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## Calculating Initial Amount given Rate and k
Simaran Dosanjh 3E
Posts: 26
Joined: Fri Sep 25, 2015 3:00 am
### Calculating Initial Amount given Rate and k
Given the rate and k and the molar mass of an unknown species, how do I find the initial amount of the species in grams?
We are also given that he added the initial amount in grams to a 500 ml reaction flask.
Heerali Patel 3A
Posts: 26
Joined: Fri Sep 25, 2015 3:00 am
### Re: Calculating Initial Amount given Rate and k
The units of k should tell you what the order of the reaction is. You can then find the rate law of the reaction and use the given info. to solve for the (initial) concentration of the species. The question asks for initial amount in grams so you can then use the given info. to convert this concentration to grams.
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1. ## inhomogenous eqn
Does anyone know how to solve $\frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=10exp(-x)sinx$ thnx
2. Originally Posted by oxrigby
Does anyone know how to solve $\frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=10exp(-x)sinx$ thnx
First solve the homogenious equation. The auxillary equation is
$m^2+2m+2=0 \iff m^2+2m+1=-1 \iff (m+1)^2=-1 \iff m= -1 \pm i$
So the complimentry solution is
$y_c=c_1e^{-x}\cos(x)+c_2e^{-x}\sin(x)$
Since $e^{-x}\sin(x)$ appears in the original equation the form of your particular solution is
$y_p=Axe^{-x}\cos(x)+Be^{-x}\sin(x)$
now take two derivatives to get
$y_p'=(A-Ax+Bx)e^{-x}\cos(x)+(-Ax+B-Bx)e^{-x}\sin(x)$
$y_p''=(-2Ax+2B-2Bx)e^{-x}\cos(x)+(-2A+2Ax-2B)e^{-x}\sin(x)$
Now plug all of this into the ODE and simplify to get
$-2Ae^{-x}\sin(x)+2Be^{-x}\cos(x)=10e^{-x}\sin(x)$
So we can see that $B=0,A=-5$
So $y_p=-5xe^{-x}\cos(x)$
$y=y_c+y_p$
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# Consider the following statements in respect of class intervals of grouped frequency distribution: 1. Class intervals are mutually exclusive. 2. Class
48 views
closed
Consider the following statements in respect of class intervals of grouped frequency distribution:
1. Class intervals are mutually exclusive.
2. Class intervals should be equal in width.
Which of the above statements are correct?
1. Only 1
2. Only 2
3. Both 1 and 2
4. Neither 1 nor 2
by (54.0k points)
selected
Correct Answer - Option 3 : Both 1 and 2
Concept:
Class Interval: Consider we have large data $\rm x_1, x_2, .......,x_n$, while arranging them, they are grouped into different classes to get an idea of the distribution, and the range of such class of data is called the Class Interval.
Calculation:
In grouped frequency distribution, the range of given class of data is called the class interval.
Class intervals are generally equal in width and are mutually exclusive.
Hence, option (3) is correct.
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# How to deduce a linear system's impulse response from a set of input/output signals?
I want to know how to solve those types of problems.. is it by inspection ?
Consider the linear system below. When the inputs to the system $x_1[n]$, $x_2[n]$ and $x_3[n]$, the responses of the systems are $y_1[n]$, $y_2[n]$ and $y_3[n]$ as shown.
a. Determine whether the system is time invariant or not. Just your answer.
b. What is the impulse response?
Edit: Assuming a general case where the given inputs don't contain a scaled impulse like $x_2[n]$
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Hint: Use $x_2[n]$ and $y_2[n]$ to determine what the impulse response of $T$ must be (since $x_2[n]$ is just a scaled impulse). That gives you the answer to part (b). Then, check the other two cases to see if the inputs/outputs are consistent with that impulse response (using the superposition property of a linear system) to get an answer for part (a). – Jason R Jan 3 at 14:42
That's a more difficult problem in the general case. If they are all short like this, you know an upper bound on the duration of the impulse response, and you have enough input/output pairs, then you could set up a system of linear equations that you could solve to arrive at the unknown impulse response values. – Jason R Jan 3 at 14:50
In the general case it's also quite possible that there is no FIR solution or no solution at all. Hint: check the DC values of x1[n] and y1[n]. – Hilmar Jan 3 at 15:52
Hint: What does the signal $x_2[n]-x_2[n-2]$ look like? For an LTI system, the response should be $y_2[n]-y_2[n-2]$, no? Is it? Also, note that for a discrete-time linear time-variant system, there is not one unit-pulse response but an infinitude of unit-pulse responses, one for each time instant when the unit pulse occurs. – Dilip Sarwate Jan 3 at 16:29
@DilipSarwate: I agree that this is a dreadful homework problem. However, the system does look causal. While $y_3[n]$ is nonzero for $n=-2$, so is $x_3[n]$, so the system output isn't leading the input in time. – Jason R Jan 5 at 18:15
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# Multi-Objective Covariance Matrix Adaptation Evolution Strategy¶
The multi-objective covariance matrix adaptation evolution strategy (MO-CMA-ES) is one of the most powerful evolutionary algorithms for multi-objective real-valued optimization. In Shark, we provide a reference implementation of the algorithm (see MOCMA.h).
This tutorial illustrates applying the MO-CMA-ES to the DTLZ2 benchmark function. Please note that the methods presented here apply to all multi-objective optimizers available in the Shark library. That is, applying an optimizer to an objective function requires the following steps:
• Instantiate and configure the objective function.
• Instantiate the optimizer.
• Configure the optimizer instance for the objective function instance.
• Execute the optimizer until a termination criterion is fulfilled.
First of all, the following header files are required:
// Implementation of the MO-CMA-ES
#include <shark/Algorithms/DirectSearch/MOCMA.h>
#include <shark/ObjectiveFunctions/Benchmarks/Benchmarks.h>
Next, an instance of the objective function is created and configured for a two-dimensional objective space and a three-dimensional search space, respectively:
shark::DTLZ2 dtlz2;
dtlz2.setNumberOfVariables( 3 );
Thereafter, the optimizer is instantiated and initialized for the objective function instance:
shark::MOCMA mocma;
// Initialize the optimizer for the objective function instance.
dtlz2.init();
mocma.init( dtlz2 );
Finally, we iterate the optimizer until the objective function instance has been evaluated 25000 times:
// Iterate the optimizer
while( dtlz2.evaluationCounter() < 25000 ) {
mocma.step( dtlz2 );
}
As in all optimizers, the MO-CMA keeps track of the best known solution found so far. In contrast to single objective optimization, the solution is not a single point but a pareto front approximated by a set of points. We can print the pareto front using the following snippet:
// Print the optimal pareto front
for( std::size_t i = 0; i < mocma.solution().size(); i++ ) {
for( std::size_t j = 0; j < dtlz2.numberOfObjectives(); j++ ) {
std::cout<< mocma.solution()[ i ].value[j]<<" ";
}
std::cout << std::endl;
}
Running the example and visualizing the resulting Pareto-front approximation with the help of gnuplot will give you the following graphics:
Please see the file MOCMASimple.cpp for the complete source code of this tutorial.
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000160775 001__ 160775
000160775 005__ 20181203022215.0
000160775 0247_ $$2doi$$a10.4171/JEMS/130
000160775 02470 $$2ISI$$a000257869200008
000160775 037__ $$aARTICLE 000160775 245__$$aHorocyclic products of trees
000160775 269__ $$a2008 000160775 260__$$c2008
000160775 336__ $$aJournal Articles 000160775 520__$$aLet T-1, ... , T-d be homogeneous trees with degrees q(1) + 1, ... , q(d) + 1 >= 3; respectively. For each tree, let h : Tj -> Z be the Busemann function with respect to a fixed boundary point ( end). Its level sets are the horocycles. The horocyclic product of T-1 , ... , T-d is the graph DL(q1, ... , q(d)) consisting of all d-tuples x(1) ... x(d) is an element of T-1 x ... x T-d with h(x(1)) + ... + h(x(d)) = 0, equipped with a natural neighbourhood relation. In the present paper, we explore the geometric, algebraic, analytic and probabilistic properties of these graphs and their isometry groups. If d =2 and q(1) = q(2) = q then we obtain a Cayley graph of the lamplighter group ( wreath product) 3q (sic) Z. If d = 3 and q(1) = q(2) = q(3) = q then DL is a Cayley graph of a finitely presented group into which the lamplighter group embeds naturally. In general, when d - 4 and q(1) = ... = q(d) = q is such that each prime power in the decomposition of q is larger than d 1, we show that DL is a Cayley graph of a finitely presented group. This group is of type Fd - 1, but not Fd. It is not automatic, but it is an automata group in most cases. On the other hand, when the q(j) do not all coincide, DL(q(1) , ... , q(d))is a vertex-transitive graph, but is not a Cayley graph of a finitely generated group. Indeed, it does not even admit a group action with finitely many orbits and finite point stabilizers. The l(2)-spectrum of the "simple random walk" operator on DL is always pure point. When d = 2, it is known explicitly from previous work, while for d = 3 we compute it explicitly. Finally, we determine the Poisson boundary of a large class of group-invariant random walks on DL. It coincides with a part of the geometric boundary of DL.
000160775 6531_ $$arestricted wreath product 000160775 6531_$$atrees
000160775 6531_ $$ahorocycles 000160775 6531_$$aDiestel-Leader graph
000160775 6531_ $$agrowth function 000160775 6531_$$anormal form
000160775 6531_ $$aMarkov operator 000160775 6531_$$aspectrum
000160775 6531_ $$aDiestel-Leader Graphs 000160775 6531_$$aFinitely Presented Group
000160775 6531_ $$aArc-Transitive Digraphs 000160775 6531_$$aRandom-Walks
000160775 6531_ $$aEquilateral Triangle 000160775 6531_$$aLamplighter Groups
000160775 6531_ $$aInfinite-Graphs 000160775 6531_$$aAffine Group
000160775 6531_ $$aBoundary 000160775 700__$$aBartholdi, Laurent
000160775 700__ $$aNeuhauser, Markus 000160775 700__$$aWoess, Wolfgang
000160775 773__ $$j10$$tJournal Of The European Mathematical Society$$q771-816 000160775 909C0$$xU10077$$0252369$$pSB
000160775 909CO $$pSB$$particle$$ooai:infoscience.tind.io:160775 000160775 917Z8$$xWOS-2010-11-30
000160775 937__ $$aEPFL-ARTICLE-160775 000160775 973__$$rREVIEWED$$sPUBLISHED$$aEPFL
000160775 980__ aARTICLE
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## Universal locally finite central extensions of groups.(English)Zbl 0582.20022
An interesting new class of locally finite simple groups is studied; they generalize the well-known locally finite groups introduced by P. Hall [J. Lond. Math. Soc. 34, 305-319 (1959; Zbl 0088.023)]. The starting point is a periodic abelian group A. The group G is called a universal locally finite central extension of A if (i) A lies in the centre of G; (ii) G is locally finite; (iii) if $$A\leq B\leq D$$ are groups such that A is in the centre of D and of finite index in D, then every monomorphism of B into G that fixes the elements of A can be extended to a monomorphism of D into G.
