A. what is the spring constant of the spring? The equation for Hooke's Law in my ...
1answer
354 views
### Plotting a SHO in matlab
I have no prior experience of using matlab. My teacher want me to solve this question. I have been trying for a couple of hours now with no luck, please help! The mass of 100 g hanging in a spring ... | {"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": 62, "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": 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": 0.9285756349563599, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/47323-question-invloving-fourier-series.html | # Thread:
1. ## Question invloving Fourier Series
Hi All,
I have a question that I need to do a bit of catching-up work before I can answer it. I'd like to be able to have a go at it myself before anybody puts an answer up but then check an answer for it later in the week. So if anybody can provide me with an answer to the following that would be great. Thanks.
Q. Evaluate
$\int x^N cos(\alpha x) dx$ and $\int x^N sin(\alpha x) dx$ using integration by parts, for N=1,2.
Sketch the function defined by
$f(x) = 2x - x^2$, $0 \leq x < 1$;
$f(x + 1) = x$, $-\infty < x < \infty$
and find the corresponding Fourier Series.
By considering $f(x)$ at $x = 0$ and $x = \frac{1}{2}$ deduce that
$\sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$
and
$\sum_{n = 1}^{\infty} \frac{(-1)^{(n+1)}}{n^2} = \frac{\pi^2}{12}$
2. For the first part I got
For $N = 1 \Longrightarrow$
$\int x^1 \ cos(\alpha x) \ dx = \frac{1}{\alpha} \ x \ sin(\alpha x) + \frac{1}{\alpha^2} \ cos(\alpha x)+ C$
$\int x^1 \ sin(\alpha x) \ dx = -\frac{1}{\alpha} \ x \ cos(\alpha x) + \frac{1}{\alpha^2} \ sin(\alpha x)+ C$
For $N = 2 \Longrightarrow$
$\int x^2 \ cos(\alpha x) \ dx = \frac{1}{\alpha} \ x^2 \ sin(\alpha x) + \frac{1}{\alpha^2} \ 2x \ cos(\alpha x) - \frac{1}{\alpha^3} \ 2 \ sin(\alpha x)+ C$
$\int x^2 \ sin(\alpha x) \ dx = -\frac{1}{\alpha} \ x^2 \ cos(\alpha x) + \frac{1}{\alpha^2} \ 2x \ sin(\alpha x)+ \frac{1}{\alpha^3} \ 2 \ cos(\alpha x)+ C$
3. Originally Posted by woody198403
Sketch the function defined by
$f(x) = 2x - x^2$, $0 \leq x < 1$;
$f(x + 1) = x$, $-\infty < x < \infty$
I think you mean $f(x+1)=f(x)$.
Now expand $f$ into a Fourier series.
Evaluate the series at $x=0$ and compare sides.
4. Yes, I did mean $f(x+1)=f(x)$ oops
Im still working on it but is
$a_0 = \frac{2}{3}$
and in determining $a_n \ \mbox{and} \ b_n$, would the following give me the correct values?
$a_n = 4 \int_{0}^{1} x \ cos(2 n\pi x) dx - \int_{0}^{1} x^2 \ cos(2 n\pi x) dx$
$b_n = 4 \int_{0}^{1} x \ sin(2 n\pi x) dx - \int_{0}^{1} x^2 \ sin(2 n\pi x) dx$
5. Can someone PLEASE tell me if my calculation is correct? and if not, can someone please show me where I went wrong? Thanks
$a_n =- \frac{2 \ cos(n\pi)^2}{n^2\pi^2} + \frac{2 \ sin(n\pi)}{n^3\pi^3}$
and
$b_n = \frac{2 \ cos(n\pi)^2 - 2}{n^3\pi^3} + \frac{2 sin(n\pi) \ cos(n\pi)}{n^2\pi^2}$
6. Originally Posted by woody198403
Can someone PLEASE tell me if my calculation is correct? and if not, can someone please show me where I went wrong? Thanks
$a_n =- \frac{2 \cos^2(n\pi)}{n^2\pi^2} + \frac{2 \sin(n\pi)}{n^3\pi^3}$
and
$b_n = \frac{2 \cos^2(n\pi) - 2}{n^3\pi^3} + \frac{2 \sin(n\pi) \cos(n\pi)}{n^2\pi^2}$
I don't have the patience to go through the whole calculation, but you should know that $\sin(n\pi)=0$ and $\cos(n\pi)=(-1)^n$. That simplifies the first formula to $a_n =- \frac{2}{n^2\pi^2}$. To check whether this is correct, see what happens at x=0. The periodic function has a jump from 1 to 0 at this point, and the theory tells you that the Fourier series ought to converge to a point halfway up the jump, namely 1/2. The Fourier series is ${\textstyle\frac12}a_0 + \sum_{n=1}^\infty (a_n\cos(2\pi nx) + b_n\sin(2\pi nx))$. When you put x=0 that gives $\frac12 = \frac13 - \sum_{n=1}^\infty \frac2{n^2\pi^2}$. This is in the right ballpark for showing that $\sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$, but it isn't quite right.
My guess is that you should look more carefully at those terms $\cos^2(n\pi)$ and $\sin(n\pi) \cos(n\pi)$ in your formulas for a_n and b_n. There is nothing in the integration-by-parts formulas that you gave earlier to suggest that there should be any products of trig functions in these formulas.
7. Letting $f(x)=2x-x^2,\;0\leq x \leq 1$ with $f(1+x)=f(x)$, I get:
$a_0=2\int_0^1 f(x)dx=4/3$
$a_n=2\int_0^1 f(x)\cos(2n\pi x)dx=\frac{-2n\pi+(1+2n^2\pi^2)\sin(2n\pi)}{2n^3\pi^3};\quad n\geq 1$
$b_n=2\int_0^1 f(x)\sin(2n\pi x)dx=\frac{1-(1+2n^2\pi^2)\cos(2n\pi)}{2n^3\pi^3}$
The plot below is the first 20 terms in the interval $(0,3)$. Red is the function $f(x)$, Gray is the Fourier series for this function. I'm not real familiar with Fourier Series so not sure about this. Also, as Opalg stated above, at the points of discontinuities, the Fourier series converges to the average of the end points. The plot however, only shows a sharp vertical line due to plotting limitations. If I drew only a set of points for the plot, a single point would be between the upper and lower limits at the points of discontinuities.
Attached Thumbnails
8. Originally Posted by shawsend
The plot below is the first 20 terms in the interval $(0,3)$. Red is the function $f(x)$, Gray is the Fourier series for this function.
That plot shows a nice illustration of the Gibbs phenomenon, with the Fourier series overshooting on either side of the discontinuity.
9. but you should know that $\sin(n\pi)=0$ and $\cos(n\pi)=(-1)^n$. That simplifies the first formula to $a_n =- \frac{2}{n^2\pi^2}$.
Ahhhhh, of course! I am so embarrassed that I missed that
Thank-you both for the replies, but shawsend maybe its because Im tired but isnt $a_0 = \frac{2}{3} \$ ? I am almost certain of it but since you've posted it here I am paranoid I've left something out.
$a_0 = \frac{1}{2L} \ \int_{-L}^{L} f(x) \ dx$
10. Originally Posted by woody198403
Ahhhhh, of course! I am so embarrassed that I missed that
Thank-you both for the replies, but shawsend maybe its because Im tired but isnt $a_0 = \frac{2}{3} \$ ? I am almost certain of it but since you've posted it here I am paranoid I've left something out.
$a_0 = \frac{1}{2L} \ \int_{-L}^{L} f(x) \ dx$
I'm not sure. As I said, I'm not real good with these and also I was unable to obtain the two summation formulas you posed using my results.
11. Originally Posted by woody198403
isnt $a_0 = \frac{2}{3} \$ ? I am almost certain of it but since you've posted it here I am paranoid I've left something out.
$a_0 = \frac{1}{2L} \ \int_{-L}^{L} f(x) \ dx$ Really? I would have thought it was $\color{red} a_0 = \frac{1}{L} \ \int_{-L}^{L} f(x) \ dx$
The constant term in the Fourier series should always be the average value of the function over the interval, in this case $\int_0^1(2x-x^2)dx = 2/3$. The constant term is also (1/2)a_0, which suggests that a_0 should be 4/3 here.
12. Thank you all for your help but I have one last question.
So finally,
$a_0 = \frac{2}{3} \ , \ a_n = -\frac{1}{\pi^2n^2} \ ,\ b_n = -\frac{1}{2\pi n}$
so my Fourier Series is
$f(x) = \frac{2}{3} - \frac{1}{\pi^2} * \sum_{n=1}^{\infty} \ \frac{1}{n^2} cos(2\pi n x) - \frac{1}{\pi} * \sum_{n=1}^{\infty} \ \frac{1}{n^2} sin(2\pi n x)$
which I know is correct because I was able to prove that at x=0,
$\sum_{n=1}^{\infty} \ \frac{1}{n^2} = \frac{\pi^2}{6}$
(since the series converges at a $\frac{1}{2}$)
but can someone show me how to prove that at x= 1/2,
$\sum_{n=1}^{\infty} \ \frac{(-1)^{n + 1}}{n^2} = \frac{pi^2}{12}$
I dont know what it is that Im doing wrong but I keep getting the wrong answer
13. Originally Posted by woody198403
so my Fourier Series is
$f(x) = \frac{2}{3} - \frac{1}{\pi^2} * \sum_{n=1}^{\infty} \ \frac{1}{n^2} cos(2\pi n x) - \frac{1}{\pi} * \sum_{n=1}^{\infty} \ \frac{1}{n^2} sin(2\pi n x)$
Have I mis-understood what the function is? When I plot the first 50 terms of your expression above I get the plot below which I've superimposed over what I thought the original function is. That does not seem to be converging to it in the way one normally thinks of a Fourier series converging (with Gibbs phenomenon) to a function. Here's my Mathematica code in case some are interested:
Code:
```f[x_] := 2*x - x^2;
h[x_] := 2/3 - (1/Pi^2)*
Sum[Cos[2*Pi*n*x]/n^2, {n, 1, 150}] -
(1/Pi)*Sum[Sin[2*Pi*n*x]/n^2, {n, 1, 150}]
fplot1 = Plot[f[x], {x, 0, 1},
PlotStyle -> {Thickness[0.001], Red}]
fplot2 = Plot[f[x - 1], {x, 1, 2},
PlotStyle -> {Thickness[0.001], Red}]
fplot3 = Plot[f[x - 2], {x, 2, 3},
PlotStyle -> {Thickness[0.001], Red}]
p2 = Plot[h[x], {x, 0, 3}]
Show[{fplot1, fplot2, fplot3, p2},
PlotRange -> {{0, 3}, {0, 1}}]```
Attached Thumbnails
14. Woody, I think you made a typo up there. Should it not be:
$f(x) = \frac{2}{3} - \frac{1}{\pi^2} * \sum_{n=1}^{\infty} <br /> \frac{1}{n^2} cos(2\pi n x) - \frac{1}{\pi} * \sum_{n=1}^{\infty} <br /> \frac{1}{n} sin(2\pi n x)<br />$
and the deal with the constant term is I'm using $a_0/2$ which is your $2/3$ term. I now believe this expression is correct. The code above with 150 terms is converging nicely to the periodic extension of the function.
15. Originally Posted by woody198403
So finally,
$a_0 = \frac{2}{3} \ , \ a_n = -\frac{1}{\pi^2n^2} \ ,\ b_n = -\frac{1}{2\pi n}$
so my Fourier Series is
$f(x) = \frac{2}{3} - \frac{1}{\pi^2} * \sum_{n=1}^{\infty} \ \frac{1}{n^2} cos(2\pi n x) - \frac{1}{\pi} * \sum_{n=1}^{\infty} \ \frac{1}{n^2} sin(2\pi n x)$ If your formula for b_n is correct, that should be $\color{red}\frac{1}{n} sin(2\pi n x)$.
which I know is correct because I was able to prove that at x=0,
$\sum_{n=1}^{\infty} \ \frac{1}{n^2} = \frac{\pi^2}{6}$
(since the series converges at a $\frac{1}{2}$)
but can someone show me how to prove that at x= 1/2,
$\sum_{n=1}^{\infty} \ \frac{(-1)^{n + 1}}{n^2} = \frac{pi^2}{12}$
I dont know what it is that Im doing wrong but I keep getting the wrong answer
(The value of b_n won't affect this answer because the sin terms will all be 0 at x=1/2.)
Put x=1/2 in the equation $2x-x^2 = \frac{2}{3} - \frac{1}{\pi^2} * \sum_{n=1}^{\infty} \ \frac{1}{n^2} \cos(2\pi n x) - \frac{1}{\pi} * \sum_{n=1}^{\infty} \ \frac{1}{n} \sin(2\pi n x)$:
$1-\frac14 = \frac{2}{3} - \frac{1}{\pi^2} * \sum_{n=1}^{\infty} \ \frac{1}{n^2}(-1)^n - 0$.
That simplifies to $\frac{1}{\pi^2} * \sum_{n=1}^{\infty} \ \frac{(-1)^n}{n^2} = \frac23-\frac34 = -\frac1{12}$.
Now all you have to do is to multiply both sides by $-\pi^2$. | {"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": 80, "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": 0.9555311799049377, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/28063/highly-connected-compact-complex-manifolds/28068 | ## Highly connected, compact complex manifolds
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Here are four remarks about the homology and homotopy type of a compact, complex manifold $M$:
1. If $M$ is Kähler, then it is symplectic and thus $H^2(M,\mathbb{R}) \ne 0$. (Also, as explained in a blog posting by David Speyer, you still have $H^2(M,\mathbb{R}) \ne 0$ even if $M$ is non-projective but algebraic.)
2. An interesting first example of a non-Kähler manifold is a Hopf manifold, by definition $(\mathbb{C}^n\setminus 0)/\Gamma_r$, where $\Gamma_r$ is a rescaling by $r$ with $|r| \ne 0,1$. This example has $H^1(M,\mathbb{R}) \ne 0$.
3. On the other hand, even-dimensional, compact Lie groups have left-invariant complex structures. If $M$ is such a manifold and is simply connected, then it is also 2-connected. $H^1(M,\mathbb{Z}) = H^2(M,\mathbb{Z}) = 0$ and $M$ is manifestly not Kähler. On the other hand, no such example is 3-connected and you always have $H^3(M,\mathbb{R}) \ne 0$.
4. There is (or was) a long-standing conjecture that no even-dimensional sphere other than $S^2$ has a complex structure.
So, question: Is there for each $n$, a compact, complex manifold $M$ which is $n$-connected?
-
## 2 Answers
E. Calabi, B. Eckmann, A class of compact, complex manifolds which are not algebraic. Ann. of Math. (2) 58, (1953). 494–500.
From Chern's MR review (MR0057539):
This paper defines on the topological product $S^{2p+1} \times S^{2q+1}$ of two spheres of dimensions $2p+1$ and $2q+1$ respectively, $p$ > 0, a complex analytic structure. The complex manifold so obtained ... admits a complex analytic fibering, with two-dimensional tori as fibers and having as base space the product $\mathbb{P}^p \times \mathbb{P}^q$ of complex projective spaces of (complex) dimensions $p$ and $q$ respectively.
-
A slam dunk for the question! I did do a few Google searches (including Google Books and Google Scholar) and I just didn't see it. – Greg Kuperberg Jun 13 2010 at 21:36
1
There's also a Wikipedia page, with a construction that's very similar to that of a Hopf manifold. en.wikipedia.org/wiki/Calabi%2dEckmann_manifold – Greg Kuperberg Jun 13 2010 at 21:53
My googling was directed by your Lie group examples, among which there are certain products of spheres! (BTW, when you described the Hopf manifolds, I expect you meant to delete zero.) – Tim Perutz Jun 13 2010 at 23:07
Yeah, thanks, I fixed the typo. – Greg Kuperberg Jun 14 2010 at 0:06
### 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.
As shown by Calabi and Eckmann, products of odd-dimensional spheres admit complex structures. See Anns of Maths 58, 1953, 494-500.
- | {"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": 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": 0.9253824949264526, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/75071/list | ## Return to Question
8 added 669 characters in body
The set of all smooth maps $S^1\to M^n$ ($M$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space).
I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space?
EDIT: it is clear that answer is no (see answer below). So, I rewrite question: is it true that set of singular loops is a collection of submanifolds different codimension(set of loops with one selfintersection has codimension $dim M-2$, loops with zero of derivative has codimension $dim M -3$ and so on. Set of loops with infinite number of singularities should have infinite codimension... Let's forget about them.)
So, the question is about local situation: for example, let's consider a loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one self-intersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)?
EDIT 2. I reformulate question. $map(S^1 \to \mathbb R^3)$ is a functional space, so we can apply a technique of singularity theory. Generic map $f$ with one self-intersection has one-parameter versal deformation $V:[0,1]\times S^1\to \mathbb R^3$ - and any deformation are induced from $V$. Does it imply that $D$(set of singular loops) near the $f$ is a submanifold of codimension 1 in sense of Fréchet or Frolicher?
ADDED. Andrew Stacey explains in his answer and in http://ncatlab.org/nlab/show/on+the+manifold+structure+of+singular+loops why a stratum of a loop space is not a submanifold (the reason is tha same as in a standart smooth injection of line to plane where image is not a submanifold).
But locally, as Ryan said, each stratum is a submanifold (in sense of Frechet). And the kast question is:
For a smooth submanifold $X⊂Y$ of codimension 2, for a general point $x∈X$ we always have a map from small neghbourhood $U$ of $x$ to $D^2$ ($x∈U⊂Y,f:U→D^2$) such that $U\cap X= f^{−1}(0)$. I belive that in situation of space of loops there is no such map... Is it true?
7 added 1 characters in body; added 3 characters in body
The set of all smooth maps $S^1\to M^n$ ($M$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space).
I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space?
EDIT: it is clear that answer is no (see answer below). So, I rewrite question: is it true that set of singular loops is a collection of submanifolds different codimension(set of loops with one selfintersection has codimension $dim M-2$, loops with zero of derivative has codimension $dim M -3$ and so on. Set of loops with infinite number of singularities should have infinite codimension... Let's forget about them.)
So, the question is about local situation: for example, let's consider a loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one self-intersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)?
EDIT 2. I reformulate question. $map(S^1)\to map(S^1 \to \mathbb R^3$ R^3)$is a functional space, so we can apply a techic technique of singularity theory. Generic map$f$with one self-intersection has one-parameter versal deformation$V:[0,1]\times S^1\to \mathbb R^3$- and any deformation are induced from$V$. Does it imply that$D$(set of singular loops) near the$f\$ is a submanifold of codimension 1 in sense of Fréchet or Frolicher?
6 added 419 characters in body; edited tags
The set of all smooth maps $S^1\to M^n$ ($M$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space).
I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space?
EDIT: it is clear that answer is no (see answer below). So, I rewrite question: is it true that set of singular loops is a collection of submanifolds different codimension(set of loops with one selfintersection has codimension $dim M-2$, loops with zero of derivative has codimension $dim M -3$ and so on. Set of loops with infinite number of singularities should have infinite codimension... Let's forget about them.)
So, the question is about local situation: for example, let's consider a loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one self-intersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)?
EDIT 2. I reformulate question. $map(S^1)\to \mathbb R^3$ is a functional space, so we can apply a techic of singularity theory. Generic map $f$ with one self-intersection has one-parameter versal deformation $V:[0,1]\times S^1\to \mathbb R^3$ - and any deformation are induced from $V$. Does it imply that $D$(set of singular loops) near the $f$ is a submanifold of codimension 1 in sense of Fréchet or Frolicher?
5 added 2 characters in body
The set of all smooth maps $S^1\to M^n$ ($M$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space).
I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space?
EDIT: it is clear that answer is no (see answer below). So, I rewrite question: is it true that set of singular loops is a collection of submanifolds different codimension(set of loops with one selfintersection has codimension $dim M-2$, loops with zero of derivative has codimension $dim M -3$ and so on. Set of loops with infinite number of singularities should have infinite codimension... Let's forget about them.)
So, the question is about local situation: for example, let's consider a loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one self-intersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)?
4 edited title; edited title
# I was wondering if the set of singular loops is a (somewhere) submanifold of loop space?
3 added 117 characters in body; added 1 characters in body
The set of all smooth maps $S^1\to M^n$ ($M$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space).
I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space?
EDIT: it is clear that answer is no (see answer below). So, I rewrite question: is it true that set of singular loops is a collection of submanifolds different codimension(set of loops with one selfintersection has codimension $dim M-2$, loops with zero of derivative has codimension $dim M -3$and 3\$ and so on.on. Set of loops with infinite number of singularities should have infinite codimension... Let's forget about them.)
So, the question is about local situation: for example, let's consider loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one selfintersection self-intersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)?
2 added 488 characters in body; added 85 characters in body
The set of all smooth maps $S^1\to M$ M^n$($M\$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space).
I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space?
EDIT: it is clear that answer is no (see answer below). So, I rewrite question: is it true that set of singular loops is a collection of submanifolds different codimension(set of loops with one selfintersection has codimension $dim M-2$, loops with zero of derivative has codimension $dim M -3$and so on.) So, the question is about local situation: for example, let's consider loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one selfintersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)?
1
# I was wondering if the set of singular loops is a submanifold of loop space?
The set of all smooth maps $S^1\to M$ ($M$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space).
I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space? | {"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": 51, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.911068856716156, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2008/05/27/ | # The Unapologetic Mathematician
## Dual Spaces
Another thing vector spaces come with is duals. That is, given a vector space $V$ we have the dual vector space $V^*=\hom(V,\mathbb{F})$ of “linear functionals” on $V$ — linear functions from $V$ to the base field $\mathbb{F}$. Again we ask how this looks in terms of bases.
So let’s take a finite-dimensional vector space $V$ with basis $\left\{e_i\right\}$, and consider some linear functional $\mu\in V^*$. Like any linear function, we can write down matrix coefficients $\mu_i=\mu(e_i)$. Notice that since our target space (the base field $\mathbb{F}$) is only one-dimensional, we don’t need another index to count its basis.
Now let’s consider a specially-crafted linear functional. We can define one however we like on the basis vectors $e_i$ and then let linearity handle the rest. So let’s say our functional takes the value ${1}$ on $e_1$ and the value ${0}$ on every other basis element. We’ll call this linear functional $\epsilon^1$. Notice that on any vector we have
$\epsilon^1(v)=\epsilon^1(v^ie_i)=v^i\epsilon^1(e_i)=v^1$
so it returns the coefficient of $e_1$. There’s nothing special about $e_1$ here, though. We can define functionals $\epsilon^j$ by setting $\epsilon^j(e_i)=\delta_i^j$. This is the “Kronecker delta”, and it has the value ${1}$ when its two indices match, and ${0}$ when they don’t.
Now given a linear functional $\mu$ with matrix coefficients $\mu_j$, let’s write out a new linear functional $\mu_j\epsilon^j$. What does this do to basis elements?
$\mu_j\epsilon^j(e_i)=\mu_j\delta_i^j=\mu_i$
so this new transformation has exactly the same matrix as $\mu$ does. It must be the same transformation! So any linear functional can be written uniquely as a linear combination of the $\epsilon^j$, and thus they form a basis for the dual space. We call $\left\{\epsilon^j\right\}$ the “dual basis” to $\left\{e_i\right\}$.
Now if we take a generic linear functional $\mu$ and evaluate it on a generic vector $v$ we find
$\mu(v)=\mu_j\epsilon^j(v^ie_i)=\mu_jv^i\epsilon^j(e_i)=\mu_jv^i\delta_i^j=\mu_iv^i$
Once we pick a basis for $V$ we immediately get a basis for $V^*$, and evaluation of a linear functional on a vector looks neat in terms of these bases.
Posted by John Armstrong | Algebra, Linear Algebra | 26 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”).
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": 35, "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": 0.8944145441055298, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/91827?sort=votes | ## Application of polynomials with non-negative coefficients
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Question 1: Are there any deeper applications (in any field of mathematics) of polynomials (with possibly more than one variable) over the real numbers whose coefficients are non-negative? So far I have found only some vague mentions of control theory and maybe some approximations of functions, but don't know the details (e.g. www.math.ttu.edu/~barnard/poly.pdf).
Question 2: There is a similar notion, namely "positive polynomials", but they are defined as "polynomial functions that are non-negative". Does anybody know some connection between these two kinds of polynomials? Thanks.
