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http://pediaview.com/openpedia/Four_color_theorem | # Four color theorem
Example of a four-colored map
A four-coloring of a map of the states of the United States (ignoring lakes).
In mathematics, the four color theorem, or the four color map theorem states that, given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors are required to color the regions of the map so that no two adjacent regions have the same color. Two regions are called adjacent if they share a common boundary that is not a corner, where corners are the points shared by three or more regions.[1] For example, in the map of the United States of America, Utah and Arizona are adjacent, but Utah and New Mexico, which only share a point that also belongs to Arizona and Colorado, are not.
Despite the motivation from coloring political maps of countries, the theorem is not of particular interest to mapmakers. According to an article by the math historian Kenneth May (Wilson 2002, 2), “Maps utilizing only four colours are rare, and those that do usually require only three. Books on cartography and the history of mapmaking do not mention the four-color property.”
Three colors are adequate for simpler maps, but an additional fourth color is required for some maps, such as a map in which one region is surrounded by an odd number of other regions that touch each other in a cycle. The five color theorem, which has a short elementary proof, states that five colors suffice to color a map and was proven in the late 19th century (Heawood 1890); however, proving that four colors suffice turned out to be significantly harder. A number of false proofs and false counterexamples have appeared since the first statement of the four color theorem in 1852.
The four color theorem was proven in 1976 by Kenneth Appel and Wolfgang Haken. It was the first major theorem to be proved using a computer. Appel and Haken's approach started by showing that there is a particular set of 1,936 maps, each of which cannot be part of a smallest-sized counterexample to the four color theorem. Appel and Haken used a special-purpose computer program to confirm that each of these maps had this property. Additionally, any map (regardless of whether it is a counterexample or not) must have a portion that looks like one of these 1,936 maps. Showing this required hundreds of pages of hand analysis. Appel and Haken concluded that no smallest counterexamples existed because any must contain, yet not contain, one of these 1,936 maps. This contradiction means there are no counterexamples at all and that the theorem is therefore true. Initially, their proof was not accepted by all mathematicians because the computer-assisted proof was infeasible for a human to check by hand (Swart 1980). Since then the proof has gained wider acceptance, although doubts remain (Wilson 2002, 216–222).
To dispel remaining doubt about the Appel–Haken proof, a simpler proof using the same ideas and still relying on computers was published in 1997 by Robertson, Sanders, Seymour, and Thomas. Additionally in 2005, the theorem was proven by Georges Gonthier with general purpose theorem proving software.
## Precise formulation of the theorem
The intuitive statement of the four color theorem, i.e. 'that given any separation of a plane into contiguous regions, called a map, the regions can be colored using at most four colors so that no two adjacent regions have the same color', needs to be interpreted appropriately to be correct. First, all corners, points that belong to (technically, are in the closure of) three or more countries, must be ignored. In addition, bizarre maps (using regions of finite area but infinite perimeter) can require more than four colors.[2]
Example of a map of Azerbaijan with non-contiguous regions
Second, for the purpose of the theorem every "country" has to be a simply connected region, or contiguous. In the real world, this is not true (e.g., Alaska as part of the United States, Nakhchivan as part of Azerbaijan, and Kaliningrad as part of Russia are not contiguous). Because the territory of a particular country must be the same color, four colors may not be sufficient. For instance, consider a simplified map:
In this map, the two regions labeled A belong to the same country, and must be the same color. This map then requires five colors, since the two A regions together are contiguous with four other regions, each of which is contiguous with all the others. If A consisted of three regions, six or more colors might be required; one can construct maps that require an arbitrarily high number of colors. A similar construction also applies if a single color is used for all bodies of water, as is usual on real maps.
An easier to state version of the theorem uses graph theory. The set of regions of a map can be represented more abstractly as an undirected graph that has a vertex for each region and an edge for every pair of regions that share a boundary segment. This graph is planar (it is important to note that we are talking about the graphs that have some limitations according to the map they are transformed from only): it can be drawn in the plane without crossings by placing each vertex at an arbitrarily chosen location within the region to which it corresponds, and by drawing the edges as curves that lead without crossing within each region from the vertex location to each shared boundary point of the region. Conversely any planar graph can be formed from a map in this way. In graph-theoretic terminology, the four-color theorem states that the vertices of every planar graph can be colored with at most four colors so that no two adjacent vertices receive the same color, or for short, "every planar graph is four-colorable" (Thomas 1998, p. 849; Wilson 2002).
## History
### Early proof attempts
The conjecture was first proposed on October 23, 1852 [3] when Francis Guthrie, while trying to color the map of counties of England, noticed that only four different colors were needed. At the time, Guthrie's brother, Frederick, was a student of Augustus De Morgan at University College. Francis inquired with Frederick regarding it, who then took it to De Morgan (Francis Guthrie graduated later in 1852, and later became a professor of mathematics in South Africa). According to De Morgan:
"A student of mine [Guthrie] asked me to day to give him a reason for a fact which I did not know was a fact — and do not yet. He says that if a figure be any how divided and the compartments differently coloured so that figures with any portion of common boundary line are differently coloured — four colours may be wanted but not more — the following is his case in which four colours are wanted. Query cannot a necessity for five or more be invented… " (Wilson 2002, p. 18)
"F.G.", perhaps one of the two Guthries, published the question in The Athenaeum in 1854,[4][5] and De Morgan posed the question again in the same magazine in 1860.[6] Another early published reference by Arthur Cayley (1879) in turn credits the conjecture to De Morgan.
There were several early failed attempts at proving the theorem. One proof was given by Alfred Kempe in 1879, which was widely acclaimed; another was given by Peter Guthrie Tait in 1880. It was not until 1890 that Kempe's proof was shown incorrect by Percy Heawood, and 1891 Tait's proof was shown incorrect by Julius Petersen—each false proof stood unchallenged for 11 years (Thomas 1998, p. 848).
In 1890, in addition to exposing the flaw in Kempe's proof, Heawood proved the five color theorem (Heawood 1890) and generalized the four color conjecture to surfaces of arbitrary genus—see below.
Tait, in 1880, showed that the four color theorem is equivalent to the statement that a certain type of graph (called a snark in modern terminology) must be non-planar.[7]
In 1943, Hugo Hadwiger formulated the Hadwiger conjecture (Hadwiger 1943), a far-reaching generalization of the four-color problem that still remains unsolved.
### Proof by computer
During the 1960s and 1970s German mathematician Heinrich Heesch developed methods of using computers to search for a proof. Notably he was the first to use discharging for proving the theorem, which turned out to be important in the unavoidability portion of the subsequent Appel-Haken proof. He also expanded on the concept of reducibility and, along with Ken Durre, developed a computer test for it. Unfortunately, at this critical juncture, he was unable to procure the necessary supercomputer time to continue his work (Wilson 2002).
Others took up his methods and his computer-assisted approach. While other teams of mathematicians were racing to complete proofs, Kenneth Appel and Wolfgang Haken at the University of Illinois announced, on June 21, 1976,[8] that they had proven the theorem. They were assisted in some algorithmic work by John A. Koch (Wilson 2002).
If the four-color conjecture were false, there would be at least one map with the smallest possible number of regions that requires five colors. The proof showed that such a minimal counterexample cannot exist, through the use of two technical concepts (Wilson 2002; Appel & Haken 1989; Thomas 1998, pp. 852–853):
1. An unavoidable set contains regions such that every map must have at least one region from this collection.
2. A reducible configuration is an arrangement of countries that cannot occur in a minimal counterexample. If a map contains a reducible configuration, then the map can be reduced to a smaller map. This smaller map has the condition that if it can be colored with four colors, then the original map can also. This implies that if the original map cannot be colored with four colors the smaller map can't either and so the original map is not minimal.
Using mathematical rules and procedures based on properties of reducible configurations, Appel and Haken found an unavoidable set of reducible configurations, thus proving that a minimal counterexample to the four-color conjecture could not exist. Their proof reduced the infinitude of possible maps to 1,936 reducible configurations (later reduced to 1,476) which had to be checked one by one by computer and took over a thousand hours. This reducibility part of the work was independently double checked with different programs and computers. However, the unavoidability part of the proof was verified in over 400 pages of microfiche, which had to be checked by hand (Appel & Haken 1989).
Appel and Haken's announcement was widely reported by the news media around the world, and the math department at the University of Illinois used a postmark stating "Four colors suffice." At the same time the unusual nature of the proof—it was the first major theorem to be proven with extensive computer assistance—and the complexity of the human-verifiable portion, aroused considerable controversy (Wilson 2002).
In the early 1980s, rumors spread of a flaw in the Appel-Haken proof. Ulrich Schmidt at RWTH Aachen examined Appel and Haken's proof for his master's thesis (Wilson 2002, 225). He had checked about 40% of the unavoidability portion and found a significant error in the discharging procedure (Appel & Haken 1989). In 1986, Appel and Haken were asked by the editor of Mathematical Intelligencer to write an article addressing the rumors of flaws in their proof. They responded that the rumors were due to a "misinterpretation of [Schmidt's] results" and obliged with a detailed article (Wilson 2002, 225–226). Their magnum opus, Every Planar Map is Four-Colorable, a book claiming a complete and detailed proof (with a microfiche supplement of over 400 pages), appeared in 1989 and explained Schmidt's discovery and several further errors found by others (Appel & Haken 1989).
### Simplification and verification
Since the proving of the theorem, efficient algorithms have been found for 4-coloring maps requiring only O(n2) time, where n is the number of vertices. In 1996, Neil Robertson, Daniel P. Sanders, Paul Seymour, and Robin Thomas created a quadratic time algorithm, improving on a quartic algorithm based on Appel and Haken’s proof (Thomas 1995; Robertson et al. 1996). This new proof is similar to Appel and Haken's but more efficient because it reduced the complexity of the problem and required checking only 633 reducible configurations. Both the unavoidability and reducibility parts of this new proof must be executed by computer and are impractical to check by hand (Thomas 1998, pp. 852–853). In 2001, the same authors announced an alternative proof, by proving the snark theorem (Thomas; Pegg et al. 2002).
In 2005, Benjamin Werner and Georges Gonthier formalized a proof of the theorem inside the Coq proof assistant. This removed the need to trust the various computer programs used to verify particular cases; it is only necessary to trust the Coq kernel (Gonthier 2008).
## Summary of proof ideas
The following discussion is a summary based on the introduction to Appel and Haken's book Every Planar Map is Four Colorable (Appel & Haken 1989). Although flawed, Kempe's original purported proof of the four color theorem provided some of the basic tools later used to prove it. The explanation here is reworded in terms of the modern graph theory formulation above.
Kempe's argument goes as follows. First, if planar regions separated by the graph are not triangulated, i.e. do not have exactly three edges in their boundaries, we can add edges without introducing new vertices in order to make every region triangular, including the unbounded outer region. If this triangulated graph is colorable using four colors or fewer, so is the original graph since the same coloring is valid if edges are removed. So it suffices to prove the four color theorem for triangulated graphs to prove it for all planar graphs, and without loss of generality we assume the graph is triangulated.
Suppose v, e, and f are the number of vertices, edges, and regions. Since each region is triangular and each edge is shared by two regions, we have that 2e = 3f. This together with Euler's formula v − e + f = 2 can be used to show that 6v − 2e = 12. Now, the degree of a vertex is the number of edges abutting it. If vn is the number of vertices of degree n and D is the maximum degree of any vertex,
$6v - 2e = 6\sum_{i=1}^D v_i - \sum_{i=1}^D iv_i = \sum_{i=1}^D (6 - i)v_i = 12.$
But since 12 > 0 and 6 − i ≤ 0 for all i ≥ 6, this demonstrates that there is at least one vertex of degree 5 or less.
If there is a graph requiring 5 colors, then there is a minimal such graph, where removing any vertex makes it four-colorable. Call this graph G. G cannot have a vertex of degree 3 or less, because if d(v) ≤ 3, we can remove v from G, four-color the smaller graph, then add back v and extend the four-coloring to it by choosing a color different from its neighbors.
Kempe also showed correctly that G can have no vertex of degree 4. As before we remove the vertex v and four-color the remaining vertices. If all four neighbors of v are different colors, say red, green, blue, and yellow in clockwise order, we look for an alternating path of vertices colored red and blue joining the red and blue neighbors. Such a path is called a Kempe chain. There may be a Kempe chain joining the red and blue neighbors, and there may be a Kempe chain joining the green and yellow neighbors, but not both, since these two paths would necessarily intersect, and the vertex where they intersect cannot be colored. Suppose it is the red and blue neighbors that are not chained together. Explore all vertices attached to the red neighbor by red-blue alternating paths, and then reverse the colors red and blue on all these vertices. The result is still a valid four-coloring, and v can now be added back and colored red.
This leaves only the case where G has a vertex of degree 5; but Kempe's argument was flawed for this case. Heawood noticed Kempe's mistake and also observed that if one was satisfied with proving only five colors are needed, one could run through the above argument (changing only that the minimal counterexample requires 6 colors) and use Kempe chains in the degree 5 situation to prove the five color theorem.
In any case, to deal with this degree 5 vertex case requires a more complicated notion than removing a vertex. Rather the form of the argument is generalized to considering configurations, which are connected subgraphs of G with the degree of each vertex (in G) specified. For example, the case described in degree 4 vertex situation is the configuration consisting of a single vertex labelled as having degree 4 in G. As above, it suffices to demonstrate that if the configuration is removed and the remaining graph four-colored, then the coloring can be modified in such a way that when the configuration is re-added, the four-coloring can be extended to it as well. A configuration for which this is possible is called a reducible configuration. If at least one of a set of configurations must occur somewhere in G, that set is called unavoidable. The argument above began by giving an unavoidable set of five configurations (a single vertex with degree 1, a single vertex with degree 2, ..., a single vertex with degree 5) and then proceeded to show that the first 4 are reducible; to exhibit an unavoidable set of configurations where every configuration in the set is reducible would prove the theorem.
Because G is triangular, the degree of each vertex in a configuration is known, and all edges internal to the configuration are known, the number of vertices in G adjacent to a given configuration is fixed, and they are joined in a cycle. These vertices form the ring of the configuration; a configuration with k vertices in its ring is a k-ring configuration, and the configuration together with its ring is called the ringed configuration. As in the simple cases above, one may enumerate all distinct four-colorings of the ring; any coloring that can be extended without modification to a coloring of the configuration is called initially good. For example, the single-vertex configuration above with 3 or less neighbors were initially good. In general, the surrounding graph must be systematically recolored to turn the ring's coloring into a good one, as was done in the case above where there were 4 neighbors; for a general configuration with a larger ring, this requires more complex techniques. Because of the large number of distinct four-colorings of the ring, this is the primary step requiring computer assistance.
Finally, it remains to identify an unavoidable set of configurations amenable to reduction by this procedure. The primary method used to discover such a set is the method of discharging. The intuitive idea underlying discharging is to consider the planar graph as an electrical network. Initially positive and negative "electrical charge" is distributed amongst the vertices so that the total is positive.
Recall the formula above:
$\sum_{i=1}^D (6 - i)v_i = 12.$
Each vertex is assigned an initial charge of 6-deg(v). Then one "flows" the charge by systematically redistributing the charge from a vertex to its neighboring vertices according to a set of rules, the discharging procedure. Since charge is preserved, some vertices still have positive charge. The rules restrict the possibilities for configurations of positively-charged vertices, so enumerating all such possible configurations gives an unavoidable set.
As long as some member of the unavoidable set is not reducible, the discharging procedure is modified to eliminate it (while introducing other configurations). Appel and Haken's final discharging procedure was extremely complex and, together with a description of the resulting unavoidable configuration set, filled a 400-page volume, but the configurations it generated could be checked mechanically to be reducible. Verifying the volume describing the unavoidable configuration set itself was done by peer review over a period of several years.
A technical detail not discussed here but required to complete the proof is immersion reducibility.
## False disproofs
The four color theorem has been notorious for attracting a large number of false proofs and disproofs in its long history. At first, The New York Times refused as a matter of policy to report on the Appel–Haken proof, fearing that the proof would be shown false like the ones before it (Wilson 2002). Some alleged proofs, like Kempe's and Tait's mentioned above, stood under public scrutiny for over a decade before they were exposed. But many more, authored by amateurs, were never published at all.
The map (left) has been colored with five colors, and it is necessary to change at least four of the ten regions to obtain a coloring with only four colors (right).
Generally, the simplest, though invalid, counterexamples attempt to create one region which touches all other regions. This forces the remaining regions to be colored with only three colors. Because the four color theorem is true, this is always possible; however, because the person drawing the map is focused on the one large region, they fail to notice that the remaining regions can in fact be colored with three colors.
This trick can be generalized: there are many maps where if the colors of some regions are selected beforehand, it becomes impossible to color the remaining regions without exceeding four colors. A casual verifier of the counterexample may not think to change the colors of these regions, so that the counterexample will appear as though it is valid.
Perhaps one effect underlying this common misconception is the fact that the color restriction is not transitive: a region only has to be colored differently from regions it touches directly, not regions touching regions that it touches. If this were the restriction, planar graphs would require arbitrarily large numbers of colors.
Other false disproofs violate the assumptions of the theorem in unexpected ways, such as using a region that consists of multiple disconnected parts, or disallowing regions of the same color from touching at a point.
## Generalizations
By joining the single arrows together and the double arrows together, one obtains a torus with seven mutually touching regions; therefore seven colors are necessary
This construction shows the torus divided into the maximum of seven regions, every one of which touches every other.
The four-color theorem applies not only to finite planar graphs, but also to infinite graphs that can be drawn without crossings in the plane, and even more generally to infinite graphs (possibly with an uncountable number of vertices) for which every finite subgraph is planar. To prove this, one can combine a proof of the theorem for finite planar graphs with the De Bruijn–Erdős theorem stating that, if every finite subgraph of an infinite graph is k-colorable, then the whole graph is also k-colorable Nash-Williams (1967). This can also be seen as an immediate consequence of Kurt Gödel's compactness theorem for First-Order Logic, simply by expressing the colorability of an infinite graph with a set of logical formulae.
One can also consider the coloring problem on surfaces other than the plane (Weisstein). The problem on the sphere or cylinder is equivalent to that on the plane. For closed (orientable or non-orientable) surfaces with positive genus, the maximum number p of colors needed depends on the surface's Euler characteristic χ according to the formula
$p=\left\lfloor\frac{7 + \sqrt{49 - 24 \chi}}{2}\right\rfloor,$
where the outermost brackets denote the floor function.
Alternatively, for an orientable surface the formula can be given in terms of the genus of a surface, g:
$p=\left\lfloor\frac{7 + \sqrt{1 + 48g }}{2}\right\rfloor$ (Weisstein).
This formula, the Heawood conjecture, was conjectured by P.J. Heawood in 1890 and proven by Gerhard Ringel and J. T. W. Youngs in 1968. The only exception to the formula is the Klein bottle, which has Euler characteristic 0 (hence the formula gives p = 7) and requires 6 colors, as shown by P. Franklin in 1934 (Weisstein).
For example, the torus has Euler characteristic χ = 0 (and genus g = 1) and thus p = 7, so no more than 7 colors are required to color any map on a torus. The Szilassi polyhedron is an example that requires seven colors.
A Möbius strip requires six colors (Weisstein) as do 1-planar graphs (graphs drawn with at most one simple crossing per edge) (Borodin 1984). If both the vertices and the faces of a planar graph are colored, in such a way that no two adjacent vertices, faces, or vertex-face pair have the same color, then again at most six colors are needed (Borodin 1984).
There is no obvious extension of the coloring result to three-dimensional solid regions. By using a set of n flexible rods, one can arrange that every rod touches every other rod. The set would then require n colors, or n+1 if you consider the empty space that also touches every rod. The number n can be taken to be any integer, as large as desired. Such examples were known to Fredrick Guthrie in 1880 (Wilson 2002). Even for axis-parallel cuboids (considered to be adjacent when two cuboids share a two-dimensional boundary area) an unbounded number of colors may be necessary (Reed & Allwright 2008; Magnant & Martin (2011)).
## See also
Graph coloring
the problem of finding optimal colorings of graphs that are not necessarily planar.
Grötzsch's theorem
Hadwiger–Nelson problem
how many colors are needed to color the plane so that no two points at unit distance apart have the same color?
List of sets of four countries that border one another
Contemporary examples of national maps requiring four colors
Apollonian network
The planar graphs that require four colors and have exactly one four-coloring
## Notes
1. Georges Gonthier (December, 2008). "Formal Proof---The Four-Color Theorem". Notices of the AMS 55 (11): 1382–1393.From this paper: Definitions: A planar map is a set of pairwise disjoint subsets of the plane, called regions. A simple map is one whose regions are connected open sets. Two regions of a map are adjacent if their respective closures have a common point that is not a corner of the map. A point is a corner of a map if and only if it belongs to the closures of at least three regions. Theorem: The regions of any simple planar map can be colored with only four colors, in such a way that any two adjacent regions have different colors.
2. Hud Hudson (May, 2003). "Four Colors Do Not Suffice". The American Mathematical Monthly 110 (5): 417–423. JSTOR 3647828.
3. Donald MacKenzie, Mechanizing Proof: Computing, Risk, and Trust (MIT Press, 2004) p103
4. .
5. Brendan D. McKay (2012). "A note on the history of the four-colour conjecture". arXiv:1201.2852.
6. . As cited by Wilson, John (1976), "New light on the origin of the four-color conjecture", Historia Mathematica 3: 329–330, doi:10.1016/0315-0860(76)90106-3, MR 0504961.
7. Tait, P. G. (1880), "Remarks on the colourings of maps", Proc. R. Soc. Edinburgh 10: 729
8. Gary Chartrand and Linda Lesniak, Graphs & Digraphs (CRC Press, 2005) p221
## References
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http://stochastix.wordpress.com/2012/10/02/deterministic-finite-automata-in-haskell/ | # Rod Carvalho
## Deterministic Finite Automata in Haskell
Consider the deterministic finite automaton [1] illustrated below
[ state diagram courtesy of Michael Sipser [1] ]
Henceforth, we will call this automaton $M_1$. Note that $M_1$ has three states, labeled $q_1$, $q_2$, and $q_3$. The start state is $q_1$ (note the arrow coming from nowhere) and the accept state is $q_2$ (note the double circle). There are two possible inputs, labeled $0$ and $1$, and, depending on the input, the automaton will jump from one state to another. In the diagram above these state transitions are depicted using arrows connecting two states.
Suppose that the automaton $M_1$ receives an input string such as $1101$ (which the automaton reads from left to right). Since the start state is $q_1$ and the first input symbol is $1$, the automaton will jump to state $q_2$. The second input symbol is $1$, so the automaton will remain at $q_2$. The automaton reads the third input symbol, which is $0$, and then jumps from state $q_2$ to state $q_3$. The last input symbol is $1$ and thus the automaton jumps from state $q_3$ to state $q_2$. Hence, the state sequence corresponding to the input string $1101$ is the following
$q_1 \xrightarrow{1} q_2 \xrightarrow{1} q_2 \xrightarrow{0} q_3 \xrightarrow{1} q_2$
After reading the last symbol in the input string, $M_1$ produces an output. Since the final state $q_2$ is the accept state, we have that the automaton produces the output “accept”. Were the final state not $q_2$, the automaton would produce the output “reject”. We conclude that this automaton accepts the input string $1101$. What other input strings does $M_1$ accept? Michael Sipser answers this question in [1]:
Experimenting with this machine on a variety of input strings reveals that it accepts the strings $1$, $01$, $11$, and $0101010101$. In fact, $M_1$ accepts any string that ends with a $1$, as it goes to its accept state $q_2$ whenever it reads the symbol $1$. In addition, it accepts strings $100$, $0100$, $110000$, and $0101000000$, and any string that ends with an even number of $0$s following the last $1$. It rejects other strings, such as $0$, $10$, $101000$.
A set of strings is called a language [1]. The set of all input strings accepted by the deterministic finite automaton $M_1$ is a language which we denote by $L (M_1)$.
__________
Formal definition
A deterministic finite automaton (DFA) consists of a finite set of states $Q$, a finite input alphabet $\Sigma$ that tells us what the allowed input symbols are, a transition function $\delta : Q \times \Sigma \to Q$ that tells us how to jump from one state to another, a start state $q_0 \in Q$, and a set of accept states $A \subseteq Q$. A deterministic finite automaton (DFA) is thus a $5$-tuple of the form
$(Q, \Sigma, \delta, q_0, A)$
The deterministic finite automaton $M_1$ which we discussed earlier in this post is defined formally as follows
$M_1 := (\{q_1, q_2, q_3\}, \{0,1\}, \delta, q_1, \{q_2\})$
where the transition function $\delta: \{q_1, q_2, q_3\} \times \{0,1\} \to \{q_1, q_2, q_3\}$ is defined enumeratively as
$\delta (q_1, 0) = q_1$, $\delta (q_2, 0) = q_3$, $\delta (q_3, 0) = q_2$
$\delta (q_1, 1) = q_2$, $\delta (q_2, 1) = q_2$, $\delta (q_3, 1) = q_2$
Alternatively, we could view each state transition $q_i \xrightarrow{\sigma} q_j$ as an ordered triple $(q_i, \sigma, q_j)$, which might be easier to implement in some programming languages.
__________
Computing state sequences
Given an input string $\sigma = \sigma_1 \sigma_2 \dots \sigma_n$ over a given alphabet $\Sigma$, how do we obtain the corresponding state sequence? I solved that problem last January. I will not repeat myself here, but the crux of the matter is that the final state can be obtained using a left-fold
$\text{foldl} \, \delta \, q_0\, [\sigma_1, \sigma_2, \dots, \sigma_{n}]$
whereas the entire state sequence can be computed using a left-scan
$\text{scanl} \, \delta \, q_0\, [\sigma_1, \sigma_2, \dots, \sigma_{n}]$
Lacking a better name, I called this procedure the “scanl trick“, as I used the Haskell function scanl to implement it. Please let me know if you find a more sophisticated name for this “trick”.
__________
Haskell implementation of the DFA
Without further ado, here is a Haskell script:
```data State = Q1 | Q2 | Q3 deriving (Read, Show, Eq)
type Input = Integer
-- define state-transition function
delta :: State -> Input -> State
delta Q1 0 = Q1
delta Q1 1 = Q2
delta Q2 0 = Q3
delta Q2 1 = Q2
delta Q3 0 = Q2
delta Q3 1 = Q2
delta _ _ = error "Invalid input!"
-- define initial state
initialstate :: State
initialstate = Q1
-- create list of accept states
acceptstates :: [State]
acceptstates = [Q2]
-- create infinite list of input sequences
inputss :: [[Input]]
inputss = concat $ iterate g [[]]
where g = concat . map (\xs -> [ xs ++ [s] | s <- [0,1]])
-- create accept predicate
isAccepted :: [Input] -> Bool
isAccepted inputs = finalstate `elem` acceptstates
where finalstate = foldl delta initialstate inputs
-- compute language recognized by the DFA
language :: [[Input]]
language = filter isAccepted inputss```
• Sets are represented by lists. Strings are represented by lists, too. The latter is more natural than the former. Sets of strings become lists of lists.
• A new data type is created to represent the states, which we denote by Q1, Q2, and Q3. The input symbols are integers.
• Note that Input is a type synonym, inputs is a list of inputs symbols (i.e., an input string), and inputss is a list of lists of inputs (i.e., a list of input strings). Yes, it is a bit confusing.
• Note also that inputss is generated using the procedure I blogged about yesterday where the alphabet is $\Sigma_2 := \{0, 1\}$.
• The final state is computed using the higher-order function foldl (left-fold). We check if the final state is an accept state using the list membership predicate elem.
If you have any objections to this script, please do let me know. Let us now test it! We load it into GHCi and voilà:
```*Main> -- check list of input strings
*Main> take 31 $ inputss
[[],[0],[1],
[0,0],[0,1],[1,0],[1,1],
[0,0,0],[0,0,1],[0,1,0],[0,1,1],
[1,0,0],[1,0,1],[1,1,0],[1,1,1],
[0,0,0,0],[0,0,0,1],[0,0,1,0],[0,0,1,1],
[0,1,0,0],[0,1,0,1],[0,1,1,0],[0,1,1,1],
[1,0,0,0],[1,0,0,1],[1,0,1,0],[1,0,1,1],
[1,1,0,0],[1,1,0,1],[1,1,1,0],[1,1,1,1]]
*Main> -- compute the language of the automaton
*Main> -- (let us extract some 40 input strings only)
*Main> take 40 language
[[1],[0,1],[1,1],
[0,0,1],[0,1,1],[1,0,0],[1,0,1],[1,1,1],
[0,0,0,1],[0,0,1,1],[0,1,0,0],[0,1,0,1],
[0,1,1,1],[1,0,0,1],[1,0,1,1],[1,1,0,0],
[1,1,0,1],[1,1,1,1],[0,0,0,0,1],[0,0,0,1,1],
[0,0,1,0,0],[0,0,1,0,1],[0,0,1,1,1],[0,1,0,0,1],
[0,1,0,1,1],[0,1,1,0,0],[0,1,1,0,1],[0,1,1,1,1],
[1,0,0,0,0],[1,0,0,0,1],[1,0,0,1,1],[1,0,1,0,0],
[1,0,1,0,1],[1,0,1,1,1],[1,1,0,0,1],[1,1,0,1,1],
[1,1,1,0,0],[1,1,1,0,1],[1,1,1,1,1],[0,0,0,0,0,1]]```
The list of input strings is exactly the same I posted yesterday. The input strings in the language $L (M_1)$, or, to put it more precisely, the input strings in $L (M_1)$ that are displayed above either end with a $1$, or end “with an even number of $0$s following the last $1$“, as mentioned by Sipser in [1]. Why is that? The last $1$ in the input string puts $M_1$ in state $q_2$, and an even number of $0$s after that $1$ lead to transitions from $q_2$ to $q_3$ and back to $q_2$, e.g.,
$q_2 \xrightarrow{0} q_3 \xrightarrow{0} q_2$
or
$q_2 \xrightarrow{0} q_3 \xrightarrow{0} q_2 \xrightarrow{0} q_3 \xrightarrow{0} q_2$
or
$q_2 \xrightarrow{0} q_3 \xrightarrow{0} q_2 \xrightarrow{0} q_3 \xrightarrow{0} q_2 \xrightarrow{0} q_3 \xrightarrow{0} q_2 \xrightarrow{0} q_3 \xrightarrow{0} q_2$
It would also be interesting to take a look at the state sequences corresponding to the input strings in $L(M_1)$. We compute these state sequences using the infamous “scanl trick“:
```*Main> -- create list of sequences of states
*Main> let statess = map (\xs -> scanl delta initialstate xs) language
*Main> -- take the "first" 20 state sequences
*Main> take 20 statess
[[Q1,Q2],[Q1,Q1,Q2],[Q1,Q2,Q2],
[Q1,Q1,Q1,Q2],[Q1,Q1,Q2,Q2],[Q1,Q2,Q3,Q2],
[Q1,Q2,Q3,Q2],[Q1,Q2,Q2,Q2],[Q1,Q1,Q1,Q1,Q2],
[Q1,Q1,Q1,Q2,Q2],[Q1,Q1,Q2,Q3,Q2],[Q1,Q1,Q2,Q3,Q2],
[Q1,Q1,Q2,Q2,Q2],[Q1,Q2,Q3,Q2,Q2],[Q1,Q2,Q3,Q2,Q2],
[Q1,Q2,Q2,Q3,Q2],[Q1,Q2,Q2,Q3,Q2],[Q1,Q2,Q2,Q2,Q2],
[Q1,Q1,Q1,Q1,Q1,Q2],[Q1,Q1,Q1,Q1,Q2,Q2]]```
Note that all state trajectories end in state Q2, as we expected.
__________
References
[1] Michael Sipser, Introduction to the Theory of Computation (2nd edition), Thomson Course Technology, 2006. | {"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": 88, "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": 0.888286292552948, "perplexity_flag": "head"} |
http://mathhelpforum.com/discrete-math/156312-number-iterations-print.html | # Number of Iterations
Printable View
• September 15th 2010, 02:49 PM
aarnold
Number of Iterations
How many times does the innermost loop execute:
Code:
for k = 1 to n
for j = 1 to k
for i = 1 to j
#the inner loop
end for
end for
end for
I did the first few by hand to see if I noticed anything.
Looking at Pascal's triangle I saw the numbers appear in the diagonal. Eventually I came up with the answer:
(n + 2) choose (n - 1)
But I don't know how I got there. Does this have anything to do with the Principle of Inclusion and Exclusion?
• September 15th 2010, 03:16 PM
yeKciM
Quote:
Originally Posted by aarnold
How many times does the innermost loop execute:
Code:
for k = 1 to n
for j = 1 to k
for i = 1 to j
#the inner loop
end for
end for
end for
I did the first few by hand to see if I noticed anything.
Looking at Pascal's triangle I saw the numbers appear in the diagonal. Eventually I came up with the answer:
(n + 2) choose (n - 1)
But I don't know how I got there. Does this have anything to do with the Principle of Inclusion and Exclusion?
that can't be (n+2) or (n-1) iterations in that loop with "i" :D:D:D
i have question for you :D why don't compile this ? and is it ( k=1 ; k<= n or you need k<n ; k++ )
for n=2 there are 6 , for n=3 there is 18 , for n=4 there is 40 ... repetitions :D
lol for n = 50 there is 63 750 of executions of that loop :D
• September 15th 2010, 03:39 PM
aarnold
Hi, I'm assuming that I am calculating for k <= n.
A quick check with python gives the following results:
Code:
>>> def iterate(n):
c = 0
for k in range(1, n + 1):
for j in range(1, k + 1):
for i in range(1, j + 1):
c = c + 1
print c
>>> iterate(1)
1
>>> iterate(2)
4
>>> iterate(3)
10
>>> iterate(4)
20
>>> iterate(5)
35
>>> iterate(6)
56
and here's Pascal's triangle:
Code:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
If the row and column's are 0-based, look at the n+2 row and the n-1 column and you get the number of iterations.
So if n = 1, look at row 3, column 0: 1
n = 2, look at row 4, column 1: 4
n = 3, row 5, column 2: 10
etc.
I determined this to be (n+2) choose (n-1).
However, I don't think this is a mathematical enough answer.
• September 15th 2010, 04:00 PM
yeKciM
hm...
Code:
cin>> n ;
int iter = 0 ;
for (int k = 1 ; k<= n ; k++ ) {
for( int j= 1 ; j <= k ; j++ ) {
for( int i =1 ; i< =j i ++ ) {
iter++ ;
cout << << "No." << iter << "." << i << endl;
}
}
}
and get totally something else :D
• September 15th 2010, 04:16 PM
undefined
Quote:
Originally Posted by aarnold
How many times does the innermost loop execute:
Code:
for k = 1 to n
for j = 1 to k
for i = 1 to j
#the inner loop
end for
end for
end for
I did the first few by hand to see if I noticed anything.
Looking at Pascal's triangle I saw the numbers appear in the diagonal. Eventually I came up with the answer:
(n + 2) choose (n - 1)
But I don't know how I got there. Does this have anything to do with the Principle of Inclusion and Exclusion?
$1+2+3+...+n = t_n =$ nth triangle number $=\binom{n+1}{2}$
$<br /> t_1 + t_2 + t_3 + ... + t_n =$ nth tetrahedral number $=\binom{n+2}{3}$
The pattern continues. Notice that n+2-3 = n-1. (Combine this with the knowledge that $\binom{n}{k}=\binom{n}{n-k}$.)
You can justify the triangular number formula by considering the handshake problem; given a group of n people, if everyone shakes everyone else's hand, how many handshakes took place? So the first person shakes n-1 hands. Add to that n-2 for the second person -- because if you added n-1 you would re-count the handshake with the first person. So you get (n-1) + n + ... + 1, which is the (n-1)st triangle number.
We can generalize as follows: Consider a set of n points, n > 2, such that no three points are collinear. How many triangles can we form by choosing vertices from these n points? Start with one point, and ask how many triangles can form with it. You get the (n-2)nd triangle number. (Because you are choosing any two points out of the n-1 remaining points, which is a problem we just solved above.) For the next point you get the (n-3)rd triangle number, etc.
There may be a cleaner way to state it, but that at least does the job.
• September 16th 2010, 05:23 AM
CaptainBlack
Quote:
Originally Posted by aarnold
How many times does the innermost loop execute:
Code:
for k = 1 to n
for j = 1 to k
for i = 1 to j
#the inner loop
end for
end for
end for
I did the first few by hand to see if I noticed anything.
Looking at Pascal's triangle I saw the numbers appear in the diagonal. Eventually I came up with the answer:
(n + 2) choose (n - 1)
But I don't know how I got there. Does this have anything to do with the Principle of Inclusion and Exclusion?
For any value of $$$ j$ the inner loop executes $$$ j$ times.
For any value of $$$ k$ the inner look executes $\sum_{j=1}^k j$
So the total number of executions of the inner code is $\sum_{k=1}^n \left[\sum_{j=1}^k j\right]$
With a bit of jiggery-pokery this becomes: $\dfrac{n(n+1)(n+2)}{6}$
CB
• September 16th 2010, 06:54 AM
MathoMan
Here's that jiggery-pokery stuff:
$\sum_{k=1}^n \left[\sum_{j=1}^k j\right]=\sum_{k=1}^n\frac{k(k+1)}{2}=$ $=\frac{1}{2}\left(\sum_{k=1}^n k^2+\sum_{k=1}^n k\right)=\frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\ frac{n(n+1)}{2}\right)=\dfrac{n(n+1)(n+2)}{6}$
All times are GMT -8. The time now is 03:26 PM. | {"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": 13, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9528090953826904, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/67265/list | ## Return to Question
3 added 270 characters in body
It is well known that $4$ general points in $\mathbb{P}^2$ are complete intersection of two conics. On the other hand, if $d \geq 3$, $d^2$ general points are not a complete intersection of two curves of degree $d$. More precisely, if $d =3$ there is only one cubic passing through $9$ general points, whereas if $d \geq 4$ there is no curve of degree $d$ passing through $d^2$ general points.
While investigating some questions about factoriality of singular hypersurfaces of $\mathbb{P}^n$, I ran across the following problem, which seems quite natural to state.
Let $d \geq 3$ be a positive integer and let $Q \subset \mathbb{P}^3$ be a subset made of $d^2$ distinct points, with the following property: for a general projection $\pi \colon \mathbb{P}^3 \to \mathbb{P}^2$, the subset $\pi(Q) \subset \mathbb{P}^2$ is the complete intersection of two plane curves of degree $d$.
Is it true that $Q$ itself is contained in a plane (and is the complete intersection of two curves of degree $d$)?
If not, what is a counterexample?
Any answer or reference to the existing literature will be appreciated. Thank you.
EDIT. Dimitri's answer below provides a counterexample given by $d^2$ points on a quadric surface. Are there other configurations of points with the same property? It is possible to classify them up to projective transformations (at least for small values of $d$)?
2 deleted 26 characters in body
It is well known that $4$ general points in $\mathbb{P}^2$ are complete intersection of two conics. On the other hand, if $d \geq 3$, $d^2$ general points are not a complete intersection of two curves of degree $d$. More precisely, if $d =3$ there is only one cubic passing through $9$ general points, whereas if $d \geq 4$ there is no curve of degree $d$ passing through $d^2$ general points.
While investigating some questions about factoriality of singular hypersurfaces of $\mathbb{P}^n$, I ran across the following problem, which seems quite natural to statebut I have not been able to solve.
Let $d \geq 3$ be a positive integer and let $Q \subset \mathbb{P}^3$ be a subset made of $d^2$ distinct points, with the following property: for a general projection $\pi \colon \mathbb{P}^3 \to \mathbb{P}^2$, the subset $\pi(Q) \subset \mathbb{P}^2$ is the complete intersection of two plane curves of degree $d$.
Is it true that $Q$ itself is contained in a plane (and is the complete intersection of two curves of degree $d$)?
If not, what is a counterexample?
Any answer or reference to the existing literature will be appreciated. Thank you.
1
# When is a general projection of $d^2$ points in $\mathbb{P}^3$ a complete intersection?
It is well known that $4$ general points in $\mathbb{P}^2$ are complete intersection of two conics. On the other hand, if $d \geq 3$, $d^2$ general points are not a complete intersection of two curves of degree $d$. More precisely, if $d =3$ there is only one cubic passing through $9$ general points, whereas if $d \geq 4$ there is no curve of degree $d$ passing through $d^2$ general points.
While investigating some questions about factoriality of singular hypersurfaces of $\mathbb{P}^n$, I ran across the following problem, which seems natural to state but I have not been able to solve.
Let $d \geq 3$ be a positive integer and let $Q \subset \mathbb{P}^3$ be a subset made of $d^2$ distinct points, with the following property: for a general projection $\pi \colon \mathbb{P}^3 \to \mathbb{P}^2$, the subset $\pi(Q) \subset \mathbb{P}^2$ is the complete intersection of two plane curves of degree $d$.
Is it true that $Q$ itself is contained in a plane (and is the complete intersection of two curves of degree $d$)?
If not, what is a counterexample?
Any answer or reference to the existing literature will be appreciated. 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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9441551566123962, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2010/05/07/adding-and-multiplying-measurable-real-valued-functions/?like=1&_wpnonce=401286028f | # The Unapologetic Mathematician
## Adding and Multiplying Measurable Real-Valued Functions
One approach to the problem of adding and multiplying measurable functions on a measurable space $X$ would be to define a two-dimensional version of Borel sets and Lebesgue measure, and to tweak the definition of a measurable function to this space $(\mathbb{R}^2,\mathcal{B}_2)$ like we did before to treat the additive identity $(0,0)$ specially. Then we could set up products (which we will eventually do) and get a map $(f,g):X\to\mathbb{R}^2$ and compose this with the Borel map $(x,y)\mapsto x+y$ or the Borel map $(x,y)\mapsto xy$. In fact, if you’re up for it, you can go ahead and try working out this approach as an exercise.
Instead, we’ll take more of a low road towards showing that the sum and product of two measurable functions are measurable. We start with a useful lemma: if $f$ and $g$ are extended real-valued measurable functions on a measurable space $(X,\mathcal{S})$ and if $c$ is any real number, then each of the sets
$\displaystyle\begin{aligned}A&=\left\{x\in X\vert f(x)<g(x)+c\right\}\\B&=\left\{x\in X\vert f(x)\leq g(x)+c\right\}\\C&=\left\{x\in X\vert f(x)=g(x)+c\right\}\end{aligned}$
has a measurable intersection with every measurable set. If $X$ is itself measurable, of course, this just means that these three sets are measurable.
To see this for the set $A$, consider the (countable) set $\mathbb{Q}\subseteq\mathbb{R}$ of rational numbers. If $f(x)$ really is strictly less than $g(x)+c$, then there must be some rational number $r$ between them. That is, if $x\in A$ then for some $r$ we have $f(x)<r$ and $r-c<g(x)$. And thus we can write $A$ as the countable union
$\displaystyle\begin{aligned}A&=\bigcup\limits_{r\in\mathbb{Q}}\left(\left\{x\in X\vert f(x)<r\right\}\cap\left\{x\in X\vert r-c<g(x)\right\}\right)\\&=\bigcup\limits_{r\in\mathbb{Q}}\left(f^{-1}\left[-\infty,r\right]\cap\left[r-c,\infty\right]\right)\end{aligned}$
By the measurability of $f$ and $g$, this is the countable union of a collection of measurable sets, and is thus measurable.
We can write $B$ as $X\setminus\left\{x\in X\vert f(x)<g(x)-c\right\}$, and so the assertion for $B$ follows from that for $A$. And we can write $C=B\setminus A$, so the statement is true for that set as well.
Anyway, now we can verify that the sum and product of two measurable extended real-valued functions are measurable as well. We first handle infinite values separately. For the product, $\left[fg\right](x)=\infty$ if and only if $f(x)=g(x)=\pm\infty$. Since the sets $f^{-1}(\{\infty\})\cap g^{-1}(\{\infty\})$ and $f^{-1}(\{-\infty\})\cap g^{-1}(\{-\infty\})$ are both measurable, the set $[fg]^{-1}(\{\infty\})$ — their union — is measurable. We can handle $[fg]^{-1}(\{-\infty\})$, $[f+g]^{-1}(\{\infty\})$, and $[f+g]^{-1}(\{-\infty\})$ similarly.
So now we turn to our convenient condition for measurability. Since we’ve handled the sets where $f(x)$ and $g(x)$ are infinite, we can assume that they’re finite. Given a real number $c$, we find
$\displaystyle\left\{x\in X\vert f(x)+g(x)<c\right\}=\left\{x\in X\vert f(x)<c-g(x)\right\}$
which is measurable by our lemma above (with $-g$ in place of $g$). Since this is true for every real number $c$, the sum $f+g$ is measurable.
To verify our assertion for the product $fg$, we turn and recall the polarization identities from when we worked with inner products. Remember, they told us that if we know how to calculate squares, we can calculate products. Something similar is true now, as we write
$\displaystyle f(x)g(x)=\frac{1}{4}\left(\left(f(x)+g(x)\right)^2-\left(f(x)-g(x)\right)^2\right)$
We just found that the sum $f+g$ and the difference $f-g$ are measurable. And any positive integral power of a measurable function is measurable, so the squares of the sum and difference functions are measurable. And then the product $fg$ is a scalar multiple of the difference of these squares, and is thus measurable.
### Like this:
Posted by John Armstrong | Analysis, Measure Theory
## 5 Comments »
1. [...] and Negative Parts of Functions Now that we have sums and products to work with, we find that the maximum of and — sometimes written or — and their [...]
Pingback by | May 7, 2010 | Reply
2. [...] and some limit inferior . If these two coincide, then the sequence has a proper limit . But one of our lemmas tells us that the set of points where any two measurable functions coincide has a nice property: [...]
Pingback by | May 10, 2010 | Reply
3. [...] function is simple — . And thus the collection of simple functions forms a subalgebra of the algebra of measurable [...]
Pingback by | May 11, 2010 | Reply
4. [...] finally we can show that converges in measure to . We can use the same polarization trick as we’ve used before. Write ; we’ve just verified that the squares converge to squares, and we know that linear [...]
Pingback by | May 21, 2010 | Reply
5. Minor mistake: you missed a $g^{-1}$ on the second line of re-writing $A$.
Comment by | February 17, 2013 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"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": 53, "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": 0.9447290897369385, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/tagged/mean-square-error+standard-deviation | # Tagged Questions
0answers
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### What is the purpose of subtracting the mean from data when standardizing?
What is the purpose of subtracting the mean from data when standardizing? and What is the purpose of dividing by the standard deviation?
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### Step by step correlation calculation
I must understand how I can calculate the correlation for the following probability variables. ...
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### Accuracy of a telescope, measurement without systematic error [closed]
I got these values from the measurement: 20.1, 20.2, 19.9, 20, 20.5, 20.5, 20, 19.8, 19.9, 20 I know that the actual distance is 20 km and the error of the measurement is not affected by the ...
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### Getting the correct error for a mean calculation
A constant k needs to be calculated including its gaussian error. $k = f_{(u,t)}$ $k_i$ can be calculated with the values and errors of $u_i$ and $t_i$ and their respective errors. Main issue 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": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "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": 0.9140844941139221, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/57076/list | ## Return to Answer
2 added 13 characters in body
The niceness condition you want is on the action, not on the space $X$. Specifically, you want to have that $X\to X/G$ is a principle $G$-bundle, so that we have a Serre spectral sequence as you describefor $G\to X\to X/G$. Of course, since you're assuming that $G$ is a finite discrete group, the singular cohomology of $G$ is free, and only in degree 0. In fact, the requirement of being a "principle $G$-bundle" is the same as $X\to X/G$ being a covering space.
The problem: though the spectral sequence looks simple, and collapses immediately at the $E_2$ page, it's not really very useful, since all the interesting data is hidden in the local coefficient system (which is absolutely not trivial unless $G=0$.)
However, we can perhaps get the relationship you want in a much easier way if you're willing to modify the coefficient ring a bit. In particular, the answer is much simpler if you use a ring in which the order of $G$ is a unit. In that case, it's not hard to show directly (using covering space theory) that `$H^*(X/G)\to H^*(X)$` is an isomorphism onto the invariants of $G$, i.e. the subring `$H^*(X)^G$` of classes which are invariant under the action of $G$. This is an exercise in Milnor's Characteristic Classes, and I believe some form of it appears in Hatcher as well.
1
The niceness condition you want is on the action, not on the space $X$. Specifically, you want to have that $X\to X/G$ is a principle $G$-bundle, so that we have a Serre spectral sequence as you describe. Of course, since you're assuming that $G$ is a finite discrete group, the cohomology of $G$ is free, and only in degree 0. In fact, the requirement of being a "principle $G$-bundle" is the same as $X\to X/G$ being a covering space.
The problem: though the spectral sequence looks simple, and collapses immediately at the $E_2$ page, it's not really very useful, since all the interesting data is hidden in the local coefficient system (which is absolutely not trivial unless $G=0$.)
However, we can perhaps get the relationship you want in a much easier way if you're willing to modify the coefficient ring a bit. In particular, the answer is much simpler if you use a ring in which the order of $G$ is a unit. In that case, it's not hard to show directly (using covering space theory) that `$H^*(X/G)\to H^*(X)$` is an isomorphism onto the invariants of $G$, i.e. the subring `$H^*(X)^G$` of classes which are invariant under the action of $G$. This is an exercise in Milnor's Characteristic Classes, and I believe some form of it appears in Hatcher as well. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3, "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": 0.968184769153595, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/15804?sort=votes | ## When does a probability measure take all values in the unit interval?
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Let $\mathbb{P}$ be a probability measure on some probability space $(\Omega,\mathcal{A})$. Are there conditions on the $\sigma$-algebra $\mathcal{A}$ such that for every real number $c\in [0,1]$ we find a set $A\in\mathcal{A}$ with $\mathbb{P}(A)=c$. It is like the intermediate value theorem for continuous functions.
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## 4 Answers
A measure space $(\mathbb{P},\Omega,\mathcal{A})$ is atomless if for all $A\in\mathcal{A}$ with $\mathbb{P}(A)>0$ there exists $B\subset A, B\in\mathcal{A}$ such that $0<\mathbb{P}(B)<\mathbb{P}(A)$. Now according to a theorem of Sierpinski, the values of an atomless measure space form an interval. In particular, for probability spaces, every value in $[0,1]$ is taken. The original source of the article can be found here (in french). For a proof in english, you can look at on 215D on page 46 in Fremlin's book Measure Theory 2.
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2
So, the condition holds on spaces where the total mass of the atoms is at most 1/2, and on spaces where the sums of the atoms have gaps of width at most the mass of the space excluding atoms. – Douglas Zare Feb 19 2010 at 22:45
### 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's a concrete example of an atomless measure. Let $f \in L^1$ be an integrable function with total mass 1 (i.e. $\int_0^1 f = 1$). Define $$\mathbb P(A) = \int_A f(x) ~dx$$ for any Borel set $A$. It is a nice exercise to show that $\mathbb P$ is an atomless measure.
Note: $f$ is called the Radon-Nikodym derivative of $\mathbb P$ with respect to Lebesgue measure, and often written $f = \tfrac{d\mathbb P}{dx}$. If a random variable $X$ has distribution $\mathbb P$, then $f$ is called its density function.
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@TG: Instead of "Exercise: Prove...", maybe "It is a nice exercise to show..." or something like that? – Pete L. Clark Feb 19 2010 at 17:01
Pete, I like your rephrasing. I'll change my post. – Tom LaGatta Feb 19 2010 at 18:06
3
Only a mathematician would start a concrete example with "Let $f \in L^1$..." – Tom LaGatta Feb 19 2010 at 20:03
This is a property of $\mu$, not that of $\mathcal A$, and it is called being atomless. It is equivalent to not having sets $A \in \mathcal A$ of positive measure such that for all $B \in \mathcal A$, $B \subseteq A$ the measure $\mu(B)$ is either 0 or $\mu(A)$.
edit: Wikipedia article, complete with the proof of the property you describe from atomlessness.
edit: yup, the comments are right and I'm wrong. The precise condition for finite measures composed entirely of atoms to have full range is $a_n \leq \sum_{j>n} a_j$ - it is clearly necessary as $a_n-\varepsilon$ has to be produced somehow, and the greedy algorithm shows sufficiency.
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7
In fact, certain atomic measures also have range [0,1], so "atomless" is sufficient but not necessary. – Gerald Edgar Feb 19 2010 at 14:48
3
If you take a measure made up purely of atoms (the atoms having measures (a_1, a_2, a_3...) in decreasing order, what are the conditions guaranteeing full range? For example, a_i <= 2a_{i+1} suffices, but may not be necessary – Kevin P. Costello Feb 19 2010 at 20:11
A necessary and sufficient condition is that every atom is no larger than the sum of all smaller atoms, plus the non-atomic part.
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1
In particular a measure consisting only of atoms of mass $1/2, 1/4, 1/8, \ldots$ just barely suffices. – Michael Lugo Feb 25 2010 at 21:45 | {"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": 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": 0.9309103488922119, "perplexity_flag": "head"} |
http://planetmath.org/TheoryOfRationalAndIrrationalNumbers | # theory of rational and irrational numbers
The following entry is some sort of index of articles in PlanetMath about the basic theory of rational and irrational numbers, and it should be studied together with its complement: the theory of algebraic and transcendental numbers. The reader should follow the links in each bullet-point to learn more about each topic. For a somewhat deeper approach to the subject, the reader should read about Algebraic Number Theory. In this entry we will concentrate on the properties of the complex numbers and the extension $\mathbb{C}/\mathbb{Q}$.
There is also a topic entry on rational numbers.
# 1 Basic Definitions
1. 1.
A number is said to be rational if it can be expressed as a quotient of integers (with non-zero denominator). The set of all rational numbers forms a field, denoted by $\mathbb{Q}$.
2. 2.
Such rational numbers, which are not integers, may be expressed as sum of partial fractions (the denominators being powers of distinct prime numbers).
3. 3.
The real numbers are the set of all possible decimal expansions (where we don’t allow any expansion to end in all $9$’s). For the formal definition please see the entry real number. The real numbers form a field, usually denoted by $\mathbb{R}$.
4. 4.
A real number is said to be irrational if it is not rational, i.e. it cannot be expressed as a quotient of integers. The decimal expansion is non-periodic for any irrational, but periodic for any rational number.
5. 5.
For example, $\sqrt{2}$ is irrational.
6. 6.
Commensurable numbers have a rational ratio. See also sine at irrational multiples of full angle.
# 2 Small Results
1. 1.
The field $\mathbb{Q}$ is, up to an isomorphism, subfield (prime subfield) in every field where no sum of unities can be 0. One may also say that $\mathbb{Q}$ is the least field of numbers.
2. 2.
$\sqrt{2}$ is irrational. Similarly $\sqrt{d}$ is irrational as long as $d\in\mathbb{N}$ is not a perfect square.
3. 3.
The sum of two square roots of positive squarefree integers is irrational.
4. 4.
Rational and irrational: the sum, difference, product and quotient of two non-zero real numbers, from which one is rational and the other irrational, is irrational.
5. 5.
There exists real functions, which are continuous at any irrational but discontinuous at any rational number (e.g. the Dirichlet’s function).
6. 6.
Every irrational (and also rational) number is a limit of sequence of rational numbers (see real numbers). An example: the sequence $(1+\frac{1}{1})^{1},\,(1+\frac{1}{2})^{2},\,(1+\frac{1}{3})^{3},\,...$ converges to the number $e$.
7. 7.
The number e is irrational (this is not as difficult to prove as it is to show that e is transcendental). In fact, if $r\in\mathbb{Q}\setminus\{0\}$ then $e^{r}$ is also irrational. There is an easier way to show that e is not a quadratic irrational.
8. 8.
Every real transcendental number (such as $e$) is irrational, but not all irrational numbers are transcendental — some (such as $\sqrt{2}$) are algebraic.
9. 9.
“Most” logarithms of positive integers are irrational (and transcendental).
10. 10.
If $a^{n}$ is irrational then $a$ is irrational (see here).
11. 11.
A surprising fact: an irrational to an irrational power can be rational.
12. 12.
# 3 BIG Results
The irrational numbers are, in general, “easily” understood. The BIG theorems appear in the theory of transcendental numbers. Still, there are some open problems: is Euler’s constant irrational? is $\pi+e$ rational?
Type of Math Object:
Topic
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## Mathematics Subject Classification
11R04 Algebraic numbers; rings of algebraic integers
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## Versions
(v15) by alozano 2013-03-22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.8928914666175842, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/8294/how-much-of-a-star-falls-into-a-black-hole | # How much of a star falls into a black hole?
http://blogs.discovermagazine.com/badastronomy/2011/04/05/astronomers-may-have-witnessed-a-star-torn-apart-by-a-black-hole/
A lot of the star in the disc, a lot of the star in the jets, precisely how much of the star actually falls into the black hole?
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## 1 Answer
There is no correct answer for all cases. Of course the upper-bound is 100% but in practice, even if nearly all of the matter would fall in, a huge amount of energy will be radiated as x-rays and $\gamma$-rays due to heating in the accretion disk.
The dynamics of in-falling matter are quite complicated and very sensitive to initial conditions. Currently super computers are required to answer this question in any detail for a given set of initial conditions.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.920260488986969, "perplexity_flag": "head"} |
http://www.physicsforums.com/showthread.php?p=3405942 | Physics Forums
## How to visualize diff eq?
I don't feel that I can truly appreciate a math without being able to visualize it in my head. Generally speaking: calculus flows into areas, trig builds shapes, and linear algebra builds spaces, but I cannot for the life of me look at a diff eq and 'see' it, so to speak. While only proficient in trig and calculus, I'm learning linear algebra and diff eq but I'm finding it harder to grasp diff eq for this very reason.
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
Quote by cowmoo32 I don't feel that I can truly appreciate a math without being able to visualize it in my head. Generally speaking: calculus flows into areas, trig builds shapes, and linear algebra builds spaces, but I cannot for the life of me look at a diff eq and 'see' it, so to speak. While only proficient in trig and calculus, I'm learning linear algebra and diff eq but I'm finding it harder to grasp diff eq for this very reason.
Trying to visualize the differential equation itself I think is very difficult. I don't know how to do it, unless it represents a physical system where I derived the differential equation from. Visualizing solutions to ODE's at least is much easier. If you can't, then plot it in MATLAB or MAPLE to help you get an idea of what the solution is saying. I suggest looking at a simple mass/spring system, such as this:
http://en.wikipedia.org/wiki/Harmonic_oscillator
there are things we can solve by mathematical equations but not visualize it...as a proof directly , example we can visuaize (a+b)^2, or (a+b)^3......but can you visualize (a+b)^4 that easily but we can get a solution thats the beauty of mathematical expressions
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## How to visualize diff eq?
Phase plots might help.
http://tutorial.math.lamar.edu/Class...hasePlane.aspx
From further down on a google search,
www.dump.com/2011/05/18/a-cool-way-to-visualize-differential-equations-video/
I don't like to visualize them. Instead, I try to understand them. For example, with the equation $$\frac{\mathrm{d}y}{\mathrm{d}x}=x,$$ I think that how fast y changes is proportional to x. If I have $$\frac{\mathrm{d}y}{\mathrm{d}x}=x+x^2,$$ I think that there are 2 factors that contribute to the change of y - x and $x^2$.
Visualisation in the phase space or as a vector field might be helpful. However, visualisation in dimensions higher than 3 is difficult & can be misleading. Have you heard this one ? - A biologist is asked what his mental picture of a dog is. He says that the picture is as follows: DOG I think this is a little more than a mere joke.
Something all are familiar with are speed and acceleration which are simple examples of diff. equations. For example: $$v=\frac{dx}{dt}$$ $$a=\frac{dv}{dt}=\frac{d²x}{dx²}$$ where the velocity v is how long distance you got after a certain time. Or in other words, difference in distance over difference in time. And same for acceleration which describes how the speed has changed after a certain time. Don't know how that helps visualize it but you can feel acceleration :)
DEs come in a variety of sizes and shapes, so it's hard to give a general answer to your question. But there are helpful visualization methods for some common classes of DEs. Strogatz describes some of them in Nonlinear Dynamics and Chaos.
I. In Diff. Eq. there are a variety of methods for a variety of equations. Something to consider: Create a summary table showing A. what form the original equation looks like. B. steps in the solution method. C. Other notes or cautions to be aware of, possible comparisons & contrasts. If you lay these out in a clear pattern, it can help one visualize the course, & understand the varieties of methods to apply & when to apply them. II. Unfortunately, when starting a Diff. Eq. the visualization options tend to be limited to computer methods. I'm more familiar with the terms "vector field" or "slope field" See sosMath.com Any college level mathematical software will have and thankfully, wolframAlpha.com may be able to graph vector fields. See Mathematica syntax III. When solved it can be very instructive to plot a "solution family" or multiple solution curves. Using a different color for each solution can be useful. wolfram example Notice the "sample solution family", where x:0->2. Bye.
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http://unapologetic.wordpress.com/2011/03/28/local-rings/?like=1&source=post_flair&_wpnonce=c1ce40f6e1 | # The Unapologetic Mathematician
## Local Rings
Sorry for the break last Friday.
As long as we’re in the neighborhood — so to speak — we may as well define the concept of a “local ring”. This is a commutative ring which contains a unique maximal ideal. Equivalently, it’s one in which the sum of any two noninvertible elements is again noninvertible.
Why are these conditions equivalent? Well, if we have noninvertible elements $r_1$ and $r_2$ with $r_1+r_2$ invertible, then these elements generate principal ideals $(r_1)$ and $(r_2)$. If we add these two ideals, we must get the whole ring, for the sum contains $r_1+r_2$, and so must contain $1$, and thus the whole ring. Thus $(r_1)$ and $(r_2)$ cannot both be contained within the same maximal ideal, and thus we would have to have two distinct maximal ideals.
Conversely, if the sum of any two noninvertible elements is itself noninvertible, then the noninvertible elements form an ideal. And this ideal must be maximal, for if we throw in any other (invertible) element, it would suddenly contain the entire ring.