All the universal locally finite general extensions of A form the class ULF(A). Theorem 3 says (much more than) that this class is non-empty; if A is countable, then ULF(A) contains a countable member which is unique up to isomorphism; if A is finite, then there are $$2^{\aleph_ 0}$$ mutually non-embeddable members of ULF(A) of cardinality $$\aleph_ 1$$, and their central quotients are also mutually non-embeddable. Theorem 1 says, inter alia, that if $$G\in ULF(A)$$, then A is the centre of G and G/A is simple; moreover G/A uniquely determines A. If A contains the direct cube of a cyclic group of prime order, then G/A is NOT a direct limit of finite simple groups. There are many more interesting results in the paper. The method relies on embedding lemmas for amalgams of locally finite groups that exploit the permutational product of groups.
Reviewer: B.H.Neumann
### MSC:
20E25 Local properties of groups 20E32 Simple groups 20F50 Periodic groups; locally finite groups 20E22 Extensions, wreath products, and other compositions of groups
Zbl 0088.023
Full Text:
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# Bash script to batch rename files
This is my homebrew bash script for batch renaming files extensions (my music library needs organizing):
#!/bin/bash
# rext.sh
# Syntax: rext.sh EXT1 EXT2
# Batch rename any file in current directory with extension of EXT1 to EXT2
EXT1=$1 EXT2=$2
for f in *.$EXT1 do fname=${f%.*}
newfname="$fname.$EXT2"
mv -v -i "$f" "$newfname"
done
It works beautifully with my music files, but is this code sufficient or can I still improve on it? Also I've seen examples using rename instead of mv, so is there any advantage in using rename?
A tiny mistake is not quoting $EXT1 in for f in *.$EXT1. If it contains a space, the script won't work as expected. I agree it's an unrealistic corner case, but it's easy enough to protect against with proper quoting: for f in *."$EXT1". The rename Perl script is famous for this purpose. Since you asked about its advantages: • Stood the test of time • You don't have to write it • It has more features (full regex support) The equivalent usage to simulate the behavior of your script would be: rename -v 's/EXT1$/EXT2/' *.EXT1
A feature I like is the -n flag for dry run, to verify the renames that would be performed before actually performing them.
• I agree it would be a corner case, but it's not unlikely to have something like some_file.timestampt\ year, where a newbie would be like "Oh, I can use my shell script to do that". It wouldn't do any harm if there's no files some_file.timestamp and no file year because that's what improper quoting would expand to, but if there are, that would probably rename files to not what user wanted, and they'd come asking where did their files go. Nice answer, by the way – Sergiy Kolodyazhnyy May 13 '17 at 22:50
Here's a few suggestions:
• always quote your variables to avoid word splitting as in *."$EXT1" • you are using globstar , which suggest that you run your script on files in current working directory. This may break if there is a filename with leading - (which will be understood by mv command as an option flag) , so it's often recommended to use ./* glob instead, and in your specific case you should makeit ./*."$EXT1".
• typical convention is to use lowercase variables to avoid overwriting environment variables, which are typically upper case; this isn't a written rule, but strongly suggested by professional sysadmins
As for rename command, a few things must be explained. There exist several rename commands. Typically when we talk about rename we are talking about the Perl script, but if you are using Debian or Ubuntu,those have two diverging versions which do the same thing - one comes with Perl installation while the other is standalone package. See this post on Unix & Linux for more info on this topic.
Korn shell and Mir Korn shell have their own, which is why that Perl script is often referenced as prename. The big difference is that Korn shell rename is a built-in and does simple renaming much like mv does. The prename is useful for couple different reasons, but mainly because you can use perl regular expressions to do some really complex renamings. See for example one of my answers on askubuntu. The trick here is that you really want to learn regular expressions first to be able to use rename efficiently and do that complex stuff.
• If the script is called without arguments (or maybe with the argument -?) a message how to use the script should be written to stderr.
• If the script is called with only one or more than 2 arguments an error message should be written to stderr
• should files be overwritten? Maybe the script should pass its option to the mv command (-i -v -u)
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# What is an intuitive view of adjoints? (version 1: category theory)
In trying to think of an intuitive answer to a question on adjoints, I realised that I didn't have a nice conceptual understanding of what an adjoint pair actually is.
I know the definition (several of them), I've read the nlab page (and any good answers will be added there), I've worked with them, I've found examples of functors with and without adjoints, but I couldn't explain what an adjunction is to a five-year-old, the man on the Clapham omnibus, or even an advanced undergraduate.
So how should I intuitively think of adjunctions?
For more background: I'm a topologist by trade who's been learning category theory recently (and, for the most part, enjoying it) but haven't truly internalised it yet. I'm fully convinced of the value of adjunctions, but haven't the same intuition into them as I do for, say, the uniqueness of ordinary cohomology.
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Here's a related question on Maths-SX with a very nice answer in terms of posets: math.stackexchange.com/q/25455/2907 – Andrew Stacey Sep 26 '11 at 8:52
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## 5 Answers
The example I would give a five-year old is the following. Take the category R whose objects are real numbers (or perhaps rational numbers for the five-year old) and a single morphism x--> y whenever x is less than or equal to y. Let Z be the full subcategory consisting of the integers. The inclusion i:Z --> R has a left and a right adjoint: one is the floor function, the other the ceiling function. I think the five-year old will agree that these are approximations so I would then say that left and right adjoints are just jazzed up versions of approximations.
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OOPS - I forgot to post my name - Paul Smith – user1421 Jan 10 '11 at 15:10
That's going to be one amazing five-year-old. – KConrad Jan 10 '11 at 19:55
Actually, it's the other way around: the left adjoint is the ceiling and the right adjoint is the floor. If $r \in \mathbb{R}$ and $n \in \mathbb{Z}$ then $\lceil r \rceil \le n$ if and only if $r \le n$, and dually $n \le \lfloor r \rfloor$ if and only if $n \le r$. – Qiaochu Yuan Apr 7 at 6:40
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I like Wikipedia's motivation for an adjoint functor as a formulaic solution to an optimization problem (though I'm biased, because I helped write it). In short, "adjoint" means most efficient and "functor" means formulaic solution.
Here's a digest version of the discussion to make this more precise:
An adjoint functor is a way of giving the most efficient solution to some optimization problem via a method which is formulaic ... For example, in ring theory, the most efficient way to turn a rng (like a ring with no identity) into a ring is to adjoin an element '1' to the rng, adjoin no unnecessary extra elements (we will need to have r+1 for each r in the ring, clearly), and impose no relations in the newly formed ring that are not forced by axioms. Moreover, this construction is formulaic in the sense that it works in essentially the same way for any rng.
The intuitive description of this construction as "most efficient" means "satisfies a universal property" (in this case an initial property), and that it is intuitively "formulaic" corresponds to it being functorial, making it an "adjoint" "functor".
In this asymmetrc interpretation, the theorem (if you define adjoints via universal morphisms) that adjoint functors occur in pairs has the following intuitive meaning:
"The notion that F is the most efficient solution to the (optimization) problem posed by G is, in a certain rigorous sense, equivalent to the notion that G poses the most difficult problem which F solves."
Edit: I like the comment below emphasizing that an adjoint functor is a globally defined solution. If
$G:C\to D$, it may be true that terminal morphisms exist to some $C$'s but not all of them; when they always exist, this guarantees that they extend to define a unique functor $F:D\to C$ such that $F \dashv G$. This result could have the intuitive interpretation "globally defined solutions are always formulaic".
Compare this for example to the basic theorem in algebraic geometry that a global section (of the structure sheaf) of $\mathrm{Spec} (A)$ is always defined by a single element of $A$; the global sections functor is an adjoint functor representable by the formula $Hom(-,\mathrm{Spec}( \mathbb{Z}))$, so this is actually directly related.
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Just to emphasize on part of this response, note that an adjoint pair provides a global solution (category-wide) as opposed to a local solution provided by universal maps. So universality is more general, in that if the universal map exists for every object in the category, the family of universal maps can be used to define an adjoint. – marc Nov 23 '09 at 19:34
In the last section, should it be $\mathrm{Spec}(\mathbb{Z}[t])$ instead? – QcH Jan 2 '11 at 3:49
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Suppose that $F\colon C\to D$ is a functor. Then there are many situations in which thinking of finding left and right adjoints to $F$ as solving approximation problems is very good intuition. So these would constitute functorial ways to approximate objects in $D$ relative to the image of $F$ by objects in $C$ either on the right or the left. I'm not sure I've really managed to word this in a way that conveys what I have in my head but here are some examples (which I have picked because they have a particularly 'approximationy' flavour but I do think that this works reasonably well in general anyway, I think my selection bias here is more skewed toward what I think about regularly).
Torsion theories are very good examples of this principal. For instance the notion of localization with respect to a homology theory in the homotopy category of spectra or more generally the approximating triangles coming from the acyclization and localization functors of a semi-orthogonal decomposition of a triangulated category. Another nice example along these lines is say the standard t-structure on the derived category of modules over some ring. Here we again have two pairs of adjoints and we can think of one as a right approximation by a bounded below complex coming from the unit and the other as a left approximation by a bounded above complex via the counit.
One can also view resolutions in the derived category in this way. For instance we have a right adjoint to the canonical map from $K(Inj R) \to D(R)$ for a ring $R$ where the first category is the homotopy category of complexes of injective $R$-modules which is taking K-injective resolutions. Similarly other sorts of resolutions, envelopes, and covers can be realised by adjunctions.
All of these examples are particularly nice in the sense that we get triangles or short exact sequences describing the object we start with in terms of our complementary approximations (by complementary I mean that there is orthogonality floating around in all of these examples so we have in some sense decomposed our category).
I think things like the adjoint functor theorem and Brown reprensentability become very reasonable from this point of view. One can loosely interpret them as saying provided things are "small enough" to be manageable and one has enough limits/colimits then one can build universal approximations (i.e. adjunctions) by taking coarse approximations and refining them.
I think this philosophy works well with the one given on the wiki page that Andrew Critch linked to.
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This intuition is the intuition behind Freyd's adjoint functor theorem. – Spice the Bird Jan 30 '13 at 16:07
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Another intuitive notion for adjoint functors comes from the string diagram notation for $2$-categories. Functors are $1$-morphisms in the $2$-category $Cat$. Functors $D\xleftarrow L C$ and $C\xleftarrow R D$ are adjoint $L\dashv R$ iff there are natural transformations (i.e. 2-morphisms) $R\circ L\xleftarrow\eta 1_C$ and $1_D\xleftarrow\epsilon L\circ R$ such that the composites $$L\xleftarrow{\epsilon\circ 1_L}L\circ R\circ L\xleftarrow{1_L\circ\eta}L$$ $$R\xleftarrow{1_R\circ\epsilon}R\circ L\circ R\xleftarrow{\eta\circ 1_R}R$$ are equal to the identities $1_L,1_R$ respectively.
As mentioned before, this looks particularly nice in string diagram notation. A 1-morphism $B\xleftarrow f A$ is drawn as a vertical line labeled by $f$ bisecting a square with the left and right regions labeled by $B,A$ respectively. In the case that the 1-morphism is $1_A$ then you can remove the line and have a square labeled by $A$. Composition of $1$-morphisms $g\circ f$ is denoted by drawing 2 parallel vertical lines labelled $g,f$ respectively, trisecting a square, labeling regions according to source and targets of $C\xleftarrow g B\xleftarrow f A$ respectively. $2$-morphisms are drawn by putting a dot on a line bisecting a square. The dot is labeled by the $2$-morphism, the lower and upper lines it connects by its source and target $1$-morphisms respectively and the left and right regions by their source and target objects. Horizontal composition is drawn by trisecting the square as before and vertical composition by stacking your squares on top of each other. In the case that the $2$-cell is the identity $1_f$, it is drawn exactly like the $1$-cell $f$.