All comments and suggested answers here (thanks for them:)) have certainly something to do with the mentioned type of polynomials. To specify more what was my intention - a desired answer (to Question 1) should fullfil the following: "Is the application in you proposed area of such a kind that it forces an (independent) research of polynomials with non-negative coefficients on themselves?"
-
Proof of Cauchy Kovaleskaya in PDE theory as done in Evans' book. – Helge Nov 9 at 4:26
## 5 Answers
One answer to question 2 is Polya's theorem on forms positive on an orthant: Let a form (i.e. homogeneous polynomial) in several variables be given which is (strictly) positive whenever evaluated on non zero tuples of nonnegative reals. Then you can multiply it with a high power of the sum of the variables such that you obtain a form with all coefficients nonnegative (actually all coefficients of the "right" degree are positive).
You can also prove a lower bound on the exponent required, see: Powers, Reznick: A new bound for Polya’s Theorem with applications to polynomials positive on polyhedra
This theorem can be used in representation theorems involving sums of squares (cf. Patricia's answer), see my article: An algorithmic approach to Schmüdgen’s Positivstellensatz
-
### 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 you know that the coefficients are non-negative and also integral, then the polynomial can be completely determined by the values of $p(1)$ and $p(p(1))$. There might be a way to extend this to rational coefficients, but I'm not sure.
-
1
Why is this statement true? – Igor Rivin Mar 22 2012 at 1:57
1
@Igor: $q=p(1)$ gives the sum of the coefficients. Now think of $p(p(1))=p(q)$ written in base $q$; one sees that the "digits" are exactly the coefficients of $p$. The only possible ambiguity comes if $p(q)=q^n$ for some $n$, but since the coefficients sum to $q$, one sees that $p=qx^{n−1}$ in this case. – ARupinski Mar 22 2012 at 2:33
The previous proof works if $p(1)\neq 1$ (i.e. $p(x)=x^n$). But one can just take $p(2)$ and $p(p(2))$ for instance. – Miroslav Korbelar Mar 22 2012 at 14:49
@Miroslav One could also take the maximum of the coefficient, say $m$ and evaluate $p(m+1)$ in base $m+1$. – user11000 Mar 29 2012 at 6:56
@Marvis: you have to know $m$ in advance for that trick to work. If you know nothing about the polynomial, I think that the best you can do is asking for $q=p(1)$, and then for either $p(p(1))$ if $q\neq 1$ or $p(2)$ instead. This way you can always reconstruct the polynomial by asking for two evaluations. – Federico Poloni Nov 9 at 8:49
show 2 more comments
Regarding question 1: An example from combinatorics (or conceivably topology, depending on how one classifies these things): if a finite simplicial complex is Cohen-Macaulay (defined below), then its $h$-polynomial (also defined below) has nonnegative coefficients. So this is one example of a situation where you can detect that an object of some sort lacks some nice property by observing that an associated polynomial does not have nonnegative coefficients.
Terminology review, as promised above: given a finite $(d-1)$-dimensional simplicial complex $\Delta$, let $f_i(\Delta )$ be the number of $i$-dimensional faces in $\Delta$, making the convention that the empty set is a $(-1)$-dimensional face. The vector $(f_{-1}(\Delta ),f_0(\Delta ),\dots )$ is called the $f$-vector of $\Delta$. One may encode the same data with another vector called the $h$-vector of $\Delta$ given by $h_k(\Delta ) = \sum_{i=0}^k (-1)^{k-i}{d-i\choose k-i}f_{i-1}(\Delta )$ for each $0\le k\le d$, so for instance $h_d(\Delta )$ is the reduced Euler characteristic of $\Delta$. The $h$-polynomial of $\Delta$ is the polynomial $h(x) = h_0 x^d + h_1 x^{d-1} + \cdots + h_d$, while the $f$-polynomial of $\Delta$ is $f(x) = f_{-1}x^d + f_0x^{d-1} + \cdots + f_{d-1}$. Now an equivalent way (to above) of describing the relationship between $h$-vectors and $f$-vectors is via $\sum_{i=0}^d f_{i-1}(t-1)^{d-i} = \sum_{i=0}^d h_it^{d-i}$, or in other words as $h(x)=f(x-1)$. Now a finite $d$-dimensional simplicial complex $\Delta$ is said to be Cohen-Macaulay over the integers if all its reduced homology groups satisfy $\tilde{H_i}(\Delta ,\mathbf{Z})=0$ for $i < d$ and likewise the link $lk_{\Delta }(\sigma )$ of any face $\sigma \in \Delta$ has reduced homology groups satisfying $\tilde{H_i}(lk_{\Delta }(\sigma )) = 0$ for $i < dim (lk_{\Delta }(\sigma ))$. Cohen-Macaulay complexes have the property that their $h$-vector is nonnegative, hence $h$-polynomial has nonnegative coefficients. A good reference for this and related topological combinatorics is the book "Combinatorics and commutative algebra" by Richard Stanley.
Regarding question 2: An important way of constructing positive polynomials is as sums of squares. Many examples of positive polynomials with nonnegative coefficients may be constructed this way -- if the smaller polynomials being squared have nonnegative coefficients too. However, not all positive polynomials are sums of squares by any means, let alone these special ones. A good reference on this is the paper:
G. Blekherman, Nonnegative polynomials and sums of squares, Jour. Amer. Math. Soc. 25 (2012), 617-635
which proves that there are many more nonnegative polynomials than sums of squares asymptotically by showing that the volumes of cross sections of the cone of nonnegative polynomials grow much faster as total degree is increased than the corresponding cross sections of the cone of sums of squares.
-
People who haven't seen $h$-vectors before might find the definition somewhat unnatural. Since you're interested in the polynomials, not just the vectors, it might be worth mentioning the nice relationship between the $h$- and $f$-polynomials, shifting the argument by 1. – Andreas Blass Nov 9 at 0:43
@Andreas: That's a good idea! I will add that some time soon. – Patricia Hersh Nov 9 at 1:12
@Patricia: Sums of squares don't necessarily have all coefficients nonnegative. – Markus Schweighofer Nov 9 at 7:13
@Markus: I got mixed up and was using the wrong (too restrictive) definition. Thanks for catching that. – Patricia Hersh Nov 9 at 7:59
How about Rook polynomials? http://en.wikipedia.org/wiki/Rook_polynomial Or if you like examples in several variables, Schur polynomials, http://en.wikipedia.org/wiki/Schur_polynomial
Or a plentiful of other polynomials with combinatorial connections. The Schur polynomials have for example applications in representation theory.
-
In fact I had some more systematic usage in mind, not only the fact that some polynomials are just of this kind. Or does one really use this property in the examples you mentioned? – Miroslav Korbelar Mar 22 2012 at 9:18
1
Well, each coefficient counts something, so if it was negative (or non-integer), it would make no sense. – Per Alexandersson Mar 22 2012 at 11:58
Thanks to all for your comments and suggestions. The original motivation for this question was a connection between polynomials with non-negative coefficients and commutative semirings. After some search we found some papers asking for determining the least degree of a polynomial with non-negative coefficients that is divisible by a given (general) polynomial. Suprisingly similar questions were investigated repeatedly and independetly but without any deeper motivation. We deceided hence to make an overview, improvements of some results and suggested a few conjectures. The result (made before the latest updates of this webpage) can be found here http://arxiv.org/abs/1210.6868. (Suggestions and comments are welcome.)
- | {"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": 61, "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": 0.9326912760734558, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/250499/isomorphism-between-2-quotient-spaces | # Isomorphism between 2 quotient spaces
Let $M,N$be linear subspaces $L$ then how can we prove that the following map $$(M+N)/N\to M/M\cap N$$ defined by $$m+n+N\mapsto m+M\cap N$$ is surjective? Originally, I need to prove that this map is bijection but I have already proven that this map is injective and well defined,but having hard time to prove surjectivity,please help.
-
What does a typical element of $M/(M\cap N)$ look like; i.e., how do you write its general form? Given such an element, can you find an element of $(M+N)/N$ that is sent there under your map? Incidentally, note that $n+N=N$, so $m+n+N=m+N$ (assuming here that $n\in N$). – Jonas Meyer Dec 4 '12 at 5:43
the typical element is shown above,well your 2nd question is obviously equivalent to the surjectivity of f,i.e what I need!So is there an element that would preimage of arbitrarily taken element of $M/M\cap N$ – Pilot Dec 4 '12 at 5:46
Given $x=m+M\cap N$, what element of $(M+N)/N$ might go to $x$? – Jonas Meyer Dec 4 '12 at 5:47
$m+n+N$ with n being arbitrary vector in N? – Pilot Dec 4 '12 at 5:50
Yes, which can be written as $m+N$. – Jonas Meyer Dec 4 '12 at 5:51
show 1 more comment
## 2 Answers
Given an arbitrary element $x=m+M\cap N$ of $M/(M\cap N)$, note that $m+N\in (M+N)/N$ is mapped to $x$.
-
Define $T: M \to (M+N)/N$ by $m \mapsto m+N$. Show that it is linear and onto. Check the $\ker T$ and is $M\cap N$ by the first isomorphism theorem $f:M/(M\cap N) \to (M+N)/N$ is an isomorphism.
-
For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Dec 18 '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": 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": 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": 0.9437313675880432, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/12092/is-there-triangulated-category-version-of-barr-becks-theorem | ## Is there triangulated category version of Barr-Beck’s theorem?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
It is well known that Beck's theorem for Comonad is equivalent to Grothendieck flat descent theory on scheme.
There are several version of derived noncommutative geometry. I wonder whether someone developed the triangulated version of Beck's theorem. And What does it mean,if exists?
-
## 2 Answers
There isn't a descent theory for derived categories per se - one can't glue objects in the derived category of a cover together to define an object in the base. (Trying to apply the usual Barr-Beck to the underlying plain category doesn't help.)
But I think the right answer to your question is to use an enriched version of triangulated categories (differential graded or $A_\infty$ or stable $\infty$-categories), for which there is a beautiful Barr-Beck and descent theory, due to Jacob Lurie. (This is discussed at length in the n-lab I believe, and came up recently on the n-category cafe (where I wrote basically the same comment here..) This is proved in DAG II: Noncommutative algebra. In the comonadic form it goes like this. Given an adjunction between $\infty$-categories (let's call the functors pullback and pushforward, to mimic descent), if we have
1. pullback is conservative (it respects isomorphisms), and
2. pullback respects certain limits (namely totalizations of cosimplicial objects, which are split after pullback)
then the $\infty$-category downstairs is equivalent to comodules over the comonad (pullback of pushforward). (There's an opposite monadic form as well) This can be verified in the usual settings where you expect descent to hold. In other words if you think of derived categories as being refined to $\infty$-categories (which have the derived category as their homotopy category), then everything you might want to hold does.
So while derived categories don't form a sheaf (stack), their refinements do: you can recover a complex (up to quasiisomorphism) from a collection of complexes on a cover, identification on overlaps, coherences on double overlaps, coherences of coherences on triple overlaps etc. More formally: define a sheaf as a presheaf $F$ which has the property that for an open cover $U\to X$, defining a Cech simplicial object $U_\bullet={U\times_X U\times_X U\cdots\times_X U}$, then $F(X)$ is the totalization of the cosimplicial object $F(U_\bullet)$. Then enhanced derived categories form sheaves (in appropriate topologies) as you would expect. This is of course essential to having a good theory of noncommutative algebraic geometry!
-
### 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 is a Beck's theorem for Karoubian triangulated categories, proposed by Konstevich and Rosenberg in July 2004, which is proved by using the Verdier's abelianization functor and graded monads; see page 36 of A. Rosenberg, Topics in noncommutative algebraic geometry, homological algebra and K-theory, preprint MPIM Bonn 2008-57, pdf.
In fact they got simulatenously (July 15, 2004) both versions: A-infinity and triangulated. The reference to the triangulated is above: while for A-infinity there is no write up, but Kontsevich gave a talk (I think Nov 2004, van den Bergh birthday conference) where he formulated and used the result; one of nice applications was to glue certain ordinary commutative schemes to get certain formal schemes. I remember very well the weeks preceding the result when we discussed possible shape of the result seeked at IHES. Later at the conference in Split, Kontsevich gave a talk where he gave some usages in noncommutative algebraic geometry.
I disagree with the statement: "Beck's theorem for comonad is equivalent to Grothendieck flat descent theory on scheme". Namely Grothendieck gave both the flat descent theory for quasicoherent sheaves (SGA I.8.1) which is a special case of Beck's theorem (though it has some symmetries which general noncommutative case does not have), but also the (stronger) flat descent theorem for affine schemes (SGA I.8.2) and for morphisms (cf. SGA I.8.5), which unlike the descent for quasicoherent sheaves, does not generalize to the noncommutative algebras and consequently to categorical setup either.
-
Thanks Zoran! The link to Rosenberg's pdf didn't work for me. I'm very curious to see how the triangulated theorem is used - I understand how you'd use the $A_\infty$ version (which is presumably subsumed now by Lurie) but I'm wondering if there are some settings where the triangulated version is useful eg for gluing. – David Ben-Zvi Mar 9 2010 at 13:01
@David and Skoda: Now I saw Rosenberg used this triangulated version of Beck's theorem for induction theorem in triangulated category. Because in triangulated version, we do not require the exactness of functors,so the induction theorem works very smooth in this settings. For induction theorem, I mean cohomological induction. Rosenberg then used t-structures to turn this triangulated picture to abelian picture. I will elaborate this in an additional answers – Shizhuo Zhang Mar 18 2010 at 8:00
David, I corrected the link, for some reason MPI changed the format of the URL in the meantime and www at the beginning is now mandatory. – Zoran Škoda Mar 23 2010 at 16:11
Yes, in the usual geometric situations with qcoh sheaves on open neighborhoods the descent for derived categories does not work as we know from elementary counterexamples. But there are lots of examples with Cohn localization which is non-flat, and these are examples in which if we would restrict to noncommutative subvarieties which are close to the commutative the descent would suddenly become flat, and everything would work at the abelian level already. You should push Sasha & Maxim to finish their "secret" preprint started in 1999, whose delay was initially due to lack of similar tools. – Zoran Škoda Mar 23 2010 at 16:12
2
Just before Prop. 3.4, Rosenberg says on p.36, that given a monad F on a triangulated category, the category of F-modules is triangulated in such a way that the forgetful functor is exact. This would imply in particular that modules over ring spectra up to homotopy form a triangulated category. Etc, etc. Do you believe this? Specifically, I don't see how to put a module structure on the cone of a morphism of modules, unless some assumption is made about the monad (like being separable, in which case it works). – Paul Balmer Mar 23 2011 at 22:10 | {"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": 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": 0.9401208162307739, "perplexity_flag": "middle"} |
http://mathhelpforum.com/latex-help/94005-solved-sans-serif-font-vector-space-notation.html | # Thread:
1. ## [SOLVED] sans serif font for vector space notation
Hi, all.
My linear algebra text uses a sans serif font for vector spaces. Here is a sample image where it talks about arbitrary vector space "V". Notice the font:
(Sorry I couldn't find something more zoomed-in.)
Is there some way to do this in latex?
Thanks!
2. try this [tex]\vee [ /math]
$\vee$
3. Is there some way to do this in latex?
\mathsf{V} gives $\mathsf{V}$. Is that what you are looking for ?
4. Originally Posted by flyingsquirrel
\mathsf{V} gives $\mathsf{V}$. Is that what you are looking for ?
It sure was. Thank you! | {"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": false, "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": 0.8227525949478149, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/17526-some-questions-proving-convergence-tests.html | # Thread:
1. ## Some questions on proving convergence tests
I was reading proofs on some convergence tests, but I'm stuck on a few parts here:
1. Prove every cauchy sequence is bounded.
Proof: Suppose the sequence {An} is cauchy.
Let E = 1, pick N in Natural Numbers such that n >= N and m >= N
We have |An - Am| < 1, further, we have |An-AN| < 1.
This is the part that I don't understand, how do we know that|An - AN| < 1?
2. For a number r such that |r| < 1, then the series sum {from k=0} r^k = 1/(1-r).
Proof: sum {from k=0 to n} r^k = 1 + r + r^2 + r^3 + . . . + r^n = (1 - r^{n+1}) / (1 - r)
Now, how do we know that equals?
Thank you, y'all!
KK
2. Theorem: If $(s_n)$ is a Weierstrauss sequence then $s_n$ is bounded.
Proof: We now that for any $\epsilon$ we can choose $N\in \mathbb{N}$ so that $n,m>N\implies |s_n-s_m|<\epsilon$. So choose $\epsilon = 1$ and we have $|s_n - s_m|<1$. That means
$|s_n|-|s_m|\leq|s_n - s_m|<1 \mbox{ thus }|s_n|< 1+ |s_{N+1}|$ (by choose $m=N+1$).
Now the above inequality if true for only $n>N$.
So,
$|s_n|<1+|s_{N+1}|$ for $n>N$.
But what if $n=1,2,..,N$?
No problem, let,
$M= \max \{ s_1,s_2,...,s_n , 1+|s_{N+1}| \}$.
And so,
$|s_n|\leq M \mbox{ for all } n\in \mathbb{N}$.
3. Originally Posted by tttcomrader
2. For a number r such that |r| < 1, then the series sum {from k=0} r^k = 1/(1-r).
If $|r|<1$ then,
$1+r+r^2+...+r^n = \frac{1-r^{n+1}}{1-r}\to \frac{1}{1-r} \mbox{ as }n\to \infty \mbox{ since }|r|<1$
4. I understand the first proof now, thank you.
But in the second one, why does
$<br /> 1+r+r^2+...+r^n = \frac{1-r^{n+1}}{1-r}<br />$
true?
Other than that, I understand the rest, thank you!
5. Geometric series formula + the following theorem:
$\lim \ a^n = 0 \mbox{ if } |a|<1$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "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": 0.9139114022254944, "perplexity_flag": "middle"} |
http://en.m.wikipedia.org/wiki/Alternating_decision_tree | # Alternating decision tree
An alternating decision tree (ADTree) is a machine learning method for classification. It generalizes decision trees and has connections to boosting.
## History
ADTrees were introduced by Yoav Freund and Llew Mason.[1] However, the algorithm as presented had several typographical errors. Clarifications and optimizations were later presented by Bernhard Pfahringer, Geoffrey Holmes and Richard Kirkby.[2] Implementations are available in Weka and JBoost.
↑Jump back a section
## Motivation
Original boosting algorithms typically used either decision stumps or decision trees as weak hypotheses. As an example, boosting decision stumps creates a set of $T$ weighted decision stumps (where $T$ is the number of boosting iterations), which then vote on the final classification according to their weights. Individual decision stumps are weighted according to their ability to classify the data.
Boosting a simple learner results in an unstructured set of $T$ hypotheses, making it difficult to infer correlations between attributes. Alternating decision trees introduce structure to the set of hypotheses by requiring that they build off a hypothesis that was produced in an earlier iteration. The resulting set of hypotheses can be visualized in a tree based on the relationship between a hypothesis and its "parent."
Another important feature of boosted algorithms is that the data is given a different distribution at each iteration. Instances that are misclassified are given a larger weight while accurately classified instances are given reduced weight.
↑Jump back a section
## Alternating decision tree structure
An alternating decision tree consists of decision nodes and prediction nodes. Decision nodes specify a predicate condition. Prediction nodes contain a single number. ADTrees always have prediction nodes as both root and leaves. An instance is classified by an ADTree by following all paths for which all decision nodes are true and summing any prediction nodes that are traversed. This is different from binary classification trees such as CART (Classification and regression tree) or C4.5 in which an instance follows only one path through the tree.
### Example
The following tree was constructed using JBoost on the spambase dataset[3] (available from the UCI Machine Learning Repository).[4] In this example, spam is coded as $1$ and regular email is coded as $-1$.
The following table contains part of the information for a single instance.
An instance to be classified
Feature Value
char_freq_bang 0.08
word_freq_hp 0.4
capital_run_length_longest 4
char_freq_dollar 0
word_freq_remove 0.9
word_freq_george 0
Other features ...
The instance is scored by summing all of the prediction nodes through which it passes. In the case of the instance above, the score is calculate as
| | | | | Iteration | Instance values | Prediction |
|--------|----------------|---------------|-------------|---------------|-------------------|---------------|
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| N/A | .08 < .052 = f | .4 < .195 = f | 0 < .01 = t | 0 < 0.005 = t | N/A | .9 < .225 = f |
| -0.093 | 0.74 | -1.446 | -0.38 | 0.176 | 0 | 1.66 |
The final score of $0.657$ is positive, so the instance is classified as spam. The magnitude of the value is a measure of confidence in the prediction. The original authors list three potential levels of interpretation for the set of attributes identified by an ADTree:
• Individual nodes can be evaluated for their own predictive ability.
• Sets of nodes on the same path may be interpreted as having a joint effect
• The tree can be interpreted as a whole.
Care must be taken when interpreting individual nodes as the scores reflect a re weighting of the data in each iteration.
↑Jump back a section
## Description of the algorithm
The inputs to the alternating decision tree algorithm are:
• A set of inputs $(x_1,y_1),\ldots,(x_m,y_m)$ where $x_i$ is a vector of attributes and $y_i$ is either -1 or 1. Inputs are also called instances.
• A set of weights $w_i$ corresponding to each instance.
The fundamental element of the ADTree algorithm is the rule. A single rule consists of a precondition, a condition, and two scores. A condition is a predicate of the form "attribute <comparison> value." A precondition is simply a logical conjunction of conditions. Evaluation of a rule involves a pair of nested if statements:
1 if(precondition)
2 if(condition)
3 return score_one
4 else
5 return score_two
6 end if
7 else
8 return 0
9 end if
Several auxiliary functions are also required by the algorithm:
• $W_+(c)$ returns the sum of the weights of all positively labeled examples that satisfy predicate $c$
• $W_-(c)$ returns the sum of the weights of all negatively labeled examples that satisfy predicate $c$
• $W(c) = W_+(c) + W_-(c)$ returns the sum of the weights of all examples that satisfy predicate $c$
The algorithm is as follows:
1 function ad_tree
2 input Set of $m$ training instances
3
4 $w_i = 1/m$ for all $i$
5 $a = 1/2 \textrm{ln}\frac{W_+(true)}{W_-(true)}$
6 $R_0 =$ a rule with scores $a$ and $0$, precondition "true" and condition "true."
7 $\mathcal{P} = \{true\}$
8 $\mathcal{C} =$ the set of all possible conditions
9 for$j = 1 \dots T$
10 $p \in \mathcal{P}, c \in \mathcal{C}$ get values that minimize $z = 2 \left( \sqrt{W_+(p \wedge c) W_-(p \wedge c)} + \sqrt{W_+(p \wedge \neg c) W_-(p \wedge \neg c)} \right) +W(\neg p)$
11 $\mathcal{P} += p \wedge c + p \wedge \neg c$
12 $a_1=\frac{1}{2}\textrm{ln}\frac{W_+(p\wedge c)+1}{W_-(p \wedge c)+1}$
13 $a_2=\frac{1}{2}\textrm{ln}\frac{W_+(p\wedge \neg c)+1}{W_-(p \wedge \neg c)+1}$
14 $R_j =$ new rule with precondition $p$, condition $c$, and weights $a_1$ and $a_2$
15 $w_i = w_i e^{ -y_i R_j(x_i) }$
16 end for
17 return set of $R_j$
The set $\mathcal{P}$ grows by two preconditions in each iteration, and it is possible to derive the tree structure of a set of rules by making note of the precondition that is used in each successive rule.
↑Jump back a section
## Empirical results
Figure 6 in the original paper[1] demonstrates that ADTrees are typically as robust as boosted decision trees and boosted decision stumps. Typically, equivalent accuracy can be achieved with a much simpler tree structure than recursive partitioning algorithms.
↑Jump back a section
## References
1. ^ a b Yoav Freund and Llew Mason. The Alternating Decision Tree Algorithm. Proceedings of the 16th International Conference on Machine Learning, pages 124-133 (1999)
2. Bernhard Pfahringer, Geoffrey Holmes and Richard Kirkby. Optimizing the Induction of Alternating Decision Trees. Proceedings of the Fifth Pacific-Asia Conference on Advances in Knowledge Discovery and Data Mining. 2001, pp. 477-487
3. "UCI Machine Learning Repository". Ics.uci.edu. Retrieved 2012-03-16.
4. "UCI Machine Learning Repository". Ics.uci.edu. Retrieved 2012-03-16.