Why do we care? Well, it turns out that for any manifold $M$ and point $p\in M$ the algebra $\mathcal{O}_p$ of germs of functions at $p$ is a local ring. And in fact this is pretty much the reason for the name “local” ring: it is a ring of functions that’s completely localized to a single point.
To see that this is true, let’s consider which germs are invertible. I say that a germ represented by a function $f:U\to\mathbb{R}$ is invertible if and only if $f(p)\neq0$. Indeed, if $f(p)=0$, then $f$ is certainly not invertible. On the other hand, if $f(p)\neq0$, then continuity tells us that there is some neighborhood $V$ of $p$ where $f(p)\neq0$. Restricting $f$ to this neighborhood if necessary, we have a representative of the germ which never takes the value zero. And thus we can define a function $g(q)=\frac{1}{f(q)}$ for $q\in V$, which represents the multiplicative inverse to the germ of $f$.
With this characterization of the invertible germs in hand, it should be clear that any two noninvertible germs represented by $f_1$ and $f_2$ must have $f_1(p)=f_2(p)=0$. Thus $f_1(p)+f_2(p)=0$, and the germ of $f_1+f_2$ is again noninvertible. Since the sum of any two noninvertible germs is itself noninvertible, the algebra $\mathcal{O}_p$ of germs is local, and its unique maximal ideal $\mathfrak{m}_p$ consists of those functions which vanish at $p$.
Incidentally, we once characterized maximal ideals as those for which the quotient $R/I$ is a field. So which field is it in this case? It’s not hard to see that $\mathcal{O}_p/\mathfrak{m}_p\cong\mathbb{R}$ — any germ is sent to its value at $p$, which is just a real number.
## 2 Comments »
1. [...] this would mean evaluating a function at a point, yes, but here we interpret it in terms of the local ring structure of . Given a germ there is a projection , which we write as [...]
Pingback by | March 29, 2011 | Reply
2. very useful. might be good to include brief word on the local subrings of Q.
Comment by | February 23, 2013 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"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": 36, "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": 0.9345376491546631, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/78250?sort=oldest | ## Can taking the projective closure of an affine variety increase the degrees of its ideal generators?
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Say we have some equations $f_1(x)=0, \ldots f_k(x)=0$ defining a variety $X$ in ${\mathbb C}^n$ (not necessarily a minimal number of generators, and not necessarily of minimal degree), and suppose we want to know the ideal $\bar{I}$ of its closure $\overline{X}$ in ${\mathbb P}^n$. A naive question:
If all the original generators $f_i$ were of degree at most $d$, can we generate $\bar{I}$ using polynomials of degree at most $d$?
If not, what degree bound $d'$ can we give for a minimal generating set of $\bar{I}$?
Motivation, in short: Ideal saturation is an extremely slow operation, but solving a linear system to identify vanishing forms up to a certain degree is polynomial time, which would give a more practical way to compute $\bar{I}$ in high dimensions.
(Sorry, I'm new to computational algebra, and maybe people already know a fast way to compute projective closure...)
Motivation, in long: To compute $\bar{I}$, it's not enough to simply homogenize the generators with respect to a new variable $z$. Denote such homogenizations by $\hat{f}$, their ideal by $\hat{I}$, and their variety by $\hat{X}$. The problem is that $\hat{X}$ might contain irreducible components on the hyperplane at infinity $(z=0)$, so we need to saturate it, $\bar{I}=(\hat{I}:z^\infty)$, to eliminate these components and get the ideal of $\bar{X}$. But saturation is extremely slow! (I think it's at least doubly exponential time in the number of variables.) So I'd like to find $\bar{I}$ instead by solving a linear system on the vector space of homogeneous polynomials of degree at most $d$ to see which ones vanish, and declare that they cut out the variety $\bar{X}$, yay! But do they? If searching up to degree $d$ is not enough, how far do I have to go?
A simple example: Say $f_1 = x$ and $f_2 = y-x^2$ in ${\mathbb C}^2$, so $X$ is just the origin $(0,0)$. Suppose we first homogenize the generators $f_i$. The parabola $\hat{f}_2=yz-x^2=0$ is tangent to the line at infinity at $(x:y:z)=(0:1:0)$, which also lies on $\hat{f}_1=0$. So together they cut out $\hat{X} = { (0:0:1) \cup (0:1:0)}$, and we don't get $\bar{X}=(0:0:1)$ until we saturate $\hat{I}=\langle x,yz-x^2\rangle$ by $z$ to eliminate the component at infinity.
In this example, we already know the answer is the origin, and saturation is very fast for such a small system anyway. But for many variables, we don't already know what the variety looks like, and the saturation might not finish in a human lifetime, so it would be nice to solve a linear system instead.
Thanks for any help on this question!
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Just to be clear: Is $I$ radical? And, do you want to compute the radical of $\bar I$, or any ideal $J$ such that $\sqrt{J} = \sqrt{\bar I}$, or precisely $\bar I$? – auniket Oct 17 2011 at 5:22
My intention was not to take radicals of anything but simply to homogenize and saturate, but it turns out not to matter to the answer: in Manoj's answer below, all the ideals are radical, so the answer to any interpretation of the question is "no, the generator degrees can indeed go up." – Andrew Critch Oct 17 2011 at 6:36
## 1 Answer
Consider $R = \Bbbk[x_1, \ldots, x_5]$ and $I = (x_1x_2^2-x_3^2, x_1x_4^2-x_5^2)$. When we homogenize w.r.t $z$, we get $(x_1x_2^2-zx_3^2, x_1x_4^2-zx_5^2)$ whose saturation w.r.t $z$ contains the binomial $x_3^2x_4^2-x_2^2x_5^2$ of degree $4$. The original generators are of degree at most $3$.
Look at Section 15.10 of Eisenbud's commutative algebra book; one should homogenize a Groebner basis of $I$. Therefore, bound for $d'$ in terms of $d$ might turn out to be quite large. (This is just my feeling!)
PS: in the `Simple Example', do you mean that $\hat{f_2} = x^2-yz$? As stated the ideal $\hat{I}$ is already saturated w.r.t. $z$ and does not have a component at $[0:1:0]$.
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Thanks, Manoj! To add to your answer in response to auniket's comment, your ideal $I$ is in fact radical, so its (saturated) homogenization $\bar{I}$ with respect to $z$ is also radical. It turns out to be minimally generated by the three polynomials $x_1x_4^2-x_5^2z$, $x_1x_2^2-x_3^2z$, $x_3^2x_4^2-x_2^2x_5^2$, so any generating set must contain at least one quartic. Thanks also for the correction to my typo, which I've now fixed :) – Andrew Critch Oct 17 2011 at 6:50 | {"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": 65, "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": 0.9453034996986389, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/49993/distinguishing-between-valid-and-fallacious-arguments-propositional-calculus | # Distinguishing between valid and fallacious arguments (propositional calculus)
I am having some difficulties understanding logical arguments. I was taught that the notion of a valid argument is formalized as follows:
"An argument $P_1, P_2,\cdots , P_n ⊢ Q$ is said to be valid if $Q$ is true whenever all the premises $P_1, P_2,\cdots , P_n$ are true. An argument which does not satisfy this condition is a fallacy."
But (according to my book) the argument $$p \rightarrow q,\quad q\;⊢ \;p\;$$ is a fallacy, but I don't see why. If we construct a truth table
$p$ | $q\;$ | $p \rightarrow q$ |
T | T | T |
T | F | F |
F | T | T |
F | F | T |
we see that when $p$ and $q$ are true, $p \rightarrow q$ is also true (line 1). So then, why isn't this valid? I do know that for an argument to be valid, with premises $P_1, P_2,\cdots , P_n$, the proposition $$(P_1 \land P_2 \land \cdots \land P_n) \rightarrow Q$$ should be a tautology, but that doesn't dispel my confusion.
What am I missing?
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All valid arguments in propositional logic can be mechanically identified via truth tables. – André Nicolas Jul 7 '11 at 2:17
## 3 Answers
I think one the important points is that truth-functionality is too coarse for certain applications/issues, in the sense that the inner-structure of the statements does not matter, e.g., the coarseness does not prevent statements of this sort:
All Men are Mortal:=P
Adam is not Mortal:=Q
Since we can just assign to P the value T , independent of its actual meaning, same for Q.
More formally, the issue of material implication guarantees that if Q is true in P->Q, then (P->Q) itself is true, so that:
($(P\rightarrow Q)$ /\Q) is always true when Q is. In particular, this last argument is true when Q is true and P is false. This assignment, then, of P:=T and Q is false is one which makes your premises true and your conclusion false.
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In the third line of the truth table, $q$ and $p \to q$ are both true and $p$ is false. So $q, p\to q \vdash p$ is not a valid argument.
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To perhaps clarify (or perhaps just belabor) Carl's response, the premises of your argument are (p->q) and q, while the conclusion is p. Thus by the third line the argument is invalid. – Doug Spoonwood Jul 8 '11 at 15:00
To expand upon what Carl Mummert said,an argument is invalid if there is an instance where all of the premises are true and the conclusion is false. As stated this happens on the third line of your truth table, as 'p' is your conclusion.
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http://math.stackexchange.com/questions/174669/a-sum-of-terms-like-1-sin2x-1 | # A sum of terms like $(1 + \sin^2x)^{-1}$
How to prove that $$\frac12\frac1{1+\sin^2 x} + \frac12\frac1{1+\cos^2 x} + \frac12\frac1{1+\sec^2 x}+ \frac12\frac1{1+\csc^2 x} = 1?$$ Some genius please help me I have been stuck at this for one whole day.
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Please be clearer about the equation. Some would read 1/2(1+sin^2 x) as $\frac 1{2(1+\sin^2 x)}$ or at least worry that you meant that. Either (1+sin^2 x)/2, (1/2)(1+sin^2 x), or (best) $\LaTeX$. Trying to solve the problem, I see I had it wrong, so please 1/(2(1+sin^2 x)), or $\LaTeX$ – Ross Millikan Jul 24 '12 at 14:23
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I would start by getting rid of $\sec$ and $\csc$, replacing them with $\frac 1\cos$ and $\frac 1\sin$. Then you must have identities for $1+\sin^2 x$ and $1 + \cos^2 x$. Where does that take you? – Ross Millikan Jul 24 '12 at 14:30
## 2 Answers
Forget about the $2$'s for a while. We have $$\frac{1}{1+\sec^2 x}=\frac{1}{1+\frac{1}{\cos^2 x}}=\frac{\cos^2 x}{\cos^2 x+1}.$$ Now add $\frac{1}{1+\cos^2 x}$. We get something very simple. Continue.
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Observe that for any $\alpha> 0$ we have $$\frac{1}{1+\alpha}+\frac{1}{1+1/\alpha}=\frac{1+1/\alpha+1+\alpha}{(1+\alpha)(1+1/\alpha)}= 1$$ so $$\frac{1}{1+\sin^2x}+\frac{1}{1+\csc^2x} = 1$$ wherever $\csc x$ is defined and similarly $$\frac{1}{1+\cos^2x}+\frac{1}{1+\sec^2x} = 1$$
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http://math.stackexchange.com/questions/243972/probability-question-on-picking-letters/243974 | # Probability question on picking letters
5 letters: a, b, c, d, e I randomly pick 3 letters, whats the chance of having a and b in those 3 letters
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## 1 Answer
You have a total of $5$ letters and you are choosing $3$ from them. Hence, the total number of ways of choosing three letters is $\dbinom{5}3$.
Number of ways in which you have $a$ and $b$ as two of your three letters is $\dbinom{3}1$, since the third letter you can choose from any of the remaining three letters $c$, $d$ and $e$.
Hence, the probability is _____.
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I presume you can also derive '3 choose 1' by saying that, for example, there is one choice for the first selection, one for the second and three for the third and so by the multi. Principle there is 3 ways. Why don't we have to divide by 3! here though if you do it this way? – CAF Nov 24 '12 at 23:06
@CAF Yes. In fact that is how I have derived. However, I don't see why you want to divide by $3!$. – user17762 Nov 24 '12 at 23:38
Does the method I describe not mean that I have ordered the choosing of the letters? (I.e one choice for the first letter implies this one must be 'a', for example) I think I meant multiplying by 3! neglects the order? – CAF Nov 25 '12 at 9:19 | {"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": 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": 0.9643949866294861, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/39577/list | ## Return to Question
4 tag added
3 added 12 characters in body
Given a pair of Koszul dual algebras, say `$S^*(V)$` and `$\bigwedge^*(V^*)$` for some vector space $V$, one obtains an a triangulated equivalence between their bounded derived categories of finitely-generated graded modules.
Given a pair of Koszul dual operads, say the Lie and commutative operads, what is the precise analogue of a derived equivalence between their categories of algebras?
2 +operad tag
1
# koszul duality and algebras over operads
Given a pair of Koszul dual algebras, say `$S^*(V)$` and `$\bigwedge^*(V^*)$` for some vector space $V$, one obtains an equivalence between their bounded derived categories of finitely-generated graded modules.
Given a pair of Koszul dual operads, say the Lie and commutative operads, what is the precise analogue of a derived equivalence between their categories of algebras? | {"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": 0.8794468641281128, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/28158/a-learning-roadmap-request-from-high-school-to-mid-undergraduate-studies/28175 | ## A Learning Roadmap request: From high-school to mid-undergraduate studies
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Dear MathOverflow community,
In about a year, I think I will be starting my undergraduate studies at a Dutch university. I have decided to study mathematics. I'm not really sure why, but I'm fascinated with this subject. I think William Dunham's book 'Journey through Genius ' has launched this endless fascination.
I can't wait another whole year, however, following the regular school-curriculum and not learning anything like the things Dunham describes in his book. Our mathematics-book at school is a very 'calculus-orientated' one, I think. I don't think it's 'boring', but it's not a lot of fun either, compared to the evalutation of $\zeta(2)$, for example. Which is why I took up a 'job' as as a tutor for younger children to help them pass examinations. I wanted to make money (I've gathered about 300 euros so far) to buy some new math-books. I have already decided to buy the book ' Introductory Mathematics: Algebra and Analysis' which should provide me with some knowledge on the basics of Linear Algebra, Algebra, Set Theory and Sequences and Series. But what should I read next? What books should I buy with this amount of money in order to acquire a firm mathematical basis? And in what order? (The money isn't that much of a problem, though, I think my father will provide me with some extra money if I can convince him it's a really good book). Should I buy separate books on Linear Algebra, Algebra and a calculus book, like most university web-pages suggest their future students to buy?
Notice that it's important for me that the books are self-contained, i.e. they should be good self-study books. I don't mind problems in the books, either, as long as the books contain (at least a reasonable portion) of the answers (or a website where I can look some answers up).
I'm not asking for the quickest way to be able to acquire mathematical knowledge at (graduate)-university level, but the best way, as Terence Tao once commented (on his blog): "Mathematics is not a sprint, but a marathon".
Last but not least I'd like to add that I'm especially interested in infinite series. A lot of people have recommended me Hardy's book 'Divergent Series' (because of the questions I ask) but I don't think I posess the necessary prerequisite knowledge to be able to understand its content. I'd like to understand it, however!
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Please take what Harry says with a grain of salt,Max.He thinks anything with motivation is not mathematics and that's not good for beginners. – Andrew L Jun 14 2010 at 21:10
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Max, I think that the sooner you stop worrying about questions like "But isn't Topology more of a graduate subject?", the better. Most Bourbaki's books do not make good first reading for the subject, that's true, but there are topology books that can, and should be, read while still undergraduate. There is no such thing as undergraduate/graduate subject, there is mathematics and something else. – Vladimir Dotsenko Jun 14 2010 at 21:34
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1. Some of the answers you have received are a little odd, IMO. I see some fairly difficult graduate level books being recommended for a high school student who knows some calculus. (Homological algebra? Really?) I honestly don't know what to make of this. 2. Generic advice that may or may not be something you need to hear: Don't get discouraged. Math is hard for everybody. Be persistent, but if you have a book you can't make progress on, don't feel the least bit of shame in turning to a more elementary treatment or going back to learn prerequisite topics or whatever it takes. – Mike Benfield Jun 14 2010 at 22:10
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Aack! If you want italics, on MO put underscores `_` or asterisks `*` on either side, not dollar signs. In TeX, use `{\em` text `}`. Dollar signs make the computer process whatever's inside as math, as if you had all those variables to multiply together. The classic example is $difference$ versus difference — notice the spacing around the `f` s. (In the default TeX font, the correct look is `$\textit{difference}$`.) The spacing is even weirder for words with `ffi`: `$spiffier$`, `$\textit{spiffier}$`, spiffier. – Theo Johnson-Freyd Jun 14 2010 at 22:10
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I looked at the table of contents for the book the OP linked to ("Introductory Mathematics:...") and they define things like sets, functions, injective, bijective, complex numbers, vector spaces, etc. I think it would be considerate if people kept this in mind when suggesting books. Some people have not done this, and I would agree with Mike Benfield that you shouldn't be discouraged if a randomly chosen book from the answers is too difficult to understand right now. Many of them will still be difficult after several years of studying math. – Peter Samuelson Jun 15 2010 at 0:33
show 18 more comments
## 31 Answers
I'm not a big fan of full roadmaps and reading lists. Exploring mathematics is something that can be totally different depending on where and who you are. Any serious roadmap needs to be flexible and take account of the course: reading maths is a skill (one you seem well on your way to learning btw! but still...), initially you may find actual teaching easier to grasp- and your reading should work along side that. So here's my attempt at a flexible roadmap:
1) Buy some very carefully chosen books and read them cover to cover: There's a lot of baffling books out there- even some that look really UG friendly can have you weeping by page 5 in your first year- and you only want 4-5 to start with (any more will just be too expensive and you won't get round to reading all of them- top up via the library). My recommendations are: 'Naive set theory- Halmos', 'Finite dimensional vector spaces- Halmos', 'Principles of mathematical analysis- Rudin' and 'Proofs from the book- Aingler and Zeigler'.
[These, ostensibly, cover the exact same material as the book you have decided to buy- but to develop quickly I urge you to buy more mathematical texts like these: Halmos' and Rudin's writing styles are very clear but technical (in a way that the book you are interested in will not be), and will make you a better mathematician faster than any book that tries to 'bridge the gap' ever would. I also seconded Owen's call for proofs from the book: it is simultaneously inspiring and useful as a way of seeing 'advanced' topics in action- it's something you'll keep coming back to, right up to your third year!]
2) Do all of the excercises: Or as much as you can bear to- even if it looks like it's beneath you (if you're half decent- a lot of first year will!) you will be surprised as to how much it helps with your mathematical development (and the crucial high mark you'll need for a good PhD placement). This applies to classes and your 4-5 text books.
3) Ask your tutor about doing some modules from the year above: If you've read all of those textbooks and done all of the excercises, you will be ready. Get some advice from your tutor about what would be best and roll with it (most unis won't make you take the exam, so if you don't feel comfortable you're fine). Taking something like metric spaces or group theory in your first year will put you top of the pile.
4) Keep doing all of these things: Immerse yourself in maths- keep on MO, meet likeminded people and no matter how slow the course seems to be moving, no matter the allure of apathy: keep at it. Advice for later books would be pointless now, but there will be people who can give it to you there and then (use maths forums if you want). Oh, and never rule out an area- you never know where intrigue will come from...
Best of luck, Tom
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I would recommend Courant and Robbins' "What is Mathematics." It is quite inexpensive, and gives a meaningful introduction to many areas of mathematics. I think you can browse the table of contents on Amazon.com to see what is included.
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There are lots of good answers here, so I'm not going to add any additional book recommendations. I just want to warn you of one misconception I had when I was in your position. It is best illustrated by example but don't worry if you don't understand all the terms. That's part of the point.
A vector space is mathematical structure defined in terms of another called a field: for example the real numbers are a field and the plane is a vector space. Now, a field is a special case of another structure called a commutative ring. A field is just a commutative ring in which you can do division; the integers are an example of a commutative ring. Now, commutative rings are built out of abelian groups, which are themselves a certain kind of group.
My reaction to seeing such definitions was to assume that the best way to learn about the ones at the top of the hierarchy (e.g. vector spaces) was to develop a solid understanding of those at the bottom (e.g. groups). This seems natural because math is supposed to be a very methodical thing and logically if B is defined in terms of A you might expect you'd want to understand A first.
It turns out that this is for the most part wrong. The reason is that there are all sorts of crazy groups out there, but the abelian ones are some of the simplest and easiest to understand. This type of reasoning can be applied at each level, and when you get all the way up to vector spaces, you get a family of objects which behave very nicely, having eliminated some complicated behavior at each stage.
Of course, you won't be able to appreciate quite how nice the situation is until you later on learn what can go wrong when you take a few steps down the hierarchy. But generally speaking, it is easier to learn about objects with lots of structure than those which have very little.
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Definitely agree. But, at the same time, there is a sense in which additional generality makes things easier to understand (e.g. defining Lebesgue integration via abstract measures instead of using step functions and monotone convergence as it was done before abstract measure theory was developed). So I think it goes both ways. – Akhil Mathew Jun 15 2010 at 23:39
Hi Max,
I would suggest starting with group theory, analysis(real and complex), elementary number theory, algebraic number theory, analytic number theory and then working your way up to algebraic geometry, differential geometry and so on.
I found the books published by the Mathematical Association of America very useful to get some intuition.
You may also want to browse through the American Math Monthly. It has lots of tutorial-style articles on various subjects. Search mathscinet for articles which have won expository writing awards. For example, I found this article quite useful
MR0401754 (53 #5581) Abhyankar, Shreeram S. Historical ramblings in algebraic geometry and related algebra. Amer. Math. Monthly 83 (1976), no. 6, 409--448.
Make an attempt to connect the problems which occur across various areas of math. See DIEUDONNE's Panorama of pure mathematics for a good overview of various areas of math.
MR0662823 (83e:00003) Dieudonné, Jean Alexandre A panorama of pure mathematics. As seen by N. Bourbaki. Translated from the French by I. G. Macdonald. Pure and Applied Mathematics, 97. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London, 1982. x+289 pp. ISBN: 0-12-215560-2 00A05
Hope you have an interesting journey...!
This figure connecting all math may be helpful in your future research Sourced from http://space.mit.edu/home/tegmark/toe.gif
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All math? I think not! – Andrew Stacey Jun 15 2010 at 9:51
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I agree with Andrew Stacey. That's a cool diagram, but it's from the viewpoint of physics as the only end goal. – Chris Phan Jun 15 2010 at 10:56
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Buying access to a university library is a good idea, but may not take the place of buying some texts. In the university libraries of my experience, the classic texts discussed in the other answers would be perpetually checked out. – Nate Eldredge Jun 15 2010 at 23:46
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Max,
I am a current undergraduate student and remember the feeling of wanting to know more before starting my degree. My university sent out a letter recommending that we all buy a certain analysis text before the start of the course, and that if we wanted to do some work before term that this was the book to read... Suffice it to say that the book was awful, and not even particularly useful once I had started at the university... a complete waste of my time and money.
Whilst this is slightly biased towards my own interests, if you are interested in combinatorics at all then I fully recommend 'Combinatorics and Graph Theory' by Harris, Hirst and Mossinghoff. The book is certainly readable for an enthusiastic school student, and really covers the basics of combinatorial enumeration and selected topics in graph theory. What's more the book is written with great clarity, and a good sense of humour. It starts at a basic level, but works up to material that there was not room to fit into a term long second year module.
Another text that I cant speak more highly of is 'Proofs from the Book' by Aigner and Ziegler. This is a compendium of interesting and simple proofs taken from many areas of mathematics. The title refers to a line of Paul Erdos. To Erdos, a book proof was a beautiful, elegant solution to a problem... so that is exactly what you find in here. Not all of it will be accessible immediately, but a lot will, and by the time you finish your first year you should be equiped to understand all of it.
Finally, from a softer approach: Martin Gardner's columns from Scientific American have been collected into several volumes which are readily available. Gardner often wrote about pure mathematics but also sometimes about philosophy of science, or physics. The articles are written so that the reader needs no technical knowledge, but I still read through the books just to find out a little bit about something that I haven't studied. Even if it is something that I do know about, reading Gardner is still a great exercise in seeing how good teachers really work: and there can be no doubt that Gardner was an excellent educator.
Best of luck.
Owen.
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I second the recommendation of Proofs from the Book. – Chris Phan Jun 15 2010 at 10:59
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@Owen I'm VERY glad someone finally recognized the wonderful text by Harris,Hirst and Mossingoff on a very baffling subject. It is simply one of the most readable and perfectly structured introductions to any subject I've EVER seen. After reading this book and absorbing it's contents,the student will be more then prepared to move on to more advanced texts on the subject like Gross and Yellen's GRAPH THEORY WITH APPLICATIONS (the bible of the subject,in my opinion) and Bona's texts on combinatorics. – Andrew L Jun 15 2010 at 13:46
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Here are a few introductory books that I've found useful. I'm in a somewhat similar situation as you, I suppose--a year older, but having had to learn whatever mathematics that I know mostly independently (though I've been lucky to have been able to talk to several mathematicians on various occasions as well as take a couple of courses). For more advanced ones, I'd certainly be the wrong person to ask. Anyway, here are some of the books that I've used (or am currently using).
Algebra: I think Herstein's Topics in Algebra is often recommended as a good introductory-level undergraduate textbook (which I learned material from). Lang's Algebra is a nice follow-up, but I think a lot of people (certainly including myself) would be turned away from the subject if Lang were the first thing they opened.
Analysis: Rudin's Principles of Mathematical Analysis is great, but the real fun is in Real and Complex Analysis (you don't have to read all of the former to get into the latter). Also, Peter Lax's Functional Analysis is very enjoyable reading.
Differential geometry: I personally found it very hard to read Spivak's Calculus on Manifolds because there wasn't much motivation (like, why would one define a tangent space when you're just working with open subsets of $\mathbb{R}^n$?), but volume 1 of his A Comprehensive Introduction to Differential Geometry has similar material with much more explanation and motivation.
Number theory: My first introduction to number theory was from Niven and Zuckerman's book. Serre's A Course in Arithmetic is also very interesting, but I would consider that more of a second or third read (I tend to find Serre's books rather terse and consequently struggle with them, but this is likely just a personal failing). For instance, you need to know what a projective limit is to read it.
I wish I knew a really good introduction to algebraic number theory. The ones I've seen have tended to be somewhat difficult (i.e., presupposing a fair bit of material in Lang's algebra such as familiarity with localization, noetherian rings, etc.--topics that might be omitted in a first course on abstract algebra). I think Lang's Algebraic Number Theory is a great book, but it took me an enormous amount of time before any of it started to make any sense at all. And there are other "elementary" books on algebraic number theory that never get anywhere interesting.
Ireland-Rosen's A Classical Introduction to Modern Number Theory has lots of fun stuff in it and is a great book to read after one understands elementary abstract algebra (hat tip to Vladimir Dotsenko for pointing this out; I had read the book and then forgotten about it). There is also a little bit of algebraic number theory in it.
Linear algebra: I learned this from Hoffman and Kunze's book, but I think the subject is nicer in a more abstract context (e.g. after one has talked about rings).
Computer science: Wait, hold on! I think I'd be remiss if I didn't mention Sipser's An Introduction to the Theory of Computation, which has to be one of the most enjoyable books I've ever seen---and it is basically mathematics.
Logic: Ebbinghaus, Frum, and Thomas have a very nice book on mathematical logic in the UTM series.
Topology: Dugundji's book is a bit old, but I found the exposition very crisp and enjoyable. At the same time, I suppose it's not good for geometric intuition. I have heard great things about the books by Munkres but have not (yet) read them myself.
The books in the Carus mathematical monograph series are all accessible, pithy, and enjoyable; Krantz's Complex Analysis: The Geometric Viewpoint is one that I'm enjoying looking at right now.
Another bit of advice: You don't have to finish a book to "graduate" to another one! Skipping around is something many mathematicians do, and one never "really" understands an area of mathematics (at least, not as an undergraduate), so it's more efficient to move on to other things. Nor do you have to know everything. (This isn't me giving advice; it's me regurgitating something that a very respected mathematician told me a while back, and was quite a revelation for me.)
I obviously don't know whether you're near a university library and have borrowing privileges; I happen to live near three (albeit ones from liberal arts schools with no graduate math department). If not, I strongly recommend buying used books, since math textbooks tend to be ridiculously overpriced for some reason. Fortunately, there are many good resources on the web: James Milne's site is excellent, for instance.
Anyway, if you need more, the bibliography on my blog has a longer list (but the ones here should probably keep you busy for at least a little while). You could also try contacting a professor at a nearby university to see if he or she is willing to mentor you for a research project; there may be programs through which this is possible (though admittedly I have no idea how it works in Europe). The benefit of this is that you'll end up absorbing a lot of new mathematics along the way as well as better understanding what you already know.
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There's an old book by Hasse on algebraic number theory, I think, which is very well written. Also, "A Classical Introduction to Modern Number Theory" by Ireland and Rosen is a good read. – Vladimir Dotsenko Jun 14 2010 at 22:50
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Of course! How did I forget Ireland-Rosen? – Akhil Mathew Jun 14 2010 at 23:00
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Samuel's book "Introduction to the algebraic theory of numbers" is an excellent first book on algebraic number theory: rather self-contained, good algebra technique, excellent examples/applications worked out, proves some real finiteness theorems, and nice (but slightly small) selection of exercises at the end. – Boyarsky Jun 15 2010 at 0:19
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I second Boyarsky - we used it for a semester undergraduate course in algebraic number theory (we finished the book about 2/3 through the semester, then talked about L-functions and proved the Chevotarev Density Theorem). Another good book with slightly more problems and material than Samuel is Marcus's Number Fields. And if you really feel that both of those "never get anywhere interesting," I recommend Number Fields by Janusz as a second read. You can skip most of the first four chapters if you've done Marcus, and Janusz discusses class field theory. – David Corwin Jun 16 2010 at 14:54
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Thanks again Akhil (my comment was removed for some mysterious reason). Perhaps we meet each other one day, as 'true' mathematicians ;). – Max Muller Sep 24 2010 at 23:19
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Calculus: Micheal Spivak's Calculus(the definitive theoretical presentation of calculus),Donald Estep's Practical Analysis In One Variable,and Gilbert Strang's Calculus (the latter 2 for applications). Strang is available online for free.
Algebra/Linear Algebra: A Course In Algebra by E.B. Vinberg (gives an integrated introduction to both linear and abstract algebra with lots of applications,especially to geometry:VERY readable and reaches a very high level of coverage by the end), Michael Artin's Algebra (more intense,but similar in spirit,a second edition coming out later this year,can't wait for it)
Category Theory: Category Theory by Steven Awodey(the definitive introduction for beginners-I wouldn't recommend anything else for a beginner)
Topology: A First Course In Topology:Continuity And Dimension by John McCleary (the best introduction to the subject)
Overview of Mathematics: Mathematics:It's Content,Methods And Meaning by A.N.Kolomogrov,etc. (The great classic by 3 Russian masters of mathematics,will give you the best bird's eye view of the subject)
That should get you started and give you the tools needed to go forward. Good Luck and welcome to the sorcerer's guild!
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Max, if you like pictures, check out "A topological picturebook" by Francis, it's quite good and informative! – Vladimir Dotsenko Jun 14 2010 at 21:54
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@Max: EVERY mathematics book for beginners should have pictures and don't let anyone try and make you feel stupid for asking that. Human learning is largely spacial in nature. A picture is not a proof-but a proof or concept can be made much more understandable with a picture accompanying it. This is ESPECIALLY true in topology and geometry. – Andrew L Jun 14 2010 at 22:01
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Re: pictures. I would say that not just "EVERY mathematics book for beginners" but indeed every mathematics book should be replete with pictures. Humans think well visually. But note that pictures have at various times been antithetical to mathematical writing. Euclid and the other Greeks certainly drew pictures when figuring out their work, but their writing are completely devoid of them, because a picture is not a proof, which should apply to any (theoretically perfect) geometric construction, not to the (necessarily imperfect) representation in a given picture. – Theo Johnson-Freyd Jun 14 2010 at 22:14
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Re whether every math book should be replete with pictures: Pictures often help, but I would not go so far as to say that they always help. Sometimes the movie adaptation of a novel is better than the novel. But sometimes the novel is better -- it might, for example, "leave more to the imagination". Re whether a picture is a proof: Try asking Rob Kirby whether a picture is a proof. – Kevin Lin Jul 20 2010 at 9:10
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Some books I enjoyed as a first year undergraduate:
Abstract Algebra by Dummit and Foote: I have heard many people speak ill of this text. However, from the perspective of one who is interested in picking up the fundementals of algebra, in my opinion this is the text.
Introduction to Topological Manifolds by Lee: Don't let the word manifold scare you off. This book is essentially an introductory topology book with all of the boring bits left out. I know it is a GTM but I was able to pick it up during my first year of Uni and learn a lot of topology from it.
Munkres topology: This is "the" book for learning point set topology. It is a bit gentler than Lee's and goes through all the relevant set theory you will ever need (unless you become a set theorist). I picked this book up during my first year at Uni and have been opening it regularly ever since.
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I also recommend Principles of Mathematical analysis by Rudin. It has already been mentioned though. It is an amazing book. It is also the only book which i have ever really studied from start to finish. – Daniel Barter Jun 14 2010 at 22:16
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I don't think Max here is an ordinary student. He seems like he is interested in becoming a mathematician, and he seems very bright. In this light, even if he is not able to handle Lee's topological manifolds straight away, I would be very surprised if he cant handle Rudin's Principles of Mathematical analysis. Once he has mastered this book, he should be able to move onto Lee's book easily. – Daniel Barter Jun 14 2010 at 22:29
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@ Andrew L: Rudin was my first introduction to analysis. In hindsight, i knew what a derivative was (but not to much more) and i could mechanically evaluate integrals from high school. Max seems to know a lot more than I did when I first picked up rudin, and I didn't have to much trouble working my way through it. I am not telling him that he should read rudin, I am just telling him that I think he would be able to handle it if he wanted to and I think it is a good book. – Daniel Barter Jun 14 2010 at 22:35
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@Andrew L: I get the feeling that you've never even read Rudin or Bourbaki. – Harry Gindi Jun 15 2010 at 5:02
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Andrew, chill out. Different people have different tastes. – Kevin Lin Jul 18 2010 at 8:01
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What follows is mostly a list of book that I was recommended as best books in the respective area by those who were teaching me mathematics in Moscow (with some additions I came across later).
Generally great introductory (but infinitely far from pointless) books:
"What is mathematics?" by Courant and Robbins
"Math talks for undergraduates" by Serge Lang
Real analysis: "Principles of Mathematical Analysis" by Walter Rudin
"Calculus on manifolds" by Spivak
Complex analysis: "Elementary Theory of Analytic Functions of One or Several Complex Variables" by Henri Cartan
Linear algebra: Lectures on Linear Algebra by I.M.Gelfand and Linear algebra and geometry by A.I.Kostrikin and Yu.I.Manin
Great introduction to various ideas of algebra and topology: Abel's theorem in Problems and Solutions by Alekseev. Another advantage of this book, as well as Halmos "Linear Algebra Problem Book" mentioned elsewhere in answers is that it gives a balanced set of problems - and one of the most efficient ways of learning new things is through problem solving.
Algebra: "Algebra" by Serge Lang
Differential equations: "Ordinary Differential Equations" by V.I.Arnold
Geometry/mechanics: "Mathematical Methods of Classical Mechanics" by V.I.Arnold
Topology: "First concepts of topology" by Steenrod and Chinn
Category Theory: partly from the abovementioned "Algebra" by Lang, partly from Conceptual mathematics: a first introduction to categories by Lawvere and Schanuel
More Algebra: "Associative Algebras" by Richard Peirce
Homological Algebra: "Methods of homological algebra" by S.I.Gelfand and Yu.I.Manin
Homotopic topology: "Homotopic Topology" by Fuchs, Fomenko, and Gutenmacher
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@Vladimir Lang's Algebra is WAY too difficult for a beginner even at Moscow State. I'm sure most students that tried to use it struggled. Frankly,most of the Russian trained mathematicians I know learned algebra from Kostrikin's books. – Andrew L Jun 14 2010 at 21:58
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Andrew, I believe that those Russian trained algebraists whom I know who did not learn from Lang (and they are minority), learned algebra from Vinberg's book. In any case, there is a possibility that we know different mathematicians, and don't intend to question your statement. I hope that you don't intend to question mine either. – Vladimir Dotsenko Jun 14 2010 at 22:05
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I think that for someone who has not studied abstract algebra, the book by Dummit and Foote is much better than Lang. They make a much more concerted effort to explain and motivate concepts, which I think is important when first learning a subject. – Peter Samuelson Jun 15 2010 at 0:25
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@ Vladimir I'm not-and I probably know mathematicians from a previous generation. Vinberg's text is my algebra reference of choice and I am SO glad it's available now to an English-speaking audience! – Andrew L Jun 15 2010 at 4:07
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There are (today) many popular books on mathematics. I don't know any very good ones. Kac and Ulam, Mathematics and Logic, is an older style of popular book, pretty good I think, and on Google Books. (I'd be embarrassed if I didn't at least mention Timothy Gowers, Mathematics: A Very Short Introduction.)
There are plenty of textbooks, and books put together from lecture notes. These are mostly not so exciting.
You really need a book on analysis, one on algebra, and one on geometry or topology. These are for core ideas. Fourier Analysis by Thomas Körner is mostly on Google Books: Körner is the right kind of writer for your requirements. I learnt a great deal from Serge Lang, Algebra, an early edition. But his writing now seems strange to me - bias towards number theory, which is OK really, and later editions are very different. Geometry and topology are hard. Maybe Penrose's book on everything might help with concepts (bias towards physics). Any book that helps you grasp what a Lie group is. None of these writers are "orthodox".
Classics: Hardy's Divergent Series you mention, but it is too hard to read, maybe few really can these days. Weyl's Classical Groups has always been impossible? Serre, A Course in Arithmetic, yes.
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@Charles Halmos is a wonderful author and you made me realize the one book I should recommend to everyone thinking a career in math:I WANT TO BE A MATHEMATICIAN:AN AUTOMATHOGRAPHY. A MUST READ.To be honest,though-I think between the 2 texts on linear algebra he wrote;FINITE DIMENSIONAL VECTOR SPACES and A LINEAR ALGEBRA PROBLEM BOOK,the latter would be much easier and enjoyable for a beginner. FDVS is very hard and abstract. – Andrew L Jun 14 2010 at 22:06
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The following texts were very successful in helping me make the transition from high school mathematics to university level mathematics. When I left high school in the UK system (A-levels) I had a reasonably thorough grasp of calculus, trigonometry, geometry, school level algebra, statistics and mechanics (these were covered in the A-level maths and further maths syllabi, if you're familiar with these at all).
General reasoning:
1. Copi and Cohen, 'Introduction to Logic'. A basic introduction to logical reasoning. Despite having studied as much mathematics as I could at high school, I was never taught to understand the logical structure of proofs. Reading this book helped me to begin reading proofs in undergraduate/graduate mathematics books.
Analysis:
1. Bartle and Sherbert, 'Introduction to Real Analysis'. This is useful for picking up the basics of real analysis at an elementary level and is also useful in learning to read proofs of elementary results. I found that this phase of my education was a bit tedious and that I was wanting to get ahead to a more advanced text. It is a useful complement to item 3 below.
2. Kolmogorov and Fomin, 'Introductory Real Analysis'. This is an excellent book for learning analysis and was used in some honors level analysis courses for students who were familiar with reading and writing proofs. I would emphasize that reading Copi and Cohen first is an imperative.
3. Walter Rudin, 'Principles of Mathematical Analysis'. Harder to read than Kolmogorov and Fomin, but it contains beautifully written proofs and is a great text to model your own answers on.
4. H.L. Royden, 'Real Analysis'. An excellent book to learn from on measure theory, integration and basic functional analysis (classical L^p spaces, etc).
5. T. Gamelin, 'Complex Analysis'. A good book on complex analysis with lots of motivating examples.
Algebra:
1. Allan Clark, 'Elements of Abstract Algebra'. This is a Dover publication and is rather cheap (probably around \$15 or so now). It consists of a hundred or more articles; you are given definitions and the proofs of a few important theorems. Everything else is an exercise. I believe this book did the most for me in helping to build intuition for abstract algebra.
2. Hoffman and Kunze, 'Linear Algebra'. A classic book on linear algebra which I do not think has been surpassed. It provides thorough and well written proofs. I believe this is a preferable text to Axler's book, unless you are somewhat heavily inclined towards analysis and/or do not enjoy doing computations.
3. Serge Lang, 'Algebra'. This was my second book in abstract algebra, after Clark's. It is beatifully written, which is why I prefer it to Hungerford's book. As another answerer mentioned, the typesetting in Hungerford's book is also somewhat off-putting, and Lang's book does not suffer from this defect.
Topology:
1. Munkres, 'Topology'. A clearly written text with a good supply of examples.
Lastly, I would second the idea of purchasing access to a university library with a good collection of mathematics textbooks.
Best of luck.
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I recommend a single book which doesn't require any knowledge about mathematics:
Ebbinghaus et. al.: "Numbers"
Full list of authors: Heinz-Dieter Ebbinghaus (Author), Hans Hermes (Author), Friedrich Hirzebruch (Author), Max Koecher (Author), Klaus Mainzer (Author), Jürgen Neukirch (Author), Alexander Prestel (Author), Reinhold Remmert (Author), John H. Ewing (Editor), H.L.S. Orde (Translator), K. Lamotke (Introduction))
It is a translation from the german book "Zahlen", which I read with pleasure during my first year at the university. All authors are german mathematicians known for their wonderful writing.
The book teaches you a lot about different number systems, like the complex numbers, the Cayley numbers, nonstandard numbers and so on. It also connects these topics to other topics in mathematics, so it provides a beautiful example-based introduction to mathematics. It doesn't stop at basic material, so you can revisit this book after one or two years of university courses, and with a new perspective, learn something new again!
I have to warn you that it's not a book about number theory, although it touches elementary number theory preliminaries at various places and I guess you should read something like it before reading any serious number theory.
The book is full of historical remarks and motivational paragraphs. Nevertheless, it's written in a clear and formally correct way (not the sometimes difficult "colloquial style" found in American math introduction books).
If you're interested in infinite series, this book might be especially good for you.
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Here is my list. I tried to make it more practical by supplying links to electronic versions whenever possible.
Rudin: Principles of mathematical analysis
Electronic version: http://libgen.org/get?nametype=orig&md5=0AB81110FEE9FBF5C218F772AB137601
Kostrikin, Manin: Linear Algebra and Geometry
http://libgen.org/get?nametype=orig&md5=796855351C245310B4FAD3C5947C846A
Cartan: Elementary theory of analytic functions of one or several complex variables
French original is available in electronic form: http://libgen.org/get?nametype=orig&md5=6E9E2B89719AD44D885B6428D8B9D077
I couldn't find the translated book in electronic form, but Dover offers an inexpensive paperback edition for \$9.
Shafarevich: Basic notions of algebra
Electronic version: http://libgen.org/get?nametype=orig&md5=11154CB5CF3714C07D0D20FB3C79D803
Milnor: Topology from the differentiable viewpoint
Electronic version: http://libgen.org/get?nametype=orig&md5=E78E64CD53429CC8DD94D7282E2BDA27
Hatcher: Algebraic topology
Electronic version: http://www.math.cornell.edu/~hatcher/AT/ATpage.html
Helemskii: Lectures and exercises on functional analysis
Electronic version: http://libgen.org/get?nametype=orig&md5=A18C3A9EC500745D563F9D3816892E3B
Milnor: Morse theory
Electronic version: http://libgen.org/get?nametype=orig&md5=ACD9C232FDFD205E937583F301F20058
Serre: A course in arithmetic
Electronic version: http://libgen.org/get?nametype=orig&md5=C00F38F10D80A59AF2A64B3D6D427CFC
Edit: I rearranged the list so that books appear more or less in order of increasing difficulty and prerequisites of every book precede it in the list.
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I think some of these books are at a bit high of a level. I can't imagine anyone opening "A course in arithmetic" without a good grounding in Both Algebra and Complex analysis. – Daniel Barter Jun 14 2010 at 22:09
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I looked at Milnor's "Morse Theory" and there's a good chance that the following words, which appear on page 1 but are not defined, will be foreign to someone who hasn't studied much beyond caluclus: manifold, torus, tangent, homeomorphic, 2-cell, cylinder, compact, genus, and boundary. This book is clearly not appropriate. I would have similar reservations about recommending several of the other books. – Peter Samuelson Jun 15 2010 at 0:15
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"I think it's worth mentioning that reading all of these books would take most people most of their time as an undergrad." - you must be kidding. I mean, it can be true if you include in "most" all people who have no intention of becoming a mathematician... – Vladimir Dotsenko Jun 15 2010 at 18:29
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Hmm... I wonder how many people "who have intention of becoming a mathematician" have completed Hatcher's Algebraic Topology. Milnor's Morse theory is much more elementary than that, not to mention less monumental. – Victor Protsak Jul 18 2010 at 14:37
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@Ben Crowell: Disservice to whom? I fail to see how such information could be relevant to the original question (“what should I read next?”). There is no way to classify these links according to you scheme without emailing each author separately, not to mention that Rudin, Kostrikin, and Cartan are no longer alive, hence for their books it's impossible to find out the answer. Finally, it seems that the value of such classification is close to zero anyway (how one could potentially use it?). Also, being in copyright (2 vs 1 and 3) depends on the jurisdiction, of which there are about 200. – Dmitri Pavlov Oct 23 at 17:24
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This is a tricky question to answer because I'm not sure what your current mathematical background is and what exactly you want to achieve. Are you familiar on how to write proofs? Given a statement, can you take its contrapositive? Or its negation? Do you know modular arithmatic? (I'm a bit confused with what you're asking for because the first book you want to purchase is an introductory text. Then you ask for graduate level reading material.) Anyways, if not, then you might want to try a little book by Peter Eccles called "Introduction to Mathematical Reasoning." It's a book that is designed to help transition students into their first higher mathematics course. He goes over all the fundamentals: basic logic, truth tables, set theory, sequences and series, divisbility, modular arithmatic, elementary combinatorics. It's self-contained and very readable. He also provies solutions to many of the exercises as well. I'd definitely start with that text if you want to learn how to write proofs.
After that, then a nice text is the two volume series "Mathematical Analysis" by Vladimir Zorich. It's a very comprehensive and well-written text on real analysis. A nice aspect is that he gives the historical motivations for a lot of constructions so it's easy to understand how the mathematical ideas arose.
If the "two volume" series of text book scares you off, then Stephen Abbott has a nice book called "Understanding Analysis" that I would suggest checking out.
Of course, there are a lot of good suggestions out there. It'd help to have a more concrete idea of what you're looking for.
Best of luck, Bman
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In high school, I was never really exposed to the kind of books that mathematicians use in their mathematics education, the kind of material that practicing mathematicians assume that "everybody" knows. The resource that I found most helpful for identifying these "canonical" books used in the top American mathematics programs was the Chicago Undergraduate Mathematics Bibliography. It hasn't been updated in some time and was written by people who graduated about 10 years ago, so it is missing some great newer titles like Lee's Introduction to Topological Manifolds and Hatcher's Algebraic Topology. However, it does have a great list of books from the high school level to early graduate level and short reviews of each, all on one page.
As a motivated student interested in mathematics, I can think of no better way to get started now than to read some of the reviews, get the books that sound interesting, and work through them. Most (if not all) of the books listed are classics and thus are familiar to nearly everybody doing research in that particular field. Furthermore, this has the added benefit that these textbooks are often cited for background in papers/monographs/lecture notes and will likely be used as references for classes that you take at university.
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Thanks (I think) for the link to the CUMB -- rereading it just now was an interesting experience. Based on a few context clues, most of the text was probably written in late 1997 or early 1998. Perhaps the most dated point is the belief that Massey's GTM is, for better or worse, the standard introductory text in algebraic topology. Nowadays the canonical choice seems to be Hatcher's book, and rightfully so. – Pete L. Clark Jun 15 2010 at 0:00
Fundamentals of University Mathematics by C. M. McGregor, J. J. C. Nimmo, and W. W. Stothers is an excellent book for the transition from high-school to university mathematics. Definitely pitched at an appropriate level and worth a look.
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I don't know how much sense it makes to study books on the topics of the usual first courses at university since this is stuff one learns anyhow at university. This means not that I recommend to study material on higher courses. I mean that one perhaps should study elementary mathematics, which is not usually covered so much at university, but is worth knowing.
A very fine book on elementary geometry is Coxeter's Geometry Revisited. It covers on 200 pages a lot of beautiful mathematics.
Besides the already mentioned Proofs from the BOOK, I would recommend to study elementary number theory, where one gets with little formal effort quite deep theorems. Books I used (and liked) include A Classical Introduction to Modern Number Theory by Ireland and Rosen and An Introduction to the Theory of Numbers by Niven and Zuckerman, but there are a lot more: just read the amazon reviews. They are usually quite informative. Andreescu has also some quite nice problem based books.
There is also another book I would like to recommend, but hesitate since it is not really easy. It is Thurston's Three-Dimensional Geometry and Topology (Volume I). But on the other hand, it is a book one can learn very much from and it is written in a quite informal way, bringing much ideas, and begins with elementary ideas from two-dimensional geometry. I've given a short series of talks to (bright) high school students on some topics of the first chapter and I think, it was successful, so if you want to try and like to use your geometric imagination, then try.
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Niven, Zuckerman is very nice, but there exist so many other, different good books that I recommend to visit a good math library with an experienced mathematician and select in a kind of 'group-browsing' what fits best. – Thomas Riepe Jun 17 2010 at 12:44
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Some may protest that it's not really a math book, but I have to recommend Gödel, Escher, Bach, by Douglas Hofstadter. It is, among other things, a beautifully written and extremely self-contained introduction to three fascinating subjects:
1. Formal systems. What does it mean to prove a theorem from a set of axioms? How do we translate mathematical statements into statements about the real world? Is it possible to do math without thinking about the real world?
2. Computability theory. How do you unambiguously describe a computational procedure? What are the basic tools of computation---and what price do you pay for using the most powerful ones? Can the values of a mathematical function always be computed?
3. Gödel's incompleteness theorems. Can every true statement in the language of number theory be proven from the axioms of number theory? How do we know that the axioms of number theory don't subtly contradict each other? Can a mathematical theory be used to study itself?
Winding its way through these weighty topics, GEB manages to find time for a little mathematical sightseeing, discussing some classic proofs, problems, and pieces of history.
Most importantly, it's also tons of fun to read! ^_^ You'll have plenty of time for dry, boring books at university; in the meantime, you should just read what you enjoy. ~_^
p.s. GEB is a massive book, and I do not recommend trying to read it straight through. If you want to focus on the mathematical parts, you might try concentrating on these chapters:
• I -- IV
• V (starting at "Diagram G and Recursive Sequences")
• VII -- VIII
• IX (starting at "From Mumon to the MU-puzzle")
• XIII -- XV (including the dialogue, "Aria with Diverse Variations", that comes before chapter XIII)
• XVII (you'll have to wade through a lot of history and philosophy to find the math)
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Studying mathematics for the sake of learning is a feat worthy of praise; studying mathematics to fulfill a particular goal (such as gaining an academic job, solving problems in industry and society, or using it to further research in other disciplines) is a feat that is worthy not only of praise, but of asking others for financial support to help one do it. You can budget time for doing the first as long as you leave time for earning the money to pay the bills. You can budget a little more time for doing the second as long as you leave time for getting the monetary support, and it is important that you pick a good goal: a goal that others are willing to support.
Even if you do not have to support yourself, it is good to have a goal and a plan around mathematical study. One plan might be to spend x amount of time reading reviews and abstracts, and then use that to pick papers and books to spend y amount of time beginning to read. If you develop questions and sufficient maturity, some well placed questions here on Math Overflow may supplement your reading. Part of doing mathematics well is asking the right questions.
I recommend using as many free resources as you can before spending money on books. The Internet, your prospective university library, and friends and acquaintances who have books of interest are just three sources. Don't worry about the sophistication or the prerequisites needed for understanding a book or article: you are compiling a list of things for future reading and enjoyment. Buy something only for adding to your "permanent library", the library you will always carry with you or will always devote time and energy to store and maintain, or buy something only when you know you will give to someone else but you will get to read it for yourself first.
Finally, supplement your study with attending lectures on other topics. The engineering or computer science departments (or other academic departments even) will sponsor lectures which may supplant your current interest in mathematics. Things get exciting when you discover an application of what you learned to another field, but you won't see the applications without seeing something of the other fields.
Good luck in finding a good goal.
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I do not understand why nobody mentioned Schaum's Outline of Theory and Problems. These books offer a lot of practice in different branches of math. From algebra to calculus to differential equations,logic, topology,...
It has always been a great idea to practice my math skills with the problems there.
Another thing, I learned calculus I to III before getting in to univ from the book:
CALCULUS: An introduction to Applied Mathematics by H.P. Greenspan (Author), D.J. Benney You will definitely find it in most university.
Also I learned Group theory (again before getting in to univ) from:
Algebra (2nd Edition) Michael Artin
My final remark: You will get high-quality math ebooks for free from the internet, a great repository of knowledge for everyone.
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Since you live in the Netherlands, you might want to look at what is currently being offered in the Springer Yellow Sale 2010 (http://www.springer.com/sales?SGWID=5-40289-0-0-0). You'll find Lee's Introduction to Topological Manifolds there, for example, or Abbott's Understanding Analysis.
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Do you live near a university? Maybe someone in the university's math dept. would be helpfull? Anyway, I'd suggest to get access to it's library, take a few days to find a handfull of texts which look accessable and make you most curious, then work through them. E.g. Kintchin's "Three pearls of number theory" may be an idea, it was written for someone in your situation, but is not easy. (some thoughts on learning etc.: 1 ,2 )
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If you are looking for a break from Calculus and don't want to dive into Category Theory right away, my favorite book is Alan Beardon's Algebra and Geometry. What follows is an enhanced Amazon review I wrote as an undergrad. The Cambridge Schedules provide a wonderful guide to further study so do have a look.
Beardon's book has become a bible of sorts to first-year students studying mathematics at Cambridge University (please refer to the Schedules for a beautiful play-by-play of topics and books by perhaps the best foundational math curriculum the world-over). Its quality as a text cannot be doubted, although its usefulness for further years of algebra is limited. This is precisely the book to study from if you are doing vector calculus and differential equations, but still aren't sure about doing mathematics seriously. If you have not taken a course in linear algebra or abstract algebra, buy the paperback copy (~\$50) of this book and start reading right away. Beardon starts with (what I believe is the best way) the study of permutations (think about shuffling a deck of cards) to develop an intuition of the basic notions of a group. From here the fundamentals for further study in mathematics is laid. I won't repeat the table of contents here, as you can look for yourself, but believe me when I say that mastering the concepts in this book will serve you very well.
I truly wish I had a course which devoted itself to the complete digestion of this book. I used it for self-study and found that it served me very well. It does not fall easily into the structure of most American math sequences, as these departments are often forced to "modularize" mathematics into semester-bite-sized pieces. I believe that this often has a negative impact on the appreciation of mathematics as a whole, especially at the nexus between doing basic calculus and appreciating proof, rigor and beauty in mathematical structures.
The book may not go into the same depth as, say, Artin's "Algebra", but rather the foundational concepts for the study of algebra and geometry are emphasized in a variety of settings. This is very important as the study of "abstract algebra" is precisely that if you do not have a wide-selection of examples and contexts to draw from. This book has plenty of exercises of varying difficulty, and everything in this book is accessible to the beginning student of mathematics.
Bottom line: If you are someone interested in learning linear algebra, geometry, group theory, Mobius transformations, complex variables all in a rigorous yet introductory level, this is the book for you. Developing a robust mental model for mathematics requires building several thin layers at a time. This means not going too deep too quickly, but rather snorkeling around the entire reef, before you gear up for further exploration.
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Beardon's book to me is one of those "best intentions" books that doesn't quite deliver. While it presents a wealth of beautiful material that all math students should know and isn't covered in most traditional sources,the lack of pictures really hurts the presentation. It also doesn't have anywhere near enough examples or exercises. Elmer Rees' classic Notes on Geometry is much better in this regard. Here's hoping that Beardon one day gives us an enlarged second edition that fixes both those defects-if he does,we'll have a bona-fide classic. – Andrew L Jul 18 2010 at 20:26
EDIT: some people didn't like what I've written below. Rather than change it, I've added comments below instead. Why should I change it anyway? If you don't agree, that's your opinion; but these are mine, so why shouldn't Max see a fair range of differing views? One of the things I was trying to say is that if you want to have a successful career ACTUALLY USING PROPER MATHEMATICS beyond just, say, calculating percentages or putting numbers into a formula, then your chances really are very slim unless you're EXCEPTIONALLY good. The number of good jobs with genuine mathematical content is far, far smaller than the number of mathematicians with reasonable ability (at least in England). The sooner Max knows this, the better he can prepare properly and take precautions.
Hi Max, it's nice to see some youngsters with interest!
Firstly, Hardy's Divergent Series is a great, fun book, but probably too hard and of little use for your future career (it's very old-fashioned and not likely ever to come back into fashion - but I could be wrong).
My personal opinions for what they're worth (i.e., not very much); some (many?) people would disagree!
The most important thing by far is to go to the very best university you can. I assume that you have genuine mathematical talent (you must be easily the best at mathematics in your school) and that you want to go to the top university in Holland; don't waste your time at second best places. (If not, then improving your school grades or whatever you need is the top priority!) It's far better to be in the middle at the best university than at the top in a poor university.
If you can't decide which is the "best", ask your teachers, see which ones have Fields Medallists/Nobel Prize winners, look at league tables, etc. (I'm English and don't know how the Dutch system works, so apologies if this sounds like nonsense). As an absolute minimum, it MUST be a university which offers Ph.D. degrees in all areas of mathematics.
At good universities, you have nothing to worry about; your lectures/university library will provide everything you need.
Almost certainly, you will buy the wrong books (either far too easy or far too difficult) that you'll never read (or will quickly finish and want to throw away). Almost certainly, your mathematical tastes will change significantly over the next few years.
Sad, but true: many (perhaps even "most" or "almost all") maths books are boring, tedious, unsuitable and impossible to read (and different people will disagree about which books are which, so advice is not much use!)
Personally, almost every mathematical book I bought as an undergraduate or before was a total waste of money; with more experience you'll know better in future what to look for; but now is the worst possible time for you to buy books.
Until you have spent at least one or two years at university, SAVE YOUR MONEY!!! GO TO LIBRARIES!! DO NOT BUY MATHEMATICS BOOKS!!!!! (Unless you are rich or someone else is paying...)
If you do buy books, only do so after you have looked at them for several weeks or months from a library!
Only buy books with LOTS of exercises!(Apart from rare exceptions, e.g. Hardy and Wright: An Introduction to the Theory of Numbers.)
Doing fun research problems of your own, either directly or indirectly inspired by books, is just as important as pure reading; and try not to read too much right now. You won't discover anything new, but that's not the purpose (and the internet is too big - don't look up the answers too quickly, or you'll never develop).
Don't listen to everything physicists say when they're talking about mathematics...! Physicists and mathematicians are very different.
Practice basic calculations: calculus, infinite series, Fourier series, complex numbers, differential equations, basic number theory,... and whatever else interests you. It will be very useful in future.
Try looking through things on Wikipedia and the links, and just following wherever your interests take you. You'll learn a surprising amount this way, and enjoy it more too. (But be very careful of the Internet; almost everything online, if not written by proper university mathematicians, is rubbish! MathOverflow and Wikipedia are rare exceptions, but even these have a small amount of rubbish on them).
And finally...
BE FLEXIBLE! Many things you are interested in now might not be to your taste in the future (mathematics is very, very large and no-one can do all parts of it); it is even possible (what a hideous thought!) that you might lose your interest in Mathematics and prefer Physics, Computer Science, Engineering or something else instead; be prepared to change! (Very few people survive up to Ph.D. level and beyond; if you don't become a mathematician, you will get much better jobs after graduation if you do these subjects at university instead...!)
If you still want to study mathematics at university: be prepared for lifetime social exclusion, poverty and unemployment! (This is more for Ph.D. than B.Sc.; perhaps I exaggerate slightly, but you must prepare for the dangers!)
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"...lifetime social exclusion,povery and unemployment"?!? Just about everyone I know with a PHD in mathematics has a good paying job and lots of friends,Zen. You won't get rich by any means,but you're making it sound like he's going to end up a hermit mopping floors! Not only is this simply incorrect-unless the mathematician's area is something very abstract and isolated-his life will not be so desolate and deplorable! Now if you refuse to get a job unless it's being on faculty at a university-like some unreasonably stubborn academics-that's a different story..... – Andrew L Jul 17 2010 at 2:57
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I am disgusted by ZH's answer, really. Not only many things are downright wrong ("Don't listen to physicists...", "Prepare for poverty..."), but the very tone of a frustrated rant (with, of course, obligatory exclamations marks and boldface) is anything but inspiring to a beginner. – Michal Kotowski Jul 17 2010 at 10:15
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That said, Zen, you should really reconsider your last statement. People here (myself included) might not just disagree about your definition of poverty and social exclusion, but find your flippant use of them rather ill-mannered. – Yemon Choi Jul 18 2010 at 5:14
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The reason I feel the need to say all this (this is for the benefit of Max and other youngsters) is that what many careers advisors say at school and university, when talking about mathematicians, is just completely wrong. By all means, do Mathematics because you have a lot of interest and ability; but don't do it to get a good job! To get a permanent position, a Ph.D. mathematician needs YEARS of travelling around the world doing various postdocs, only a year or two at a time, with little control over where they go next. It's impossible to have any kind of financial or social stability. – Zen Harper Jul 19 2010 at 17:43
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Michael: this is getting off the subject of the original question, but what do you mean by "school math", "no extracurricular learning",...I don't understand? I'm wondering what one should do when you've completed a Ph.D. in Pure Mathematics and amassed a modest collection of published papers, but the postdocs/lecturing positions have all dried up and you've left the university system (or are approaching the end of your current position), with no new university job offer: what should you do? In fact, maybe I will ask this question myself... – Zen Harper Jul 19 2010 at 23:16
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Hello Max,
let me provide you with yet another opinion:
Imho the best way (and probably also the most fun way) of learning mathematics is not so much by learning mathematical theories from books (you will learn enough theory during your studies anyway) but by actually doing mathematics and by that I mean solving mathematical problems. A good oppertunity to do so is to compete in mathematical competitions (Check out this list from wikipedia http://en.wikipedia.org/wiki/List_of_mathematics_competitions). Not only will it provide you with challenging problems but also it is a good oppertunity in the later stages of the competitions to meet other guys who have the same passion about mathematics as you. It is a great feeling to know that one is not "the only one".