In general, a $2$-cell $g_1\circ\cdots\circ g_n\xleftarrow \varphi f_1\circ\cdots\circ f_m$ is drawn by having $n,m$ lines connecting the bottom or top of a square respectively to a dot labeled by $\varphi$ and labeling lines and regions accordingly. In the case of our $2$-morphism $R\circ L\xleftarrow\eta 1_C$, we have three lines emanating from the dot, however $1_C$ was to be drawn without a line so we have only a two lines connecting at a dot in the middle of the square. We drop the dot and simply draw it as a cup with appropriate labels. Similarly, we can draw $\epsilon$ as a cap. Finally, one can see the identity above is simply a planar isotopy of curves, something intuitive from topology.
See the catsters for a better explanation: http://www.youtube.com/watch?v=pmvVE8AGAEA.
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For a "man on the Clapham omnibus" gloss on it, I think you could do worse than the Stanford Encyclopedia of Philosophy's entry for Category Theory. It describes adjoints as "conceptual inverses", and elaborates on how to see them that way in some of the standard examples.
I guess this is most probably a lower level than you were really asking for. But I think it articulates pretty well one of the less immediately obvious core points of the intuition (at least, my intuition) of what an adjunction is.
Putting this more precisely/abstractly: when we think of generalising isomorphism between objects of a 1-category to something between objects in a 2-category, we might usually think first of isomorphism and equivalence, but adjunction is also such a generalisation.
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# Malus Law - Cosine squared term?
by Dart82
Tags: cosine, malus, squared, term
P: 291 I simply tell my students that once upon a time, they wrote it as (sinx)^2, but lazy students kept leaving off the parenthesis and wrote sinx^2, sometimes meaning to take the sine of x, then square that answer; other times meaning to square the x first, then take the sine. To eliminate confusion, when they want the sin value to be squared, they put the squared symbol right next to sin $$sin^{2}x$$ So, to shorten $$(sinx)^2$$ write $$sin^{2}x$$ and to shorten $$sin(x^2)$$ write $$sinx^2$$ (although some people prefer those parenthesis are left in the latter case.)
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# English or French translation of Gauss' “Summatio Quarumdam Serierum Singularium”
I'm interested in looking at the details of Gauss' method of determining the sign of the Gauss sum in his "Summatio Quarumdam Serierum Singularium", and I was wondering if anyone knew if there was an English or French translation available at all? I've tried searching on the internet and haven't found anything. Also, are there any modern presentations of the Gauss' method that you would recommend?
Thanks for your help.
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D M Bressoud, On the value of Gaussian sums, J Number Theory 13 (1981) 88–94, MR 82b:10046, "gives a new proof, closely paralleling that of Gauss...." – Gerry Myerson Nov 17 '10 at 4:58
Thanks-this looks like it will be really helpful – Lea M Nov 17 '10 at 5:28
"The determination of Gauss sums" by Berndt and Evans (Bull. Amer. Math. Soc., Vol. 5, Number 2 (1981), 107-129.) contains an exposition of the original proof due to Gauss. It also includes a short historical outline of various classical and modern proofs due to Dirichlet, Cauchy and others.
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Thanks for linking to this article – Lea M Nov 17 '10 at 5:29
You're welcome. – Andrey Rekalo Nov 17 '10 at 5:43
there is a German translation: pages 463-495 in: http://www.amazon.co.uk/Untersuchungen-hohere-arithmetik-Chelsea-Publishing/dp/0821842137
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(English + French)/2 = German? – Adam Hughes Dec 4 '10 at 1:04
actually, i would say that English $-$ French $\to$ German. – Suvrit Dec 4 '10 at 1:52
Unfortunately I only speak a tiny bit of German (I've just taken a very basic introductory course), but thanks for pointing out that there is a German translation. – Lea M Dec 7 '10 at 2:07
An English translation of Baumgart's exposition of Gauss's fourth proof can be found here.
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That looks great-thanks so much for providing the link. – Lea M Dec 7 '10 at 2:05
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# Probably simple C++ question
This topic is 3920 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I an trying to input a variable with type double but don't know how to check to make sure the input is double and not something else. Could someone tell me how to do this?
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Quote:
Original post by PCosmin89What about if(variable%1!=0)
Umm, no. Modular arithmetic only works on integral types (int, short, long, etc) in C++, and even if that was allowed, what if the value was 10.0?
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Quote:
% is only valid for integers. This doesn't answer the question.
Quote:
Could someone tell me how to do this?
Presuming you are using std::cin for input, then just:
double d = 0.0;std::cin >> d;
Now, if you want to make sure the user enters something that is, in fact, a double (in the above code if they enter somebody bogus you will get unexpected results), you must first read the input as a string, then parse it. boost::lexical_cast is great for this. If you don't have Boost, get it. Or use something like
bool is_double(const std::string &input){ std::istringstream marshaller(input); double dummy; marshaller >> dummy; return !marshaller.fail();}
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Quote:
Original post by jpetrieOr use something likebool is_double(const std::string &input){ std::istringstream marshaller(input); double dummy; marshaller >> dummy; return !marshaller.fail();}
This is actually no different from
double x;if (cin >> x) { // ok}else { // bad}
In that both will seem ok if the user enters something like "42asdf" -- x = 42, but "asdf" will still be left in the buffer. If you really need to verify that what the user entered is a real number and absolutely nothing else, that's kinda difficult. You can use atof, but that returns 0.0 in case of an error. So you'd have to compare it to the stringstream conversion result, to make sure that the user didn't actually enter 0.
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// Evaluates whether a "C" string is a real.bool isReal(const char ch[]){ unsigned i = 0; bool has_point = false, has_sign = false; while ( ch != '\0' ) { // Look for a digit. if ( !std::isdigit(ch) ) { switch (ch) { case '.': if ( has_point == true ) { // Oops, allowed one occurrence. return false; } else { // Found the first point. has_point = true; } break; case '-': if ( has_sign == true ) { // Oops, allowed one occurrence. return false; } else { // Found the first sign. has_sign = true; } break; case '+': if ( has_sign == true ) { // Oops, allowed one occurrence. return false; } else { // Found the first sign. has_sign = true; } break; default: // Oops, no number, no point, no minus, no plus. return false; } } ++i; } // Has point, sign and at least one number. if ( has_point && has_sign && i > 2 ) return true; // Has point and at least one number. if ( has_point && i > 1 ) return true; // Has number, but nothing else. return false;}
Rules:
1. Can have either a + or - sign or none, but not both signs, or it fails.
2. Must have decimal point, or else assumed to be an integer, and it fails.
3. Must have at least one digit, or it fails.
4. If it has anything other than +/-, digits and a point, then it fails.
--random
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Quote:
Original post by random_thinkerHow about...*** Source Snippet Removed ***Rules:1. Can have either a + or - sign or none, but not both signs.2. Must have decimal point, or else assumed to be an integer.3. Must have at least one number.4. If it has anything other than +/-, digits and a point, then it fails.--random
Consider:
45.+0
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Quote:
Original post by drakostar
Quote:
Original post by jpetrieOr use something likebool is_double(const std::string &input){ std::istringstream marshaller(input); double dummy; marshaller >> dummy; return !marshaller.fail();}
This is actually no different from
*** Source Snippet Removed ***
In that both will seem ok if the user enters something like "42asdf" -- x = 42, but "asdf" will still be left in the buffer. If you really need to verify that what the user entered is a real number and absolutely nothing else, that's kinda difficult. You can use atof, but that returns 0.0 in case of an error. So you'd have to compare it to the stringstream conversion result, to make sure that the user didn't actually enter 0.
It is quite different. Your one tries to interpret a double straight from std::cin, which if it fails you must reset the streams state and possibly ignore characters from the stream. I prefer to read in a string and then do the conversion as a separate step.
@random_thinker
what about "3E-3". Also, why not std::string?
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If you want to check that there are no additional characters left in the string after reading the double, you can add a check to jpetrie's solution:
bool is_double(const std::string &input) { std::istringstream marshaller(input); double dummy; marshaller >> dummy; return !marshaller.fail() && (marshaller.rdbuf()->in_avail() == 0);}
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Quote:
Consider:45.+0
Quote:
what about "3E-3". Also, why not std::string?
Good points, could alter the logic to allow signs only at the beginning of the sequence to solve this potential problem. However, it is a lightweight and reliable approach even though it's very C-like. I couldn't be bothered to permit 'E-3' type constructs.
Could use std::string if you wanted, with a few alterations. I have used this primarily with std::getline() and feed the resulting string.c_str() to it.
--random
What about something like:
// Declare a double.double d=0;// Reference character set.const std::string double_chars("0123456789Ee-+.");// Make a cache.std::string cache; // Fill the cache.std::getline(std::cin,cache);// Check for any non-double characters.if (cache.find_first_not_of(double_chars) == std::string::npos){ // All characters are in reference set. d=strtod(cache.c_str(),0);}else{ // Fail.}
[Edited by - random_thinker on September 28, 2007 5:56:17 PM]
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1. ## Related rates problem
Hello, i'm not exactly sure how to do this one.
Water is draining for a conical filter into a cylindrical cup at the rate of 10 cubic centimeters per minute. The conical filter measures 10 cm tall and 10 cm across its top. The diameter of the cylindrical cup is 15 cm.
(a) How fast is the level of water in the filter falling when the water in the filter is 8 cm deep?
(b) At the same time as in part a, how fast is the level of water in the cup rising?
I can somewhat get it, but I need a start and some explanation. Thanks a lot.
2. First you can get by triangle analogies that the radius of the water cone is always half the height. Let h be the heigth of this water cone.
$V = \frac{\pi h^3}{4}$
$\frac{dV}{dh} = \frac{3 \pi h^2}{4}$
$\frac{dh}{dV} = \frac{4}{3 \pi h^2}$
You are given $\frac{dV}{dt} = 10$
$\frac{dh}{dV}\frac{dV}{dt} = \frac{dh}{dt} = 10\frac{4}{3 \pi h^2}$
Replace h by what value you want and be careful that your units are consistent.
Exercice b is quite easier than exercice a
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Utku Olcar
is open to work
Job titles
Robotics Engineer · Robotics Software Engineer · Robotics Specialist · Mechatronics Engineer · Embedded Software Engineer
Start date
Immediately, I’m actively applying
Job types
Full-time (I can work remotely)
Details
I have a virtually created Stewart Platform that is designed with Solidworks software. I have created to explain how does inverse kinematics works with parallel robots.
The main purpose of the controller's embedded software of this device is that moving the end effector plate to the desired position and the rotation. So, in this study, our inputs will be ;
• The center position of the end effector plate concerning the base plate.
• The Euler Angles (Roll, Pitch, Yaw) of the rotation of the end effector plate, again concerning the base plate's rotation.
If we name the six actuators from A to F, the links are called as
Links = \begin{bmatrix} L { \atop A} \\[1em] L { \atop B} \\[1em] L { \atop C}\\[1em] L { \atop D} \\[1em] L { \atop E} \\[1em] L { \atop F} \\[1em] \end{bmatrix}
Our outputs will be the link lengths of the linear actuators. The length parameters are shown with "Len" ;
Input = \begin{bmatrix} pos X \\[1em] pos Y \\[1em] pos Z \\[1em] \phi \\[1em] \theta \\[1em] \psi \\[1em] \end{bmatrix}
Output = \begin{bmatrix} Len { \atop A} \\[1em] Len { \atop B} \\[1em] Len { \atop C} \\[1em] Len { \atop D} \\[1em] Len { \atop E} \\[1em] Len { \atop F} \\[1em] \end{bmatrix}
The base plate's frame will be frame 0 and the end effector plate frame will be frame 1.