↑Jump back a section | {"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": 39, "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": 0.9054344892501831, "perplexity_flag": "middle"} |
http://cs.stackexchange.com/questions/tagged/np-complete | # Tagged Questions
Questions about the hardest problems in NP, i.e. of those that can be solved in polynomial time by nondeterministic Turing machines.
1answer
51 views
### Converting math problems to SAT instances
What I want to do is to turn a math problem I have to solve into a Boolean Satisfiability problem and then solve it using a SAT Solver. I wonder if someone knows a manual, guide or anything that will ...
1answer
80 views
### Is “Find the shortest tour from a to z passing each node once in a directed graph” NP-complete?
Given a directed graph with the following attributes: - a chain from node $a$ to node $z$ passing nodes $b$ to $y$ exists and is unidirectional. - additionally a set of nodes having bidirectional ...
1answer
31 views
### Reducing from Hamiltonian Cycle problem to the Graph Wheel problem cannot be proved vice versa [closed]
I saw a proof by Saeed Amiri, We will add one extra vertex v to the graph G and we make new graph G′, such that v is connected to the all other vertices of G. G has a Hamiltonian cycle if and only if ...
1answer
89 views
### Strategic vertex labeling
We are given a graph $G=(V,E)$ with positive edge weights $w_{i}$ and numerical {0,1,-1} labels $l$ for all vertices . We know that $G$ has a subset $G'$ with all vertices labeled 0. The problem is to ...
2answers
101 views
### Wheel subgraph problem [duplicate]
In the following two threads I specified the question in the wrong way (easier to solve that way). Proving that finding wheel subgraphs is NP-complete Reducing from Hamiltonian Cycle problem to the ...
1answer
25 views
### Doubts related to set cover NP-complete problem
I have some doubts related to the set cover ${\sf NP}$-complete problem. I am trying to show that a problem is ${\sf NP}$-complete, so I am trying to transform the Set Cover problem to it. I am ...
1answer
146 views
### Reducing from Hamiltonian Cycle problem to the Graph Wheel problem [duplicate]
EDIT: This question is different from the other in a sense that unlike it this one goes into specifics and is intended to solve the problem. In the previous post, the only answer was a hint. In this ...
1answer
32 views
### Reduction from Vertex Cover to an Independent Set problem
Assume there exists some algorithm that solves vertex cover problem in time polynomial in terms of $n$ and exponential for $k$ with the run time that looks like this $O(k^2 55^k n^3)$. Can we claim ...
1answer
91 views
### NP complete problems that are solvable in polynomial time if the input (e.g. number of variables) is fixed?
I have seen some problems that are NP-hard but polynomially solvable in fixed dimension. Examples, I think, are Knapsack that is polynomial time solvable if the number of items is fixed and Integer ...
1answer
30 views
### reducing Max3SAT to Max2sat
I want to reduce $MAX3SAT$ to $MAX2SAT$ ... MAX-n-SAT : given $\phi$ n-CNF formula and number k does $\phi$ has an assignment that satisfy k clauses?
2answers
59 views
### k-path problem - P, NP or NPC?
I need to determine which complexity class this problem belongs to: Given a graph $G(V, E)$, two vertices $u$ and $v$ and a natural number $k$, does a path of length $k$ exist between thesee two ...
1answer
72 views
### How to reduce INDEPENDENT SET to INDEPENDENT SET SIZE?
Suppose you are given a polynomial-time algorithm for the following problem related to INDEPENDENT SET: INDEPENDENT SET VALUE Input: An undirected graph G. Output:The size of the largest ...
1answer
66 views
### Prove NP-completeness of deciding satisfiability of monotone boolean formula
I am trying to solve this problem and I am really struggling. A monotone boolean formula is a formula in propositional logic where all the literals are positive. For example, \$\qquad (x_1 \lor x_2) ...
1answer
41 views
### Reduction from 3-SAT to a graphe problem
I have a question, i was trying to reduce 3-SAT to a particular graph problem and i'm not quite sure about a thing i used in the reduction. In fact the reduction build a bipartite graph, the edge ...
2answers
150 views
### Proving that finding wheel subgraphs is NP-complete
Can you help me with this problem ? Given an undirected graph $G$ and an integer $n$, prove that determining whether the graph has wheel on $n$ vertices $W_{n}$ (a wheel $W_{i}$ is such that $i$ ...
0answers
50 views
### Convert undirected graph to a directed graph [duplicate]
is it possible to convert G to a directed graph by assigning directions to each of its edges so that every node in C has indegree 0 or outdegree 0, and every other node in G has indegree at least 1? ...
1answer
57 views
### Prove NP-completeness of deciding whether there is an edge-tour of at most a given length
We are given a graph G, integer b < |E|, and subset F in E. The problem is to detect whether there is a cycle in the graph with length at most b and includes each edge in F. Prove that this is NP ...
0answers
31 views
### In a directed graph, the indegree of a node is the number of incoming edges and the outdegree is the number of outgoing edges [duplicate]
In a directed graph, the indegree of a node is the number of incoming edges and the outdegree is the number of outgoing edges. Show that the following problem is NP-complete. Given an undirected graph ...
1answer
103 views
### Pebbling Problem
Pebbling is a solitaire game played on an undirected graph $G$ , where each vertex has zero or more pebbles. A single pebbling move consists of removing two pebbles from a vertex $v$ and adding ...
2answers
85 views
### How to show a problem is NP-complete [closed]
So I'm pretty clueless when it comes to NP complete problems. I'm having a hard time understanding how 3SAT's applied to reduce a problem. Could someone enlighten me a bit? EDIT: It seems like I ...
1answer
59 views
### Asymptotic bounds on number of 3SAT formulas with unique solutions
A set is sparse if it contains polynomially bounded number of strings of any given string length $n$ otherwise it is dense. All known NP-complete sets are dense. It was proven that P=NP if and only if ...
1answer
16 views
### Set cover problem and the existence of such cover
In the set cover problem we want to find in the $\mathbb{S} \subset 2^\mathbb{U}$ the subset $\{s_i\}_{1..k}$, such that $\cup s_i = \mathbb{U}$ for given $K$, where $k \le K$. But how to reduce the ...
1answer
83 views
### How does the problem of having a coffee-machine close relate to vertex cover?
Meeting rooms on university campuses may or may not contain coffee machines. We would like to ensure that every meeting room either has a coffee machine or is close enough to a meeting room ...
2answers
95 views
### Show that the following problem is NP-complete
In a directed graph, the indegree of a node is the number of incoming edges and the outdegree is the number of outgoing edges. Show that the following problem is NP-complete. Given an undirected graph ...
1answer
35 views
### Is the 0-1 Knapsack problem where value equals weight NP-complete?
I have a problem which I suspect is NP-complete. It is easy to prove that it is NP. My current train of thought revolves around using a reduction from knapsack but it would result in instances of ...
1answer
170 views
### NP-Complete Proof [closed]
$n$ people live in a house and wish to share their expenses equally. Their respective expenses before settling are $x_1, x_2, \ldots, x_n$. Assume that all of these are greater than 0. They agree to ...
2answers
108 views
### Can one show NP-hardness by Turing reductions?
In the paper Complexity of the Frobenius Problem by Ramírez-Alfonsín, a problem was proved to be NP-complete using Turing reductions. Is that possible? How exactly? I thought this was only possible by ...
1answer
140 views
### Do any decision problems exist outside NP and NP-Hard?
This question asks about NP-hard problems that are not NP-complete. I'm wondering if there exist any decision problems that are neither NP nor NP-hard. In order to be in NP, problems have to have a ...
0answers
95 views
### Fastest known algorithm for 3-Partition problem
3-Partition problem is $\mathsf{NP}$-Complete in a strong sense meaning there is no pseudo-polynomial time algorithm for it unless $\mathsf{P}=\mathsf{NP}$. I'm looking for the fastest known exact ...
1answer
71 views
### How to analyze the Steiner tree problem?
I have a problem where I am supposed to analyze the Steiner tree problem by doing the following 3 steps. 1) Look up what the Steiner tree problem is. 2) Find a ...
2answers
74 views
### Polynomial time reductions using binary search
There are many NP-complete decision problems that ask the question whether it holds for the optimal value that OPT=m (say bin packing asking whether all items of given sizes can fit into m bins of a ...
1answer
113 views
### How hard is a variant of Sudoku puzzle?
Sudoku is well known puzzle which is known to be NP-complete and it is a special case of more general problem known as Latin squares. A correct solution of the $N \times N$ square consists of filling ...
1answer
119 views
### Proving NP Completeness of a subset-sum problem - how?
So I'm trying to understand P/NPC problems. The one I'm trying to tackle now is subset sum (we have a collection of integers $S$ and a $k$ param: is there a subset of $S$ that sum of all it's elements ...
1answer
56 views
### Prove Matrix Correspondence is NP-complete
Consider the following problem. Given a $m \times n$ integer matrix $A$ and a $p \times q$ integer matrix $B$, do there exist one-to-one functions $$r:\{1,2,...,m\} \rightarrow \{1,2,...,p\}$$ ...
1answer
125 views
### Prove finding a near clique is NP-complete
An undirected graph is a near clique if adding an additional edge would make it a clique. Formally, a graph $G = (V,E)$ contains a near clique of size $k$ where $k$ is a positive integer in $G$ if ...
4answers
79 views
### Does the complexity of strongly NP-hard or -complete problems change when their input is unary encoded?
Does the difficulty of a strongly NP-hard or NP-complete problem (as e.g. defined here) change when its input is unary instead of binary encoded? What difference does it make if the input of a ...
2answers
103 views
### NP-complete and polynomial time reduction
A decision problem is NP-complete if it is in NP and all other problems in NP can be reduced to it by a reduction that runs in polynomial time. Why it is important to require that the reduction runs ...
1answer
98 views
### $1+\epsilon$ approximation for inapproximable problems
I am currently confused by the following situation: 1) The metric $k$-center problem is inapproximable in polynomial time within $2-\epsilon$ unless $P=NP$. 2) The metric $k$-center problem can ...
1answer
69 views
### min-cut with extra condition
I have a undirected graph with no edge costs. A subset of the nodes are labeled $c_1, c_2, ..., c_k$ and one node is labeled $K$. I want to find the minimum cut of the graph with the extra condition ...
1answer
59 views
### Why is MAX-2SAT in NP?
Max-2-SAT is defined as follows. We are given a 2-CNF formula and a bound k, and asked to find an assignment to the variables that satisfies at least k of the clauses. I can understand the ...
3answers
159 views
### Is there a efficient test for if an NFA accepts a subset of another NFA?
So, I know that testing if a regular language $R$ is a subset of regular language $S$ is decidable, since we can convert them both to DFAs, compute $S \cap \bar{R}$, then test if this language is ...
2answers
168 views
### Doron ZEILBERGER's P = NP computer proof
In 2009 Doron has published a paper stating "Using 3000 hours of CPU time on a CRAY machine, we settle the notorious P vs. NP problem in the affirmative, by presenting a “polynomial” time algorithm ...
2answers
129 views
### Is 2-DNF is NP-complete?
I want to know whether the 2-DNF problem is NP-complete or not? If it is NP-complete, can anyone provide a proof?
1answer
118 views
### Showing that minimal vertex deletion to a bipartite graph is NP-complete
Consider the following problem whose input instance is a simple graph $G$ and a natural integer $k$. Is there a set $S \subseteq V(G)$ such that $G - S$ is bipartite and $|S| \leq k$? I would ...
2answers
330 views
### Are there subexponential-time algorithms for NP-complete problems?
Are there NP-complete problems which have proven subexponential-time algorithms? I am asking for the general case inputs, I am not talking about tractable special cases here. By sub-exponential, I ...
1answer
68 views
### How do I explain that a polynomial time reduction is in fact polynomial time?
I have as an assignment question to show that $QuadSat=\{\langle\phi\rangle\mid\phi$ is a satisfiable 3CNF formula with at least 4 satisfying assignments$\}$ is $\sf NP$-Complete. My solution is as ...
1answer
40 views
### Polynomial time reductions
I'm having a very hard time understanding what's what. $$L_{1}\leq_{p}L_{2}$$ If $L_2$ is stated to be in $\textbf{NP}$, is it necessarily true that $L_1$ is $\textbf{NP}$-Complete? I need to show ...
1answer
62 views
### Hardness of Approximating 0-1 Integer Programs
Given a $0,1$ (binary) integer program of the form: \begin{array}{lll} \text{min} & f(x) & \\ \text{s.t.} &A\vec{x} = \vec{b} & \quad \forall i\\ &x_i\ge 0 & \quad \forall ...
0answers
36 views
### Experimental Survey on Different Heuristics for Knapsack Problem
I am looking for a good survey/study of experimental results of heuristics for Knapsack problem (or implemented libraries in java/c++). Any help is appreciated!
2answers
89 views
### Weak and strong completeness
What does a pseudo-polynomial algorithm tell us about the problem it solves? I don't see how running time improves if the algorithm is exponential in the input length and polynomial in the input ... | {"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": 77, "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": 0.9364688992500305, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/22288/conversion-of-motion-equation-from-cartesian-to-polar-coordinates-is-covariant | # Conversion of motion equation from Cartesian to Polar coordinates: Is covariant differentiation necessary?
I have earlier posted the same question here on `math stackexchange` but without any answer. As the question concerns tensors, I guess that I have come to the right place i.e. to physicists.
Say I have the following equation of motion in the `Cartesian` coordinate system for a typical mass spring damper system:
$$M \; \ddot{x} + C \; \dot{x} + K \; x = 0$$
where the dot $^\dot{}$ represents differentiation with respect to time.
Now I would like to convert this equation to `Polar` coordinates. So I introduce
$$x=r \; \cos{\theta}$$ to obtain
$$\dot{x}=\dot{r} \; \cos{\theta} - r \; \dot{\theta} \sin{\theta}$$
and $$\ddot{x}=\ddot{r} \; \cos{\theta}-2 \; \dot{r} \; \dot{\theta} \; \sin{\theta}-r \; \dot{\theta}^2 \; \cos{\theta}- r \; \ddot{\theta} \; \sin{\theta}$$
I can insert $x, \; \dot{x} \; \text{and} \; \ddot{x}$ in my original equation in the `Cartesian` coordinate system to yield
$$M \; (\ddot{r} \; \cos{\theta}-2 \; \dot{r} \; \dot{\theta} \; \sin{\theta}-r \; \dot{\theta}^2 \; \cos{\theta}- r \; \ddot{\theta} \; \sin{\theta}) + C \; (\dot{r} \; \cos{\theta} - r \; \dot{\theta} \sin{\theta}) + K \; (r \; \cos{\theta}) = 0$$
Note: I am just showing the equation and derivatives in the x-direction. But the full system has both $x$ and $y$ components.
I wonder if the above way of thinking is right. I am very new to tensors and I after reading about covariant derivatives, I am now thinking that one should include consider the basis vectors of the `Polar`coordinate system (a `non-Cartesian`coordinate system) also since unlike the basis vectors of the `Cartesian` coordinate system which do not change direction in the 2D space, `Polar` coordinate basis vectors change direction depending on the angle $\theta$.
I am thinking about `covariant derivatives` as the conversion process includes differentiation with respect to the bases. For example if $x=r \; \cos{\theta}$, then
$$\dot{x}=\frac{dx}{dt}=\frac{\partial{x}}{\partial{r}} \cdot \frac{dr}{dt} + \frac{\partial{x}}{\partial{\theta}} \cdot \frac{d\theta}{dt}$$
So we have terms like $\frac{\partial{x}}{\partial{r}}$ and $\frac{\partial{x}}{\partial{\theta}}$ that concern basis vectors both in the `Cartesian` and in the `Polar` coordinate systems.
Hope that someone can shed some light on this.
-
## 1 Answer
1) The first derivation is correct. When you make the second one for $\dot x$ written as $\dot x = \dfrac{d x}{d t}$ you make an error when using full derivatives instead of partial ones. In the latter case you would have obtained: $$\dot x=\dfrac{d x}{d t}=\dfrac{\partial x}{\partial \theta}\dfrac{d \theta}{d t} +\dfrac{\partial x}{\partial \phi}\dfrac{d \phi}{d t}+\dfrac{\partial x}{\partial r}\dfrac{d r}{d t}$$
Then, as $x=r \cos \theta$, you can insert $\dfrac{\partial x}{\partial \phi}=0, \dfrac{\partial x}{\partial \theta}=-r \sin \theta$, $\dfrac{\partial x}{\partial r}=\cos \theta$ into the last formula to obtain the same result you had before.
2) The derivatives you are using here are not covariant derivatives, they are always partial. The derivative $\dfrac{d}{d t}$ is expressed through partial derivatives, in contrast to full derivative $\dfrac{D}{D t}$, which is expressed through covariant derivatives. As you are actually not using covariant derivatives in your derivation, you don't have to worry about the basis vectors.
3) You could make you derivation alternatively by rewriting your original equations in a covariant form $M\dfrac{D^2 x^i}{D t^2}+C\dfrac{D x^i}{D t}+Kx^i=0$, where $x^i$ stands for $(x,y,z)$, stating that the equations of motion do not depend on the choice of coordinates, rewriting the equation in spherical coordinates as $M\dfrac{D^2 \tilde x^i}{D t^2}+C\dfrac{D \tilde x^i}{D t}+K \tilde x^i=0$, where $\tilde x = (r, \theta, \phi)$, and expanding all the derivatives $\dfrac{D}{D t}$ through partial derivatives and Christoffel symbols in spherical coordinates. You would have to take care of the basis, as the result would be in natural basis, whereas the equations of motion are usually written in orthonormal basis.
4) Another powerful way to transform coordinates would be by using lagrangian form of the equations of motion. This would be possible for the case of conservative systems only (no damping terms). In this case you could just rewrite the lagrangian of your system in spherical coordinates and write lagrangian equations from it in a more or less straightforward manner.
-
1
As a coda to this, the critical place where covariant differentiation shows up irrevocably is in the case where you have spatial derivatives. – Jerry Schirmer Mar 13 '12 at 4:29
@Alexey Thank you. I am working in Polar coordinates with $r$ and $\theta$ and you have illustrated the case for spherical coordinates, thus the additional azimuth angle $\phi$ in you equations. 1 vote up. – yCalleecharan Mar 13 '12 at 8:09
@Jerry Schirmer Thanks. But we do have spatial derivatives of $x$ w.r.t. $r$ and so on as indicated in my post. 1 vote up. – yCalleecharan Mar 13 '12 at 8:11
@yCalleecharan: I mean, in the original Lagrangian. If you tried to convert something like the wave equation $\frac{1}{c^{2}}\frac{\partial^{2}\phi}{\partial t^{2}} - \left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)\phi$ into polar coordinates, you'd find that the transformed equation would contain terms that only have first spatial derivatives of $\phi$. The coefficients of these terms are your nonzero $\Gamma_{ij}{}^{k}$ – Jerry Schirmer Mar 14 '12 at 5:25
@Alexey I have corrected my post and put the partial derivatives where necessary. – yCalleecharan Mar 17 '12 at 19:46 | {"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": 29, "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": 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": 0.9380297660827637, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/57850-differential-equations-problem.html | # Thread:
1. ## Differential Equations Problem
A two-mode laser produces two di erent kinds of photons. The number of photons in the laser eld of each type are n1 and n2:
These vary in time and are governed by the DE system
n*1(n with dot on top) = G1*N*n1 - k1*n1
n*2(n with dot on top) = G2*N*n2 - k2*n2
where N(t) = N0 - alpha1*n1 - alpha2*n2
For a given experiment the parameters are given by G1 = 2; G2 = 1; alpha1 =
1 ;alpha2 = 2;k1 = 2;k2 = 2 and N0 = 4
(a) Find and classify all the equilibrium points and sketch the solutions in
the phase plane.
(b) What is the long term behaviour of the laser?
2. flaming . . . you know how to get started with this? Let me give you some tips:
(1) Write the system as pretty as possible else some will not want to bother interpreting your writing. Here's what I did after a lot of processing:
$n'_1=6n_1-2n_1^2-4n_1n_2$
$n'_2=2n_2-2n_2^2-n_1n_2$
See. That makes a lot of difference. Now shake and bake:
(a) Find the equilibrium points by setting the right side to zero. Flat out, I'm not going to do that by hand (just use Solve in Mathematica). Maybe you want to. Anyway they are: $e_1=(0,0),\; e_2=(0,1),\; e_3=(3,0)$
(b) Linearize it by calculating the Jacobian matrix for each equilibrium point.
(c) Find the eigenvalues for each matrix.
(d) Based on the sign of the eigenvalues, determine the stability of the equilibrium points.
(e) Finally, rely heavily on Mathematica to plot the results. You can use VectorFieldPlot to draw the vector field easily. Use NDSolve to solve the system for some select initial conditions and superimpose these solutions in the vector field to illustrate how the solutions follow the field. Use Point to show where the eq. points are and by plotting a lot of solutions, show how the solutions are affected by the stability of the eq. points.
See. Poke-a-poke | {"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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8978000283241272, "perplexity_flag": "middle"} |
http://mathhelpforum.com/differential-equations/165424-inverse-operators.html | # Thread:
1. ## Inverse operators
Hey I need to solve this problem by means of inverse operators:
y''-3y'+2y=x
Thank you!
2. Originally Posted by deltafee
Hey I need to solve this problem by means of inverse operators:
y''+3y'+2y=x
Thank you!
Just FYI your questions are very vague.
I think this is what you mean...
$\displaystyle y''+3y'+2y=x \iff (D^2+3D+2)y=x$
Solving for y gives
$\displaystyle y=\frac{1}{(D+1)(D+2)}[x]$
Since $x$ is a polynomial we can express each operator as a geometric series, but 1st partial fractions gives
$\displaystyle \frac{1}{(D+1)(D+2)}=\frac{1}{D+1}+\frac{-1}{D+2}$
Then
$\displaystyle \frac{1}{1+D}=\sum_{n=0}^{\infty}(-1)^nD^n$
and
$\displaystyle \frac{1}{2+D}=\frac{1}{2}\frac{1}{1+\frac{D}{2}}=\ sum_{n=0}^{\infty}(-1)^n\frac{D^n}{2^n}$
Combing the sums gives
$\displaystyle \frac{1}{(D+1)(D+2)}=\sum_{n=0}^{\infty} \left( \frac{2^n+1}{2^n}\right)(-1)^nD^n$
Note that since the right hand side is only a degree 1 polynomial we only need two terms from the series
$\displaystyle 1-\frac{3}{2}D$
$\displaystyle \left( 1-\frac{3}{2}D\right)x=x-\frac{3}{2}$
So this gives $\displaystyle y_p=x-\frac{3}{2}$
You still need to use the eigenfunctions to find the complimentary solution.
3. Very elegant! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9306107759475708, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/57958/microlocal-geometry-a-theorem-of-verdier | ## Microlocal geometry - A theorem of Verdier
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
(1) In "Geometrie Microlocale", Verdier states the following theorem.
Theorem: Let $E$ be a vector space and $F$ a constructible complex on $E$. Then for $\ell$ a linear form on $E$, we have a morphism between the fibers at $0$: $$i_0^*\psi_\ell (F) \longrightarrow i_0^*\psi_\ell(\nu_0(F))$$ which is an isomorphism for generic $\ell$ (same thing for the vanishing cycles $\phi_\ell$).
Here $\nu_0: D^b_c(E) \to D^b_{mon}(T_0E)$ is Verdier specialization functor to the tangent cone $T_0 E = E$, i.e. the nearby cycles of $p^*F$ along the deformation of to the normal cone $t:D_0 E \to \mathbb{A}^1$.
The theorem says that when computing $\psi_\ell(F)_0$, one can usually suppose $F$ is monodromic.
Verdier doesn't prove the theorem. Instead he just gives an indication of what generic means: consider the blow-up $p:\tilde{E} \to E$ at $0$, take a stratification for which $p^*F$ is constructible and such that $p^{-1}(0)$ is reunion of strata. The condition is that the strict transform of the hyperplane ${ \ell = 0}$ should be transverse to the strata contained in $p^{-1}(0)$.
I don't understand how to prove the theorem based on this.
Question 1: Blowing-up a strata to get things transverse in order to compute nearby/vanishing cycles seems not uncommon. Is there some kind of machinery that makes it a standard trick?