But you were asking for books so I give you my all time favourite:
László Lovász - Combinatorial Problems and Exercises
This books contains nothing but mathematical problems from rather easy ones you can figure out in a quarter of an hour to really challenging ones which could take you up to a week or more to solve. As background you only need basic linear algebra, probability and calculus. Solving all of these problems (of course without looking into the solution before) would probably teach you enough math to start tackling real unsolved problems in the beautiful field of discrete mathematics afterwards, i.e. to do real mathematical research. Sounds tempting doesn't it? ;)
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Beste Max,
Have you ever been to the Mathematics Institute in Leiden? You are always welcome. I will show you around (including the library :)).
As far as I understand, in principle, you can attend some lectures even in the coming year. I would advise the Algebra course given by Hendrik Lenstra. The course material is available in Dutch at http://websites.math.leidenuniv.nl/algebra/
In my opinion, you should not concentrate so much on "doing things properly", but rather read and think about some nice elementary (which is not synonymous to easy) mathematical problem. There are quite a few such problems. In my last year in high school, I was attending a "course" on Pell's equation, which was really nice and served the purpose of doing something outside the standard school curriculum.
Anyway, do not hesitate to contact me.
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First of all, why not go ahead and check out the math library of your university. Go there, sit on the floor and take books of the shelves and just skim through them.
Also, I noticed that you don't mind if there are problems in the books. That's good because you are wasting your time, to be honest, if you don't work through every problem on your own and treat doing the problems as more important than reading the sections.
Alright moving onto specific recomendations:
Conjecture and Proof, M. Lackzovich (this is a great into to doing proofs, has problems, and is wonderful for self study at an amatuer level)
A survey of modern algebra, birkhoff and mclane (I strongly recommend this book as an intro to modern algebra, it's very clear and easy to work through! Plus it will help you with proofs in a gentle way)
Differential Geometry, stoker (this is what i learned the subject from, I could not possibly recommend this book more for self study of this subject)
Principles of Analysis, rudin (this is a 100% no brainier, everyone else recommended it too)
Mechanics, L.D. Landau (course of theoretical physics vol 1)
Introduction to Analysis, whittaker (first edition, the second and third editions are not as good!)
Complex Variables, Fischer
Princeton Lectures on Analysis Vol 1 and 2, Shakarchi and Stein
Introduction to analytic number theory, apostle
Theory of Numbers, serpinski (this guy is from the older school, but its written really really well, and it is always nice to study the masters)
Lectures on Ramanujan, G H Hardy
Disquisitions Arithmatic, Gauss (just try to get as far as you can!)
Abstract Set Theory, Fraenkel (this author has the axioms of set theory named after him, and this book is a really fast and fun read)
that should be enough to keep you busy for awhile =)
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Hi! This is a very useful question. Getting the right books is very important. But Math books are expensive, so I suggest you to buy international edition (you can look on abebooks.com) or check your library. Below is a list of books which will definitely prepare you for a rigorous graduate program in Mathematics. It's divided into subfields of Math and within each subfield, it's sorted from easy to hard (from beginning university level to finishing university).
• Analysis:
• Principles of Mathematical Analysis (Rudin)
• Calculus on Manifolds (Spivak)
• Fourier Analysis, Complex Analysis, and Real Analysis (3 book sequence by Stein and Shakarchi)
• For Complex and Real Analysis, you could use Rudin's Complex and Real Analysis as well.
• Functional Analysis (Rudin)
• Differential Geometry:
• Differential Geometry and Riemmanian Geometry (Do Carmo)
• Introduction to Topological Manifolds (John M. Lee)
• Introduction to Smooth Manifolds (John M. Lee)
• Riemannian Manifolds: An Introduction to Curvature (John M. Lee)
• Topology:
• Topology (Munkres) -- very good introduction.
• From Calculus to Cohomology (Madsen and Tornehave) -- a good introduction to Algebraic Topology through differential forms; can be read after Calculus on Manifolds by Spivak.
• Algebraic Topology (Hatcher) -- very geometric, but you need to know some algebra first (see the algebra list below).
• Algebra:
• Abstract algebra (Dummit and Foote) -- Easy to digest, but I much prefer the following
• Algebra (Lang, GTM) -- very amazing introduction, a bit terse.
• Commutative Algebra (Atiyah and Macdonald).
• Basic Algebraic Geometry (Shafarevich), or Algebraic Curves (Fulton, free online) -- good intro to algebraic geometry.
• Algebraic Geometry (Hartshorne) -- difficult book, but worth the effort if you want to study algebraic geometry.
• Number Theory:
• Algebraic Number Theory (Neukirch) -- good book, but you need to at least read the first 4 chapters of Lang's algebra before starting. The section on Analytic Number theory also requires a good understanding of Complex Analysis.
Of course, you don't need to read everything up there. What you need to learn depends on what math you want to do in the future. I think it's good to see a professor and talk to him in person to see what you really want to do.
I hope that helps!
Edit: added suggestions by Andrew L.
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Hi there Max,
I too am of the same age, and I too am fascinated with math. :)
Now that I think about it, I was inspired by two teachers and two books.
1. The first book was a pretty old one called "Elementary Mathematics" by three Russian authors called G.Dorofeev,Potapov and Rozov. These books were published by a publishing company called Mir Publishers. They used to print extremely good books which were available for a pittance a couple of decades back, but they have shut down now after the collapse of the Soviet Union.
2. The second one was an Indian book called "Challenges and Thrills of Pre-College mathematics". You might be able to order that if you try hard enough.
Another one I recommend is "Men of Mathematics" by E.T.Bell. This is a well-known classic, and I'm reading it now.
All the best for your future and I hope we meet someday in the math world! :)
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I found "On Knots" by Louis Kauffman to be very inspiring when I was in high school. It's not a book to be read linearly, but rather you should hop around from section to section. As your mathematical career progresses you'll be able to understand more and more of it.
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http://www.physicsforums.com/showthread.php?t=507141 | Physics Forums
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## Faster than light time travel
Hello, i'm a high school senior interested in studying physics in college. I was listening to a podcast by Neil deGrasse Tyson in which he stated that if one could travel faster than the speed of light, it might be possible to travel back in time. He didn't really expand on that so I was wondering if someone here could explain this to me.
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: Gold Member Science Advisor Staff Emeritus Either you misunderstood him or he made a mistake. Material objects can only travel at less than the speed of light.
I think he was speaking theoretically, and that if you plug in a time faster than the speed of light into Einstein's equations time starts to flow backwards.
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## Faster than light time travel
Quote by abledpilot Hello, i'm a high school senior interested in studying physics in college. I was listening to a podcast by Neil deGrasse Tyson in which he stated that if one could travel faster than the speed of light, it might be possible to travel back in time. He didn't really expand on that so I was wondering if someone here could explain this to me.
Things that have happened cannot be undone. no matter the frame.
Simularly, I can add numbers to 299,792,458m/s, however that doesn't impact the reality. What point would it be to see SR/GR equations compute time reversal at FTL speeds, it's beyond the predictability of those equations I'm sure (let alone reality).
I don't know in what context Neil deGrasse Tyson was speaking from. It seems a very odd statement for SR/GR.
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Quote by abledpilot I think he was speaking theoretically, and that if you plug in a time faster than the speed of light into Einstein's equations time starts to flow backwards.
No, that's incorrect, so either he made a mistake or you misunderstood him.
-Ben
Well, as you tend to the limit of traveling at the speed of light time dilations increases. That is, time appears to 'slow down' (on the outside of your spaceship). If you extrapolate this tendency just for the fun of it you would say that time stays 'still' at the speed of light, and taking this to a further step would lead you to say that time goes 'backwards' at higher speeds. So its not crazy talk, just a baseless or "hypotetical" speculation. You can't even travel at the speed of light, much less at a higer speed*. *at least as our observations appear to confirm us so far
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Quote by AndyUrquijo Well, as you tend to the limit of traveling at the speed of light time dilations increases. That is, time appears to 'slow down' (on the outside of your spaceship). If you extrapolate this tendency just for the fun of it you would say that time stays 'still' at the speed of light, and taking this to a further step would lead you to say that time goes 'backwards' at higher speeds.
No, that's incorrect. If you plug v>c into the equation $\gamma=1/\sqrt{1-v^2/c^2}$, you don't get a negative result, you get an imaginary one.
-Ben
Recognitions: Homework Help In special relativity, causality is limited to one direction because of the speed limit of c. (It is possible to reverse the order of two events, by looking at them through a different reference frame, but this requires that the two events must be separated such that one event could not have caused the other). But, if there was a wormhole between the two events, then you could in principle reverse the order of two causally connected events. But of course, a wormhole doesn't mean travelling faster than c. Nothing can travel faster than c. So you'd be more correct to say that in certain situations, the curvature of spacetime could lead to the reversal of the cause-and-effect of two events.
Quote by bcrowell No, that's incorrect. If you plug v>c into the equation $\gamma=1/\sqrt{1-v^2/c^2}$, you don't get a negative result, you get an imaginary one. -Ben
Yeah, my bad. I didn't mean to say that was correct. I just think this was probably the (naive) reasoning that went on to make such claim. You are right of course. And you get an undefined result when you use v=c. Which is irrelevant since you can't get to that speed to begin with (if you have non-zero mass).
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Quote by abledpilot Hello, i'm a high school senior interested in studying physics in college. I was listening to a podcast by Neil deGrasse Tyson in which he stated that if one could travel faster than the speed of light, it might be possible to travel back in time. He didn't really expand on that so I was wondering if someone here could explain this to me.
It's possible he was referring to the tachyon, a hypothetical particle which, if it existed, would always travel faster than light and would never go slower. There are many good reasons for believing that tachyons don't exist, but nevertheless we can consider how they might behave if they did.
If a tachyon is travelling faster than light forwards in time relative to one inertial observer, it is travelling backwards in time relative to some other inertial observers. If you could relay a message via at least two tachyons, you could send a message into your own past. See Tachyonic_antitelephone for details. That would give rise to all sorts of grandfather paradox-like problems, one reason that we suspect tachyons don't exist -- there are other reasons, too.
Quote by DrGreg It's possible he was referring to the tachyon, a hypothetical particle which, if it existed, would always travel faster than light and would never go slower. There are many good reasons for believing that tachyons don't exist, but nevertheless we can consider how they might behave if they did. If a tachyon is travelling faster than light forwards in time relative to one inertial observer, it is travelling backwards in time relative to some other inertial observers. If you could relay a message via at least two tachyons, you could send a message into your own past. See Tachyonic_antitelephone for details. That would give rise to all sorts of grandfather paradox-like problems, one reason that we suspect tachyons don't exist -- there are other reasons, too.
Why could virtual particles travel faster than light but yet i don't hear them violating causality.
Quote by DrGreg ... If you could relay a message via at least two tachyons, you could send a message into your own past. See [url=http://en.wikipedia.org/wiki/Tachyonic_antitelephone]...
Oh ok, so FTL message transfer is paradoxic because formula says so :)))) I don't mean to offend - But do you realize how hilarious that sounds to someone who doesn't rely on formula doctrine? In just another thread on this forum ("A photon's time") it is agreed that photon's frame can not be expressed with Lorentz transformations which are used in the Tachyonic Antitelephone example. It's the same as saying that GR is wrong because newtonian formulas give wrong results at relativistic speeds.
Does anyone have a better thought experiment that doesn't involve formulas that are likely not accounting for FTL? I'm sure there has to be something.
And FYI, I am not saying that FTL is possible... All I am saying is that formulas are not useful answers to reply to thoughts outside of things that we can experiment with.
Quote by bcrowell No, that's incorrect, so either he made a mistake or you misunderstood him.
If, by FTL, we mean the ability to travel between two spacelike-separated points, then in that sense there is no distinction between faster-than-light travel and time travel, because one can easily draw a spacelike curve which ends within the past lightcone of its origin. (Assuming Lorentz invariance; if there's a preferred foliation which restricts such trajectories, then that's another matter.)
Actually a few things travel faster than light but are not helpful or useful. For example, the EPR parardox that information between 2 electrons is shared instantly and faster than lightspeed, however it proves no use if the information cannot be controlled and only random.
Quote by abaio Actually a few things travel faster than light but are not helpful or useful. For example, the EPR parardox that information between 2 electrons is shared instantly and faster than lightspeed, however it proves no use if the information cannot be controlled and only random.
Yes, but I'm supposing we don't really know that the other entangled particle changes state instantly upon the 1st being observed, even though we might casually assume such. There's no proof it's instant, yes?
GrayGhost
I believe we do know its instant. Thats why Einstein himself referred to the event as "spooky". Either way, there are still other ways to conquer the speed of light.
Coming from a very pure math background, I see why travelling at or faster than light screws up all the math formulas, but physically why can't an object travel at (or faster) than light? What fundamental law of the universe could you be breaking? Thanks.
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http://physics.stackexchange.com/questions/tagged/inertia | # Tagged Questions
The inertia tag has no wiki summary.
2answers
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### Two Different Sorts of Inertia: Inertial Mass and Moment of Inertia
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### What is the present state of Mach's Principle amongst physicists? [duplicate]
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1answer
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2answers
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### Why do objects with different masses fall at the same rate? [duplicate]
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### Fictitious Forces $\overset{?}{⇔}$ Constraint Forces (re: D'Alembert's Principle)
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1answer
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### Why are there some inconsistencies with the underlying principle of center of gravity and rotational inertia?
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5answers
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### Inertia in an empty universe
I was reading a recent article on Mach's Principle. In it, the author talks about inertia in an empty universe. I'll quote some lines from the article: Imagine a single body in an otherwise empty ... | {"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": 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": 0.9324100613594055, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/52094/exciton-energy-splitting-in-indirect-semiconductor/52459 | # Exciton energy splitting in indirect semiconductor
Let's take an indirect bulk semiconductor and imagine the exciton that build up at the indirect gap. Since we have a bulk semiconductor we should find three p-like orbitals: $p_x,p_y,p_z$. Would their energy levels be degenerate due to the mass anisotropy which we have at the indirect gap? My intuition screams yes. But if I try to bring it on the paper I actually find the opposite. Any suggestions, ideas or helpful tips?
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## 1 Answer
Just in case someone was interested, I'll give myself an answer :)
Yes they do split due to mass anisotropy. One can see that when solving the Wannier-equation for an indirect semiconductor. For the special cases that the wave vector lies in the $x$, $y$ or $z$ direction one finds that $x$ and $y$ will give the same integral and $z$ will differ from that. One of course has to define the band structure correspondingly: the masses in $x$ and $y$ are the same and in $z$ is different.
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http://mathoverflow.net/questions/84120?sort=oldest | ## Fubini’s theorem and unique mean value
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Following the terminology of Rosenblatt, I will say that a bounded function $f:\mathbb Z\rightarrow\mathbb R$ has a unique mean value if for every pair of finitely additive translation invariant probability measures $\lambda_1,\lambda_2$ on $\mathbb Z$, one has $\int fd\lambda_1=\int fd\lambda_2$.
I will say that $f$ has Fubini's property if for all finitely additive probability measures $\mu,\nu$ on $\mathbb Z$, one has $\int\int f(x+y)d\mu d\nu=\int\int f(x+y)d\nu d\mu$.
Question: Is it true that the two properties above are equivalent?
It is obvious that if $f$ has Fubini's property, then it has also a unique mean value, but the converse is not clear to me. I don't have a real evidence why the two properties should be the same.. let's say, that I am interested in studying the relation between these two properties and I was not able to find a function with a unique mean value which does not have Fubini's property.
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## 1 Answer
Let $g$ be the characteristic function on the positive odd numbers and let $f(x) = g(x) - g(x + 1)$, so that $f$ has a unique mean value. If $\mu$ is non-principle and supported on the positive odd numbers and $\nu$ is non-principle and supported on the negative even numbers then we have $\iint f(x + y) \; d\mu d\nu = 1$, while $\iint f(x + y) \; d\nu d\mu = 0$.
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http://www.physicsforums.com/showpost.php?p=3788761&postcount=2 | View Single Post
Recognitions: Gold Member Science Advisor Staff Emeritus No, amazingly, you are wrong and your teacher is right! Since that integrand does not exist at x= 0, you have to use the definition: $$\int_{-1}^1 \frac{dx}{x}= \lim_{\alpha\to 0^-}\int_{-1}^\alpha \frac{dx}{x}+ \lim_{\beta\to 0^+}\int_\beta^1\frac{dx}{x}$$ and those limits do not exist. You cannot just evaluate the anti-derivative at 1, -1, and then subtract- that's ignoring the whole problem of what happens at 1. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9424735307693481, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/29247/catenary-curve-under-non-uniform-gravitational-field/29523 | Catenary curve under non-uniform gravitational field
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The catenary curve is the shape of a chain hanging between two equal-height poles under the influence of gravity. But the derivation of the (hyperbolic cosine) curve equation from the physics traditionally assumes a uniform gravitational field. Suppose instead one uses the non-uniform gravitational field that diminishes with distance from the center of the Earth. (Perhaps this would be relevant for a very long chain that sags significantly.) Does this lead to an interesting curve, known in some closed form? Or just to a differential equation that can only be solved numerically?
I ask this primarily out of curiosity, so please interpret in that spirit!
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1
Your first sentence, describing the traditional catenary, would remain correct without the "equal height" restriction. – Andreas Blass Sep 5 at 13:38
5 Answers
Well, the next physically interesting situation is the skipping rope equation, where the gravitational field is proportional to the distance from a fixed axis. Clebsch, 1860. You may find a lot of material googoling "skipping rope equation" ("courbe de la corde à sauter" or "Springseilkurve" should you read French or German).
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Very interesting! The skipping rope differential equation solves in term of Jacobi’s elliptic sinus amplitudinis function. Here is a paper that solves it: iopscience.iop.org/0143-0807/28/2/009/pdf/…. Here is a link on the sn function: mymathlib.webtrellis.net/functions/…. – Joseph O'Rourke Jun 23 2010 at 21:03
I search for "skipping rope equation" (in quotes) on Google, and got exactly ONE hit. Without quotes was better. :-) – Hans Lundmark Jun 26 2010 at 9:14
mathcurve.com/courbes2d/cordeasauter/… is an interesting (French) page on this. (They link to the European Journal of Physics article linked to by Joseph.) – J. M. Nov 1 2010 at 11:05
@Hans: Then you googled ""skipping rope equation"". – Marcos Cossarini May 10 2012 at 23:13
@Marcos: It took me a while to understand what you mean by """skipping rope equation""" – Pietro Majer May 11 2012 at 7:36
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You might look at Robert Osserman's article The mathematics of the gateway arch, which discusses many generalizations of the catenary problem. I'm not sure that it does exactly what you want, however.
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Great reference, great paper--Thanks! He explores versions of a "weighted catenary," where the chain's density varies along its length. He does not seem to mention the natural condition of using the inverse-square law of gravity. But it seems likely that one could select a density variation to simulate the effect of true gravity... – Joseph O'Rourke Jun 25 2010 at 17:16
This started out as a comment, but it wouldn't fit in the box:
If you are interested in deviation from the catenary caused by variation in the strength of gravity, you should be almost as interested in deviation caused by variation in its direction. Imagine a suspension bridge with a span of length s and towers of height h. These towers will be further apart at the top, by a factor of 1 in R/h, where R is the radius of the Earth. (For a concrete example, the Akashi Kaikyō Bridge, the longest in the world, has s=1991m and h=283m, giving a factor of about 1 in 22000, or 9cm.) The difference in the force of gravity between top and bottom is R2/(R+h)2, or 1 in R/2h.
So the directional effect is half as strong as the height-induced effect. Whether they affect the shape of the curve in the same proportion is another question.
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Yes, you are absolutely correct! Surely some classical scholar worked out the chain curve under true gravity, both in magnitude and direction... – Joseph O'Rourke Jun 27 2010 at 1:34
The skipping rope equation is known as the troposkien. It is well known to anyone who designs Darrieus style vertical axis wind turbines. It is to the best of my knowledge solved only iteratively.
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See also Pietro Majer's answer on the skipping-rope equation. – Joseph O'Rourke May 11 2012 at 12:13
If one uses the functions $f_d(t) = (\cos (d t))^{\frac 1d}$ (which were implicitly introduced by Colin Maclaurin) one can compute the catenaries for central gravitational fields which are proportional to a power of the distance to the Sun. These are curves of the form $rf(\theta)=1$ where $f$ is a function of the above type (or a dilation and rotation thereof).
Interestingly, one can also get catenaries for parallel forces from the same functions. In fact, parametrised curves of the form $(F(t),f(t))$ where $f$ is one of the above functions and $F$ is a primitive, give the catenaries for gravitational fields which are proportional to powers of the distance to the $X$-axis. These curves, the Maclaurin catenaries, are new.
Not relevant to the question but worth mentioning is that if one rotates them about the $X$-axis, they provide examples of surfaces for which all six parameters in the fundamental forms ($E$, $F$, $G$, $L$, $M$, and $N$) are proportional to powers of the above distance.
Details in arXiv 1102.1579
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http://meetings.aps.org/Meeting/MAR11/Event/142863 | # Bulletin of the American Physical Society
## APS March Meeting 2011 Volume 56, Number 1
Monday–Friday, March 21–25, 2011; Dallas, Texas
### Session W29: Symmetric Discrete Structures for Finite Dimensional Quantum Systems
11:15 AM–2:15 PM, Thursday, March 24, 2011
Room: C148
Sponsoring Unit: GQI
Chair: Christopher Fuchs, Perimeter Institute for Theoretical Physics
Abstract ID: BAPS.2011.MAR.W29.5
### Abstract: W29.00005 : Isotropic States in Discrete Phase Space
1:39 PM–2:15 PM
Preview Abstract MathJax On | Off Abstract
#### Author:
William Wootters
(Williams College)
An energy eigenstate of a harmonic oscillator is isotropic in phase space, in the sense that the state looks the same along any ray emanating from the origin. It is possible to extend this notion of isotropic'' to quantum systems with finite-dimensional state spaces---the rays are then rays in discrete phase space. In this talk I present examples of discrete isotropic states and discuss their properties. One can show that every isotropic state minimizes a specific information-theoretic measure of uncertainty with respect to a complete set of mutually unbiased bases. Numerical results on a certain class of isotropic state vectors suggest that their components, in any of those same mutually unbiased bases, exhibit a semicircular distribution when the dimension of the state space is large.
To cite this abstract, use the following reference: http://meetings.aps.org/link/BAPS.2011.MAR.W29.5 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.8135775327682495, "perplexity_flag": "middle"} |
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# Risk Management for Enterprises and Individuals, v. 1.0
by Etti Baranoff, Patrick L. Brockett, and Yehuda Kahane
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## 7.4 Appendix: Modern Loss Reserving Methods in Long Tail Lines
The actuarial estimation in loss reserving is based on data of past claim payments. This data is typically presented in the form of a triangle, where each row represents the accident (or underwriting) period and each column represents the development period. Table 7.11 "A Hypothetical Loss Triangle: Claim Payments by Accident and Development Years ($Thousand)" represents a hypothetical claims triangle. For example, the payments for 2006 are presented as follows:$13 million paid in 2006 for development year 0, another $60 million paid in development year 1 (i.e., 2007 = [2006+1]), and another$64 million paid during 2008 for development year 2. Note, that each diagonal represents payments made during a particular calendar period. For example, the last diagonal represents payment made during 2008.
Table 7.11 A Hypothetical Loss Triangle: Claim Payments by Accident and Development Years (\$ Thousand)
Development Year
Accident Year 0 1 2 3 4 5 6
2002 9,500 50,500 50,000 27,500 9,500 5,000 3,000
2003 13,000 44,000 53,000 33,500 11,500 5,000
2004 14,000 47,000 56,000 29,500 15,000
2005 15,000 52,000 48,000 35,500
2006 13,000 60,000 64,000
2007 16,000 47,000
2008 17,000 ?
The actuarial analysis has to project how losses will be developed into the future based on their past development. The loss reserve is the estimate of all the payments that will be made in the future and is still unknown. In other words, the role of the actuary is to estimate all the figures that will fill the blank lower right part of the table. The actuary has to “square the triangle.” The table ends at development year 6, but the payments may continue beyond that point. Therefore, the actuary should also forecast beyond the known horizon (beyond development year 6 in our table), so the role is to “rectanglize the triangle.”
The actuary may use a great variety of triangles in preparing the forecast: the data could be arranged by months, quarters, or years. The data could be in current figure or in cumulative figures. The data could represent numbers: the number of reported claims, the number of settled claims, the number of still pending claims, the number of closed claims, and so forth The figures could represent claim payments such as current payments, payments for claims that were closed, incurred claim figures (i.e., the actual payments plus the case estimate), indexed figures, average claim figures, and so forth.
All actuarial techniques seek to identify a hidden pattern in the triangle, and to use it to perform the forecast. Some common techniques are quite intuitive and are concerned with identifying relationships between the payments made across consecutive developing years. Let us demonstrate it on Table 7.11 "A Hypothetical Loss Triangle: Claim Payments by Accident and Development Years ($Thousand)" by trying to estimate the expected payments for accident year 2008 during 2009 (the cell with the question mark). We can try doing so by finding a ratio of the payments in development year 1 to the payments in development year 0. We have information for accident years 2002 through 2007. The sum of payments made for these years during development year 1 is$300,500 and the sum of payments made during development year 0 is $80,500. The ratio between these sums is 3.73. We multiply this ratio by the$17,000 figure for year 2008, which gives an estimate of \$63,410 in payments that will be made for accident year 2008 during development year 1 (i.e., during 2009). Note that there are other ways to calculate the ratios: instead of using the ratio between sums, we could have calculated for each accident year the ratio between development year 1 and development year 0, then calculated the average ratio for all years. This would give a different multiplying factor, resulting in a different forecast.
In a similar way, we can calculate factors for moving from any other development period to the next one (a set of factors to be used for moving from each column to the following one). Using these factors, we can fill all other blank cells in Table 7.11 "A Hypothetical Loss Triangle: Claim Payments by Accident and Development Years ($Thousand)". Note that the figure of$63,410 that we inserted as the estimate for accident year 2008 during development year 1 is included in estimating the next figure in the table. In other words, we created a recursive model, where the outcome of one step is used in estimating the outcome of the next step. We have created a sort of “chain ladder,” as these forecasting methods are often referred to.
In the above example, we used ratios to move from one cell to the next one. But this forecasting method is only one of many we could have utilized. For example, we could easily create an additive model rather than a multiplicative model (based on ratios). We can calculate the average difference between columns and use it to climb from one cell to the missing cell on its right. For example, the average difference between the payments for development year 1 and development year 0 is $36,667 (calculated only for the figures for which we have data on both development years 0 and 1, or 2002 through 2007). Therefore, our alternative estimate for the missing figure in Table 7.11 "A Hypothetical Loss Triangle: Claim Payments by Accident and Development Years ($ Thousand)", the payments that are expected for accident year 2008 during 2009, is $53,667 ($17,000 plus \$36,667). Quite a different estimate than the one we obtained earlier!
We can create more complicated models, and the traditional actuarial literature is full of them. The common feature of the above examples is that they are estimating the set of development period factors. However, there could also be a set of “accident period factors” to account for the possibility that the portfolio does not always stay constant between years. In one year, there could have been many policies or accidents, whereas in the other year, there could have been fewer. So, there could be another set of factors to be used when moving between rows (accident periods) in the triangle. Additionally, there could also be a set of calendar year factors to describe the changes made while moving from one diagonal to the other. Such effects may result from a multitude of reasons—for example, a legal judgment forcing a policy change or inflation that increases average payments. A forecasting model often incorporates a combination of such factors. In our simple example with a triangle having seven rows, we may calculate six factors in each direction: six for the development periods (column effects), six for the accident periods (row effects), and six for the diagonals (calendar or payment year effects). The analysis of such a simple triangle may include eighteen factors (or parameters). A larger triangle (which is the common case in practice) where many periods (months, quarters, and years) are used involves the estimation of too many parameters, but simpler models with a much smaller number of factors can be used (see below).
Although the above methods are very appealing intuitively and are still commonly used for loss reserving, they all suffer from major drawbacks and are not ideal for use. Let us summarize some of the major deficiencies:
• The use of factors in all three directions (accident year, development year, and payment year) may lead to contradictions. We have the freedom to determine any two directions, but the third is determined automatically by the first two.
• There is often a need to forecast “beyond the horizon”—that is, to estimate what will be paid in the development years beyond year six in our example. The various chain ladder models cannot do this.
• We can always find a mathematical formula that will describe all the data points, but it will lack good predictive power. At the next period, we get new data and a larger triangle. The additional new pieces of information will often cause us to use a completely different set of factors—including those relating to previous periods. The need to change all the factors is problematic, as it indicates instability of the model and lack of predictive power. This happens due to overparameterization (a very crucial point that deserves a more detailed explanation, as provided below).
• There are no statistical tests for the validity of the factors. Thus, it is impossible to understand which parameters (factors) are statistically significant. To illustrate, it is clear that a factor (parameter) based on a ratio between only two data points (e.g., a development factor for the sixth year, which will be based on the two figures in the extreme right corner of the Table 7.11 "A Hypothetical Loss Triangle: Claim Payments by Accident and Development Years (\$ Thousand)" triangle) is naturally less reliable, although it may drastically affect the entire forecast.
• The use of simple ratios to create the factors may be unjustified because the relationships between two cells could be more complicated. For example, it could be that a neighboring cell is obtained by examining the first cell, adding a constant, and then multiplying by a ratio. Studies have shown that most loss reserves calculated with chain ladder models are suffering from this problem.
• Chain ladder methods create a deterministic figure for the loss reserves. We have no idea as to how reliable it is. It is clear that there is zero probability that the forecast will exactly foretell the exact future figure. But management would appreciate having an idea about the range of possible deviations between the actual figures and the forecast.
• The most common techniques are based on triangles with cumulative figures. The advantage of cumulative figures is that they suppress the variability of the claims pattern and create an illusion of stability. However, by taking cumulative rather than noncumulative figures, we often lose much information, and we may miss important turning points. It is similar to what a gold miner may do by throwing away, rather than keeping, the little gold nuggets that may be found in huge piles of worthless rocks.
• Many actuaries are still using triangles of incurred claim figures. The incurred figures are the sum of the actually paid numbers plus the estimates of future payments supplied by claims department personnel. The resulting actuarial factors from such triangles are strongly influenced by the changes made by the claims department from one period to the other. Such changes should not be included in forecasting future trends.
There are modern actuarial techniques based on sophisticated statistical tools that could be used for giving better forecasts while using the same loss triangles. Let us see how this works without engaging in a complicated statistical discussion. The purpose of the discussion is to increase the understanding of the principles, but we do not expect the typical student to be able to immediately perform the analysis. We shall largely leave the analysis to actuaries that are better equipped with the needed mathematical and statistical tools.
A good model is evaluated by its simplicity and generality. Having a complex model with many parameters makes it complicated and less general. The chain ladder models that were discussed above suffer from this overparameterization problem, and the alternative models that are explained below overcome this difficulty.
Let us start by simply displaying the data of Table 7.11 "A Hypothetical Loss Triangle: Claim Payments by Accident and Development Years (\$ Thousand)" in a graphical form in Figure 7.3 "Paid Claims (in Thousands of Dollars) by Development Year". The green dots describe the original data points (the paid claims on the vertical axis and the development years on the horizontal axis). To show the general pattern, we added a line that represents the averages for each development year. We see that claim payments in this line of business tend to increase, reach a peak after a few years, then decline slowly over time and have a narrow “tail” (that is, small amounts are to be paid in the far future).
Figure 7.3 Paid Claims (in Thousands of Dollars) by Development Year
We can immediately see that the entire claims triangle can be analyzed in a completely different way: by fitting a curve through the points. One of these tools to enable this could be regression analysis. Such a tool can give us a better understanding of the hidden pattern than does the chain ladder method. We see that the particular curve in our case is nonlinear, meaning that we need more than two parameters to describe it mathematically. Four parameters will probably suffice to give a mathematical function that will describe the pattern of Figure 7.3 "Paid Claims (in Thousands of Dollars) by Development Year". The use of such methods can reach a level of sophistication that goes beyond the scope of this book. It is sufficient to say that we can get an excellent mathematical description of the pattern with the use of only four to six parameters (factors). This can be measured by a variety of statistical indicators. The coefficient of correlation for such a mathematical formula is above 95 percent, and the parameters are statistically significant.
Such an approach is simpler and more general than any chain ladder model. It can be used to forecast beyond the horizon, it can be statistically tested and validated, and it can give a good idea about the level of error that may be expected. When a model is based on a few parameters only, it becomes more “tolerant” to deviations: it is clear that the next period payment will differ from the forecast, but it will not force us to change the model. From the actuary’s point of view, claim payments are stochastic variables and should never be regarded as a deterministic process, so why use a deterministic chain ladder analysis?
It is highly recommended, and actually essential, to base the analysis on a noncumulative claims triangle. The statistical analysis does not offer good tools for cumulative figures; we do not know their underlying statistical processes, and therefore, we cannot offer good statistical significance tests. The statistical analysis that is based on the current, noncumulative claim figures is very sensitive and can easily detect turning points and changing patterns.
One last point should be mentioned. The key to regression analysis is the analysis of the residuals, that is, the differences between the observed claims and the figures that are estimated by the model. The residuals must be spread randomly around the forecasted, modeled, figures. If they are not randomly spread, the model can be improved. In other words, the residuals are the compass that guides the actuary in finding the best model. Traditional actuarial analyses based on chain ladder models regard variability as a corrupt element and strive to get rid of the deviations to arrive at a deterministic forecast. By doing so, actuaries throw away the only real information in the data and base the analysis on the noninformative part alone! Sometimes the fluctuations are very large, and the insurance company is working in a very uncertain, almost chaotic claims environment. If the actuary finds that this is the case, it will be important information for the managers and should not be hidden or replaced by a deterministic, but meaningless, forecast.
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http://unapologetic.wordpress.com/2008/08/ | # The Unapologetic Mathematician
## Convergence of Power Series
Now that we’ve imported a few rules about convergent series of complex numbers we can talk about when the series we get from evaluating power series converge or not.
We’ll just consider our power series to be in $\mathbb{C}[[X]]$, because if we have a real power series we can consider each coefficient as a complex number instead. Now we take a complex number $z$ and try to evaluate the power series $\sum\limits_{k=0}^\infty c_kX^k$ at this point. We get a series of complex numbers
$\displaystyle\sum\limits_{k=0}^\infty c_kz^k=\lim\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nc_kz^k$
by evaluating each polynomial truncation of the power series at $z$ and taking the limit of the sequence. For some $z$ this series may converge and for others it may not. The amazing fact is, though, that we can always draw a circle in the complex plane — $|z|=R$ — within which the series always converges absolutely, and outside of which it always diverges. We’ll say nothing in general about whether it converges on the circle, though.
The tool here is the root test. We take the $n$th root of the size of the $n$th term in the series to find $\sqrt[n]{|c_nz^n|}=|z|\sqrt[n]{|c_n|}$. Then we can pull the $z$-dependance completely out of the limit superior to write $\rho=|z|\limsup\limits_{n\rightarrow\infty}\sqrt[n]{|c_n|}$. The root test tells us that if this is less than ${1}$ the series will converge absolutely, while if it’s greater than ${1}$ the series will diverge.
So let’s define $\lambda=\limsup\limits_{n\rightarrow\infty}\sqrt[n]{|c_n|}$. The root test now says that if $|z|<\frac{1}{\lambda}$ we have absolute convergence, while if $|z|>\frac{1}{\lambda}$ the series diverges. Thus $\frac{1}{\lambda}$ is the radius of convergence that we seek.
Now there are examples of series with all sorts of behavior on the boundary of this disk. The series $\sum\limits_{k=0}^\infty z^k$ has radius of convergence ${1}$ (as we can tell from the above procedure), but it doesn’t converge anywhere on the boundary circle. On the other hand, the series $\sum\limits_{k=1}^\infty\frac{1}{k^2}z^k$ has the same radius of convergence, but it converges everywhere on the boundary circle. And, just to be perverse, the series $\sum\limits_{k=1}^\infty\frac{1}{k}z^k$ has the same radius of convergence but converges everywhere on the boundary but the single point $z=1$.
Posted by John Armstrong | Analysis, Calculus, Power Series | 6 Comments
## Convergence of Complex Series
Today, I want to note that all of our work on convergence of infinite series carries over — with slight modifications — to complex numbers.
First we have to get down an explicit condition on convergence of complex sequences. In any metric space we can say that the sequence $z_n$ converges to a limit $z$ if for every $\epsilon>0$ there is some $N$ so that $d(z_n,z)<\epsilon$ for all $n>N$. Of course, here we’ll be using our complex distance function $|z_n-z|=\sqrt{(z_n-z)\overline{(z_n-z)}}$. Now we just have to replace any reference to real absolute values with complex absolute values and we should be good.
Cauchy’s condition comes in to say that the series $\sum\limits_{k=0}^\infty a_k$ converges if and only for every $\epsilon>0$ there is an $N$ so that for all $n\geq m>N$ the sum $\left|\sum\limits_{k=m}^na_k\right|<\epsilon$.
Similarly, we say that the series $\sum\limits_{k=0}^\infty a_k$ is absolutely convergent if the series $\sum\limits_{k=0}^\infty|a_k|$ is convergent, and this implies that the original series converges.
Since the complex norm is multiplicative, everything for the geometric series goes through again: $\sum\limits_{k=0}^\infty c_0r^k=\frac{c_0}{1-r}$ if $|r|<1$, and it diverges if $|r|>1$. The case where $|r|=1$ is more complicated, but it can be shown to diverge as well.
The ratio and root tests are basically proven by comparing series of norms with geometric series. Since once we take the norm we’re dealing with real numbers, and since the norm is multiplicative, we find that the proofs go through again.
Posted by John Armstrong | Analysis, Calculus, Power Series | 1 Comment
## Evaluating Power Series
Now that we’ve got some topological fields to use as examples, let’s focus in on power series over $\mathbb{R}$ or $\mathbb{C}$.
Remember that a power series is like an infinite polynomial. In fact, we introduced a topology so we could see in any power series a sequence of polynomials that converged to it. To be explicit, we write the series as a limit
$\displaystyle S=\sum\limits_{k=0}^\infty c_kX^k=\lim\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nc_kX^k$
where the $c_k$ are coefficients selected from our base field.
Now evaluation of power series is specified by two conditions: it should agree with evaluation of polynomials when we’ve got a power series that cuts off after a finite number of terms, and it should be continuous.
The first condition says that each of our approximating polynomials should evaluate just the same as it did before. That is, if we cut off after the degree-$n$ term and evaluate at the point $x$ in the base field, we should get $\sum\limits_{k=0}^nc_kx^k$.
The second condition says that evaluation should preserve limits. And we’ve got a sequence right here: the $n$th term is the evaluation of the $n$th approximating polynomial! So the power series should evaluate to the limit $\lim\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nc_kx^k$. If this limit exists, that is. And that’s why we need a topological field to make sense of evaluations.
Now we’re back in the realm of infinite series, and taking the limit of a sequence of partial sums. The series in question has as its $n$th term the evaluated monomial $c_nx^n$. We can start using our old techniques to sum these series.
Posted by John Armstrong | Analysis, Calculus, Power Series | 4 Comments
## Some Topological Fields
Okay, hopefully now that I’m back in Kentucky and I’ve gotten the first day of all three of my classes under my belt, I can get back into the groove.
In a little while we’re going to want to talk about “evaluating” a power series like we did when we considered polynomials as functions. But when we try to map into our base field, we get a sequence of values and we ask questions about convergence. That means we need to have a topology on our field! Luckily there are a few hanging around.
The real numbers have a topology. In fact, that’s really their main characteristic. The rational numbers have a topology too, but the whole motivation of the construction of the real numbers is that rational numbers have a lot of holes that sequence limits fall through. Since we’re talking about sequences here we really need the tighter weave that the real numbers provide.
What else can we use? The complex numbers form a two-dimensional vector space over the reals, which means that as a vector space we have the isomorphism $\mathbb{C}\cong\mathbb{R}\times\mathbb{R}$. So let’s just use the product metric, along with the topology it gives on the complex numbers.
Let’s be a little explicit here: the product metric depends on picking the basis $\{1,i\}$ for $\mathbb{C}$ as a vector space over $\mathbb{R}$. We get the “size” of a complex number $|a+bi|=\sqrt{a^2+b^2}$, and then we define the distance between two complex numbers as the size of their difference.
I said before that there may be many different distance functions that give the same topology, so why do we really want to use this one? Well, it turns out that this formula can actually be written in a really neat way in complex number language. If we have a complex number $z=a+bi$ we also have its conjugate $\bar{z}=a-bi$. Then we can calculate the product
$z\bar{z}=(a+bi)(a-bi)=(a^2+b^2)+0i$
This is just the square of our distance function, written as a complex number! But also notice that this formula is symmetric between a complex number and its conjugate. That is, conjugation preserves the size of complex numbers, as we should expect because there’s no a priori difference between the two.
Now we need to prove that the field operations are continuous. For multiplication, for example, we need to ask that if $z_2$ is close to $z_1$ and $w_2$ is close to $w_1$, then $z_2w_2$ is close to $z_1w_1$. We write $z_2=z_1+\Delta z$ and $w_2=w_1+\Delta w$, and multiply out
$z_2w_2=z_1w_1+z_1\Delta w+\Delta zw_1+\Delta z\Delta w$
The condition is that for any real $\epsilon>0$ we can find real $\delta_z$ and $\delta_w$ so that $\delta_z>|\Delta z|$ and $\delta_w>|\Delta w|$ together imply that $|z_1\Delta w+\Delta zw_1+\Delta z\Delta w|<\epsilon$. From here it’s a bunch of mucking around with our formula, but none of it is very difficult.
At the end of the day, we’ve got two topological fields to work with — the real numbers $\mathbb{R}$ and the complex numbers $\mathbb{C}$ — and we can talk about evaluating power series from $\mathbb{R}[[X]]$ or $\mathbb{C}[[X]]$.
UPDATE: I forgot to mention that it’s also easy to see that the norm is multiplicative. That is, $|zw|=\sqrt{zw\overline{zw}}=\sqrt{z\bar{z}}\sqrt{w\bar{w}}=|z||w|$.
Posted by John Armstrong | Algebra, Ring theory | 5 Comments
## On the road again
Today is my last long drive for a while. I’ll try to post a Sunday Sample tonight, or maybe tomorrow. But for now, here’s a problem to chew on.
In a multiple choice test, one question was illegible, but the following choice of answers was clearly printed. Which must be the right answer?
1. All of the below
2. None of the below
3. All of the above
4. One of the above
5. None of the above
6. None of the above
Posted by John Armstrong | Uncategorized | 15 Comments
## Zero-Knowledge Sudoku
Scientific American has been doing a lot about privacy, since that’s the theme of its latest issue. One of its posts today linked to this one from two years ago at a weblog that my old teacher Bill Gasarch has something to do with. It’s a pretty cogent explanation of zero-knowledge proofs, with a sudoku puzzle as the centerpiece.
A zero-knowledge proof is a way of convincing you that I know something without you getting any information about what that thing is. I know, it sounds like LitCrit, but this actually makes sense.
The classic motivating example is a cave with two entrances. Way back deep in the cave there’s a door with a combination lock I tell you I can open. I go into one of the cave entrances (you don’t see which, but if I can’t open the door I’m stuck in whichever one I picked). Then you come to the mouth(s) of the cave and yell out which entrance I should come out of.
If I can open the door, I can come out either entrance, so I come out the one you want. If I can’t open the door, then I can only come out the one you want if that’s the one I went in. Juggling the numbers a bit we can find that the probability I don’t know the combination given that I came out the correct entrance is $\frac{1}{2}$.
This means that half the time I could have gotten the right answer by luck rather than by skill. That’s not very convincing, but we can repeat the experiment, and each run is independent of all the others. So if we go through the motions $n$ times and I’ve met your challenge every single time, then the probability that I’ve just been lucky is $\frac{1}{2^n}$, which gets very small very quickly. After a while you’ll be sufficiently convinced that I know the answer.
In a way it’s more like a physics “proof” than the ones mathematicians are used to. The hypothesis can be falsified at any step, and each confirmation just increases the probability that it’s accurate.
Posted by John Armstrong | Cryptography | 4 Comments
## Products of Metric Spaces
Shortly we’re going to need a construction that’s sort of interesting in its own right.
We know about products of topological spaces. We can take products of metric spaces, too, and one method comes down to us all the way from Pythagoras.
The famous Pythagorean theorem tells us that in a right triangle the length $c$ of the side opposite the right angle stands in a certain relation to the lengths $a$ and $b$ of the other two sides: $c^2=a^2+b^2$. So let’s say we’ve got metric spaces $(M_1,d_1)$ and $(M_2,d_2)$. For the moment we’ll think of them as being perpendicular and define a distance function $d$ on $M_1\times M_2$ by
$d((x_1,x_2),(y_1,y_2)=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}$
The quantity inside the radical here must be nonnegative, since it’s the sum of two nonnegative numbers. Since the result needs to be nonnegative, we take the unique nonnegative square root.
Oops, I don’t think I mentioned this before. Since the function $f(x)=x^2$ has $f'(x)=2x$ as its derivative, it’s always increasing where $x$ is positive. And since we can eventually a square above any real number we choose, its values run from zero all the way up to infinity. Now the same sort of argument as we used to construct the exponential function gives us an inverse sending any nonnegative number to a unique nonnegative square root.
Okay, that taken care of, we’ve got a distance function. It’s clearly nonnegative and symmetric. The only way for it to be zero is for the quantity in the radical to be zero, and this only happens if each of the terms $d_1(x_1,y_1)$ and $d_2(x_2,y_2)$ are zero. But since these are distance functions, that means $x_1=y_1$ and $x_2=y_2$, so $(x_1,x_2)=(y_1,y_2)$.
The last property we need is the triangle inequality. That is, for any three pairs $(x_1,x_2)$, $(y_1,y_2)$, $(z_1,z_2)$ we have the inequality
$d((x_1,x_2),(z_1,z_2))\leq d((x_1,x_2),(y_1,y_2))+d((y_1,y_2),(z_1,z_2))$
Substituting from the definition of $d$ we get the statement
$\sqrt{d_1(x_1,z_1)^2+d_2(x_2,z_2)^2}\leq\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}+\sqrt{d_1(y_1,z_1)^2+d_2(y_2,z_2)^2}$
The triangle inequalities for $d_1$ and $d_2$ tell us that $d_1(x_1,z_1)\leq d_1(x_1,y_1)+d_1(y_1,z_1)$ and $d_2(x_2,z_2)\leq d_2(x_2,y_2)+d_1(y_2,z_2)$. So if we make these substitutions on the left, it increases the left side of the inequality we want. Thus if we can prove the stronger inequality
$\begin{aligned}\sqrt{d_1(x_1,y_1)^2+2d_1(x_1,y_1)d_1(y_1,z_1)+d_1(y_1,z_1)^2+d_2(x_2,y_2)^2+2d_2(x_2,y_2)d_2(y_2,z_2)+d_2(y_2,z_2)^2}\\\leq\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}+\sqrt{d_1(y_1,z_1)^2+d_2(y_2,z_2)^2}\end{aligned}$
we’ll get the one we really want. Now since squaring preserves the order on the nonnegative reals, we can find this equivalent to
$\begin{aligned}d_1(x_1,y_1)^2+2d_1(x_1,y_1)d_1(y_1,z_1)+d_1(y_1,z_1)^2+d_2(x_2,y_2)^2+2d_2(x_2,y_2)d_2(y_2,z_2)+d_2(y_2,z_2)^2\\\leq d_1(x_1,y_1)^2+d_2(x_2,y_2)^2+2\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}\sqrt{d_1(y_1,z_1)^2+d_2(y_2,z_2)^2}+d_1(y_1,z_1)^2+d_2(y_2,z_2)^2\end{aligned}$
Some cancellations later:
$\begin{aligned}d_1(x_1,y_1)d_1(y_1,z_1)+d_2(x_2,y_2)d_2(y_2,z_2)\\\leq \sqrt{d_1(x_1,y_1)^2d_1(y_1,z_1)^2+d_1(x_1,y_1)^2d_2(y_2,z_2)^2+d_2(x_2,y_2)^2d_1(y_1,z_1)^2+d_2(x_2,y_2)^2d_2(y_2,z_2)^2}\end{aligned}$
We square and cancel some more:
$2d_1(x_1,y_1)d_1(y_1,z_1)d_2(x_2,y_2)d_2(y_2,z_2)\leq d_1(x_1,y_1)^2d_2(y_2,z_2)^2+d_2(x_2,y_2)^2d_1(y_1,z_1)^2$
Moving these terms around we find
$\begin{aligned}0\leq\left(d_1(x_1,y_1)d_2(y_2,z_2)\right)^2-2\left(d_1(x_1,y_1)d_2(y_2,z_2)\right)\left(d_2(x_2,y_2)d_1(y_1,z_1)\right)+\left(d_2(x_2,y_2)d_1(y_1,z_1)\right)^2\\=\left(d_1(x_1,y_1)d_2(y_2,z_2)-d_2(x_2,y_2)d_1(y_1,z_1)\right)^2\end{aligned}$
So at the end of the day, our triangle inequality is equivalent to asking if a certain quantity squared is nonnegative, which it clearly is!
Now here’s the important thing at the end of all that calculation: this is just one way to get a metric on the product of two metric spaces. There are many other ones which give rise to different distance functions, but the same topology and the same uniform structure. And often it’s the topology that we’ll be most interested in.
In particular, this will give us a topology on any finite-dimensional vector space over the real numbers, but we don’t want to automatically equip that vector space with this norm unless we say so very explicitly. In fact, we don’t even want to make that same assumption about the two spaces being perpendicular to each other. The details of exactly why this is so I’ll leave until we get back to linear algebra, but I want to be clear right now that topology comes for free, but we may have good reason to use different “distances”.
## Power Series
Prodded by some comments, I think I’ll go even further afield from linear algebra. It’s a slightly different order than I’d originally thought of, but it will lead to some more explicit examples when we’re back in the realm of linear algebra, so it’s got its own benefits.
I’ll note here in passing that mathematics actually doesn’t proceed in a straight line, despite the impression most people get. The lower-level classes are pretty standard, yes — natural-number arithmetic, fractions, algebra, geometry, calculus, and so on. But at about this point where most people peter out, the subject behaves more like an alluvial fan — many parallel rivulets carry off in different directions, but they’re all ultimately part of the same river. So in that metaphor, I’m pulling a bit of an avulsion.
Anyhow, power series are sort of like polynomials, except that the coefficients don’t have to die out at infinity. That is, when we consider the algebra of polynomials $\mathbb{F}[X]$ as a vector space over $\mathbb{F}$ it’s isomorphic to the infinite direct sum
$\displaystyle\mathbb{F}[X]\cong\bigoplus\limits_{k=0}^\infty\mathbb{F}X^k$
but the algebra of power series — written $\mathbb{F}[[X]]$ — is isomorphic to the infinite direct product
$\displaystyle\mathbb{F}[[X]]\cong\prod\limits_{k=0}^\infty\mathbb{F}X^k$
It’s important to note here that the $X^i$ do not form a basis here, since we can’t write an arbitrary power series as a finite linear combination of them. But really they should behave like a basis, because they capture the behavior of every power series. In particular, if we specify that $\mu(X^m,X^n)=X^{m+n}$ then we have a well-defined multiplication extending that of power series.
I don’t want to do all the fine details right now, but I can at least sketch how this all works out, and how we can adjust our semantics to talk about power series as if the $X^i$ were an honest basis. The core idea is that we’re going to introduce a topology on the space of polynomials.
So what polynomials should be considered “close” to each other? It turns out to make sense to consider those which agree in their lower-degree terms to be close. That is, we should have the space of tails
$\displaystyle\bigoplus\limits_{k=n+1}^\infty\mathbb{F}X^k$
as an open set. More concretely, for every polynomial $p$ with degree $n$ there is an open set $U_p$ consisting of those polynomials $q$ so that $X^{n+1}$ divides the difference $q-p$.
Notice here that any power series defines, by cutting it off after successively higher degree terms, a descending sequence of these open sets. More to the point, it defines a sequence of polynomials. If the power series’ coefficients are zero after some point — if it’s a polynomial itself — then this sequence stops and stays at that polynomial. But if not it never quite settles down to any one point in the space. Doesn’t this look familiar?
Exactly. Earlier we had sequences of rational numbers which didn’t converge to a rational number. Then we completed the topology to give us the real numbers. Well here we’re just doing the same thing! It turns out that the topology above gives a uniform structure to the space of polynomials, and we can complete that uniform structure to give the vector space underlying the algebra of power series.
So here’s the punch line: once we do this, it becomes natural to consider not just linear maps, but continuous linear maps. Now the images of the $X^k$ can’t be used to uniquely specify a linear map, but they will specify at most one value for a continuous linear map! That is, any power series comes with a sequence converging to it — its polynomial truncations — and if we know the values $f(X^k)$ then we have uniquely defined images of each of these polynomial truncations since each one is a finite linear combination. Then continuity tells us that the image of the power series must be the limit of this sequence of images, if the limit exists.
## Factoring Real Polynomials
Okay, we know that we can factor any complex polynomial into linear factors because the complex numbers are algebraically closed. But we also know that real polynomials can have too few roots. Now, there are a lot of fields out there that aren’t algebraically closed, and I’m not about to go through all of them. But we use the real numbers so much because of the unique position it holds by virtue of the interplay between its topology and its algebra. So it’s useful to see what we can say about real polynomials.
We start by noting that since the real numbers sit inside the complex numbers, we can consider any real polynomial as a complex polynomial. If the polynomial has a real root $r$, then the complex polynomial has a root $r+0i$. So all the real roots still show up.
Now, we might not have as many real roots as the degree would indicate. But we are sure to have as many complex roots as the degree, which will include the real roots. Some of the roots may actually be complex numbers like $a+bi$. Luckily, one really interesting thing happens here: if $a+bi$ is a root, then so is its complex conjugate $a-bi$.
Let’s write out our polynomial $c_0+c_1X+...+c_nX^n$, where all the $c_i$ are real numbers. To say that $a+bi$ is a root means that when we substitute it for ${X}$ we get the equation
$c_0+c_1(a+bi)+...+c_n(a+bi)^n=0$
Now we can take the complex conjugate of this equation
$\overline{c_0+c_1(a+bi)+...+c_n(a+bi)^n}=\overline{0}$
But complex conjugation is a field automorphism, so it preserves both addition and multiplication
$\overline{c_0}+\overline{c_1}\left(\overline{a+bi}\right)+...+\overline{c_n}\left(\overline{a+bi}\right)^n=\overline{0}$
Now since all the $c_i$ (and ${0}$) are real, complex conjugation leaves them alone. Conjugation sends $a+bi$ to $a-bi$, and so we find
$c_0+c_1(a-bi)+...+c_n(a-bi)^n=0$
So $a-bi$ is a root as well. Thus if we have a (complex) linear factor like $\left(X-(a+bi)\right)$ we’ll also have another one like $\left(X-(a-bi)\right)$. These multiply to give
$\begin{aligned}\left(X-(a+bi)\right)\left(X-(a-bi)\right)=X^2-(a-bi)X-(a+bi)X+(a+bi)(a-bi)\\=X^2-(2a)X+(a^2+b^2)\end{aligned}$
which is a real polynomial again.
Now let’s start with our polynomial $p$ of degree $n$. We know that over the complex numbers it has a root $\lambda$. If this root is real, then we can write
$p=(X-\lambda)\tilde{p}$
where $\tilde{p}$ is another real polynomial which has degree $n-1$. On the other hand, if $\lambda=a+bi$ is complex then $\bar{\lambda}$ is also a root of $p$, and so we can write
$p=(X-(a+bi))(X-(a-bi))\tilde{p}=(X^2-(2a)X+(a^2+b^2))\tilde{p}$
where $\tilde{p}$ is another real polynomial which has degree $n-2$.
Either way, now we can repeat our reasoning starting with $\tilde{p}$. At each step we can pull off either a linear term or a quadratic term (which can’t be factor into two real linear terms).
Thus every real polynomial factors into the product of a bunch of linear polynomials and a bunch of irreducible quadratic polynomials, and the number of linear factors plus twice the number of quadratic factors must add up to the degree of $p$. It’s not quite so nice as the situation over the complex numbers, but it’s still pretty simple. We’ll see many situations into the future where this split between two distinct real roots and a conjugate pair of complex roots (and the border case of two equal real roots) shows up with striking qualitative effects.
## Additive Functions
I’ve got internet access again, but I’m busy with a few things today, like assembling my furniture. Luckily, Tim Gowers has a post on “How to use Zorn’s lemma”. His example is the construction of additive (not linear) functions from $\mathbb{R}$ to itself.
In practice, as he points out, this is equivalent to defining linear functions (not just additive) from $\mathbb{R}$ to itself, if we’re considering $\mathbb{R}$ as a vector space over the field $\mathbb{Q}$ of rational numbers! This is a nifty little application within the sort of stuff we’ve been doing lately, and it really explains how we used Zorn’s Lemma when we showed that every vector space has a basis.
Posted by John Armstrong | Algebra, Linear Algebra | 9 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": 172, "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": 0.9313035011291504, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/45253/almost-linear-ode-how-node-becomes-a-spiral | ## Almost linear ODE: how node becomes a spiral
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Most introductory ODE books contain a discussion of almost linear systems, and there are two cases when the behavior of an almost linear system near an equilbrium point can differ from the behaviour of a linearlized system. One of these two cases is when the linearlized system has a repeated nonzero real eigenvalue; then the equilibrium solution of the linearlized system is a node, while the original almost linear system could be a node or a spiral, and I wish to see an explicit example of when the spiral occurs.
To give you an idea of what I need, let's discuss the other case when the linearized system is a center, but the almost linear system could be a center or a spiral, and the example when the spiral occurs is $x^\prime=y+x(x^2+y^2)$ and $y^\prime=-x+y(x^2+y^2)$.
UPDATE: I now feel much better about the question because it turned out not as silly as I initially feared. Apparently, given a $2\times 2$ almost linear system $X^\prime=F(X)$ if the solution $X=0$ of the linearized system $X^\prime=AX$ is a node, then it is also a node for $X^\prime=F(X)$, as long as $F$ is $C^2$. If $F$ is merely $C^1$ there is a counterexample indicated in the comments. In cartezian coordinates the counterexample is $x^\prime=-x-\frac{2y}{\ln(x^2+y^2)}$ and $y^\prime=-y+\frac{2x}{\ln(x^2+y^2)}$; the right hand side is not $C^2$.
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Maybe I don't understand, but does $x'= -x + y^3$, $y'= -y - x^3$ qualify? However, the spiral and the node are always conjugated, so I don't see quite clearly what you mean by differ (when there are zero eigenvalues this is clear, since you can get, as in your example, really different behaviour). – rpotrie Nov 8 2010 at 10:03
In a small neighborhood of the equilibrium point $p$, the spiral goes around $p$ infinitely many times, the node does not, so the behaviour is quite different. However, I found an example that does the job. In polar coordinates it looks like $r^\prime=-r$ and $\theta^\prime=1/ln(r)$. – Igor Belegradek Nov 8 2010 at 12:39
@rpotrie, your example looks like a node in a small neigborhood of the origin. In a larger neigborhood it starts looking like a spiral but my question was about local begavior. Still it is a nice example, thanks! – Igor Belegradek Nov 8 2010 at 18:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.18, "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": 0.9430640935897827, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/68100/what-are-the-requirements-for-a-rotation-matrix?answertab=oldest | # What are the requirements for a rotation matrix?
Generally speaking, what are the necessary and sufficient properties of a matrix to make it a rotation matrix?
Is det(A) = 1 enough?
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## 3 Answers
No, not by far. The matrix needs to be orthogonal, which means that $A^tA=I$ where $A^t$ is the transposed matrix -- and then it also has to have determinant 1.
(You can think of orthogonality by considering how the matrix acts on the standard basis vectors -- since they were orthogonal to each other and had length 1 before the rotation, this must also be true after the rotation. But the vectors after-the-rotation are just the columns of $A$, and $A^tA=I$ is then a compact way to take the dot product of each pair of new vectors and see if they come out right, all in one operation).
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That $A^tA=I$ doesn't guarantee it's an orthogonal matrix unless it's also a square matrix, i.e. the number of rows is the same as the number of columns. – Michael Hardy Sep 28 '11 at 3:12
It is easy to understand why orthogonal matrices preserve the vector length. Can you explain why $\det A=1$ means rotation and $\det A=-1$ means reflection? – Shiyu Sep 28 '11 at 3:22
@Shiyu, not right off the cuff. Intuitively it's because $SO(n)$ is connected, but I don't have a quick demonstration of that ready. Also, I don't even think I'd be comfortable calling something like $\pmatrix{0.6&0.8&0\\-0.8&0.6&0\\0&0&-1}$ a "reflection". – Henning Makholm Sep 28 '11 at 3:44
– J. M. Sep 28 '11 at 9:36
Not it is not. A rotation matrix is a square matrix with orthonormal columns and determinant 1. The set of all $n\times$n such matrices is commonly referred to as the special orthogonal group, and is denoted $SO(n)$.
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In $\mathbb{R}^3$, one can prove that the set of all rotation matrices is precisely $SO_3(\mathbb{R})$.
Assuming this fact, a rotation matrix $A$ with respect to the standard basis would then have the properties
1. $AA^{T} = I$
2. $\langle Ax, Ay \rangle = \langle x,y \rangle$, where $\langle,\rangle$ is just the standard inner product in $\mathbb{R}^3$ and $x,y$ any vectors in $\mathbb{R}^3$.