The joint points of the links are named for base joints :
Base Joints = \begin{bmatrix} A{ \atop 0} \\[1em] B { \atop 0} \\[1em] C { \atop 0} \\[1em] D { \atop 0} \\[1em] E { \atop 0} \\[1em] F { \atop 0} \\[1em] \end{bmatrix}
End Effector Joints = \begin{bmatrix} A{ \atop 1} \\[1em] B { \atop 1} \\[1em] C { \atop 1} \\[1em] D { \atop 1} \\[1em] E { \atop 1} \\[1em] F { \atop 1} \\[1em] \end{bmatrix}
Let's begin to solve the problem. To find the link lengths, we need the 3-dimensional position differences of two joint of the links. To find them, we need all calculated positions concerning the base frame.
Output = Link Lengths = \begin{bmatrix} Len { \atop A} =\begin{vmatrix} A{0 \atop 1} - A{0 \atop 0} \end{vmatrix} \\[1em] Len { \atop B} =\begin{vmatrix} B{0 \atop 1} - B{0 \atop 0} \end{vmatrix} \\[1em] Len { \atop C} =\begin{vmatrix} C{0 \atop 1} - C{0 \atop 0} \end{vmatrix} \\[1em] Len { \atop D} =\begin{vmatrix} D{0 \atop 1} - D{0 \atop 0} \end{vmatrix} \\[1em] Len { \atop E} =\begin{vmatrix} E{0 \atop 1} - E{0 \atop 0} \end{vmatrix} \\[1em] Len { \atop F} =\begin{vmatrix} F{0 \atop 1} - F{0 \atop 0} \end{vmatrix} \\[1em] \end{bmatrix}
We know the positions of points concerning their own frame (base point concerning base frame, end effector points concerning end effector frame). All we need is the 3D positions of the end effector joints concerning the base frame.
The positions of the end effector joints with respect to the end effector frame are constant.
A{1 \atop 1} = \begin{pmatrix} x{ \atop a1} ,& y{ \atop a1} ,& z{ \atop a1} \end{pmatrix}
The positions of base joints with respect to the base frame are constant too.
A{0 \atop 0} = \begin{pmatrix} x{ \atop a0} ,& y{ \atop a0} ,& z{ \atop a0} \end{pmatrix}
We need the transformation of these two frames from 0 to 1. To describe the transformation matrix, we need the rotation matrix and the position of the end effector frame with respect to the base.
T{0 \atop 1} = \begin{bmatrix} R{0 \atop 1} & P{0 \atop 1} \\[1em] (0,0,0) & 1 \\[1em] \end{bmatrix}
The position is given as input. We put that input directly to the right column.
P{0 \atop 1} = \begin{bmatrix} pos X \\[1em] pos Y \\[1em] pos Z \\[1em] . \end{bmatrix}
The rotation matrix will be generated;
R{0 \atop 1} = R{ \atop z}(\psi) R{ \atop y}(\theta) R{ \atop x}(\phi) = \begin{bmatrix} \cos \psi & -\sin \psi & 0 \\[1em] \sin \psi & \cos \psi & 0 \\[1em] 0 & 0 & 1 \\[1em] \end{bmatrix} \begin{bmatrix} \cos \theta & 0 & \sin \theta \\[1em] 0 & 1 & 0 \\[1em] -\sin \theta & 0 & \cos \theta \\[1em] \end{bmatrix} \begin{bmatrix} \cos \phi & -\sin \phi & 0 \\[1em] \sin \phi & \cos \phi & 0 \\[1em] 0 & 0 & 1 \\[1em] \end{bmatrix} \\[1em]
If we multiply all matrix, the result would be ;
R{0 \atop 1} = \begin{bmatrix} \cos \psi \cos \theta & \cos \psi \sin \theta \sin \phi - \sin \psi \cos \phi & \cos \psi \sin \theta \cos \phi + \sin \psi \sin \phi \\[1em] \sin \psi \cos \theta & \sin \psi \sin \theta \sin \phi +\cos \psi \cos \phi & \sin \psi \sin \theta \cos \phi - \cos \psi \sin \phi \\[1em] - \sin \theta & \cos \theta \sin \phi & \cos \theta \cos \phi \\[1em] \end{bmatrix} \\[1em]
So the Transformation Matrix has been described :
T{0 \atop 1} = \begin{bmatrix} \cos \psi \cos \theta & \cos \psi \sin \theta \sin \phi - \sin \psi \cos \phi & \cos \psi \sin \theta \cos \phi + \sin \psi \sin \phi & pos X \\[1em] \sin \psi \cos \theta & \sin \psi \sin \theta \sin \phi +\cos \psi \cos \phi & \sin \psi \sin \theta \cos \phi - \cos \psi \sin \phi & pos Y \\[1em] - \sin \theta & \cos \theta \sin \phi & \cos \theta \cos \phi & pos Z \\[1em] 0 & 0 & 0 & 1 \\[1em] \end{bmatrix} \\[1em]
Okey. Let's calculate the positions for link joints. The calculations before this step were the same for all the parallel robot systems. Starting with this step, we will calculate the Stewart Platform's geometry. Every Stewart Platform has some similarities about the geometry. So we will calculate the positions with these common assumptions.
pos A { \atop 0} = ( X{ \atop A0}, Y{ \atop A0}, 0 )
pos B { \atop 0} = ( X{ \atop B0}, Y{ \atop B0}, 0 )
pos C { \atop 0} = ( X{ \atop C0}, Y{ \atop C0}, 0 )
pos D { \atop 0} = ( X{ \atop D0}, Y{ \atop D0}, 0 )
pos E { \atop 0} = ( X{ \atop E0}, Y{ \atop E0}, 0 )
pos F { \atop 0} = ( X{ \atop F0}, Y{ \atop F0}, 0 )
Fill the X, Y data with the "r" and "Ω" variables ;
pos A {0 \atop 0} = ( r \sin \Omega, r \cos\Omega, 0 )
pos B{0 \atop 0} = ( r \sin (120-\Omega),r \cos(120-\Omega) , 0 )
pos C {0 \atop 0} = (r \sin (120+\Omega), r \cos(120+\Omega), 0 )
pos D{0 \atop 0} = ( r \sin (240-\Omega), r \cos (240-\Omega), 0 )
pos E{0 \atop 0} = ( r \sin (240+\Omega), r \cos(240+\Omega), 0 )
pos F{0 \atop 0} = ( r \sin (360-\Omega), r \cos(360-\Omega), 0 )
End Effector :
posA{1 \atop 1} = ( r \sin \Sigma, r \cos\Sigma, 0 )
pos B{1 \atop 1}= ( r \sin (120-\Sigma),r \cos(120-\Sigma) , 0 )
pos C{1 \atop 1} = (r \sin (120+\Sigma), r \cos(120+\Sigma), 0 )
pos D {1 \atop 1} = ( r \sin (240-\Sigma), r \cos (240-\Sigma), 0 )
pos E{1 \atop 1} = ( r \sin (240+\Sigma), r \cos(240+\Sigma), 0 )
pos F{1 \atop 1} = ( r \sin (360-\Sigma), r \cos(360-\Sigma), 0 )
These positions are calculated with respect to the end effector frame. We need these positions with respect to the base frame.
P{0 \atop1} = T {0 \atop 1} P{1 \atop 1}
So we can calculate these using the Transformation Matrix easily using this formula.
For example :
pos A{0 \atop 1} = \begin{bmatrix} \cos \psi \cos \theta & \cos \psi \sin \theta \sin \phi - \sin \psi \cos \phi & \cos \psi \sin \theta \cos \phi + \sin \psi \sin \phi & pos X \\[1em] \sin \psi \cos \theta & \sin \psi \sin \theta \sin \phi +\cos \psi \cos \phi & \sin \psi \sin \theta \cos \phi - \cos \psi \sin \phi & pos Y \\[1em] - \sin \theta & \cos \theta \sin \phi & \cos \theta \cos \phi & pos Z \\[1em] 0 & 0 & 0 & 1 \\[1em] \end{bmatrix} \begin{bmatrix} r \sin \Sigma \\[1em] r \cos\Sigma \\[1em] 0 \\[1em] 1 \\[1em] \end{bmatrix}
Len B = magnitude of ( pos B{0 \atop 1} - pos B{0 \atop 0} )
Len C = magnitude of ( pos C{0 \atop 1} - pos C{0 \atop 0} )
Len D = magnitude of ( pos D {0\atop 1} -pos D {0 \atop 0} )
Len E = magnitude of ( pos E {0 \atop 1} - pos E{0 \atop 0} )
Len F = magnitude of ( pos F {0 \atop 1} - pos F{0 \atop 0} )
Reference:
• Mark ,S, Seth, H. and M., Vidyasagar, 2005. Robot Modeling and Control. 1th ed. Wiley
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• Februray 2, 2017
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• November 3, 2017
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# Latex Beamer Madrid theme - how to get smoothbars in the footer not header?
How do I create a custom beamer style where my header contains only the title of the slide, and my footer - the smoothbars "dots" style of how far the presentation has progressed, i.e.
(((TITLE)))
Frame content
(((what you get above the title if you use \useoutertheme[subsection=false]{smoothbars}, but I want it in the footer)))
I'm using Madrid as my basis style at the moment, but I'm happy to use another style if it can't be customized.
-
## migrated from stackoverflow.comJul 14 '12 at 21:07
This question came from our site for professional and enthusiast programmers.
Starting from the Madrid theme, one could redefine the footline template (taking immediately into account to suppress the subsections, as it is possible to see from \useoutertheme[subsection=false]{smoothbars}).
Code:
\documentclass[compress]{beamer}
\usepackage{lmodern}
\makeatletter
\setbeamertemplate{footline}
{%
\vskip-9ex%
\begin{beamercolorbox}{}
\end{beamercolorbox}%
\end{beamercolorbox}%
}%
\makeatother
\title{My title}
\author{My name}
\institute{My institute}
\begin{document}
\section{First section}
\subsection{a}
\begin{frame}{Title}
hello
\end{frame}
\subsection{b}
\begin{frame}{Another title}
hello
\end{frame}
\section{Second section}
\subsection{a}
\begin{frame}{Title}
hello
\end{frame}
\subsection{b}
\begin{frame}{Another title}
hello
\end{frame}
\end{document}
which gives:
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# Let’s make a DQN: Double Learning and Prioritized Experience Replay
### Introduction
Last time we implemented a Full DQN based agent with target network and reward clipping. In this article we will explore two techniques, which will help our agent to perform better, learn faster and be more stable – Double Learning and Prioritized Experience Replay.
### Double Learning
One problem in the DQN algorithm is that the agent tends to overestimate the Q function value, due to the max in the formula used to set targets:
$Q(s, a) \xrightarrow{} r + \gamma max_a Q(s', a)$
To demonstrate this problem, let’s imagine a following situation. For one particular state there is a set of actions, all of which have the same true Q value. But the estimate is inherently noisy and differs from the true value. Because of the max in the formula, the action with the highest positive error is selected and this value is subsequently propagated further to other states. This leads to positive bias – value overestimation. This severe impact on stability of our learning algorithm1.