(2) Here are my thoughts. We have a triangle $$F \to e_* e^* F \to \bigoplus_{q=1}^{d-1} i_* i^* F(-q)[-2q] \to +1$$ The theorem is trivial for $i_* i^* F$ so it is enough to prove that $$i^* e_* \psi_{\tilde{\ell}} (\tilde{F}) = i^* \psi_\ell (e_* e^* F) \longrightarrow i^* \psi_{d\ell} \nu_0(e_* e^* F) = i^* (Ce)_* \psi_{d\tilde{\ell}} \nu_D (\tilde{F})$$ is an isomorphism (with $\tilde{F} = e^* F$, $\tilde{l} = l\circ e$). But $\tilde{l}^{-1}(0) = \tilde{H} \cup D$ where $\tilde{H}$ is the strict transform of $H = {\ell = 0}$ and $D = e^{-1}(0)$ is the exceptionnal divisor. Outside of $H$, $\tilde{\ell}$ is an equation for $D$ and we always have $\psi_f (F) = \psi_{df} \nu_{f=0}(F)$ (in the topological setting this is proved in Kashiwara-Schapira, I don't know any reference for the étale setting) so $\psi_{\tilde{\ell}} (\tilde{F}) \to \psi_{d\tilde{\ell}} \nu_D (\tilde{F})$ is an isomorphism outside of $\tilde{H}\cap D$. The tricky part is
Question 2 How do we prove that
$$\psi_{\tilde{\ell}}(\tilde{F}) \to \psi_{d\tilde{\ell}}\nu_D (\tilde{F})$$ is an isomorphism on $\tilde{H}\cap D$, the only hypothesis being that $\tilde{H}$ is transverse to the strata of $D$?
(3) I have (almost) entirely checked that Question 2 cand be answered by using resolution of singularities and proper base change to reduce to the case where $\tilde{F}$ is constructible w/r to the stratification defined a normal crossing divisor, the strict transform of $H$ being one of its components and then using the explicit computation of the nearby cycles in terms of the nearby cycles relative to each component.
But it seems that this line of reasoning is overly convoluted and that there should be a simpler, more natural argument. My main evidence is that stating the computation of nearby cycles in the normal crossing case is already painful. The proof is even worse and it doesn't make things much clearer. Since we don't really need the full computation here, I'd like to avoid it. Plus, the idea that things shouldn't change under specialization if the characteristic directions are separated seems very natural. For example, I think this could be obvious to someone who understands (stratified) Morse theory if I was able to translate it in that topological language.
Question 3 I think Verdier's proof was based on a "standard" dévissage.
Is there a way not to use resolution of singularity and/or explicit computation?
Am I over-optimistic in thinking that there should be a proof as simple as the statement of the theorem?
(4) Any advice or reference (on what you think Verdier's proof is, general techniques of dévissage, the non-characteristic hypotheses etc...) would be truly appreciated. Thanks.
- | {"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": 38, "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": 0.9472013711929321, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/164559-geometric-series.html | # Thread:
1. ## Geometric Series
Find a specific geometric series which sums to 30
2. Originally Posted by thamathkid1729
Find a specific geometric series which sums to 30
Can you solve $\dfrac{1}{1-r}=30~?$
3. Thank you
Is it possible to find a geometric series summing to -1/3?
4. A geometric series converges if and only if $|r|<1$.
5. So since r = 4, the answer is NO?
6. Originally Posted by thamathkid1729
Thank you
Is it possible to find a geometric series summing to -1/3?
$\displaystyle -\left(\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + ... \right) = -\frac{1}{3}$
7. Can only rational numbers be sums of geometric series? Precisely which real numbers can be sums of geometric series?
8. Originally Posted by thamathkid1729
Can only rational numbers be sums of geometric series? Precisely which real numbers can be sums of geometric series?
the sum can be any real number ... depends on the first term and the common ratio.
9. The series $\sum_{n=0}^\infty ar^n$ sums to $\frac{a}{1- r}$. Whether that is a rational number or an irrational number depends upon a and r. For example, given any rational number, u, take a= 1 and solve $\frac{1}{1- r}= u$. Given any irrational number, v, take a= v/2 and a= 1/2. | {"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": 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": 0.8579142093658447, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/62958/do-bistellar-flips-preserve-shellability/62960 | ## Do bistellar flips preserve shellability?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I notice there is a strong connection between shellability of simplicial complexes and bistellar flips on these complexes; in particular, adding in a new facet of a shelling induces a bistellar flip on the boundary.
Is it always the case that bistellar flips preserve shellability of a complex? In other words, if I apply a bistellar flip to a shellable simplicial complex, is the resulting simplicial complex also shellable? If so, how can this be seen easily?
-
## 1 Answer
Obviously not, because there exist non-shellable combinatorial spheres, but any combinatorial $n$-sphere is bistellar-equivalent to the boundary of the $(n+1)$-simplex.
The observation you mentioned is also in the very end of Lickorish's paper
- | {"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": 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": 0.887788712978363, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/6062/does-it-take-brute-force-to-find-a-pair-of-plaintext-and-ciphertext-that-each-fo/6067 | # Does it take brute force to find a pair of plaintext and ciphertext that each follow a certain condition, given an AES encryption key?
Suppose that I have an AES key $K$, and I'm instructed to find a plaintext such that the first 32 bits of the plaintext are some string of bits $S_1$, and the last 32 bits of the ciphertext once the plaintext has been encrypted with $K$ are another string of bits $S_2$.
Is this difficult? Is there a known attack against this that's faster than just fixing one of the $S$'s and brute-force searching for a text with the other $S$?
-
## 2 Answers
Well, it's moderately difficult; you couldn't use that as a security assumption; however, it would be too difficult to expect someone to solve during an active protocol. There's no nonobvious trick to it; the two methods at your disposal would be:
• Select random plaintexts with the first 32 bits as $S_1$; encrypt them with key $K$, and check to see if the last 32 bits of the ciphertext happen to be $S_2$
• Select random ciphertexts with the last 32 bits as $S_2$; decrypt them with key $K$, and check to see if the first 32 bits of the plaintext happen to be $S_1$
Either approach will take an expected $2^{32}$ random trials before success; at 1 $\mu$sec per AES operation (quite conservative; modern CPUs typically can do several times as fast as that), we're looking at perhaps an hour or two.
Perhaps you were hoping there was some clever way to take advantage of the partly known plaintext/ciphertext; it doesn't work out. For example, you might be hoping to translate the AES cipher into a large series of boolean operations (with the known key, and known 32 bits of plaintext and ciphertexts), and solve the resulting set of equations. However, AES has a fairly quick 'avalanche'; very quickly (within two rounds), all the internal bits will depend on all the unknown bits in subtle ways; the resulting set of equations will not have an easier solution than just trying various combinations of the 96 bits on one side (which is effectively what the straight-forward solutions do)
-
That's what I was looking for; thanks. I was wondering whether there was any method easier than those two you listed. If there aren't, that's perfect. – Joe Z. Jan 22 at 11:57
"Perhaps you were hoping there was some clever way to take advantage of the partly known plaintext/ciphertext; it doesn't work out." Actually, I was hoping there wasn't such a way and that somebody would verify the fact. :P – Joe Z. Jan 22 at 12:07
Could I use it as a security assumption if I upped the bit specification to 64? – Joe Z. Jan 22 at 17:29
@JoeZeng: well, I suppose you could, depending on the adversary. Someone could solve with with around $2^{64}$ AES operations; there are real world organizations which could feasibly do that much for for something they really wanted. Of course, your potential adversaries might not fall into that camp -- that's a call you would need to make. Personally, I prefer solutions that require $2^{128}$ operations (or more) to defeat; that means I don't have to worry if a TLA would be interested. – poncho Jan 22 at 20:56
1
@Joe: Note that, if you fix 64 bits of the plaintext and 64 bits of the ciphertext of a 128-bit block cipher, then there's about a $e^{-1}\approx 0.37$ chance that no such pair will exist. (More generally, if you fix $j\gg0$ out of $n$ bits of the plaintext and $k\gg0$ out of $n$ bits of the ciphertext, then the probability of there being at least one matching pair is about $1-e^{-2^{n-j-k}}$.) – Ilmari Karonen Jan 23 at 12:19
show 2 more comments
For an ideal 128-bit block cipher (a family of random permutations on the set of 128-bit blocks), there's no better way to find such pairs than the brute force method described by poncho. If a significantly more efficient method did exist for AES, that would allow efficiently distinguishing AES from an ideal block cipher, which would contradict the security assumptions of AES.
As no such attack has been published, despite the considerable amount of cryptanalytic research on AES, we can be fairly confident that there's no easily discoverable method to find such pairs that would be more efficient than brute force.
- | {"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": 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": 0.9699374437332153, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/66084/open-problems-with-monetary-rewards/66101 | ## Open problems with monetary rewards
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Since the old days, many mathematicians have been used to attaching monetary rewards to problems they admit are difficult. Their reasons could be to draw other mathematicians' attention, to express their belief in the magnitude of the difficulty of the problem, to challenge others, "to elevate in the consciousness of the general public the fact that in mathematics, the frontier is still open and abounds in important unsolved problems.1", etc.
Current major instances are
Other problems with money rewards
-Kimberling's list of problems
Question: What others are there? To put some order into the answers, let's put a threshold prize money of 100 USD. I expect there are more mathematicians who have tucked problems in their web-pages with some prizes.
What this question does not intend to achieve:
-once offered but then collected or withdrawn offers
-new pledges of sums of money just here
P.S. Some may be intersted in the psychological aspects of money rewards. However, to keep the question focused, I hope this topic won't be ignited here. One more, I understand that mathematician's do not work merely for money.
-
3
I parse the title as "Difficulties arising by offering money for solutions to questions." Perhaps you might disambiguate the title? Gerhard "Offer Drinks Instead Of Money" Paseman, 2011.05.26 – Gerhard Paseman May 26 2011 at 18:38
6
I see some votes to close this post. I have started a meta asking why here: meta.mathoverflow.net/discussion/1055/… So, please all discussion concerning this should happen there. – To be cont'd May 27 2011 at 9:08
7
@Thierry -- I've often felt that even for the Clay million-dollar problems, that's got to be one of the hardest ways to earn a million dollars! I should think just the glory of solving such a problem would be, by far, a greater motivation. (BTW: I think it is small.) – Todd Trimble May 27 2011 at 9:40
12
Well, this question already has 4 votes to close. To whoever is thinking of pulling the plug: I also thought of voting to close at first, but since the question has already generated a couple of useful answers (including some problems I would probably never have heard of otherwise), I think it should stay open. – algori May 28 2011 at 1:57
11
I agree, I think that four votes to close is four too many. – Greg Kuperberg May 29 2011 at 13:42
show 7 more comments
## 19 Answers
Two which are for food rather than cash:
Let $f = t^{2d} + f_1 t^{2d-1} + f_2 t^{2d-2}+ \cdots f_d t^d + \cdots+ f_2 t^2 +f_1 t + 1$ be a palindromic polynomial, so the roots of $f$ are of the form $\lambda_1$, $\lambda_2$, ..., $\lambda_d$, $\lambda_1^{-1}$, $\lambda_2^{-1}$, ..., $\lambda_d^{-1}$. Set $r_k = \prod_{j=1}^d (\lambda_j^k-1)(\lambda_j^{-k} -1)$.
Conjecture: The coefficients of $f$ are uniquely determined by the values of $r_1$, $r_2$, ... $r_{d+1}$.
Motivation: When computing the zeta function of a genus $d$ curve over $\mathbb{F}_q$, the numerator is essentially of the form $f$. (More precisely, it is of the form $q^d f(t/\sqrt{q})$ for $f$ of this form.) Certain algorithms proceed by computing the $r_k$ and recovering the coefficients of $f$ from them. Note that you have to recover $d$ numbers, so you need at least $r_1$ through $r_d$; it is known that you need at least one more and the conjecture is that exactly one more is enough.
Reward: Sturmfels and Zworski will buy you dinner at Chez Pannise if you solve it.
Consider the following probabilistic model: We choose an infinite string, call it $\mathcal{A}$, of $A$'s, $C$'s, $G$'s and $T$'s. Each letter of the string is chosen independently at random, with probabilities $p_A$, $p_C$, $p_G$ and $p_T$.
Next, we copy the string $\mathcal{A}$ to form a new string $\mathcal{D}_1$. In the copying process, for each pair $(X, Y)$ of symbols in ${ A, C, G, T }$, there is some probability $p_1(X \to Y)$ that we will miscopy an $X$ as a $Y$. (The $16$ probabilities stay constant for the entire copying procedure.)
We repeat the procedure to form two more strings $\mathcal{D}_2$ and $\mathcal{D}_3$, using new probability matrices $p_2(X \to Y)$ and $p_3(X \to Y)$.
We then forget the ancestral string $\mathcal{A}$ and measure the $64$ frequencies with which the various possible joint distributions of `$\{ A, C, G, T \}$` occur in the descendant strings $(\mathcal{D}_1, \mathcal{D}_2, \mathcal{D}_3)$.
Our procedure depended on $4+3 \times 16$ inputs: the $(p_A, p_C, p_G, p_T)$ and the $p_i(X \to Y)$. When you remember that probabilities should add up to $1$, there are actually only $39$ independent parameters here, and we are getting $63$ measurements (one less than $64$ because probabilities add up to $1$). So the set of possible outputs is a semialgeraic set of codimension $24$.
Conjecture: Elizabeth Allman has a conjectured list of generators for the Zariski closure of the set of possible measurements.
Motivation: Obviously, this is a model of evolution, and one which (some) biologists actually use. Allman and Rhodes have shown that, if you know generators for the ideal for this particular case, then they can tell you generators for every possible evolutionary history. (More descendants, known sequence of branching, etc.) There are techniques in statistics where knowing this Zariski closure would be helpful progress.
Reward: Elizabeth Allman will personally catch, clean, smoke and ship an Alaskan Salmon to you if you find the generators. (Or serve it to you fresh, if you visit her in Alaska.)
-
2
Thanks for the nice exposition of the problems. – To be cont'd May 27 2011 at 8:45
Do you mean, independently at random rather than uniformly at random? – Greg Kuperberg May 27 2011 at 20:28
Corrected, thanks. – David Speyer May 27 2011 at 21:06
Is there some genericity condition on the $\lambda_i$ in the first conjecture? For instance if $\lambda_1 = 1$ then all the $r_k$ are zero, and that's obviously not enough information to reconstruct $f$? – Abhinav Kumar Jul 26 at 21:19
Yes, the conjecture is for generic $\lambda_i$. I can't seem to find an original print source for it, but you can find it restated as Conjecture 1.2 in arxiv.org/abs/math.AG/0411414 – David Speyer Jul 27 at 1:02
### 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.
Addendum: There is a paper by Fan Chung similar to this book by Chung and Graham, Open problems of Paul Erdős in graph theory. She says there, "In November 1996, a committee of Erdős' friends decided no more such awards will be given in Erdős' name." But the same article says that Chung and Graham decided to still sponsor questions in graph theory, and this article in Science Magazine implies that they are still sponsoring the Erdős problems in general.
Some 19 years ago I collected a list of Erdős prize problems and posted them to Usenet. The problems were from "A Tribute to Paul Erdős" (1990) and "Paths, Flows, and VLSI Layout" (1980). I can repeat the problems here, although I have no idea which ones may have been solved.
$\$10000$. (T4N) Consecutive primes are often far apart. Conjecture: For every real number$C$, the difference between the$n\$'th prime and n+1'st prime exceeds
$$C \log(n) \log(\log(n))\log(\log(\log(\log(n))))/\log(\log(\log(n)))^2$$
infinitely often. (The wording in the source does not clearly indicate that the money will be awarded if the conjecture is disproved, only if it is proved.)
$\$3000\$. (T3N) Divergence implies arithmetic progressions. If the sum of the reciprocals of a set of positive integers is infinite, must the set contain arbitrarily long finite arithmetic progressions?
$\$1000$. (T2N) Unavoidable sets of congruences. A set of congruences$n = a_1 \bmod b_1$,$n = a_2 \bmod b_2$,... is unavoidable if each$n$satisfies at least one of them. Is there an$N$such that every unavoidable set of congruences either has two equal moduli$b_i$and$b_j$or some modulus$b_i$less than$N\$?
$\$1000$. (T1C) Three-petal sunflowers. Is there an integer$C$such that among$C^n$sets with$n$elements, there are always three whose mutual intersection is the same as each pairwise intersection? (Problem P2 is the same, except that Erdos asks about$k$-petal sunflowers for every$k$but then says he would be satisfied with$k=3\$.)
$\$500$. (T7N) Asymptotic bases of order 2 (I). Consider an infinite set of positive integers such that every sufficiently large integer is the sum of two members of the set. Can there be an$N$such that no positive integer is the sum of two members of the set in more than$N\$ ways?
$\$500$. (T8N) Asymptotic bases of order 2 (II). In the context of the previous problem, let$f(n)$be the number of ways that n is the sum of two members of the set. Can$f(n)/\log(n)$converge to a finite number as$n\$ goes to infinity?
$\$500$. (T9N) Evenly distributed two-colorings. Given a black-white coloring of the positive integers, let$A(n,k)$be the number of blacks minus the number of whites among the first$n$multiples of$k$. Can the range of$A\$ be bounded on both sides?
$\$500$. (T4C) Friendly collections of half-sized subsets. Given$1+(\binom{4n}{2n} - \binom{2n}{n}^2)/2$distinct, half-sized subsets of a set with$4n\$ elements, must there be two subsets which intersect only in one element? (As problem P1, 250 pounds is offered.)
$\$500$. (T1G) Uniformity of distance in the plane (I). Is there a real number$c$such that n points in the plane always determine at least$cn/\sqrt{\log(n)}\$ distinct distances?
$\$500$. (T1G) Uniformity of distance in the plane (II). Is there a real number$c$such that given n points in the plane, no more than$n^{(1+c/\log(\log(n)))}\$ pairs can be unit distance apart?
$\$500$. (P2) Sets with distinct subset sums. Is there a real number$c$such that, given a set of n positive integers whose subsets all have distinct sums, the largest element is at least$c2^n\$? (As in problem T1N, no prize is mentioned.)
$\$250$. (P4) Collections of sets not represented by smaller sets. Is there a real number$c$such that for infinitely many positive integers$n$, there exists$cn$or fewer sets with n elements, no two of which are disjoint, and every$(n-1)\$-element set is disjoint from at least one of them?
$\$250/\$100$. (P15) Slowly increasing Turan numbers. If H is a (simple) graph, the Turan number $T(n,H)$ is the largest number of edges a graph with $n$ vertices can have without containing a copy of $H$. Conjecture: the function $f(n) = T(n,H)/n^{3/2}$ is bounded above if and only if every connected subgraph of $H$ has a vertex of valence 1 or 2. The larger award would be granted for a proof.
$\$100/\$25000$. (T6N) Consecutive early primes. An early prime is one which is less than the arithmetic mean of the prime before and the prime after. Conjecture: There are infinitely many consecutive pairs of early primes. The larger award would be granted for a disproof.
$\$100$. (T8G) Quadrisecants in the plane. Given an infinite sequence of points in the plane, no five of which are collinear, let$r(n)$be the number of lines that pass through four points among the first$n$. Can it happen that$r(n)/n^2\$ does not converge to zero?
-
@Greg, thank you very much. Your collection of the Erdos prize problems were linked to previously but this organized answer makes the question really interesting. – To be cont'd May 28 2011 at 5:47
Of course mathematicians do not work merely for money! They are motivated by higher things like access to moderation tools at 10000 reputation level on Mathoverflow!
I've heard from Ron Graham that the bookkeeping for Erdos rewards is difficult because many people frame the check and never cash it.
There is always the chance of earning \$327.68 from Donald Knuth. It is stretching things more than a bit to include that in and of itself, but the linked article is amusing and the general considerations are pertinent.
The EFF offers large rewards for large primes.
-
10
I've heard that the cleverer people who win Erdos money photocopy the check, frame the copy, and keep the original. – Gerry Myerson May 27 2011 at 5:37
6
Knuth says his cheques are much more often cached than cashed. – Max Jun 21 2011 at 19:09
Paul Erdős was famous for offering money for solutions to math problems. My understanding is that those prizes are still being administered by Ron Graham, even though Erdős passed away several years ago.
One of his most famous (and biggest money) problems is the following:
Conjecture: If $S$ is a set of positive integers such that the series $\sum_{s \in S} \frac{1}{s}$ diverges, then $S$ contains arbitrarily large arithmetic progressions.
I believe that \$3000 is offered for a proof.
Several graph theory related problems, with prize money listed for many of them, are collected in Erdős on graphs, by Chung and Graham.
-
@Matthew, thank you. – To be cont'd May 26 2011 at 18:20
7
With a solution of that problem a Field medal would also come, unless you are over 40. – Maurizio Monge May 26 2011 at 18:48
1
See "Does there exist a comprehensive compilation of Erdos open problems?" mathoverflow.net/questions/27716 – Joseph O'Rourke May 26 2011 at 20:52
3
A partial list of problems with monetary awards is available at: math.niu.edu/~rusin/known-math/93_back/prizes.erd – Ben Linowitz May 27 2011 at 1:16
9
@Maurizio: I'm wondering what makes you so sure of that? – Todd Trimble May 27 2011 at 21:47
show 2 more comments
See http://people.math.jussieu.fr/~talagran/prizes.pdf for a total of $\$7000$offered by Michel Talagrand for a solution of three problems. In particular$\$5000$ for the "Bernoulli conjecture". You may also be interested in the following picture of Mazur awarding Per Enflo a live goose as promised for the solution of the approximation problem. http://en.wikipedia.org/wiki/File:MazurGes.jpg
-
John Conway has prize money out for various problems, such as \$1000 for the thrackle conjecture.
-
It was pointed out by Randall Munroe that by proving the inconsistency of logic you can earn quite a lot of money:
(Source)
-
Ex falso quodlibet! So, I doubt this creates that many errors. – quid Jul 26 at 12:00
9
I should have mentioned that the hidden text is : "Dear Reader: Enclosed is a check for ninety-eight cents. Using your work, I have proven that this equals the amount you requested." – Adrien Jul 26 at 12:21
Oh, so you suggest to prove just this, and then cash in essentially all the things mentioned so far. Nice plan ;D – quid Jul 26 at 15:07
J.-B. Zuber offered respectively 1, 2 and 3 bottles of Champagne for the classification of finite quantum subgroups of $SU_q(3)$, $SU_q(4)$, and $SU_q(5)$, respectively.
Ocneanu solved the first two cases, but as far as I know the problem for $SU_q(5)$ is still open.
-
I am not sure if he would still be willing to pay out, but Vaughan Pratt offers 2000 smackers for the solution to the following:
http://thue.stanford.edu/puzzle.html
Let D be a set of subsets of a set A with the following three properties.
(i) D contains A and the empty set.
(ii) Let C be any set of pairs (a,b), a,b in A, such that for all a in A, the sets {b | (a,b) is in C} and {b | (b,a) is in C} are in D. Then {b | (b,b) is in C} is in D.
(iii) (T1) For any two distinct elements a,b of A, D contains a set containing a but not b, and another containing b but not a.
The set D consisting of all subsets of A evidently satisfies (i)-(iii). Does any other set of subsets of A do so?
-
1
If smackers is a synonym for kisses, I'm not so sure I want to work in the same field as this problem. Gerhard "Choosy About Who He Kisses" Paseman, 2011.05.26 – Gerhard Paseman May 27 2011 at 0:05
According to thefreedictionary.com, smacker = \$1 or £1. Which is a relief, 'cos my first thought was that it had something to do with smack. Which would be scandalous, of course. ;) – Nikita Sidorov May 27 2011 at 1:25
The free dictionary got it right. – Steven Gubkin May 27 2011 at 1:40
HAHAHA! This made me laugh. – Koundinya Vajjha May 27 2011 at 5:16
A. Bressan has advertised two monetary rewards of \$500 each for solutions to problems on mixing flows and blocking problems.
The first problem was unsolved as of Jan. 15, 2011, although progress in relevant directions is noted in the linked announcement. The second problem was announced on Jan. 19, 2011.
-
The $3n+1$ Conjecture has some money assigned to it.
Define $T(n) = n/2$ when $n$ is even and $3n+1$ when $n$ is odd.
For any positive integer $n$ does there exist a positive integer $N$, such that $T^N(n) = 1$?
The origin of this precise question seems to be obscure, although Lothar Collatz made similar conjectures during the 1930s1. For example, Bryan Thwaites claims to have been the first to make this conjecture in 19522, and this does not seem to have been decisively refuted. (The 1937 dates in the Wikipedia and Mathworld articles are missing citations - the Wikipedia edit dates to 7 September 2004.)