3. $||Ax|| = ||x||$ for any vector $||x||$.
In fact the properties above hold not just for the rotation matrices but for matrices in $O_3$ (and in general $O_n$ ) as well.
It remains now to prove why the rotation matrices are exactly the matrices of $SO_3$. To prove it you need at least two facts:
1. Every linear operator on a real vector space $V$ of odd dimension has a (real) eigenvalue.
2. If $W$ is a subspace of $V$ stable the restriction of a linear operator $T$ to $W$, then $W^{\perp}$ is stable too.
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http://mathoverflow.net/questions/66584/optimal-tax-rate/66589 | ## Optimal tax Rate
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Assume you have two countries A and B, with a tax rates $T_A$ and $T_B$. The tax is redistributed to each people equally. Hence if you live in A and you make $I$ as income then you will finally receive
$$I*(1-T_A) + \overline{I}*T_A$$
where $\overline{I}$ is the average income in $A$.The country A wants to choose an optimal rate, in order to do it the decision is taken by the median income. But the people can migrate if the new rate makes them poorer than if they were living in $B$. Of course this migration to B as a cost $M$, hence if the median income choose as new rate $T$ the people in A such that
$$I(1-T) + \overline{I}\ T < I\ (1-T_B) + \overline{I}\ T_B -M$$
will leave A to B. And symmetrically the people in B such that
$$I\ (1-T_B) + \overline{I}\ T_B < I\ (1-T) + \overline{I}\ T -M$$
will leave B to A. Which changes the configuration of incomes in A and hence the decision of the median income since his income depends on the average income.
My question is how can find the taxe rate which will optimize the income of the median income after migration?
I have think to a dynamical approach, but it looks hard to show that we converge to an equilibrium. Is there is general tools for this kind of problem?
I hope, i have been clear enough.
P.S: I have already ask this question on Math.stackexchange, but i think it is in fact a research problem since i have find nothing in the literature except a a paper of Stéphane Rossignol and Emmanuelle Taugourdeau :Asymmetric social protection systems with migration in J Popul Econ 19:481–505 (2006). But they study an asymmetric case.
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4
I don't think this is a research question in economics- and certainly not in mathematics. It is also underspecified. What is the overall income distribution? What is the initial income distribution? And there is nothing dynamic here. – Michael Greinecker May 31 2011 at 19:19
We can assume the income distribution is gaussian, for example. For me it looks like a non-cooperative game with $N$ player, isn't? and the question is to find the Nash equilibrium, which is a question of "applied" mathematics, no?. But I am not a specialist of game theory, that why i ask to be enlightened. – Raphael May 31 2011 at 19:30
I cannot figure out what median income you're trying to maximize. Country A currently contains x people. Then the tax rates are set, migration happens, and now Country A contains y people. Are we trying to maximize the median income of the the original population in Country A? Or of the final population in Country A? Or of that fraction of the original population of Country A that remains in Country A? Or perhaps of the total population of the two countries? And does "income" mean income net of moving costs? These are the first of many things I find unclear. – Steven Landsburg May 31 2011 at 20:35
There is three steps: first the median income choose the tax rate second people migrate third they pay their tax and receive the redistributed income. That's my problem, because the median income try to maximize is income but this one depends on the migration which depends on his choice. The fact that the choice is made by the median income is a consequence of the median voter theorem: en.wikipedia.org/wiki/Median_voter_theorem – Raphael May 31 2011 at 21:32
@Raphael: you may get more answers at the sister site devoted to math finance quant.stackexchange.com – Andrey Rekalo May 31 2011 at 21:34
show 1 more comment
## 2 Answers
A response to the OP's comment than to the original question. The income distribution is definitely NOT gaussian -- a lot of thought has been devoted to figuring out what exactly it is, which thought has led to the creation of the "fractal" view of the world. Check out
The Misbehavior of Markets: A Fractal View of Financial Turbulence [Paperback] Benoit Mandelbrot (Author), Richard L. Hudson
The pivotal point in the book is Mandelbrodt's talk at Harvard, where he was about to talk about the distribution of incomes, but saw the curves from his talk on his host's blackboard (they came from some questions on variation of commodity prices). The distribution is a power law of some sort.
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Thanks for your answer. I will see this talk of Mandelbrot. But in fact we can choose whatever distribution to start. – Raphael May 31 2011 at 21:23
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For person $i$ with income $I_i$ let $w_i$ be 1 or -1 depending on whether said person lives in $A$ or $B$. Then the condition that the person is living in the right place is a linear constraint on $w_i$ and the other $w$. To make this tractable we can allow $w_i$ to be in the interval $[-1,1]$, since the optium will still be at the endpoints (you can also interpret this is a probability that a person with tht income woul dbe in $A$ or $B$. This gives a linear system for equilibrium. Unfortunately when $M>0$ there will not be a single equilibrium (as an extreme, when $M$ is sufficiently large any state is an equilibrium). If $M=0$ you get a unique equilibrium (or if you somehow pick a canonical equilibrium) and you could try to do a numerical search on $T$ (ternary search?) Another way is to pick a target median income, form a linear constraint that the median is at least this target, and do a binary search to find the best feasible median.
A discussion of the applicability of this to economics is beyond the scope of this post :)
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http://physics.stackexchange.com/questions/1513/why-does-my-watch-act-like-a-mirror-under-water/1514 | # Why does my watch act like a mirror under water?
I have a digital watch, rated to go underwater to $100 \rm m$. When it is underwater it can be read normally, up until you reach a certain angle, then suddenly, it becomes almost like a mirror, reflecting all the light that hits it, almost perfectly (you cannot read the digits, the entire surface becomes reflective.)
I've seen this happen with other waterproof watches too, so I don't think it's unique to mine or the specific model. I'm wondering what causes this? My physics teacher was stumped when I told him about this (we're doing lenses and imaging in physics right now.) I think it has something to do with internal refraction. I haven't been able to measure the angle it becomes reflective accurately, I estimate about 30-40 degrees. Near this critical point, it can be half and half reflective, where only half becomes a mirror, but it's always either reflective or not - never in between being reflective and non-reflective.
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## 1 Answer
What you are observing is total internal reflection. Snell's law tells you that, for a ray transmitting through a surface $n_{1}\sin\theta_{1} = n_{2}\sin\theta_{2}$, where $\theta$ represents the angle of reflection from the surface, $n$ represents the index of refraction of the substance in question, and the labels 1 and 2 represent the source medium and the destination medium.
If $n_{1}>n_{2}$, as would be the case for light leaving water ($n\approx 1.33$)and entering the air ($n\approx 1$)inside of your watch, simple algebra will tell you that there is a range of $\theta_{1}$ at which you will find that Snell's law predicts $\sin \theta_{2}>1$. For this range of angles, since you can't solve for $\theta_{2}$, light cannot be transmitted, and must be reflected. So the watch looks like a mirror. In fact, if you flip over, and look at the surface of the water, you will find a portion of the surface of the water looks like a mirror, too!
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Good guess, but you've got your $n$'s mixed up. There is no water-air surface here, because there's glass or plastic (the watchplate) in between the water and the air! The index of plastics and glasses is usually higher than water, so unless this watchplate happens to have $n\approx 1.2$, then there must be something else going on. – ptomato Dec 1 '10 at 15:56
Alternately, if the index of refraction of the glass/plastic is similar to that of water, then the dominant reflection event would happen at the glass/air interface, and the argument above would map over. The qualitative behaviour described above leads me to believe that this has to be total internal reflection at some surface. The only other plausible mechanism is thin-film diffraction, but that effect is wavelength dependent, and you would see several maxima as you rotated the watch, not just one critical angle. – Jerry Schirmer Dec 1 '10 at 16:33
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@ptomato, @jerry : Actually , you have 3 indexes $n_w, n_a, n_g$, for water, glass and air. Applying the refraction law at each inreface, we have $n_w\sin\theta_w=n_g\sin\theta_g=n_a\sin\theta_a$ which imply the relation given by Jerry. In other words, you can neglect the glass if there is no total reflection at the first interface. – Frédéric Grosshans Dec 1 '10 at 17:20
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@Frédéric, @Jerry: if the total reflection happens at the glass-air surface, then you don't even need to be underwater to see the effect. I wanted to try some experiments, but I didn't have a watch handy. I'll see if I can find one this evening. – ptomato Dec 1 '10 at 18:18
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I tried it myself. Underwater the faceplate turned opaque at an angle of about 30 degrees like in the original post. I tried to observe the effect in air, but couldn't - I saw Fresnel reflection, but up to an angle of about 80 degrees I could still see the digits under the faceplate at least faintly. I couldn't observe all the way to 90 because the watch housing sticks up around the faceplate a little. – ptomato Dec 1 '10 at 22:40
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": 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": 0.9587823748588562, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/36961/list | Return to Answer
2 capitalization
There is a direct connection between Hall's marriage theorem (combinatorics) and Linear programming (linear inequalities). Of course, the latter is about finite dimensions, but prominently features duality and convexity, two important tools in functional analysis.
To elaborate on Hall's marriage theorem, consider the following picture:
The dots on the left represent men, the dots on the right represent women and the connecting lines indicate whether this man and woman like each other. The question is whether it is possible to arrange simultaneous, monogamous marriages such that everyone marries someone he or she likes. Hall's theorem gives a necessary and sufficient condition for that: for every subset $M$ of men, the set $$W = \lbrace w \text{ woman}\ |\ w \text{ likes } m, m\in M\rbrace$$ of women liked by these men must fulfill $|M| \le |W|$.
This problem is also known as perfect Matching matching in a bipartite graph. It turns out that it is equivalent to a maximum flow problem, for which we have the min-cut max-flow theorem, which is equivalent to the duality theorem for linear programming. The details of this equivalence are not very difficult and can be found here: http://web.mit.edu/k_lai/www/6.046/r11-handout.pdf .
Unfortunately, I haven't found a ready-made proof of Hall's theorem from the duality theorem, you'd have to work that out yourself for your lecture. The intermediate reformulations are bit long, I don't think it's worth spending more than a cursory remark on them; I'd jump right to the reformulation as linear program.
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There is a direct connection between Hall's marriage theorem (combinatorics) and Linear programming (linear inequalities). Of course, the latter is about finite dimensions, but prominently features duality and convexity, two important tools in functional analysis.
To elaborate on Hall's marriage theorem, consider the following picture:
The dots on the left represent men, the dots on the right represent women and the connecting lines indicate whether this man and woman like each other. The question is whether it is possible to arrange simultaneous, monogamous marriages such that everyone marries someone he or she likes. Hall's theorem gives a necessary and sufficient condition for that: for every subset $M$ of men, the set $$W = \lbrace w \text{ woman}\ |\ w \text{ likes } m, m\in M\rbrace$$ of women liked by these men must fulfill $|M| \le |W|$.
This problem is also known as perfect Matching in a bipartite graph. It turns out that it is equivalent to a maximum flow problem, for which we have the min-cut max-flow theorem, which is equivalent to the duality theorem for linear programming. The details of this equivalence are not very difficult and can be found here: http://web.mit.edu/k_lai/www/6.046/r11-handout.pdf .
Unfortunately, I haven't found a ready-made proof of Hall's theorem from the duality theorem, you'd have to work that out yourself for your lecture. The intermediate reformulations are bit long, I don't think it's worth spending more than a cursory remark on them; I'd jump right to the reformulation as linear program. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9579049944877625, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/47499/list | 2 edited tags
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# multiplicity under specialization
Let $C$ be a projective non-singular curve defined over a field $K$ with the characteristic zero. Let $y,z$ be non-constant rational functions defined $K$ such that $y$ is defined at all poles and zeros of $z$ and gives an injective mapping of this set into the algebraic closure of $K.$ Let $(y_1,y_2,...,y_m)$ be all conjugates of $y$ over $K(z)$ and $r$ be the largest of the multiplicities of the zeros of $z.$ If $(y'_1,...,y'_m)$ is a specialization of $(y_1,y_2,...,y_m)$ over $z \rightarrow 0.$ \ Is this true that the multiplicity of any $y'_k$ in $(y'_1,...,y'_m)$ is less than $r ?$ | {"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": 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": 0.8886575698852539, "perplexity_flag": "head"} |
http://en.wikipedia.org/wiki/Molecular_mass | # Molecular mass
Assuming hydrogen and oxygen are standard weights in this image (as opposed to deuterium oxide) the molecular mass should be 18.01528 u.
The molecular mass (ma) is the mass of a molecule.
## Units
The SI unit of mass is kg. However, for molecular mass more commonly the atomic mass unit (symbol u or Da) is used. For example, the molecular mass of water is approximately
ma(H2O) ≈ ≈ 3×10−26 kg.
## Related quantities
### Molecular weight
Molecular mass is sometimes called molecular weight. However, this is incorrect because mass and weight are different (see Mass versus weight). Another possible source of confusion is that some older textbooks use the term "molecular weight" to mean the molar mass.[1] This terminology is not used in modern literature.
### Relative molecular mass
The numerical value of the absolute molecular mass expressed in unified atomic mass units is equivalent to the dimensionless relative molecular mass Mr[2]
Mr (H2O) ≈ 18.
Thus, it is incorrect to express relative molecular mass (molecular weight) in daltons (Da).
### Molar mass
Main article: molar mass
The molar mass of a substance is the mass of 1 mol (the SI unit for the basis SI quantity amount of substance, having the symbol n) of the substance. This has a numerical value which is the average molecular mass of the molecules in the substance multiplied by Avogadro's constant (approximately 6.022×1023). The most common units of molar mass are g/mol because in those units the numerical value equals the average molecular mass in units of u.
Many chemists use molecular mass as a synonym of molar mass,[3] differing only in units (see average molecular mass below). A stricter interpretation does not equate the two, as the mass of a single molecule is not the same as the average of an ensemble. Because a mole of molecules may contain a variety of molecular masses due to natural isotopes, the average mass is usually not identical to the mass of any single molecule. The actual numerical difference can be very small when considering small molecules and the molecular mass of the most common isotopomer in which case the error only matters to physicists and a small subset of highly specialized chemists; however it is always more correct, accurate and consistent to use molar mass in any bulk stoichiometric calculations. The size of this error becomes much larger when considering larger molecules or less abundant isotopomers. The molecular mass of a molecule which happens to contain heavier isotopes than the average molecule in the sample can differ from the molar mass by several mass units.
### Average molecular mass
The average molecular mass is another variation on the use of the term molecular mass. The average molecular mass is the abundance weighted mean (average) of the molecular masses in a sample. This is often closer to what is meant when "molecular mass" and "molar mass" are used synonymously and may have derived from shortening of this term. The average molecular mass and the molar mass of a particular substance in a particular sample are in fact numerically identical and may be inter-converted by Avogadro's constant. It should be noted, however, that the molar mass is almost always a computed figure derived from the Relative atomic masss, whereas the average molecular mass, in fields that need the term, is often a measured figure specific to a sample. Therefore, they often vary since one is theoretical and the other is experimental. Specific samples may vary significantly from the expected isotopic composition due to real deviations from earth's average isotopic abundances.
## Computation
The molecular mass can be calculated as the sum of the individual isotopic masses (as found in a table of isotopes) of all the atoms in any molecule. This is possible because molecules are created by chemical reactions which, unlike nuclear reactions, have very small binding energies compared to the rest mass of the atoms ($<$ 10−9) and therefore create a negligible mass defect. The use of average atomic masses derived from the standard atomic masses found on a standard periodic table will result in an average molecular mass, whereas the use of isotopic masses will result in a molecular mass consistent with the strict interpretation of the definition, i.e. that of a single molecule. However, any given molecule may contain any given combination of isotopes, so there may be multiple molecular masses for each chemical compound.
## Measurement
The molecular mass can also be measured directly using mass spectrometry. In mass spectrometry, the molecular mass of a small molecule is usually reported as the monoisotopic mass, that is, the mass of the molecule containing only the most common isotope of each element. Note that this also differs subtly from the molecular mass in that the choice of isotopes is defined and thus is a single specific molecular mass of the many possible. The masses used to compute the monoisotopic molecular mass are found on a table of isotopic masses and are not found on a typical periodic table. The average molecular mass is often used for larger molecules since molecules with many atoms are unlikely to be composed exclusively of the most abundant isotope of each element. A theoretical average molecular mass can be calculated using the standard atomic weights found on a typical periodic table, since there is likely to be a statistical distribution of atoms representing the isotopes throughout the molecule. This however may differ from the true average molecular mass of the sample due to natural (or artificial) variations in the isotopic distributions.
The basis for determination of molecular mass according to the Staudinger method (since replaced by the more general Mark-Houwink equation[4]) is the fact that relative viscosity of suspensions depends on volumetric proportion of solid particles.
There are several companies that manufacture accurate instruments for determining absolute molecular mass quickly and easily. These companies include Brookhaven Instruments, Malvern Instruments, and Horiba
## Isotopic considerations
The average atomic mass of natural hydrogen is 1.00794 u and that of natural oxygen is 15.9994 u; therefore, the molecular mass of natural water with formula H2O is (2 × 1.00794 u) + 15.9994 u = 18.01528 u. Therefore, one mole of water has a mass of 18.01528 grams. However, the exact mass of hydrogen-1 (the most common hydrogen isotope) is 1.00783, and the exact mass of oxygen-16 (the most common oxygen isotope) is 15.9949, so the mass of the most common molecule of water is 18.01056 u. The difference of 0.00472 u or 0.03% comes from the fact that natural water contain traces of water molecules containing, oxygen-17, oxygen-18 or hydrogen-2 (Deuterium) atoms. Although this difference is trivial in bulk chemistry calculations, it can result in complete failure in situations where the behavior of individual molecules matters, such as in mass spectrometry and particle physics (where the mixture of isotopes does not act as an average).
There are also situations where the isotopic distributions are not typical such as with heavy water used in some nuclear reactors which is artificially enriched with Deuterium. In these cases the computed values of molar mass and average molecular mass, which are ultimately derived from the standard atomic weights, will not be the same as the actual molar mass or average molecular mass of the sample. In this case the mass of deuterium is 2.0136 u and the average molecular mass of this water (assuming 100% deuterium enrichment) is (2 × 2.0136 u) + 15.9994 u = 20.0266 u. This is a very large difference of ~11% error from the expected average molecular mass based on the standard atomic masses. Furthermore the most abundant molecular mass is actually slightly less than the average molecular mass since oxygen-16 is still the most common. (2 × 2.0136 u) + 15.9949 u = 20.0221 u. Although this is an extreme artificial example, natural variation in isotopic distributions do occur and are measurable. For example, the atomic mass of lithium as found by isotopic analysis of 39 lithium reagents from several manufacturers varied from 6.939 to 6.996.[5]
## References
1. For instance, see Hill, Nora E. et al. (1969). Dielectric properties and molecular behaviour. London: Van Nostrand Reinhold. ISBN 0442034113. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9134964346885681, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/94378/on-continuous-markov-chains-statistics-of-recurrent-states | On continuous Markov chains: statistics of recurrent states
Given a continuous Markov chains (and given the transition rates between the states) I would like to know the following:
1. mean time of permanence for all states.
2. higher order moments (i.e., variance and, possibly, CDF?)
In particular, I am also interested about computing the above quantities in the case when the system has to stay at least for a fixed amount of time in every state. For example, suppose there are for states {A, B, C, D}. When the system is in state A, it has to stay there for at least $T_A$ seconds, and the same for state B ($T_B$ seconds) and so on for all remaining states. In this case, is it still possible to determine the quantities in the list above, namely, mean permanence time for all states, variance and CDF?
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1 Answer
If it has to stay at least a fixed amount of time in a state, it's no longer a continuous Markov chain: the future depends not only on the present but also on the past (namely how long you've been in the current state). However, if by "permanence time" you mean the amount of time it is in the state $A$ from when it enters until it first leaves that state, that will simply be $T_A + T$ where $T$ has an exponential distibution with rate $v_A$ (the rate at which the system makes transitions out of state $A$ after the initial period is up).
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do you know how to compute the time it takes on average to come back to a certain state when the system leaves it? – Bob Dec 27 '11 at 16:18
Let $u_i$ be the expected time to reach state $A$ after , leaving state $i$. By a "first-step analysis", you get a system of linear equations for the $u_i$, which you can solve. – Robert Israel Dec 27 '11 at 17:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 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": 0.9533785581588745, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/125396/solve-int-infty-infty-fraceqiy-k-sqrt-lambda-a-iy-sqrt-lambda?answertab=active | # solve $\int_{-\infty}^{\infty}\frac{e^{qiy -K(\sqrt{\lambda -a-iy}-\sqrt{\lambda})}}{a+iy}$?
Is there any way to solve this integral?
$$\int_{-\infty}^{\infty}\frac{e^{qiy -K(\sqrt{\lambda -a-iy}-\sqrt{\lambda})}}{a+iy} dy$$ where $K,\lambda, a$ and $q$ are real numbers and $K>0$, $a>0$, $\lambda > 0$ and $q<0$
I have tried the standard contour approaches, but the branch cut makes it complicated on the lower half plane, and the integrand grows unbounded on the upper half plane.
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No assumptions on $a$ and $\lambda$? If we use the branch cut of the square root along the negative real axis, I suppose $\lambda-a$ had better have a positive real part. (And I find it easier to think of this as an integral along the imaginary axis wrt $z=iy$.) – Harald Hanche-Olsen Mar 28 '12 at 14:33
You are right, otherwise it will be even nastier. I forgot to write the assumptions for $\lambda$ and $\alpha$, I added them now. – Johnmimo Mar 28 '12 at 15:02 | {"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": 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": 0.9297432899475098, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/47905/integers-in-a-triangle-and-differences/47947 | ## Integers in a triangle, and differences
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I read about the following puzzle thirty-five years ago or so, and I still do not know the answer.
One gives an integer $n\ge1$ and asks to place the integers $1,2,\ldots,N=\frac{n(n+1)}{2}$ in a triangle according to the following rules. Each integer is used exactly once. There are $n$ integers on the first row, $n-1$ on the second one, ... and finally one in the $n$th row (the last one). The integers of the $j$th row are placed below the middle of intervals of the $(j-1)$th row. Finally, when $a$ and $b$ are neighbours in the $(j-1)$th row, and $c$ lies in $j$-th row, below the middle of $(a,b)$ (I say that $a$ and $b$ dominate $c$), then $c=|b-a|$.
Here is an example, with $n=4$. $$\begin{matrix} 6 & & 10 & & 1 & & 8 \\ & 4 & & 9 & & 7 \\ & & 5 & & 2 & & \\ & & & 3 & & & \end{matrix}$$
Does every know about this ? Is it related to something classical in mathematics ? Maybe eigenvalues of Hermitian matrices and their principal submatrices.
If I remember well, the author claimed that there are solutions for $n=1,2,3,4,5$, but not for $6$, and the existence was an open question when $n\ge7$. Can anyone confirm this ?
Trying to solve this problem, I soon was able to prove the following.
If a solution exists, then among the numbers $1,\ldots,n$, exactly one lies on each line, which is obviously the smallest in the line. In addition, the smallest of a line is a neighbour of the highest, and they dominate the highest of the next line.
The article perhaps appeared in the Revue du Palais de la Découverte.
Edit. Thanks to G. Myerson's answer, we know that these objects are called Exact difference triangles in the literature.
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This reminds me of graceful labeling of a graph: en.wikipedia.org/wiki/Graceful_labeling . It seems your problem could be rephrased as gracefully labeling a particular graph. Not clear this would yield any insights, though... – Joseph O'Rourke Dec 1 2010 at 14:19
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The problem reminds me of Ducci sequences : see en.wikipedia.org/wiki/Ducci_sequence – François Brunault Dec 1 2010 at 18:47
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in all examples in every row the values never grow/fall twice. Can one proof this or is it just wrong ? – HenrikRüping Dec 2 2010 at 0:15
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@HenrikRüping: the n=5 example with first row [6,14,15,3,13] shows this doesn't hold always. – ndkrempel Dec 2 2010 at 1:08
## 4 Answers
This is the first problem in Chapter 9 of Martin Gardner, Penrose Tiles to Trapdoor Ciphers. In the addendum to the chapter, he writes that Herbert Taylor has proved it can't be done for $n\gt5$. Unfortunately, he gives no reference.
There may be something about the problem in Solomon W Golomb and Herbert Taylor, Cyclic projective planes, perfect circular rulers, and good spanning rulers, in Sequences and their applications (Bergen, 2001), 166–181, Discrete Math. Theor. Comput. Sci. (Lond.), Springer, London, 2002, MR1916130 (2003f:51016).
See also http://www.research.ibm.com/people/s/shearer/dts.html and the literature on difference matrices and difference triangles.
EDIT. Reading a little farther into the Gardner essay, I see he writes,
The only published proof known to me that the conjecture is true is given by G. J. Chang, M. C. Hu, K. W. Lih and T. C. Shieh in "Exact Difference Triangles," Bulletin of the Institute of Mathematics, Academia Sinica, Taipei, Taiwan (vol. 5, June 1977, pages 191- 197).
This paper can be found at http://www.math.sinica.edu.tw/bulletin/bulletin_old/51/5120.pdf and the review is MR0491218 (58 #10483).
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
For small values of $n$, there is a relatively small state space to search.
In the most naive way possible, I found the following (showing the top row of triangle only):
````1: 1 way: [1]
2: 4 ways: [1,3], [2,3], [3,1], [3,2]
3: 8 ways: [1,6,4], [2,6,5], [4,1,6], [4,6,1], [5,2,6], [5,6,2], [6,1,4], [6,2,5]
4: 8 ways: [6,1,10,8], [6,10,1,8], [8,1,10,6], [8,3,10,9], [8,10,1,6], [8,10,3,9] [9,3,10,8], [9,10,3,8]
5: 2 ways: [6,14,15,3,13], [13,3,15,14,6]
6: no ways
````
In particular, it is possible for $n = 5$, but not possible for $n = 6$.
The computation for $n = 7$ seems entirely feasible, and I'm happy to carry it out.
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Ok, using a slightly less naive search, I've shown there are none for n = 7 either. – ndkrempel Dec 1 2010 at 22:08
Update: none for n = 8. That took about half an hour (using Ruby 1.9), so a more clever approach or faster language/computer may be required soon. – ndkrempel Dec 1 2010 at 22:48
Thanks. I mixed up 5 and 6. I make a correction. – Denis Serre Dec 2 2010 at 0:42
Since I set it running before more definitive answers came in, I may as well report there are none for n = 9 either, although that took about 6 hours (without improving the method). – ndkrempel Dec 2 2010 at 5:56
Noting, that the biggest number must always be placed in the first row, the second biggest number in the first row or below a bigger number etc. should speed up the program a lot (if u didn't use that already). – HenrikRüping Dec 2 2010 at 10:14
show 2 more comments
Not an answer, but a question about constraints mod 2. There you are just taking differences, and the number 0s and 1s must be the same for n that is 0 or 3 mod 4 (one different in the other two cases). On the face of it the left edge of the triangle could be any binary sequence. Does this give anything?
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It does indeed. For instance, for $n=5$, I got only two distinct (up to rotation) $(0,1)$ patterns. – Denis Serre Dec 2 2010 at 0:44
From what I remember there is now an actual proof by 6 authors (one is Andy Liu); I think the paper was presented at g4g9. The paper might have not been published yet.
It is my understanding that Herbert Taylor was true but way to complicated, I think that Liu & et used his basic idea but simplified a lot the proof, which is now at an elementary level.
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For the non-cognoscenti, g4g9 is the 9th Gathering for Gardner, held earlier this year. I couldn't find anything about this paper on their webpage, nor on Andy Liu's. I'm not suggesting Nick S is wrong, just reporting on the difficulty of finding anything. – Gerry Myerson Dec 2 2010 at 22:50
Yea I couldn't find anything on Andy's page either and the g4g9 page doesn't contain any info on the speakers (they have the list from last year). But Andy gave a talk on this during the October 21st Celebration of Mind (see the Edmonton event). – Nick S Dec 3 2010 at 6:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9263166189193726, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/120322/distribute-n-objects-to-k-boxes-such-that-no-box-has-more-than-c-objects-i | # Distribute $N$ objects to $K$ boxes such that no box has more than $c$ objects in it
I'm trying to find a way to calculate a problems such as this: if you have $n$ objects and $k$ indistinguishable boxes, how do you put in $n$ objects such that each box has no more than $C$, where $C \le k$.
For example, I have $6$ objects that need to be placed in $4$ boxes, where no box can have more than $4$ objects in it. I tried $\displaystyle\binom{n+k-1} {k-1}$. I got $84$ ways. But that is too big. When I do it by hand, I get $74$.
Would like actual explanation.
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## 2 Answers
Added: The answer below is based on the assumption that your calculation with $\binom{n+k-1}{k-1}$ was relevant. However, this is the case if and only if the it is the objects that are indistinguishable, and the boxes can be distinguished, which (I now note) contradicts your statement of the problem. If in fact the boxes are indistinguishable, then the problem is altogether different. If the objects are distinguishable, the answer will involve Stirling numbers of the second kind; if not, it will involve partitions of an integer and be even messier.
You start with the $84$ ways that you calculated, but then you have to subtract the distributions that violate the upper limit. This is a standard inclusion-exclusion argument; the actual result for essentially this problem can be found in this question and answer.
In more detail: A first approximation to the desired result is, as you tried, $\dbinom{n+k-1}{k-1}$. However, this counts all possible distributions, not just that have at most $C$ objects in each box, so it’s necessary to subtract the distributions that violate this limit. We’ll first count the distributions that violate it for the first box. Those all have at least $C+1$ objects in the first box, so we can count them by first putting $C+1$ objects in Box 1 and then distributing the remaining $n-C-1$ objects freely amongst the $k$ boxes. This can be done in $$\binom{n-C-1+k-1}{k-1}=\binom{n-C+k-2}{k-1}$$ ways. There are just as many distributions that violate the limit on the second box, just as many again that violate it on the third box, and so on through the $k$ boxes, so there are (to a first approximation) $$k\binom{n-C+k-2}{k-1}$$ distributions that violate it for at least one box. That leaves us with $$\binom{n+k-1}{k-1}-k\binom{n-C+k-2}{k-1}\tag{0}$$ acceptable distributions as a second approximation.
Unfortunately, we’ve now subtracted too much, if $n$ is large enough: a distribution that violates the limit for two boxes has been subtracted twice. Thus, we must add back in all distributions that violate the limit for two boxes. Pick a pair of boxes; how many distributions violate the limit for both boxes of that pair? Those are the distributions that have $C+1$ objects in each of the two boxes and the remaining $n-2C-2$ objects distributed arbitrarily amongst the $k$ boxes. There are $$\binom{n-2C-2+k-1}{k-1}=\binom{n-2C+k-3}{k-1}$$ such distributions. And there are $\binom{k}2$ pairs of boxes, so altogether we must add back in $$\binom{k}2\binom{n-2C+k-3}{k-1}$$ distributions to get a third approximation of $$\binom{n+k-1}{k-1}-k\binom{n-C+k-2}{k-1}+\binom{k}2\binom{n-2C+k-3}{k-1}$$ acceptable distributions.
This time we’ve potentially added too much: distributions that violate the limit for three boxes have been added back in more than once, so we have to subtract a correction. And then we’ll have to correct for having subtracted too much in the case of distributions that go over the limit for four boxes. And so on.
The correction for distributions violating the limit for $i$ boxes must be made for every one of the $\binom{k}i$ sets of $i$ boxes, so it will have a factor of $\binom{k}i$. As you can see from the pattern up to this point, it will be a positive correction if $i$ is even and a negative correction if $i$ is odd; this can be handled with a factor of $(-1)^i$. Finally, the number of distributions that go over the limit on a particular set of $i$ boxes is
$$\binom{n-iC-i+k-1}{k-1}=\binom{n-iC+k-(i+1)}{k-1}\;,$$
so the correction term is $$(-1)^i\binom{k}i\binom{n-iC+k-(i+1)}{k-1}\;.\tag{1}$$
The original approximation of $\dbinom{n+k-1}{k-1}$ is simply $(1)$ with $i=0$, so the final answer is $$\sum_{i\ge 0}(-1)^i\binom{k}i\binom{n-iC+k-(i+1)}{k-1}\;.$$ Of course all terms with $i>k$ are $0$, so this is a finite sum.
In general there isn’t a simple closed form. To get that, you have to know something about $C$ and $n$. For example, you can’t exceed the upper limit on more than one box if $n<2C+2$; if that’s the case, only one correction term is non-zero, and you can simplify the answer to $(0)$.
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@user9915: If you simply want to be able to compute the result correctly, you should be able to plug into the formula that I gave in the other answer. Yes, you’ll have to change the names of the quantities, but that’s a triviality. If you want a full explanation, it would have been helpful to have said so in the question, rather than simply asking for ‘a way to calculate’ the result. And if you got no help from the cited article on inclusion-exclusion, it would be helpful if you said just what you don’t understand. – Brian M. Scott Mar 15 '12 at 2:53
Yes, I want a full explanation. Also, from linked answer: "Place 10 identical balls in 5 distinct containers with no more than 4 balls in any container. We see that if left unconstrained, 1 or 2 containers may violate the restriction, so we allot 5 balls each to 1 or 2 containers, and correct for such configurations to get..." Why do only 1 or 2 containers violate the restriction? – user9915 Mar 15 '12 at 2:55
The inclusion - exclusion argument is what I don't understand. Only one guy said anything remotely about it, and it didn't make any sense. – user9915 Mar 15 '12 at 3:10
(n−C+1+k−1k−1)=(n−C+k−2k−1) ... mmm... where is this coming from. And is the first 1, supposed to be -1 or +1? – user9915 Mar 15 '12 at 3:46
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@user9915: Without a clearer idea of just what it is that you don’t understand, I doubt that I can help much; I’ve already included explanation of where every part of the formula comes from. – Brian M. Scott Mar 15 '12 at 14:10
show 2 more comments
First, I want to be clear about your problem. There are, in general, 4 types of such problems: you may place distinguishable/indistinguishable objects into distinguishable/indistinguishable boxes.
Consider this example: you have 3 objects and 2 boxes. If neither objects nor boxes are distinguishable, then you have 2 cases only: either put all three objects into one box, or put one in a box and put two others in the other box. If the objects are distinguishable, then you have 1+3=4 cases: when you put one object separately, you have 3 choices to decide which one. If the boxes are distinguishable but objects are not, then you have 2+2=4 different combinations (this case is where your expression $\binom{n+k-1}{k-1}$ belongs to). Finally, if everything is distinguishable, then you have 2+2*3=8 combinations.
Then, there are some variations of each type of the problem, like the minimum number of objects in each box (e.g. the boxes must be non-empty), or the maximum number of objects (your problem) etc.
So, I think your expression $\binom{n+k-1}{k-1}$ is for the case when there is no restriction ($C\ge n$), and, more importantly, the boxes are distinguishable, while in your original question your clearly say that the boxes are indistinguishable. Although, I have no idea how you got 74...
This little difference leads to a lot of trouble.
If objects and boxes are indistinguishable, then this problem is closely related to the the one of partitioning of integer numbers. Consider the following question: what is the number of ways one can represent a positive integer number $n$ as a sum of positive integer numbers? This question must be MUCH EASIER to address: first, there is no restriction $C$ on how big the terms of the sum could be (i.e. the case $C\ge n$), and, second, there is no restriction on the number of terms in the sum (which corresponds to no restriction on number of boxes in your problem, i.e. the case $k\ge n$).
As an example on how the problems are related: suppose you have $n=C=k=4$, then you can either put all 4 objects into one box, or put one in a box and three others in another, or put 2 in a box and two others in another, or put 2 and 1 and 1, or all four in different boxes. This corresponds to the partitioning of 4 as a sum: 4=4=3+1=2+2=2+1+1=1+1+1+1.
Once again, this is an easier question, and often referred as the number of unrestricted partitions. Yet, this has no closed form solution, only generating function (thanks to Euler) and recursive expressions.
Here is the sequence of the number of partitions of $n$ into at most $k$ parts (still, no restriction by $C$): OEIS A026820. You can also start reading about the partition function, for example, here: Wolfram MathWorld on Partition Function.
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http://math.stackexchange.com/questions/17087/why-is-the-space-of-surjective-operators-open | # Why is the space of surjective operators open?
Suppose $E$ and $F$ are given Banach spaces. Let $A$ be a continuous surjective map. Why is there a small ball around $A$ in the operator topology, such that all elements in this ball are surjective?
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Perhaps it would be more profitable to ask why the nonsurjective operators form a closed set. – Grumpy Parsnip Jan 11 '11 at 14:07
@Jim: No. The solution is a corollary of the open mapping theorem and can be for instance found in Lang's book on functional analysis – OrbiculaR Jan 11 '11 at 16:59
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If you know where to find the solution what's your motivation for asking? – Rasmus Jan 11 '11 at 17:21
Well, I found the reference afterwards... Sorry! – OrbiculaR Jan 11 '11 at 20:06
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If you have a solution, for sake of completeness I think it'd be wise if you post an answer, so the next time someone comes to ask this question there can be a reference to direct them to (or they could find it on their own). – Asaf Karagila Jan 11 '11 at 20:36
## 2 Answers
Since this question bugged me, I decided to write down the proof (I don't have access to Lang's book, so I hope my argument is not much more complicated than necessary). The idea is the same as in the proof of the Banach-Schauder theorem.
By the open mapping theorem we may scale the norm on $E$ in such a way that $A$ maps the unit ball of $E$ onto the unit ball of $F$, that is $B_{\leq 1} F \subset A(B_{\leq 1}E)$. Since $A$ is linear we have $B_{\leq r} F \subset A(B_{\leq r}F)$ for all $r > 0$.
Claim. If $B: E \to F$ is such that $\alpha := \|A - B\| < 1$ then $B$ is onto.
Proof. Let $f \in F$. We want to show that there is $e$ such that $f = Be$. For convenience, we put $f_{0} = f$ and assume $\|f_{0}\| \leq 1$.
Choose $e_{0}$ with $\|e_{0}\| \leq 1$ such that $Ae_{0} = f_{0}$. Define $f_{1} = f_{0} - Be_{0}$ and observe $\|f_{1}\| = \|(A - B) e_{0}\| \leq \alpha$, so we may choose $e_{1}$ with $\|e_{1}\| \leq \alpha$ such that $Ae_{1} = f_{1}$. Now $f_{2} = f_{1} - Be_1$ has norm $\|f_{2}\| = \|(A - B)e_{1}\| \leq \alpha^{2}$, so we obtain by induction two sequences $\{f_{n}\}_{n=0}^{\infty}$ and $\{e_{n}\}_{n=0}^{\infty}$ having the following properties:
• $\|e_{n}\|, \|f_{n}\| \leq \alpha^{n}$ for all $n$.
• $e_{n}$ is such that $f_{n} = A(e_{n})$,
• $f_{n+1} = f_{n} - B(e_{n}) = (A-B)(e_{n})$.
Finally $e = \sum_{n = 0}^{\infty} e_{n}$ has norm $\|e\| \leq \sum_{n=0}^{\infty} \alpha^{n} = \frac{1}{1-\alpha}$ and, moreover, $$B(e) = \sum_{n=0}^{\infty} B(e_{n}) = \sum_{n=0}^{\infty} (f_{n} - f_{n+1}) = f_{0} = f,$$ as we wanted to show.
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Among the many Banach-Schauder theorems the one I have in mind is the following: Let $p: E \to F$ have norm at most one. If $p(B_{\leq 1}E)$ is dense in $B_{\leq 1}F$ then $p$ is onto and the map $E/\text{Ker}\,p \to F$ is isometric. – t.b. Jan 12 '11 at 0:33
This is essentially the proof given by Lang. The proof reminds me of the proof that the space of right- or left-invertible operators is open (usually the proof uses a series, but one can also write down a sequential versions via partial sums - yielding the proof you just gave) – OrbiculaR Jan 12 '11 at 7:21
@OrbiculaR: Thanks for this info. Yes, there is an entire family of results whose proof follows this `telescopic' pattern, but I'm unable to put my finger on exactly what their statements have in common (up to the fact that it always has to do with special classes of surjective maps). It definitely is a nice argument and probably worth remembering. – t.b. Jan 12 '11 at 10:33
A bounded linear operator $A: E \to F$, where $E$ and $F$ are Banach spaces, is surjective if and only if there is $c > 0$ such that for all $\phi \in F^*$, $\|A^* \phi\| \ge c \|\phi\|$ (see e.g. Rudin, "Functional Analysis", Theorem 4.15). Now if $A$ is such an operator, so is $B$ for $\|A-B\| < c$, since $$\|B^* \phi\| \ge \|A^* \phi\| - \|A^* - B^*\| \|\phi\| \ge (c - \|A-B\|) \|\phi\|$$
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That's very nice. I had a feeling that there must be a slicker way than the proof I gave... On the other hand, it uses much less machinery than yours. – t.b. Apr 6 '12 at 23:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 53, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": 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": 0.9528842568397522, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/61315/lie-group-examples/62023 | Lie group examples
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I'm looking for interesting applications of Lie groups for an introductory Lie groups graduate course. In particular I'd like to hear of non-standard examples that at first sight do not seem to be related to Lie groups (so please don't suggest well-known things like Clifford algebras or triality that appear in standard Lie groups texts such as Fulton and Harris). Here are some examples of the sorts of things I'm looking for:
*The cohomology of a compact Kaehler manifold is a representation of SL2, so the Hopf manifold cannot be Kaehler.
*q-binomial coefficients are unimodal, as they are characters of representations of SL2
*Hilbert's theorem on the finite generation of rings of invariants can be proved using invariant integration on compact Lie groups.
*Holomorphic modular forms are really highest weight vectors of discrete series representations of certain Lie groups.
*Most closed 3-manifolds are quotients of SL2(C) by discrete subgroups.
*Bessel functions cannot be expressed using elementary functions and indefinite integration. (Differential Galois theory was one of Lie's original motivations, but seems to have been eliminated from texts on Lie theory.)
*Classifying manifolds up to cobordism uses orthogonal groups.
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Does the fact that any principal $G$-bundle ($G$ simply connected) over a $3$-manifold is trivializable count? It does involve Lie group although it's a nice starting point of Chern-Simons theory. – Somnath Basu Apr 11 2011 at 21:16
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Lie Groups can be used to solve differential and difference equations, see: amazon.com/… and amazon.com/… . – Max Muller Apr 11 2011 at 21:24
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I'd just like to advertise Graeme Segal's section of the book 'Lectures on Lie groups and Lie algebras' (books.google.co.uk/books?id=3gRjcfsGu_EC). It is a beautiful exposition covering a somewhat unusual collection of topics. – Neil Strickland Apr 12 2011 at 8:35
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Community wiki for a list, without a single right answer? – Allen Knutson Apr 13 2011 at 12:58
7 Answers
Here are three examples.
1. A finite field extension of `$\mathbb R$` must be quadratic. For if `$\mathbb R^n$` carries the structure of a field, then its group of units `$\mathbb R^n - \{0\}$` is an abelian Lie group. But `$\mathbb R^n - \{0\}$` is simply connected if `$n>2$`, which means that `$\exp$` gives us an isomorphism of groups `$\mathbb R^n - \{0\} \cong \text{Lie}(\mathbb R^n - \{0\}) = \mathbb R^n$`, which is absurd.
2. The complex structure on the complex grassmannian `$Gr(d,\mathbb C^n)$` is locally rigid. Roughly what this means is that if you have a smoothly varying family $X_t$ of complex manifolds, where the index $t$ takes values in an open connected subset of `$\mathbb{C}^N$` that contains $0$, and if `$X_0 = Gr(d,\mathbb C^n)$`, then one can find a neighborhood $U$ of $0$ such that $X_t$ is isomorphic to $X_0$ as a complex manifold for all $t \in U$. A theorem of Frölicher and Nijenhuis states that the complex structure of a compact complex manifold $X$ is locally rigid if `$H^1(X,T_X)=0$`, where `$T_X$` is the holomorphic tangent bundle of $X$. Using representation theory, Bott showed that `$H^q(X,T_X)=0$` for all $q\geq1$ if `$X=G/P$` for $G$ a complex semisimple Lie group and $P$ a parabolic subgroup, i.e., if $X$ is a "generalized flag variety." This establishes the local rigidity of the complex structure of generalized flag varieties, such as `$Gr(d,\mathbb C^n)$`.
3. It's easy to believe that the combinatorics of integer partitions and Young diagrams is related to the representation theory of the symmetric group `$S_n$` (over `$\mathbb C$`, say). Schur--Weyl duality relates the latter to the representation theory of `$GL_m(\mathbb C)$`. This in turn can be related to the geometry of the flag varieties of `$GL_m(\mathbb C)$`. With these observations one can, for example, relate the combinatorics of Young diagrams to the multiplication in the cohomology ring of the grassmannian, which of course carries some kind of geometric information. This is quite remarkable, in my opinion.
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Doesn't 1) prove the fundamental theorem of algebra..? I guess the computation of the fundamental group of C* is enough to prove it (it's in May's Algebraic Topology), but I would count your proof as "essentially different". – Piotr Pstrągowski May 20 2012 at 17:57
Piotr: Yes, 1) essentially proves the fundamental theorem of algebra. – Faisal Jun 13 at 23:03
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The sum identity
$$\frac{1}{k} \sum_{i=0}^{k-1} \left( 2 \cos \frac{2 \pi i}{k} \right)^{2n} = {2n \choose n}$$
(where $k > 2n$) can be explained combinatorially as follows: the adjacency matrix $A_k$ of the cycle graph of size $k$ has eigenvalues $2 \cos \frac{2 \pi i}{k}, 0 \le i \le k-1$, and the sum of the $2n^{th}$ powers of the eigenvalues of the adjacency matrix is the total number of closed walks of length $2n$ on the graph. For $k > 2n$ this is easily seen to be $k {2n \choose n}$ (where the coefficient of $k$ comes from the choice of starting vertex), and dividing by $k$ we get the identity above.
Letting $k \to \infty$ the above becomes a Riemann sum and we obtain the integral identity
$$\int_0^1 (2 \cos 2 \pi x)^{2n} \, dx = {2n \choose n}$$
which is fairly straightforward to prove but not quite as straightforward to interpret directly, since it's not obvious that the argument above about adjacency matrices generalizes.
This identity can be explained combinatorially using the representation theory of the circle group $\text{SO}(2)$. Associated to (say) any compact Lie group $G$ and representation $V$ of $G$ there is a graph $\Gamma_G(V)$, the principal graph, whose vertices are the irreducible representations of $G$, and where the number of edges from a representation $A$ to a representation $B$ is $\dim \text{Hom}_G(A \otimes V, B)$.
The principal graph of the standard representation of $\text{SO}(2)$ is precisely the Cayley graph of $\mathbb{Z}$ with generators $\pm 1$, which one can think of as the "limit" of the cycle graphs above (the Cayley graphs of the finite, rather than infinite, cyclic groups) in some appropriate sense. It follows that the number of closed walks from the origin to itself on $\mathbb{Z}$ of length $2n$ is, on the one hand, clearly ${2n \choose n}$ and, on the other hand, is $\dim \text{Hom}_G(V^{\otimes n}, 1)$, or the dimension of the invariant subspace of $V^{\otimes n}$, and this quantity can be computed by character theory in a way that exactly generalizes the eigenvalue computation above.
The principal graph of the standard representation of $\text{SU}(2)$ is similar, but is infinite in one direction rather than two. This gives a corresponding integral identity for the Catalan numbers, and in order to get the Riemann sum version of this integral identity one must pass to the representation theory of quantum groups at roots of unity, as I learned in the linked MO question.
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Let $C$ be a finite-dimensional field extension of the real numbers, so $(C \setminus 0)/{\mathbb R}_+$
is a compact abelian Lie group, and a sphere of dimension $\dim_{\mathbb R} C - 1$. If $C \neq {\mathbb R}$ so this group is connected, then it's a $K(\pi,1)$ (proved using the exponential map, which is a surjective group homomorphism). In particular, it can't be a sphere of dimension $\geq 2$, so $\dim C = 2$.
(Anybody know who this proof is due to? I believe I heard Mazur or Gross, but I'm not sure.)
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Allen: Gross had discovered a proof of the fundamental theorem of algebra via Lie theory, although the same basic argument goes back earlier to Witt. See the answer by curious at math.stackexchange.com/questions/2300/…. – KConrad May 20 2012 at 21:42
Integrable system theory might be an example, however the examples I can propose requires some technicality level (which probably undesired)...
The line of application is the following
Questions (seemingly not related to Lie groups): Consider the differential operator - $H = \sum \partial_i^2 + \sum exp(x_i - x_{i+1} )$
Qeust 1: Can you find some differential operators which commute with H? Quest 2: Can you find eigenvectors for H ?
(This is called Toda quantum system, Calogero and some other can be considered as well)
Lie groups comes into the game like this:
The main idea is that this differential operator comes from the Casimir of gl_n, so higher Casimirs (i.e. Z(U(gl_n)) will provide the commiting operators. What we need to do to obtain this operator from standard Casimir - is to make some reduction (integration) over some some subgroup. Eigenfucntions can be obtained by intgration of some characters of the representations, which comes from the fact that Casimirs acts by scalars on any irreps...
The sl(2) example is simple technically and for me it was quite a beatiful, when it was explained to me... In sl(2) case we get Bessel functions as eigenfunctions, some properties like recurrent formulas for different "n" in Bessel can be derived from tensor decomposition of corresponding representations...
If necessary I can provide refrences...
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These are "quantum integrable" systems, similar can be done for classical ones - one can obtain solutions of diffurs.
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What could be a nicer example than Milnor invariants? At first sight, these "higher linkage" invariants seem to have nothing to do with Lie groups at all; you can even define them using Habiro moves to make any connection to Lie algebras even more mysterious. Or, you can speak of the Baguenaudier puzzle, whose solution involves Habiro moves in disguise, as discussed by Przytycki and Sikora.
And yet, the natural home for Milnor invariants is the group $D(H)$, which is the kernel of the left bracketting map $L(H)\otimes H \to L(H)$, where $L(H)$ denotes the free Lie algebra over $H$, the first homology of the link complement.
The group $D(H)$ also comes up in other contexts which at first sight don't seem to have anything to do with free Lie algebras. It's related to the rational homology of the outer automorphism groups of free groups, as first observed by Kontsevich. Morita, and Conant-Vogtmann, took this idea and ran with it. The group $D(H)$ was also used by Dennis Johnson to study the relative weight filtration of the mapping class group of a surface.
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Classifying of fiber bundles is dependent on its structure group, which is a Lie group, essentially it is how those fibers are pasted together. For any Lie Group $G$, there is a classifying space $BG$ associated to this group. And the theorem says, bundles over a space $X$ (good enough) with given fiber $F$ (with good enough action) is one to one correspondence with the homotopy classes from $X$ to $BG$. This also leads to the definition of Characteristic classes, which is in some sense, just the pull back of the generator of the cohomology Ring for $BG$.
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There is a jet group: the group of $\sum_{i>0}a_ix^i$ ($a_1\ne 0$) under the operation of composition. For example you can easily find every pair of f(x) and g(x) such that f(g(x))=g(f(x)) (for $a_1=1$ it is very easy)
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http://physics.stackexchange.com/questions/18088/what-symmetry-causes-the-runge-lenz-vector-to-be-conserved/18093 | # What symmetry causes the Runge-Lenz vector to be conserved?
Noether's theorem relates symmetries to conserved quantities. For a central potential $V \propto \frac{1}{r}$, the Laplace-Runge-Lenz vector is conserved. What is the symmetry associated with the conservation of this vector?
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– Dan Dec 21 '11 at 7:19
## 4 Answers
1) Problem. The Kepler problem has Hamiltonian
$$H~:=~ \frac{p^2}{2m}- \frac{k}{q},$$
where $m$ is the 2-body reduced mass. The Laplace–Runge–Lenz vector is (up to an irrelevant normalization)
$$A^j ~:=~a^j + km\frac{q^j}{q}, \qquad a^j~:=~({\bf L} \times {\bf p})^j~=~{\bf q}\cdot{\bf p}~p^j- p^2~q^j,\qquad {\bf L}~:=~ {\bf q} \times {\bf p}.$$
2) Action. The Hamiltonian Lagrangian is
$$L_H~:=~ \dot{\bf q}\cdot{\bf p} - H,$$
and the action is
$$S[{\bf q},{\bf p}]~=~ \int {\rm d}t~L_H .$$
The non-zero fundamental canonical Poisson brackets are
$$\{ q^i , p^j\}~=~ \delta^{ij}.$$
3) Inverse Noether's Theorem. Quite generally in the Hamiltonian formulation, given a constant of motion $Q$, then the infinitesimal variation
$$\delta~=~ \varepsilon \{Q,\cdot\}$$
is a global off-shell symmetry of the action $S$ (modulo boundary terms). Here $\varepsilon$ is an infinitesimal global parameter, and $X_Q=\{Q,\cdot\}$ is a Hamiltonian vector field with Hamiltonian generator $Q$. The full Noether current is (minus) $Q$, see e.g. my answer to this question. (The words on-shell and off-shell refer to whether the equations of motion are satisfied or not.)
4) Variation. Let us check that the three Laplace–Runge–Lenz components $A^j$ are Hamiltonian generators of three continuous global off-shell symmetries of the action $S$. In detail, the infinitesimal variations $\delta= \varepsilon_j \{A^j,\cdot\}$ read
$$\delta q^i ~=~ \varepsilon_j \{A^j,q^i\} , \qquad \{A^j,q^i\} ~ =~ 2 p^i q^j - q^i p^j - {\bf q}\cdot{\bf p}~\delta^{ij},$$ $$\delta p^i ~=~ \varepsilon_j \{A^j,p^i\} , \qquad \{A^j,p^i\}~ =~ p^i p^j - p^2~\delta^{ij} +km\left(\frac{\delta^{ij}}{q}- \frac{q^i q^j}{q^3}\right),$$ $$\delta t ~=~0,$$
where $\varepsilon_j$ are three infinitesimal parameters.
5) Notice for later that
$${\bf q}\cdot\delta {\bf q}~=~\varepsilon_j({\bf q}\cdot{\bf p}~q^j - q^2~p^j),$$
$${\bf p}\cdot\delta {\bf p} ~=~\varepsilon_j km(\frac{p^j}{q}-\frac{{\bf q}\cdot{\bf p}~q^j}{q^3})~=~- \frac{km}{q^3}{\bf q}\cdot\delta {\bf q},$$
$${\bf q}\cdot\delta {\bf p}~=~\varepsilon_j({\bf q}\cdot{\bf p}~p^j - p^2~q^j )~=~\varepsilon_j a^j,$$
$${\bf p}\cdot\delta {\bf q}~=~2\varepsilon_j( p^2~q^j - {\bf q}\cdot{\bf p}~p^j)~=~-2\varepsilon_j a^j~.$$
6) The Hamiltonian is invariant
$$\delta H ~=~ \frac{1}{m}{\bf p}\cdot\delta {\bf p} + \frac{k}{q^3}{\bf q}\cdot\delta {\bf q}~=~0,$$
showing that the Laplace–Runge–Lenz vector $A^j$ is classically a constant of motion
$$\frac{dA^j}{dt} ~\approx~ \{ A^j, H\}+\frac{\partial A^j}{\partial t} ~=~ 0.$$
(We will use the $\approx$ sign to stress that an equation is an on-shell equation.)
7) The variation of the Hamiltonian Lagrangian $L_H$ is a total time derivative
$$\delta L_H~=~ \delta (\dot{\bf q}\cdot{\bf p})~=~ \dot{\bf q}\cdot\delta {\bf p} - \dot{\bf p}\cdot\delta {\bf q} + \frac{d({\bf p}\cdot\delta {\bf q})}{dt}$$ $$=~ \varepsilon_j\left( \dot{\bf q}\cdot{\bf p}~p^j - p^2~\dot{q}^j + km\left( \frac{\dot{q}^j}{q} - \frac{{\bf q} \cdot \dot{\bf q}~q^j}{q^3}\right)\right)$$ $$- \varepsilon_j\left(2 \dot{\bf p}\cdot{\bf p}~q^j - \dot{\bf p}\cdot{\bf q}~p^j- {\bf p}\cdot{\bf q}~\dot{p}^j \right) - 2\varepsilon_j\frac{da^j}{dt}$$ $$=~\varepsilon_j\frac{df^j}{dt}, \qquad f^j ~:=~ A^j-2a^j,$$
and hence the action $S$ is invariant off-shell up to boundary terms.
8) Noether current. The bare Noether current $j^k$ is
$$j^k~:=~ \frac{\partial L_H}{\partial \dot{q}^i} \{A^k,q^i\}+\frac{\partial L_H}{\partial \dot{p}^i} \{A^k,p^i\} ~=~ p^i\{A^k,q^i\}~=~ -2a^k.$$
The full Noether current $J^k$ (which takes the total time-derivative into account) becomes (minus) the Laplace–Runge–Lenz vector
$$J^k~:=~j^k-f^k~=~ -2a^k-(A^k-2a^k)~=~ -A^k.$$
$J^k$ is conserved on-shell
$$\frac{dJ^k}{dt} ~\approx~ 0,$$
due to Noether's first Theorem. Here $k$ is an index that labels the three symmetries.
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Is this from a textbook? If so, which one? – Dan Jan 22 at 17:28
@Dan: No, I derived it from scratch, relying mainly on the links that I provided in the answer. – Qmechanic♦ Jan 22 at 18:02
Ah, I was a bit thrown off by the 1) Problem at the beginning. – Dan Jan 22 at 19:36
While Kepler second law is simply a statement of the conservation of angular momentum (and as such it holds for all systems described by central forces), the first and the third laws are special and are linked with the unique form of the newtonian potential $-k/r$. In particular, Bertrand theorem assures that only the newtonian potential and the harmonic potential $kr^2$ give rise to closed orbits (no precession). It is natural to think that this must be due to some kind of symmetry of the problem. In fact, the particular symmetry of the newtonian potential is described exactly by the conservation of the RL vector (it can be shown that the RL vector is conserved iff the potential is central and newtonian). This, in turn, is due to a more general symmetry: if conservation of angular momentum is linked to the group of special orthogonal transformations in 3-dimensional space $SO(3)$, conservation of the RL vector must be linked to a 6-dimensional group of symmetries, since in this case there are apparently six conserved quantities (3 components of $L$ and 3 components of $\mathcal A$). In the case of bound orbits, this group is $SO(4)$, the group of rotations in 4-dimensional space.
Just to fix the notation, the RL vector is:
\begin{equation} \mathcal{A}=\textbf{p}\times\textbf{L}-\frac{km}{r}\textbf{x} \end{equation}
Calculate its total derivative:
\begin{equation}\frac{d\mathcal{A}}{dt}=-\nabla U\times(\textbf{x}\times\textbf{p})+\textbf{p}\times\frac{d\textbf{L}}{dt}-\frac{k\textbf{p}}{r}+\frac{km(\textbf{p}\cdot \textbf{x})}{r^3}\textbf{x} \end{equation}
Make use of Levi-Civita symbol to develop the cross terms:
\begin{equation}\epsilon_{sjk}\epsilon_{sil}=\delta_{ji}\delta_{kl}-\delta_{jl}\delta_{ki} \end{equation}
Finally:
\begin{equation} \frac{d\mathcal{A}}{dt}=\left(\textbf{x}\cdot\nabla U-\frac{k}{r}\right)\textbf{p}+\left[(\textbf{p}\cdot\textbf{x})\frac{k}{r^3}-2\textbf{p}\cdot\nabla U\right]\textbf{x}+(\textbf{p}\cdot\textbf{x})\nabla U \end{equation}
Now, if the potential $U=U(r)$ is central:
\begin{equation} (\nabla U)_j=\frac{\partial U}{\partial x_j}=\frac{dU}{dr}\frac{\partial r}{\partial x_j}=\frac{dU}{dr}\frac{x_j}{r} \end{equation}
so
\begin{equation} \nabla U=\frac{dU}{dr}\frac{\textbf{x}}{r}\end{equation}
Substituting back:
\begin{equation}\frac{d\mathcal A}{dt}=\frac{1}{r}\left(\frac{dU}{dr}-\frac{k}{r^2}\right)[r^2\textbf{p}-(\textbf{x}\cdot\textbf{p})\textbf{x}]\end{equation}
Now, you see that if $U$ has exactly the newtonian form then the first parenthesis is zero and so the RL vector is conserved.
Maybe there's some slicker way to see it (Poisson brackets?), but this works anyway.
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it can be shown that the RL vector is conserved iff the potential is central and newtonian. Is this done by showing that the corresponding poisson bracket involving the Hamilton function vanishes? Could You expand this a bit in Your answer, please (I am just interested in this) ;-) ? – Dilaton Dec 10 '11 at 14:01
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Comment to v1: Good to mention the somewhat hidden $SO(4)$ symmetry. Since OP mentions Noether's theorem, chances are that OP is really asking a slightly different question (which @Christoph addresses in his answer; I hope he puts it back up), namely, what is the explicit expression for the (off-shell) symmetry of the action $S$, that via Noether's theorem generates the conservation law for the RL vector. – Qmechanic♦ Dec 10 '11 at 15:05
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@Qmechanic: I'll put my answer back up after I've had time to do some facts checking - in particular, Emmy Noether's paper does include the case of velocity/momentum-dependant symmetries (as you mentioned in a comment), even though the formulation of the 'classical' Noether's theorem one finds in textbooks often does not; also, I've found another source for the symmetry corresponding to the LRL vector, which - at least at first glance - does not agree with the one from Wikipedia... – Christoph Dec 10 '11 at 22:36
The symmetry is an example of an open symmetry, i.e. a symmetry group which varies from group action orbit to orbit. For bound trajectories, it's SO(4). For parabolic ones, it's SE(3). For hyperbolic ones, it's SO(3,1). Such cases are better handled by groupoids.
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I was asking for the symmetry transformation, not the symmetry group, but +1 anyway. – Dan Dec 22 '11 at 9:20
(This post may be old, but we can add some precisions) The conservation of the RL vector is not trifling, it goes with the fact that you consider a central force, lead here by a Newtonian potential $\frac{1}{r}$ which has the property to be invariant under rotations (as $\frac{1}{r^n}$ but it works only for $n=1$ as shown by @quark1245).