A solution to this problem was proposed by Hado van Hasselt (2010)2 and called Double Learning. In this new algorithm, two Q functions – $Q_1$ and $Q_2$ – are independently learned. One function is then used to determine the maximizing action and second to estimate its value. Either $Q_1$ or $Q_2$ is updated randomly with a formula:
$Q_1(s, a) \xrightarrow{} r + \gamma Q_2(s', argmax_a Q_1(s', a))$
or
$Q_2(s, a) \xrightarrow{} r + \gamma Q_1(s', argmax_a Q_2(s', a))$
It was proven that by decoupling the maximizing action from its value in this way, one can indeed eliminate the maximization bias2.
When thinking about implementation into the DQN algorithm, we can leverage the fact that we already have two different networks giving us two different estimates $Q$ and $\tilde{Q}$ (target network). Although not really independent, it allows us to change our algorithm in a really simple way.
The original target formula would change to:
$Q(s, a) \xrightarrow{} r + \gamma \tilde{Q}(s', argmax_a Q(s', a))$
Translated to code, we only need to change one line to get the desired improvements:
t[a] = r + GAMMA * pTarget_[i][ numpy.argmax(p_[i]) ]
The Deep Reinforcement Learning with Double Q-learning1 paper reports that although Double DQN (DDQN) does not always improve performance, it substantially benefits the stability of learning. This improved stability directly translates to ability to learn much complicated tasks.
When testing DDQN on 49 Atari games, it achieved about twice the average score of DQN with the same hyperparameters. With tuned hyperparameters, DDQN achieved almost four time the average score of DQN. Summary of the results is shown in a table in the next section.
### Prioritized Experience Replay
One of the possible improvements already acknowledged in the original research2 lays in the way experience is used. When treating all samples the same, we are not using the fact that we can learn more from some transitions than from others. Prioritized Experience Replay3 (PER) is one strategy that tries to leverage this fact by changing the sampling distribution.
The main idea is that we prefer transitions that does not fit well to our current estimate of the Q function, because these are the transitions that we can learn most from. This reflects a simple intuition from our real world – if we encounter a situation that really differs from our expectation, we think about it over and over and change our model until it fits.
We can define an error of a sample S = (s, a, r, s’) as a distance between the Q(s, a) and its target T(S):
$error = |Q(s, a) - T(S)|$
For DDQN described above, T it would be:
$T(S) = r + \gamma \tilde{Q}(s', argmax_a Q(s', a))$
We will store this error in the agent’s memory along with every sample and update it with each learning step.
One of the possible approaches to PER is proportional prioritization. The error is first converted to priority using this formula:
$p = (error + \epsilon)^\alpha$
Epsilon $\epsilon$ is a small positive constant that ensures that no transition has zero priority.
Alpha, $0 \leq \alpha \leq 1$, controls the difference between high and low error. It determines how much prioritization is used. With $\alpha = 0$ we would get the uniform case.
Priority is translated to probability of being chosen for replay. A sample i has a probability of being picked during the experience replay determined by a formula:
$P_i = \frac{p_i}{\sum_k p_k}$
The algorithm is simple – during each learning step we will get a batch of samples with this probability distribution and train our network on it. We only need an effective way of storing these priorities and sampling from them.
##### Initialization and new transitions
The original paper says that new transitions come without a known error3, but I argue that with definition given above, we can compute it with a simple forward pass as it arrives. It’s also effective, because high value transitions are discovered immediately.
Another thing is the initialization. Remember that before the learning itself, we fill the memory using random agent. But this agent does not use any neural network, so how could we estimate any error? We can use a fact that untrained neural network is likely to return a value around zero for every input. In this case the error formula becomes very simple:
$error = |Q(s, a) - T(S)| = |Q(s, a) - r - \gamma \tilde{Q}(s', argmax_a Q(s', a))| = | r |$
The error in this case is simply the reward experienced in a given sample. Indeed, the transitions where the agent experienced any reward intuitively seem to be very promising.
##### Efficient implementation
So how do we store the experience and effectively sample from it?
A naive implementation would be to have all samples in an array sorted according to their priorities. A random number s, $0 \leq s \leq \sum_k p_k$, would be picked and the array would be walked left to right, summing up a priority of the current element until the sum is exceeded and that element is chosen. This will select a sample with the desired probability distribution.
But this would have a terrible efficiency: O(n log n) for insertion and update and O(n) for sampling.
A first important observation is that we don’t have to actually store the array sorted. Unsorted array would do just as well. Elements with higher priorities are still picked with higher probability.
This releases the need for sorting, improving the algorithm to O(1) for insertion and update.
But the O(n) for sampling is still too high. We can further improve our algorithm by using a different data structure. We can store our samples in unsorted sum tree – a binary tree data structure where the parent’s value is the sum of its children. The samples themselves are stored in the leaf nodes.
Update of a leaf node involves propagating a value difference up the tree, obtaining O(log n). Sampling follows the thought process of the array case, but achieves O(log n). For a value s, $0 \leq s \leq \sum_k p_k$, we use the following algorithm (pseudo code):
def retrieve(n, s):
if n is leaf_node: return n
if n.left.val >= s: return retrieve(n.left, s)
else: return retrieve(n.right, s - n.left.val)
Following picture illustrates sampling from a tree with s = 24:
With this effective implementation we can use large memory sizes, with hundreds of thousands or millions of samples.
For known capacity, this sum tree data structure can be backed by an array. Its reference implementation containing 50 lines of code is available on GitHub.
##### Results
Tests performed on 49 Atari games showed that PER really translates into faster learning and higher performance3. What’s more, it’s complementary to DDQN.
Following table summarizes the results for 49 Atari games benchmark. Values are taken from Schaul et al. (2015)3 and rescaled. Because the DQN+PER value for proportional PER was not available, the provided value is for similar, but different rank-based PER. For specifics of how these tests were performed, look into the paper3.
DQN DQN+PER DDQN DDQN+PER
100% 291% 343% 451%
### Implementation
An implementation of DDQN+PER for an Atari game Seaquest is available on GitHub. It’s an improvement over the DQN code presented in last chapter and should be easy to understand.
The DQN architecture from the original paper4 is implemented, although with some differences. In short, the algorithm first rescales the screen to 84×84 pixels and extracts luminance. Then it feeds last two screens as an input to the neural network. This ensures that the algorithm is also aware of a direction of movement of the game elements, something which would not be possible with only a current screen as input. Experience is stored in a sum tree with capacity of 200 000 samples. Neural network uses three convolutional layers and one dense hidden layer with following parameters:
Convolution2D(32, 8, 8, subsample=(4,4), activation='relu')
Convolution2D(64, 4, 4, subsample=(2,2), activation='relu')
Convolution2D(64, 3, 3, activation='relu')
Dense(output_dim=512, activation='relu')
Dense(output_dim=actionCount, activation='linear')
These hyperparameters were used:
Parameter Value
memory capacity 200000
batch size 32
γ 0.99
exploration εmax 1.00
exploration εmin 0.10
final exploration frame 500000
PER α 0.6
PER ε 0.01
RMSprop learning rate 0.00025
target network update frequency 10000
It is possible to run this program on a regular computer, however it is very resource demanding. It takes about 12 GB of RAM and fully utilizes one core of CPU and whole GPU to slowly improve. In my computer it runs around 20 FPS. After about 12 hours, 750 000 steps and 700 episodes, it reached an average reward of 263 (mean reward of a random agent is 87). You can see it in action here:
To get better results, you have to run it for at least tens of millions of steps. The following graph shows that the improvement is indeed very slow:
### Conclusion
We addressed Double Learning and Prioritized Experience Replay techniques that both substantially improve the DQN algorithm and can be used together to make a state-of-the-art algorithm on the Atari benchmark (at least as of 18 Nov 2015 – the day Prioritized Experience Replay3 article was published).
This articles finishes the Let’s make a DQN series. It was meant as a simplified tutorial for those who don’t want to read whole research articles but still want to understand what DQN is and what it does. I also hope that it sparkled your interest in this interesting direction of AI and that you will want to learn even more now.
I hope you enjoyed these articles at least as me writing them and that you learned something. If you have any questions or have anything to add, please feel free to leave a comment or contact me at author@jaromiru.com.
1. Hado van Hasselt, Arthur Guez, David Silver – Deep Reinforcement Learning with Double Q-learning, arXiv:1509.06461, 2016
2. Hado van Hasselt – Double Q-learning, Advances in Neural Information Processing Systems, 2010
3. Schaul et al. – Prioritized Experience Replay, arXiv:1511.05952, 2015
4. Mnih et al. – Human-level control through deep reinforcement learning, Nature 518, 2015
## 17 thoughts on “Let’s make a DQN: Double Learning and Prioritized Experience Replay”
1. Donal Byrne says:
Hi Jaromiru, absolutely love the ddqn series, definitely some of the best material on the subject that I have found. I was wondering if there would be a noticeable difference if you use 4 stacked images in the CNN instead of 2? I tried this on my laptop but it seemed to slow everything own a lot. This makes sense as the CNN has to do twice the work. I was wondering if it would be worth the extra training time or if the extra stacks would make a noticeable difference? Thanks
1. ヤロミル says:
Yes. In fact, four stacked images were used in the original DQN paper and this was definitely done for a reason. Therefore we can assume, that four stacked images would work better.
2. Yuhang Cao says:
I have some problems. In cartpole environment, what is the transition model or transition probability, should it be 1, that is P(s’|s, a) = 1? Can you give me a MDP define for cartpole and breakout? Really thanks to it.
1. Anonymous says:
In the OpenAI gym cartpole environment the state transition probability matrix is going to be 1 because the physics model (i.e. the set of equations in the cartpole environment) are deterministic. But you don’t typically care to know the state transition probability matrix explicitly when doing Q learning since the network learns those through experience that it’s sampled.
3. Yuhang Cao says:
Awesome post, it really helps me, thanks very much!
4. seraphli says:
It’s not a thing about difficulties. You are using Q learning, which is depended on the assumption about MDP. If you don’t use history stack, the agent can not know the moving direction of a particle on the screen, so MDP is not satisfied. Anyway, you use two images in one state in your code. But this will consume about (84*84*2*2+2)*4*200000 Bytes = 21GB memory, not 12GB memory.
1. ヤロミル says:
Samples in the memory are duplicated. Say a sample (s0, r, a, s1) and (s1, r, a, s2) both share s1, which occupies only one part of the *physical* memory.
So the formula is 84×84 x 4 (float32) x 2 (per state) x 200 000 ~ 10.5 GB
Thinking about it, you can actually decouple the images in the state itself. Changing line 237:
s_ = numpy.array([s[1], processImage(img)]) #last two screens
from numpy array to python list would half the memory requirements (some code change needed later, possible performance degradation).
2. ヤロミル says:
About the MDP property – you’re right. I have limited resources and so I decided to reduce the frame count in a state from 4 to 2, halving the memory requirements, while still retaining the agent’s ability to understand motion.
I’m not trying to replicate the papers here, rather explain the concept.
5. Hi. I notice that your implement is not using history stack. Because using history stack will use much more memory. That’s different from the original article.
1. ヤロミル says:
What do you mean by history stack? If you mean the last 4 frames stacked together, you are right. But that’s easy to implement and I didn’t want to introduce more difficulties here.