Rewards offered to date include 1000 UK pounds from Bryan Thwaites, 500 US dollars from Paul Erdos, and 50 dollars (Canadian?) from H.S.M. Coxeter1.
1. Lagarias, The $3x+1$ problem and its generalizations Am. Math. Monthly 92 (1985) 3-23.
2. Bryan Thwaites, Two conjectures or how to win £1100. Math. Gazette 80 (1996) 35-36.
-
3
The statement is not correct as $N$ must be allowed to depend on $n$. Is there documentation for Halmos' offer? Did the offer expire with the professor? Not that I expect anyone to collect on it any time soon.... – Gerry Myerson May 27 2011 at 5:41
As the conjecture is stated, False as $\forall N$, $T^N(2^{N+1}) = 2 \neq 1$ but I doubt I'll get \\$500 for stating that. – qwerty1793 May 27 2011 at 9:45
2
500 double dollars. I feel like I just walked into a trigun episode. – Steven Gubkin May 27 2011 at 14:09
@Gerry Thanks for pointing out my errors. – Daniel Parry May 27 2011 at 18:42
1
@Daniel, ordinarily I'd say, if Lagarias says so, it must be true, but I'm disappointed that he doesn't give any reference. – Gerry Myerson May 28 2011 at 5:49
show 2 more comments
Gerard Cornuejols offers \$5000 for the first proof (or refutation) of each of the 18 conjectures in his 2001 book "Combinatorial Optimization: Packing and Covering". Six of the conjectures have been resolved so far, five - by Maria Chudnovsky, Paul Seymour and coauthors.
-
Guy, Unsolved Problems In Number Theory, 3rd edition, A12 (page 45) says "Selfridge, Wagstaff & Pomerance offer 500.00 + 100.00 + 20.00 for a composite $n\equiv3{\rm\ or\ }7\pmod{10}$ which divides both $2^n-2$ and the Fibonacci number $u_{n+1}$ or 20.00 + 100.00 + 500.00 for a proof that there is no such $n$." John Selfridge having left us, I do not know whether his part of the offer still applies.
-
1
"John Selfridge having left us..." Oh dear, I didn't know that. He was my office-mate for a while when I was at Macquarie, and we had a lot of fun together. RIP... – Todd Trimble May 28 2011 at 11:11
I remember reading in Havil's book Gamma that supposidly Hardy was willing to offer his Savilian Chair at Oxford University to anyone who could prove that the Euler Mascheroni constant is irrational. I wonder if this offer still stands?
-
2
Page 52 of Havil's book, but no reference given there. Did Hardy really make this offer, even in jest? – Gerry Myerson May 30 2011 at 0:28
Bill Gasarch is offering \$289 for a 4-coloring of a 17-by-17 grid such that there is no rectangle with all four corners the same color. (No reward for proving one doesn't exist.)
Edit: The above problem, of 4-coloring grids, has since been solved completely; see the comments for the 12x21 case.
Scott Aaaronson offers $200 for an oracle relative to which BQP is not contained in PH, or$100 for an oracle relative to which BQP is not contained in AM.
-
Recently Ian Morrison issued a 100 dollar prize for the construction of an effective divisor on $\overline M_g$ with slope less than 6 (See the recent preprint of Chen, Farkas and Morrison).
-
As it is pointed out in a footnote in A. Zorich's survey "Flat surfaces", Anatole Katok promised (on behalf of the Center for Dynamics and Geometry of Penn State University) a prize of 10,000 euros for the solution of the problem of finding periodic orbits and describing the behavior of generic orbits of billiards in (all/almost all) triangular tables. See page 13 of the ArXiv version of the survey for more comments.
-
Stepan Holub is offering €200 for a solution to the following problem:
Do there exist $n\ge 2$ and nonempty words $u_1,\ldots,u_n$ such that
$(u_1 \ldots u_n)^2 = u_1^2 \ldots u_n^2$ and
$(u_1 \ldots u_n)^3 = u_1^3 \ldots u_n^3$
but the $u_i$ don't all commute with one another?
(I guess the $n\ge 2$ and nonemptiness requirement are technically redundant.)
Source: https://www.student.cs.uwaterloo.ca/~cs462/openproblems.html
-
I'm a computer scientist by trade, but I really just enjoy working on open mathematical problems in my free time. Kimberling's page is pretty nice as you mentioned, I was able to knock one of them out (a minor one, the Swappage Problem), and always look forward to when new ones are posted.
One of the best places I have found for open problems is probably open problem garden. The site is frequently updated with new problems ranging from graph theory, theoretical computer science, algebra, etc. The nice thing is that the problems are also ranked by relative difficulty.
As for cash: A number of the problems DO offer some type of cash bounty (as clearly indicated in the summary section for the problem next to "Prize" text if it exists). Problems such as The Erdos-Turan conjecture on additive bases offer cash incentives for solving.
There are also other problems listed that offer monetary compensation and are posted periodically throughout the site. However, sifting through the problems can be time consuming as many of them do not offer cash incentives.
-
4
If you can edit the answer so that it properly answers the question, please do so. Other MathOverflow questions talk about open problem lists; this one talks about those with a cash bounty (historical or present). Gerhard "Email Me About System Design" Paseman, 2011.06.20 – Gerhard Paseman Jun 21 2011 at 4:29
While not all the problems listed here offer cash bounty, a number of them do. I'll rephrase the post accordingly. – Vincent Russo Jun 21 2011 at 16:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 158, "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": 0.9418177008628845, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/155380/why-accept-the-axiom-of-infinity | # Why accept the axiom of infinity?
According to my readings, Russell showed that a principle Frege used to reduce Peano arithmetic to logic lead to a contradiction. So, Russell tried to reduce mathematics to logic a different way but realized that he needed things such as an infinity axiom to get his reduction off the ground. But of course, that there are infinite collections is not just a matter of logic. So it seems that Russell just had to stipulate an infinity axiom. (This is just background.)
So, in modern set theory, is the axiom of infinity just stipulated? Or is there an argument for its truth?
Some directions:
G Boolos derived the ZFC axioms from the iterative conception of set, and thus gave a motivation or argument in favor of the axiom of infinity.
Or, someone might think, as Cantor did, that all consistent mathematical results have (material?) instantiations in nature. Much of mathematics is dependent on the natural numbers, the real numbers, etc., and thus there is reason to accept axioms of infinity.
There are some similar threads to mine:
-
2
Do you want induction? Do you want to be able to speak about the set of natural numbers in order to state induction? (This isn't a completely rhetorical question. Some people reject the full strength of induction precisely because they don't believe the natural numbers ought to exist as a "completed infinity.") – Qiaochu Yuan Jun 7 '12 at 23:21
5
I'm a little confused, because all axioms are, by definition, stipulated. – Neal Jun 7 '12 at 23:23
11
I don't think the existence of anything follows from logic alone. – Michael Greinecker Jun 7 '12 at 23:33
1
@William: well, a model of PA is an infinite set even if PA doesn't talk directly about infinite sets. So if you want set theory to be capable of constructing models of PA... – Qiaochu Yuan Jun 7 '12 at 23:54
3
@pichael The axiom of infinity is great because it allows you to consider more interesting aspect of ordinals and cardinals. However, plenty of math can be done in first order arithmetics or ZF - Inf which has model the hereditarily finite sets. Here you just have to take the finitist view that HF or $\mathbb{N}$ is just a abbreviation and not a real object within the system. – William Jun 8 '12 at 0:54
show 8 more comments
## 6 Answers
A finitist, who rejects the axiom of infinity, will be denied both the Dedekind and Cauchy constructions of real numbers. The problem? The reals are uncountable (Cantor showed this), and in a finitist's universe, the universe itself is countable.
$V_0=\varnothing$
$V_{n+1}= \mathscr P (V_n)$ where $\mathscr P$ denotes the powerset.
The universe if we accept the negation of the axiom of infinity is $V_\omega=\bigcup_{x\in\omega} V_x$, a countable union of at most countable sets.
-
1
Note that the problem with the Cauchy and Dedekind Cut construction in finitistic mathematics come well before Cantors uncountability argument. Finistist have $\omega$, as an abbreviation, but $\omega$ is not a real object. However in the Cauchy Construction, your objects are equivalent class of sequence. Sequences are function on $\omega$. So you need $\omega$ to exists for this construction. Similary, Dedekind Cut construction are pairs of infinite sets of rational numbers, which again do not exists in finite math. – William Jun 8 '12 at 6:36
1
I am not sure if a finitist would even believe in the Cantor Diagonalization argument since it require you to assume the existence of a bijection from $\omega \rightarrow \mathbb{R}$ and they do not believe that either exists as a formal object. – William Jun 8 '12 at 6:39
@William: It doesn’t assume the existence of such a bijection; it’s a machine that when given a function from $\omega$ to $\Bbb R$ constructs a real number not in the range of the function. (This may be just as objectionable to a finitist, of course.) – Brian M. Scott Jun 8 '12 at 6:59
@BrianM.Scott What is a machine? Finistic certainly believe in Turing Machines. However, how will you represent a real number. Finistic believe in anything that can be encoded as a natural number. But if you encode a real number as decimal expansion, binary expansion, equivalence class of Cauchy Sequence, Dedekind cut, you will formally require the existence of at least $\omega$. Functions on $\omega$ are also infinite objects. – William Jun 8 '12 at 7:04
2
@William: There is absolutely no need to assume a bijection. The diagonal argument shows directly, without any appeal to contradiction, that no function from $\omega$ to $\Bbb R$ is surjective. – Brian M. Scott Jun 8 '12 at 7:49
show 6 more comments
It is possible and interesting to have as our axiom system the usual ZF, but with the Axiom of Infinity replaced by its negation. The resulting theory, which one might call the theory of (hereditarily) finite sets, turns out to be bi-interpretable with first-order Peano arithmetic.
-
If you believe in the set of natural numbers you already accept the axiom of infinity. For most mathematician existence of the set of natural numbers is an intuitively clear fact that doesn't need an argument so mathematicians are typically not bothered with the axiom. In addition lots of classical mathematics depends on such infinite concepts.
The question of accepting or rejecting such an axiom is mainly interesting for philosophers not mathematicians. One can reject the axiom of infinity (such people are often called finitists) but most mathematicians do not. They believe in the existence of the set of natural numbers and therefore see the axiom of infinity as a trivially true fact.
-
As a formalist, I echo the comment of Neal that says "axioms are, by definition, stipulated". The real question is whether the structure so axiomatized is interesting, or worth study.
As many mathematicians make arguments that involve a set of natural numbers, it follows that a good universe of sets should include a set of natural numbers. Whether its existence is an axiom or a theorem is not really important; that's merely matter of exposition. Exposition is important of course, but I wanted to emphasize that's all it is.
Now, I also like to observe the close analogy between logic and set theory. IMO, that's really the main reason why set theory has attained the importance it has: it's really just a systematization of the things we like to do with logic.
For example, if I decide to play at being a finitist and consider finite set theory, I can still meaningfully say the word "natural number", and consider predicates like "$x$ is a natural number". This lets me say things about the class of natural numbers, consider class functions on the natural numbers like $f(x) = x+1$, consider predicates on the class of natural numbers like "E(x) := $x$ is even", quantify predicates over the natural numbers such as $\forall x \in \mathbb{N} : E(x) \leftrightarrow \neg E(x+1)$, and so forth.
And because I can consider predicates like E(x), this means I can also talk about the class of even natural numbers.
If I can consider one variable denoting a natural number, I can consider two variables denoting natural numbers: I can talk about the class $\mathbb{N} \times \mathbb{N}$.
If one is of the sort to focus on such things, one can talk about mechanically translating all such statements into equivalent statements in an untyped language of finite set theory, and keep thinking in the back of one's mind that one is actually working with these untyped statements rather than the more suggestive ones the notation indicates.
While such a translation is possible, and it's a good thing to know, I think the way of thinking is unjustified -- the fact that translation is possible means that you should have no qualms about thinking in terms of the new and suggestive ideas rather than handicapping yourself with a restrictive thought process.
So really, when I'm playing at being a finitist, I still have access to a limited amount of set theory that includes a set of natural numbers. NBG set theory codifies this into having "sets" and "classes". NBG with anti-infinity has two sorts of objects: sets, and classes. The natural numbers would be a (proper) class. I assume that in the presence of anti-infinity, NBG is still 'equivalent' to ZFC.
If I further allow myself to consider second-order logic, I can do more. My objects now include first-order predicates. I can consider the second order predicate "$\varphi(P) := \forall x: P(x) \implies x \in \mathbb{N}$". In other words, I can now talk about subsets of the natural numbers.
If I don't stop there and go all the way up to higher order logic, then I can do this for every type. In the logic-set theory analogy, I now consider power sets. I assert that this means that all of the types I can talk about are organized into a Boolean topos with natural number object -- in other words, that the higher order logic of finite set theory is an instance of the first-order logic of bounded Zermelo set theory (with infinity). Bounded means that I'm never allowed to just say $\forall x: \cdots$ instead, $x$ must be bound to some set/type, as in $\forall x \in T: \cdots$.
Because of all of these things, I am thoroughly unpersuaded when a finitist rejects the notion of a set of natural numbers. I could understand being restrictive in what sorts of things you can do with said set, but rejecting it outright is, IMO, a silly notion which I ascribe more to simply having a contrary attitude than substantive content.
-
I've been asking a lot of question about Skolem's Paradox here. You're post interested me for the following reason. You post mentioned the usefulness of higher-order logics. Now, I've heard (1) that countability is relative in first order logic whereas in second order logic it is not and (2) that people (mathematicians/philosophers) don't reject the relativity of countability on the grounds that second order logic is "better," but for other reasons. Do you know what these "other" (probably more philosophical) reasons are? Do you yourself have a view? Thanks for the great post. – pichael Jun 10 '12 at 1:45
Just as side note, without enough assumptions about natural numbers one cannot manipulate symbols, so formalist already assume syntax manipulations which is similar to manipulation of natural numbers. Also using unbounded quantifiers is typically not considered finitism. – Kaveh Jun 10 '12 at 20:09
1
@pichael: I have no problem with relativity of countability myself. I do not maintain an external, absolute notion of countability but instead accept that each suitable sort of structure -- e.g. a universe of sets -- can define its own meaning, and various different structures simply may or may not agree in situations where they can both speak about an object. There's also another point of view (I'm not sure what I think about it) that cardinality is more about complexity than counting. So the countable model in Skolem's paradox really just lacks the complexity to construct a bijection. – Hurkyl Jun 11 '12 at 5:49
@Kaveh: For the record, throughout the construction, I'm thinking things like 'Cartesian category' or 'topos' in the back of my mind, and the internal logic of such things is indeed bounded. I'm correct in assuming a finitist is allowed to quantify over the natural numbers? – Hurkyl Jun 11 '12 at 5:52
@Hurkyl, the problem is that finitism is not just one specific philosophy. What I remember is that unbounded quantification is not considered finitistic because it is expressing a non-finite statement, there is not much difference between the set of natural numbers satisfying the formula and the formula itself, one doesn't need to consider the extension a set, simply assuming that the formula has a truth value for each particular assignment (of natural numbers to its free variables) would go over what a finitist is ready to accept. Take for example Hilbert's work in logic and proof theory – Kaveh Jun 11 '12 at 15:40
show 1 more comment
Frege's set theory leads to contradictions and we aim to move away from it by using Zermelo set theory or NBG set theory, however to move into these more developed (or more constricted) types, we need to define the universe very well in how our classes and sets behave.
Recall that a class is a basic universe if it follows 6 specific axioms, for the class to be transitive, swelled, to include the empty set and the power set of any x in the class, to include any union of any x in the class and to include {x,y} for any x and y in the class.
We cannot move onto Zermelo set theory or NBG by the definition of the universe I have just given; we add in a new axiom, we accept that the set of natural numbers can be included in our universe, which is now no longer basic and thus a Zermelo universe.
I would argue that the axiom of infinity guarantees the existence of infinite sets, however arguments such as a large amount of work being placed on the natural and real numbers are poor ones for the support and supposed truth of the axiom of infinity, say we find out that this axiom causes trouble to some framework; our arguments for placing huge emphasis on the real and natural numbers do us no favour at this point, rather we use the axiom to construct set theory further.
-
The Axiom of Infinity in ZFC, gives you a successor function $S(x)=x\cup \{ x \}$ from which you can construct the equivalent of the infinite set of natural numbers.
If you find that a bit strange, as I do, another approach is to simply define the natural numbers by means of some equivalent of Peano's Axioms which describe the required properties of a successor function on the set of natural numbers in set-theoretic terms, as you would, say, for the group or ring axioms.
Still another approach is to construct the set of natural numbers starting from the axioms for a complete ordered field (the real numbers).
-
You ought to be able to create the naturals without the reals. Most (all) constructions of the reals rely on the naturals. The flow is in the other direction. – Ross Millikan Jun 13 '12 at 3:25
– Dan Christensen Jun 13 '12 at 3:47
BTW, the Axiom of Infinity has nothing to do with resolving Russell's Paradox. To get around RP, ZFC has the Axiom of Regularity and the Axiom Schema of Separation. – Dan Christensen Jun 13 '12 at 3:57
1
The successor function has nothing to do with the axiom of infinity. It is well definable without it. With or without this axiom the function $x\mapsto S(x)$ is a proper class as it can be applied to all sets in the universe. The axiom of infinity says that there is a set which contains $\varnothing$ and is closed under $S$. Without the axiom of infinity this would simply amount to a class, which may be a set or a proper class depending on the universe. – Asaf Karagila Jun 13 '12 at 15:37
The axiom of infiting simply says there an infinite set, nothing more, no direct relation to successor or natural numbers. – Kaveh Jun 13 '12 at 16:38
show 1 more comment | {"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": 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": 0.9430180191993713, "perplexity_flag": "head"} |
http://wiki.math.toronto.edu/DispersiveWiki/index.php/Hamiltonian | # Hamiltonian
### From DispersiveWiki
A Hamiltonian equation takes the form
$u_t = \nabla_\omega H(u)$
where H is the Hamiltonian, and $\nabla_\omega$ is the (formal) gradient with respect to a symplectic form. | {"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": 2, "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": 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": 0.8726546168327332, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/242199/i-dont-understand-why-x-sum-i-1n-x-i-and-y-sum-i-1n-y-i/242209 | # I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
I was reading the solution to this problem and noticed it used $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$. I think I understand all other parts except this one. Would you please explain why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$?
-
## 2 Answers
$X_k$ and $Y_k$ are always either $0$ or $1$. $X_k=1$ precisely when the $k$-th die shows a $1$, and $Y_k=1$ precisely when the $k$-th die shows a $2$. When you sum the $X_k$’s, you add $1$ for each die that shows a $1$ and $0$ for each die that shows something else, so the total is simply the number of dice showing a $1$. Similarly, when you sum the $Y_k$’s, you add $1$ for each die that shows a $2$ and $0$ for each die that shows something else, so the total is simply the number of dice showing a $2$.
For example, suppose that $n=8$, and the $5$ dice show $1,3,1,2,6,4,2,2$; then $X_1=X_3=1$, and all of the other $X_k$ are $0$, while $Y_4=Y_7=Y_8=1$, and the other five $Y_k$ are all $0$. Thus,
$$\begin{align*} X&=X_1+X_2+X_3+X_4+X_5+X_6+X_7+X_8\\ &=1+0+1+0+0+0+0+0\\ &=2\;, \end{align*}$$
which is indeed the number of dice showing a $1$, and
$$\begin{align*} Y&=Y_1+Y_2+Y_3+Y_4+Y_5+Y_6+Y_7+Y_8\\ &=0+0+0+1+0+0+1+1\\ &=3\;, \end{align*}$$
the number of dice showing a $2$.
-
i've been told that capital X and Y are random variables, I never knew you can obtain the actual value of X and Y by summing $X_k$ and $Y_k$. This summation method to obtain X and Y would not work if X and Y are, say, binomially distributed correct? – user133466 Nov 21 '12 at 19:32
@user133466: The distributions of $X$ and $Y$ are completely determined by those of the $X_k$ and the $Y_k$, respectively; if they’re binomial, they’re binomial, and if not, not. The random variables $X$ and $Y$ are defined to be these sums of other random variables. You can always define random variables in such fashion. – Brian M. Scott Nov 21 '12 at 19:34
say we have $X$ ~ $Bin(n,p)$ what would big $X$ equal to in this case? – user133466 Nov 21 '12 at 19:38
@user133466: I don’t understand the question. As I said, the distribution of $X=\sum_{k=1}^nX_k$ depends entirely on the distributions of the random variables $X_k$, and you’ve not said anything about them. – Brian M. Scott Nov 21 '12 at 19:40
1
@user133466: We define $X$ to be $\sum_{i=1}^nX_i$, so of course it’s equal to $\sum_{i=1}^nX_i$. I’m really not sure what you’re trying to ask. If the random variables $X_i$ are $1$ when something happens and $0$ when it doesn’t, then $X$ will count the number of successes, regardless of the distribution(s) of the $X_i$’s. – Brian M. Scott Nov 21 '12 at 20:08
show 2 more comments
Here $X$ is the number of $1$'s when you roll a fair die $n$ times.
Recall that the random variables $X_1, X_2, X_3, \dots, X_n$ were defined as follows. $X_1=1$ if we get a $1$ on the first roll, and $X_1=0$ otherwise.
$X_2=1$ if we get a $1$ on the second roll, and $X_2=0$ otherwise.
$X_3=1$ if we get a $1$ on the third roll, and $X_3=0$ otherwise.
In general, $X_i=1$ if we get a $1$ on the $i$-th roll, and $X_i=0$ otherwise.
Then $$X_1+X_2+X_3+\cdots+X_n$$ is the total number of times you got a $1$, so it is the number of $1$'s that you got. But $X$ is by definition the number of $1$'s ou got.
Every time you get a $1$ on the die you write down a $1$, every time you don't get a $1$, you write down a $0$. So the sum of the numbers you write down is just the total number of $1$'s that you rolled.
Exactly the same idea works for $Y$. You write down a $1$ if you roll a $2$, and write down a $0$ if you don't roll a $2$. The sum of all the numbers you wrote down is the total number of $2$'s that you rolled.
This trick of using what are called indicator random variables $X_i$ is useful in many calculations.
-
i've been told that capital X and Y are random variables, I never knew you can obtain the actual value of X and Y by summing $X_k$ and $Y_k$. This summation method to obtain X and Y would not work if X and Y are, say, binomially distributed correct? – user133466 Nov 21 '12 at 19:35
Actually, $X$ and $Y$ are binomially distributed random variables. Each time, for $X$, the probability of "success" is $1/6$. Every binomially distributed random variable is the sum of indicator random variables. For example, toss a coin $10$ times in a row. Let $X_i=1$ if on the $i$-th toss you get a head, and $X_i=0$ if you get a tail. Then $X_1+X_2+\cdots+X_{10}$ is the total number of heads. This as you know has binomial distribution. – André Nicolas Nov 21 '12 at 19:40
– user133466 Nov 21 '12 at 19:44
No, sorry. They are used in many ways, sometimes very complicated. The simplest is for finding the mean and variance of the binomial. Since $X=X_1+\cdots+X_n$, we have $E(X)=E(X_1+\cdots+X_n)$. But expectation of a sum is sum of the expectations, and $E(X_i)$ is very easy to find. A not much harder argument gives you the variance of the binomial. – André Nicolas Nov 21 '12 at 19:49
I understand $E(X)=E(X_1+...+X_n)$ but I did not know $X = X_1 + ... +X_n$. Does that mean if we have $X_1,X_2,...X_n$ are independent Poisson, $X$ would still equal to $\sum_{i=1}^n X_i$? – user133466 Nov 21 '12 at 19:58
show 5 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": 110, "mathjax_display_tex": 3, "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": 0.9226962327957153, "perplexity_flag": "head"} |
http://electronics.stackexchange.com/questions/29740/signal-reflections-at-the-driver/29742 | # Signal reflections at the driver?
This question is related to How do I use directional couplers on a differential signal?
I understand that if I have an impedance mismatch at the end of a signal cable, E.G. from a terminator which doesn't match the cable impedance, then I can expect to get reflections back.
What I don't understand is exactly what happens when you have an impedance mismatch at the driver end of the cable.
If a very short part of the cable at the driver end has an unknown impedance, is this a problem? Will is cause noticeable reflections? My instinct would be 'no' because the length is so short.
I mean, this must happen in all situations anyway. The impedance of the driver IC's legs are not impedance matched, but they are so short that it's not a problem. Obviously, the length in that case would be very short indeed. If that's the case, then the question would be: "How long is too long?"