Therefore, the S0(3) which has not 6 conserved quantities as said before but 3, the 3 generators of the symmetry $J_i$, i=1..3 such that the symmetry transformation under an infinitesimal change $x \rightarrow x + \epsilon$ is given in the canonical formalism by $$\delta_i X = \{X, J_i(\epsilon) \}$$ and the algebra is $$\{ J_i, J_j \} = \epsilon_{ij}^k J_k.$$ They are conserved because, at least for the Kepler problem, the system is invariant w.r.t a time translation, and the Hamiltonian is also conserved, and the calculations show that $$\{H,J_i\}= 0.$$
Before their redefinition as shown on Wikipedia to see that the previous algebra is fulfilled, the generators of the rotations are : one is the angular momentum $L$ which shows that the movement is planar, therefore invariant under rotation around $L$, one is the RL vector which is in the plan, therefore perpendicular to $L$ and parallel to the major axis of the ellipse, and the third one has a name I don't remember, but is parallel to the minor axis.
We can see that their are only 3 degrees of freedom if we take place in the referential such that $\vec{J}_1 = \vec{L} = (0,0,L_z)$, then the planar generators are $A = (A_x,0,0)$ and $B = (0,B_y,0)$.
It has been shown that they can be constructed from the Killing-Yano tensors (which mean symmetry), and it works also at dimensions greater than 3. A nice review about the LRL vector derivation can be found in HeckmanVanHaalten
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http://en.wikipedia.org/wiki/Numerically_stable | # Numerical stability
(Redirected from Numerically stable)
In the mathematical subfield of numerical analysis, numerical stability is a desirable property of numerical algorithms. The precise definition of stability depends on the context, but it is derived from the accuracy of the algorithm.
An opposite phenomenon is instability. Typically, algorithms would approach the right solution in the limit, if there were no round-off or truncation errors, but depending on the specific computational method, errors can be magnified, instead of damped, causing the error to grow exponentially.
Sometimes a single calculation can be achieved in several ways, all of which are algebraically equivalent in terms of ideal real or complex numbers, but in practice when performed on digital computers yield different results. Some calculations might damp out approximation errors that occur; others might magnify such errors. Calculations that can be proven not to magnify approximation errors are called numerically stable. One of the common tasks of numerical analysis is to try to select algorithms which are robust – that is to say, have good numerical stability among other desirable properties.
## Example
As an example of an unstable algorithm, consider the task of adding an array of 100 numbers. To simplify things, assume our computer only has two significant figures (for example, numbers can be represented as 2.3, 77, 100, 110, 120, etc., but not 12.3 or 177).
The naive way to do this would be the following:
``````function sumArray(array) is
let theSum = 0
for each element in array do
let theSum = theSum + element
end for
return theSum
end function```
```
That looks reasonable, but suppose the first element in the array was 1.0 and the other 99 elements were 0.01. In exact arithmetic, the answer would be 1.99. However, on our two-digit computer, once the 1.0 was added into the sum variable, adding in 0.01 would have no effect on the sum, and so the final answer would be 1.0 – not a very good approximation of the real answer. Furthermore, we see that the algorithm depends on the ordering of elements within the array, in contrast to the exact arithmetic.
An example of a stable algorithm in this specific case is one that first sorts the array by the absolute values of the elements in ascending order and then sums them up using the pseudo code given above. This ensures that the numbers closest to zero will be taken into consideration first. Once that change is made, all of the 0.01 elements will be added, giving 0.99, and then the 1.0 element will be added, yielding a rounded result of 2.0 – a much better approximation of the real result.
However, for a larger array or for a computer with worse accuracy, sorting the array before adding the numbers together may not be enough. Consider for example the same task as above but with an array consisting of 1000 numbers instead of 100, and where all numbers have the value 1. In this case, sorting the numbers before summing them together will not have any effect since the numbers are all equally large. Once the calculated sum has reached 100, adding another number to it will no longer have any effect since the addition would be truncated down to 100 again. The calculated sum will therefore stop at 100 which is a very bad approximation of the actual sum which is 1000.
Instead, a stable algorithm for solving this more general problem can for example be a divide and conquer algorithm where the array is recursively split into two parts for which the sum is calculated respectively, and where these two sums then are summed together to give the final sum.
## Forward, backward, and mixed stability
There are different ways to formalize the concept of stability. The following definitions of forward, backward, and mixed stability are often used in numerical linear algebra.
Diagram showing the forward error Δy and the backward error Δx, and their relation to the exact solution map f and the numerical solution f*.
Consider the problem to be solved by the numerical algorithm as a function f mapping the data x to the solution y. The result of the algorithm, say y*, will usually deviate from the "true" solution y. The main causes of error are round-off error and truncation error. The forward error of the algorithm is the difference between the result and the solution; in this case, Δy = y* − y. The backward error is the smallest Δx such that f (x + Δx) = y*; in other words, the backward error tells us what problem the algorithm actually solved. The forward and backward error are related by the condition number: the forward error is at most as big in magnitude as the condition number multiplied by the magnitude of the backward error.
In many cases, it is more natural to consider the relative error
$\frac{|\Delta x|}{|x|}$
instead of the absolute error Δx.
The algorithm is said to be backward stable if the backward error is small for all inputs x. Of course, "small" is a relative term and its definition will depend on the context. Often, we want the error to be of the same order as, or perhaps only a few orders of magnitude bigger than, the unit round-off.
Mixed stability combines the concepts of forward error and backward error.
The usual definition of numerical stability uses a more general concept, called mixed stability, which combines the forward error and the backward error. An algorithm is stable in this sense if it solves a nearby problem approximately, i.e., if there exists a Δx such that both Δx is small and f (x + Δx) − y* is small. Hence, a backward stable algorithm is always stable.
An algorithm is forward stable if its forward error divided by the condition number of the problem is small. This means that an algorithm is forward stable if it has a forward error of magnitude similar to some backward stable algorithm.
## Error growth
Comparing the linear error growth of a stable algorithm and the exponential error growth of an unstable algorithm used to solve the same problem, with the same initial data.
Suppose that Ei > 0 denotes an initial error and En represents the magnitude of an error after n subsequent operations. If En ∼ C∙n∙Ei, where C is a constant independent of n, then the growth of the error is said to be linear. If En ∼ Cn∙Ei, for some C > 1, then the growth of the error is called exponential.
## Stability in numerical differential equations
The above definitions are particularly relevant in situations where truncation errors are not important. In other contexts, for instance when solving differential equations, a different definition of numerical stability is used.
In numerical ordinary differential equations, various concepts of numerical stability exist, for instance A-stability. They are related to some concept of stability in the dynamical systems sense, often Lyapunov stability. It is important to use a stable method when solving a stiff equation.
Yet another definition is used in numerical partial differential equations. An algorithm for solving a linear evolutionary partial differential equation is stable if the total variation of the numerical solution at a fixed time remains bounded as the step size goes to zero. The Lax equivalence theorem states that an algorithm converges if it is consistent and stable (in this sense). Stability is sometimes achieved by including numerical diffusion. Numerical diffusion is a mathematical term which ensures that roundoff and other errors in the calculation get spread out and do not add up to cause the calculation to "blow up". von Neumann stability analysis is a commonly used procedure for the stability analysis of finite difference schemes as applied to linear partial differential equations. These results do not hold for nonlinear PDEs, where a general, consistent definition of stability is complicated by many properties absent in linear equations.
## References
• Nicholas J. Higham, Accuracy and Stability of Numerical Algorithms, Society of Industrial and Applied Mathematics, Philadelphia, 1996. ISBN 0-89871-355-2.
• Richard L. Burden and J. Douglas Faires, Numerical Analysis 8th Edition, Thomson Brooks/Cole, U.S., 2005. ISBN 0-534-39200-8 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9066302180290222, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/28422/does-the-poincare-series-of-a-coxeter-group-always-describe-a-flag-variety | ## Does the Poincare series of a Coxeter group always describe a “flag variety”?
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Let $W$ be a Coxeter group and let $P_W(q) = \sum_{w \in W} q^{\ell(w)}$ be its Poincare series. When $W$ is the Weyl group of a simple algebraic group $G$ (hence $W$ is finite), $P_W(q)$ is the generating function describing the cells in the Bruhat decomposition of the flag variety $G/B$, $B$ a Borel.
What happens when $W$ is infinite, e.g. when $W$ is an affine Weyl group or a hyperbolic Coxeter group? Can $P_W$ still be associated to an infinite-dimensional "flag variety"?
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## 2 Answers
I think that Shrawan Kumar's book "Kac-Moody groups, their flag varieties, and representation theory" will contain the flag varieties (which are really ind-varieties in the non-finite case) that you are looking for.
A crystallographic Coxeter group is one of the form $\langle s_1,\ldots,s_n| s_i^2=(s_i s_j)^{m_{ij}}=1\rangle$ where each mij is equal to 2,3,4,6 or infinity. Such Coxeter groups are precisely the Weyl groups associated to arbitrary Kac-Moody algebras. In this case, there is a corresponding Kac-Moody group, together with an associated flag variety and Schubert cells, which seems to me to be what you want.
It is this geometry that is the starting point of the geometric interpretation of Kazhdan-Lusztig polynomials in the crystallographic case. As is to be expected, the finite case is easier than the affine case, which again is easier than the arbitrary KM case.
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My understanding was that the definition of crystalographic is more subtle than this. The condition is that you can work over the integers. Your condition is necessary. It is also sufficient for finite type and affine type. My recollection is that there is a condition along the lines of: every cycle has an even number of 3's. – Bruce Westbury Jun 17 2010 at 6:31
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An example to explain why crystallographic is more subtle than this: Take the Coxeter group on three generators, with $p^2=q^2=r^2=1$ and $(pq)^4=(pr)^4=(qr)^4=1$. This is a valid Coxeter group, but is NOT crystallographic. If there were an associated root system, then, for each pair of $\alpha_p$, $\alpha_q$ and $\alpha_r$, one element of the pair would have length either $1/2$ or $2$ times the other. This is impossible. – David Speyer Jun 17 2010 at 9:30
However, Peter is right that Poincare series of crystallographic Coxeter groups are always Betti numbers of Kac-Moody flag varieties, and that Kumar's book covers this very well. – David Speyer Jun 17 2010 at 9:31
@David Your example can be considered as the Weyl group associated to the Cartan Matrix a_ij=-2,2 or -1 according to whether i is less than, equal to or greater than j resp. This is not symmetrisable as your argument shows. I had thought that this flag variety construction did not require the KM group to arise from a symmetrisable Cartan datum, but I am not sufficiently versed in this subject to make a definite claim either way on the matter. – Peter McNamara Jun 19 2010 at 19:05
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Coxeter groups form a very large class of groups defined by generators and relations, whose Poincare series are unlikely to have a common geometric interpretation. However, the Poincare series of an affine Weyl group has been important in work of Bott on topology of Lie groups, as well as in work of Iwahori-Matsumoto and others on p-adic groups, etc. References are given in section 8.9 of my 1990 Cambridge Press book Reflection Groups and Coxeter Groups, while the seminal 1965 IHES paper by Iwahori-Matsumoto is available online at www.numdam.org (search for article):
Iwahori, Nagayoshi; Matsumoto, Hideya, On some Bruhat decomposition and the structure of the Hecke rings of $p$-adic Chevalley groups. Publications Mathématiques de l'IHÉS, 25 (1965), p. 5-48 Full entry | Full text djvu | pdf | Reviews MR 32 #2486 | Zbl 0228.20015
There are Kac-Moody groups in some generality, with associated "Weyl groups" and "Bruhat decompositions"; but Bruhat cells may have finite dimension in some cases, finite codimension in others. It gets complicated.
It's possible that work of Michael Davis and Ruth Charney on hyperbolic Coxeter groups would be relevant to your question, but I don't know enough about that.
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Are we assuming $W$ is crystalographic? – Bruce Westbury Jun 16 2010 at 19:48
I think the original question as well as the special cases I've mentioned all involve "crystallographic" Coxeter groups only, though the precise definition of that term in Kac-Moody theory may not be the usual one in combinatorial geometry. Other Coxeter groups, especially in the finite case, are intriguing but not obviously close to concepts coming from Lie theory. – Jim Humphreys Jun 16 2010 at 20:42
@Jim: Thanks for the answer! I was secretly hoping for a nice answer involving Kac-Moody algebras, but I guess the situation is complicated. @Bruce: Not necessarily - I'd be interested in any case for which an answer of some form exists. – Qiaochu Yuan Jun 16 2010 at 21:15 | {"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": 21, "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": 0.9235432147979736, "perplexity_flag": "head"} |
http://mathhelpforum.com/math-topics/25946-physics1.html | # Thread:
1. ## Physics1
A hockey puck hits the board with a velocity of 10m/s at an angle of 20deg to the board. It is deflected with a velocity of 8m/s at 24deg to the board. If the time of impact is 0.03s, what is the average acceleration of the puck?
$(2.3 x 10^2m/s^2$ out from the board at an angle of 107deg)
Thank you!
2. Acceleration is the rate of change of velocity in time. It asks for average acceleration. The formula of average acceleration is:
$a = \frac{\Delta V}{\Delta t}$
$\text{Attention !}:$ you can use this formula only for average acceleration, don't even think of using it for instantaneous acceleration.
So, we have to find $\Delta V$, which means the change of velocity. Remember that velocity is a vector. We have to find it using vector subtraction. You can't say $\Delta V = V_1 - V_0 = 8 - 10$. This only would be true if the velocities were in the same direction. In here, they have different angles.
O.K, here are the velocity vectors.
I leave you the subtraction work. After doing that, find the velocity of the subtracted vector, this will be our $\Delta V$. You know $\Delta t = 0.03 sec$. So use them in the average acceleration formula to finish.
3. It turns out to be the exact same calculation, but I'll bet you were supposed to use the "Impulse-Momentum" theorem, though I agree that simply using the acceleration definition is easier.
$\bar{F} \Delta t = m \Delta v$
The average force can be assumed to be a constant net force over the time interval so $\bar{F} = ma$
$ma \Delta t = m \Delta v$
$a \Delta t = \Delta v$
$a = \frac{\Delta v}{\Delta t}$
-Dan | {"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": 12, "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": 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": 0.9456757307052612, "perplexity_flag": "head"} |
http://mathhelpforum.com/geometry/120307-area-region.html | # Thread:
1. ## Area of Region
On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is contructed. The interiors of the square and the 12 triangles have no points in common. Let R be the region formed by the union of the square and all the trianlges, and let S be the smallest convex polygon that contains R. What is the area of the region that is inside S but outside R?
I tried drawing the square and triangles several times but I must be misunderstanding the problem. Could i get some help pls?
Vicky.
2. Hello, Vicky!
On each side of a unit square, an equilateral triangle of side length 1 is constructed.
On each new side of each equilateral triangle, another equilateral triangle of side length 1 is contructed.
The interiors of the square and the 12 triangles have no points in common.
Let R be the region formed by the union of the square and all the trianlges,
and let S be the smallest convex polygon that contains R.
What is the area of the region that is inside S but outside R?
Code:
B
*-------*-------o
*:::::*:*:::::* \
* *:::*:::*:::* o C
|::* *:*:::::*:* *::|
|:::::*-------o:::::|
|::*::| A|::*::|
*:::::| |:::::*
|::*::| |::*::|
|:::::*-------*:::::|
|::* *:*:::::*:* *::|
* *:::*:::*:::* *
*:::::*:*:::::*
*-------*-------*
The region $R$ is comprised of a unit square and 12 equilateral triangles.
The area of $R$ is: . $A_R \;=\;1 + 12\left(\tfrac{\sqrt{3}}{4}\right) \;=\;1 + 3\sqrt{3}$
$\Delta ABC$ has two sides $AB = AC = 1$ and included angle $\angle BAC - 30^o$
Its area is: . $\tfrac{1}{2}(1^2)\sin30^o \:=\:\tfrac{1}{4}$
The area of $S$ is the area of $R$ plus four of those triangles.
. . $A_S \;=\;(1 + 3\sqrt{3}) + 4\left(\tfrac{1}{4}\right) \:=\:2 + 3\sqrt{3}$
The area inside $S$ and outside $R$ is: . $(2+3\sqrt{3}) - (1 + 3\sqrt{3}) \;=\;1$
3. Originally Posted by Soroban
Hello, Vicky!
Code:
B
*-------*-------o
*:::::*:*:::::* \
* *:::*:::*:::* o C
|::* *:*:::::*:* *::|
|:::::*-------o:::::|
|::*::| A|::*::|
*:::::| |:::::*
|::*::| |::*::|
|:::::*-------*:::::|
|::* *:*:::::*:* *::|
* *:::*:::*:::* *
*:::::*:*:::::*
*-------*-------*
The region $R$ is comprised of a unit square and 12 equilateral triangles.
The area of $R$ is: . $A_R \;=\;1 + 12\left(\tfrac{\sqrt{3}}{4}\right) \;=\;1 + 3\sqrt{3}$
$\Delta ABC$ has two sides $AB = AC = 1$ and included angle $\angle BAC - 30^o$
Its area is: . $\tfrac{1}{2}(1^2)\sin30^o \:=\:\tfrac{1}{4}$
The area of $S$ is the area of $R$ plus four of those triangles.
. . $A_S \;=\;(1 + 3\sqrt{3}) + 4\left(\tfrac{1}{4}\right) \:=\:2 + 3\sqrt{3}$
The area inside $S$ and outside $R$ is: . $(2+3\sqrt{3}) - (1 + 3\sqrt{3}) \;=\;1$
Thanks!!!!!!
Now I clearly understand what I was doing wrong.
Vicky. | {"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": 26, "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": 0.9174787402153015, "perplexity_flag": "head"} |
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http://en.wikipedia.org/wiki/Unitarity | # Unitary operator
(Redirected from Unitarity)
For unitarity in physics, see Unitarity (physics).
In functional analysis, a branch of mathematics, a unitary operator (not to be confused with a unity operator) is a bounded linear operator U : H → H on a Hilbert space H satisfying
$U^*U=UU^*=I \!$
where U∗ is the adjoint of U, and I : H → H is the identity operator. This property is equivalent to the following:
1. U preserves the inner product 〈 , 〉 of the Hilbert space, i.e., for all vectors x and y in the Hilbert space, $\langle Ux, Uy \rangle = \langle x, y \rangle$, and
2. U is surjective.
It is also equivalent to the seemingly weaker condition:
1. U preserves the inner product, and
2. the range of U is dense.
To see this, notice that U preserves the inner product implies U is an isometry (thus, a bounded linear operator). The fact that U has dense range ensures it has a bounded inverse U−1. It is clear that U−1 = U∗.
Thus, unitary operators are just automorphisms of Hilbert spaces, i.e., they preserve the structure (in this case, the linear space structure, the inner product, and hence the topology) of the space on which they act. The group of all unitary operators from a given Hilbert space H to itself is sometimes referred to as the Hilbert group of H, denoted Hilb(H).
The weaker condition U∗U = I defines an isometry. Another condition, U U∗ = I, defines a coisometry.[1]
A unitary element is a generalization of a unitary operator. In a unital *-algebra, an element U of the algebra is called a unitary element if
$U^*U=UU^*=I$
where I is the identity element.[2]:55
## Examples
• The identity function is trivially a unitary operator.
• Rotations in R2 are the simplest nontrivial example of unitary operators. Rotations do not change the length of a vector or the angle between 2 vectors. This example can be expanded to R3.
• On the vector space C of complex numbers, multiplication by a number of absolute value 1, that is, a number of the form ei θ for θ ∈ R, is a unitary operator. θ is referred to as a phase, and this multiplication is referred to as multiplication by a phase. Notice that the value of θ modulo 2π does not affect the result of the multiplication, and so the independent unitary operators on C are parametrized by a circle. The corresponding group, which, as a set, is the circle, is called U(1).
• More generally, unitary matrices are precisely the unitary operators on finite-dimensional Hilbert spaces, so the notion of a unitary operator is a generalization of the notion of a unitary matrix. Orthogonal matrices are the special case of unitary matrices in which all entries are real. They are the unitary operators on Rn.
• The bilateral shift on the sequence space $\ell^2$ indexed by the integers is unitary. In general, any operator in a Hilbert space which acts by shuffling around an orthonormal basis is unitary. In the finite dimensional case, such operators are the permutation matrices. The unilateral shift is an isometry; its conjugate is a coisometry.
• The Fourier operator is a unitary operator, i.e. the operator which performs the Fourier transform (with proper normalization). This follows from Parseval's theorem.
• Unitary operators are used in unitary representations.
## Linearity
The linearity requirement in the definition of a unitary operator can be dropped without changing the meaning because it can be derived from linearity and positive-definiteness of the scalar product:
$\langle \lambda\cdot Ux-U(\lambda\cdot x), \lambda\cdot Ux-U(\lambda\cdot x) \rangle$
$= \| \lambda \cdot Ux \|^2 + \| U(\lambda \cdot x) \|^2 - \langle U(\lambda\cdot x), \lambda\cdot Ux \rangle - \langle \lambda\cdot Ux, U(\lambda\cdot x) \rangle$
$= |\lambda|^2 \cdot \| Ux \|^2 + \| U(\lambda \cdot x) \|^2 - \overline{\lambda}\cdot \langle U(\lambda\cdot x), Ux \rangle - \lambda\cdot \langle Ux, U(\lambda\cdot x) \rangle$
$= |\lambda|^2 \cdot \| x \|^2 + \| \lambda \cdot x \|^2 - \overline{\lambda}\cdot \langle \lambda\cdot x, x \rangle - \lambda\cdot \langle x, \lambda\cdot x \rangle$
$= 0$
Analogously you obtain $\langle U(x+y)-(Ux+Uy), U(x+y)-(Ux+Uy) \rangle = 0$.
## Properties
• The spectrum of a unitary operator U lies on the unit circle. That is, for any complex number λ in the spectrum, one has |λ|=1. This can be seen as a consequence of the spectral theorem for normal operators. By the theorem, U is unitarily equivalent to multiplication by a Borel-measurable f on L²(μ), for some finite measure space (X, μ). Now U U* = I implies |f(x)|² = 1 μ-a.e. This shows that the essential range of f, therefore the spectrum of U, lies on the unit circle.
## Footnotes
1. Doran, Robert S.; Victor A. Belfi (1986). Characterizations of C*-Algebras: The Gelfand-Naimark Theorems. New York: Marcel Dekker. ISBN 0-8247-7569-4.
## References
• Lang, Serge (1972). Differential manifolds. Reading, Mass.–London–Don Mills, Ont.: Addison-Wesley Publishing Co., Inc.
• Halmos, Paul (1982). A Hilbert space problem book. Springer. | {"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": 10, "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": 0.8948893547058105, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/5/how-can-you-prove-that-the-square-root-of-two-is-irrational/16544 | # How can you prove that the square root of two is irrational?
I have read a few proofs that $\sqrt{2}$ is irrational.
I have never, however, been able to really grasp what they were talking about.
Is there a simplified proof that $\sqrt{2}$ is irrational?
-
I can't edit yet, could someone edit "grash" to "grasp"? Thanks. – Coltin Jul 20 '10 at 19:53
– LLN BBK Aug 10 '10 at 19:17
11
$\sqrt{2}$ isn't integer (it's strictly in between 1 and 2). So if it's rational, it's equal to an irreducible fraction $p/q$. Then the fraction $p^2 / q^2$ is also irreducible, but it is equal to 2, which is an integer! – Alexei Averchenko Apr 3 '11 at 1:16
Plato said I can't call myself human unless I can prove this. – Mike Jones May 7 '11 at 19:48
## 7 Answers
You use a proof by contradiction. Basically, you suppose that $\sqrt{2}$ can be written as $p/q$. Then you know that $2q^2 = p^2$. However, both $q^2$ and $p^2$ have an even number of factors of two, so $2q^2$ has an odd number of factors of 2, which means it can't be equal to $p^2$.
-
Thanks! This is quite clear. – John Gietzen Jul 22 '10 at 14:51
1
This is by far my favorite proof of $\sqrt{2}$ irrational. I believe it is due to Chaitin (at least I think it's his -- it's in his book Meta Math! (p. 98), and he does not attribute it to anyone else). Of course, this depends on unique prime factorization, but it's still quite elementary. The descent method in the standard proof is, of course, hidden in the prime factorization proof, but that's a fine place for it. Note that the original poster couldn't grasp the popular proof, and I bet the descent with contradiction is the obstacle -- I've seen that with many students. – David Lewis Apr 3 '11 at 10:53
This proof easily generalizes to any exponent k and ratio b >= 2 which is not a perfect power of k, as follows (not in Chaitin's book, but it ain't so hard)... Assume $m^k = b n^k$ Then the unique prime factorizations of $m^k$ and $n^k$ must have all exponents that are multiples of k, and that must also therefore be true of b. But that means b is a perfect k-th power, $b = c^k$ for some integer c. The case k = b = 2 is the classical theorem, with 2 not a perfect square. – David Lewis Apr 3 '11 at 10:54
It is interesting that the Greeks (and I guess most everybody since) missed this proof, because they had unique prime factorization (cf. Euclid's algorithm), and this proof makes clear that that the irrationality of non-perfect roots is intimately related to it. – David Lewis Apr 3 '11 at 10:55
1
@Bill -- thanks. Makes sense that this proof is not due to Chaitin. It's just that I had never seen it, and it is so attractive (IMHO) that I assumed it must be recent, or it would be better known. He does not attribute it, but I guess that is normal with ancient, folklore proofs. I wonder if anyone does know where or from whom it did originate -- did the Greeks know it? As for not requiring unique prime factorization, you are correct, mathematically. But I was thinking more pedagogically -- it's feasible to introduce this proof in school when prime factorization has been taught. – David Lewis Mar 14 '12 at 17:27
show 3 more comments
Consider this proof by contradiction:
Assume that $\sqrt{2}$ is rational. Then there exists some rational $R=\sqrt{2}=\frac{Q}{D}$, where $Q$ and $D$ are positive integers and relatively prime (since $R$ can be expressed in simplified form).
Now consider $R^2 = 2 = \frac{Q^2}{D^2}$. Since $Q$ and $D$ are relatively prime, this means that only $Q^2$ can have $2$ in its prime decomposition, and the exponent must be one. Thus, $Q^2 = 2^1 x$, for some odd integer $x$. But $Q^2$ is a square, and thus the exponents for all of its prime factors must be even. Here we have a contradiction.
Thus, $\sqrt{2}$ must be irrational.
-
If you are implicitly using uniqueness of prime factorizations then you need to explicitly state that, and state how it applies to yields your deduction. This is essential for proofs at this level. – Gone Apr 14 '12 at 0:19
The continued fraction proof in Aryabhata's answer can be recast into an elementary form that requires no knowledge of continued fractions. Below is a variant of such that John Conway (JHC) often mentions, followed by my (WGD) reinterpretation of it to highlight the key role played by the principality of (denominator) ideals in $\:\mathbb Z\:$ (which I call unique fractionization).
THEOREM (JHC) $\quad \rm r = \sqrt{n}\ \:$ is integral if rational,$\:$ for $\:\rm n\in\mathbb{N}$
Proof $\ \ \$ Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B>0\:.\;$ $\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A.\ \:$ Taking fractional parts yields $\rm\displaystyle\ \frac{b}B = \frac{a}A\$ for $\rm\ 0 \le b < B\:.\$ But $\rm\displaystyle\ B\nmid A\ \Rightarrow\:\ b\ne 0\ \:\Rightarrow\ \frac{A}B = \frac{a}b\$ contra $\rm B$ least. $\:$ QED
Abstracting out the Euclidean descent at the heart of the above proof yields the following
THEOREM (WGD) $\quad \rm r = \sqrt{n}\ \:$ is integral if rational,$\:$ for $\:\rm n\in\mathbb{N}$
Proof $\ \$ Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B>0\:.\;$ $\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A\ \Rightarrow\ B\:|\:A\$ by this key result:
Unique Fractionization $\$ The least denominator $\rm\:B\:$ of a fraction divides every denominator.
Proof $\rm\displaystyle\ \ \frac{A}B = \frac{C}D\ \Rightarrow\ \frac{D}B = \frac{C}A \:.\$ Taking fractional parts $\rm\displaystyle\ \frac{b}B = \frac{a}A\$ where $\rm\ 0 \le b < B\:.\$ But
$\rm\displaystyle\ \:B\nmid D\ \Rightarrow\ b\ne 0\ \Rightarrow\ \frac{A}B = \frac{a}b\ \$ contra leastness of $\rm\:B\:.\quad\quad$ QED
Thus JHC's proof essentially "inlines" the above proof - which is better viewed as principality of (denominator) ideals in $\mathbb Z\:,$ cf. my post here. See also this discussion between John Conway and I.
-
or...suppose gcd(a,b) = 1, and $\sqrt{n} = \frac{a}{b}$ for integers a and b. then $a^2 = b^2n$. since $b^2$ divides $b^2n$, it must be the case that $b^2|a^2$. if p is a prime that divides b then p divides $a^2$, hence p divides a contradicting gcd(a,b) = 1. thus, there can be no such prime that divides b, so b = -1 or b = 1, that is, a/b is an integer. – David Wheeler Mar 14 '12 at 9:36
Another method is to use continued fractions (which was used in one of the first proofs irrationality of $\displaystyle \pi$).
Instead of $\displaystyle \sqrt{2}$, we will consider $\displaystyle 1 + \sqrt{2}$.
Now $\displaystyle v = 1 + \sqrt{2}$ satisfies
$$v^2 - 2v - 1 = 0$$
i.e
$$v = 2 + \frac{1}{v}$$
This leads us to the following continued fraction representation
$$1 + \sqrt{2} = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \dots}}$$
Any number with an infinite simple continued fraction is irrational and any number with a finite simple continued fraction is rational and has at most two such simple continued fraction representations.
Thus it follows that $\displaystyle 1 + \sqrt{2}$ is irrational, and so $\displaystyle \sqrt{2}$ is irrational.
Exercise: Show that the Golden Ratio is irrational.
-
If $\sqrt 2$ were rational, we could write it as a fraction $a/b$ in lowest terms. Then $$a^2 = 2 b^2.$$ Look at the last digit of $a^2$. It has to be $0$, $1$, $4$, $5$, $6$ or $9$. Now look at the last digit of $2b^2$. It has to be $0$, $2$ or $8$. As $a^2$ and $2b^2$ are the same number, its last digit must be $0$. But that's only possible if $a$ ends in $0$ and $b$ ends in $0$ or $5$. Either way both $a$ and $b$ are multiples of $5$ contradicting $a/b$ being in lowest terms.
-
Can you explain how you got the numbers, `0, 1, 4, 5, 6, or 9` – Tyler Hilton Aug 10 '10 at 19:59
The last digit of the square of a number depends only on the last digit of the number. To see this, just think about how you usually multiply two numbers (by hand) and focus on what can contribute to the 1's column. From here, you just compute 0^2, 1^2, 2^2,..., 9^2 and record the last digits to get 0,1,4,9,6,5,6,9,4,1, which, not counting multiples, is 0,1,4,5,6, or 9. – Jason DeVito Aug 16 '10 at 2:20
You can also use the rational root test on the polynomial equation $x^2-2=0$ (whose solutions are $\pm \sqrt{2}$). If this equation were to have a rational solution $\frac{a}{b}$, then $a \vert 2$ and $b \vert 1$, hence $\frac{a}{b}\in \{\pm 1, \pm 2\}$. However, it's straightforward to check that none of $1,-1,2,-2$ satisfy the equation $x^2-2=0$. Therefore the equation has no rational roots and $\sqrt{2}$ is irrational.
-
Here are some of my favorite (sketches) of proofs for the irrationality of $\sqrt{2}$.
• Using Newton's method to approximate roots of the polynomial $f(x) = x^2 - 2$, then showing that the sequence does not converge to a rational number.
• Proof by contradiction, assume that $\sqrt{2} = \frac{n}{m}$ for some $n,m \in \mathbb{Z}$ with $m \neq 0$, then $2m^{2} = n^2$, hence $n$ must be even and we can let $n = 2k$ for some $k \in \mathbb{Z}$, but then $m^2 = 2k^2$ will also be even, which is impossible if $\frac{n}{m}$ is reduced. Therefore, $\sqrt{2}$ cannot be expressed as a ratio of integers.
• Since $f(x) = x^2 -2$ is irreducible over $\mathbb{Q}[x]$, its roots must lie in some finite extension field $\mathbb{Q}(\sqrt{2})$ over the rationals.
[Reposted from closed topicProve the square root of 2 is irrational
-
1
It would help to give further details about the first and third methods. I know a lot about these topics yet I cannot be sure precisely what you have in mind. – Gone Nov 7 '12 at 20:15 | {"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": 120, "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": 0.9619242548942566, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/12453/list | ## Return to Answer
1 [made Community Wiki]
I would suggest the approach Tom Apostol takes in his linear algebra book. In chapter 1, after introducing abstract vector spaces, he goes on to Gram-Schmidt, and then immediately to best approximations. At the end of the first chapter, he solves questions like: "find the polynomial of degree three $p(x)$ which approximates $\sin(x)$ best over $[2,3]$ in the sense of minimizing the error $\int_2^3 (sin(x)-p(x))^2 dx$.
When I first read this, I was amazed. Prior to this, I only knew high school mathematics plus basic calculus - no abstract math at all. The problem of approximating one function by another seemed completely unsolvable given the mathematics I knew at the time. And yet here it had a simple solution.
Even more amazingly, the solution was right in front of me all along. If you had asked me how to approximate the vector $(1,2,3)$ by a vector whose last coordinate was $0$ - I would have immediately said $(1,2,0)$. I knew a little bit about geometry problems, and the problem of finding the closest point in a plane seemed "easy" and "natural" to me. And yet this this is all the solution of this problem required - all I needed was just to think about "vectors" or "points" a little more abstractly. I was completely sold on the benefits of the abstract approach. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 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": 0.9774447679519653, "perplexity_flag": "head"} |
http://mathforum.org/mathimages/index.php?title=Four_Color_Theorem_Applied_to_3D_Objects&oldid=33869 | # Four Color Theorem Applied to 3D Objects
### From Math Images
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Four Color Theorem
Field: Graph Theory
Image Created By: [[Author:| ]]
Website: Chroma
Four Color Theorem
This picture shows an example of the extension of the four color theorem to non-flat surfaces using a bumpy 3D shape.
# Basic Description
The Four Color Theorem states that any planar map needs at most four colors to be colored so that no bordering regions are the same color. So when coloring a flat map with four colors, same color regions only ever need to touch at their vertices. They will never need to share an edge. To clear things up, vertices are points on a graph. Edges are the sides of a region. A map is a collection of points. And Graph Theory is the study of graphs. Also, a planar graph is a graph in which no edges overlap each other.
The Four Color Theorem only applies explicitly to maps on flat, 2D surfaces, but as I'll be talking about, the theorem holds for the surfaces of many 3D shapes as well.
In the picture, a 3D surface is shown colored with only four colors: red, white, blue, and green. This picture is demonstrating the Four Color Theorem because not one object is touching another object of the same color. There are, however, two blocks that are not colored in. But we could easily complete the four-coloring of this map by coloring the upper block green and the lower block blue, demonstrating the Four Color Theorem.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Graph Theory
[Click to view A More Mathematical Explanation]
Chart of My Findings
[[Image:Sean Is Cool.png|Graph of My Finding [...]
[Click to hide A More Mathematical Explanation]
Chart of My Findings
Graph of My Findings
Nets
In order to figure which 3D surfaces the Four Color Theorem applies to, I started with a trial and error experiment.
## How I Conducted My Research
I conducted my experiment of applying the Four Color Theorem to 3D objects by drawing nets of seven different 3D objects. Then I colored the nets as if they were just flat, 2D shapes. Then I did the same thing, but I colored the nets as if they were the actual 3D objects, keeping in mind that each face touches many other faces. In both cases I made sure not to color any two regions touching by edges the same color. Then I put my work into the chart and graph above.
The picture of the nets shows the minimum number of colors used to cover each shape. The first one in each set is a coloring of the 2D shape, and the second coloring in each set is done as if on the actual 3D object.
The chart above was the beginning of my research. The chart shows the difference between the minimum number of colors used to cover the 2D net of a 3D object, as well as the minimum number of colors used to cover the actual 3D object without any colors touching each other, unless by vertices (which is acceptable in the Four Color Theorem).
## Observations
After doing my experiment, I noticed a lot of different things within my work. I also connected my results to graph theory, cycle graphs, and topology. This will show up later. What I noticed is as follows:
• All the shapes without a curved side have a higher minimum number of colors used to cover the 3D object than the 2D surface.
• The cylinder is the only object that has the same minimum number of colors used to cover the 3D object as well as the 2D surface (the net of the 3D object).
• I think the cylinder uses the same minimum number of colors to cover the 3D and 2D object/ surface because the curvature of the cylinder mimics how a 3D objects' faces touch multiple other sides.
• The two circular parts of the cylinder are also completely opposite each other in the 3D format, and as I noticed in the other 3D objects, opposite sides can always be colored the same color. However, there are not always multiple pairs of opposite sides.
• All the 2D nets are colored in at most 2 colors.
## Connection to Graph Theory
Picture of Bipartite Graph
Now lets get into graph theory. It was basically explained earlier, but I will get into Bipartite graphs which are the basis of my results applying the Four Color Theorem to 3D objects.
A Bipartite graph is a set of points that can be separated into two independent sets. The vertices within each set are not connected by edges. Only between the two sets are vertices connected.
As you can see, no points in the U set are connected by edges, which is the same for the V set. Edges do connect the two sets together, however. And if you were to spread the points out randomly but kept the same connections between the vertices, you would notice how the blue points can share edges with the green ones but not each other. For vertices of the same color to be connected, edges would have to cross, which would mean the graph could not be planar as is the overhead graph used in this project.
For a graph to be Bipartite it
• Has an even cycle (A set of points that create a shape like a square or a hexagon that have an even number of vertices).
• Is colored in two or less colors
• (Look into Graph Theory for other, more complex reasons for a graph to be called Bipartite)
Cycle Graph
This leads me into a discovery I made on my own before looking into Bipartite graphs. It was that if the base number of any shape is odd, more colors are needed to cover the 3D object than if the base number were even. A base number is the number of vertices in the base of the object. For example, in the pentagon to the right the base number is five. There are less colors needed when the base number is even because every other vertex can be colored the same color, which means that only 2 two colors need be used. If you have a shape like a square with an even base number, the colors of the vertices in the cycle graph would go: red, yellow, red, yellow (two colors). If you have a shape like a pentagon with an odd base number, the coloring would have to go: red, yellow, red, yellow, green (three colors). A cycle graph, by the way, is a a set of vertices with only one closed chain connecting the points (much like what a 2D shape looks like on paper).
## Connection to topology
Just as this mug and torus can be morphed into each other, spheres can be deformed into cubes, cylinders, and many other 3-D shapes. As it turns out, the four color theorem survives this transformation.
These results also make sense in the context of topology.
Topology is the study of properties of objects that survive when the objects are distorted. This type of property is very useful in figuring out the minimum number of colors needed for maps on non-flat surfaces, like the surfaces of 3D objects. If the key properties of a flat map for which the four color theorem applies survive when the map is distorted to cover the surface of a 3-D object, then we can still apply the four color theorem.
The four color theorem tells us that no more than four colors are needed to color a map on a flat, 2D surface so that no two bordering regions are the same color. So the property that matters here is adjacency, or whether two regions border each other.
Now think of the difference between a curved map of the world on the surface of a globe and a flat map of the world. The sizes of some countries are distorted in the flat map, but countries adjacent on the globe stay adjacent on the flat map. We know the four color theorem applies to the flat version of the map, and since adjacency is preserved, we now know it applies to the globe version of the map as well. In fact, the four color theorem applies to any map on the surface of a sphere.
Using this fact, we can extend the four color theorem to the surfaces of even more 3-D objects. In topology, when an object can be stretched and bent into another object the two objects are said to be homeomorphic. Homeomorphic objects share many properties, including adjacency patterns on their surfaces. Spheres are homemorphic to cones, cylinders, and any convex polyhedron, including the cube, the pyramid and the other shapes mentioned above. Since the sphere is deformable into any of these objects, the four color theorem applies to all of their surfaces as well.
# Why It's Interesting
The topic in and of itself is interesting. If you think about it, it would seem that you would need more than four colors to cover even a flat map. But really, only four colors are ever needed. If you think about any planar map with any amount of regions within it, it only needs four colors to cover it without any edges touching. That is interesting within itself.
## More than four color theorems
Not all maps are planar, and not all surfaces can be colored with four or less colors. Maps on the surface of a Torus, a shape that looks like a donut, may need as many as 7 colors. The Mobius Strip can require as many as six colors. These surfaces which we can't easily imagine deforming into spheres require a different color theorem.
While maps on a flat surface, cylinder, sphere, etc. require no more than 4 colors, maps on a torus, Mobius Strip, or related shape can require more.
A formula called the Heawood Conjecture can be used to figure out the maximum possible number of colors needed to color a map on the surface of these other kinds of shape so that no two bordering regions are the same color. The formula is
$\gamma \left ( \chi \right ) = \bigg\lfloor \dfrac{7+\sqrt{49-24\chi}}{2} \bigg\rfloor$
where $\gamma \left ( \chi \right )$ is the maximum possible number of necessary colors for a surface with an Euler Characteristic of $\chi$. An Euler Characteristic is a topological property of a shape, a number that describes its structure regardless of how it is stretched or bent. Spheres and all the shapes they are homeomorphic to have an Euler Characteristic of 2. A torus has an Euler characteristic of 0. Applying Heawood's formula to the torus, we see that
$\gamma \left ( 0 \right ) = \bigg\lfloor \dfrac{7+\sqrt{49-0}}{2} \bigg\rfloor = \bigg\lfloor \dfrac{7+\sqrt{49}}{2} \bigg\rfloor = \bigg\lfloor \dfrac{7+7}{2} \bigg\rfloor = \bigg\lfloor \dfrac{14}{2} \bigg\rfloor = \big\lfloor 7 \big\rfloor$
Which means that, as mentioned earlier, a map on the surface of a Torus can require as many as 7 colors.
The only known exception to this formula is a surface called the Klein Bottle. The formula gives a maximum of 7 colors, when in actuality only 6 are ever needed to color a map the Klein Bottle.
# How the Main Image Relates
My main image relates to the topics discussed here because the surface of the figure is covered with only four colors. No regions of the same color are touching each other, unless by a corner. This is an example of an extension of the Four Color Theorem to a 3D shape, which is what I looked into and experimented with.
It also relates because this is a picture of a game where the player puts cubes of four different colors in the board. The idea is to get the cubes of the same color to be spread out enough that one color block doesn't touch another block of the same color. This relates to my topics because it uses both 3D objects and the Four Color Theorem.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# Related Links
### Additional Resources
This is a link to the page on the torus. The torus is a counterexample to the Four Color Theorem's extension to many 3D objects. It is an interesting topic that shares the same ideas as my initial project.
This is another link to the Four Color Theorem Page. This is a good link for a little bit of background information.
# Future Directions for this Page
If anyone can find an actual equation to work with my findings, that would be great. I have very little faith there is one to work with what I have, but if one is found, that would be greatly appreciated.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
[[Category:]]
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http://en.wikipedia.org/wiki/Randomized_algorithm | # Randomized algorithm
Probabilistic
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Randomized algorithm
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A randomized algorithm is an algorithm which employs a degree of randomness as part of its logic. The algorithm typically uses uniformly random bits as an auxiliary input to guide its behavior, in the hope of achieving good performance in the "average case" over all possible choices of random bits. Formally, the algorithm's performance will be a random variable determined by the random bits; thus either the running time, or the output (or both) are random variables.
One has to distinguish between algorithms that use the random input to reduce the expected running time or memory usage, but always terminate with a correct result in a bounded amount of time, and probabilistic algorithms, which, depending on the random input, have a chance of producing an incorrect result (Monte Carlo algorithms) or fail to produce a result (Las Vegas algorithms) either by signalling a failure or failing to terminate.
In the second case, random performance and random output, the term "algorithm" for a procedure is somewhat questionable. In the case of random output, it is no longer formally effective.[1] However, in some cases, probabilistic algorithms are the only practical means of solving a problem.[2]
In common practice, randomized algorithms are approximated using a pseudorandom number generator in place of a true source of random bits; such an implementation may deviate from the expected theoretical behavior.
## Motivation
As a motivating example, consider the problem of finding an ‘a’ in an array of n elements.
Input: An array of n≥2 elements, in which half are ‘a’s and the other half are ‘b’s.
Output: Find an ‘a’ in the array.
We give two versions of the algorithm, one Las Vegas algorithm and one Monte Carlo algorithm.
Las Vegas algorithm:
```findingA_LV(array A, n)
begin
repeat
Randomly select one element out of n elements.
until 'a' is found
end
```
This algorithm succeeds with probability 1. The running time is random (and arbitrarily large) but its expectation is upper-bounded by $O(1)$. (See Big O notation)
Monte Carlo algorithm:
```findingA_MC(array A, n, k)
begin
i=1
repeat
Randomly select one element out of n elements.
i = i + 1
until i=k or 'a' is found
end
```
If an ‘a’ is found, the algorithm succeeds, else the algorithm fails. After k iterations, the probability of finding an ‘a’ is:
$\Pr[\mathrm{find~a}]=1-(1/2)^k$
This algorithm does not guarantee success, but the run time is fixed. The selection is executed exactly k times, therefore the runtime is $O(k)$.
Randomized algorithms are particularly useful when faced with a malicious "adversary" or attacker who deliberately tries to feed a bad input to the algorithm (see worst-case complexity and competitive analysis (online algorithm)) such as in the Prisoner's dilemma. It is for this reason that randomness is ubiquitous in cryptography. In cryptographic applications, pseudo-random numbers cannot be used, since the adversary can predict them, making the algorithm effectively deterministic. Therefore either a source of truly random numbers or a cryptographically secure pseudo-random number generator is required. Another area in which randomness is inherent is quantum computing.
In the example above, the Las Vegas algorithm always outputs the correct answer, but its running time is a random variable. The Monte Carlo algorithm (related to the Monte Carlo method for simulation) completes in a fixed amount of time (as a function of the input size), but allow a small probability of error. Observe that any Las Vegas algorithm can be converted into a Monte Carlo algorithm (via Markov's inequality), by having it output an arbitrary, possibly incorrect answer if it fails to complete within a specified time. Conversely, if an efficient verification procedure exists to check whether an answer is correct, then a Monte Carlo algorithm can be converted into a Las Vegas algorithm by running the Monte Carlo algorithm repeatedly till a correct answer is obtained.
## Computational complexity
Computational complexity theory models randomized algorithms as probabilistic Turing machines. Both Las Vegas and Monte Carlo algorithms are considered, and several complexity classes are studied. The most basic randomized complexity class is RP, which is the class of decision problems for which there is an efficient (polynomial time) randomized algorithm (or probabilistic Turing machine) which recognizes NO-instances with absolute certainty and recognizes YES-instances with a probability of at least 1/2. The complement class for RP is co-RP. Problem classes having (possibly nonterminating) algorithms with polynomial time average case running time whose output is always correct are said to be in ZPP.
The class of problems for which both YES and NO-instances are allowed to be identified with some error is called BPP. This class acts as the randomized equivalent of P, i.e. BPP represents the class of efficient randomized algorithms.
## History
Historically, the first randomized algorithm was a method developed by Michael O. Rabin for the closest pair problem in computational geometry.[3] The study of randomized algorithms was spurred by the 1977 discovery of a randomized primality test (i.e., determining the primality of a number) by Robert M. Solovay and Volker Strassen. Soon afterwards Michael O. Rabin demonstrated that the 1976 Miller's primality test can be turned into a randomized algorithm. At that time, no practical deterministic algorithm for primality was known.
The Miller-Rabin primality test relies on a binary relation between two positive integers k and n that can be expressed by saying that k "is a witness to the compositeness of" n. It can be shown that
• If there is a witness to the compositeness of n, then n is composite (i.e., n is not prime), and
• If n is composite then at least three-fourths of the natural numbers less than n are witnesses to its compositeness, and
• There is a fast algorithm that, given k and n, ascertains whether k is a witness to the compositeness of n.
Observe that this implies that the primality problem is in Co-RP.
If one randomly chooses 100 numbers less than a composite number n, then the probability of failing to find such a "witness" is (1/4)100 so that for most practical purposes, this is a good primality test. If n is big, there may be no other test that is practical. The probability of error can be reduced to an arbitrary degree by performing enough independent tests.
Therefore, in practice, there is no penalty associated with accepting a small probability of error, since with a little care the probability of error can be made astronomically small. Indeed, even though a deterministic polynomial-time primality test has since been found (see AKS primality test), it has not replaced the older probabilistic tests in cryptographic software nor is it expected to do so for the foreseeable future.
## Applications
### Quicksort
Quicksort is a familiar, commonly used algorithm in which randomness can be useful. Any deterministic version of this algorithm requires O(n2) time to sort n numbers for some well-defined class of degenerate inputs (such as an already sorted array), with the specific class of inputs that generate this behavior defined by the protocol for pivot selection. However, if the algorithm selects pivot elements uniformly at random, it has a provably high probability of finishing in O(n log n) time regardless of the characteristics of the input.
### Randomized incremental constructions in geometry
In computational geometry, a standard technique to build a structure like a convex hull or Delaunay triangulation is to randomly permute the input points and then insert them one by one into the existing structure. The randomization ensures that the expected number of changes to the structure caused by an insertion is small, and so the expected running time of the algorithm can be upper bounded. This technique is known as randomized incremental construction.[4]
### Verifying matrix multiplication
Main article: Freivalds' algorithm
Input: Matrix A ∈ Rm × p, B ∈ Rp × n, and C ∈ Rm × n.
Output: True if C = A · B; false if C ≠ A · B
We give a Monte Carlo algorithm to solve the problem.[5]
``` begin
i=1
repeat
Choose r=(r1,...,rn) ∈ {0,1}n at random.
Compute C · r and A · (B · r)
if C · r ≠ A · (B · r)
return FALSE
endif
i = i + 1
until i=k
return TRUE
end
```
The running time of the algorithm is $O(kn^2)$.
Theorem: The algorithm is correct with probability at least $1-(\frac{1}{2})^k$.
We will prove that if $A \cdot B \neq C$ then $Pr[A \cdot B \cdot r=C \cdot r]\leq 1/2$.
If $A \cdot B\neq C$, by definition we have $D=A \cdot B-C \neq 0$. Without loss of generality, we assume that $d_{11} \neq 0$.
On the other hand, $Pr[A \cdot B \cdot r=C \cdot r] = Pr[(A \cdot B-C) \cdot r=0] = Pr[D \cdot r=0]$.
If $D \cdot r=0$, then the first entry of $D \cdot r$ is 0, that is
$\sum_{j=1}^n d_{1j}r_j=0$
Since $d_{11} \neq 0$, we can solve for $r_1$:
$r_1=\frac{-\sum_{j=2}^n d_{1j}r_j}{d_{11}}$
If we fix all $r_j$ except $r_1$, the equality holds for at most one of the two choices for $r_1\in \{0,1\}$. Therefore,$Pr[ABr=Cr]\leq 1/2$.
We run the loop for k times. If $C=A \cdot B$, the algorithm is always correct; if $C\neq A \cdot B$, the probability of getting the correct answer is at least $1-(\frac{1}{2})^k$.
### Min cut
Main article: Karger’s algorithm
Figure 2: Successful run of Karger’s algorithm on a 10-vertex graph. The minimum cut has size 3 and is indicated by the vertex colours.
Input: A graph G(V,E)
Output: A cut partitioning the vertices into L and R, with the minimum number of edges between L and R.
Recall that the contraction of two nodes, u and v, in a (multi-)graph yields a new node u ' with edges that are the union of the edges incident on either u or v, except from any edge(s) connecting u and v. Figure 1 gives an example of contraction of vertex A and B. After contraction, the resulting graph may have parallel edges, but contains no self loops.
Figure 1: Contraction of vertex A and B
Karger's [6] basic algorithm:
``` begin
i=1
repeat
repeat
Take a random edge (u,v)∈ E in G
replace u and v with the contraction u'
until only 2 nodes remain
obtain the corresponding cut result Ci
i=i+1
until i=m
output the minimum cut among C1,C2,...,Cm.
end
```
In each execution of the outer loop, the algorithm repeats the inner loop until only 2 nodes remain, the corresponding cut is obtained. The run time of one execution is $O(n)$, and n denotes the number of vertices. After m times executions of the outer loop, we output the minimum cut among all the results. The figure 2 gives an example of one execution of the algorithm. After execution, we get a cut of size 3.
Lemma 1: Let k be the min cut size, and let C = {e1,e2,...,ek} be the min cut. If, during iteration i, no edge e ∈ C is selected for contraction, then Ci = C.
Proof: If G is not connected, then G can be partitioned into L and R without any edge between them. So the min cut in a disconnected graph is 0. Now, assume G is connected. Let V=L∪ R be the partition of V induced by C : C={ {u,v} ∈ E : u ∈ L,v ∈ R } (well-defined since G is connected). Consider an edge {u,v} of C. Initially, u,v are distinct vertices. As long as we pick an edge f ≠ e, u and v do not get merged. Thus, at the end of the algorithm, we have two compound nodes covering the entire graph, one consisting of the vertices of L and the other consisting of the vertices of R. As in figure 2, the size of min cut is 1, and C = {(A,B)}. If we don't select (A,B) for contraction, we can get the min cut.
Lemma 2: If G is a multigraph with p vertices and whose min cut has size k, then G has at least pk/2 edges.
Proof: Because the min cut is k, every vertex v must satisfy degree(v) ≥ k. Therefore, the sum of the degree is at least pk. But it is well known that the sum of vertex degrees equals 2|E|. The lemma follows.
Analysis of algorithm
The probability that the algorithm succeeds is 1 − the probability that all attempts fail. By independence, the probability that all attempts fail is
$\prod_{i=1}^m \Pr(C_i\neq C)=\prod_{i=1}^m(1-\Pr(C_i=C)).$
By lemma 1, the probability that Ci = C is the probability that no edge of C is selected during iteration i. Consider the inner loop and let Gj denote the graph after j edge contractions, where j ∈ {0,1,...,n − 3}. Gj has n − j vertices. We use the chain rule of conditional possibilities. The probability that the edge chosen at iteration j is not in C, given that no edge of C has been chosen before, is $1-\frac{k}{|E(G_j)|}$. Note that Gj still has min cut of size k, so by Lemma 2, it still has at least $\frac{(n-j)k}{2}$ edges.
Thus, $1-\frac{k}{|E(G_j)|}\geq 1-\frac{2}{n-j}=\frac{n-j-2}{n-j}$.
So by the chain rule, the probability of finding the min cut C is $Pr[C_i=C] \geq (\frac{n-2}{n})(\frac{n-3}{n-1})(\frac{n-4}{n-2})\ldots(\frac{3}{5})(\frac{2}{4})(\frac{1}{3}).$
Cancellation gives $\Pr[C_i=C]\geq \frac{2}{n(n-1)}$. Thus the probability that the algorithm succeeds is at least $1-(1-\frac{2}{n(n-1)})^m$. For $m=\frac{n(n-1)}{2}\ln n$, this is equivalent to $1-\frac{1}{n}$. The algorithm finds the min cut with probability $1-\frac{1}{n}$, in time $O(mn)=O(n^3\log n)$.
## Derandomization
Randomness can be viewed as a resource, like space and time. Derandomization is then the process of removing randomness (or using as little of it as possible). From the viewpoint of computational complexity, derandomizing an efficient randomized algorithm is the question, is P = BPP ?
There are also specific methods that can be employed to derandomize particular randomized algorithms:
• the method of conditional probabilities, and its generalization, pessimistic estimators
• discrepancy theory (which is used to derandomize geometric algorithms)
• the exploitation of limited independence in the random variables used by the algorithm, such as the pairwise independence used in universal hashing
• the use of expander graphs (or dispersers in general) to amplify a limited amount of initial randomness (this last approach is also referred to as generating pseudorandom bits from a random source, and leads to the related topic of pseudorandomness)
## Where randomness helps
When the model of computation is restricted to Turing machines, it is currently an open question whether the ability to make random choices allows some problems to be solved in polynomial time that cannot be solved in polynomial time without this ability; this is the question of whether P = BPP. However, in other contexts, there are specific examples of problems where randomization yields strict improvements.
• Based on the initial motivating example: given an exponentially long string of 2k characters, half a's and half b's, a random access machine requires at least 2k−1 lookups in the worst-case to find the index of an a; if it is permitted to make random choices, it can solve this problem in an expected polynomial number of lookups.
• In communication complexity, the equality of two strings can be verified using $\log n$ bits of communication with a randomized protocol. Any deterministic protocol requires $\Theta(n)$ bits.
• The volume of a convex body can be estimated by a randomized algorithm to arbitrary precision in polynomial time.[7] Bárány and Füredi showed that no deterministic algorithm can do the same.[8] This is true unconditionally, i.e. without relying on any complexity-theoretic assumptions.
• A more complexity-theoretic example of a place where randomness appears to help is the class IP. IP consists of all languages that can be accepted (with high probability) by a polynomially long interaction between an all-powerful prover and a verifier that implements a BPP algorithm. IP = PSPACE.[9] However, if it is required that the verifier be deterministic, then IP = NP.
• In a chemical reaction network (a finite set of reactions like A+B → 2C + D operating on a finite number of molecules), the ability to ever reach a given target state from an initial state is decidable, while even approximating the probability of ever reaching a given target state (using the standard concentration-based probability for which reaction will occur next) is undecidable. More specifically, a Turing machine can be simulated with arbitrarily high probability of running correctly for all time, only if a random chemical reaction network is used. With a simple nondeterministic chemical reaction network (any possible reaction can happen next), the computational power is limited to primitive recursive functions.
• The inherent randomness of algorithms such as Hyper-encryption, Bayesian networks, Random neural networks and Probabilistic Cellular Automata was harnessed by Krishna Palem et al. to design highly efficient hardware systems using Probabilistic CMOS or PCMOS technology that were shown to achieve gains that are as high as a multiplicative factor of 560 when compared to a competing energy-efficient CMOS based realizations.[10]
## Notes
1. "Probabilistic algorithms should not be mistaken with methods (which I refuse to call algorithms), which produce a result which has a high probability of being correct. It is essential that an algorithm produces correct results (discounting human or computer errors), even if this happens after a very long time." Henri Cohen (2000). A Course in Computational Algebraic Number Theory. Springer-Verlag, p. 2.
2. "In testing primality of very large numbers chosen at random, the chance of stumbling upon a value that fools the Fermat test is less than the chance that cosmic radiation will cause the computer to make an error in carrying out a 'correct' algorithm. Considering an algorithm to be inadequate for the first reason but not for the second illustrates the difference between mathematics and engineering." Hal Abelson and Gerald J. Sussman (1996). . MIT Press, section 1.2.
3. Smid, Michiel. Closest point problems in computational geometry. Max-Planck-Institut für Informatik|year=1995
4. A. A. Tsay, W. S. Lovejoy, David R. Karger, Random Sampling in Cut, Flow, and Network Design Problems, Mathematics of Operations Research, 24(2):383–413, 1999.
5. Dyer, M.; Frieze, A.; Kannan, R. (1991), "A random polynomial-time algorithm for approximating the volume of convex bodies", 38 (1): 1–17, doi:10.1145/102782.102783
6. Füredi, Z.; Bárány, I. (1986), "Computing the volume is difficult", , New York, NY: ACM, pp. 442–447, doi:10.1145/12130.12176, ISBN 0-89791-193-8
7. Shamir, A. (1992), "IP = PSPACE", Journal of the ACM 39 (4): 869–877, doi:10.1145/146585.146609
8.
## References
• Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. MIT Press and McGraw–Hill, 1990. ISBN 0-262-03293-7. Chapter 5: Probabilistic Analysis and Randomized Algorithms, pp. 91–122.
• Jon Kleinberg and Éva Tardos. Algorithm Design. Chapter 13: "Randomized algorithms".
• Don Fallis. 2000. "The Reliability of Randomized Algorithms." British Journal for the Philosophy of Science 51:255–71.
• M. Mitzenmacher and E. Upfal. Probability and Computing : Randomized Algorithms and Probabilistic Analysis. Cambridge University Press, New York (NY), 2005.
• Rajeev Motwani and P. Raghavan. Randomized Algorithms. Cambridge University Press, New York (NY), 1995.
• Rajeev Motwani and P. Raghavan. Randomized Algorithms. A survey on Randomized Algorithms.
• Christos Papadimitriou (1993), Computational Complexity (1st ed.), Addison Wesley, ISBN 0-201-53082-1 Chapter 11: Randomized computation, pp. 241–278.
• M. O. Rabin. (1980), "Probabilistic Algorithm for Testing Primality." Journal of Number Theory 12:128–38.
• A. A. Tsay, W. S. Lovejoy, David R. Karger, Random Sampling in Cut, Flow, and Network Design Problems, Mathematics of Operations Research, 24(2):383–413, 1999. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 38, "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": 0.8882690072059631, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2008/07/30/ | # The Unapologetic Mathematician
## Roots of Polynomials I
When we consider a polynomial as a function, we’re particularly interested in those field elements $x$ so that $p(x)=0$. We call such an $x$ a “zero” or a “root” of the polynomial $p$.
One easy way to get this to happen is for $p$ to have a factor of $X-x$. Indeed, in that case if we write $p=(X-x)q$ for some other polynomial $q$ then we evaluate to find
$p(x)=(X-x)q(x)=0$
The interesting thing is that this is the only way for a root to occur, other than to have the zero polynomial. Let’s say we have the polynomial
$p=c_0+c_1X+c_2X^2+...+c_nX^n$
and let’s also say we’ve got a root $x$ so that $p(x)=0$. But that means
$0=c_0+c_1x+c_2x^2+...+c_nx^n$
This is not just a field element — it’s the zero polynomial! So we can subtract it from $p$ to find
$\begin{aligned}p=\left(c_0+c_1X+c_2X^2+...+c_nX^n\right)-\left(c_0+c_1x+c_2x^2+...+c_nx^n\right)\\=c_1(X-x)+c_2(X^2-x^2)+...+c_n(X^n-x^n)\end{aligned}$
Now for any $k$ we can use the identity
$X^k-x^k=(X-x)(X^{k-1}+X^{k-2}x+...+Xx^{k-2}+x^{k-1})$
to factor out $(X-x)$ from each term above. This gives the factorization we were looking for.