6. Most likely I’m getting something wrong but may I ask:
In your mountain car DQN implementationt this is line 170:
[a] = r + GAMMA * p_[i][ numpy.argmax(p[i]) ]
Shouldn’t it be numpy.argmax(p_[i])? The max Q-Value predicted from the Online Network for the newstate instead of the current state?
1. jaromiru says:
Nice catch! You are referring to a file open_gym/MountainCar-v0.py in my other repository, I assume.
It should be:
=====
p = agent.brain.predict(states)
p_ = agent.brain.predict(states_, target=False)
pTarget_ = agent.brain.predict(states_, target=True)
====
t[a] = r + GAMMA * pTarget_[i][ numpy.argmax(p_[i]) ] # double DQN
===
The solution you proposed leads to a regular DQN.
7. Vincent says:
Very interesting read! This makes it very easy to understand. Are you planning. Are you planning on writing posts about more RL subjects? I would love to read more intuitive articles about policy gradients methods. Good luck with your PhD!
1. ヤロミール says:
Thank you! More articles are definitely possible, although not in the very near future.
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# Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..)
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Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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Originally posted by EgmatQuantExpert on 13 Jul 2018, 06:28.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:04, edited 2 times in total.
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Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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Updated on: 18 Jul 2018, 12:09
3
EgmatQuantExpert wrote:
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
Let the tank = 60 liters.
Since P takes 3 hours to fill the 60-liter tank, P's rate $$= \frac{60}{3} = 20$$ liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate $$= \frac{60}{4} = 15$$ liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate $$= \frac{60}{-5} = -12$$ liters per hour.
Since R works to EMPTY the tank, R's rate is negative.
To MINIMIZE the time required to fill the tank, we must MAXIMIZE the time for P, since P's input rate is greater than Q's input rate.
Since P must close for $$\frac{1}{4}$$ hour after each hour of work, P's time will be maximized if -- over a 5-hour period -- P works for 4 hours and takes 4 quarter-hour breaks, with Q working for $$\frac{1}{4}$$ hour during each break.
Thus, the input rate for each 5-hour period = (4 hours of work for P) + (1 hour of work for Q) - (5 hours of work for R) $$= (4*20) + (1*15) - (5*12) = 35$$ liters for every 5 hours of work.
Since the volume increases by 35 liters every 5 hours, the hourly rate $$≈ \frac{35}{5} ≈ 7$$ liters per hour.
(The hourly rate is an approximation because it increases when P works but decreases when Q works.)
Since the hourly rate is about 7 liters per hour, the approximate time to fill the 60-liter tank $$= \frac{60}{7} ≈ 8.5$$ hours.
10am + about 8.5 hours ≈ 6:30pm.
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Originally posted by GMATGuruNY on 18 Jul 2018, 03:43.
Last edited by GMATGuruNY on 18 Jul 2018, 12:09, edited 1 time in total.
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Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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13 Jul 2018, 08:09
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EgmatQuantExpert wrote:
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
To get the minimum time, the faster that is P should function the most......
so the functioning should be P for 1 hr and then Q for 15 min and then P for 1 hour and so on
two cases:-
1) when P and R function for 1 hour.
$$\frac{1}{3}-\frac{1}{5}=\frac{2}{15}$$...
2) when Q and R function for 15 min or 1/4 hr
$$\frac{1}{4}-\frac{1}{5}=\frac{1}{20}$$ in one hour, so $$\frac{1}{20}*\frac{1}{4} = \frac{1}{80}$$ in 15 min
so 1 hr 15 min or 5/4 hr work = $$\frac{2}{15}+\frac{1}{80}=\frac{35}{240}$$ and thus 1 hr work = $$\frac{35}{240}*\frac{4}{5}=\frac{7}{60}$$
and the work will get completed in $$\frac{60}{7}$$ = 8 hr 35 min approx
time wise it will finish at 10:00 + 8 hr 35 min = 18:35 or 06:35 pm ~ 06:30 pm
C
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Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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Updated on: 12 Aug 2018, 23:41
Hey everyone,
We will post the solution very soon. Till then, try it one more time and post your analysis.
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Originally posted by EgmatQuantExpert on 18 Jul 2018, 01:16.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:41, edited 1 time in total.
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Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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18 Jul 2018, 03:01
1
Let total capacity of tank be = 60 litres.
P's one hour work = 60/3 = +20 litres/hour
Q's one hour work = 60/4 = +15 litres/hour
R's one hour work = 60/5 = -12 litres/hour
We would require P to work for maximum time as P+R work = 8 l/hour which is higher than Q+R = 3 l/hour.
P would work for 1 hour remain closed for 15 minutes when Q will work and the P will resume. Q+R 15 minute work = 3/4 =0.75 litres/hour
tank filled in 1 hour 15 minutes (75 minutes) = 8.75 litres.
Number of such cycles required = 60/8.75 = 6000/875
Minutes required = 6000*75/875
Hours required = 6000*75/(875*60) = 8.5 hours approx
Thus 10:00 AM + 8.5 hours = 6:30 PM
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Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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18 Jul 2018, 12:05
GMATGuruNY wrote:
Let the tank = 60 liters.
Since P takes 3 hours to fill the 60-liter tank, P's rate $$= \frac{60}{3} = 20$$ liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate $$= \frac{60}{4} = 15$$ liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate $$= \frac{60}{-5} = -12$$ liters per hour.
Since R works to EMPTY the tank, R's rate is negative.
To MINIMIZE the time required to fill the tank, we must MAXIMIZE the time for P, since P's input rate is greater than Q's input rate.
Since P must close for $$\frac{1}{4}$$ hour after each hour of work, P's time will be maximized if -- over a 5-hour period -- P works for 4 hours and takes 4 quarter-hour breaks, with Q working for $$\frac{1}{4}$$ hour during each break.
Thus, the rate for each 5-hour period = (4 hours of work for P) + (1 hour of work for Q) - (5 hours of work for R) $$= (4*20) + (1*15) - (5*12) = 35$$ liters per hour.
Since the volume increases by 35 liters every 5 hours, the hourly rate $$≈ \frac{35}{5} ≈ 7$$ liters per hour.
(The hourly rate is an approximation because it increases when P works but decreases when Q works.)
Since the hourly rate is about 7 liters per hour, the approximate time to fill the 60-liter tank $$= \frac{60}{7} ≈ 8.5$$ hours.
10am + about 8.5 hours ≈ 6:30pm.
Dear GMATGuruNY
I believe the highlighted should be modified as 35 liter per 5 hrs. Correct?
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Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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18 Jul 2018, 12:17
Mo2men wrote:
I believe the highlighted should be modified as 35 liter per 5 hrs. Correct?
Good catch.
I've corrected the typo.
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Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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18 Jul 2018, 22:58
It is understood that for least time to fill up the tank, P should be opened for the maximum time.
=> P should be opened for 1 hour after 10 am
R is open at all times
In this one hour, the proportion of tank which is has been filled is (1/3 - 1/5) = 2/15
After the 1 hour, Q is opened for a mandatory 15 mins while P is closed
In these 15 mins, the proportion of tank which has been filled up is (1/16 - 1/20) = 1/80
So, for a cycle of 1 hours 15 mins, tank filled is (2/15) + (1/80) = 35/240
The above cycle has to be repeated till the tank is filled.
It can be seen that this will run for 6 complete cycles, and the tank will be filled in the 7th cycle when P is opened with R (Since 35/240*6 is 210/40, and 35*7/240>1)
6 cycles of 1 hours 15 mins yields 7 hours 30 mins. => Time after 6 cycles is 17:30 hours
Tank remaining empty is 1 - 35*6/240 = 1/8
1/8 tank will be filled by P and R open
(1/8)/(2/15) = approximately 1 hour.
Time at the time of finishing 18:30.
I think assuming hourly rates of a cycle works only when rates of P and Q are comparable and the size of the cycle is not too big. Otherwise, the approximation may result into wrong answer.
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Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
### Show Tags
Updated on: 12 Aug 2018, 23:37
Solution
Given:
• Three pipes P, Q, and R are attached to a tank
• P and Q can fill the tank in 3 hours and 4 hours respectively.
• R can empty the tank in 5 hours.
• P is opened at 10 am and Q is opened at 11 am, while R is open all the time.
• P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour.
To find:
• Time when the tank will be full.
Approach and Working:
Let us assume that the capacity of tank is 60 Litres. Hence:
• In 1 hour, P can fill= 20 L
• In 1 hour, Q can fill= 15 L
• In 1 hour, R can empty= 12 L
Since, we are asked the earliest time to fill the tank, P should be opened for as much time as it can.
Now, from 10 AM to 11 AM:
• P and Rare open.
• Hence, amount of water in the tank= 20-12= 8 L
From 11 AM to 11:15AM:
• Q and R are open.
• Hence, amount of water in the tank now=$$8+(\frac{15}{4}-\frac{12}{4})=8+\frac{3}{4}$$
From 11:15 AM to 12:15 PM:
• P and R are open.
• Hence, amount of water in the tank= $$(8+\frac{3}{4})+8$$
From 12:15 PM to 12:30 PM:
• Q and R are open.
• Hence, amount of water in the tank now= $$(8+\frac{3}{4})+8+\frac{3}{4}= (8+\frac{3}{4})+(8+\frac{3}{4})= 2(8+\frac{3}{4})$$
From 12:30 PM to 1:30 PM
• The amount of water in the tank= $$2(8+\frac{3}{4}) +8$$
From 1:30 PM to 1:45 PM
• The amount of water in the tank=$$3(8+\frac{3}{4})$$
So, if you can understand then:
• In every 1 hour 15 min or $$\frac{5}{4}$$ hour, $$8+\frac{3}{4} or \frac{35}{4} l$$ water is getting filled.
• Hence, in 1 hour= $$\frac{35}{4}* \frac{4}{5}= 7$$ litres.
• Thus, time taken to fill 60 litres= $$\frac{60}{7}≈ 8.5$$ hours
• Hence, Time= 10AM+8h 30 min= 6:30 PM
Hence, the correct answer is option C.
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Originally posted by EgmatQuantExpert on 22 Jul 2018, 13:18.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:37, edited 1 time in total.
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Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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25 Jul 2018, 05:36
EgmatQuantExpert wrote:
Solution
Given:
• Three pipes P, Q, and R are attached to a tank
• P and Q can fill the tank in 3 hours and 4 hours respectively.
• R can empty the tank in 5 hours.
• P is opened at 10 am and Q is opened at 11 am, while R is open all the time.
• P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour.
To find:
• Time when the tank will be full.
Approach and Working:
Let us assume that the capacity of tank is 60 Litres. Hence:
• In 1 hour, P can fill= 20 L
• In 1 hour, Q can fill= 15 L
• In 1 hour, R can empty= 12 L
Since, we are asked the earliest time to fill the tank, P should be opened for as much time as it can.
Now, from 10 AM to 11 AM:
• P and Rare open.
• Hence, amount of water in the tank= 20-12= 8 L
From 11 AM to 11:15AM:
• Q and R are open.
• Hence, amount of water in the tank now=$$8+(\frac{15}{4}-\frac{12}{4})=8+\frac{3}{4}$$
From 11:15 AM to 12:15 PM:
• P and R are open.
• Hence, amount of water in the tank= $$(8+\frac{3}{4})+8$$
From 12:15 PM to 12:30 PM:
• Q and R are open.