This cable will be carrying 100Mb/s LVDS (which stands for Low Voltage Differential Signal). This is my estimation of the spectrum of the signal.
Which is based on this wave shape:
-
## 1 Answer
I don't think you would get significant reflections. If the load impedance is matched to the transmission line, there would be no reflections on that transmission line. In reality there's a tiny transmission line between the driver load and the long transmission line, so there's an opportunity for reflections there.
Think of it this way: "looking into" the transmission line, you will "see" a certain impedance (the wave impedance). Likewise, "looking into" the driver impedance from the driver itself, you will "see" another impedance.
Your driver transmission line is 15mm. The relative permittivity of most metals is 1, so we can assume the propagation speed is $c$. Your signal is 100 Mb/s, so let's assume a signal frequency of 100 MHz. The wavelength is thus:
$\lambda=c/f=(3 \times 10^{8} \mathrm{m/s})/(100 \times 10^{6} \mathrm{Hz})=3 \mathrm{m}$
What matters is the length of the tiny transmission line in multiples of the wavelength:
$l/\lambda=0.015/3=0.005\lambda$
This is too small to cause significant reflections. While you might get reflections due to the possible mismatch (as explained here, the reflection depends on the impedance values and not the transmission line length), the line is too small for the standing wave pattern to change.
The only thing you have to worry about is the voltage divider. Since the main transmission line is matched, you'll "see" it as a 100 ohm load: this will form a voltage divider with the driving load. For optimal power transfer to the load you'll want to minimize the driver impedance.
EDIT: If you want further reading, I'd recommend a textbook on the subject. I've been reading Fundamentals of Applied Electromagnetics by Ulaby et al., it goes over reflections and matching in detail in Chapter 2.
EDIT 2: I'm not in the field, I'm an EE student and we just happen to be studying this subject at the moment. I'm sure that someone with professional experience could give a more practical answer.
-
Thank you. This is a very helpful answer. – Rocketmagnet Apr 11 '12 at 21:55
– Kortuk♦ Apr 11 '12 at 22:17
As not to be all negative, Thanks for taking the time to visit the site and become brave enough to write an answer. I hope you continue to visit us. – Kortuk♦ Apr 11 '12 at 22:18
Not negative at all, I appreciate the comments. – Jeff E Apr 11 '12 at 22:27
@Kortuk I would love to know your thoughts on this question. – Rocketmagnet Apr 12 '12 at 10:04
show 4 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": 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": 0.9417088627815247, "perplexity_flag": "middle"} |
http://mathhelpforum.com/trigonometry/126452-angular-velocity-problem.html | # Thread:
1. ## Angular velocity problem
A planet takes 60 hours to rotate and the radius of the equator is 7007.910 miles and the circumference of the equator is 14015.820 miles. How would I find the angular velocity of the equator? Is it just 2pi/60hours since one complete rotation is 2pi?
2. Originally Posted by princesasabella
A planet takes 60 hours to rotate and the radius of the equator is 7007.910 miles and the circumference of the equator is 14015.820 miles. How would I find the angular velocity of the equator? Is it just 2pi/60hours since one complete rotation is 2pi?
the angular velocity of any point on the planet (except its poles) is just $2\pi$ radians per 60 hours | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.8802372813224792, "perplexity_flag": "middle"} |
http://mathhelpforum.com/pre-calculus/49373-inequalities-print.html | # inequalities
Printable View
• September 16th 2008, 04:44 PM
xian791
inequalities
2(x-3)-3(x+2)<5
i seriously don't know where to start. I was thinking of putting 5 on the other side, but i would be stuck from there.
P.S. if i do find the numbers for instance, 3,-3, 8. I would have to graph them on a number line and see if its a positive or negative which i know how to do but i am confused on how to figure out the final answer from that
• September 16th 2008, 04:59 PM
11rdc11
$2(x-3)-3(x+2)<5$
1) Distribute
$2x-6 -3x-6<6$
2) Combine like terms
$-x<18$
3) Divide by -1 and remember to change your sign
$x>-18$
• September 16th 2008, 05:04 PM
xian791
so there is only 1 answer? in the questions my teacher showed us there were mutiply answers like x^2(x-16)(x-3)<=0 and the answer was 3<=x<=16
and how you get <6 when it was <5?
• September 16th 2008, 05:09 PM
11rdc11
The amount of answers depend on the degree of the exponent. So in this case it was to the degree of 1 so you had 1 answer. Now $x^2$ is to the degree of 2 so you will have at most 2 solutions
All times are GMT -8. The time now is 04:54 PM.
Copyright © 2005-2013 Math Help Forum. All rights reserved.
Copyright © 2005-2013 Math Help Forum. All rights reserved. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9651302695274353, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?p=4229546 | Physics Forums
## What is a tensor?
I'm a bit confused not just on what exactly a tensor is but also how I should be thinking about a tensor. I figure it's just a vector's bigger brother, but I'm just trying to wrap my head around it.
Any explanations are appreciated.
Thanks in advance.
I figure it's just a vector's bigger brother
A vector's bigger brother is actually a matrix. A tensor is like the bad uncle that used to bully you and try to make you think that you were stupid because you couldn't tell the difference between co-variance and contra-variance.
Quote by DiracPool A vector's bigger brother is actually a matrix. A tensor is like the bad uncle that used to bully you and try to make you think that you were stupid because you couldn't tell the difference between co-variance and contra-variance.
I'll try harder, uncle Paul -- I promise!
But really, this tensor business is boggling me.
Mentor
## What is a tensor?
If your teacher has told you something like "it's something that transforms under rotations according to the tensor transformation law", it's understandable that you're confused, because what does that statement even mean? It's been almost 20 years since someone tried to explain tensors to me that way, and it still really irritates me when I think about it.
I will explain one way to interpret that statement for the case of "covariant vectors" and "contravariant vectors". (I really hate those terms).
Consider a function T that takes coordinate systems to ordered triples of real numbers. We can choose to denote the members of the triple corresponding to a coordinate system S by T(S)i, or by T(S)i. If S' is another coordinate system, that can be obtained from S by a rotation, then it might be useful to have formulas for T(S')i in terms of the T(S)i, and for T(S')i in terms of the T(S)i. Those formulas can be thought of as describing how (T(S)1, T(S)2, T(S)3) and (T(S)1, T(S)2, T(S)3) "transform" when the coordinate system is changed from S to S'. It's possible that neither of those formulas looks anything like the tensor transformation law, but it's also possible that one of them looks exactly like the tensor transformation law.
If that's the case with the formula for T(S')i, then some people call the triple (T(S)1, T(S)2, T(S)3) a "contravariant vector". And if it's the case with the other formula, then some people call the triple (T(S)1, T(S)2, T(S)3) a "covariant vector". I think that terminology is absolutely horrendous. The information about how the triple will "transform" isn't present in the triple itself. It's in the function T and the choice about where to put the indices. So if anything here should be called a "contravariant vector", it should be something like the pair (T,"upstairs"). I also don't like that this defines "vector" in a way that has nothing to do with the standard definition in mathematics: A vector is a member of a vector space.
If you understand this, then it's not too hard to understand how this generalizes to other tensors. It's just hard to explain it.
I consider this entire approach to tensors obsolete. A much better approach is to define tensors as multilinear functions that take tuples of tangent vectors and cotangent vectors to real numbers. This post and the ones linked to at the end explain some of the basics. (If you read it, skip the first two paragraphs. You can also ignore everything that mentions the metric if you want to. What you need is an understanding of the terms "manifold", "tangent space", "cotangent space" and "dual basis").
Quote by Fredrik Let V be a finite-dimensional vector space, and V* its dual space. A tensor of type (n,m) is a multilinear map $$T:\underbrace{V^*\times\cdots\times V^*}_{\text{$n$ factors}}\times\underbrace{V\times\cdots\times V}_{\text{$m$ factors}}\rightarrow\mathbb R.$$ The components of the tensor in a basis $\{e_i\}$ for V are the numbers $$T^{i_1\dots i_n}{}_{j_1\dots j_m}=T(e^{i_1},\dots,e^{i_n},e_{j_1},\dots,e_{j_m}),$$ where the $e^i$ are members of the dual basis of $\{e_i\}$. The multilinearity of T ensures that the components will change in a certain way when you change the basis. The rule that describes that change is called "the tensor transformation law".
My explanation is that a tensor is an n dimensional grid of numbers. Since a scalar is a 0d grid, a vector is a 1d grid, a matrix is a 2d grid, we can say that a tensor is any of these, plus 3d grids, 4d grids etc. But it isn't just a grid of numbers, it also has a few operations on it, such as addition and multiplication, which work like matrix algebra in the 2d case, and real number algebra in the 0d case, etc.
Also, you might want to consider the analog to the Cartesian product of vector spaces (or linear spaces) to get a "multi"-linear object (hence the term multi-linear algebra which is another word for tensors).
Recognitions:
Gold Member
Science Advisor
Staff Emeritus
Quote by DiracPool A vector's bigger brother is actually a matrix. A tensor is like the bad uncle that used to bully you and try to make you think that you were stupid because you couldn't tell the difference between co-variance and contra-variance.
Quote by TGlad My explanation is that a tensor is an n dimensional grid of numbers. Since a scalar is a 0d grid, a vector is a 1d grid, a matrix is a 2d grid, we can say that a tensor is any of these, plus 3d grids, 4d grids etc. But it isn't just a grid of numbers, it also has a few operations on it, such as addition and multiplication, which work like matrix algebra in the 2d case, and real number algebra in the 0d case, etc.
These are both quite a bit misleading. In a given coordinate system, we can represent a vector as a "list" of numbers and we can represent a tensor by a rectanguar grid of numbers or matrix. But the crucial point of both vectors and tensors (technically, a vector is a tensor) is that they have an existence independent of any given coordinate system. And the change as we change from one coordinate system to another is "linear homogeneous". One reason that is important is that if a tensor happens to be all 0s in one coordinate system, it is all 0s in any coordinate system. So if we have an equation, say A= B where A and B are tensors, in one coordinate system, we can write that as A- B= 0 so that A- B= 0, i.e. A= B in any coordinate system: tensor equations are independent of the coordinate system.
Mentor
Quote by TGlad My explanation is that a tensor is an n dimensional grid of numbers. Since a scalar is a 0d grid, a vector is a 1d grid, a matrix is a 2d grid, we can say that a tensor is any of these, plus 3d grids, 4d grids etc. But it isn't just a grid of numbers, it also has a few operations on it, such as addition and multiplication, which work like matrix algebra in the 2d case, and real number algebra in the 0d case, etc.
You left out the most important thing, which is that one of these "grids" isn't enough. There must be one for each coordinate system. And the relationship between the "grids" associated with two different coordinate systems must be given by the tensor transformation law.
But the crucial point of both vectors and tensors is that they have an existence independent of any given coordinate system.
So is calling a real a number misleading? since reals have an existence independent of coordinate system (which is just scale in 1d).
There must be one for each coordinate system. And the relationship between the "grids" associated with two different coordinate systems must be given by the tensor transformation law
I did mention that tensors come with a set of operators that have specific rules, which are basically like real, vector and matrix algebra in the 0d, 1d and 2d cases. I didn't suggest that they are just data structures.
Are you suggesting that you need something more for tensors that you don't need for say just matrices? Matrices can be transformed between coordinate systems.
Mentor
Quote by TGlad Are you suggesting that you need something more for tensors that you don't need for say just matrices?
Yes. A matrix is not a tensor. You can however define a tensor by specifying an n×n matrix, a coordinate system and the tensor type (i.e. if you want the tensor to be a map from V×V into ℝ, from V*×V into ℝ, from V×V* into ℝ, or from V*×V* into ℝ), because there's exactly one tensor of each type that has components in the chosen coordinate system that are equal to the components of the matrix.
Quote by TGlad Matrices can be transformed between coordinate systems.
Right, ##A'=RAR^{-1}##. You can define a type (1,1) tensor by saying that that we associate each rotation matrix with a coordinate system, and that we for each rotation matrix R, associate the matrix ##RAR^{-1}## with the coordinate system associated with R. But you can also define a type (2,0) tensor by instead associating the matrix ##RAR## with the coordinate system associated with R.
Thread Tools
| | | |
|----------------------------------------|------------------------------|---------|
| Similar Threads for: What is a tensor? | | |
| Thread | Forum | Replies |
| | Calculus & Beyond Homework | 2 |
| | Differential Geometry | 6 |
| | General Physics | 2 |
| | Special & General Relativity | 5 |
| | Calculus & Beyond Homework | 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": 3, "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": 0.9591342806816101, "perplexity_flag": "middle"} |
http://gravityandlevity.wordpress.com/2010/01/30/the-universe-is-a-giant-energy-minimization-machine/ | ## A blog about the big ideas in physics, plus a few other things
Physics, at its worst, can be tremendously complicated. Sometimes learning physics (or any other branch of science) can seem having to memorize a giant catalogue of phenomena, each discovered by some guy who got an equation named after him. It can seem like if I want to solve a problem, then I need to begin by preparing a huge cheat sheet of possible relations and dependencies that have to be taken into account. For example, if I were solving a problem in atmospheric physics, I might feel compelled to make a list like this:
1. Remember that charged particles drift in crossed electric and magnetic fields (…formula…)
2. Remember the coriolis acceleration (…formula…)
3. Remember relativistic effects (…formula…)
4. Remember that ionization depends exponentially on temperature (…formula…)
5. etc.
Apparently, solving a problem in atmospheric physics requires me to remember and understand all these effects, plus dozens of others, and decide how each will affect the problem. What an ugly mess! I would never have chosen to become a scientist if I knew it would be like that. And sometimes, for sure, it is. I have definitely read some ugly scientific papers, in which a bunch of different effects (each with its own formula) get patched together to form some awful and unintelligible Frankenstein theory.
That’s not to say that all the different effects in physics aren’t interesting. They usually are, and discovering a new phenomenon is the rush that everyone studies physics for. But no one wants to approach every problem by making a long list of things they have to remember and then making sure that each one gets its say in the answer.
Luckily, physics doesn’t actually have to be like this. Strangely enough, I find that the more I study physics and expand my “catalogue of effects”, the more I end up approaching every problem the exact same way. That is, instead of adhering to a big messy list, I am guided by a single principle:
• Remember that the universe is a giant energy minimization machine
$\hspace{10mm}$
I’ve talked before about how energy is the most important concept in physics. But I’ve realized lately that the degree to which I rely on it for solving problems is kind of astonishing. In fact, these days I approach just about every problem using the same three steps:
1. Draw a big picture and write out all the things that can possibly happen (usually by defining some mathematical variables).
2. Decide how much energy the various possibilities require (usually by writing down some equations).
3. Find out which possibility has the lowest energy (usually by minimizing those equations). This is what will actually happen.
Now, these steps can be deceptively hard, but the approach is remarkably simple. If you want to know what will happen to some object or set of objects in the future, you just have to remember this: of all the options available, the universe will choose the one with the lowest (free) energy. It may be hard to describe how that thing will happen, but you can rest assured that eventually it will.
In this post and the next one, I want to give an example or two of surprising realizations you can reach by adhering to the doctrine that the universe will somehow find a way to minimize its total energy.
$\hspace{10mm}$
$\hspace{10mm}$
Example #1: a mysterious force of suction
Imagine performing the following experiment: you take two flat plates of material and give them equal and opposite electric charges (if you want, you can do it the way they did it in the 1600s: by rubbing a piece of glass against a piece of resin). Once you’ve done that, take your plates to a big reservoir of water, and hold them parallel to each other so that their bottom edges just barely touch the surface. Like so:
The lines here indicate the presence of electric field: since there are positive charges next to negative charges, there is an electric field going from positive to negative. If the plates are fairly large relative to the area between them, then the electric field lines will be pretty much parallel and horizontal.
This is the end of the experimental setup. What will happen?
At first, it looks like nothing should happen at all. The plates attract each other, but you are holding them in place so that they can’t move. The only other possibility is that maybe the water could be pushed around by the electric field. But it doesn’t seem like this should happen either, since the electric field goes in straight lines and apparently doesn’t even touch the water.
But try approaching this from the perspective I advocated in the introduction. Don’t ask “what forces are pushing things around?”. Instead, ask “how could the universe lower its energy?”. The answer to that question is a little more interesting.
In this case, there is some energy to be saved by bringing water in between the charged plates. That’s because water has a high dielectric constant (about 80 times larger than air), so if there is water in between the plates then the strength of the electric field is diminished. And electric field is a type of energy (with energy density $\vec{E}^2/2$), so if you reduce its magnitude then you lower the energy of the universe. You can also think of it this way: water molecules are polar objects, with a positive end and a negative end. Each of them would gain some energy by being in the presence of en electric field, where they can point their positive ends toward the negative charge and their negative ends toward the positive charge.
Apparently, then, there is some energy to be gained if the water spontaneously jumps up in between the two plates. And, in fact, it does, like this:
This is a conclusion that we might not have reached from the beginning. There don’t seem to be any strong forces pushing the water upward. But the water gets sucked up nonetheless, because the energy of the universe can be lowered when it does.
If you want to figure out how high the water level rises in between the plates, then you have to balance this electrostatic energy savings with the cost of picking up all that water against the earth’s gravity. You can do so by writing down the total energy as a function of the water height $h$ and then finding out which value of $h$ minimizes the total energy. For anyone following along at home, I’ll give the final answer:
$h = \frac{79}{160} L \frac{Q^2 d}{2 \epsilon_0 L^2 g \rho A^2}$.
Here, $L$ is the height of the plates, $Q$ is the charge on them, $d$ is the distance between them, $\epsilon_0$ is the vacuum permitivity, $g = 9.81 m/s^2$ is the acceleration due to gravity, and $\rho$ is the density of water.
$\hspace{10mm}$
You may find this answer pretty unsatisfying; certainly I did when I first encountered it. There seems to be a contradiction: at the beginning of the problem there is no force acting on the water molecules, but somehow they rise up to fill the space between the plates. How is that possible?
The answer is that there is, in fact, a force on the water molecules. My description above of the electric field — that it goes in straight lines and exists only between the two plates — isn’t really correct. The electric field bends a little bit at the edges and at the boundary of the water surface. This bending is enough to provide a force that pushes the water upward and holds it there once it has risen. This a subtle point. But, amazingly, we didn’t need to know it in order to figure out what would happen (or even to calculate the magnitude of the force). Just knowing the general behavior of the electric field, and how much energy it stored, was enough to figure everything out.
$\hspace{10mm}$
$\hspace{10mm}$
For those of my readers with a background/inclination toward economics, I pose the following question: do you think about economics problems in a similar way? Do you approach them with a similar guiding principle, like “a person or population will do whatever results in the maximum income?”, even when it’s not clear how or why they should do that thing?
When I taught introductory physics, I liked to state the parallels between energy and money explicitly: it isn’t created or destroyed, and everyone is trying to get as much (or for energy, as little) of it as possible.
### Like this:
from → physics
17 Comments leave one →
1. January 30, 2010 1:02 pm
As an economist I like the idea of finding an analogous ‘law of energy minimization’. My tuppence worth for discussion:
‘Money flows to scarcity’
or
‘All invisible costs want to become visible’
or
‘The optimum choice minimises opportunity cost’
… thanks for the inspiration!
2. January 30, 2010 1:09 pm
Energy is for statics. The universe is dynamic, and seeks to extremize action!
• gravityandlevity *
January 30, 2010 11:55 pm
I really do need to write a post about the least action principle. It ties in closely to all my talk about “the quantum field”.
But I have to admit that I still struggle with it a little bit from a philosophical standpoint.
3. January 30, 2010 6:49 pm
This is interesting. Yes, economists tend to think of competitive markets as automatically accomplishing the efficient outcome (generating the most total good)…but there are caveats and caveats and caveats. Would you say that physics is free of these, or just free enough that the energy minimization principle is usually a pretty good guide?
We could compare the ways markets fail to their analogs in the physical world. For one, I suppose the universe doesn’t need to worry about whether property rights are well-defined, and other human constructs. Perfect information isn’t a problem in the universe either; gravity just acts between two bodies, unconditionally. Another way markets can fail is if there are too few agents (e.g. monopoly is inefficient). Do you ever run into issues in physical situations where there aren’t so many zillions of molecules bouncing around? Maybe you can confirm or correct my physics speculations above.
It seems that physicists are in pretty good shape. But it occurs to me that there’s one really fortunate area we’ve got you beat. If it costs a company \$100 up front to make a product it can turn around and sell for \$1,000, you can bet it’s going to do it. But because physical objects are not forward-looking, they can’t get over the activation energy hump by themselves. The reason it’s fortunate is that it’s _good_ for society when the firm creates \$900 of surplus, whereas on reflection, a universe that truly minimized energy in this way would have…shall we say…some problems.
• gravityandlevity *
January 31, 2010 12:10 am
There are a lot of good talking points here.
First, I should probably say that in practice the law of decrease in energy is primarily a statistical one. When there is no thermal energy (zero temperature), it is strictly obeyed. But at finite energy, it is only true because of the law of large numbers, because the average energy of a zillion interacting particles becomes very sharply defined. In the study of a small number of particles moving around, the behavior has to be described in a less deterministic, and more probabilistic, way.
Your second point, about humans being forward-looking, is also a pretty interesting one. I would say that quantum mechanics does in fact allow for particles to be “forward-looking”. That is, a particle trapped in a potential well can borrow energy from the environment in which it sits in order to jump over any barriers and achieve a lower energy. In fact, philosophically, quantum mechanics predicts that every such barrier will eventually be overcome. But the rate at which a particle can jump (tunnel) through the barrier depends exponentially on how much energy it has to borrow from the universe in order to do so. What’s more, all the energy it borrows must be returned.
Out of curiosity, has anyone ever noticed a similar relationship in economics? If a person can only make money by first borrowing some large sum, does the rate at which they succeed decay exponentially with the amount they are required to borrow?
• January 31, 2010 7:01 pm
That’s interesting about the quantum borrowing. I don’t know that exponential decay makes a lot of sense in the general context of success rates when borrowing money etc., but I wouldn’t be surprised if it showed up in some particular, related context.
Speaking of zillions of molecules, here’s an excerpt from the preface of an old textbook by Donald McCloskey:
“The governing image of ersatz price theory is that of one person cheating another, taking ‘unfair’ advantage; that of real price theory is of many people trying to cheat all the others, but in fact helping them. The unintended consequence of selfish behavior is altruistic; the apparent chaos of competition, unplanned by moral or civil law, leads to orderly social change; direct attempts to help this or that person are thwarted by the logic of economic events. Such are the paradoxes in which economists delight. The key to the paradoxes is that each person’s behavior is constrained by all others’ behavior. The person is a molecule in a social gas bumping against other molecules, unable to move in a selfish straight line. The theory of price is based on methodological individualism, adding up the bumping molecules.”
4. January 30, 2010 7:07 pm
The late Paul Samuelson was express about the influence of minimization principles on his work. I will try to find a quote.
The concept of spontaneous symmetry breaking in far from equilibrium systems has been a guide for me in noticing many macro and micro dynamics that others ignore.
I have on my blog pointed out a few places where post-Feynman work on minimzation in far-from-equilibrium systems might be applied fruitfully in economics. Using the Ising Model to understand how leverage can result in synchronized market-wide buying/selling, for example.
I think minimization principles generalize to social systems. The concept of a general equilibrium in neoclassical economics is predicated on extremization of aggregate utility –a kind of functional.
• gravityandlevity *
January 31, 2010 12:13 am
Wow, that’s interesting. I hadn’t realized that economists had adapted these kinds of ideas from physics to such sophisticated levels.
And if you do find the quote, please post it. I’d be interested to hear.
5. January 30, 2010 7:30 pm
@Gerard: “Money flows to scarcity”? That seems to contradict the observation that large concentrations of money are used to buy influence over markets which serves to further concentrate money. If anything, money behaves more like gravity than an ideal gas, in that existing inequalities are magnified over time.
@Xan: I think the difference is, as I stated before, physics is dynamic. That is, when everything is stable, you can minimize energy, and ideal markets will reach steady-states in efficient outcomes. But the real work is what happens when the world changes? It’s the process of getting to the steady states that’s interesting, and there are a lot more hooks in human interactions than in physics that prevent the transition from one state to another from proceeding smoothly.
• Jonathan Gardner
February 1, 2010 12:08 am
#1 Your perception of markets is wrong then. It’s much cheaper and profitable to “go with the flow” than to try and manipulate entire markets. Successful businesses do not need to manipulate the market, except to satiate demand and increase supply.