## 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": 18, "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": 0.9177030920982361, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2008/07/24/the-inclusion-exclusion-principle/?like=1&source=post_flair&_wpnonce=4ca8fc8c32 | # The Unapologetic Mathematician
## The Inclusion-Exclusion Principle
In combinatorics we have a method of determining the cardinality of unions in terms of the sets in question and their intersection: the Inclusion-Exclusion Principle. Predictably enough, this formula is reflected in the subspaces of a vector space. We could argue directly in terms of bases, but it’s much more interesting to use the linear algebra we have at hand.
Let’s start with just two subspaces $V$ and $W$ of some larger vector space. We’ll never really need that space, so we don’t need to give it a name. The thing to remember is that $V$ and $W$ might have a nontrivial intersection — their sum may not be direct.
Now we consider the sum of the subspaces. Every vector in $V+W$ is the sum of a vector in $V$ and a vector in $W$. This tells us that there’s a linear map $f_1:V\oplus W\rightarrow V+W$ defined by $f_1(v,w)=v+w$. Note carefully that this is linear as a function of the pair $(v,w)$. It’s not bilinear — linear in each of $v$ and $w$ separately — which would mean bringing in the tensor product.
This function is always surjective, but it may fail to be injective. Specifically, if $V$ and $W$ have a nontrivial intersection then it gives us a nontrivial kernel. If $x\in V\cap W$ then $f_1(x,-x)=0$. But rather than talk about elements, let’s use the intersection to parametrize the kernel! We’ve got another linear map $f_2:V\cap W\rightarrow V\oplus W$, defined by $f_2(x)=(x,-x)$.
Now it’s clear that $f_2$ has a trivial kernel. On the other hand, its image is precisely the set of pairs that $f_1$ kills off. That means we’ve got an exact sequence:
$\mathbf{0}\rightarrow V\cap W\xrightarrow{f_2}V\oplus W\xrightarrow{f_1}V+W\rightarrow\mathbf{0}$
Taking the Euler characteristic we find that
$\dim\left(V\cap W\right)-\dim\left(V\oplus W\right)+\dim\left(V+W\right)=0$
Some simple juggling turns this into
$\dim\left(V+W\right)=\dim\left(V\right)+\dim\left(W\right)-\dim\left(V\cap W\right)$
which is the inclusion-exclusion principle for two subspaces.
What about three subspaces $U$, $V$, and $W$? We start off again with
$U\oplus V\oplus W\xrightarrow{f_1}U+V+W\rightarrow\mathbf{0}$
defined by $f_1(u,v,w)=u+v+w$. Then the kernel contains triples like $(0,x,-x)$, $(-y,0,y)$, and $(z,-z,0)$. We use the pairwise intersections to cover the kernel by the map
$\left(V\cap W\right)\oplus\left(W\cap U\right)\oplus\left(U\cap V\right)\xrightarrow{f_2}U\oplus V\oplus W$
defined by $(x,y,z)=(z-y,x-z,y-x)$. The image of $f_2$ is exactly the kernel of $f_1$, but this time it has a nontrivial kernel itself! Now we need another map
$\mathbf{0}\rightarrow U\cap V\cap W\xrightarrow{f_3}\left(V\cap W\right)\oplus\left(W\cap U\right)\oplus\left(U\cap V\right)$
defined by $f_3(t)=(t,t,t)$. Now $f_3$ is injective, and its image is exactly the kernel of $f_2$. We stick these maps all together to have the long exact sequence
$\begin{aligned}\mathbf{0}\rightarrow U\cap V\cap W\xrightarrow{f_3}\left(V\cap W\right)\oplus\left(W\cap U\right)\oplus\left(U\cap V\right)\\\xrightarrow{f_2}U\oplus V\oplus W\xrightarrow{f_1}U+V+W\rightarrow\mathbf{0}\end{aligned}$
Taking the Euler characteristic and juggling again we find
$\begin{aligned}\dim\left(U+V+W\right)=\dim\left(U\right)+\dim\left(V\right)+\dim\left(W\right)\\-\dim\left(U\cap V\right)-\dim\left(U\cap W\right)-\dim\left(V\cap W\right)\\+\dim\left(U\cap V\cap W\right)\end{aligned}$
We include the dimensions of the individual subspaces, exclude the dimensions of the pairwise intersections, and include back in the dimension of the triple intersection.
For larger numbers of subspaces we can construct longer and longer exact sequences. The $k$th term consists of the direct sum of the intersections of the subspaces, taken $k$ at a time. The $k$th map consists of including each $k$-fold intersection into each of the $(k-1)$-fold intersections it lies in, with positive and negative signs judiciously sprinkled throughout to make the sequence exact.
The result is that the dimension of a sum of subspaces is the alternating sum of the dimensions of the $k$-fold intersections. First add the dimension of each subspace, then subtract the dimension of each pairwise intersection, then add back in the dimension of each triple intersection, and so on.
From here it’s the work of a moment to derive the combinatorial inclusion-exclusion principle. Given a collection of sets, just construct the free vector space on each of them. These free vector spaces will have nontrivial intersections corresponding to the nonempty intersections of the sets, and we’ve got canonical basis elements floating around to work with. Then all these dimensions we’re talking about are cardinalities of various subsets of the sets we started with, and the combinatorial inclusion-exclusion principle follows. Of course, as I said before we could have proved it from the other side, but that would require a lot of messy hands-on work with bases and such.
The upshot is that the combinatorial inclusion-exclusion principle is equivalent to the statement that exact sequences of vector spaces have trivial Euler characteristic! This little formula that we teach in primary or secondary school turns out to be intimately connected with one of the fundamental tools of homological algebra and the abstract approach to linear algebra. Neat!
### Like this:
Posted by John Armstrong | Algebra, Linear Algebra
## 13 Comments »
1. Beyond neat! Satisfies the true purpose of Mathematics: enlightenment.
Comment by | July 24, 2008 | Reply
2. What is also good about this observation is that this form of inclusion-exclusion holds in lattices more general than distributive lattices (which were the types of lattice emphasized in the post you linked to) — subspace lattices satisfy the weaker property of modularity.
Comment by | July 24, 2008 | Reply
3. Indeed they do. That’s why I said as much last time when I talked about subspace sums.
Comment by | July 24, 2008 | Reply
4. Sorry, I don’t know sequences, but it seems either you have a major error or I completely don’t understand your post.
dim(U+V+W)=dim(U)+dim(V)+dim(W)-dim(UV)-dim(UW)-dim(WV)+dim(UVW)
Consider R^2 as a vector space over R and let:
U={(x,y):x=0}
V={(x,y):y=0}
W={(x,y):x=y}
These are clearly subspaces. U+V=R^2 and their intersection is {0}. Same for U,W and V,W.
So the left side is dim(U+V+W)=dim(R^2)=2 and the right side is dim(U)+dim(V)+dim(W)=3 because the intersections vanish.
The problem is that in general (U+V) intersection W is not equal to (U intersection W) + (V intersection W). Take these 3 mentioned spaces as a counterexample. Once my class received a task to determine whether PIE is valid for n vector spaces. Of course everybody gave a “proof” of it.:)
Best, –k.g.
(Feel free to add TeX or correct my English; don’t know how to use it)
Comment by k.g. | July 26, 2008 | Reply
5. You’re right.. I’d implicitly assumed that the subspaces were in what we topologists like to call “general position”. I can see two remedies.
First, turn your reason why my proof failed into an algebraic statement about subspaces and their maps, then establish a new sequence (or show how to modify existing ones) to handle this degeneracy.
Alternately, scrap the given sequences altogether and give new ones which handle the degenerate case as well as the generic.
I’m not in a position to type either solution out myself on my iPhone from a restaurant in Meriden, MS. But good catch there.
Comment by | July 27, 2008 | Reply
• Dear Professor John:
I am very eager to know what you mean by “general position”,
When your formular is true.
Can you email me the answer through Email [email protected] ?
Comment by mingming | March 10, 2010 | Reply
• Essentially, if you wiggle the subspaces a bit, the dimensions of their intersections don’t change.
Comment by | March 10, 2010 | Reply
6. [...] inclusion-exclusion principle tells us [...]
Pingback by | February 19, 2009 | Reply
7. [...] inclusion-exclusion principle tells us [...]
Pingback by | April 8, 2009 | Reply
8. You learned the inclusion-exclusion principle in primary school? I don’t think anybody I know who didn’t do math contests in high school has the faintest idea what it is.
Anyway, I’ve always found the use of the free vector space construction in combinatorics extremely mysterious. For example, in poset theory one can use certain linear operators to show that certain graded posets are rank-unimodal and Sperner, and some of the consequences of this method don’t have known “direct” proofs. So does the usefulness of linear algebra in combinatorics ultimately stem from categorical and “topological” properties of the category of vector spaces? What makes this category unique for combinatorial purposes?
Comment by Qiaochu Yuan | June 7, 2009 | Reply
9. Sure. It’s part of the remnants of the “New Math” hanging around the curriculum.
Comment by | June 7, 2009 | Reply
10. [...] have a nonempty intersection, which leads us to think that maybe this has something to do with the inclusion-exclusion principle. We’re “overcounting” the intersection by just adding, so let’s subtract it [...]
Pingback by | December 23, 2009 | Reply
11. [...] little more about characteristic functions, let’s see how they can be used to understand the inclusion-exclusion principle. Our first pass was through a very categorified lens, which was a neat tie-in to Euler [...]
Pingback by | December 28, 2009 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"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": 47, "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": 0.920688271522522, "perplexity_flag": "head"} |
http://mathhelpforum.com/algebra/45287-trinomial-factoring-help.html | # Thread:
1. ## Trinomial factoring help
Hello,
I have been learning how to factor trinomials and I am stuck..
2x^2+x-15 = (2x-5)(x+3)
3a^2-2a-5 = (3a-5)(a+1)
Fc^2+17c+14 = (fc+7)(c+2)
To factor these types of problems, is the only way real to just plug in factors and do a full test with foil?? I mean some of these could have 100 possible factor pairs.. do you have to plug them in and keep going until it works? Am I missing a shortcut?
thanks!
2. when you start out learning factoring you normally do it the harder way.
but there's some logic with factoring.
given $ax^2 + bx + c$, if the a is greater than one, you can do what's called the AC method.
For $2x^2{\color {blue}{+x}}-15$ you first multiply the a and c.
That's -30. Since the linear term has a coefficient of +1, you need to find two numbers that multiply to give you -30 and add up to 1.
That's 6 and -5.
So rewrite this as: $2x^2 + 6x -5x - 15$
Notice how I keep the -5x near the -15 since they have a common term.
Pull out a greatest common factor between terms: $2x(x + 3) -5(x + 3)$
Factor by grouping: $\boxed{(2x - 5)(x + 3)}$
I might've went too fast, but we'll explain something more detailed if you ask the question.
3. Pull out a greatest common factor between terms:
Factor by grouping:
will you always get a similar as in the (X+3) in the part? what happens if its different, or will it not be different?
thanks
4. You can just use the quadratic formula to factor trinomials. You may find this easier.
For example, factor $2x^{2}+x-15$.
Use quadratic formula,
$x= {\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}} = {\dfrac{-(1) \pm \sqrt{(1)^{2}-4(2)(-15)}}{2(2)}}$.
So $x = -3$ or $x=\frac{5}{2}$.
Now take $x = -3$ and add 3 to both sides of the equation to get: $(x+3)=0$.
Now take $x=\frac{5}{2}$ and multiply both sides of the equation by 2 to get $2x=5$. Then subtract both sides by 5 to get $(2x-5)=0$.
Multiplying the two quantities,
$(2x-5)(x+3)=0$
And you're done.
5. Hey I never thought of that before! That was spectacular Pn0yS0ld13r
6. Originally Posted by blip911
Pull out a greatest common factor between terms:
Factor by grouping:
will you always get a similar as in the (X+3) in the part? what happens if its different, or will it not be different?
thanks
it might be different sometimes, you'll have to group the terms differently. | {"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": 15, "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": 0.9430229067802429, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/76093?sort=oldest | ## An inequality on Difference of Entropies
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Hi,
I have the following problem that came up. It is not a homework problem or something similar. I did my simulations and it seems to hold but i was unable to prove it.## Heading ##
Let $P$ and $Q$ be two discrete probability distributions on the alphabet ${1,2,\dots n}$. Prove that:
$H(P)-H(Q) \leq \sum\limits_{i=1}^n \big [ (P_i-Q_i)\log(\frac{1}{\frac{P_i}{e}+(1-\frac{1}{e})Q_i}) \big ]$, where $e$ is the base of the natural log. All entropies are measured in nats.
Thank you very much for your help! Any ideas would be very helpful.
-
## 2 Answers
If $P=(P_i)$ and $Q=(Q_i)$ are two distributions then $\sum_i P_i\log \frac{1}{Q_i}=H(P)+D(P\|Q)$.
Hence $\sum_i (P_i-Q_i)\log\left(\frac{1}{\frac{P_i}{e}+(1-\frac{1}{e})Q_i}\right)=H(P)-H(Q)+D(P\|R)-D(Q\|R)$, where $R=\frac{1}{e}P+(1-\frac{1}{e})Q$.
So, if we can show that $D(P\|R)-D(Q\|R) \ge 0$, we are done. I hope this can be true from the Pythagorean property of relative entropy of Csiszar.
-
Notice, (if I am not mistaken), that nothing in this requires that $e$ is the base of natural log. Rather, what is crucial is just that $R$ is a convex combination of $P$ and $Q$. – R Hahn Sep 22 2011 at 7:15
This does not hold for any convex combination of $P$ and $Q$. For example if we choose $R$ to be very close to $P$ then the first term is close to zero whereas the second is always positive. Therefore the number $e$ plays an important role. I have corroborated these with simulations. Number $\frac{1}{e}$ seems to be the largest number for which this holds for all the $P$ and $Q$ I have tested. (if we choose something less than $\frac{1}{e}$ then we get "closer" to $Q$ and this difference can become negative.) – Kostas Sep 22 2011 at 7:44
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EDIT: This is wrong -- careless mistake as noted in the comments. I thought I had deleted it, but here it still is.
Working with the RHS of your inequality we have
\begin{eqnarray}\sum_i (P_i - Q_i) \log{\left(\frac{1}{\frac{P_i}{e} + (1-\frac{1}{e})Q_i}\right)} &=& \sum_i (P_i - Q_i)\log{\left(\frac{e}{P_i + (e-1)Q_i}\right)}\\ & = & \sum_i (P_i - Q_i) (1 - \log{(P_i + (e-1)Q_i)})\\ & = & \sum_i (P_i - Q_i) + \sum_i (Q_i - P_i)\log{(P_i + (e-1)Q_i)}\\ & = & 1 - 1 + \sum_i (Q_i - P_i)\log{(P_i + (e-1)Q_i)}\\\ & = & \sum_i Q_i \log{(P_i + (e-1)Q_i)} - \sum_i P_i \log{(P_i + (e-1)Q_i)}\\\ & \geq & \sum Q_i \log{(Q_i)} - \sum_i P_i \log{(P_i)}\\ & =& -\mbox{H}(Q) + \mbox{H}(P). \end{eqnarray} The inequality follows from $\log{(P_i)} \leq \log{(P_i + (e-1)Q_i)}$ and $\log{(Q_i)} \leq \log{(P_i + (e-1)Q_i)}$.
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Are you assuming $Q_i\geq P_i$ for all $i$? :) – Gjergji Zaimi Sep 22 2011 at 16:33
Nope, but I'm not seeing it -- where is that required? – R Hahn Sep 22 2011 at 16:42
Could you elaborate on the last sentence then? I see the inequality is equivalent to ∑Pi(log(Pi+(e−1)Qi)−log(Pi))≤∑Qi(log(Pi+(e−1)Qi)−log(Qi)) – Gjergji Zaimi Sep 22 2011 at 17:29
In the fourth line I assume that the {P_i} and the {Q_i} sum to one. I guess I'm not 100% sure that is what the OP intended. – R Hahn Sep 22 2011 at 18:32
Hi, in the 5th to the 6th line you use the same inequality for both the positive and negative term of the formula. For the positive term you can lower bounded by the inequality that you give. Yet the for the negative term it does not hold. – Kostas Sep 22 2011 at 18:46
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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9201187491416931, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/255181/irreducible-subvariety-not-cut-out-by-one-equation | # Irreducible subvariety not cut out by one equation
Let $X$ be an irreducible nonsingular algebraic variety. Let $Y$ be an irreducible codimension 1 closed subvariety. An important theorem states that locally $Y$ is cut out by a single equation regular in the neighborhood of a given point. What would be an example where this is not true globally?
In other words,
What is an example of $X,Y$ as above in which no rational function on $X$ has precisely $Y$ as a zero set?
(If $X$ is affine and $k[X]$ is a UFD then $Y$ is always globally cut out by one equation. Similarly if $X$ is projective and the homogeneous coordinate ring is a UFD. So I know I need a failure of unique factorization to get the example I want. That said, so far I haven't constructed it, so I humbly turn to your assistance.)
-
## 1 Answer
Consider the blowup $X=\operatorname{Bl}_{(0,0)}\Bbb A^2$ of the affine plane at the origin, and let $Y\subseteq X$ denote the exceptional divisor. Note that $X$ and $\Bbb A^2$ are birational (and are irreducible and nonsingular), hence $k(X)\cong k(\Bbb A^2)=k(x,y).$ Thus, a rational function cutting out $Y$ globally on $X$ must correspond to a rational function $\dfrac{f(x,y)}{g(x,y)}$ cutting out $(0,0)$ on $\Bbb A^2,$ which is of course impossible, since the origin has codimension two.
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this is a very nice example ! – QiL'8 Dec 10 '12 at 23:13
@QiL, thank you! – Andrew Dec 10 '12 at 23:18 | {"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": 0.18, "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": 0.9360435009002686, "perplexity_flag": "head"} |
http://www.physicsforums.com/showthread.php?t=547728 | Physics Forums
## What is an unbiased estimator ??
I do not really understand what is an unbiased estimator during my statistic studies ~~ THANKS ~~
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
This should be in the statistics forum, but since I can answer it... Remember that estimators are random variables; an estimator is "unbiased" if its expected value is equal to the true value of the parameter being estimated. To use regression as an example... Suppose you measured two variables x and y where the true (linear) relationship is given by...$$y = 5x + 2$$Of course, any sample that you draw will have noise in it, so you have to estimate the true values of the slope and intercept. Suppose that you draw a thousand samples of x,y and calculate the least squares estimators for each sample (assuming that the noise is normally distributed). As you do that, you'll notice two things... 1) All of your estimates are different (because the data is noisy) 2) The mean of all of those estimates starts to converge on the true values (5 and 2) The second occurs because the estimator is unbiased.
Mentor Thread moved.
## What is an unbiased estimator ??
any sample that you draw will have noise in it, >---- what does the noise mean by ??
mean like a One-to-One function ??
Quote by Voilstone I do not really understand what is an unbiased estimator during my statistic studies ~~ THANKS ~~
Hey Voilstone and welcome to the forums.
Lets say you have a parameter, for simplicity lets say its the mean.
Now the mean has a distribution based on a sample. Using your sample, you are trying to estimate a parameter: this is why we call these things estimators because they are estimating something.
Unbiased estimators have the property that the expectation of the sampling distribution algebraically equals the parameter: in other words the expectation of our estimator random variable gives us the parameter. If it doesn't, then the estimator is called unbiased.
Of course, we want estimators that are unbiased because statistically they will give us an estimate that is close to what it should be.
Also the key thing is that the estimate stays the same even when the sample grows. You will learn that an estimator should be consistent which basically means that the variance of the estimator goes to zero as the sample size goes to infinity.
Quote by chiro Hey Voilstone and welcome to the forums. Of course, we want estimators that are unbiased because statistically they will give us an estimate that is close to what it should be.
Thanks , i am still new here .
since unbiased estimator = mean/paramete , we want it to be close to the mean for what purposes ??
Quote by Voilstone Thanks , i am still new here . since unbiased estimator = mean/paramete , we want it to be close to the mean for what purposes ??
The estimator is a function of a sample. Since each observation in the sample comes from the same distribution, we consider each observation to be the realization of a random variable that corresponds to the true distribution. We also consider that each observation is independent: this simplifies many things like variance because independent samples have no covariance terms which means we can add variances very easily (You will see results like this later).
So our estimator is a random variable that is a function of other random variables. Now our estimator random variable is the actual distribution for the parameter we are estimating.
When you start off, we look at estimating things like means and variances, but we can create estimators that are really complicated if we want to: it uses the same idea as the mean and the variance but it measures something else of interest.
But with regards to unbiasedness, if our estimator was unbiased, then if we used statistical theory, we may not get the right intervals for the actual parameter.
Just think about if you had an estimator, and you did a 95% confidence interval that was really unbiased (lets say five standard deviations away from estimator mean): it wouldn't be useful using that estimator would it? Might as well not use an estimator at all if that is all you had.
So yeah in response to being close to the parameter, yes that is what we want. We want the mean for the estimator random variable to be the same, and to be the parameter of interest no matter what the sample is and no matter how big the sample is.
Recognitions:
Science Advisor
Quote by Voilstone I do not really understand what is an unbiased estimator during my statistic studies ~~ THANKS ~~
First, do you understand what an estimator is? In particular, do you understand that an estimator is a random variable ? (It is not generally a single number like 2.38. )
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http://mathematica.stackexchange.com/questions/11535/stationary-distribution-of-a-transition-matrix | # stationary distribution of a transition matrix
How can I solve the stationary distribution of a finite Markov Chain? In other words, how can I estimate the eigenvectors of a transition matrix?
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2
Welcome to Mathematica.SE, whynot! Please let us know what you tried. In addition, may I suggest: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the FAQs! 3) When you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign – Verbeia♦ Oct 4 '12 at 7:46
## 1 Answer
The `Eigensystem[ ]` command would be the way to go. Say you have a transition matrix:
````trans = Transpose[{{1/6, 1/6, 4/6}, {0, 3/4, 1/4}, {1/10, 1/10, 8/10}}]
````
Then you would get the eigenvalues and eigenvectors as:
````{eVals, eVecs} = Eigensystem[trans]
````
You can interpret these using
````eVals // MatrixForm
````
and
````eVecs // MatrixForm
````
In this case, for the transition matrix above, the eigenvector corresponding to the eigenvalue $1$ is the first row of the eVecs matrix, which is $\{ 0.12, 0.48, 1.\}$. You can check that this is true by evaluating
````trans.{0.12, 0.48, 1.}
````
which indeed returns $\{ 0.12, 0.48, 1.\}$. To get the actual steady state distribution, you would need to normalize this, i.e., divide the vector by the sum of the elements
````{0.12, 0.48, 1.}/Total[{0.12, 0.48, 1.}]
````
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nice answer! Another way to do the last step would be `Normalize[{0.12, 0.48, 1.}, Total]`. – Thies Heidecke Oct 4 '12 at 13:46
yes, it works, thanks. why do you need to transpose? – whynot Oct 4 '12 at 19:58
As commonly defined, a stochastic matrix has row sum equal to $1$, and has a left eigenvector corresponding to the unit eigenvalue. Mathematica computes right eigenvectors by default. So what the above does is to calculate the right eigenvector of $A'$, which is the desired left eigenvector of $A$. – bill s Oct 5 '12 at 5:59
lang-mma | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8766573667526245, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/38472/need-help-understanding-a-topos-theory-proof-any-topos-generated-by-subobjects-o/38483 | ## Need help understanding a topos theory proof (any topos generated by subobjects of 1 in whose subobject lattices are complete and Boolean satisfies AC)
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Hi all, I'm reading Mac Lane & Moerdijk's book "Sheaves in Geometry and Logic" and I don't understand a proof; sorry if this is the wrong place to ask, if there's somewhere better please let me know.
The proof in question is of Proposition VI.1.8 (page 276 in the paperback), which states:
Let $\mathcal{E}$ be a topos which is generated by subobjeccts of $1$, and moreover has the property that for each object $E$, $\mathrm{Sub}(E)$ is a complete Boolean algebra. Then $\mathcal{E}$ satisfies the axiom of choice.
Here the axiom of choice is the statement that any epimorphism $p: X \to I$ has a section $s: I \to X$, i.e. $ps = 1_I$. The proof starts as follows:
Let $p: X \to I$ be an epimorphism in $\mathcal{E}$. By completeness of $\mathrm{Sub}(I)$, we can apply Zorn's lemma and find a maximal subobject $m: M \to I$ such that $p$ has a section $s: M \to X$, i.e., $ps = m$.
I don't see how to show that Zorn's lemma applies here. Presumably I need to show that, given a linearly ordered set of subobjects $m_i: M_i \to I$ of $I$, $m_i = m_j k_{ij}$ for some $k_{ij}: M_i \to M_j$ whenever $m_i \leq m_j$, together with a set of sections $s_i: M_i \to X$ such that $p s_i = m_i$, $s_j k_{ij} = s_k$, there exists a section $s: M \to X$ such that $ps= m$, where $m: M \to I$ is the least upper bound in $\mathrm{Sub}(I)$ of the $m_i$'s, and such that the restriction of $s$ to each $M_i$ is $s_i$. It seems plausible that $m: M \to I$ will turn out to be a colimit in $\mathcal{E}/I$ of the collection $m_i: M_i \to I$, from which the required result would follow easily, but I don't know how to prove this. Can anybody help?
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## 2 Answers
Your approach is absolutely right: apply Zorn's lemma to the poset of pairs $(M,s)$ where $M \subseteq I$ is a subobject and $s: M \to X$ is a partial section of $p$.
As you say, to see that this poset is chain-complete, we just need to show that the least upper bound in $\newcommand{\Sub}{\mathrm{Sub}} \Sub(I)$ of a chain ${M_i}$ is in fact a colimit in $\mathcal{C}/I$, and hence that the sections also extend.
The key point here is that as long as the chain is inhabited, this is a filtered colimit. Filtered colimits are computed in $\Sub(I)$ just as colimits in $\newcommand{\C}{\mathcal{C}}\C$ (Charles Rezk's answer has just appeared, and shows this nicely); and all colimits in $\C/I$ are computed just as colimits in $\C$; so filtered colimits in $\Sub(I)$ are colimits in $\C/I$, and we're done.
…at least modulo the question of the empty chain! but this is easy to fix, in (at least) two ways. The simplest way is just to look at the case of the empty chain separately, and see that it's trivial. A nicer way (to my taste) is to state Zorn's lemma as “any (inhabited chain)-complete poset has a maximal element above any given element”. Neither of these feels quite right to me here — both seem a little ad hoc — but they do at least both work :-)
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Thanks for your reply. I can see how showing that filtered colimits in $\mathrm{Sub}(I)$ coincide with those in $\mathcal{C}$ gives me the result I want. However I do not see how Charles's answer shows this, can you explain? Regarding your last paragraph, the version of Zorn's lemma I know says that a non-empty poset in which every chain has an upper bound has a maximal element; since any element is an upper bound for the empty chain, this case is automatic for any non-empty poset. – Phil Wild Sep 12 2010 at 17:39
Ah… like Charles, I had thought we were working with Grothendieck not elementary toposes and so could assume $\mathcal{E}$ was complete. Hmmm… – Peter LeFanu Lumsdaine Sep 13 2010 at 3:15
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I'll try to show that a filtered colimit of a diagram of subobjects is still a subobject, which I suppose should include your linearly ordered diagram as a special case.
A map $X\to I$ is a subobject (i.e., a monomorphism), if and only if the projection maps $X\times_I X\to X$ from the fiber product to either of the factors are isomorphisms. If I have a family $X_i\to I$ of such monomorphisms indexed by a filtered category $C$, then `$\mathrm{colim}_{i\in C} X_i \to \mathrm{colim}_{i\in C} I$` has to be monomorphism, since filtered colimits preserve finite limits, and because $\mathrm{colim}_{i\in C}I \approx I$ since $i\mapsto I$ is a constant diagram on the filtered category $C$.
(Possibly, when I say "filtered", I really mean "cofiltered". Can't figure out which is which.)
Added. I was thinking about a Grothendieck topos, which is cocomplete. But it seems you (and Moerdijk and Mac Lane) want a proof which works in an elementary topos, and I don't know what to do in that case.
The key line in the proof seems to be "By completeness of $\mathrm{Sub}(I)$ ..." Presumably, if $\mathrm{Sub}(I)$ is complete, we can form arbitrary "intersections" of subobjects; so $\mathrm{colim} X_i$ should be the intersection of all subobjects $Y$ of $I$ which contain all the $X_i$'s.
I don't know why $\mathrm{Sub}(I)$ should be complete in an elementary topos. It is mentioned (for instance, on p. 491) that in a cocomplete topos $\mathrm{Sub}(I)$ is a Heyting algebra, so is complete. So the proof should work in that case.
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Sorry if I'm being thick, but I don't follow your reply. Are the colimits you refer to those in the original topos $\mathcal{E}$? If so then how do I know that they exist? The completeness of the subobject lattices shows that the corresponding colimits exist in $\mathrm{Sub}(I)$; I don't see how it follows that $\mathcal{E}$ itself has all relevant colimits. – Phil Wild Sep 12 2010 at 17:46
Ah, I was thinking of a Grothendieck topos. So "topos" means "elementary topos" here? – Charles Rezk Sep 12 2010 at 19:02
Yes, I think the proposition is supposed to apply to elementary topoi. I don't understand your comment that you "don't know why \$\mathrm{Sub}(I) should be complete [...]" - I don't think it's supposed to be true in general, it's just one of the hypotheses of the proposition. – Phil Wild Sep 12 2010 at 23:10
Oh, it's a hypothesis! I missed that. – Charles Rezk Sep 13 2010 at 3:49 | {"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": 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": 0.9586145877838135, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/211353-optimization-transformation-target-function-conditions.html | # Thread:
1. ## Optimization: transformation of target function and conditions
In a recent microeconomics lecture I was confronted with the following problem:
maxx1, x2, x3 2*sqrt(x1)
s.t.
2*sqrt(x2y)=64
y=2*sqrt(2x3)
x1+x2+x3=112
the professor reformulated the problem without explanation to
maxx1, x2, x3 4x1
s.t.
16x22y22=644
y2=8x3
x1+x2+x3=112
and then solved by plugging the conditions into the target function. How can such a transformation be determined without changing the problem at hand? what is the logic or the reason behind this?
thanks for any clue
2. ## Re: Optimization: transformation of target function and conditions
If $2\sqrt{x_1}$ is at a maximum, then so is $4x_1=(2\sqrt{x_1})^2$. The changes made to the conditions are similar - taking the 4th power of both sides and squaring both sides.
- Hollywood | {"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": 0.8817735314369202, "perplexity_flag": "middle"} |
http://stats.stackexchange.com/questions/3276/regularization-and-mean-estimation | # Regularization and Mean Estimation
Suppose I have some i.i.d. data $x_1, \ldots, x_n \sim N(\mu, \sigma^2)$, where $\sigma^2$ is fixed and $\mu$ is unknown, and I want to estimate $\mu$.
Instead of simply giving the MLE of $\mu = \bar{x}$, one could estimate
(1) $\mu = \lambda \mu_0 + (1 - \lambda) \bar{x},$
for some "prior best guess" $\mu_0$. This also has a nice Bayesian interpretation: we place a prior $\mu \sim N(\mu_0, \sigma^2_0)$ on $\mu$, and $\lambda$ is the weighted precision.
I seem to recall that this also has an explicit L2 regularization-interpretation (i.e., we choose some penalty and minimize the squared loss to get the above estimate), similar to things like lasso and ridge regression, but I can't remember how it goes. Can anyone explain what the L2 regularization-interpretation of (1) is?
[More general answers, where the data isn't necessarily normal distributed, are welcome as well.]
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Similar question at stats.stackexchange.com/questions/2957/…. A nice review paper on the topic is webee.technion.ac.il/Sites/People/YoninaEldar/Download/… – Robby McKilliam Oct 3 '10 at 7:30
## 2 Answers
Sure, it would be equivalent to the following ridge-like optimization problem:
$\underset{\mu\in\mathbb{R}|\mu_0,\lambda\geq0}{\min} ||x_i-\mu-\mu_0||_2+\lambda\mu^2$
For $\lambda=0$, $\mu+\mu_0$ goes to the OLS solution (i.e. $\bar{x}$), for $\lambda=\infty$, it shrinks to $\mu_0$.
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Ridge regression (Hoerl and Kennard, 1988) was initially developed to overcome singularities when inverting $X^tX$ (by adding $\lambda$ to its diagonal elements). Thus, the regularization in this case consists in working with a vc matrix $(X^tX-\lambda I)^{-1}$. This L2 penalization leads to "better" predictions than with usual OLS by optimizing the compromise between bias and variance (shrinkage), but it suffers from considering all coefficients in the model. The regression coefficients are found to be
$$\hat\beta=\underset{\beta}{\operatorname{argmin}}\|Y-X\beta\|^2 + \lambda\|\beta\|^2$$
with $\vert\vert\beta\vert\vert^2 = \sum_{j=1}^p\beta_j^2$ (L2-norm).
From a bayesian perspective, you can consider that the $\beta$'s must be small and plug them into a prior distribution. The likelihood $\ell (y,X,\hat\beta,\sigma^2)$ can thus be weighted by the prior probability for $\hat\beta$ (assumed i.i.d. with zero mean and variance $\tau^2$), and the posterior is found to be
$$f(\beta|y,X,\sigma^2,\tau^2)=(y-\hat\beta^tX)^t(y-\hat\beta^tX)+\frac{\sigma^2}{\tau^2}\hat\beta^t\hat\beta$$
where $\sigma^2$ is the variance of your $y$'s. It follows that this density is the opposite of the residual sum of squares that is to be minimized in the Ridge framework, after setting $\lambda=\sigma^2/\tau^2$.
The bayesian estimator for $\hat\beta$ is thus the same as the OLS one when considering the Ridge loss function with a prior variance $\tau^2$. More details can be found in The Elements of Statistical Learning from Hastie, Tibshirani, and Friedman (§3.4.3, p.60 in the 1st ed.). The second edition is also available for free.
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http://stats.stackexchange.com/questions/4485/is-this-the-correct-way-to-calculate-mse-from-ss-texttreatments-df | # Is this the correct way to calculate $MSE$ from $SS_{\text{treatment(s)}}$, $df_{\text{treatment(s)}}$, $F$?
As described in a few of my previous questions (here and there), I am interested in deriving summary statistics from statistics reported in the literature.
I would very much appreciate any advice into the validity or errors found in the following calculations.
To solve for $MSE$ given $F$, $df_{\text{group}}$, and $SS$.
This is required when a partial anova table is provided.
Given: \begin{equation}\label{eq:f} F = MS_g/MS_e \end{equation}
Where $g$ indicates the group, or treatment. Rearranging this equation gives: $$MS_e=MS_g/F$$
Given
$$MS_x = SS_x/df_x$$
Substitute $SS_g/df_g$ for $MS_g$ in the first equation
$$F=\frac{SS_g/df_g}{MS_e}$$
Then solve for $MS_e$
\begin{equation}\label{eq:mse} MS_e = \frac{SS_g}{df_g\times F} \end{equation}
Example from table 3 in Starr 2008.
The results are from one (two?) factor ANOVA with repeated measures, with treatment and week as the factors and no replication. "Effects of treatment on individual species were analyzed using Repeated Measures ANOVA in the statistical package Superanova (Abacus Concepts, Berkeley)."
We will calculate MSE from the $SS_{\text{treatment}}$ df_{\text{treatment}}, and $F$-value given in the table; these are $109.58$, $2$, and $0.570$, respectively; $df_{\text{weeks}}$ is given as $10$.
For the 1997 \textit{Eriphorium vaginatum}, the mean $A_{max}$ in table 4 is $13.49$.
Calculate $MS_e$:
$$MS_e = \frac{109.58}{0.57 \times 2} = 96.12$$
If this is the correct way, then shouldn't $MS_e$ be the same calculated based on factor a or factor b?
This is only a first draft, but my first attempt at presenting this detailed of a mathematical derivation. Feedback on the writing and presentation would be much appreciated.
Thanks!
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## 1 Answer
Your derivation is perfectly fine for regular (non-repeated meaures) ANOVA. For repeated measures ANOVA the F-statistics does not always equal $MS_g/MS_e$. Assuming weeks is the repeated measure, this formula is correct for the weeks and weeks*treatmeant terms (note that they will give the same MSE), but not for the treatment term. There is another term that is not shown in this table: subjects within treatment, and the MSE you found actually corresponds to that term.
If you are interested in complicated ANOVA tables, I would recommend a rather old book, that has a lot of these formulas worked out:
O. J. Dunn and V.A. Clark: Applied Statistics: Analysis of Variance and Regression, Wiley
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Hi @Aniko, Thanks for the feedback on this and for the reference. If my goal is to estimate the variance that would be measured on a single day in a single treatment, which term would be approapriate? – David Nov 12 '10 at 21:44 | {"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": 24, "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": 0.9275451898574829, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/100615/given-a-die-what-is-the-probability-that-the-second-roll-of-a-die-will-be-less/100742 | # Given a die, what is the probability that the second roll of a die will be less than the first roll?
If you are given a die and asked to roll it twice. What is the probability that the value of the second roll will be less than the value of the first roll?
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13
It makes me very sad to see that everyone here is assuming a d6 – TehShrike Jan 21 '12 at 3:13
1
Not everyone...I upvoted the answer that showed the generalized form. But, given the conciseness of Pete's answer (and the fact that I can generalize it myself) I accepted it as the best answer – Salman Paracha Jan 21 '12 at 3:17
me too! thanks! – CutieKrait Feb 14 at 0:17
Isn't it too easy-- 50 votes surprising.People love symmetry.. – Halil Duru Apr 2 at 17:42
## 8 Answers
There are various ways to answer this. Here is one:
There is clearly a $1$ out of $6$ chance that the two rolls will be the same, hence a $5$ out of $6$ chance that they will be different. Further, the chance that the first roll is greater than the second must be equal to the chance that the second roll is greater than the first (e.g. switch the two dice!), so both chances must be $2.5$ out of $6$ or $5$ out of $12$.
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4
You're assuming a six-sided die, which is probably the most common, not the only type. Is there a way to generalize this for a die with `N` sides where `N > 3` (because a die with three or less sides couldn't exist, obviously). – casperOne Jan 20 '12 at 20:00
16
Isn't a 2-sided die called a "coin"? – AShelly Jan 20 '12 at 20:27
6
@casperOne: There are also "3-sided" dice, in that each of 3 rolls are equally likely. You can use triangular-prisms, or a sphere with 3 corners cut off. Or just a normal 6-sided die, with two of each number. – BlueRaja - Danny Pflughoeft Jan 20 '12 at 20:40
9
Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$. – Pete L. Clark Jan 20 '12 at 21:22
3
@Pureferret: If $x$ and $y$ are mutually exclusive events of equal probability such that the probability that either $x$ or $y$ occurs is $p$, then the probability that $x$ occurs is $\frac{p}{2}$. In writing up my answer I chose to assume that the OP would understand this, at least intuitively. Fortunately, this assumption turned out to be correct. I'm sorry you didn't like my writeup, but in any case there are plenty of other answers... – Pete L. Clark Jan 21 '12 at 18:41
show 12 more comments
Here another way to solve the problem $$P(\textrm{second} > \textrm{first}) + P(\textrm{second} < \textrm{first}) + P(\textrm{second} = \textrm{first}) = 1$$ Because of symmetry $P(\text{second} > \text{first}) = P(\text{second} < \text{first})$ $$P(\text{second} > \text{first}) = \frac{1 - P(\text{second} = \text{first})}{2}$$ $$P(\text{second} > \text{first}) = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12}$$
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+1. And, I suggest that you'll TeXify your math here! – user21436 Jan 19 '12 at 23:51
1
+1 for mathematical formula rather than a long text. – SidCool Jan 20 '12 at 20:01
@SidCool This is biased, one should compare a mathematical formula with a short text (and my vote would definitely go to the latter). – Did Nov 4 '12 at 10:18
Looking at your reputation, I solemnly agree :) – SidCool Nov 5 '12 at 16:58
It might help to draw a picture:
$$\begin{array}{c|cccccc} &1&2&3&4&5&6 \\ \hline \\ 1&=&<&<&<&<&< \\ 2&>&=&<&<&<&< \\ 3&>&>&=&<&<&< \\ 4&>&>&>&=&<&< \\ 5&>&>&>&>&=&< \\ 6&>&>&>&>&>&= \\ \end{array}$$
Here, the $<$ signs mark the outcomes where the row number is less than the column number, and the $>$ signs mark those where to row number is greater than the column number. It's easy to see from the picture that the number of $<$ (or $>$) signs is $5+4+3+2+1=15$ out of $6^2 = 36$.
In fact, if you look at the picture a bit longer, you might realize that there's an even easier way to count the $<$ signs: the total number of $<$ and $>$ signs equals the total number of all signs ($6^2 = 36$) minus the number of $=$ signs ($6$), and the number of $<$ signs is half of that. Thus, there are $(36 - 6)/2 = 30/2 = 15$ out of $36$ $<$ signs in the table.
Once you've noticed that, it's easy to generalize the result: if you roll two $n$-sided dice, there are $n^2$ possible outcomes, out of which in $(n^2-n)/2$ the second roll will be less than the first. Thus, the probability of the second roll being less than the first is $$\frac{n^2-n}{2n^2} = \frac{n-1}{2n}.$$
For six-sided dice, this works out to $\frac{30}{72} = \frac{5}{12} = 0.41666\ldots$
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If the first roll is n, the chance that the second roll will be less is $\frac{n-1}{6}$. Summation over all possible values of n and multiplying by the chance for each value of n gives
$$\sum _{n=1}^6 \frac{1}{6}*\frac{(n-1)}{6}=\frac{5}{12}$$
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An anonymous user wanted to edit this answer. The edit was about adding a general formula for $m$-sided dice. Looked good to me, but I feel that editing somebody else's answer that way is not best. IMHO such a generalization is better placed as a comment to this answer. The OP is, of course, welcome to edit this answer. – Jyrki Lahtonen Feb 20 '12 at 5:07
If the:
````first roll is a 6 odds are: 5/6
first roll is a 5 odds are: 4/6
first roll is a 4 odds are: 3/6
first roll is a 3 odds are: 2/6
first roll is a 2 odds are: 1/6
first roll is a 1 odds are: 0/6
````
Therefore the total odds are the average of all those roll possibilities so: $$\frac{\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} + \frac{0}{6}}{6} = \frac{\frac{15}{6}}{6} = \frac{15}{36} = \frac{5}{12} = \frac{2.5}{6}$$
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1
this is by far the easiest to understand :) – Christian Jan 20 '12 at 14:39
One way to answer this question is to count the total number of pairs of results and number of pairs $(i, j)$ where $i < j$. The former is just $n^2$, and the latter is just $\binom{n}{2}$ where $n$ is the number of possible results of rolls. Here $n = 6$, so our answer is $$\frac{\binom{6}{2}}{6^2} = \frac{5}{12}$$ The same idea applies if we wanted to count the probability of an increasing sequence of rolls of length $k$. $$\frac{\binom{n}{k}}{n^k}$$
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1
+1 Nice generalization. – dot dot Dec 24 '12 at 17:37
The number of total possibilities when two dice are rolled is 36. The sample space for the experiment can be described as the set of ordered pairs in the following sense:
$$\Omega=\{(x, y)|1 \leq x,y \leq 6\}$$
Your question boils down to be able to count the number of ordered pairs where the second co-ordinate is less than the first co ordinate.
So, the answer is $\dfrac{15}{36}=\dfrac{5}{12}$
EDITED TO ADD PETE's COMMENTS:
How do you count?
The number of ordered pairs, where the $2^{nd}$ co-rdinate is $6$ and the $1^{st}$ co-ordinate is more than $6$ is $0$. Similarly, the number of ordered pairs, where the $2^{nd}$ co-ordinate is $5$ and and the $1 ^{st}$ co-ordinate is more than $5$ is $1$. Continuing this way, the number of pairs will be $0+1+2+3+4+5=15$
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You didn't show how you counted the ordered pairs. If you do this explicitly, I believe you will find $0 + 1 + 2 + 3 + 4 + 5 = 15$ of them, not $12$, for a probability of $\frac{15}{36} = \frac{5}{12}$. – Pete L. Clark Jan 19 '12 at 23:38
Okay, you fixed the typo in your answer. I'll leave the above comment since it hints on how to count other than pure brute force. (Added: what I hint at is done more fully in wnvl's answer.) – Pete L. Clark Jan 19 '12 at 23:39
I soon realised that I made a mistake. That's the reason why I edited it within few minutes of posting it. Thanks any way! – user21436 Jan 19 '12 at 23:40
I think some of the explanations, though correct, are unnecessarily complex. Out of a total of 36 combinations (6*6), how many are success cases?
If the result of first die throw is 1, we have 0 success cases as it doesn't matter what the second throw is.
If the result of first die throw is 2, there is 1 success case, where second throw is 1
If the result of first die throw is 3, there are 2 success cases, where second throw is 1 or 2
If the result of first die throw is 4, there are 3 success cases, where second throw is 1,2 or 3
If the result of first die throw is 5, there are 4 success cases, where second throw is 1,2,3 or 4
If the result of first die throw is 6, there are 5 success cases, where second throw is 1,2,3,4 or 5
Total # of success cases = 0+1+2+3+4+5 = 15. Probability is 15/36 or 5/12
Easy to test this in many languages like python, haskell. At the command prompt of haskell if you type
```
[(x,y) | x <- [2..6], y <- [1..x-1]]
```
you will get [(2,1),(3,1),(3,2),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5)]
If you type
```
length $ [(x,y) | x <- [2..6], y <- [1..x-1]]
```
you will get 15
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+1 for Haskell Codes. I sincerely hope among those that give "hard" arguments, mine is not one. I will be glad if I am proved wrong – user21436 Jan 21 '12 at 17:23
## protected by Qiaochu YuanJan 21 '12 at 4:28
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site. | {"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": 10, "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": 0.9282670021057129, "perplexity_flag": "middle"} |
http://en.wikipedia.org/wiki/Differential_equations | # Differential equation
(Redirected from Differential equations)
Not to be confused with Difference equation.
Differential equations
Scope
Classification
Operations
Attributes of variables
Relation to processes
• Difference (discrete analogue)
• stochastic
• Delay
Solutions
Solution topics
Mathematicians
Visualization of heat transfer in a pump casing, created by solving the heat equation. Heat is being generated internally in the casing and being cooled at the boundary, providing a steady state temperature distribution.
A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. Differential equations play a prominent role in engineering, physics, economics, and other disciplines.
Differential equations arise in many areas of science and technology, specifically whenever a deterministic relation involving some continuously varying quantities (modeled by functions) and their rates of change in space and/or time (expressed as derivatives) is known or postulated. This is illustrated in classical mechanics, where the motion of a body is described by its position and velocity as the time value varies. Newton's laws allow one (given the position, velocity, acceleration and various forces acting on the body) to express these variables dynamically as a differential equation for the unknown position of the body as a function of time. In some cases, this differential equation (called an equation of motion) may be solved explicitly.
An example of modelling a real world problem using differential equations is the determination of the velocity of a ball falling through the air, considering only gravity and air resistance. The ball's acceleration towards the ground is the acceleration due to gravity minus the deceleration due to air resistance. Gravity is considered constant, and air resistance may be modeled as proportional to the ball's velocity. This means that the ball's acceleration, which is a derivative of its velocity, depends on the velocity (and the velocity depends on time). Finding the velocity as a function of time involves solving a differential equation.
Differential equations are mathematically studied from several different perspectives, mostly concerned with their solutions —the set of functions that satisfy the equation. Only the simplest differential equations admit solutions given by explicit formulas; however, some properties of solutions of a given differential equation may be determined without finding their exact form. If a self-contained formula for the solution is not available, the solution may be numerically approximated using computers. The theory of dynamical systems puts emphasis on qualitative analysis of systems described by differential equations, while many numerical methods have been developed to determine solutions with a given degree of accuracy.
## Directions of study
The study of differential equations is a wide field in pure and applied mathematics, physics, meteorology, and engineering. All of these disciplines are concerned with the properties of differential equations of various types. Pure mathematics focuses on the existence and uniqueness of solutions, while applied mathematics emphasizes the rigorous justification of the methods for approximating solutions. Differential equations play an important role in modelling virtually every physical, technical, or biological process, from celestial motion, to bridge design, to interactions between neurons. Differential equations such as those used to solve real-life problems may not necessarily be directly solvable, i.e. do not have closed form solutions. Instead, solutions can be approximated using numerical methods.
Mathematicians also study weak solutions (relying on weak derivatives), which are types of solutions that do not have to be differentiable everywhere. This extension is often necessary for solutions to exist, and it also results in more physically reasonable properties of solutions, such as possible presence of shocks for equations of hyperbolic type.
The study of the stability of solutions of differential equations is known as stability theory.
## Nomenclature
The theory of differential equations is well developed and the methods used to study them vary significantly with the type of the equation.
### Ordinary and partial
• An ordinary differential equation (ODE) is a differential equation in which the unknown function (also known as the dependent variable) is a function of a single independent variable. In the simplest form, the unknown function is a real or complex valued function, but more generally, it may be vector-valued or matrix-valued: this corresponds to considering a system of ordinary differential equations for a single function.
Ordinary differential equations are further classified according to the order of the highest derivative of the dependent variable with respect to the independent variable appearing in the equation. The most important cases for applications are first-order and second-order differential equations. For example, Bessel's differential equation
$x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - \alpha^2)y = 0$
(in which y is the dependent variable) is a second-order differential equation. In the classical literature a distinction is also made between differential equations explicitly solved with respect to the highest derivative and differential equations in an implicit form. Also important is the degree, or (highest) power, of the highest derivative(s) in the equation (cf. : degree of a polynomial). A differential equation is nonlinear if its degree is not one (a sufficient but unnecessary condition).
• A partial differential equation (PDE) is a differential equation in which the unknown function is a function of multiple independent variables and the equation involves its partial derivatives. The order is defined similarly to the case of ordinary differential equations, but further classification into elliptic, hyperbolic, and parabolic equations, especially for second-order linear equations, is of utmost importance. Some partial differential equations do not fall into any of these categories over the whole domain of the independent variables and they are said to be of mixed type.
### Linear and non-linear
Both ordinary and partial differential equations are broadly classified as linear and nonlinear.
• A differential equation is linear if the unknown function and its derivatives appear to the power 1 (products are not allowed) and nonlinear otherwise. The characteristic property of linear equations is that their solutions form an affine subspace of an appropriate function space, which results in much more developed theory of linear differential equations. Homogeneous linear differential equations are a further subclass for which the space of solutions is a linear subspace i.e. the sum of any set of solutions or multiples of solutions is also a solution. The coefficients of the unknown function and its derivatives in a linear differential equation are allowed to be (known) functions of the independent variable or variables; if these coefficients are constants then one speaks of a constant coefficient linear differential equation.
• There are very few methods of solving nonlinear differential equations exactly; those that are known typically depend on the equation having particular symmetries. Nonlinear differential equations can exhibit very complicated behavior over extended time intervals, characteristic of chaos. Even the fundamental questions of existence, uniqueness, and extendability of solutions for nonlinear differential equations, and well-posedness of initial and boundary value problems for nonlinear PDEs are hard problems and their resolution in special cases is considered to be a significant advance in the mathematical theory (cf. Navier–Stokes existence and smoothness). However, if the differential equation is a correctly formulated representation of a meaningful physical process, then one expects it to have a solution.[1]
Linear differential equations frequently appear as approximations to nonlinear equations. These approximations are only valid under restricted conditions. For example, the harmonic oscillator equation is an approximation to the nonlinear pendulum equation that is valid for small amplitude oscillations (see below).
### Examples
In the first group of examples, let u be an unknown function of x, and c and ω are known constants.
• Inhomogeneous first-order linear constant coefficient ordinary differential equation:
$\frac{du}{dx} = cu+x^2.$
• Homogeneous second-order linear ordinary differential equation:
$\frac{d^2u}{dx^2} - x\frac{du}{dx} + u = 0.$
• Homogeneous second-order linear constant coefficient ordinary differential equation describing the harmonic oscillator:
$\frac{d^2u}{dx^2} + \omega^2u = 0.$
• Inhomogeneous first-order nonlinear ordinary differential equation:
$\frac{du}{dx} = u^2 + 1.$
• Second-order nonlinear ordinary differential equation describing the motion of a pendulum of length L:
$L\frac{d^2u}{dx^2} + g\sin u = 0.$
In the next group of examples, the unknown function u depends on two variables x and t or x and y.
• Homogeneous first-order linear partial differential equation:
$\frac{\partial u}{\partial t} + t\frac{\partial u}{\partial x} = 0.$
• Homogeneous second-order linear constant coefficient partial differential equation of elliptic type, the Laplace equation:
$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.$
• Third-order nonlinear partial differential equation, the Korteweg–de Vries equation:
$\frac{\partial u}{\partial t} = 6u\frac{\partial u}{\partial x} - \frac{\partial^3 u}{\partial x^3}.$
## Related concepts
• A delay differential equation (DDE) is an equation for a function of a single variable, usually called time, in which the derivative of the function at a certain time is given in terms of the values of the function at earlier times.
• A stochastic differential equation (SDE) is an equation in which the unknown quantity is a stochastic process and the equation involves some known stochastic processes, for example, the Wiener process in the case of diffusion equations.
• A differential algebraic equation (DAE) is a differential equation comprising differential and algebraic terms, given in implicit form.
## Connection to difference equations
See also: Time scale calculus
The theory of differential equations is closely related to the theory of difference equations, in which the coordinates assume only discrete values, and the relationship involves values of the unknown function or functions and values at nearby coordinates. Many methods to compute numerical solutions of differential equations or study the properties of differential equations involve approximation of the solution of a differential equation by the solution of a corresponding difference equation.
## Universality of mathematical description
Many fundamental laws of physics and chemistry can be formulated as differential equations. In biology and economics, differential equations are used to model the behavior of complex systems. The mathematical theory of differential equations first developed together with the sciences where the equations had originated and where the results found application. However, diverse problems, sometimes originating in quite distinct scientific fields, may give rise to identical differential equations. Whenever this happens, mathematical theory behind the equations can be viewed as a unifying principle behind diverse phenomena. As an example, consider propagation of light and sound in the atmosphere, and of waves on the surface of a pond. All of them may be described by the same second-order partial differential equation, the wave equation, which allows us to think of light and sound as forms of waves, much like familiar waves in the water. Conduction of heat, the theory of which was developed by Joseph Fourier, is governed by another second-order partial differential equation, the heat equation. It turned out that many diffusion processes, while seemingly different, are described by the same equation; the Black–Scholes equation in finance is, for instance, related to the heat equation.
## Notable differential equations
### Physics and engineering
• Newton's Second Law in dynamics (mechanics)
• Euler–Lagrange equation in classical mechanics
• Hamilton's equations in classical mechanics
• Radioactive decay in nuclear physics
• Newton's law of cooling in thermodynamics
• The wave equation
• Maxwell's equations in electromagnetism
• The heat equation in thermodynamics
• Laplace's equation, which defines harmonic functions
• Poisson's equation
• Einstein's field equation in general relativity
• The Schrödinger equation in quantum mechanics
• The geodesic equation
• The Navier–Stokes equations in fluid dynamics
• The Diffusion equation in stochastic processes
• The Convection–diffusion equation in fluid dynamics
• The Cauchy–Riemann equations in complex analysis
• The Poisson–Boltzmann equation in molecular dynamics
• The shallow water equations
• Universal differential equation
• The Lorenz equations whose solutions exhibit chaotic flow.
### Biology
• Verhulst equation – biological population growth
• von Bertalanffy model – biological individual growth
• Lotka–Volterra equations – biological population dynamics
• Replicator dynamics – found in theoretical biology
• Hodgkin–Huxley model – neural action potentials
## See also
• Complex differential equation
• Exact differential equation
• Integral equations
• Linear differential equation
• Picard–Lindelöf theorem on existence and uniqueness of solutions
• Numerical methods
• Recurrence relation, also known as 'Difference Equation'
## References
• P. Abbott and H. Neill, Teach Yourself Calculus, 2003 pages 266-277
• P. Blanchard, R. L. Devaney, G. R. Hall, Differential Equations, Thompson, 2006
• E. A. Coddington and N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hill, 1955
• E. L. Ince, Ordinary Differential Equations, Dover Publications, 1956
• W. Johnson, A Treatise on Ordinary and Partial Differential Equations, John Wiley and Sons, 1913, in University of Michigan Historical Math Collection
• A. D. Polyanin and V. F. Zaitsev, Handbook of Exact Solutions for Ordinary Differential Equations (2nd edition), Chapman & Hall/CRC Press, Boca Raton, 2003. ISBN 1-58488-297-2.
• R. I. Porter, Further Elementary Analysis, 1978, chapter XIX Differential Equations
•
• D. Zwillinger, Handbook of Differential Equations (3rd edition), Academic Press, Boston, 1997.
1. Boyce, William E.; DiPrima, Richard C. (1967). Elementary Differential Equations and Boundary Value Problems (4th ed.). John Wiley & Sons. p. 3. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "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": 0.9130606055259705, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/47762/list | ## Return to Answer
2 fixed a typo noted by a reader, used markdown instead of abusing MathJax
I
$\DeclareMathOperator{\Hom}{Hom}\newcommand{\amod}{\mathscr{A}\text{-}{\bf Mod}}\newcommand{\scrA}{\mathscr{A}}\newcommand{\scrE}{\mathscr{E}}\newcommand{\Ab}{\mathbf{Ab}}\DeclareMathOperator{\Lex}{\mathbf{Lex}}\DeclareMathOperator{\coker}{Coker}$I only know one proof of the embedding theorem- the —the expositions differ heavily in terminology but the approaches all are equivalent, as far as I can tell. I think the proof in Swan's book on K-theory makes the relation between the Freyd-Mitchell approach and Gabriel's approach pretty clear. Let me just say that there is no cheap way of getting the Freyd-Mitchell embedding theorem as since there is a considerable amount of work you need to invest in order to get all the details straight. On the other hand, if you manage to feel comfortable with the details, you will have learned quite a good amount of standard tools of homological algebra, so I think it's well worth the effort.
Think of the category of functors $\mathscr{A} \scrA \to {\bf Ab}$ \Ab$as$\mathscr{A}$-modules, \scrA$-modules, that's why the notation $\mathscr{A}-{\bf Mod}$ \amod$is fairly common. The Yoneda embedding$A \mapsto \hom(A,{-})$Hom(A,{-})$ even yields a fully faithful contravariant functor $y: \mathscr{A} \to y\colon \mathscr{A}-{\bf Mod}$.scrA\to\amod\$.
The category $\mathscr{A}-{\bf Mod}$ \amod$inherits a bunch of nice properties of the category${\bf Ab}$\Ab$ of abelian groups:
• The functors $\hom(A,{-})$ \Hom(A,{-})$are injective and$\prod_{A \in\mathscr{A}} in\scrA} \hom{(A,{-})}$Hom{(A,{-})}$ is an injective cogenerator.
• Since there is an injective cogenerator, the category ${\scr A}-{\bf Mod}$ \amod\$ is well-powered.
• However, the Yoneda embedding is ${\it not}$ not exact: If $0 \to A' \to A \to A'' \to 0$ is a short exact sequence, we only have an exact sequence
$\hom(A'',{-})$0 \to \hom(A,{-}) Hom(A',{-}) \to \hom(A',{-}) Hom(A,{-}) \to 0$\Hom(A'',{-})$\$
in $\mathscr{A}-{\bf Mod}$.\amod\$.
It turns out that the functor $Q = \ker{(\hom(A'',{-}) coker{(\Hom(A,{-}) \to \hom(A,{-}))}$ Hom(A'',{-}))}$is "weakly effaceable", “weakly effaceable”, so we would want it to be zero in order to get an exact functor. How can we achieve this? Well, just$\it force$force them to be zero: say a morphism$f: f\colon F \to G$in${\scr A}-{\bf Mod}$\amod$ is an isomorphism if both its kernel and its cokernel are weakly effaceable. If this works then a weakly effaceable functor $E$ is isomorphic to zero because $E \to 0$ has $E$ as kernel. Now the full subcategory $\scr E$ \scrE$of weakly effaceable functors is a Serre subcategory, so we may form the Gabriel quotient$\scr A-\bf Mod/\scr E$. \amod/\scrE$. By its construction, isomorphisms in the Gabriel quotient have precisely the description above.
On the other hand, the category ${\bf Lex}(\mathscr{A},{\bf Ab})$ \Lex(\scrA,\Ab)$of left exact functors$\mathscr{A} \scrA \to {\bf Ab}$\Ab$ is abelian. This is far from obvious when you start from the definitions. However, ${\bf Lex}(\mathscr{A},{\bf Ab})$ \Lex(\scrA,\Ab)$sits comfortably inside the abelian category$\scr A- \bf Mod$. \amod$. The inclusion has an exact left adjoint (!) (= "sheafification"), “sheafification”), so again ${\bf Lex}({\scr A}, {\bf Ab})$ \Ab)$inherits many useful properties from${\scr A}-{\bf Mod}$. \amod$. Moreover, the kernel of the left adjoint can be identified with the weakly effaceable functors, and that's why ${\bf Lex}{({\scr A}, {\bf Ab})} \Lex{(\scrA, \Ab)} = {\scr A}-{\bf Mod}/{\scr E}$.\amod/\scrE\$.
All this work shows that $A \mapsto \hom{(A,{-})}$ Hom{(A,{-})}$is a fully faithful and$\it exact$exact embedding of$\scr A$\scrA$ into ${\bf Lex}{({\scr A}, {\bf Ab})}$, {\Lex}{(\scrA, \Ab)}\$, so it remains to show that the latter can be embedded into a category of modules. This is well described in Weibel's or Swan's books, so I won't elaborate on that point and content myself by saying that you simply need to look at the endomorphism ring of an injective generatorcogenerator.