• Hence, amount of water in the tank now= $$(8+\frac{3}{4})+8+\frac{3}{4}= (8+\frac{3}{4})+(8+\frac{3}{4})= 2(8+\frac{3}{4})$$
From 12:30 PM to 1:30 PM
• The amount of water in the tank= $$2(8+\frac{3}{4}) +8$$
From 1:30 PM to 1:45 PM
• The amount of water in the tank=$$3(8+\frac{3}{4})$$
So, if you can understand then:
• In every 1 hour 15 min or $$\frac{5}{4}$$ hour, $$8+\frac{3}{4} or \frac{35}{4} l$$ water is getting filled.
• Hence, in 1 hour= $$\frac{35}{4}* \frac{4}{5}= 7$$ litres.
• Thus, time taken to fill 60 litres= $$\frac{60}{7}≈ 8.5$$ hours
• Hence, Time= 10AM+8h 30 min= 6:30 PM
Hence, the correct answer is option C.
We have to decide ...which combination will us maximum rate of filling.
combinations are
1. P alone with 15 minutes breaks
2. P, then 15 mins break, then again P....continue
3. P for 1 hour then 15 mins break, during 15 mins break run Q.....stop Q in 15 mins and then again run P.
Combination three will give maximum filling rate. solution is as given by Chetan sir
Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) &nbs [#permalink] 25 Jul 2018, 05:36
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# Abstract Cauchy problem
The condensed formulation of a Cauchy problem (as phrased by J. Hadamard) in an infinite-dimensional topological vector space. While it seems to have arisen between the two World Wars (F. Browder in [a2], Foreword), it was apparently introduced as such by E. Hille in 1952, [a2], Sec. 1.7.
Narrowly, but loosely speaking, the abstract Cauchy problem consists in solving a linear abstract differential equation (cf. also Differential equation, abstract) or abstract evolution equation subject to an initial condition. More precise explanations slightly differ from textbook to textbook [a2], [a5]. Following A. Pazy [a5], given a linear operator $A$ on a Banach space $X$ with domain $D ( A )$ and given an element $x _ { 0 } \in X$, one tries to solve
\begin{equation*} x ^ { \prime } ( t ) = A x ( t ) , t > 0 ; \quad x ( 0 ) = x 0, \end{equation*}
i.e., one looks for a continuous function $x$ on $[ 0 , \infty )$ such that $x$ is differentiable on $( 0 , \infty )$, $x ( t ) \in D ( A )$ for all $t > 0$, and $( d / d t ) x ( t ) = A x ( t )$ for all $t \in ( 0 , \infty )$.
Since $x$ is required to be continuous at $0$, the Cauchy problem can only be solved for $x _ { 0 } \in \overline { D ( A ) }$.
A Cauchy problem is called correctly set if the solution $x$ is uniquely determined by the initial datum $x _ { 0 }$. It is called well-posed (properly posed) if, in addition, the solution $x$ depends continuously on the initial datum $x _ { 0 }$, i.e., for every $\tau > 0$ there exists some constant $c > 0$ (independent of $x _ { 0 }$) such that
\begin{equation*} \| x ( t ) \| \leq c \| x _ { 0 } \| \text { for all } \, t \in [ 0 , \tau ], \end{equation*}
and all $x _ { 0 }$ for which a solution exists. Sometimes it is also required that solutions exist for a subspace of initial data which is large enough in an appropriate sense, e.g., dense in $X$.
The notion of a Cauchy problem can be extended to non-autonomous evolution equations [a2], [a5] and to semi-linear [a5], quasi-linear [a5], or fully non-linear evolution equations [a1], [a4]. In this process it may become necessary to replace classical solutions by more general solution concepts (mild solutions [a1], limit solutions [a4], integral solutions (in the sense of Ph. Bénilan; [a4]) in order to keep the problem meaningful. See [a1] and the references therein.
Well-posedness of linear Cauchy problems is intimately linked to the existence of $C _ { 0 }$-semi-groups of linear operators (cf. also Semi-group of operators), strongly continuous evolution families [a2], [a5] and related more general concepts like distribution semi-groups, integrated semi-groups, convoluted semi-groups, and regularized semi-groups, while the well-posedness of non-linear Cauchy problems is linked to the existence of non-linear semi-groups (the Crandall–Liggett theorem and its extensions) or (semi-) dynamical systems [a1], [a4], and to (evolutionary) processes and skew product flows [a3].
#### References
[a1] P. Benilan, P. Wittbold, "Nonlinear evolution equations in Banach spaces: Basic results and open problems" K.D. Bierstedt (ed.) A. Pietsch (ed.) W.M. Ruess (ed.) D. Vogt (ed.) , Functional Analysis , Lecture Notes Pure Appl. Math. , 150 , M. Dekker (1994) pp. 1–32 [a2] H.O. Fattorini, "The Cauchy problem" , Addison-Wesley (1983) [a3] J.K. Hale, "Asymptotic behavior of dissipative systems" , Amer. Math. Soc. (1988) [a4] V. Lakshmikantham, S. Leela, "Nonlinear differential equations in abstract spaces" , Pergamon (1981) [a5] A. Pazy, "Semigroups of linear operators and applications to partial differential equations" , Springer (1983)
How to Cite This Entry:
Abstract Cauchy problem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Abstract_Cauchy_problem&oldid=50297
This article was adapted from an original article by H. Thieme (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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# Tag Info
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Some history here might be useful; one of the oldest cosmological theories that we have was developed by Ionian philosophers which was an atomic theory; in that theory uncertainty as in random motion was taken as something fundamental (they called it the clinamen which is usually translated as swerve). This shows that pure determinism, physically, is ...
1
The Heisenberg Uncertainty Principle (HUP) holds for special observables, as energy and time, space and momentum, .. To every observable there corresponds a quantum mechanical operator. Quantum mechanical operators either commute or not commute, and are seen in the commutation relationships. Observables that do not commute are what the HUP is about. Is ...
-1
First and foremost we have to understand that if we are deriving laws of nature then our primary assumption is that the natural phenomenons are not random. If they are random, it would be impossible to say anything about them. Now let's come back to Quantum Mechanics. There is nothing random about the motion of electrons or any other subatomic particles ...
-1
There is something going on. Which is that certain observables are fundamentally incompatible. That means firstly that you can do an experiment for one observable or for another observable, but you can't do an experiment for both observables at the same time. And what's worse if you did an experiment for A then one for A again and then one for B the two ...
0
Out of the many quantum-mechanically possible states of an oscillator (be it a mechanical one or light waves), the ones we almost exclusively observe are the coherent states. In a way, they are the states where uncertainty is evenly distributed, such that every uncertain quantity scales as $\sqrt{N}$ for $N$ quanta (e.g. photons or energy quanta in an ...
1
A photon is both a particle and a wave is utterly incorrect. A photon is the carrier of the electromagnetic interaction and does not appear in the description of quantum mechanics until you introduce quantum field theory. a particle doesn't have both speed and momentum defined values at the same time Maybe you mean position and momentum (velocity ...
0
My issue here is it seems as though if his experiment were to be performed, it would violate known laws of physics. The picture and words in the section of the video you cited seem a little too vague to uniquely specify a specific experiment. But in the whole first 40 minutes of the video you see that all he is claiming is that we don't see departures ...
1
1
In nonrelativistic quantum mechanics the wavefunction is defined on configuration space. So for $n$ particles there is a $3n$ dimensional configuration space. And the quantum wavefunction is a function from $\mathbb R^{3n}$ into the complex numbers (or into a joint spin state if you have spin). There could be regions where the wave is zero or very small, ...
-3
As a quantum theorist, this area of physics perfectly intersects my area of knowledge. A simple Google search can lead you to my favourite site for philosophical inquiry. The Many-Worlds Interpretation (MWI) of quantum mechanics holds that there are many worlds which exist in parallel at the same space and time as our own. The existence of the other ...
1
I don't think that my terminology is suitable, because the term hidden realm carries the connotation of hidden variable theories, which is something else. I don't see them as very different. A hidden variable theory could take as the hidden variable the spinor or the wavefunction as a hidden variable. Many do. Events in the measurable realm ...
1
A priori, they could have been different things. The Equivalence Principle - the hypothesis that they are actually the same - is a core input the General Relativity. To the extent that General Relativity is empirically validated, we have evidence that these really are the same. There's no complete theory of quantum gravity, so I think we'd have to say ...
1
There are classical and quantum descriptions of the world. One of the differences of quantum description is paying attention to the process of measurement and how it affects the measured system. Description of measurements is an integral part of quantum description. Splitting this is into "realms" doesn't make much sense.
0
It is wrong to describe the equipment as "faulty" . It is just different and when you carry out the experiment with the "different" equipment you will get a different answer.The scientific method requires that experiments be repeatable.So if you can repeat the experiment with the "faulty" equipment then you have just carried out a new experiment.
3
No one thinks the wavefunction of nonrelativistic quantum mechanics is a wave in space and time. No one that wants to agree with observations that is. If someone is a realist about the wavefunction of nonrelativistic quantum mechanics then they start out by making a mathematical model by having the model include a wavefunction. Which when there are $n$ ...
0
Imagine a simpler set up. You have a double slit and you purchase a super fancy which way device and out it next to the left slit. But you forget to remove the wrapper it came with. So it just goes off randomly at random times based on some thermal properties of say the wall current you plug it into. You might incorrectly think that almost all the ...
0
Whether the MWI allows communication depends on what definition of communication you want to use. Communication in everyday life is something like this. I have information I want to send to you. I can arrange for you to have access to one copy while I have access to another. For example, you will be able to see this answer to your question on your computer ...
0
It's my opinion that the reason is as follows. Time evolution is just an illusion, the future versions of you just exist out there. What we refer to as the time evolution according to the Schrödinger equation is not really a time evolution in the sense that things change. Nothing changes, we only ever have the same static eternal multiverse. All you are ...
0
We don't need to worry about the whole universe here-- we can start with a single quantum coin flip in a sealed laboratory and take it from there. If you think you can consider a single particle in isolation then you fail to grasp the entire method by which separate worlds form in the MWI. It would be like it you tried to use the Copenhagen ...
1
Your example is bad. The reason your example is wrong, as covered by Emilio Pisanty, is that measurement of your state simply results in getting some random $(n, f(n))$ pair, which is very easy to do classically; in the theory-framework of a non-deterministic Turing machine we would say that it certainly contains a "probabilistic state" which has to ...
3
In principle there is nothing in the MWI which prevents communication between the different branches of the wavefunction. Indeed a (fairly controversial) paper was published in Foundations of Physics in 1997 with a proposal for an experiment to test this very possibility arXiv link. There are other proposals including several David Deutch which exploit the ...
3
No. More explicitly, there's nothing particularly quantum about your scheme. This is easy to see because it fails the crucial test of replacing superpositions with mixed states. That means that you can replicate exactly the same protocol using the density matrix $$\rho=\frac{1}{2^m}\sum_x|x⟩⟨x|\otimes|f(x)⟩⟨f(x)|.$$ If you measure the output register in ...
-1
Let's say we have a quantum computer with two registers taking in $m$ and $n$ qubits respectively, with $m$, $n$ suitably large. Let $f:\{0,1\}^m \rightarrow \{0,1\}^n$ be a one-way function. According to wikipedia the existence of a one way function on the natural numbers would prove $P\neq NP$ which I believe means the Clay institute would give you a ...