#2 Regarding the “dynamic” nature of physics, consider that time is just another variable along with position, and suddenly, nothing is dynamic. Rather dynamic simply means, “Now add in the variable ‘t’ and solve again”. A whole great piece of physics is the study of static “motion” and how it directly applies to dynamic motion.
Also, consider that ultimately, all energy is turned into heat thanks to entropy, and the neat little patterns of motion that particles make soon ends up being a pretty consistent porridge. A very large class of problems are accurately solved by G&L’s method, which is extraordinarily simple and easy to apply in the real world. Use it to understand why your gas mileage diminishes on an upward climb, or diminishes as you drive faster, etc… and you will quickly see its power.
• February 1, 2010 11:27 am
Maybe you can explain what the adage means? Your response sounds nonsensical given the plain reading of the words. I had a feeling that it was jargony, which is why I said “seems”.
As for statics vs. dynamics, that works fine for high school and intro college stuff. Throw in some general relativity, some quantum mechanics, and suddenly you’re talking Cauchy data and time is very different in practice.
But even within basic classical mechanics, when you include time it’s not energy that’s extremized, but action.
6. February 1, 2010 8:03 am
Here is the simplest possible example from economics (followed by a few words on more sophisticated examples):
Suppose a world with one lettuce seller and one lettuce buyer. The cost (to the seller) of producing x heads of lettuce is C(x). The value (to the buyer) of x heads of lettuce is V(x).
Given a price p, the seller chooses a quantity x to maximize his profit, which is px-C(x). The buyer choose a quantity x to maximize his surplus, which is V(x)-px. The economy is in equilibrium if the seller wants to sell the amount the buyer wants to buy.
One way to compute the equilibrium is to write down the first order conditions for the seller’s and the buyer’s problems. If x1 and x2 solve those problems, the conditions are: C’(xs)=p for the seller and V’(xd)=p for the buyer.
So to find the equilibrium, we must solve three equations in the three unknowns xs, xd, and p, namely:
1) C’(xs)=p
2) V’(xd)=p
3) xs=xd
But a faster way to find the equilibrium quantities is to pose a different problem: The “planner’s problem” is to maximize social welfare, defined as V(x)-C(x). The value of x that solves this problem is equal to the common value of xs and xd in equilibrium.
This is an extremely simple example (though it already has profound implications about the desirable properties of competitive equilibrium). But in more sophisticated examples, incorporating infinite numbers of goods, goods delivered at different times, decision making in the face of uncertainty, etc., it can be essentially impossible to compute the competitive directly—and in these cases, a common strategy is to formulate an appropriate “planner’s problem”, prove that every competitive equilibrium is a solution to the planner’s problem, and then solve the planner’s problem.
What you’ve done with your water example is nearly perfectly analogous to this standard technique in economics.
• gravityandlevity *
February 1, 2010 10:12 am
Thanks Steven, I like that example a lot, and the paradigm of the “planner’s problem”. I had a feeling that there must be some analogue of “solving by force equilibration” and “solving by energy minimization” in economics, and I’m glad to see that there is.
7. February 1, 2010 4:58 pm
Sorry I can’t post a full excerpt. See page 21 of Foundations of Economic Analysis by Samuelson
Although the analogy is not complete, one could call Hal Varian’s nonparametric tests of the Weak Axiom Revealed Preferences in the early ’80s a kind of calculus of variations. Unlike an exact analysis, those tests only establish that WARP is not violated by aggregate data about consumer behavior, making WARP a kind of mean-field approximation to the behavior of individual consumers within the aggregate.
On my view, all the action in economics will be in making perturbative corrections to those mean-field approximations. The behavioralist critique, which goes after the rational hypothesis itself, is built up from a wholly different set of data, and ill-suited to correcting the approximations of neoclassical theory.
Almost all of what I just wrote is speculative and personal. If you find another person who understands both sets of theories (i.e., both field theory and neoclassical economic theory), and they disagree, I’d like to know.
8. February 12, 2010 5:25 pm
This is a particularly good review article.
http://rsta.royalsocietypublishing.org/content/368/1914/1175
9. Lee
March 6, 2010 9:58 am
Are you missing a factor of the relative permittivity in your formula for h? It seems that odd that the whole effect is based on the reduction of energy caused by relative permittivity and I don’t see it in your expression. It is this energy gain that must be balanced with the energy loss due to the increased gravitational potential.
Or is that already used in computing the front coefficient 79/160. It could be (e-1)/2e, assuming e = 80.
• gravityandlevity *
March 9, 2010 8:38 am
Yes, that’s right. My 79/160 is actually (e – 1)/2e, where e is the dielectric constant of water. I suppose it would have been easy enough to write that out, but the equation was kind of messy already.
%d bloggers like this: | {"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": 16, "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": 0.9413204789161682, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/6708/proof-of-the-standard-pseudorandom-generator-xor-encryption-scheme-in-goldreic | # Proof of the standard pseudorandom generator + XOR encryption scheme in Goldreich
Reading Goldreich's Foundations of Cryptography II, I found this proof for the security of the common pseudorandom generator + XOR encryption scheme (Proposition 5.2.12 in the book):
Assume you have a secure pseudorandom generator $g \colon \lbrace 0,1 \rbrace^n \to \lbrace 0,1 \rbrace^m$, and you construct your encryption scheme in the standard way by XOR'ing the output of $g$ to a message of length $m$. To prove that this is one-message-IND-secure, Goldreich proceeds as follows:
1. Assume, for the sake of contrdiction, that there is an adversary $\mathcal{A}$ (really a poly-sized circuit in the book, but that is not important here) against this encryption scheme, such that for $x, y \in \lbrace 0,1 \rbrace^m$, $U_n \in \lbrace 0,1 \rbrace^n$ and polynomial $p$ we have: $$|Pr[\mathcal{A}(x \oplus g(U_n)) = 1] - Pr[\mathcal{A}(y \oplus g(U_n)) = 1] > \frac{1}{p(n)}.$$
2. Observe that for $U_m \in \lbrace 0,1 \rbrace^m$ we have: $$Pr[\mathcal{A}(x \oplus U_m) = 1] = Pr[\mathcal{A}(y \oplus U_m) = 1].$$
3. Then, WLOG: $$|Pr[\mathcal{A}(x \oplus g(U_n)) = 1] - Pr[\mathcal{A}(x \oplus U_m) = 1]| > \frac{1}{2p(n)},$$ which contradicts the security of $g$ (since we can distinguish its output from random).
My question
How do you get to that last inequality in step 3? In particular, how do you derive the term $\frac{1}{2p(n)}$?
-
## 1 Answer
The inequality is obtained by a distance argument.
Consider two points $X,Y$ on the real line. Taking another point $Z$, you have $|X-Z| + |Y-Z| \geq |X-Z+Z-Y| = |X-Y|$.
Applying this "triangle" inequality to your equality 1, we have for any $z \in \mathbb{R}$, $\begin{array}{l} \bigl\lvert\Pr[A(x\oplus g(U_n))=1] - z\bigr\rvert + \bigl\lvert \Pr[A(y\oplus g(U_n))=1]- z\bigl\rvert \\ \geq \bigl\lvert\Pr[A(x\oplus g(U_n))=1] - z + z - \Pr[A(y\oplus g(U_n))=1]\bigr\rvert \\ = \bigl\lvert\Pr[A(x\oplus g(U_n))=1]-\Pr[A(y\oplus g(U_n))=1]\bigr\rvert \\ > 1/p(n). \\ \end{array}$
By setting $z = \Pr[A(x \oplus U_m)=1] = \Pr[A(y \oplus U_m) = 1]$, we have $\begin{array}{l} \bigl\lvert\Pr[A(x\oplus g(U_n))=1] - \Pr[A(x \oplus U_m)=1]\bigr\rvert \\ + \bigl\lvert \Pr[A(y\oplus g(U_n))=1]-\Pr[A(y \oplus U_m) = 1]\bigr\rvert \\ > 1/p(n). \\ \end{array}$
The one of two absolute values is bigger than $1/(2p(n))$. Therefore, WLOG, you obtain your equation 3.
-
Exactly the answer I was looking for, thx! It's funny though: I had done exactly the same thing as you, including the consideration of the "triangle" inequality. But, when I got to your last equation, I couldn't figure out where that $1/2$ factor should come from. However, its good too see that I wasn't to far of :) – hakoja Mar 16 at 15:35
Suppose the two absolute values are at most 1/(2p), then you get 1/p < |...| + |...| <= 2 * 1/(2p) = 1/p, that is, contradiction. – xag Mar 20 at 16:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 3, "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": 0.930199146270752, "perplexity_flag": "middle"} |
http://mathhelpforum.com/statistics/2185-please-help.html | # Thread:
1. ## Please help
Delete
2. Originally Posted by Jon
need help ASAP Six dwarfs: Sleepy, Happy, Bashful, Doc, Grumpy, and Wishbone are available for a behavior experiment. In the experiment, 4 of therm are selcted and arranged in a row. Find the probability for the following events:
Doc is first
Wishbone is last
Bashful is first and Happy is last
Happy is first, Grumpy is second , Doc is thrid and Wishbone is last
Sleepy is first, second or third but not last
Wishbone is not selected
Doc is first:
He needs to be selected AND he would be the first. The probability that he is selected is $\frac{_5C_3\cdot _1C_1}{_6C_4}=\frac{2}{3}$
The probability that he is first is $1/4$ thus, the answer is $\frac{1}{4}\cdot \frac{2}{3}=\frac{1}{6}$
Wishbone:
Same thing as in the first problem, same answer.
Bashful is first, Happy is last:First they need to be selected out of 6 people which is $\frac{_2C_2\cdot _4C_2}{_6C_4}=\frac{2}{5}$
Now bashful is first which is $1/4$ and hapy is last which is $1/3$ because only 3 poeple remain after bashful. Thus,
$\frac{2}{5}\cdot \frac{1}{4}\cdot \frac{1}{3}=\frac{1}{30}$
For the next problem, you need to select all 4 which is $\frac{_4C_4}{_6C_4}=\frac{1}{15}$, Now you need them is proper order which is for the first one $1/4$ for the second $1/3$ for the third $1/2$ and for the last $1/1$ Thus,
$\frac{1}{15}\cdot \frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{1}=\frac{1}{360}$
Sleepy Not last: Add that he is first or second or third. Which is $\frac{1}{6}$ as in the first problem. Thus, $\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$
Wishbone not selected: If the probability of selecting him is $\frac{2}{3}$ then the probability of not selecting him is $\frac{1}{3}$
Hope this helps, and also that I did not erred.
3. ## Thanks
I did come up with some of the answers, but on others I see where I made my mistake. Thanks.
4. Originally Posted by Jon
I did come up with some of the answers, but on others I see where I made my mistake. Thanks.
Welcome, no need to delete the post let other members see if there are any other ways of answering it or maybe they want to see the 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": 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": 0.9762764573097229, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2008/02/16/ftoc-flame-war-wrap-up/?like=1&source=post_flair&_wpnonce=a082172354 | # The Unapologetic Mathematician
## FToC Flame War Wrap-Up
Well, we’ve certainly had a lively time the last few days. Regular commentercomplainer Michael Livshits kicked it off by noting that I presented
The same pathetic proof that mixes apples and oranges, and makes the reader believe that MVT has anything to do woth FTC!
Then came some back-and-forth. I argued that there are many approaches, and due to different motivations we’ve chosen different ones. Michael argued that I was part of some “Church of Limitology” which “indoctrinates” calculus students, and that my proofs “suck”, are “trashy”, and are “in bad taste”. My point that this is not the approach I actually take in a classroom setting was ignored.
However, he did have some points. One of them was that we can weaken the theorem to only assume that the function $f$ is continuous at the point $x$, and my proof assumed way too much to say the function is continuous everywhere. But let’s consider where I actually used continuity. First it shows up in invoking the Integral Mean Value Theorem, but there I really only need it to say there’s a maximum and a minimum, so the Darboux sums work out. An integrable function still manages to satisfy this condition. Then I use continuity to show that $\lim\limits_{c\rightarrow x}f(c)=f(x)$, which really only needs continuity at $x$. In fact, my proof already works in Michael’s extended context.
He also tried presenting his own proof of the crucial step. He argues by continuity at $x$ that for any $\epsilon$ there is a $\delta$ so that $|f(x+\Delta x)-f(x)|<\epsilon$ when $|\Delta x|<\delta$. This fact then shows that $\int_x^{x+\Delta x}f(t)-f(x)dt<\epsilon|\Delta x|$, and so the limit (there’s that awful word again!) that I claimed in the post works out.
But what does his proof really mean? Go on, try and draw a picture. It’s saying that the difference in area we add by integrating from $x$ to $\Delta x$ from the area we’d add if we just used a constant height of $f(x)$ by less than any constant multiple of the width $\Delta x$. And that means… it’s obscure to me, at least.
On the other hand, my proof says that the area function changes by some amount as we go from $x$ to $x+\Delta x$, which means there’s some average (“mean”) rate of change over that interval. At some point along the way, the derivative actually attains that mean value, and as we contract the interval we push that middle point down to $x$. Now that makes sense to me.
Now, the real genius came later, after I sewed the two parts of the FToC together. Eventually, Michael said:
By MVT I meant the one for continuous function, that it hits the zero if it changes sign. Is it the one you were talking about, or you were talking abouth the Lagrange theorem (= MVT for the derivative)? I’m a bit confused. Well, either way it’s not too important.
And here it all runs off the rails. This whole time he hasn’t actually been reading a single thing about my proof, and evidently he hasn’t read the proofs in the calculus textbooks he so despises. He doesn’t even know which theorem I’m invoking! And it is important, because the different theorems say vastly different things.
Now it’s plain as day that Michael is a crank, pushing his pet theories while remaining so embittered to the “system” that “indoctrinates” students against him. Either that or he’s been trolling. The actual merits of my own proof — which I hope I’ve shown above to meet his tests — never mattered at all. I do hope he will set up his own weblog to present all his work in his own space, and then interested readers can judge for themselves the merits and demerits of different proofs.
As for this discussion, it’s closed. I have a proof, and Michael has a proof, and they both work. Our proofs emphasize different aspects of the theorem, and we choose between them depending on what we want to highlight for our current audience. Despite all his ranting, neither one is “the right way” or “the wrong way”, independent of context. I’m glad to hear alternative approaches here, since they might highlight points that I missed. But as a word to future ranters: don’t even try to use my weblog as your soapbox. That sort of behavior really is trashy, and in bad taste.
Besides, I’m the Dennis Miller around here.
[UPDATE]: I’ve come to a decision, since the war seems to rage on unabated, and Mr. Livshits refuses to take the olive branches of equanimity I’ve been offering since the beginning. As of midnight (Central Standard Time) tonight, this is over. Mr. Livshits goes in the kill file, and I wash my hands of the whole business. I’m sure he’ll cry foul, and oppression, and maybe he’s right. However, this whole mess just distracts from my work here, and I’m sick of it. From sideline conversations with numerous non-commenting readers, I’m not the only one.
I’ve made my case, and tried over and over to say that ultimately the whole debate comes down to aesthetics. His approach has its merits, as does mine. He really dislikes my approach, so much so that he’s willing to fight tooth and nail. Ultimately, I really don’t care to fight this any more. But since I can’t seem to continue this project without having my “mathematical taste” insulted left and right, I’m using my authority as the owner of this space to cut off debate. This does not continue here.
If Mr. Livshits wants to continue his tirades, he’s free to set up his own weblog, as I’ve encouraged him to do time and again. He can even continue to read and post to his own space in parallel to my coverage. If he’s right and a significant majority of my readers want to hear his side, he’ll have a built-in audience ready and waiting, and he’s welcome to it. Just like the sky, there’s a lot of blogosphere out there. Of course, he’ll eventually have to come up with something to fill his space, since I’m not spending the rest of my life here on calculus and elementary analysis.
### Like this:
Posted by John Armstrong | Analysis, Calculus
## 1 Comment
1. [...] has been the subject of immense amounts of educational material, ranging from textbooks to blog posts. Unfortunately, that is now all obsolete. The definitive presentation of calculus is here: [...]
Pingback by | February 20, 2008
Sorry, the comment form is closed at this time.
« Previous | Next »
## 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": 0.9629624485969543, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/188354-lengths-sides-right-angled-triangle-pythagoras-theorem.html | # Thread:
1. ## The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Hi guys,
Question: The lengths of the sides of a right angled triangle are $(2x+1)$ cm, $6x$cm and $(5x+3)$ cm. Calculate the value of x and hence find the area of the triangle
I tried using Pythagoras Theorem but if I were to expand the algebra given, the answers would be completely different? How should I go around to solve it?
Thanks
2. ## Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Originally Posted by FailInMaths
Hi guys,
Question: The lengths of the sides of a right angled triangle are $(2x+1)$ cm, $6x$cm and $(5x+3)$ cm. Calculate the value of x and hence find the area of the triangle
I tried using Pythagoras Theorem but if I were to expand the algebra given, the answers would be completely different? How should I go around to solve it?
Thanks
There are two possibilities for the hypotenuse that you need to consider the 6x and the 5x+3 these could both be the hypotenuse, 6x if x>3 and 5x+3 if x<3.
You need to look at both of these cases.
CB
3. ## Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Originally Posted by CaptainBlack
There are two possibilities for the hypotenuse that you need to consider the 6x and the 5x+3 these could both be the hypotenuse, 6x if x>3 and 5x+3 if x<3.
You need to look at both of these cases.
CB
Hmm, the answer stated here is 2. Can I assume that this question is unclear? I mean, like what you said, there are 2 hypotenuse
4. ## Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
If the answer is 2 then consider like CB has said $5x+3$ as the hypothenuse, that means if we use Phytagoras theorem:
$(5x+3)^2=(6x)^2+(2x+1)^2$
Solve this equation.
5. ## Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Originally Posted by Siron
If the answer is 2 then consider like CB has said $5x+3$ as the hypothenuse, that means if we use Phytagoras theorem:
$(5x+3)^2=(6x)^2+(2x+1)^2$
Solve this equation.
Yeah, but I think the question itself is unclear. Without the answer, how would we have know that (5x+3) is the hypotenuse
6. ## Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Originally Posted by FailInMaths
Yeah, but I think the question itself is unclear. Without the answer, how would we have know that (5x+3) is the hypotenuse
By doing the calculations in both cases, you will find only one solution consistent with x>0 and x<3 for the 6x hypotenuse and x>3 for the 5x+3 hypotenuse.
So you compute all four solutions and only one is consistent with the constraints.
CB
7. ## Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Originally Posted by CaptainBlack
By doing the calculations in both cases, you will find only one solution consistent with x>0 and x<3 for the 6x hypotenuse and x>3 for the 5x+3 hypotenuse.
So you compute all four solutions and only one is consistent with the constraints.
CB
Ok, I get it now, thanks | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9357152581214905, "perplexity_flag": "middle"} |
http://www.mathplanet.com/education/algebra-2/discrete-mathematics-and-probability/probabilities | Probabilities
One cannot discern outcomes in all situations, for example whether we will get heads or tails when tossing a coin. We cannot foresee how the coin will land as it is decided by chance. However, if we were to toss the coin infinite times, then half of the results will be heads and half will be tails.
One may say that the probability of achieving heads is 0.5 and the probability for tails is 0.5.
One designates probability by the letter P, and writes the probability of achieving heads as: P(heads)=0,5 and tails as: P(tails)=0,5 for tails.
The probability of an event always lies between 0 (0%) and 1 (100%) and is calculated by the following formula:
$Probability=\frac{number \; of\; favorable\; outcomes}{number \;of\; possible\; outcomes}$
Example
What is the probability of throwing a 5 with a die?
The number of favorable outcomes = 1 (there is only one 5 on a die)
The number of possible outcomes = 6 (a die has 6 sides)
$P(5)=\frac{1}{6}\approx 0.167$
In order to calculate this probability one multiplies the probability of the one event with that of the other one:
$P(A\;and\;B)=P(A)\cdot P(B)$
Example
If one has two dice, what is the probability of throwing a 5 with the first die and a 6 with the other die?
P(5)=1/6
P(6)=1/6
$P(5\;and\;6)=P(5)\cdot P(6)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$
In order to add two probabilities we have to determine whether or not they are mutually exclusive or inclusive events. Our events are considered to be mutually exclusive if they cannot happen at the same time.
Example
We toss a coin, either heads or tails might turn up, but not heads and tails at the same time.
If A and B are mutually exclusive events then we determine the probability of A happening or the probability of B happening with the following formula:
$P(A\;or\;B )=P(A)+P(B)$
If the events are inclusive, e.g. both can happen but not at the same time, then we use the following formula to determine the probability that either A or B occurs:
$P(A\;or\;B )=P(A)+P(B)-P(A\; and\; B)$
Videolesson: what is the probability of not throwing a 6 when throwing a die?
Next Class: Trigonometry, Trigonometric functions
• Pre-Algebra
• Algebra 1
• Algebra 2
• Geometry
• Sat
• Act | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 6, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "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": 0.9336081743240356, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/87050/linearization-of-singular-foliation-in-the-plane | ## Linearization of singular foliation in the plane
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello,
I would like to obtain a smooth model around a singular point of a foliation (in my case the model should be the linearization of the foliation). It seems that the answer could be hidden in Grobmann-Hartman, and I apologize for not having read the proof.
Here is my problem:
I have a singular foliation on $\mathbb{R}^2$ given by the smooth $1$-form
$$\beta = a(x,y) dx + b(x,y) dy$$
and I assume that $(0,0)$ is a singular point.
Additionally I suppose that the matrix $$\begin{pmatrix} \partial_x b & \partial_y b \\ - \partial_x a & -\partial_y a \end{pmatrix}$$ does not have purely imaginary eigenvalues. (In fact, I have $d\beta \ne 0$ so that in the worst case, I could have a 0 eigenvalue, but I assume for now that this is not the case).
Since the kernel of $\beta$ is spanned by the vector field $$X = b \partial_x - a \partial_y,$$ I can use the Hartman-Grobmann theorem to say that my foliation is homeomorphic to its linearization, but I would like to have a diffeomorphism.
I have seen that there exist counter-examples for vector fields to be smoothly conjugated to its linearization, but on the other hand, I'm not interested in the time parameter of the trajectories and I would only like to map the foliation onto the linearized foliation. Could it be possible that I still get a diffeomorphism in my situation? I would bet that there must be a reference, where this has already been worked out, but googling only confused me (it spoke of resonance conditions and other things).
Thank you very much for any response.
Best Klaus
- | {"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": 3, "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": 0.9643194079399109, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/10034-functions-inverses.html | # Thread:
1. ## Functions and Inverses
just 2. I got the rest down but these ones are driving me insane.
Find inverse of functions
1. h(x) = 2x/(x+10)
2. f(x) = x^2 + 6x + 11
Thanks.
2. Originally Posted by Freaky-Person
1. h(x) = 2x/(x+10)
$y=\frac{2x}{x+10}$
Substitute x for y and solve,
$x=\frac{2y}{y+10}$
$x(y+10)=2y$
$xy+10x=2y$
$2y-xy=10x$
$y(2-x)=10x$
$y=\frac{10x}{2-x}$
2. f(x) = x^2 + 6x + 11
It has no inverse.
Horizontal line test, a parabola can be passes twice or more with a horizontal line.
3. It has no inverse.
Horizontal line test, a parabola can be passes twice or more with a horizontal line.
I know it stops being a function, but there should still be an inverse right? Cause you can still draw y=x and flip the original perabola along that line to get the shape, even if it's not a function.
Like the inverse of y = X^2 is y = +-sqrtX, it's still an inverse even if it's not a function....
4. Originally Posted by ThePerfectHacker
$y=\frac{2x}{x+10}$
Substitute x for y and solve,
$x=\frac{2y}{y+10}$
$x(y+10)=2y$
$xy+10x=2y$
$2y-xy=10x$
$y(2-x)=10x$
$y=\frac{10}{2-x}$
It has no inverse.
Horizontal line test, a parabola can be passes twice or more with a horizontal line.
Minor error on #1; did not carry over the x.
Should be (10x)/(x-2)
5. Originally Posted by Freaky-Person
I know it stops being a function, but there should still be an inverse right? Cause you can still draw y=x and flip the original perabola along that line to get the shape, even if it's not a function.