As for references, I think you cannot can't do much better than Freyd's book. Don't be too intimidated by Swan's exposition in his K-theory book. If you're really interested in understanding this proof, I think it's worth reading the two expositions (first Freyd, then Swan). There also is a proof in volume 2 of Borceux's Handbook of categorical algebra with a more "hands “hands on" ” approach.
1
I only know one proof of the embedding theorem - the expositions differ heavily in terminology but the approaches all are equivalent. I think the proof in Swan's book on K-theory makes the relation between the Freyd-Mitchell approach and Gabriel's approach pretty clear. Let me just say that there is no cheap way of getting the Freyd-Mitchell embedding theorem as there is a considerable amount of work you need to invest in order to get the details straight. On the other hand, if you manage to feel comfortable with the details, you will have learned quite a good amount of standard tools of homological algebra, so I think it's well worth the effort.
Think of the category of functors $\mathscr{A} \to {\bf Ab}$ as $\mathscr{A}$-modules, that's why the notation $\mathscr{A}-{\bf Mod}$ is fairly common. The Yoneda embedding $A \mapsto \hom(A,{-})$ even yields a fully faithful contravariant functor $y: \mathscr{A} \to \mathscr{A}-{\bf Mod}$.
The category $\mathscr{A}-{\bf Mod}$ inherits a bunch of nice properties of the category ${\bf Ab}$ of abelian groups:
• It is abelian.
• It is complete and cocomplete ((co-)limits can be computed pointwise on objects)
• The functors $\hom(A,{-})$ are injective and $\prod_{A \in\mathscr{A}} \hom{(A,{-})}$ is an injective cogenerator.
• Since there is an injective cogenerator, the category ${\scr A}-{\bf Mod}$ is well-powered.
• etc.
However, the Yoneda embedding is ${\it not}$ exact: If $0 \to A' \to A \to A'' \to 0$ is a short exact sequence, we only have an exact sequence
$\hom(A'',{-}) \to \hom(A,{-}) \to \hom(A',{-}) \to 0$
in $\mathscr{A}-{\bf Mod}$.
It turns out that the functor $Q = \ker{(\hom(A'',{-}) \to \hom(A,{-}))}$ is "weakly effaceable", so we would want it to be zero in order to get an exact functor. How can we achieve this? Well, just $\it force$ them to be zero: say a morphism $f: F \to G$ in ${\scr A}-{\bf Mod}$ is an isomorphism if both its kernel and its cokernel are weakly effaceable. If this works then a weakly effaceable functor $E$ is isomorphic to zero because $E \to 0$ has $E$ as kernel. Now the full subcategory $\scr E$ of weakly effaceable functors is a Serre subcategory, so we may form the Gabriel quotient $\scr A-\bf Mod/\scr E$. By its construction, isomorphisms in the Gabriel quotient have precisely the description above.
On the other hand, the category ${\bf Lex}(\mathscr{A},{\bf Ab})$ of left exact functors $\mathscr{A} \to {\bf Ab}$ is abelian. This is far from obvious when you start from the definitions. However, ${\bf Lex}(\mathscr{A},{\bf Ab})$ sits comfortably inside the abelian category $\scr A- \bf Mod$. The inclusion has an exact left adjoint (!) (= "sheafification"), so again ${\bf Lex}({\scr A}, {\bf Ab})$ inherits many useful properties from ${\scr A}-{\bf Mod}$. Moreover, the kernel of the left adjoint can be identified with the weakly effaceable functors, and that's why ${\bf Lex}{({\scr A}, {\bf Ab})} = {\scr A}-{\bf Mod}/{\scr E}$.
All this work shows that $A \mapsto \hom{(A,{-})}$ is a fully faithful and $\it exact$ embedding of $\scr A$ into ${\bf Lex}{({\scr A}, {\bf Ab})}$, so it remains to show that the latter can be embedded into a category of modules. This is well described in Weibel's or Swan's books, so I won't elaborate on that point and content myself by saying that you simply need to look at the endomorphism ring of an injective generator.
As for references, I think you cannot do much better than Freyd's book. Don't be too intimidated by Swan's exposition in his K-theory book. If you're really interested in understanding this proof, I think it's worth reading the two expositions (first Freyd then Swan). There also is a proof in volume 2 of Borceux's Handbook of categorical algebra with a more "hands on" approach. | {"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": 80, "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": 0.9330196976661682, "perplexity_flag": "head"} |
http://nrich.maths.org/2192/index?nomenu=1 | There are six dots in a Braille cell. Each dot can be either raised or left blank. A black dot is raised on the paper so you can feel it (circles represent dots that have not been used). A quick calculation should tell you that there are $2^6$ possible arrangements of the six dots. However, some would be difficult to tell apart if you were blind and so no two letters use the same arrangements placed in slightly different positions, for example the letter "d" is represented in the following way: so no letter uses . | {"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": 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": 0.9573720097541809, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/51992/how-i-calculate-degree-of-the-algebraic-curve | ## How I calculate degree of the algebraic curve?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let F be algebraically closed field. Let C be a curve in F^n defined as zeroes of polynomials $p_1(x_1,\ldots,x_n),..,p_{n-1}(x_1,\ldots x_n)$. Let us define degree of the curve as $\max_S { S\cap C }$ were $S$ $n-1$ dimensional linear subspace such that ${ S\cap C }$ is finite.
How does possible to calculate this degree of the curve?
In general degree of curve should be product of degrees of $p_1,\ldots p_{n-1}$ but for example if $C=(x-f_1(z),y-f_2(z))$ were $f_1,f_2$ are of degree d then degree of $C$ is $d$ and not $d^2$.
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## 1 Answer
What's happening is that indeed the degree of your curve (when it is a curve) is the product of the degrees of the $p_i$, counted with multiplicity! In other words, the intersection scheme has that degree.
In your example in the last paragraph, the intersection is actually $d\cdot C$, so getting degree $d$ for $C$ is the correct answer. The issue is that in this example the intersection multiplicity of the defining equations is $d$ everywhere along the intersection.
Here is a specific example to see what's happening: Let's say that $d=2$, $f_1(z)=z^2+l_1(z)$ and $f_2(z)=z^2+l_2(z)$ where the $l_i$ are linear polynomials in $z$. Then the ideal generated by $x-f_1(z)$ and $y-f_2(z)$ contains $x-y+l_1(z)-l_2(z)$, a linear polynomial and you get the same ideal with the generators $x-f_1(z)$ and $x-y+l_1(z)-l_2(z)$. If you take the intersection of these two hypersurfaces, then you get the correct degree $d=2$. You can easily see that if you take a local ring of the ambient space at a point of the intersection curve, then the original defining equations are both in the square of the maximal ideal, so their intersection multiplicity has to be (at least) $2$.
By the way, intersection theory should be really done in the projective space.
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Thanks for you answer. How can I calculate this degree for a given set of polynomials. One specific example is what will be degree if I have two random polynomials $p(x,y)$ of degree $d_1$ and $g(x_1,x_2,x_3,x_4)$ of degree $d_2$ what will be typical degree of $C=<p(x_1,x_2),p(x_3,x_4), g(x_1,x_2 x_3,x_4)>$ – Klim Efremenko Jan 13 2011 at 20:31
1
I think in this case you should expect a degree $d_1^2 d_2$ curve. Basically, that's your degree, divided by the multiplicity of the curve. This is of course assuming that the curve is irreducible. If $C$ is reducible, you can get different multiplicities along different components. In any case, what you need to determine is the intersection multiplicities along each irreducible component of the intersection. In the case of three polynomials as in your comment, you would probably be best off by doing it first for the intersection of just two of them and then intersect with the third. – Sándor Kovács Jan 13 2011 at 21:01
ps: the reason I would expect degree $d_1^2d_2$ in the particular case you are asking about is because the two occurrences of $p$ happens in separate variables, so their zero loci are transversal. Then if $g$ is general, then it should be transversal to the intersection. – Sándor Kovács Jan 13 2011 at 21:04
But, this still does not answer the question: How can I calculate degree of the curve? Say given two polynomials $p(x,y,z), g(x,y,z)$ Does there exist algorithm for calculating degree of intersection? Or what is the multiplicity of the intersection? – Klim Efremenko Jan 16 2011 at 8:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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": 0.9478799700737, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/285882/group-automorphisms-of-the-non-zero-elements-of-a-field | Group automorphisms of the non-zero elements of a field
I would like to know what is $Hom_{Groups}(K^*,K^*)$, at least in the case $K$ is a complete non-archimedean (valued) field. Is this $\mathbb{Z}$?
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1 Answer
Much bigger than $\mathbb Z$, at least if (as is standard) you mean by “$K^*$” the set of nonzero elements of $K$. (Your title suggests something different.)
I will say something about the automorphisms of a local field of mixed characteristic, i.e. a finite extension of $\mathbb Q_p$. The question is certainly much more difficult if you accept discontinuous automorphisms, and I will not consider such. What you need to do is decompose $K^*$ as direct sum of simpler groups, and you find that $$K^*\cong \mathbb Z\oplus W\oplus(\mathbb Z_p)^n\>,$$ where $W$ is the (finite) group of roots of unity of $K$, and $n=[K\colon\mathbb Q_p]$. This decomposition is not unique. Now to see all automorphisms of this group you need to look not only at the automorphisms of the summands but also at the homomorphisms from one summand to another.
- | {"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": 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": 0.9512078762054443, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/283636/linear-optimization-of-line-under-points | # Linear optimization of line under points.
I have a set of points in two-dimensional space. I'd like to find the line of best fit that minimizes the distance between the line and the points such that the line is below all points. How can I find the slope and intercept of a line that minimizes this distance?
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Yes, I know OLS, but I don't want a line of best fit though the points. I want to find a line below all points. – user1728853 Jan 21 at 18:19
Thanks for the help. Unfortunately, I don't know how to do this. Can you provide a simple explanation? – user1728853 Jan 21 at 18:25
you can find the line with least squares and then translate this line so that he pass over the lowest point. At this point you can say that the line is below all points (except for one) – the_candyman Jan 21 at 18:44
1
@the_candyman: Are you sure the translated line will be optimal among all possible lines that lie below all points? – Rahul Narain Jan 21 at 22:28
## 1 Answer
Just solve the following optimization problem for $(a,b)$: \begin{align*} \text{Minimize}&\quad \sum_{i=1}^n (y_i - (a \, x_i + b))^k \\ \text{such that}& \quad a \, x_i + b \le y_i \end{align*} Here, you can choose $k = 2$ (least squares) or $k=1$.
This is a smooth, convex problem (for $k=1$ even a linear problem). Try to write down the optimality conditions and solve them.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "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": 0.9016312956809998, "perplexity_flag": "head"} |
http://cstheory.stackexchange.com/questions/16216/searching-for-the-first-item-satisfying-property-with-penalty-for-every-test | # Searching for the first item satisfying property with penalty for every test
The problem in terms of a step function on integers: A step function of integers is $0$ until $s$ (the "item" in question) and then $1$. That is, $s$ is the first integer that satisfies the property $f(s) = 1$. The only way to find out the function's value $f(x)$ is to run a test to evaluate it, where each test has a penalty $p(x)$. Given a range $[a,b]$ in which the step is known to occur, find the stepping point $s$ with minimal total penalty.
In the case I'm interested in, $p(x) = x$.
If $p(x)$ is just a constant, then binary search is the solution (I think?).
Background: There is a process that takes X minutes before a certain event occurs, always at the same time, but this time is unknown. I need to find this time. To check, start the process, wait X minutes and then run a test to find out if the event happened or not. After testing, the process restarts (so you can't just test every minute to find out the minimum time).
Edit: I forgot to mention that for simplicity's sake the distribution of $s$ is uniform in $[a,b]$, that is $P(x=s, x \in [a,b]) = 1/(b-a+1)$. A more realistic distribution in my actual problem is monotonically increasing, for example $P(x) = 2x/(b-a+1)^2$.
Update: usul's answer (with a minor tweak) describes how to calculate the value which needs minimizing, which is the expected cost given a penalty function and the probability distribution of the target value.
The question still stands: which algorithm actually minimizes that value in the case $p(x)=x$ and the two distributions discussed above (uniform and increasing)? More generally - what's an algorithm that takes the penalty function and always yields the minimum expected cost?
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2
It seems that unless you have negative penalty, it does not matter what the cost p(x) is, you just need to minimize it. The only relevant information here is the distribution. I suspect that a binary search will always be applicable, but that the "regions" will be divided by the median point of the distribution rather than by the center of the considered interval. – cody Jan 24 at 19:13
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@cody, counterexample: $p(x)$ where odd numbers have double the cost of even ones. In that case binary search is worse than binary search tuned to always pick even numbers. – sinelaw Jan 24 at 19:31
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Compute the minimum total cost for every subrange of [a,b]. – Tsuyoshi Ito Jan 24 at 20:54
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It seems like there should be a closed form for the algorithm (at least upto a constant factor). – Suresh Venkat♦ Jan 25 at 6:09
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@Suresh: I was wondering about the possibility for a closed formula, too. If you have one, I am interested to see it. – Tsuyoshi Ito Jan 25 at 6:23
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## 2 Answers
Here is a writeup following up on Tsuyoshi's pointer to use DP.
Given $p(x)$, we can write any (decent, deterministic) algorithm as a binary decision tree with nodes weighted by $p(x)$. The root of the tree is the first node the algorithm selects to test; the left child of each $x$ is the next node to test if $f(x)$ is zero and the right node is the next to test if $f(x) = 1$.
An algorithm's worst-case performance on weights $p(x)$ is its weighted depth: the maximum over any path to a leaf of the sum of the path's weights.
Therefore the optimal deterministic algorithm minimizes this value.
The "minimum maximum length" or "cost" of a path starting at node $x$ is $p(x)$ plus the cost of the tree on the nodes remaining to be searched. The nodes remaining to be searched will always be an interval. This gives the DP relation:
$cost[x, z] = \min_{y \in [x,z-1]} \left( p(y) + \max ( cost[x, y-1], cost[y, z]) \right)$
with $cost[x,x] = 0$ as a base case (except $cost[b,b] = p(b)$ unless you are promised that a $1$ occurs in the interval).
(Update)
With a probability distribution, say you want to minimize total expected cost. The expected cost for searching an interval is the cost of searching the first node $y$, plus the expected cost for searching the remainder. Using $\Pr[a,b|c,d]$ for the conditional probability of being in interval [a,b] given that we're in interval [c,d], this gives the following:
$cost[x,z] = \min_{y \in [x,z-1]} \left(p(y) + \Pr[x,y-1|x,z] cost[x,y-1] + \Pr[y,z|x,z] cost[y,z] \right)$ .
Here's what I had before, which does something ... let's say "different" instead of "wrong."
It minimizes maximum expected cost. The idea is, given an algorithm, consider the node $y$ with highest expected cost $\Pr[y] p(y)$. We want the algorithm that minimizes this value. We might view this as risk-averse, I suppose.
$cost[x,z] = \min_{y \in [x,z-1]} \left( p(y) + \max ( \Pr[x,y-1 | x,z] cost[x,y-1], \Pr[y,z | x,z] cost[y,z]) \right)$.
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(1) Your DP recurrence is a little incorrect because in this problem, a query on x only tells you whether x<s or x>=s, not the 3-way comparison between x and s. – Tsuyoshi Ito Jan 25 at 5:43
@TsuyoshiIto - Thanks, fixed! – usul Jan 25 at 5:45
(2) I was about to write a comment that the average case can be done by the same DP idea, but your updated answer already addresses this case. :) – Tsuyoshi Ito Jan 25 at 6:19
I started to write "just multiply p(x) by Pr[x]", then had to stop and think about it for 45 min... – usul Jan 25 at 6:23
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Oh, by the way, I have just realized that your recurrence in the average case is still incorrect. The max must be replaced by addition. In addition, in both recurrences, the range of y should be [x+1,z] instead of [x,z]. – Tsuyoshi Ito Jan 25 at 21:01
show 13 more comments
One greedy method would be to test the position which will minimize expected entropy (the amount of information that is unknown) weighted by the cost of testing the position.
Assuming that we have determined that $c \le s \le d$ (via previous tests), then the current amount of entropy is:
$$-\sum_{x=c}^d \ln\big(Pr(s = x \mid c \le x \le d)\big)Pr(s = x \mid c \le x \le d)$$
After testing a position, $y$, between $c$ and $d$, the entropy will be
$$-\sum_{x=c}^y \ln\big(Pr(s = x \mid c \le x \le y)\big)Pr(s = x \mid c \le x \le y)$$
with probability $Pr(c \le s \le y \mid c \le s \le d)$
and
$$-\sum_{x=y+1}^d \ln\big(Pr(s = x \mid y < x \le d)\big)Pr(s = x \mid y < x \le d)$$
with probability $Pr(y < s \le d \mid c \le s \le d)$
This makes the expected entropy:
$$-Pr(c \le s \le y \mid c \le s \le d)\sum_{x=c}^y \ln\big(Pr(s = x \mid c \le x \le y)\big)Pr(s = x \mid c \le x \le y) -Pr(y < s \le d \mid c \le s \le d)\sum_{x=y+1}^d \ln\big(Pr(s = x \mid y < x \le d)\big)Pr(s = x \mid y < x \le d)$$
Thus, to choose the test which minimizes expected entropy while penalizing cost, use the value of $y$ which minimizes:
$$-p(y)\Big( Pr(c \le s \le y \mid c \le s \le d)\sum_{x=c}^y \ln\big(Pr(s = x \mid c \le x \le y)\big)Pr(s = x \mid c \le x \le y) +Pr(y < s \le d \mid c \le s \le d)\sum_{x=y+1}^d \ln\big(Pr(s = x \mid y < x \le d)\big)Pr(s = x \mid y < x \le d) \Big)$$
Repeat.
Note, for uniform penalty and distribution, this becomes a binary search. I believe it will also work for your "penalize odds" example.
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(1) In the case of the uniform penalty and the uniform prior distribution, your method gives linear search instead of binary search. (2) You might want to fix this particular case by tweaking the answer, but more importantly, where is the proof of correctness? – Tsuyoshi Ito Jan 24 at 23:46
While I refresh my memory trying to understand your answer - you're suggesting a greedy algorithm, but why should this be optimal? As for the "quantum weirdness", the question started from a weird bug in a program that expires user's session after a period of inactivity, where the only way to test the session's validity was to perform some action - and if the session isn't expired, it would restart the timeout (hence forced reset upon observation) – sinelaw Jan 25 at 1:22
@TsuyoshiIto, that's because I messed up the entropy calculation. I've fixed it so it should yield binary search now with uniform pmf and penalty. – bbejot Jan 25 at 3:53
@sinelaw, it's no proof Thursday! That is, it isn't necessarily optimal. I took inspiration from the calculation of decision trees by minimizing expected entropy. Weighting by multiplying by the penalty just seemed like an intuitive thing to do. I've changed my opening phrase from "The aswer is" to "One greedy method would be". – bbejot Jan 25 at 4:00
"Greedy method" does not mean "an algorithm based on the greedy method which may or may not be correct," at least not in theoretical computer science. If you mean the latter, you should state so, but then I do not know what the point of posting it is in the first place. – Tsuyoshi Ito Jan 25 at 5:24
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http://mathhelpforum.com/math/27940-riemann-hypothesis-print.html | Riemann Hypothesis
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• February 10th 2008, 07:15 PM
ThePerfectHacker
Riemann Hypothesis
Today I recieved an e-mail from my mathematics advisor (not sure how to call him). He is an expert in complex analysis.
Quote:
I heard today that there will soon be an announcement of a proof of the Riemann Hypothesis!
You might be excited. But I am not. :(. It is a wonderful problem, I wish to solve it, not anybody else.
• February 10th 2008, 07:30 PM
tukeywilliams
haven't there been many proofs proposed?
• February 10th 2008, 07:40 PM
ThePerfectHacker
Quote:
Originally Posted by tukeywilliams
haven't there been many proofs proposed?
Yes. And all of them false. Hopefully this will be false too.
Interesting fact: Many people think that the Riemann Hypothesis is a number theory problem, but in fact it is really a problem from complex analysis!
• February 10th 2008, 07:42 PM
Jhevon
Quote:
Originally Posted by ThePerfectHacker
Interesting fact: Many people think that the Riemann Hypothesis is a number theory problem, but in fact it is really a problem from complex analysis!
one of the candidates for the dean of our math department described it as a "number theory" problem :D ... maybe he mumbled complex analysis, i don't remember ...
• February 10th 2008, 07:46 PM
ThePerfectHacker
Quote:
Originally Posted by Jhevon
one of the candidates for the dean of our math department described it as a "number theory" problem :D ... maybe he mumbled complex analysis, i don't remember ...
It has connections to number theory but it is certainly not a number theory problem. "If the zeta function is analytically extended to $\mathbb{C}\setminus \{0\}$ then the (non-trivial) zeros have real part equal to 1/2". That is exactly what the hypothesis says. Look at the words: zeta function, analytic continuation, $\mathbb{C}$, real part. All of those are words from complex analysis, not from number theory.
• February 10th 2008, 07:49 PM
Jhevon
Quote:
Originally Posted by ThePerfectHacker
It has connections to number theory but it is certainly not a number theory problem. "If the zeta function is analytically extended to $\mathbb{C}\setminus \{0\}$ then the (non-trivial) zeros have real part equal to 1/2". That is exactly what the hypothesis says. Look at the words: zeta function, analytic continuation, $\mathbb{C}$, real part. All of those are words from complex analysis, not from number theory.
yes. i know the theorem. just curious, what doors would it open if the theorem is actually proven? are there theorems in existence that are like "if the Riemann Hypothesis were true, then ..." like there were with the Takamuri conjecture? (i'm almost certain i spelt that wrong)
i have to read up on how Riemann came up with the hypothesis...
• February 10th 2008, 08:00 PM
ThePerfectHacker
Quote:
Originally Posted by Jhevon
yes. i know the theorem. just curious, what doors would it open if the theorem is actually proven? are there theorems in existence that are like "if the Riemann Hypothesis were true, then ..." like there were with the Takamuri conjecture? (i'm almost certain i spelt that wrong)
You mean Taniyama-Shimura conjecture (now called Taniyama-Shimura-Wiles theorem (Bow)). There is a lot of work on a concept introduced by Lejuenne Dirichlet (Bow) called L-series. In fact, one of the millenium problems, the Birch-Swinnerton-Dyer conjecture is based a little bit on the Dirichlet L-series. The problem is some of the results are conjectural because they need the Riemann hypothesis, but I am no expert on this so I am not entirely sure.
Quote:
i have to read up on how Riemann came up with the hypothesis...
How? All the history I know is that Riemann (Bow) wrote a short paper "On the number of primes less than a given magnitude", one of his few he ever wrote, but he had to introduce his new zeta function. And then he investigated its zeros. If I remember, Riemann was able to solve for some of the zeros himself, those were found scrippled around in his notes somewhere (I cannot confirm the origin of this last sentence, it needs reference).
• February 10th 2008, 08:08 PM
tukeywilliams
Ramanujan tried to prove the Riemann Hypothesis right?
• February 10th 2008, 08:13 PM
ThePerfectHacker
Quote:
Originally Posted by tukeywilliams
Ramanujan tried to prove the Riemann Hypothesis right?
I do not know. I would guess that no. Because Ramanujan was not very knowledgable. And when I say knowledgable I do not mean he was stupid, I mean he did not know very advanced concepts in math. This is coming from Hardy, who was the advisor of Ramanajun, he explains that Ramanajun was unfamilar with many advanced concepts in complex analysis. Since the Riemann hypothesis is very complicated problem I assume Ramanajun could not understand it to the level that he could solve it. However, Hardy did a breakthrough work by proving there are infinitely many zeros on the critical line.
• February 11th 2008, 01:23 PM
Aryth
Just recently... de Branges from Purdue claimed he had a proof. But, of course, he was wrong and later apologized. Another statement of the hypothesis though, is this:
$\sigma(n) \leq H_n + ln(H_n)e^{H_n}, \ for \ all \ n \ \in \ \mathbb{N}$
$\sigma(n)$ is the sum of the positive divisors of n.
$H_n$ is the nth harmonic number.
It's an elementary equivalent to the Riemann Hypothesis.
I don't know nearly enough math to begin solving it, but I am quite fascinated by the problem.
• February 11th 2008, 01:36 PM
galactus
Here's an interesting article.
K2Crypt
• February 11th 2008, 01:40 PM
wingless
Quote:
Originally Posted by ThePerfectHacker
You might be excited. But I am not. :(. It is a wonderful problem, I wish to solve it, not anybody else.
If I don't prove that it's impossible to prove the Riemann Hypothesis (like I can), why not ;p
• February 11th 2008, 06:06 PM
ThePerfectHacker
The nice thing about this problem is that it is not as difficult to understand as the other millenium problems.
• February 12th 2008, 04:06 PM
Skinner
If they proved the riemann hypothesis, wouldn't they also be able to prove loads of number theory conjectures (or disprove them)?
• February 12th 2008, 06:59 PM
ThePerfectHacker
Quote:
Originally Posted by Skinner
If they proved the riemann hypothesis, wouldn't they also be able to prove loads of number theory conjectures (or disprove them)?
You should read the previous posts (#7).
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http://math.stackexchange.com/questions/287067/y-8-cos8x-32y-with-y0-1-initial-value-problem-diffeq | $y'=(8\cos8x)/(3+2y)$ with $y(0)=-1$ Initial Value Problem: DiffEq
Im told to find the explicit form $y(x)$ from the given differential equation and its initial value. Then find where the solution $x=?$ attain a maximum. What I did was:
$3+2ydy=8\cos 8xdx$ integrated both sides
$3y+y^2=\sin8x+c$
Im stuck because I can't figure out how to get $y$ by itself.
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3 Answers
HINT: Use the quadratic formula, (Pretend the expression in $x$ is constant). Also, at the point you are at now, you can solve for the constant $c$, using your initial conditions.
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To find where the function achieves its maximum, just put $y'=0$ in the differential equation
$$y'=\frac{8\cos(8x)}{y+2} \implies 0=\frac{8\cos(8x)}{y+2} \implies \cos(8x)=0\implies \dots.$$
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I often prefer completing the square to using the explicit quadratic formula, even though they are equivalent.
Since $(y+b)^2 = y^2 + 2by + b^2$, to complete the square for $y^2+3y$, $2b=3$ or $b = 3/2$. To complete the square add $(3/2)^2 = 9/4$ to both sides getting
$$\sin 8x+c +9/4 = y^2+3y+9/4 = (y+3/2)^2.$$
You can now easily solve for $y$ and use the initial conditions.
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http://physics.stackexchange.com/questions/29985/would-the-rate-of-ascent-of-an-indestructible-balloon-increase-as-function-of-it | # Would the rate of ascent of an indestructible balloon increase as function of it's altitude?
Assume a balloon filled with Hydrogen, fitted with a perfect valve, and capable of enduring vacuum (that is to say, it would retain it's shape and so well insulated that the extremes of temperature at high altitudes and in space would have little effect) were to be launched.
As long as the balloon were in atmosphere it would ascend upwards (and also affected by various winds/currents, and gravity). As the balloon passed through increasingly rare atmosphere, would it rise faster?
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## 3 Answers
There are several things to note here. First, the atmosphere stops being buoyant around 100-150 km above the surface of Earth (I believe the mesopause is roughly where this occurs). Balloons simply cannot float beyond this point, no matter how fluffy (lower density per unit volume) they are. Second, past the tropopause (where the mixing of the lower atmosphere stops), one starts to see the atmosphere lose its heavier components such as argon and diatomic oxygen (as well as trace molecules like carbon dioxide) to lighter molecules like atomic oxygen and nitrogen. This will have some affect on the rise of the balloon.
Third, you can get a more consistent rise, if you replace the rigid balloon of your question with a zero-pressure balloon, filled with a less dense gas, which maintains the same pressure on both sides of the balloon. Your perfect balloon expands so that the displacement remains the same even though the atmosphere becomes less dense, crudely exponentially, as one rises.
This is the sort of balloon actually used in placing things in the upper atmosphere. For example, the traditional latex weather balloon, while not zero-pressure, comes close with a low pressure drop between inside and outside of the balloon. It can easily reach the upper stratosphere. Proper zero-pressure balloons (which are effectively bags of helium or hydrogen with a hole on the bottom) can get significantly higher. I think somewhere over 50km is the current altitude record which is in the lower mesosphere roughly.
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Why does the atmosphere stops being buoyant at 100-150 km ? A too long mean free path ? – Frédéric Grosshans Jun 13 '12 at 16:19
Updated the question to specify what gas I had in mind for the balloon – Everyone Jun 15 '12 at 7:36
Short Answer: NO
Long Answer: the rate of ascent of a balloon is based on his bouyancy (the link mention a fluid, in this case the air), the bouyancy is independent is the gravity field, with density being the only variable.
$$m=\rho_fV_{\mathrm{disp}}$$ reference: wikipedia
with the law of ideal gas the can obtanin the relationship with the presure, and the altitude
$$\rho=\frac{p}{R_{\mathrm{specific}}T}$$ reference: other wikipedia
$$p=p_0\cdot\left(1-\frac{Lh}{T_0}\right)^{\frac{gM}{RL}}\approx p_0\exp\left(-\frac{gMh}{RT_0}\right)$$reference: another wikipedia
this result in the following exponential graph:
Referenceengineering Toolbox
• Conclusion: As Altitude rise, the pressure lowers and proportionality the density, witch lowers the bouyancy net force.
Lastly (cite wikipedia)
"As a balloon rises it tends to increase in volume with reducing atmospheric pressure, but the balloon's cargo does not expand. The average density of the balloon decreases less , therefore, than that of the surrounding air. The balloon's buoyancy decreases because the weight of the displaced air is reduced. A rising balloon tends to stop rising. Similarly, a sinking balloon tends to stop sinking."
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1
– Alen Jun 12 '12 at 23:13
2
On a second though your question is refering to the velocity upwards, and since there is a net upward force as is explained before, in this case there is a increase in his ascent rate until the drag force equals the vertical forces, but this isn´t due to his altitude. – Alen Jun 12 '12 at 23:30
Forgive me for being so thick, would the volume increase in a rigid container too? – Everyone Jun 15 '12 at 7:41
@Everyone - no, of course not (instead, the differential pressure would rise), but that is not really the point. The buoyancy of the container decreases because of the decrease in the density of the surrounding air. An expanding balloon can counter that effect (by displacing more of a less dense fluid) to some extent. – Hanno Fietz Jan 3 at 22:07
I'll try to be more to the point than Karl Hallowell and a little more terse than Alen, but both answers should give you a picture already.
The exact answer to your question is, the rate of ascent (i. e., its velocity) increases as long as there is a net upward force acting on the balloon. This will only be the case for part of the balloon's journey.
However, that has to do only in part with its height. The balloon is accelerating upwards as long as it has positive buoyancy*, i. e. its velocity is rising as a function of time. At some point, the acceleration will be offset by the decceleration induced by the drag of the surrounding air and the velocity will remain constant for a while. Finally, the net buoyancy will decrease all the way down to zero, and drag will slow down the balloon until it stops.
Now, that last part actually is a function of the balloon's height, or, more precisely, a function of the density of the surrounding air (which decreases with height). The less dense the fluid displaced by a body, the lesser the buoyancy force exerted by the fluid. For a rigid container, that would be the whole story, for an expanding container (like a latex balloon), its increase in volume under lower ambient pressure counters that effect to some extent by displacing more of a now lighter fluid. As Karl has pointed out, this is an idealization, and a decrease occurs anyway.
Useful Wikipedia articles (as already given by Alen):
*note that I'm mixing gravity and buoyancy into "net buoyancy", which may be an oversimplification.
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http://mathoverflow.net/questions/72851/asymptotic-behaviour-of-the-solution-to-a-certain-pde | ## Asymptotic behaviour of the solution to a certain PDE
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
$\delta M=\beta(x-y)M_y+\mu x(y-1)M_x+\delta y$, where $M(x,y)=\sum_{n=0}^{\infty}{\sum_{k=0}^{\infty}{s_{n,k}x^ny^k}}$ is the generating function for a certain probability distribution ${s_{n,k}}$ (the exact formula for $s_{n,k}$ is unknown), and $\delta$, $\beta$, $\mu$ are all constants.
The problem comes out of a probability model, and data show that the distribution should have a finite mean and divergent moment ($M_{xx}(1,1)=\infty$). My question is that is there any way to get the asymptotic of $M_{xx}(x,1)$ when $x\rightarrow 1^{-}$? (by asymptotic I mean something like $M_{xx}(x,1)\sim C(1-x)^{-\zeta}$)
Thank you!
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http://math.stackexchange.com/questions/92867/testing-what-probability-model-to-use/92884 | # Testing what probability model to use
Suppose we have some random variable $X$ that ranges over some sample space $S$. We also have two probability models $F$ and $G$. Let $f(x)$ and $g(x)$ be the probability density functions for these distributions. Does the following quantity $$\log \frac{f(x)}{g(x)} = \log \frac{P(F|x)}{P(G|x)}- \log \frac{P(F)}{P(G)}$$ basically tell us how much more likely model $F$ is the true model than model $G$?
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What is the difference between $\log\frac{f(x)}{g(x)}$ and the log likelihood ratio (LLR)? The value of the LLR does not depend on the a priori probabilities $P(F)$ and $P(G)$ at all. In other words, there is cancellation on the right side of your displayed equation which makes the a priori probabilities disappear, and the LLR does not have any information about the a priori probabilities. – Dilip Sarwate Dec 20 '11 at 18:53
## 2 Answers
I am totally confused by the last comment made by Michael (the answer is ok, it is the link with logistic regression which went too far for me). Logistic regression is to be used when you have pairs of observations (X, Y) where Y is a binary variable (taking values in {0,1}) which is modeled as a Bernoulli variable $\mathcal{B}(p)$ the parameter of which depends of the value $x$ taken by $X$ : $\mathrm{logit}(p) = \beta_0 + \beta_1 x$. Here you don’t observe a variable Y taking value 1 when the model if F and 0 when it is G, the model is fixed beforehand and would not change along the observations... and you wouldn’t write $\mathrm{logit} P(F) = \beta_0 + \beta_1 x$. To me, this doesn‘t make any sense.
I will slightly reword Michael’s answer, just to give you some additionnal keywords. If you have a single observation $x$, then $f(x)$ is the likelihood of the model F, denote it by $L(F; x) = f(x)$, and $g(x)$ is the likelihood of the model G, denote it by $L(G;x) = g(x)$. As you stated, the likelihood ratio $L(F ; x)/L(G ;x) = f(x)/g(x)$ tells you how much the data support F against G.
If you have prior probabilities for F and G, denoted by P(F) and P(G) = 1 - P(F), then you can write posterior probabilities P(F|x) and P(G|x). You have $$P(F | x) = { L(F ; x) P(F) \over L(F;x) P(F) + L(G;x) P(G)},$$ $$P(G | x) = { L(G ; x) P(G) \over L(F;x) P(F) + L(G;x) P(G)},$$ and $${P(F | x) \over P(G |x) } = {P(F) \over P(G)} \times {L(F ; x) \over L(G ;x)}.$$ This is as Michael stated an application of Bayes’ theorem. The quantity P(F)/P(G) = P(F)/(1-P(F)) is called the odds of the model F. You can take the log of this last equality to get an additive statement, which is very usual (cf Michael’s answer). The quantity L(F;x)/L(G;x) is called a Bayes factor.
If you have $n$ independent observations $\mathbf{x} = x_1, \dots, x_n$, the same thing holds with $L(F ; \mathbf{x}) = \prod_i f(x_i)$ and $L(G ; \mathbf{x}) = \prod_i g(x_i)$.
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I've deleted the logistic regression remark. In logistic regression there are multiple observed instances of success or failure, each corresponding to an observed $x$ value. In this case, there is only one one binary choice, between $F$ and $G$, and its outcome is not observed, but the conditional distribution of $X$ given $F$ or $G$ is known, and the conditional probabilities of $F$ and $G$ are sought. – Michael Hardy Dec 20 '11 at 22:09
I was slightly puzzled by the notation, but I'm assuming that by $P(F)$ you mean the probability that $F$ is the right model, and $P(F\mid x)$ is the conditional probability that $F$ is the right model given the event $X=x$.
The identity you write is then a form of a special case of what is sometimes called Bayes' theorem.
If one assumes the right model must be either $F$ or $G$, but not both, then one can say that $$\operatorname{logit} P(F) = \log \frac{P(F)}{1-P(F)} = \log(\operatorname{odds}(F))$$ increases by $\log(f(x)/g(x))$ when it is observed that $X=x$.
So that quantity tells you by how much the logit of the probability increases when you observe the data.
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So $\text{logit} \ P(F) = \log \frac{P(F|x)}{P(G|x)} - \log \frac{f(x)}{g(x)}$. The first term plays the term of the intercept right? We need it to be of the form $\text{logit}\ P(F) = \beta_0+ \beta_{1}x$? – Tommy Dec 20 '11 at 3:26
Actually, $\operatorname{logit}(P(F\mid X=x))$ would be a function of $x$ that depends on what probability distributions $X$ has under $F$ and $G$. – Michael Hardy Dec 21 '11 at 15:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.952322781085968, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/239802/gauss-divergence-theorem | # Gauss divergence theorem
I'm trying to solve the following problem using the Gauss divergence theorem. I have to calculate the Flux through a sphere. The sphere is given as $$x^2 +y^2+z^2==4$$ where as the z is resticted from $$\sqrt{3} \; to \; 2$$ I determined the divergence to $$4z$$ I first tried to use spherical coordinates which resolve to:
$$\int _{\sqrt{3}}^2\int _0^{\pi /6}\int _0^{2\pi }4 *r*\text{Cos}[o]*r^2*\text{Sin}[o]dpdodr = \frac{7 \pi }{4}$$
To confirm this solution I tried to solve this problem using a regular volume integral. Which resolve to: $$\int _{\sqrt{3}}^2\int _{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}\int _{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}}(4 *z) dxdydz = \pi$$ I guess the error lies in my changed z variable. I replaced it with $$r*\text{Cos}[o]$$ Could you guys please take a look into it? 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": 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": 0.959648847579956, "perplexity_flag": "head"} |
http://cstheory.stackexchange.com/questions/tagged/chernoff-bound | # Tagged Questions
The chernoff-bound tag has no wiki summary.
2answers
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### Chernoff Bounds for settings with limited dependence
Could someone point me to a way of bounding the tail probability of sums of bernoulli variables each generated by the same distribution but the condition of independence is only partially satisfied. ...
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### Reverse Chernoff bound
Is there an reverse Chernoff bound which bounds that the tail probability is at least so much. i.e if $X_1,X_2,\ldots,X_n$ are independent binomial random variables and \$\mu=\mathbb{E}[\sum_{i=1}^n ...
1answer
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### Estimator for sum of independent and identically distributed (iid) variables
This is a repost of a question at math.stackexchange, but I was told by a reliable source that people around here might be able to help me, so I thought I'd give it a shot. Consider the Chernoff ...
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### An extension of Chernoff bound
I am looking for a reference (not a proof, that I can do) to the following extension of Chernoff. Let $X_1,..,X_n$ be Boolean random variables, not necessarily independent. Instead, it is guaranteed ...
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### Chernoff-type inequality for random variable with 3 outcomes
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### Chernoff-type Inequality for pair-wise independent random variables
Chernoff-type inequalities are used to show that the probability that a sum of independent random variables deviates significantly from its expected value is exponentially small in the expected value ...
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### Chernoff bound for weighted sums
Consider $X = \sum_i \lambda_i Y_i^2$, where lambda_i > 0 and Y_i is distributed as a standard normal. What kind of concentration bounds can one prove on X, as a function of the (fixed) coefficients ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "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": 0.9032220244407654, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?t=562548 | Physics Forums
## Please explain Dopler effect of light
How can there be any shifting of light frequency due to dopler effect? Frequency is just a fancy way of stating 'wave' or, as depicted on all explanitory diagrams, a sinusoidal wave. Each cycle of the wave represents one part of the total waves measured in a given amount of time which is the frequency of the light. To have a dopler effect the waves must either arrive farther apart or closer together than they were when they were originated. If the speed of light is universal how can there be a difference?
If the star were to blink on then off for one second you would have one second's worth of light coming from the point of origin. This would be a measurable, distinct and determinable length based on the speed of light. Within this length would be a fixed number of waves. The front of the main group of waves arrives at earth at the speed of light as does the back of the main group. Doesnt each wave within the group also arrive at that same speed? If this is true then the only other factor would be that the entire length of the main group of 1 second's worth of light is longer at arrival than when it left.
But if light speed is universal how can there be any lengthening?
tex
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See: Special Relativity
Quote by thetexan Frequency is just a fancy way of stating 'wave' or, as depicted on all explanitory diagrams, a sinusoidal wave.
While frequency is universally a proper of a wave, it is certainly not synonymous with 'wave'---and applies to any type of wave (sinusoidal or otherwise).
Frequency is the rate at which a wave oscillates, which is not the same thing as the wave's velocity. The simplest way to measure or describe frequency, is to count the number of wave-crests (or any repeating element) which pass by a given point, in a given amount of time. The frequency is related to the velocity by the equation $$v = \nu \lambda$$.
Quote by thetexan To have a dopler effect the waves must either arrive farther apart or closer together than they were when they were originated. If the speed of light is universal how can there be a difference?
Good point, and good question. While the speed of light is constant, the other 2 elements of the above equation are not: the frequency (the inverse of the 'period' or duration for a single oscillation), and the wavelength (length of the wave) are both not constant, because of special relativity (for example).
As the special relativity article (linked above) explains, an observer at rest will measure a different length and different time duration than another observer at a constant velocity with respect to the first. There are also other (similar) causes of doppler effects.
Quote by thetexan If the star were to blink on then off for one second you would have one second's worth of light coming from the point of origin. This would be a measurable, distinct and determinable length based on the speed of light. Within this length would be a fixed number of waves.
If the star 'measures' one second during which it is 'on', a moving observer would measure a longer duration of time. Similarly, if the star 'measures' the wave traveling 1 light year before it is measured; a moving observer would measure a shorter distance. These two effects conspire together to make the product of the frequency and the wavelength constant---the speed of light.
Recognitions: Gold Member Science Advisor Staff Emeritus Consider the following animations. The first shows waves of light emitted by a source stationary to the two dots. The waves expand out as circles from the source. The second shows what happens if the source moves. The waves still expand out as circles from the point of emission, but now, between the emission of the front of the wave to the back of the wave, the source has moved closer to the blue dot and further from the Red dot. Since the ends of the waves travel at c and the back of the wave has a shorter distance to travel, it takes a shorter time to reach the blue dot (and conversely, a longer time to reach the Red). Thus, to the blue observer, the wave is compressed and the for the red observer, it is stretched out. All without the speed of light changing at all.
## Please explain Dopler effect of light
Well, this is what I understand. Please tell me if Im correct.
A star, very far away, blinks on then off sending out a 7 cycle burst of light(the center image)...at the speed of light.
It arrives, after billions of years of travel, at the earth which, due to expansion, is receding from the star at a very great rate of speed. Due to the observer's speed the original burst is elongated, or appears so, which causes the entire 7 cycle burst to arrive at a lower frequency thus causing the redish color. If there were a situation where we were getting closer to a star the bottom image would be the case and the star would seem blueish. This stretching of the waves is what is refered to as Dopler. Is that right?
tex
Recognitions: Gold Member Wonderfully clear animations of the Doppler Effect! Thanks, Janus.
Recognitions:
Gold Member
Quote by thetexan A star, very far away, blinks on then off sending out a 7 cycle burst of light(the center image)...at the speed of light. It arrives, after billions of years of travel, at the earth which, due to expansion, is receding from the star at a very great rate of speed. Due to the observer's speed the original burst is elongated, or appears so, which causes the entire 7 cycle burst to arrive at a lower frequency thus causing the redish color. If there were a situation where we were getting closer to a star the bottom image would be the case and the star would seem blueish. This stretching of the waves is what is refered to as Dopler. Is that right? tex
Without going in depth, yes.
However, there is a critical piece or two that you are missing here. The standard picture of a light wave being 1 or 2 squiggly lines is NOT what light is. The up and downs of that picture represent a shift from positive to negative and back to positive of the lights electric and magnetic fields. Your statement, "A star emits a 7 cycle burst..." is not what happens. Light is only emitted or absorbed in discrete quanta called Photons. In spite of this, it still behaves extremely similar to any other wave, such as a water or sound wave, and does in fact exhibit doppler shift.
Another problem is that frequency shift due to the expansion of space is similar to, but not exactly like the Doppler Effect. (At least to my knowledge) As far as I know the expansion of space actually "stretches" out the light wave as it travels, resulting in a frequency shift. Whether there is any technical difference between the two is beyond me.
You should also look at these... Gravitational red-shift and... http://en.wikipedia.org/wiki/Gravitational_redshift
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| | Introductory Physics Homework | 5 | | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9525055885314941, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/100060/convexity-of-the-set-of-gradients-of-a-convex-function | # Convexity of the set of gradients of a convex function
If $\vec{y}_1$ and $\vec{y}_2$ are the gradients of a differentiable convex function $f(x)$ at points $\vec{x}_1$ and $\vec{x}_2$ does there exist a $\nabla f^{-1}( \alpha \vec{y}_1 + (1-\alpha)\vec{y}_2)$ in the domain of $f.$
The forward image of the domain mapped by the gradient is the domain of the Legendre conjugate so I think the above is true. If you could point me to a proof that will be very helpful.
I can see that the Legendre conjugate is convex because it is a supremum of affine functions, but am looking for a more direct intuition / construction for the forward image of the domain mapped by the gradient. What is bothering me is what happens when the supremum that is used to define the conjugate is not achieved. The value of the conjugate is set to $\infty$ but is there a way to show that the set over which the supremum is achieved is convex (even if $f$ is unbounded below).
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http://math.stackexchange.com/questions/180744/compute-int-frac-sinx-sinx-cosx-mathrm-dx/180746 | # Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$
I've got troubles in computing the below integral: $$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$$ I hope it can be expressed in elementary functions. I've tried simple substitution as $u=\sin(x)$ and $u=\cos(x)$ but not so effective the results were.
Any suggestions are welcomed. Thanks.
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I removed the display from your title. It's better to keep titles simple so that the main page does not break too much. – Mariano Suárez-Alvarez♦ Aug 9 '12 at 17:01
– clark Aug 9 '12 at 17:06
## 5 Answers
Let $I:=\int\frac{\cos x}{\cos x+\sin x}dx$ and $J:=\int\frac{\sin x}{\cos x+\sin x}dx$. Then $I+J=x+C$, and $$I-J=\int\frac{\cos x-\sin x}{\cos x+\sin x}dx=\int\frac{u'(x)}{u(x)}dx,$$ where $u(x)=\cos x+\sin x$. Now we can conclude.
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This is almost too cute for its own good :-) – Mariano Suárez-Alvarez♦ Aug 9 '12 at 17:06
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Nice proof. Simple and elegant. Thanks! – Michael Li Aug 9 '12 at 17:12
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You can make this even simpler by just writing the $\sin x$ in the numerator as ${1 \over 2}(\sin x + \cos x) - {1 \over 2}(\cos x - \sin x)$. – Zarrax Aug 9 '12 at 17:12
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@Zarrax, that magical step only helps in hiding what is going on :-) – Mariano Suárez-Alvarez♦ Aug 9 '12 at 18:09
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– Erick Wong Aug 9 '12 at 18:29
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Hint: $\sqrt{2}\sin(x+\pi/4)=\sin x +\cos x$, then substitute $x+\pi/4=z$
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Thanks. This is a new formula for me. The pity is a number of constants appears meanwhile. – Michael Li Aug 9 '12 at 17:30
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$\sin(A+B)=\sin A \cos B+\cos A \sin B$ actually. – Taxi Driver Aug 9 '12 at 17:32
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Very nice, at least if we can regard $\int \csc t \,dt$ as a "book" integral. – André Nicolas Aug 9 '12 at 18:57
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@Michael This formula generalizes to $$A\cos x+B\sin x=\sqrt{A^{2}+B^{2}}\sin \left( x+\arctan \frac{A}{B}\right) ,$$ which is very useful in (electrical) engineering. – Américo Tavares Aug 9 '12 at 20:05
You can do this without thinking: use the Weierstrass substitution to reduce the integral to a rational function, and integrate that as usual.
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We can write the integrand as $$\begin{equation*} \frac{1}{1+\cot x} \end{equation*}$$ and use the substitution $u=\cot x$. Since $du=-\left( 1+u^{2}\right) dx$ we reduce it to a rational function
$$\begin{equation*} I:=\int \frac{\sin x}{\sin x+\cos x}dx=-\int \frac{1}{\left( 1+u\right) \left( u^{2}+1\right) }\,du. \end{equation*}$$
By expanding into partial fractions and using the identities
$$\begin{eqnarray*} \cot ^{2}x+1 &=&\csc ^{2}x \\ \arctan \left( \cot x\right) &=&\frac{\pi }{2}-x \\ \frac{\csc x}{1+\cot x} &=&\frac{1}{\sin x+\cos x} \end{eqnarray*}$$
we get
$$\begin{eqnarray*} I &=&-\frac{1}{2}\int \frac{1}{1+u}-\frac{u-1}{u^{2}+1}\,du \\ &=&-\frac{1}{2}\ln \left\vert 1+u\right\vert +\frac{1}{4}\ln \left( u^{2}+1\right) -\frac{1}{2}\arctan u +C\\ &=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left( \cot ^{2}x+1\right) -\frac{1}{2}\arctan \left( \cot x\right) +C \\ &=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left( \csc ^{2}x\right) +\frac{1}{2}x+\text{ Constant} \\ &=&\frac{1}{2}x-\frac{1}{2}\ln \left\vert \sin x+\cos x\right\vert +\text{ Constant.} \end{eqnarray*}$$
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Write the numerator (here $\sin x$) as a linear combination of the denominator and the derivative of the denominator: $$A(\sin x+ \cos x) + B( \cos x- \sin x) = \sin x$$ Solve for $A$ and $B$ and split the fraction accordingly. Integrating give a linear term and an $\ln$
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http://physics.stackexchange.com/questions/2523/is-there-a-limit-to-loudness/2525 | # Is there a limit to loudness?
Is there any reason to believe that any measure of loudness (e.g. sound pressure) might have an upper boundary, similar to upper limit (c) of the speed of mass?
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## 4 Answers
Let's talk about the sound waves in the air first. Physically they are longitudinal waves of pressure. The bunch of air in one place will get compressed (in comparison with equilibrium state) and after that will expand, compressing the adjacent air and so on the wave propagates. These (single frequency) waves are essentially described by three numbers: the speed of propagation (called speed of sound; it depends on the type of the material and the temperature), the frequency of oscillation (determining the pitch), and the amplitude of oscillation.
It is the amplitude of oscillations which determines the loudness. So you are essentially asking whether there is any limit on amplitude of compression. Well, of course. At high enough pressures, the air would freeze even at normal temperatures. So this is the limit for air. Similar thing would happen with liquids: at certain pressures they would condense into solids. You could continue with phases of matter in this way and applying higher and higher pressures you would eventually end up with a black hole. That would be an ultimate limit.
But of course, in reality the limit is set by our engineering capabilities and I doubt it's possible to create sound waves that would be able to freeze air.
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You've completely missed the effects of adiabatic compressional heating. You won't liquify the air by compression, its temperature rises with pressure, so it will remain gaseous, and the soundwave transitions into a shockwave. So it becomes an issue of symantics, is a shockwave a highamplitude soundwave, or is it something different? Of course the local soundspeed scales as the squareroot of temperature, so as the compression ratio of the shock increases its velocity increases as well. – Omega Centauri Jan 5 '11 at 17:10
Of course you can liquify it, but it's true that it will heat somewhat (depending on precise process used). I had intended to discuss also temperature dependence and other phenomena but apparently I forgot to mention it for some reason. – Marek Jan 5 '11 at 18:31
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I really doubt it will liquify. Adiabatic heating is stronger than you imagine. For a perfect gas, T is proportional to density to the 2/3rds power, air has higher heat capacity, so at gamma =7/5ths it only increases at the .4 power of density. That still means a 300x increase in density implies a 10x increase in temperature. Can we get liquid phase on that curve, or will we end up much hotter than the triple point? These pressures are readily reachable via high explosives. – Omega Centauri Jan 6 '11 at 5:31
@Omega: I guess you meant critical point? Sure if the temperature got bigger than that there would no longer be a distinction between gas and liquid but the gas/liquid should still condense to a solid. But I have to admit, it's not clear to me whether at those pressures and temperatures other processes (like nuclear and eventually GR) wouldn't take over. – Marek Jan 6 '11 at 10:34
Yes - there is a sound pressure limit for undistorted sound. Over that limit we have a shock wave. It depends on the environmental pressure, but there is a theoretical limit to loudness which you can find here.
The limit is basically equal to the pressure.
Theoretical limit for undistorted sound at 1 atmosphere environmental pressure 101,325 Pa ~194.094 dB
The lower limit of audibility is therefore defined as 0 dB, but the upper limit is not as clearly defined. While 1 atm (191 dB) is the largest pressure variation an undistorted sound wave can have in Earth's atmosphere, larger sound waves can be present in other atmospheres, or on Earth in the form of shock waves.
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Well, the short answer is: there is, when the hydrodynamic approximation (that fluid is composed of small "fluid particles" i which real particles move in the reference frame of the "fluid particle" like in stationary fluid) breaks.
The upper bound can be approximated with wave amplitude equal to ambient pressure, so that the pressure is going down to 0 in wave minimas (this plus minus corresponds to cavitation); yet this corresponds to loudness of $\infty$dB.
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+1. @mbq what is "weave pressure"? – user346 Jan 5 '11 at 13:49
@space_cadet wave amplitude, which is pressure... But I'll fix it. – mbq♦ Jan 5 '11 at 15:03
lol. you meant "wave". doh. sorry, it was very early in the morning where I live! – user346 Jan 5 '11 at 21:13
"yet this corresponds to loudness of ∞dB." No, it's 194 dB SPL. – endolith May 8 at 3:59
The only practical limit to sound-pressure might be when the medium were to be compressed into a black hole. Although I do not know about the sound-propagating features of black holes. Long before that the medium would come apart, f.e. into plasma.
After all, you compare two things: an upper limit of a speed with a power. How much power can you put into a particle to speed it up with c in mind?
PS: I'd rather not guess what kind of sound-generator would be necessary. One might ask the band "Disaster Area" (https://secure.wikimedia.org/wikipedia/en/wiki/List_of_minor_The_Hitchhiker%27s_Guide_to_the_Galaxy_characters#Hotblack_Desiato)
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You'll reach a vacuum in the negative direction long before you reach a black hole in the positive direction. – endolith May 8 at 4:00 | {"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": 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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.932852566242218, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/84510/irreducibility-of-induced-representation-over-arbitrary-field/84518 | ## Irreducibility of Induced Representation over arbitrary field
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In representation theory of finite groups, there is Mackey test for irreducibility of an induced representation (Serre- Linear Representations of Finite Groups - #7.4). The author has stated it for field $\mathbb{C}$. Is it true for representations over arbitrary field whose characteristic is zero or co-prime to the order of group ?
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A good reference for this seems to be D. Bump's Lie Groups (chapter 34 and further). Bump comments on the dependence on the ground field. – Igor Rivin Dec 29 2011 at 13:31
## 3 Answers
I think so. The key property used in the proof is $$Res^G_H Ind^G_H V=\bigoplus_{s\in H\setminus G/H} Ind^H_{H_s}V_s,$$ where $V_s$ is the twist of $V$ by $s$, and $H_s=sHs^{-1}\cap H$, and its proof is more or less manifestly characteristic-free. Other than that, instead of characters and scalar products $\langle \chi_U,\chi_W\rangle$ you can use the actual spaces of intertwiners $Hom_G(U,W)$, and finally instead of saying that irreducibility of $W$ means the intertwining number $\dim Hom_G(W,W)$ is equal to 1 (which is true over an algebraically closed field of characteristic zero), you can say that irreducibility of $W$ means that $Hom_G(W,W)$ is a division algebra (which is true in the semisimple case, so under the assumption you want to make).
Let me, following a comment of Darij Grinberg, make the last step a bit more clear. There is a natural homomorphism of algebras $Hom_H(V,V)\to Hom_G(Ind^G_HV, Ind^G_HV)$. The induced representation $Ind^G_HV$ is irreducible if and only if this map is an isomorphism of algebras. However, to conclude that it is sufficient to show that it is an isomorphism of vector spaces, which we do without trouble via the Frobenius reciprocity.
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We get $\Hom_G\left(V,V\right) = \bigoplus\limits_{s\in H\backslash G / H} \Hom_H\left(\rho,\rho^s\right)$ as vector spaces. I am pretty sure this doesn't say anything about their ring structure, if the RHS has any at all; how do you see that $\Hom_G\left(V,V\right)$ is a division algebra iff $\Hom_H\left(\rho,\rho^s\right)=0$ for all $s\neq e$? – darij grinberg Dec 29 2011 at 13:36
I added a bit of a clarification. It's a kinda cheap trick with applying forgetful functors which I tend to like a lot. – Vladimir Dotsenko Dec 29 2011 at 13:56
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The isomorphism of vector spaces $Hom_H(V,V)\rightarrow Hom_G(Ind^G_H V, Ind^G_H V)$ will hold if $V$ is irreducible and $Res^H_{H\cap H_s} V$ and $Res^{H_s}_{H\cap H_s}V_s$ ($s\in G\setminus H$) do not share an irreducible component; what about converse? – joseph Jan 11 2012 at 4:23
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You are right, the converse is actually not clear at all. Probably without any assumptions on a field you get only an if, not iff, statement. – Vladimir Dotsenko Jan 11 2012 at 9:20
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(after one more minute of thinking) The natural 4d irreducible representation of the quaternionic group is induced from the nontrivial irreducible representation of the normal subgroup $\{1,-1\}$! – Vladimir Dotsenko Jan 11 2012 at 9:23
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To reinforce what Vladimir says, I'd replace his "I think so" by "Yes". The basic paper by Mackey in Amer. J. Math. 73 (1951) is retrievable online through JSTOR and might be worth looking at. His initial discussion of finite groups is done over arbitrary fields of characteristic 0, intended to generalize the work of Frobenius while providing background for the study of locally compact groups. It's only a convenience to assume that the ground field is `$\mathbb{C}$` (done by Serre in part because his audience for the early lectures consisted of people with applied interests in character theory).
By now there are multiple textbook sources (including Serre, Bump, Isaacs, Dornhoff, Grove, etc.), each with different coverage and emphases. Probably the most comprehensive treatment of induction (along with tensor products and intertwining maps) is found in Part I, Section 10, of the large treatise Methods of Representation Theory by Curtis & Reiner, which goes beyond their pioneering 1962 book. At first they can work over arbitrary commutative rings, while discussing only the general module results; this flexibility is needed later to pass to J.A. Green's work on modular representations.
But in order to get an applicable irreduciblity criterion for induced characters in (10.25), they take a subfield of `$\mathbb{C}$` which is a splitting field for the given finite group and all its subgroups. Even this kind of assumption can be weakened, since you only need a field of characteristic 0 containing sufficiently many roots of unity to ensure absolute irreducibility of the characters of the various irreducible representations involved.
If you really want to work with fields of prime characteristic, it's also a standard principle that the character theory works the same way provided only that the characteristic of the field doesn't divide the group order. To deal with absolute irreducibility of characters, you still need the splitting field assumption.
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As you have pointed out, for splitting fields, the theorem is true; no doubt about it. But I am not sure, whether it holds for non-splitting field. Is the answer "Exactly YES"? – joseph Jan 11 2012 at 4:26
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Virtually all of the classical results on characters for finite groups (Frobenius, Schur, Mackey, etc.) are developed over a splitting field for the group and sometimes for all its subgroups too when induction is involved. It's usually a subtle question to sort out further questions about irreducibility vs. absolute irreducibility. The field of complex numbers is convenient for exposition but of course much too large. In any case, it seems unrealistic to expect clean general statements for an arbitrary field (of characteristic not dividing the group order). – Jim Humphreys Jan 11 2012 at 16:04
The inclusion ${\rm End}_H (V)\longrightarrow {\rm End}_G ({\rm Ind}_H^G V)$ is much clearer at the level of Hecke algebras. If ${\mathcal H}(G,V)$ denotes the spherical Hecke algebra attached to $V$ (which is known to be isomorphic to ${\rm End}_G ({\rm Ind}_H^G V)$), then ${\rm End}_H (V)$ naturally identifies with the subalgebra of functions in the Hecke algebra with support $H$ : to $\psi$ you attach the function with support $H$ given by $h\mapsto \pi (h)\circ \psi =\psi \circ \pi (h)$. Then one has just to observe that the hypothesis of Mackey's criterion exactly says that functions in ${\mathcal H}(G,V)$ have support in $H$ !
For all of this you need to assume that the order of $G$ is prime to the caracteristics of the field of coefficients.
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http://math.stackexchange.com/questions/88581/how-do-i-prove-the-following-inequality-involving-exponential-functions | How do I prove the following inequality involving exponential functions
Let $R>2$ be a real number. Then, for any $n\geq 1$, it holds that
$$\frac{ \exp(n R)+ 1}{\exp( n R) - \exp(\pi R/2)} \leq \exp( R/n^2).$$
How do I prove this?
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For $n=1$ the inequality: $$e^R+1\leq e^R(e^R-e^{\pi R/2})$$ is false, because $e^{(1+\pi/2)R}\geq e^{2R}-e^R-1$ for $R\geq 0$. – Pacciu Dec 5 '11 at 16:19
1 Answer
The inequality as stated: $$\frac{ \exp(n R)+ 1}{\exp( n R) - \exp(\pi R/2)} \leq \exp( R/n^2)$$ can not be true for all $R>2$ and $n \geq 1$. Indeed, choose $n=2$, then for all $2 < r < r_\ast$, the inequality does not hold, where $r_\ast$ is the unique positive root of $$e^{2 r}-e^{9 r/4}+\exp\left({\frac{\pi r}{2}+\frac{r}{4}}\right)+1$$ approximately equal $r_\ast \approx 2.118953$.
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http://math.stackexchange.com/questions/228754/prove-rmcolb-subseteq-rmnulla/228758 | # Prove : $\rm{col}(B) \subseteq \rm{null}(A)$
Let $A$ be an $m\times n$ matrix and $B$ be a $n\times m$ matrix.
Prove that:
$$AB = 0 \iff \rm{col}(B) \subseteq \rm{null}(A)$$
Here is the problem in my textbook. I just have a simple solution if $B$ is transpose of $A$. But at general case, I don't have result yet.
Thanks :)
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## 3 Answers
I'll assume your field of scalars is $\mathbb{R}$ (not that it matters).
\begin{align*} &AB = 0 \\ \Leftrightarrow & (AB)x = 0 \quad \forall \, x \in \mathbb{R}^m \\ \Leftrightarrow & A (Bx) = 0 \quad \forall \, x \in \mathbb{R}^m \\ \Leftrightarrow & A y = 0 \quad \forall \, y \in \textbf{col}(B) \\ \Leftrightarrow & \textbf{col}(B) \subset \textbf{null}(A). \end{align*}
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Hints:
1. If $AB=0$ and $v \in \mbox{col}(B)$ then $v=Bx$ for some $x$, so $Av=???$.
2. If $\mbox{col}(B) \subseteq \mbox{null}(A)$ then for all $x$ we have $Bx \in ???$, which means that $ABx=???$.
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… 3. ??? 4. PROFIT! (Sorry, couldn't resist.) – K. Stm. Nov 4 '12 at 9:11
You can split a matrix product like $AB$ according to the columns of the right factor $B$: each column of $AB$ is equal to $A$ applied to the corresponding column of $B$ (this is how matrix multiplication works; you can also split the product according to the rows of $A$, but that is not helpful here).
So saying $AB=0$ means precisely that for each column $c$ of $B$ you have $Ac=0$, or $c\in\operatorname{null}A$. So you have to prove the column space of $B$ is contained in $\operatorname{null}A$ if and only if each individual column of $B$ lies in this same space $\operatorname{null}A$. You should be able to prove that yourself.
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http://crypto.stackexchange.com/questions/767/how-to-fairly-select-a-random-number-for-a-game-without-trusting-a-third-party?answertab=votes | # How to fairly select a random number for a game without trusting a third party?
Several people are playing a game with random events and require a way to produce a random number. (Such as dice rolls or a lottery.)
Can this be done such that each player has the power to be reasonably sure that the random number was fairly selected, without having to trust anyone else?
This question stems from a discussion in comments to my answer on the Bitcoin SE. I came up with a protocol but someone else spotted a flaw. This site seems more appropriate.
Please feel free to make use of my answer to that question.
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## 4 Answers
The answer you posted is actually correct (more or less, see below): have each participant commit to their random number $r_i$ by publishing, e.g., $\mathcal{H}(r_i)$ in the first round. And then in the second round, each participant opens the commitment by publishing $r_i$ and everyone checks that it matches the committed value by hashing it. The final random number is the XOR of each $r_i$.
The commenter there suggested an attack if there is a collision. However the definition of a secure hash function is that its collision-resistant. For a good modern hash function like SHA256, that is the case.
There are two subtleties.
The first is that your scheme only works for large random number, large enough that an attacker couldn't try every possible value for $r_i$, hash it, and see if it matches. For committing to small values, you need a randomized commitment function. Good randomized commitments are based on discrete logarithms but you could construct one (with reasonable safeness) using a hash function by generating a large random value $a_i$ and then computing your commitment as $\mathcal{H}(r_i||a_i)$. To open, reveal both $r_i$ and $a_i$. To change $r_i$, you'd have to find a different $a_i$ so that the hash is the same: this is hard based on collision resistance. The bitlength of both $r_i$ and $a_i$ must be predetermined so it is clear from $r_i||a_i$ where to split the value into $r_i$ and $a_i$ (otherwise you could open it to different values by simply splitting in a different spot).
The second subtlety is that the last participant to reveal his value will learn the accumulated random number before he reveals his number (he can see everyone else's number and he knows his own). This may cause him to abort if he doesn't like the result. You can do more complicated things to provide fairness, but generally if any participant refuses to open their commitment, the protocol should be restarted from the beginning with that participant excluded.
Also read this answer from @D.W. on a different question where (s)he explains the same protocol.
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Actually, there's a subtlety here. Using the XOR of the $r_i$s is not secure. An attack: Suppose there are two participants, Alice and Bob. Alice is honest, picks $r_1$ randomly, and publishes $h_1 = H(r_1)$. Bob is malicious and rather than picking his own random number and hashing it, just copies Alice's hash: i.e., Bob publishes $h_1$. In the second round, Alice reveals $r_1$. Bob reveals the same thing. Now the final random number will be all-zeros. This allows a malicious Bob to force the final "random" number to be all-zeros, so it isn't random after all. – D.W. Sep 26 '11 at 4:20
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P.S. The fix is, instead of XORing the $r_i$s, you can hash them. e.g., the final random number is $H(r_1 || r_2 || ...)$, not $r_1 \oplus r_2 \oplus ...$. – D.W. Sep 26 '11 at 4:21
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Good point! The attack is worst if you use a non-malleable commitment scheme (I had suggested Pedersen commitments; edited out now) because then you can disguise the fact that you copied another commitment (or formed a related one). – PulpSpy Sep 26 '11 at 14:07
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Another fix is to require the number of random bits be higher than (say) 256, and disallow repeated digests. Or better, to force the 2nd publisher of a repeated digest to reveal first (better because it allows you conclusively identify cheaters). – Fixee Sep 27 '11 at 1:18
The problem pointed out by JGWeissman on Bitcoin.SE is only an issue if the hash function lacks collision resistance. Admittedly, collision resistance is one of the strongest properties usually demanded of hash functions, and collision attacks have been found for some hash functions commonly used in the past, such as MD5, but still, any secure cryptographic hash function should be collision resistant.
In fact, a simpler scheme should work:
1. Each player $i$ chooses a random number $x_i \in \{0,\dotsc,n-1\}$ and a random $k$-bit string $s_i$, for some reasonably large $k$ (say, $k=128$).
2. Each player announces $h_i = H(i \,||\, s_i \,||\, x_i)$, where $H$ is a collision-resistant hash function and $||$ denotes concatenation.
3. After all players have announced their $h_i$, they now reveal their $s_i$ and $x_i$. Each player checks that the $h_j$ value announced by every other player $j$ in step 2 matches that player's $s_j$ and $x_j$.
4. The final random number $x$ is calculated as the sum of all the $x_i$ values modulo $n$.
The value $h_i$ announced in step 2 is called a commitment. By publishing $h_i$, player $i$ commits to a specific $x_i$ without actually revealing $x_i$ itself. In this way, each player $i$ can be sure that the other players chose their random numbers without knowing $x_i$.
The purpose of the salt $s_i$ is simply to prevent brute force cracking of $h_i$ by trying each possible value of $x_i$ in turn. The player identifier $i$ is included in the hash to prevent a cheating player from copying another player's commitment. I assume that the identifiers are unique and known to all players in advance, or at least before step 3. (Another way to prevent such replay attacks would be to abort the protocol in step 3 if any two commitments are identical, which, due to the inclusion of the salt, should virtually never occur by chance.)
Actually, if the length of $h_i$ is substantially greater than $k\;\log_2(n^2 - n)$, this protocol may be secure even if $H$ is not collision-resistant, simply because there may be no exploitable collisions to be found. In particular, assuming that the outputs of $H$ are uniformly distributed over the set of $m$-bit strings, with $2^m \gg 2^k$ (and $2^m \gg n^2$), the probability of there being at least one exploitable collision is approximately
$$1-\left(1-\frac{1}{2^m}\right)^{2^k\tfrac{n^2-n}{2}} \approx 1-\exp \left(-2^{k-m}\frac{n^2-n}{2}\right) \approx 2^{k-m}\frac{n^2-n}{2}.$$
However, keep in mind that you also don't want to make $k$ too small, since then the salt becomes ineffective. Increasing $m$ should always be safe, though.
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Actually, there's a subtlety here. Using the sum of the $r_i$s is not secure. Using a variant of the attack I described on @PulpSpy's protocol, a malicious participant Bob can force the sum to be even, which means that the final "random" number is not actually random. The fix is the same as I described in my comment on Pulpspy's answer: you should hash all of the $x_i$ values (rather than summing them). – D.W. Sep 26 '11 at 4:23
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@D.W.: Very good point. I should note, though, that there are at least two other ways to fix the protocol: either abort/restart the protocol if $h_i = h_j$ for any $i \ne j$ (which should be astronomically unlikely if nobody cheats), or include some kind of player identifier in the input to $H$ (so that anyone replaying another player's commitment will be caught in step 3). I'll edit the answer to implement the latter option, since it's good practice anyway. – Ilmari Karonen Sep 26 '11 at 17:39
Say $m$ persons meet physically and want to draw a positive integer $x$ less than $n$.
Each person $j$ secretly selects a positive integer $x_j$ less than $n$, writes it down on a piece of paper, and fold it to hide her choice. The $m$ folded pieces of paper are brought together then publicly unfolded, revealing the $x_j$. The outcome of the protocol is $x = (\sum x_j)\mod n$. To illustrate with $n=10$, the $m$ digits $x_j$ are added and the rightmost digit of the sum is $x$.
If any of the person participating has chosen her $x_j$ at random, then $x$ is random. Hence, no participant can complain that $x$ is not random.
It is possible to replace the pieces of papers with electronic exchanges of messages: in addition of $x_j$, each person draws a random $r_j$, and sends $H_j = Hash(x_j \| r_j)$ to all the others, which constitutes her commitment on the choice of $x_j$. When everyone has all the commitments, everyone reveals her $x_j$ and $r_j$. The outcome is computed as above, and the $H_j$ can be verified by anyone from $x_j$ and $r_j$. If the exchange of messages is reliable, no one can complain.
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The protocol with pieces of paper is secure, but the protocol with electronic exchanges of messages is not. There's a subtle attack here, which I've described in my comments on PulpSpy and Ilmari's answers: the attacker can copy someone else's commitments and thereby influence/bias the final random number. If you want to use electronic messages, you need to hash everyone's $x_j$ values, rather than summing them. – D.W. Sep 26 '11 at 4:25
@D.W.: You are of course right! As pointed out in other answers, this can be fixed by defining $H_j=Hash(x_j∥r_j∥j)$. – fgrieu Sep 27 '11 at 6:15
Two variations.
If one of the "players" is going to generate the number for others to guess (lottery) then you could do it this way:
1. Collect a token from each participant (a nonce, a random number, a favorite color, etc.)
2. Generate or select the secret number.
3. Hash the generated number with the tokens.
4. Publish the hash and the tokens.
5. Allow each other player to guess. After the guesses are complete and published you publish the secret number.
6. Everyone can verify it is the actual number by repeating the hash.
If you have multiple "players" who are each equal in the guessing then do it this way:
1. Collect a token from each players (a nonce, a random number, a favorite color, etc.)
2. Publish the tokens to all players.
3. Hash the tokens.
4. Use the hash to see a PRNG or RNG to generate the random numbers.
5. Everyone can repeat the process to generate the same random numbers. Of course as soon as you publish the tokens then all future random numbers are known.
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http://mathhelpforum.com/differential-geometry/131805-question-jordan-content.html | # Thread:
1. ## A question on Jordan content
If A is Jordan-measurable with Jordan content c(A), $B\subseteq A$ and is of content zero. Prove that A-B is also Jordan-measurable with the Jordan-content equal to c(A). Thanks!
2. I finally proved it. First we prove A-B is Jordan-measurable. This is equivalent to showing that $\partial(A-B)$ has content zero and it suffices to show $\partial(A-B)\subseteq\partial A\cup\partial B$, here $\partial$ means boundary. Let ${\bf x}\in\partial(A-B)$, then ${\bf x}\in\overline{A-B}\subseteq\bar A$. If ${\bf x}\notin{\rm Int} A$, here Int means interior, then ${\bf x}\in\bar A-{\rm Int} A=\partial A$ and we get what we want. If ${\bf x}\in {\rm Int}A$, we try to prove that ${\bf x}\in\partial B$. Supposing ${\bf x}\notin\bar B$, this means there is a ball $B({\bf x},\epsilon_1)$ not intersecting B. Because ${\bf x}\in {\rm Int}A$, there is an another ball $B({\bf x},\epsilon_2)\subseteq A$. Take $\epsilon=\min(\epsilon_1,\epsilon_2)$, then for the ball $B({\bf x},\epsilon)$, any element of it satisfies both $\in A$ and $\notin B$, that is, $\in A-B$, so ${\bf x}\in {\rm Int}(A-B)$, but this contradicts the supposition ${\bf x}\in\partial(A-B)$, so ${\bf x}\in\bar B$. On the other hand, c(B)=0 implies B contains no interior point, otherwise, there would be a ball in square metric, or a subinterval, lying completely in Int B, then $\underline J(P,B)$ would >0 and this would in turn leads to c(B)>0, a contradiction of the hypothesis. Therefore ${\bf x}\notin {\rm Int} B$, so ${\bf x}\in\bar B-{\rm Int} B=\partial B$, as desired.
Secondly let me prove c(A-B)=0. For this purpose, I introduce two lemmas:
1)If S is of Jordan-content zero, f is defined and bounded on S, then $\int_S f$ exists and equals 0.
Proof. By hypothesis, S is of measure 0, so discontinuities of f in S, as a subset of a content 0 set, is content 0 and therefore measure 0, so f is integrable on S $^{[1]}$. Since c(S)=0, there is a partition P some finite subintervals (denote it as A) of which cover $\bar S$ and the sum of measures of these subintervals < any given $\epsilon>0$. Let $g({\bf x})=\left\{ \begin{array}{ll}<br /> f({\bf x}) & {\bf x}\in S \\ <br /> 0 & {\bf x}\in I-S \\ <br /> \end{array} \right.$ where I is the closed interval containing S, then $\int_S f=\int_I g$. Assume $|f|\leq M$, then $M_k=\sup\{f({\bf x})|{\bf x}{\rm\;in\;subinterval\;}I_k\}\leq M$ and $m_k=\inf\{f({\bf x})|{\bf x}{\rm\;in\;subinterval\;}I_k\}\geq -M$. So $U(P,g)=\sum\limits_A M_k\mu(I_k)\leq M\sum\limits_A\mu(I_k)\leq M\epsilon$. So $\bar\int\leq 0$ by arbitraryness of $\epsilon$. Similar argument can be applied to get $\underline\int\geq 0$. So $\underline\int=\bar\int=0$, that is, $\int_I g({\bf x})d{\bf x}=\int_S f({\bf x})d{\bf x}=0$.
2)If S is Jordan-measurable, then $\int_S 1$ exists and equals c(S). This is an extension of [2] where S is required to be compact.
Proof. It is easy to see $\int_S 1$ exists since $c(\partial S)=0$. Let $\chi_S$ be characteristic function of S and P any partition. Since those subintervas $I_k$ contributing to $U(P,\chi_S)$ contain elements of S and therefore elements of $\bar S$, they must also contribute to $\bar J(P,S)$. But the converse is not true, since it is likely that some $I_k$ contains elements of $\bar S$ but no element of S. So $U(P,\chi_S)\leq\bar J(P,S)$. Since these two numbers are both indexed by P, we can impose inf on both sides and get $\bar\int_I\chi_S\leq\bar c(S)$. On the other hand, for those subintervals $I_k$ contributing to $\underline J(P,S)$, since they are all contained in S, $\chi_S$ assumes 1 throughout and $m_k=1$ as a result. So these subintervals also contribute to $L(P,\chi_S)$. But the converse is not true since there may be $I_k$ which is contained completely in S but contains points not belonging to Int S, in this situation, $m_k$ is still 1, but $I_k$ is not counted in $\underline J(P,S)$. So $\underline J(P,S)\leq L(P,\chi_S)$. Taking sup on both sides, we obtain $\underline c(S)\leq\underline\int_I\chi_S$. From above we have $\underline c(S)\leq\underline\int_I\chi_S\leq\bar\int_I\chi_S \leq\bar c(S)$, but $\underline c(S)=\bar c(S)$, so $\int_I\chi_S=\int_S 1=c(S)$.
Now the work becomes easy. Since A is Jordan-measurable, by Lemma 2) we have $\int_A 1$ exists. Note that A-B and B are both Jordan-measurable, by the additive property of the Riemann integral $^{[3]}$, $\int_A 1=\int_B 1+\int_{A-B} 1$, in which $\int_A 1=c(A)$ by Lemma 2), $\int_B 1=0$ by Lemma 1) and $\int_{A-B} 1=c(A-B)$. The conclusion is thus immediate.
If there is any mistakes or there is any other good proof, please kindly point out. PS: Just as it is greatly beneficial to study general topology before studying mathematical analysis, it is greatly beneficial to have some knowledge of measure theory before studying Integral Theory.
[1]Th 14.11 of Apostol's "Mathematical Analysis"
[2]Th 14.12 of Apostol's "Mathematical Analysis"
[3]Th 14.13 of Apostol's "Mathematical Analysis" | {"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": 72, "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": 0.9415428638458252, "perplexity_flag": "head"} |
http://stats.stackexchange.com/questions/30162/sampling-for-imbalanced-data-in-regression | # Sampling for Imbalanced Data in Regression
There have been good questions on handling imbalanced data in the classification context, but I am wondering what people do to sample for regression.
Say the problem domain is very sensitive to the sign but only somewhat sensitive to the magnitude of the target. However the magnitude is important enough that the model should be regression (continuous target) not classification (positive vs. negative classes). And say in this problem domain that any set of training data will have 10x more negative than positive targets.
In this scenario, I might oversample the positive-target examples to match the count of negative-target examples, and then train a model to differentiate the two cases. Obviously the training approach does badly on imbalanced data, so I need to do sampling of some sort. What would be a decent way to "undo" this oversampling when making predictions? Perhaps translating by the (negative) mean or median of the target of the natural training data?
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## 1 Answer
Imbalance is not necessarily a problem, but how you get there can be. It is unsound to base your sampling strategy on the target variable. Because this variable incorporates the randomness in your regression model, if you sample based on this you will have big problems doing any kind of inference. I doubt it is possible to "undo" those problems.
You can legitimately over- or under-sample based on the predictor variables. In this case, provided you carefully check that the model assumptions seem valid (eg homoscedasticity one that springs to mind as important in this situation, if you have an "ordinary" regression with the usuals assumptions), I don't think you need to undo the oversampling when predicting. Your case would now be similar to an analyst who has designed an experiment explicitly to have a balanced range of the predictor variables.
Edit - addition - expansion on why it is bad to sample based on Y
In fitting the standard regression model $y=Xb+e$ the $e$ is expected to be normally distributed, have a mean of zero, and be independent and identically distributed. If you choose your sample based on the value of the y (which includes a contribution of $e$ as well as of $Xb$) the e will no longer have a mean of zero or be identically distributed. For example, low values of y which might include very low values of e might be less likely to be selected. This ruins any inference based on the usual means of fitting such models. Corrections can be made similar to those made in econometrics for fitting truncated models, but they are a pain and require additional assumptions, and should only be employed whenm there is no alternative.
Consider the extreme illustration below. If you truncate your data at an arbitrary value for the response variable, you introduce very significant biases. If you truncate it for an explanatory variable, there is not necessarily a problem. You see that the green line, based on a subset chosen because of their predictor values, is very close to the true fitted line; this cannot be said of the blue line, based only on the blue points.
This extends to the less severe case of under or oversampling (because truncation can be seen as undersampling taken to its logical extreme).
````# generate data
x <- rnorm(100)
y <- 3 + 2*x + rnorm(100)
# demonstrate
plot(x,y, bty="l")
abline(v=0, col="grey70")
abline(h=4, col="grey70")
abline(3,2, col=1)
abline(lm(y~x), col=2)
abline(lm(y[x>0] ~ x[x>0]), col=3)
abline(lm(y[y>4] ~ x[y>4]), col=4)
points(x[y>4], y[y>4], pch=19, col=4)
points(x[x>0], y[x>0], pch=1, cex=1.5, col=3)
legend(-2.5,8, legend=c("True line", "Fitted - all data", "Fitted - subset based on x",
"Fitted - subset based on y"), lty=1, col=1:4, bty="n")
````
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Thanks for the answer, Peter. Would you please elaborate on what you mean by "Because this variable incorporates the randomness in your regression model"? The target is an observable in the environment, so do you means measurement error? – someben Jun 10 '12 at 1:56
Yes it is well known that it is inappropriate to sample on the values of the dependent variable. I will see if I can dig up some references. I don't quite get what the OP is doing though, so I can't say much more. – Andy W Jun 10 '12 at 2:14
1
@someben - I've elaborated and added an example. It is well described in the regression literature that you cannot sample based on the dependent variable. This should apply to other models too. A sample that is "unbalanced" is a different sort of thing and is not a problem; unless you have deliberately created it by an unjustifiable sampling strategy. It's not the balance or lack of it that is the problem, but how you get your data. – Peter Ellis Jun 10 '12 at 2:57
2
@someben, no I don't think it makes any difference. The issue is more fundamental than that. – Peter Ellis Jun 10 '12 at 4:17
1
– Andy W Jun 10 '12 at 13:24
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": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "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": 0.9220724701881409, "perplexity_flag": "middle"} |
http://www.haarp.alaska.edu/haarp/mbcalc.html | Search the Site
```
```
The Antenna System
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Page updated January 20, 2008
### Calculation of the Expected Lunar Echo Receive Signal Strength
When a radio signal is transmitted from an earth station, scattered from the moon's surface and received back on the earth's surface, the propagation can be described by the bistatic radar equation where the received signal $P$r is given by :
in this equation:
$P$t is the transmitted power in Watts. The transmitter power of the HAARP facility is 3.6 X $106$ Watts.
$G$t is the transmitter antenna gain. For the frequency used in this experiment, the HAARP transmitting antenna has a gain of 735 (28.6 dB).
$G$r is the gain of the receiving antenna over isotropic. For a dipole antenna, the gain is approximately 1.5.
σ is the scattering cross section of the moon. We use the cross section for an isotropic sphere modified by the moon's albedo which at these frequencies is approximately 0.15. The value of σ used for this calculation is 1.4 x $1012$ square meters.
λ is the wavelength of the frequency used. For this experiment, the wavelength is 42 meters.
R is the distance from the earth to the moon. For this calculation, we have taken the distance from the moon to the transmitter to be approximately the same as the distance from the moon to the receiving antenna and equal to approximately 370,000 km.
#### Results of the calculation
Using the equation above, the power delivered at the terminals of a resonant 40 meter dipole antenna is 2.5 X $10-13$ Watts or about -96 dBm. This is equivalent to about 4.4 microvolts across 70 ohms. Typically, the S meters in commercial communication receivers are calibrated such that S-9 corresponds to an input signal of 50 microvolts and each succesively lower S unit represents a decrease of about 5-6 dB in voltage. Therefore, the calculated (predicted) level of the received lunar echo is equivalent to about an S-5 signal strength.
#### Other considerations
The calculation above is for an ideal case. Here are some other potential sources of loss that would affect the signal level for the lunar echo.
1. The height of the antenna used for reception. Antennas are generally mounted as high as possible to maximize the radiation and reception of signals at low elevation angles. However, a horizontal antenna mounted 0.5 wavelength above ground (21 meters at this frequency) would have an overhead null in its pattern. Thus, an antenna that performs well for long distance communication, may not work as well when the (lunar echo) signal is arriving from a high angle.
2. The HAARP antenna beam was repositioned to track the moon during the experiment in 3 minute steps. Although the experiment took place at the low point of the solar cycle and during evening hours when ionization could be expected to be low, there could still be some ionospheric absorption and refraction possible. Absorption would cause some small decrease in the signal both going toward the moon and returning. (The HAARP riometer indicated no absorption duuring the 19 January test and a small amount during the 20 January test.) Ionospheric refraction, if present, could have affected the exact positioning of the maximum in the HAARP transmitted signal as it passed beyond the ionosphere.
3. A planar antenna such as the HAARP array, loses gain as the beam is positioned off the vertical. For this experiment, the moon was at a zenith angle of about 35 degrees as seen from HAARP. The additional loss in gain of a few dB was not included in the calculation.
4. The moon's distance from the earth varied during the test. This resulted in a small but measureable Doppler shift on the lunar echo. The amount of this frequency shift was as much as 7 Hz positive or negative, depending on the specific time during the experiment. Several of the reports that have been sent to us have reported this Doppler shift.
#### References
Reference Data for Radio Engineers, Howard W. Sams & Co., Indianapolis, IN, 1977. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9436861276626587, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/35348/the-difference-between-a-sequential-space-and-a-space-with-countable-tightness/105551 | ## The difference between a sequential space and a space with countable tightness
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi, I have recently encountered these two definitions of a sequential space and a space of countable tightness. And I seem to have difficulty understanding what is the difference between these two definitions. For example, I know that the space of ultrafilters over ,say, R or N is not weakly Frechet Urysohn so it should not be sequential. But how can one show it directly from the definition? Also, Does these spaces have countble tightness? Thanks!
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## 4 Answers
All three notions, "countably tight," "sequential," and "Frechet-Urysohn," say that each point $p$ in the closure of a set $A$ can be "approached in some countable way" by points from $A$. The difference is in the "countable ways." The strongest of the three, Frechet-Urysohn, requires $p$ to be the limit of a sequence of points in $A$. "Sequential" allows iteration of this: Take the set of limits of sequences of points in $A$; then take limits of sequences of such points; then take limits ... ; eventually you get $p$. (More formally, let `$A_0=A$`, let `$A_{\alpha+1}$` be the set of limits of sequences from `$A_\alpha$`, and for limit ordinals let `$A_\lambda$` be the union of all `$A_\alpha$`'s for $\alpha<\lambda$. Then $p$ should be in `$A_\alpha$` for some $\alpha$. The $\alpha$ here can always be taken to be countable, but that's the best bound you can get.) Finally, "countably tight" only requires $p$ to be in the closure of some countable subset of $A$; that can happen even if there are no convergent sequences of points from $A$ (except of course the eventually constant sequences).
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)
• any metric space will be Frechet-Urysohn (choose `$x_n$` in `$A$` within `$1/n$` of `$p$`); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of `$A$` and `$U_n$` where `$\{U_n\}_n$` is a countable base at the limit point);
• a sequential but not Frechet-Urysohn space is given by taking `$((\omega+1)\times\omega)\cup\{*\}$` where each copy of $\omega+1$ has the usual topology and a base for `$*$` consists of sets `$A_{m,n}=\{(m,n)|m>M,n>N_m\}$` for `$M,N_m\in\omega$` (ie cofinitely many elements of cofinitely many fibers) - then `$*$` is in the closure of `$\omega\times\omega$` but is not the limit of any sequence of points in `$\omega\times\omega$`; however, it is the limit of the sequence `$x_n=(\omega,n)$` and each `$x_n$` is the limit of a sequence of points from `$\omega\times\omega$`;
• a countably-tight but not sequential space could be given by taking `$(\omega\times\omega)\cup\{*\}$` where all points `$(m,n)$` are open and a base for `$*$` consist of sets `$A_{M,N}=\{(m,n)|m>M, n>N_m\}$` for `$M,N_m\in\omega$` - this space is trivially countably tight (it's countable) but is not sequential: `$*$` is not the limit of any sequence in `$\omega\times\omega$` (since we can always exclude any putative sequence converging to `$*$`);
• finally a non-countably-tight space is given by `$\omega_1+1$` with the usual topology: `$\omega_1$` (as a point) is in the closure of `$\omega_1$` (as a set) but any countable subset of `$\omega_1$` has bounded (countable) closure.
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Just a partial answer.
For $\beta \mathbb{N}$ (the set of all ultrafilters on $\mathbb{N}$ with the Stone topology) it is not hard to see that a sequence converges iff it is eventually constant. Hence any subset of $\beta \mathbb{N}$ is sequentially open -- and of course, $\beta \mathbb{N}$ is not discrete, so it cannot be sequential. Similarly, for the ultrafilters on $\mathbb{R}$.
If I recall correctly, this 'trivial sequential convergence' holds in all extremally disconnected spaces -- this should be an exercise in the book 'Rings of continuous functions' by Gillman and Jerison (there is also a PDF/TeX-file with all exercise solutions freely available on the web somewhere).
Also, I could be wrong, but I think $\beta \mathbb{N}$ is not countably tight since its remainder is not (since there exist weak P-points). Maybe somebody else can confirm or reject.
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Does anyone here know how to show that sequential implies pytkeev? or, where can one find a proof of it?
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This should be a separate question rather than an answer. Wikipedia claims to give a proof at en.wikipedia.org/wiki/Pytkeev_space but it involves terminology that I'm not familiar with ($\pi$-net) and too busy to look up just now, so I don't guarantee that the proof is correct. – Andreas Blass Aug 26 at 22:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "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": 0.9513883590698242, "perplexity_flag": "head"} |
http://www.nag.com/numeric/CL/nagdoc_cl23/html/S/s18eec.html | # NAG Library Function Documentnag_bessel_i_nu (s18eec)
## 1 Purpose
nag_bessel_i_nu (s18eec) returns the value of the modified Bessel function ${I}_{\nu /4}\left(x\right)$ for real $x>0$.
## 2 Specification
#include <nag.h>
#include <nags.h>
double nag_bessel_i_nu (double x, Integer nu, NagError *fail)
## 3 Description
nag_bessel_i_nu (s18eec) evaluates an approximation to the modified Bessel function of the first kind ${I}_{\nu /4}\left(x\right)$, where the order $\nu =-3,-2,-1,1,2$ or $3$ and $x$ is real and positive. For positive orders it may also be called with $x=0$, since ${I}_{\nu /4}\left(0\right)=0$ when $\nu >0$. For negative orders the formula
$I - ν / 4 x = I ν/4 x + 2 π sin πν 4 K ν/4 x$
is used.
## 4 References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
## 5 Arguments
1: x – doubleInput
On entry: the argument $x$ of the function.
Constraints:
• if ${\mathbf{nu}}<0$, ${\mathbf{x}}>0.0$;
• if ${\mathbf{nu}}>0$, ${\mathbf{x}}\ge 0.0$.
2: nu – IntegerInput
On entry: the argument $\nu $ of the function.
Constraint: $1\le \mathrm{abs}\left({\mathbf{nu}}\right)\le 3$.
3: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_INT
On entry, ${\mathbf{nu}}=〈\mathit{\text{value}}〉$.
Constraint: $1\le \mathrm{abs}\left({\mathbf{nu}}\right)\le 3$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_OVERFLOW_LIKELY
The evaluation has been abandoned due to the likelihood of overflow. The result is returned as zero.
NE_REAL_INT
On entry, ${\mathbf{x}}=〈\mathit{\text{value}}〉$, ${\mathbf{nu}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{x}}>0.0$ when ${\mathbf{nu}}<0$.
On entry, ${\mathbf{x}}=〈\mathit{\text{value}}〉$, ${\mathbf{nu}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{x}}\ge 0.0$ when ${\mathbf{nu}}>0$.
NE_TERMINATION_FAILURE
The evaluation has been abandoned due to failure to satisfy the termination condition. The result is returned as zero.
NE_TOTAL_PRECISION_LOSS
The evaluation has been abandoned due to total loss of precision. The result is returned as zero.
NW_SOME_PRECISION_LOSS
The evaluation has been completed but some precision has been lost.
## 7 Accuracy
All constants in the underlying functions are specified to approximately 18 digits of precision. If $t$ denotes the number of digits of precision in the floating point arithmetic being used, then clearly the maximum number of correct digits in the results obtained is limited by $p=\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(t,18\right)$. Because of errors in argument reduction when computing elementary functions inside the underlying functions, the actual number of correct digits is limited, in general, by $p-s$, where $s\approx \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,\left|{\mathrm{log}}_{10}x\right|\right)$ represents the number of digits lost due to the argument reduction. Thus the larger the value of $x$, the less the precision in the result.
None.
## 9 Example
The example program reads values of the arguments $x$ and $\nu $ from a file, evaluates the function and prints the results.
### 9.1 Program Text
Program Text (s18eece.c)
### 9.2 Program Data
Program Data (s18eece.d)
### 9.3 Program Results
Program Results (s18eece.r) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 34, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.6587115526199341, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/1119/randomized-oblivious-transfer?answertab=oldest | # Randomized Oblivious Transfer
If we define Oblivious Transfer as following: Alice inputs $(x_0,x_1) \in F^2$, where $F$ is a field, and Bob inputs $b\in\{0,1\}$, then Alice gets a dummy output(for which she knows nothing about b), and Bob gets $x_b$. Now if we use a Random functionality R, which requires no input but output random bits $(z_0,z_1)\in F^2$ to Alice and $(c,z_c)$ to Bob, where $c \in \{0,1\}$ is also a random bit. How to design a protocol, so that it will realize OT securely with access to R only once at the beginning of the protocol? I am thinking that the protocol should be as following: First, Alice inputs $(x_0,x_1)$, then she gets $(x_0 \oplus z_0,x_1 \oplus z_1)$ as output, thus learning nothing. Now Bob gets $(c,z_c)$, treat $c$ as his input $b$, then he can recover $x_c$ from $(x_0 \oplus z_0,x_1 \oplus z_1)$.
I don't know if it will work, or maybe I am in the wrong direction?
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## 1 Answer
No, as written, your protocol doesn't work -- the problem is that Bob is supposed to be allowed to choose $b$, your protocol selects a random one for him.
However, it is close -- here is a modification that I believe does work:
First, suppose Alice has her values $(x_0, x_1)$, and Bob has his bit $b$.
They run their Random functionality R, and so Alice gets the values $(z_0, z_1)$, and Bob gets the values $(c, z_c)$.
Now, Bob sends the Alice the bit $e$ defined as $e = b \oplus c$; Alice sends back a pair of values $(y_0, y_1)$ defined as $(y_0, y_1) = (x_0 \oplus z_e, x_1 \oplus z_{1 \oplus e})$.
Bob then computes $z_c \oplus y_b = x_b$.
I claim that Alice gets no information about the value of $b$, and that Bob gets no information about the value of $(x_{1 \oplus b})$, and thus this is a valid Oblivious Transfer function.
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Thanks, phocho, so if I do the modification for random OT as following:Alice gets random bits $(S_A, P_A)$ and Bob gets random bits $(S_B, P_B)$ with one condition that is $S_A+S_B=P_A P_B$, how should the protocol be modified to realize OT? I think with that condition $(S_A,P_A)$ and $(S_B,P_B)$ can be mapped into $(z_0,z_1)$ and $(c,z_c)$ one by one, but I couldn't really write the mapping out explicitly. Do you have any idea about that? – huyichen Nov 2 '11 at 22:03
1
Well, the obvious approach would be to notice that if Bob gets $P_B=0$, he knows the value $S_A$, and if Bob gets $P_B=1$, he knows the value $S_A+P_A$, and so you could attempt to run the above protocol with $z_0=S_A$ and $z_1=S_A+P_A$. It doesn't appear that the information that Alice gets about $S_B$ or $S_B+P_B$ allows her to deduce $b$. However, there is a subtle problem with that; the above protocol implicitly assumes that the values $z_0$, $z_1$ and $c$ are both uniformly and independently distributed. It's not clear that's true with this random generator. – poncho Nov 3 '11 at 3:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "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": 0.9515094757080078, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/265821/does-linearity-decompose-down-convex-sums | # Does linearity decompose down convex sums?
I'm doing some convex optimisation where I'm minimising sum function $f(x) = \sum g_i(x)$, where the $g$'s are convex (and hence so is $f$) and the sum is finite.
In doing so it turns out that $f$ is a linear function between any two (global) optima. Apparently $f$ being linear here implies that the $g$'s are also linear, why is this so?
-
If f is linear, then f is convex as well as concave. – TenaliRaman Dec 27 '12 at 9:45
## 1 Answer
The second derivative of the linear function $f$ is $0 = \sum_{i} g_i''(x)$ where $g_i'' \geq 0$ for every $i$.
We can do the same thing without regularity assumption: if $f$ is defined on the interval $[a,b]$, then for every $t \in[0,1]$ we write $$0 = f(ta + (1-t)b) - tf(a)-(1-t)f(b) = \sum_{i} g_i(ta + (1-t)b) - tg_i(a)-(1-t)g_i(b).$$
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9278892874717712, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/43638/list | ## Return to Question
6 added 537 characters in body
Let $\lambda$ denote a partition of size $n$. Let $$d_{\lambda}= \text{number of distinct parts of } \lambda$$ $$o_{\lambda}= \text{number of odd parts of } \lambda$$ $$f_{\lambda}= \text{number of standard Young tableau of shape } \lambda$$ Given an involution $\pi \in S_{n}$, whose insertion tableau has shape $\lambda$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that : $$o_{\lambda^{t}}= \text{number of fixed points in the involution } \pi$$ $$\sum_{\lambda \vdash n} f_{\lambda}= \text{number of involutions in } S_{n}$$
In the aforementioned formulae, $\lambda^{t}$ refers to the conjugate of the partition $\lambda$. Now, some computations I have carried out for Kronecker products of two irreducible characters of $S_{n}$ revealed the following identity in a special case: $$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n}o_{\lambda}f_{\lambda}$$
Note that the right hand side actually counts the total number of fixed points in all involutions in $S_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S_{n}$.
Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper).
Thanks!
Edit: I had a look at Sagan, which I did not have handy last night and made a minor change in saying the number of fixed points in an involution $\pi \in S_{n}$ is the number of odd columns in the insertion tableau of $\pi$.
Edit(10/27):
I thought I should put down the idea that I had. But since I am not sure if this should count as an answer, I am putting it in the body of the question. Note that $$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n+1}f_{\lambda}-\sum_{\lambda \vdash n}f_{\lambda}$$ So all that remains to be shown is the nice fact that the total number of fixed points in all the involutions of $S_{n}$ is the difference between the number of involutions in $S_{n+1}$ and the number of involutions in $S_{n}$.
5 edited body
Let $\lambda$ denote a partition of size $n$. Let $$d_{\lambda}= \text{number of distinct parts of } \lambda$$ $$o_{\lambda}= \text{number of odd parts of } \lambda$$ $$f_{\lambda}= \text{number of standard Young tableau of shape } \lambda$$ Given an involution $\pi \in S_{n}$, whose insertion tableau has shape $\lambda$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that : $$o_{\lambda^{t}}= \text{number of fixed points in the involution } \pi$$ $$\sum_{\lambda \vdash n} f_{\lambda}= \text{number of involutions in } S_{n}$$
In the aforementioned formulae, $\lambda^{t}$ refers to the conjugate of the partition $\lambda$. Now, some computations I have carried out for Kronecker products of two irreducible characters of $S_{n}$ revealed the following identity in a special case: $$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n}o_{\lambda}f_{\lambda}$$
Note that the right hand side actually counts the total number of fixed points in all involutions in $S_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S_{n}$.
Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper).
Thanks!
Edit: I had a look at Sagan, which I did not have handy last night and made a minor change in the saying the number of fixed points in an involution $\pi \in S_{n}$ is the number of odd columns in the insertion tableau of $\pi$.
4 added 331 characters in body
Let $\lambda$ denote a partition of size $n$. Let $$d_{\lambda}= \text{number of distinct parts of } \lambda$$ $$o_{\lambda}= \text{number of odd parts of } \lambda$$ $$f_{\lambda}= \text{number of standard Young tableau of shape } \lambda$$ Given an involution $\pi \in S_{n}$, whose insertion tableau has shape $\lambda$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that : $$o_{\lambda}= o_{\lambda^{t}}= \text{number of fixed points in the involution } \pi$$ $$\sum_{\lambda \vdash n} f_{\lambda}= \text{number of involutions in } S_{n}$$
In the aforementioned formulae, $\lambda^{t}$ refers to the conjugate of the partition $\lambda$. Now, some computations I have carried out for Kronecker products of two irreducible characters of $S_{n}$ revealed the following identity in a special case: $$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n}o_{\lambda}f_{\lambda}$$
Note that the right hand side actually counts the total number of fixed points in all involutions in $S_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S_{n}$.
Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper).
Thanks!
Edit: I had a look at Sagan, which I did not have handy last night and made a minor change in the saying the number of fixed points in an involution $\pi \in S_{n}$ is the number of columns in the insertion tableau of $\pi$.
3 Corrected definition of $o_{\lambda}$
Let $\lambda$ denote a partition of size $n$. Let $$d_{\lambda}= \text{number of distinct parts of } \lambda$$ $$o_{\lambda}= \text{number of odd parts of } \lambda$$ $$f_{\lambda}= \text{number of standard Young tableau of shape } \lambda$$ Given an involution $\pi \in S_{n}$, whose insertion tableau has shape $\lambda$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that : $$o_{\lambda}= \text{number of fixed points in the involution } \pi$$ $$\sum_{\lambda \vdash n} f_{\lambda}= \text{number of involutions in } S_{n}$$
Now, some computations I have carried out for Kronecker products of two irreducible characters of $S_{n}$ revealed the following identity in a special case: $$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n}o_{\lambda}f_{\lambda}$$
Note that the right hand side actually counts the total number of fixed points in all involutions in $S_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S_{n}$.
Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper).
Thanks!
2 deleted 5 characters in body
Let $\lambda$ denote a partition of size $n$. Let $$d_{\lambda}= \text{number of distinct parts of } \lambda$$ $$o_{\lambda}= \text{number of distinct odd parts of } \lambda$$ $$f_{\lambda}= \text{number of standard Young tableau of shape } \lambda$$ Given an involution $\pi \in S_{n}$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that : $$o_{\lambda}= \text{number of fixed points in the involution } \pi$$ $$\sum_{\lambda \vdash n} f_{\lambda}= \text{number of involutions in } S_{n}$$
Now, some computations I have carried out for Kronecker products of two irreducible characters of $S_{n}$ revealed the following identity in a special case: $$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n}o_{\lambda}f_{\lambda}$$
Note that the right hand side actually counts the total number of fixed points in all involutions in $S_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S_{n}$.
Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper).
Thanks!
1
# A Distinct parts/Odd parts identity for standard Young tableaux
Let $\lambda$ denote a partition of size $n$. Let $$d_{\lambda}= \text{number of distinct parts of } \lambda$$ $$o_{\lambda}= \text{number of distinct parts of } \lambda$$ $$f_{\lambda}= \text{number of standard Young tableau of shape } \lambda$$ Given an involution $\pi \in S_{n}$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that : $$o_{\lambda}= \text{number of fixed points in the involution } \pi$$ $$\sum_{\lambda \vdash n} f_{\lambda}= \text{number of involutions in } S_{n}$$
Now, some computations I have carried out for Kronecker products of two irreducible characters of $S_{n}$ revealed the following identity in a special case: $$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n}o_{\lambda}f_{\lambda}$$
Note that the right hand side actually counts the total number of fixed points in all involutions in $S_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S_{n}$.
Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper).
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": 56, "mathjax_display_tex": 37, "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": 0.942167341709137, "perplexity_flag": "head"} |
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http://mathoverflow.net/questions/109687/is-it-true-that-the-orbit-space-of-a-free-finite-group-action-on-a-cw-complex-is/109763 | ## Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex?
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Suppose a finite group G acts freely and continuously on an n-dimensional CW-complex X. Then can we conclude that the orbit space of this action is still an n-dimensional CW-complex? (or homotopy equivalent to an n-dimensional CW-complex?) In particular, we do not assume G acts cellularly on X.
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Perhaps you can perform repeated subdivisions on the cell structure of $X$ to arrive at the cellular case? I'm not claiming this is always possible, but it may be. – Mark Grant Oct 15 at 14:10
3
Here is a suggestion for proving that $X/G$ is homotopy equivalent to a CW-complex in a special case when the complex $X$ is countable and locally finite. Any metrizable ANR is homotopy equivalent to a CW-complex. If the complex $X$ is countable and locally finite, then it is a metrizable separable ANR, and I suspect metrizability and separability are inherited by $X/G$. Now a metrizable separable space that is locally an ANR is globally an ANR, so $X/G$ would then be a metrizable ANR. – Igor Belegradek Oct 15 at 14:28
1
In seeing whether $X/G$ is homeomorphic to a CW-complex, even the case when $X$ is a smooth manifold is unclear. Indeed, if the $G$-action is nonsmoothable, then $X/G$ would only be a topological manifold, and in general it is unclear to me whether $X/G$ is homeomorphic to a CW-complex. (I think the existence of a CW structure on a topological manifold is unknown in dimension 4 and also for noncompact manifolds in higher dimensions, at least the proof in Kirby-Siebenmann's book is for compact case only). – Igor Belegradek Oct 15 at 14:35
I would see if their is some sort of Borel construction. At the very least, you can take the singular chains on $X$, make the group action free with a simplicial Borel construction, then take the geometric realization of that. After this process, you will be in possession of a CW complex with the right weak homotopy type. I suspect that if you began with a CW complex, you will have a homotopy equivalence. – Spice the Bird Oct 15 at 15:27
1
@Igor: I think you should write your comments as an answer. Note that if $X$ is metrizable, so is $X/G$ (by averaging the distance function on $X$ under the group action: Sum of distance functions is again a distance function). Local finiteness passes to the quotient, separability too. – Misha Oct 15 at 18:07
## 4 Answers
Lemma If $X$ is a countable locally finite CW-complex and $G$ acts freely and properly discontinuously on $X$, then $X/G$ is homotopy equivalent to a CW-complex.
Proof Any metrizable ANR is homotopy equivalent to a CW-complex (I am not sure who proved it first but see Theorem 3.6.1 here. Since $X$ is countable and locally finite, it is a metrizable separable ANR. As Misha remarks in comments averaging the metric over the group action implies that $X/G$ is metrizable. Also a countable dense subset of $X$ projects to a countable dense subset of $X/G$. Finally, if a metrizable separable space is locally ANR, it is an ANR (see Borsuk's "Theorey of Retracts", Corollary 10.4, Chapter IV). It follows that $X/G$ is a metrizable ANR as desired.
Remark In seeing whether $X/G$ is homeomorphic to a CW-complex, even the case when $X$ is a PL manifold is unclear. The difficulty is that it seems unknown which topological manifolds are homeomorphic to CW-complexes (Kirby-Siebenmann prove this for compact manifolds of dimension $\ge 6$ (or maybe $\ge 5$?, but certainly not $4$). So there might exist manifolds not homeomorphic to CW-complexes but whose finite covers are PL.
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I wonder when $G$ is finite and the CW-complex $X$ is of dimension $n$, can we choose the CW-complex homotopic to the orbit space $X/G$ to be $n$-dimensional too? – Li Yu Oct 16 at 2:48
@Igor: I do not think you need compactness for this. The point is that you can exhaust an open topological manifold $X$ by an increasing sequence of codimension $0$ compact submanifolds with boundary $X_i$. Let $B_i=\partial X_i$. Then, inductively, the handle structure extends from the collar $B_i\times I\subset X_{i+1}$ to the rest of $X_{i+1}$ (all the existence results are relative). Incidentally, the handle decomposition of Kirby-Siebenmann works in dimensions $\ge 6$; it is extended to dimension $5$ by Frank Quinn in "Ends of Maps-III". – Misha Oct 16 at 5:00
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The 3-sphere gives an example of an action with fixed points. If one takes the solid Alexander horned sphere, then Bing proved that its double is homeomorphic to the 3-sphere. So the quotient of the involution acting on $S^3$ is the solid Alexander horned sphere. However, the solid horned sphere is not homeomorphic to a CW complex. This follows from the answer to this question on the Alexander horned sphere. If the solid Alexander horned sphere were a CW complex, then one could attach the exterior 3-ball to get a CW structure on $S^3$ with the Alexander horned sphere being the boundary of the closure of a 3-cell, which is a contradiction to the other question.
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This action is not free, as requested, the subspace of fixed points is Alexander's horned sphere (the common boundary). – Fernando Muro Oct 15 at 21:10
Ian, does not your involution have fixed points? – Igor Belegradek Oct 15 at 21:11
I also missed the assumption that the action was supposed to be free. However, the question is still a good one without that assumption and Ian's answer is instructive so I hope he leaves it up. – Neil Strickland Oct 15 at 21:24
I missed the freeness assumption, but as Neil suggests I'll leave it up. – Agol Oct 15 at 21:46
If $G$ (finite or more generally discrete) acts cellularly on $X$, i.e.
• if $\sigma$ is an open cell of $X$ then $g\sigma$ is again an open cell in $X$ for all $g \in G$
• if $g \in G$ fixes an open cell $\sigma$ (i.e. $g\sigma=\sigma$), then it fixes $\sigma$ pointwise (i.e. $gx=x$ for all $x \in \sigma$)
then $X/G$ is a CW-complex. This follows from Prop. 1.15 and Ex. 1.17(2) of tom Dieck: Transformation Groups
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2
What I want to know is exactly the case when G does not acts cellularly! – Li Yu Oct 15 at 9:15
2
Yes, but when he answered this, you had not made that clear – David White Oct 15 at 13:38
1
I am very sorry. I thought the cellularly action case is easy. So I did not mention it at the beginning. But after your answer, I realize I should emphasize where the difficulty of the question lies. – Li Yu Oct 15 at 13:56
1
@David White: your comment seems unfair. If you look in the edits, the first version of the question did not assume the action was cellular; it said "Suppose a finite group G acts freely and continuously on a CW-complex X. Then can we conclude that the orbit space of this action is still a CW-complex? (or homotopy equivalent to a CW-complex?)". In my view this is clearly stated, and it was Ralph who misread the question; why blame the OP? – Igor Belegradek Oct 15 at 16:18
1
@Igor: 1) Right, the question in its original version makes no assumption on cellularity. Hence there are two case: a) cellular action b) non-cellular action. My answer obviously treats case a). So how do you conclude I misread the question ? 2) If one is only interested in case b) then it's good style to point it out, as mentioned by David. - Anyway, now it's clear and I'm curious about the answer of this interesting problem. – Ralph Oct 15 at 17:45
show 1 more comment
This is not really an answer, but a comment about an interesting special case. Suppose that $G$ acts smoothly on $S^2$. By averaging we can choose a $G$-invariant Riemannian metric. This gives $S^2$ a conformal structure, making it a Riemann surface. Any Riemann surface homeomorphic to $S^2$ is conformally equivalent to the standard Riemann sphere. Thus, we can reduce to the case where $G$ acts on $\mathbb{C}\cup\{\infty\}$ by conformal and anticonformal maps, which must have the form $z\mapsto (az+b)/(cz+d)$ or $z\mapsto (a\overline{z}+b)/(c\overline{z}+d)$. I think it even works out here that the quotient $(\mathbb{C}\cup\{\infty\})/G$ is always either a sphere or a disc. Thus, one cannot get any local pathology in this context. This contrasts with other settings where smooth functions can generate topological pathology: for example, any closed subset of $\mathbb{R}^n$, however fractal, can be expressed as the zero set of a smooth function $f\colon\mathbb{R}^n\to\mathbb{R}$.
Along somewhat similar lines, I think one can show that when $X$ is a one-dimensional CW complex with continuous action of a finite group $G$, then $X/G$ is again a one-dimensional CW complex (up to homeomorphism, not just homotopy equivalence).
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Neil: Even more, if $G$ acts topologically freely and properly discontinuously on a topological manifold $X$ of dimension $\ne 4$ then the quotient $X/G$ is homeomorphic to a CW-complex. The same applies if $X$ is merely a simplicial complex of dimension $\le 3$ and $G$ acts topologically. – Misha Oct 15 at 22:04
Misha: is your $X$ compact? If not, would you give a reference for the first claim, say in higher dimensions? – Igor Belegradek Oct 15 at 22:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 81, "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": 0.9333794116973877, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/40832/what-are-relativistic-and-radiative-effects-in-quantum-simulation?answertab=votes | # What are relativistic and radiative effects (in quantum simulation)?
I'm reading about Quantum Monte Carlo, and I see that some people are trying to calculate hydrogen and helium energies as accurately as possible.
QMC with Green's function or Diffusion QMC seem to be the best ways to converge on the "exact" solution to Schrodinger's equation.
However, if one wants to be very exact, then the Born-Oppenheimer approximation must be removed. A lot of papers mention that the results are still not exact enough, and must be corrected for relativistic and radiative effects.
I'm pretty sure I know what relativistic effects are -- the non-relative Schrodinger's equation cannot account for GR as particles approach the speed of light (or even small but measurable effects at lower speeds). But what are radiative effects?
And I would think you would include these two things into your QMC calculation instead of applying a post-simulation correction factor if you wanted to be ultra-precise (e.g. use the Dirac equation instead for relativistic effects). So why don't most researchers do this? Does it raise the calculation time by orders of magnitude for an additional 4th decimal place of accuracy?
Finally, is there anything at a "deeper" level than relativistic and radiative effects? In other words, if I left a supercomputer running for years to compute helium's energies without the BO approximation, and with relativistic and radiative effects included in the MC calculations, would this converge on the exact experimental values?
(Actually, I just thought of one such left-out factor -- gravity ... and might you have to simulate the quarks within the protons individually? Anything else?)
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## 1 Answer
Relativistic effects are those that disappear in the non-relativistic approximation $1/c\to 0$, usually small corrections to the non-relativistic approximate results that are proportional to $1/c^2$ or higher powers of the inverse speed of light.
Let me correct a typo: "cannot account for GR" should have read "cannot account for the special theory of relativity". When we talk about relativistic corrections, we always talk about the 1905 special theory of relativity, not about GR i.e. the 1915 general theory of relativity. Corrections that have something to do with general relativity are "gravitational" or "quantum gravitational" corrections and they're typically proportional to powers of Newton's constant $G$ which makes them even more negligible.
For example, the Hydrogen atom may be described by the [=special] relativistic Dirac equation which reduces to the Pauli equation, i.e. the Schrödinger equation with the spin, in the $c\to\infty$ limit. However, there are some relativistic corrections and they actually make the energy slightly depend on the angular momentum, too. This effect seen in the Dirac equation is both due to the corrections to the $p^2/2m$ formula for the kinetic energy as well as due to the spin-orbital coupling.
Radiative corrections are corrections due to the virtual particles that can only be seen in the language of quantum field theory. The hydrogen atom, for example, may emit a photon and reabsorb it: that is the Lamb shift. Or a particle may temporarily create a positron-electron pair. Those processes are expressed by Feynman diagrams and if they have loops in the middle, their loop diagrams and all effects due to Feynman diagrams with loops are know as radiative processes or corrections. Note that this goes beyond the simple Dirac equation.
Radiative corrections are typically smaller because of an extra factor such as the fine-structure constant $\alpha=1/137.036...$. Well, it's usually corrections like $1/2\pi\alpha$ so it's about 1,000 times smaller than the "main" term. So those things are small and even if you want to incorporate them, it doesn't matter at what stage you do so. You must only do it right.
Quantum field theory – which in principle contains all relativistic and radiative corrections – is enough to explain any observable lab experiment within any realistic error margin. If you care about things that can't be measured by current or realistic technology, you need the full theory of everything including the quantum gravity effects, i.e. string/M-theory.
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http://mathoverflow.net/questions/75753?sort=votes | ## Elements of unit modulus in ring generated by root of unity
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
When thinking of an apparently unrelated problem I stumbled upon the following question, which is certainly elementary to many readers of this site. Let $\omega_l=e^{i2\pi/l}$, and let $z\in Z[\omega_l]$ have $|z|=1$. Can we conclude that $z=\omega_l^k$ for some integer $k$? Thanks in advance for any answer!
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2
Here's a cheap negative answer: $e^{2\pi i/6}=-(e^{2\pi i/3})^2$. – Anthony Quas Sep 18 2011 at 14:53
## 1 Answer
A.Quas already noted in his comment that ${\bf Z}[\omega_l]$ might contain a $2l$-th root of unity (he gave $l=3$ but even $l=1$ works...). But that's the only possibility: we show that the only algebraic integers $z$ in ${\bf Q}(\omega_l)$ satisfying $|z| = 1$ are the roots of unity in ${\bf Q}(\omega_l)$. More generally, if $K_+$ is any totally real field, and $K$ is a totally imaginary quadratic extension of $K_+$, then the only algebraic integers in $K$ satisfying $|z|=1$ are the roots of unity in $K$.
This result is surely well-known/standard, but it's easier to recite or reconstruct a proof than to track down a reference. Let $d = [K_+:{\bf Q}]$, so that $2d = [K:{\bf Q}]$; for example $d = \varphi(l)/2$ when $K = {\bf Q}[\omega_l]$. Now $|z|=1$ iff $z$ is in the kernel of the map ${\rm Nm} : z \mapsto z \bar z$ from $K^*$ to $K_+^*$. In our setting $z$ is assumed to be an algebraic integer in the kernel, so it is in the unit group of [the algebraic integers in] $K$, with inverse $\bar z$. But by the Dirichlet unit theorem both $K$ and $K_+$ have unit groups of rank $d-1$. Moreover the image of ${\rm Nm}$ has rank at least $d-1$ because it contains the squares of all units in $K_+$. Hence its rank is exactly $d-1$, and the kernel of ${\rm Nm}$ has rank zero, and therefore consists only of roots of unity, as desired.
EDIT: Thanks to David Speyer for the link to his answer to the same question on Stackexchange (http://math.stackexchange.com/questions/39856), using an alternative route via Kronecker's theorem (an algebraic integer $z$ is a root of unity iff every Galois conjugate of $z$ has absolute value $1$ in ${\bf C}$) instead of the Dirichlet unit theorem. This method too works in the general "CM" setting of a totally imaginary quadratic extension of a totally real field.
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For another proof, see math.stackexchange.com/questions/39856/… – David Speyer Sep 18 2011 at 15:36
Thanks! This is exactly what I needed. MathOverflow is great, too! – Jean-Marc Schlenker Sep 18 2011 at 17:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 44, "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": 0.9339736104011536, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/65517/useful-topology-on-space-of-smooth-structures-on-mathbb-r4 | # Useful topology on space of smooth structures on $\mathbb R^4$?
Mathoverflow is intimidating, so I thought I'd ask here first (second). If I don't get any useful answers here in a few days, I'll ask there.
$Q_0$: Is there any use for a topology on the (continuum of) smooth structures on $\mathbb{R}^4$?
$Q_1$: If so, is a useful topology known?
The only reference I have found is a paper by K. Kuga, "A note on Lipschitz distance for smooth structures on noncompact manifolds", MR1117158 (92f:57025). He apparently shows that several obvious topologies are discrete. One might be able to metrize the maximal exotic $\mathbb{R}^4$ and use some variant of the Hausdorff-Gromov pseudometric, but that's such an obvious idea that I'd be surprised if it works, given that I haven't seen it.
Edit 2: The previous edit was edited into the comments below. Edit 3: I may as well clarify "is there any use?" in the OP as the zeroth question.
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1
Useful for what purposes -- what do you want to use this topology for? For example, does Kuga answer your question, and if not, why not? – Ryan Budney Sep 18 '11 at 18:17
@Ryan: Useful for anything, really. I do not know enough to know if there is a potential use for this, aside from having a vague feeling that since all 4-manifolds are modeled on $\mathbb{R}^4$, knowing more about the smooth structures on $\mathbb{R}^4$ should give information about 4-manifolds in general. (Perhaps I should rephrase my question, "$\exists$ use for a topology on the space of smooth structures on $\mathbb{R}^4$?") – Neal Sep 18 '11 at 20:33
All I know about Kuga's paper is the review on Mathscinet; the paper doesn't appear to be on JSTOR, according to Mathscinet no paper has referenced it, and the top Google hit is this very thread. – Neal Sep 18 '11 at 20:33
– Zev Chonoles♦ Sep 18 '11 at 21:11
@Zev: Thank you! I'm registering now. We'll see if this works! – Neal Sep 18 '11 at 22:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "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": 0.945847749710083, "perplexity_flag": "middle"} |
http://mathcountsnotes.blogspot.com/ | # mathcounts notes
The best math program for middle school students
## Wednesday, May 15, 2013
### This Week's Work : Week 12 -- for Inquisitive Young Mathletes
Link to the online timed test on questions you mostly got wrong or not fast enough.
(sent through e-mail)
Common Pythagorean Triples:
3, 4, 5 and its derivatives
5, 12, 13
8, 15, 17
7, 24, 25 (at least these for SAT I and II)
9, 40, 41, (the rest for state and Nationals, so we'll learn them later)
11, 60, 61
12, 35, 37
13, 84, 85
20, 21, 29
Shoe string method in finding the area of any polygon
Heron's formula in finding the area of a triangle.
Don't mix up the "s" with the other "S" of finding
the area of an equilateral triangle -- proof and formula (You can also use 30-60-90 special right
triangle to get that.)
or
the area of a regular hexagon
In Heron's case, "s" stands for half of the perimeter.
Besides, I've noticed most of the questions, when given the sides, are best solved by using Pythagorean triples, especially in sprint round questions, so make sure to actively evaluate the question(s) at hand and use the most efficient strategy.
Here is the link to 2003 chapter #29 that most of you got wrong:
You don't need to use complementary counting for that specific question since it's equal cases either way. Make sure you understand why you need to times 3. (AA_, A_A, and _AA for team A to be chosen two out of three days).
From Mathcounts Mini: Area of irregular polygon
See if you can use shoestring method to get the same answer.
Second half is again on similar triangles, dimensional change and sometimes
Pythagorean triples.
From NOVA : Fractals - Hunting the Hidden Dimension
Posted by at 9:51 AM
### The Largest Rectangle Inscribed in Any Triangle
$$\Delta EHI\sim\Delta EFG$$ $$\rightarrow$$ $$\dfrac {a} {c}=\dfrac {d-b} {d}$$$$\rightarrow$$ $$a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}$$
We are going to find out what the largest area of a rectangle is with the side length a and b.
It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.
$$a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}$$.
From there, you know that when $$b= \dfrac {1} {2}d$$, it will give you the largest area, which is $$\dfrac {1} {4}dc$$.
$$a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c$$.
Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.
Posted by at 9:15 AM
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