Like the inverse of y = X^2 is y = +-sqrtX, it's still an inverse even if it's not a function....
f^(-1)(x) = {(y,x) | such that y = f(x)}
f will not have an inverse if its not a one-to-one function.
6. Originally Posted by AfterShock
f^(-1)(x) = {(y,x) | such that y = f(x)}
f will not have an inverse if its not a one-to-one function.
whaaaa!! Then what am I supposed to do!?! It says
For the function f(x) = x^2 + 6x + 11
a)sketch the function and its inverse on the same grid
b) Find the equation of the inverse
...
7. Originally Posted by Freaky-Person
Find inverse of functions
2. f(x) = x^2 + 6x + 11
Put:
$y=x^2+6x+11$
then:
$x^2+6x+11-y=0$,
Now use the quadratic formula to get:
$x=\frac{-6 \pm \sqrt{36-4(11-y)}}{2}$
or:
$g(y)=\frac{-6 \pm \sqrt{36-4(11-y)}}{2}$.
Which is not a function when you are looking for functions from
R to R as it fails to be single valued, but it is what you are expected to
produce, and there are interpretations under which it is a function
(such as the extension of f and g to mappings from P(R) the set of
subsets of R, to itself) .
RonL
8. Maybe he is trying to find the inverse,
$f^{-1}: f[\mathbb{R}]\to \mathbb{R}$.
Meaning from the image of the function to the set of real numbers. In that case an inverse exists.
9. Kay I got it. I'll post it up for future reference or something like that.
f(x) = x^2 + 6x + 11
y= x^2 + 6x + 11
for inverse
x = y^2 + 6y + 11
x = (y^2 + 6y + 9) + 2
x - 2 = (y + 3)^2
+-sqrt(x - 2) = y + 3
+-sqrt(x - 2) - 3 = y
BTW, anyone know any sort of tutorial for that big letter CODE writing?
10. Originally Posted by ThePerfectHacker
Maybe he is trying to find the inverse,
$f^{-1}: f[\mathbb{R}]\to \mathbb{R}$.
Meaning from the image of the function to the set of real numbers. In that case an inverse exists.
I have no idea what you just said there, but it's probably it.
11. Originally Posted by Freaky-Person
BTW, anyone know any sort of tutorial for that big letter CODE writing?
I am impressed, a teenager who actually wants to learn something! I wish more of your types existed, otherwise their philosphy is simple, "If it not on the exam we do not care, we do not need it".
http://www.mathhelpforum.com/math-he...-tutorial.html | {"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": 20, "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": 0.9097926616668701, "perplexity_flag": "middle"} |
http://mathforum.org/mathimages/index.php/Surface_Normals | # Surface Normals
### From Math Images
Surface Normals
Fields: Geometry and Calculus
Created By: Nordhr
Website: [ ]
Surface Normals
This picture is of plane segments in different rotations in space. They each have an arrow coming out of their front face at exactly $90^{^{\circ}}$ (orthogonal) to the surface.
# Basic Description
Several kinds of computation in computer graphics depend on knowing the normal to a surface at a point. Often the point is a vertex of a graphics primitive, though it may be more general. In this section we will describe two ways to compute a normal, one geometric and one analytic. While both of these techniques will create a normal to the surface, what is really important is the direction of the normal. The direction is a unit vector, so the results below must be normalized to give us the actual normal direction.
## Geometric Normal
All the graphics primitives lie in a plane, so all the normals at any point in the primitive are the same. We can get the normal by taking any two vectors in the plane and computing their cross product, with the order chosen by taking the second vector as one lying in the positive angle direction from the first with angle less than π. The easiest choice of vectors comes from taking two adjacent edges, as shown. While this is shown for a triangle (and we’ve shown that all graphics primitives are basically made up of triangles) it also works for planar quads and polygons.
Taking the cross product of any two vectors tangent to the surface, not just polygon edge vectors, will give the same result.
# A More Mathematical Explanation
[Click to view A More Mathematical Explanation]
## Analytic Normal
If you are working on a surface that is defined analytically, there are technique [...]
[Click to hide A More Mathematical Explanation]
## Analytic Normal
If you are working on a surface that is defined analytically, there are techniques from calculus you can use to get a normal. Some of these work on vertices of primitives that define the surface; some of these can work on every point in the surface.
Let’s consider surfaces where the surface is defined by a single analytic function
$F(u,v) = (f_x (u,v), f_y (u,v), f_z (u,v))$
We then have the two directional derivatives (which give tangents)
$F_u (u,v) = \frac{\partial F(u,v)}{\partial u} = (\frac{\partial f_x }{\partial u}, \frac{\partial f_y }{\partial u}, \frac{\partial f_z }{\partial u})$
and
$F_v (u,v) = \frac{\partial F(u,v)}{\partial v} = (\frac{\partial f_x }{\partial v}, \frac{\partial f_y }{\partial v}, \frac{\partial f_z }{\partial v})$
Then both $F_u (u,v)$ and $F_v (u,v)$ are vector-valued, and the cross product $F_u (u,v) \times F_v (u,v)$ of these two directional derivatives, when normalized, gives us a normal to the surface.
If you have something like Bézier patches, the parametric equation in two variables looks like
$f(u,v) = \sum_{i=0}^{3} \sum_{i=0}^{3} f_i (u)f_j (v)P_{ij}$
with the points control points $P_{ij}$ and the functions $f_i$ mapping $R^1$ to $R^3$. The partial derivatives of this function are
$\frac{\partial f(u,v)}{\partial u} = \sum_{i=0}^{3} \sum_{i=0}^{3} \frac{\partial f_i (u)}{\partial u} \times f_j(v) \times P_{ij}$ and
$\frac{\partial f(u,v)}{\partial v} = \sum_{i=0}^{3} \sum_{i=0}^{3} f_i (u) \times \frac{\partial f_j (v)}{\partial v} \times P_{ij}$
The separation of the variables in each parametric direction may make this a little simpler, or at least give you a pattern you can implement to cover a wide number of examples.
In both cases, the vertices of your graphics primitives are given by subdividing the parameter space. You can apply the cross-product technique to only the vertices to get the normals at each vertex. From there you can use different shading techniques that interpolate for each pixel to draw each primitive.
It’s worth noting, however, that you could do your interpolation in the parameter domain instead of pixel space when you draw each primitive. If you do this, then you can evaluate the directional derivatives and calculate the normal exactly for each parameter point, and this can give you an image that is superior to any pixel interpolation technique.
# Why It's Interesting
Surface normals are often used in graphics for lighting and shading. The amount of highlight created by a light source is dependent on the angle between the surface normal and the light. They can also be used to detect silhouette edges.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# References
Page written by Steve Cunningham.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
Categories: | | | | | | {"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": 14, "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": 0.8929069638252258, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-algebra/181098-isomorphisms.html | # Thread:
1. ## Isomorphisms
Is there any simple way of telling if two groups are isomorphic? For example, given a list like this and asked to partition into isomorphism classes
1. $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$
2. $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$
3. $Q_{8}$
4. $D_{8}$
5. $\mathbb{Z}_{8}$
6. $\powerset \left\{ 1,2,3 \right\}$ under symmetric set difference $\left( X \cup Y \right) \backslash \left( X \cap Y \right)$
7. Solutions of $x^{8}-1=0$ in $\mathbb{C}$
8. $\left< \left\( 24\right\), \left( 12 \right) \left( 34 \right) \right> \leq S_{4}$
9. $\left< \left[ {\begin{array}{cc} i & 0 \\ 0 & -i \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & i \\ i & 0 \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right] \right> \subseteq \mathbb{C}^{2\times2}$
10. $\mathbb{Z}_{16}^{*}$
Can I work out if they are isomorphic without listing out the elements of each, trying to find a map to from each one to another etc?
2. Originally Posted by Conn
Is there any simple way of telling if two groups are isomorphic? For example, given a list like this and asked to partition into isomorphism classes
1. $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$
2. $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$
3. $Q_{8}$
4. $D_{8}$
5. $\mathbb{Z}_{8}$
6. $\powerset \left\{ 1,2,3 \right\}$ under symmetric set difference $\left( X \cup Y \right) \backslash \left( X \cap Y \right)$
7. Solutions of $x^{8}-1=0$ in $\mathbb{C}$
8. $\left< \left\( 24\right\), \left( 12 \right) \left( 34 \right) \right> \leq S_{4}$
9. $\left< \left[ {\begin{array}{cc} i & 0 \\ 0 & -i \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & i \\ i & 0 \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right] \right> \subseteq \mathbb{C}^{2\times2}$
10. $\mathbb{Z}_{16}^{*}$
Can I work out if they are isomorphic without listing out the elements of each, trying to find a map to from each one to another etc?
I mean there is no simple way in general. So, for example $\mathbb{Z}_4\times\mathbb{Z}_2\not\cong\mathbb{Z}_ 2^3$ since the former has the element $(1,0)$ of order four and the latter has no element of order 4. It's easy that $\mathbb{Z}_4\times\mathbb{Z}_2\not\cong Q_8,D_8,$ since the latter two aren't abelian. The fact that $D_8\not\cong Q_8$ can be gotten from counting element orders.
One can easily see that $\left\{x\in\mathbb{C}:x^8=1\right\}\cong\mathbb{Z} _8$ (since it's cyclic).
You can check that $2^{\{1,2,3\}}$ with symmetric difference is isomorphic to $\mathbb{Z}_2^3$ since it's abelian and $A^2=A\Delta A=\varnothing$ for every $A\in 2^{\{1,2,3\}}$
Try a few more and see if you can work it out. Things to check are number of subgroups of a given order, number of normal subgroups of a given order, number of elements of a given order, whether the two groups are abelian, whether they are cyclic, etc. are all properties preserved under isomorphism.
3. Originally Posted by Conn
Is there any simple way of telling if two groups are isomorphic? For example, given a list like this and asked to partition into isomorphism classes
1. $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$
2. $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$
3. $Q_{8}$
4. $D_{8}$
5. $\mathbb{Z}_{8}$
6. $\powerset \left\{ 1,2,3 \right\}$ under symmetric set difference $\left( X \cup Y \right) \backslash \left( X \cap Y \right)$
7. Solutions of $x^{8}-1=0$ in $\mathbb{C}$
8. $\left< \left\( 24\right\), \left( 12 \right) \left( 34 \right) \right> \leq S_{4}$
9. $\left< \left[ {\begin{array}{cc} i & 0 \\ 0 & -i \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & i \\ i & 0 \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right] \right> \subseteq \mathbb{C}^{2\times2}$
10. $\mathbb{Z}_{16}^{*}$
Can I work out if they are isomorphic without listing out the elements of each, trying to find a map to from each one to another etc?
I think the easiest thing to do here is to sort out groups that can't be isomorphic. For example, (1) and (2) can't be iso.
since one has elements of order 4 whereas the second one hasn't, and none is iso. with (3) since this last one isn't
abelian whereas the first two are.
Continue in this fashion until you have fewer cases to check.
Tonio
4. basically, what you have to do create finer and finer equaivalences on your 10-element set. one of the coarsest filters, if you will, is to sort finite groups by order. here, all 10 groups are of order 8, so that isn't particularly helpful.
the next partition would be abelian versus non-abelian. this gives us 2 subsets {1,2,5,6,7,10} and {3,4,8,9}.
for the abelian groups, we can subdivide by cyclic versus non-cyclic: {1,2,6,10} and {5,7}. since cyclic groups of the same order are isomorphic, {5,7} is one equivalence class under isomorphism.
by considering orders of elements, we can further sub-dvide {1,2,6,10} into {1,10} and {2,6}. Drexel28 outlined how you can define an isomorphism between
P({1,2,3}) and Z2 x Z2 x Z2, so {2,6} is another equivalence class. in fact U(Z16) is isomorphic to Z4 x Z2, so {1,10} is a 3rd equivalence class (to actually prove this, you might try to exhibit an isomorphism. i suggest finding an element x of order 4 in U(Z16) and an order y of order 2 in U(Z16), and sending
(a,b) --> (x^a)(y^b) mod 16).
so that leaves us with {3,4,8,9}. considering orders of elements shows that 3 and 9 have only 1 element of order 2, while 4 and 8 have more than 1. finding an explicit isomorphism between D4 and <(2 4), (1 2)(3 4)> shouldn't be hard (some effort has been made to disguise the nature of the latter group. (1 2)(3 4)(2 4) = (1 2 3 4), which is a 4-cycle).
by a suitable assignment of i,j,k to the 3 2x2 complex matrices shown, you should be able to show the last isomorphism equivalence class is {3,9}. | {"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": 45, "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": 0.9211003184318542, "perplexity_flag": "head"} |
http://quant.stackexchange.com/questions/1947/how-to-uncluster-a-set-of-financial-data/2225 | # How to “uncluster” a set of financial data?
I am attempting to evaluate and compare the profit factor of different "test runs" of a FOREX trading strategy.
My problem is that, despite an average time between orders of 2hr+, some of these runs can have 20+ orders in a row, every 5 minutes, in the same direction. I need some way to normalize these clusters that occur without just throwing the data out.
I want to treat the cluster as 1 data point by averaging the gain/loss of each trade within the cluster.
I was thinking of doing it with a moving time window in the following manner:
````For each order:
Weight = 1/(N+1)
N = count of consecutive orders within 30 minutes of the current order.
````
But I am not sure if that is correct.
-
## 2 Answers
That is definitely not correct. Your test results will suffer from look-ahead bias. At the point in time that you will be entering your order, you will not know yet how many orders will come in the next 30 minutes, thus a proper backtest should not use that information to size the trade.
There are many factors to consider when running a proper backtest (see wikipedia), but the key factor behind all of them is to replicate as closely as possible what is realizable in live trading. In your case, that means that you will have to predict how many trades your strategy is about to enter and size all subsequent trades appropriately. Once you reach your "limit," at which point in real trading you would have run out of capital, your backtest is effectively prevented from entering new trades.
If your strategy isn't able to predict whether trades are about to cluster, you could try an incremental system, whereby the first trade is assigned a proportion $\delta$ of total capital, the second gets $\delta(1-\delta)$, etc.
-
This is not true. There is no look-ahead bias. You are right that at the point in the when I enter the order I do not know how many orders will come in the next 30 minutes - BUT - I do know how many orders came in the next 30 minutes of SIMILAR orders to the current order - in the past. THAT is what I am evaluating and what I am basing my entry size on - it is a meta test. – Mike Sep 19 '11 at 0:29
Mike, it sounds like you are trying to get around the fact that you are allowing look-ahead bias by calling it a "meta" test. I would just be very careful with that. The way your original question is phrased, it is clear that you cannot use the weight scheme you proposed. You cannot even replace N with E[N], because there will be some occasions that N>E[N], and your test will implicitly be allowing you to use more capital than you actually have available. These sort of "meta" tests are nearly worthless for anything but a quick and very rough first pass at the data. – Tal Fishman Sep 19 '11 at 2:08
How could there be look-ahead bias when everything in the meta test has already occurred? – Mike Sep 19 '11 at 2:23
– Tal Fishman Sep 19 '11 at 2:31
That's the thing - it HAS occurred at that point in time. Thats why I call it a meta test. – Mike Sep 19 '11 at 2:42
show 2 more comments
you could perhaps cluster the information in a candle stick manner and bin the price data into high, low and closing, instead of throwing some data out and keeping only its closing price.
Using the high, low and closing prices you are able to make 3 separate estimates with regards to Profit and lost. Obviously with a long position, profit calculated by entering at a high price while using a low price for its exit would be the pessimistic estimate of profit.
- | {"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": 0.9571588635444641, "perplexity_flag": "middle"} |
http://en.wikisource.org/wiki/Page:A_Treatise_on_Electricity_and_Magnetism_-_Volume_1.djvu/255 | # Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/255
From Wikisource
The images within the first sphere form a converging series, the sum of which is
$-P\frac{e^{\varpi-u}-1}{e^{\varpi}-1}.$
This therefore is the quantity of electricity on the first or interior sphere. The images outside the second sphere form a diverging series, but the surface-integral of each with respect to the spherical surface is zero. The charge of electricity on the exterior spherical surface is therefore
$P\left(\frac{e^{\varpi-u}-1}{e^{\varpi}-1}-1\right)=-P\frac{e^{\varpi}-e^{\varpi-u}}{e^{\varpi}-1}$
If we substitute for these expressions their values in terms of $OA, OB$, and $OP$, we find
charge on $A=-P\frac{OA}{OP}\frac{PB}{AB},$
charge on $B=-P\frac{OB}{OP}\frac{AP}{AB}.$
If we suppose the radii of the spheres to become infinite, the case becomes that of a point placed between two parallel planes $A$ and $B$. In this case these expressions become
charge on $A=-P\frac{PB}{AB},$,
charge on $B=-P\frac{AP}{AB}.$.
Fig. 15
172.] In order to pass from this case to that of any two spheres not intersecting each other, we begin by finding the two common inverse points $O, O'$ through which all circles pass that are orthogonal to both spheres. Then, inverting the system with respect to either of these points, the spheres become concentric, as in the first case.
The radius $OAPB$ on which the successive images lie becomes an arc of a circle through $O$ and $O'$, and the ratio of $O'P$ to $OP$ is | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 16, "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": 0.9335909485816956, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/6941/maximal-exotic-mathbbr4/6954 | ## Maximal exotic $\mathbb{R}^4$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Article Exotic $\mathbb{R}^4$ on Wikipedia says that there is at least one maximal smooth structure on $\mathbb{R}^4$, that is such an atlas on $\mathbb{R}^4$ that any other smooth $\mathbb{R}^4$ can be embedded into it. Is the construction of such a maximal exotic $\mathbb{R}^4$ explicit? Can anyone give a reference to the construction? What is a good source (or sources) with examples of exotic smooth structures on $\mathbb{R}^4$? Thanks.
-
## 3 Answers
The paper by Freedman and Taylor mentioned by Carsten Schultz (above or below) is indeed the place to find the explicit construction.
Very roughly, the idea of the construction is as follows. Recall that a Casson handle is, among other things, a smooth 4-manifold which is homeomorphic (but not diffeomorphic) to the standard open 2-handle. It turns out that a countable collection of diffeomorphism classes of Casson handles suffice for solving a certain 5-dimensional h-cobordism problem. The universal $R^4$, call it $U$, is constructed by gluing together countably many copies of each of the Casson handles in the countable collection. Given an arbitrary smooth 4-manifold homeomorphic to $R^4$, we can construct an embedding (but not a proper embedding) into $U$ using the fact that $U$ contains enough Casson handles to solve any link slice problem we might encounter along the way.
Well, that was kind of vague, but hopefully not inaccurate.
-
### 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.
Another excellent reference is the book "4-Manifolds and Kirby Calculus" by Gompf and Stipsicz. Section 9.4 is devoted to exotic R4s. It describes many constructions of exotic R4s and describes how to find the universal one.
The book gives a beautiful overview of the complexity of smooth structures in dimension 4 and I highly recommend it.
-
Google + Zentralblatt (Zbl 0586.57007) tell me:
author="Freedman, Michael H. and Taylor, Laurence R.", title="{A universal smoothing of four-space.}", journal="J. Differ. Geom. ", volume="24", pages="69-78", year="1986",
- | {"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": 0.9089996218681335, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/19142/characterization-of-a-certain-class-of-modules-broader-than-noetherian | ## Characterization of a certain class of modules-broader than Noetherian
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let R be a commutative ring with 1.
An R-module K has the 'S' property if K/T = K implies that the submodule T is trivial.
By Fitting's Lemma any Noetherian module has the 'S' property. There exist non-Noetherian modules with this property. For example the infinite product of Z_{2}xZ_{3}xZ_{5}x... running over all of the primes has the 'S' property, but is not Noetherian.
I am curious if there is a characterization of these kinds of modules out there.
-
What do you mean $K/T=K$? Do you mean $K/T$ is isomorphic to $K$? – Keenan Kidwell Mar 23 2010 at 20:31
Yes, sorry isomorphism. – Johannes Wachs Mar 23 2010 at 20:33
## 1 Answer
These are called Hopfian modules. I didn't see any particularly exciting general characterization, but there are several special case characterizations (that show up easily in a google or mathscinet search). There are also several papers devoted to giving "interesting" examples.
An exercise in Lam's Lectures on Modules and Rings asks one to prove that every finitely generated module over a commutative ring is Hopfian (so if the ring is not-noetherian, this is a generalization).
- | {"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": 0.8756048679351807, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/14489/limit-of-a-series-of-singularities/105146 | ## Limit of a series of singularities
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The $A_\infty$ and $D_\infty$ plane curve singularities have defining equations $x^2=0$ and $x^2y=0$. These equations are "clearly" natural limiting cases of the equations for $A_n$ singularities $x^2 + y^{n+1}=0$ and $D_n$ singularities $x^2y+y^{n-1}=0$ as $n \to \infty$, since large powers are small in the adic topology. So we're tempted to say that $A_\infty$ and $D_\infty$ are "limits" of the "series of singularities" ${A_n}$ and ${D_n}$. This was already observed by Arnol'd in 1981, who wrote "Although the series undoubtedly exist, it is not at all clear what a series of singularities is."
Have there been any attempts since Arnol'd to make sense out of the phrases in quotes in the previous paragraph? That is:
Are there precise definitions of a "series of singularities", and of the "limit" of a series of singularities, under which $\lim_{n\to \infty} A_n = A_\infty$ and $\lim_{n\to \infty} D_n = D_\infty$?
If the answer is Yes, here's another desideratum: does the notion of "limit" extend to modules/sheaves over the singularities? My motivation here is that the $A_n$ and $D_n$ are (almost) precisely the equicharacteristic hypersurfaces with finite Cohen-Macaulay type (i.e. only finitely many indecomposable MCM modules), while $A_\infty$ and $D_\infty$ are precisely the ones with countable or bounded CM type. I'd really like some statement that each MCM module over the "limit" "comes from" a module "at some finite stage".
-
I think the answer might be No. The reason is that varieties/singularities are parametrized by coefficients rather than by degrees. But that is my guess which won't serve a correct/formal answer to your question. – 7-adic Mar 20 2010 at 19:45
regarding the second point, note that in both cases you can put in a scaling parameter and get families $x^2 + \epsilon y^{n+1}$ and $x^2 y + \epsilon y^{n-1}$. Taking $\epsilon \to 0$ gives a map from matrix factorizations of $A_n$ to matrix factorizations of $A_\infty$, which is presumably injective and eventually gets everything, so I guess a posteriori you learn that everything over the limit comes from a finite stage. – Vivek Shende Feb 23 at 6:39
my previous comment was really meant to be a question: are you asking that the notion of limit should be compatible with the above described maps? – Vivek Shende Feb 23 at 6:43
## 2 Answers
This is not an answer, but rather a long comment (grad student level, so please don't take it seriously). I use surfaces for simplicity. The answer must yes in some form. My belief is from the moduli space theory. It is known that the normal stable surfaces admit at worst log canonical isolated singularities. This includes $xyz+x^p+y^r+z^q$ singularities. However, to have a complete moduli space of surfaces, we must include no isolated singularities of the form $xyz$, $xyz+x^p$, and $xyz+x^p+y^r$ (among others). The resemblance of the equations must be more than a coincidence. So, I can imagine we can have an isolated singularity and consider all the deformations from it to non isolated ones. Then to look for the minimal "complete" family of such degenerations.
I wish someone can say something more about all this.
-
### 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.
Large power are small in adic topology: For series of singularities let A infinity and D infinity are plane curve singularities:x^2=0 & X^2.y=0 both equations are natural limiting for A(n) singularities x^2+y^n+1=0 & D(n) singularities x^2y+y^n-1=0 as n_>infinty thats why here we are able to say that A (infinity) and D(infinity) are limits of series of singularity A(n) and D(n)
-
3
Isn't this just the first paragraph in the question? – Ketil Tveiten Aug 21 at 9:49 | {"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": 28, "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": 0.9499034881591797, "perplexity_flag": "middle"} |
http://www.reference.com/browse/Plucker+matrix | Definitions
Nearby Words
# Plucker matrix
Plucker matrices are a representation of a line used in relation to 3D homogeneous coordinates. Specifically, a Plucker matrix is a 4×4 skew-symmetric homogeneous matrix, defined as $mathbf\left\{P\right\} = AB^T - BA^T$, where $A$ and $B$ are two homogeneous coordinates on the line.
P has rank 2. Its 2 dimensional null-space is spanned by the pencil of planes with the line as axis. It has 4 degrees of freedom. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 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": 0.9061643481254578, "perplexity_flag": "head"} |
Subsets and Splits