-1
The Copenhagen interpretation (CI) doesn't say anything about the register. Or, rather, different versions of the CI may say different things about the register. The CI has a number of ingredients: (1) The CI rejects the idea that quantum mechanics (QM) is an accurate description of how the world works. (2) The CI claims that QM can nevertheless be used to ...
2
In the Faraday pilot-wave fluid droplet dynamics, the fluid wave is meant as an analogy for the wavefunction. More specifically, the experiments are constructed as physical implementations analogous to the de Broglie-Bohm theory, where a particle with discrete coordinates is 'guided' by a pilot wave which follows the Schrödinger equation. To be a bit more ...
-4
In the oil droplet experiments that suggest de Broglie’s pilot wave theory might be accurate, what does the fluid surface correspond to? The three dimensional space that particle is moving through. But might I add that you shouldn't think of the particle as the oil droplet. Instead think of the particle as something more like a hurricane. The eye of the ...
0
As a particle travels to a screen, it is traveling through 3-dimensional space. No, what happens is the configuration space of the system changes. This is essential for explaining interactions that destroy the interference. If your right slit deflected downwards and the your left slit deflected upwards then the two waves wouldn't overlap and hence when ...
Top 50 recent answers are included
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# Entropy and Fractal Dimension ¶
In the study of dynamical systems there are many quantities that identify as "entropy". Notice that these quantities are not the more commonly known thermodynamic ones, used in Statistical Physics. Rather, they are more like the to the entropies of information theory, which represents information contained within a dataset, or information about the dimensional scaling of a dataset. Based on the definition of the generalized entropy, one can calculate an appropriate dimension.
###### Summary
In the study of dynamical systems there are many quantities that identify as "entropy". Notice that these quantities are not the more commonly known thermodynamic ones, used in Statistical Physics. Rather, they are more like the to the entropies of information theory, which represents information contained within a dataset, or information about the dimensional scaling of a dataset. Based on the definition of the generalized entropy, one can calculate an appropriate dimension.
March 22, 2019
###### Recent Posts
• Lorenz Attractor We start with exploring Lorenz differential equations (also known as Lorenz attractor) using...
• Poincaré Surface of Section Poincaré surface of section (also referred to as Poincaré section or Poincaré map) is powerful...
• Delay Embedding We try to reconstruct a chaotic dynamical system from a time-series using Taken's delay...
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# Math Help - Coin problem
1. ## Coin problem
I flip a symmetrical coin n times and I'm interested to know the expectation value of event A (the most probable number of times event happens), which is defined as fliping head m times in a row (m<=n). It is important to know that, for example m=10, fliping 20heads in a row represents two A events, not 11 events, also 19heads in a row represents only 1 event.
2. the expectation value of event A (the most probable number of times event happens)
These are not the same in general (mean and mode respectively). Which did you want?
3. Originally Posted by rgep
These are not the same in general (mean and mode respectively).Which did you want?
You are right. I want expectation value.
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What is the best BASIC Pocket Computer
02-28-2019, 12:33 AM (This post was last modified: 03-01-2019 04:01 AM by Valentin Albillo.)
Post: #121
Valentin Albillo Senior Member Posts: 885 Joined: Feb 2015
RE: What is the best BASIC Pocket Computer
.
Hi, Csaba:
(02-27-2019 09:21 AM)Csaba Tizedes Wrote: This is identical to EL-5510?
Yes, it is. Most SHARP PC-xxxx were also marketed in other countries renamed as SHARP EL-yyyy. I usually own one of each, though they're otherwise identical.
Quote:Do you have English pdf manual for SHARP PC-1421?
Financial Pocket Computer SHARP PC-1421 manual (German)
Nevertheless, apart from the financial functions keywords the rest of BASIC keywords and features are exactly like most other similar Level 2 BASIC SHARP pocket computers so any manual for those would do, the keywords and syntax are the same (though there were some other dedicated models with specific words for matrix operations, double precision (20 digits) and advanced statistic functions, and so the manuals for these specific models include dedicated keywords not present in the PC-1421/EL-5510).
As for the financial keywords, they mostly have the usual standard English names such as PV, FV, IRR, etc., and thus they're pretty easy to figure out if you're familiar with financial functions in general. The rest of the functionality can be easily deduced from the German manual by simply studying the many keyboard examples and the various financial application programs featured near the end of the manual. In a pinch, Google Translate can be used for very specific details.
There's also an Applications Book for this financial model, called (in English) "Dollars and Cents Computing" full of advanced financial programs using the built-in BASIC financial keywords, which can also serve as a useful source. I've seen it but I don't own it, perhaps it can be located as a net download or in TAS.
As a final comment on also using the financial SHARP PC-1421/EL-5510 for general purpose programming, not only does its 40 Kb ROM BASIC include all trigonometric functions in degrees, radians, and grads but also hyperbolic functions, linear regression, factorial, statistic functions, cube root, polar-rectangular conversions, the works ! Compare that to the HP-12C's function set ...
The dedicated Statistics model (SHARP PC-1425) is even rarer than the Financial model, which is saying something (never located an English manual for it ). I got one of those, too !
Quote:Thank you!
You're welcome !
Regards.
V.
.
All My Articles & other Materials here: Valentin Albillo's HP Collection
04-03-2019, 10:45 AM (This post was last modified: 04-03-2019 10:45 AM by foroplus.)
Post: #122
foroplus Junior Member Posts: 27 Joined: Dec 2015
RE: What is the best BASIC Pocket Computer
(02-24-2019 11:38 PM)toml_12953 Wrote:
(02-24-2019 11:13 PM)rprosperi Wrote: Also, see this really useful website by Bill Symmes, owner of P*ROM, one of the largest VARs for Sharp Pocket Computers through the 80's and 90's, for lots of information on Sharp's full line of Pocket Computers, including lots of downloadable manuals.
http://sharppocketcomputers.com/
It has no info on my favorite Sharp! The PC-G850VS.
You can use the PC-G850V(S) user manual, translated into English by Jack W. Hsu.
There is also a thread related to this Cal somewhere on this forum.
http://basic.hopto.org/basic/manuales.php
http://basic.hopto.org
04-03-2019, 01:28 PM
Post: #123
rprosperi Super Moderator Posts: 5,475 Joined: Dec 2013
RE: What is the best BASIC Pocket Computer
(04-03-2019 10:45 AM)foroplus Wrote: You can use the PC-G850V(S) user manual, translated into English by Jack W. Hsu.
There is also a thread related to this Cal somewhere on this forum.
http://basic.hopto.org/basic/manuales.php
Yes, thanks very much, I have Jack's great manual already (I followed when he released it here last year), however I had not discovered the page you shared before, so I've found many additional manuals here.
Also: Note that the PC-G850 manual on the linked page is v2.2, but Jack released version 3.0 a bit later.
--Bob Prosperi
04-03-2019, 05:36 PM
Post: #124
EugeneNine Senior Member Posts: 303 Joined: Feb 2017
RE: What is the best BASIC Pocket Computer
(03-30-2014 03:31 AM)Namir Wrote: Hi All,
Want to ask you all to give your opinion for the best (vintage) BASIC pocket Computers, taking into account the following aspects:
1. Amount of basic and upgradable memory.
2. Number of lines in display.
3. Flexible variable names (supporting two or more alpha characters).
4. Support for SUB routines
5. Support for separate user-defined function.
I am listening!
Namiir
Raspberry Pi zero
04-03-2019, 08:45 PM
Post: #125
foroplus Junior Member Posts: 27 Joined: Dec 2015
RE: What is the best BASIC Pocket Computer
(04-03-2019 01:28 PM)rprosperi Wrote:
(04-03-2019 10:45 AM)foroplus Wrote: You can use the PC-G850V(S) user manual, translated into English by Jack W. Hsu.
There is also a thread related to this Cal somewhere on this forum.
http://basic.hopto.org/basic/manuales.php
Yes, thanks very much, I have Jack's great manual already (I followed when he released it here last year), however I had not discovered the page you shared before, so I've found many additional manuals here.
Also: Note that the PC-G850 manual on the linked page is v2.2, but Jack released version 3.0 a bit later.
Thanks. I'll update the manual.
http://basic.hopto.org
05-07-2019, 08:24 PM
Post: #126
Androsynth Junior Member Posts: 15 Joined: May 2019
RE: What is the best BASIC Pocket Computer
I was into Pocket Computers long before I got into calculators. 'Best', as always, is subjective, but here are my favorites:
For sheer options and processing power, the Sharp g850 is the best. BASIC, C, Assembler, Monitor... this thing is a really great machine. Quite a few posts about it above, so I won't go into detail, but this is a very handy machine, even today. I have a BASIC dnd stats roller on mine for rpg sessions. Great machine, and sadly, the last of an era.
What's even better is that it's completely simple to build a serial <-> usb cable for these with about \$5 in parts (arduino wires and an ebay usb signal converter) which makes transferring files a snap. My only gripe is the lack of RAM expandibility.
For portability, I really like the Sharp PC-1250/Tandy PC-3 line. Tiny, but not hamstrung like some of the other small form factor BASIC computers. I quite enjoy these, even nowadays. They are probably the best compromise of all of the smaller form factor models.
The Sharp e500 and Casio VX-4 also get honorable mention.
06-13-2019, 06:56 PM
Post: #127
Dave Britten Senior Member Posts: 2,154 Joined: Dec 2013
RE: What is the best BASIC Pocket Computer
The Casio VX-4 I ordered from Japan just arrived today. Does anybody happen to know where to find documentation (English or Japanese) for this or a similar model? The basic operation seems similar enough to other Casios I've used, but there's a lot of stuff I'm not going to be able to work out by trial and error (the C compiler and assembler, for instance).
06-13-2019, 09:18 PM
Post: #128
cdmackay Senior Member Posts: 652 Joined: Sep 2018
RE: What is the best BASIC Pocket Computer
I'd be very interested to know that too; I have put off ordering one, because I couldn't find an English translation of the manual, and feared that much may be different from the 880.
Cambridge, UK
41CL/DM41X 12/15C/16C DM15/16 71B 17B/II/II+ 28S 42S/DM42 32SII 48GX 50g 35s WP34S PrimeG2 WP43S/pilot
Casio, Rockwell 18R
06-13-2019, 09:26 PM
Post: #129
Dave Britten Senior Member Posts: 2,154 Joined: Dec 2013
RE: What is the best BASIC Pocket Computer
(06-13-2019 09:18 PM)cdmackay Wrote: I'd be very interested to know that too; I have put off ordering one, because I couldn't find an English translation of the manual, and feared that much may be different from the 880.
Yeah, the VX-4 is totally different from the 850/880, but it's apparently very similar (and possibly identical) to the 870.
I was able to find a decent BASIC reference at least:
http://luckleo.cocolog-nifty.com/pockeco...ic_jp.html
And this simple C program compiles and runs as expected:
Code:
main() { int d; printf("Hello!￥n"); printf("Enter a number: "); scanf("%d",&d); printf("%d*3=%d￥n",d,d*3); return; }
Note that you use ￥ instead of a backslash, as with many Japanese computers from this era.
I tried declaring main with an int return type, but it throws a syntax error, so who knows what standard this thing adheres to.
06-13-2019, 10:10 PM
Post: #130
thenozone Junior Member Posts: 28 Joined: Mar 2017
RE: What is the best BASIC Pocket Computer
In response to the title ( but not vintage) some might be interested in basic for android by NS-ware, they seam to keep trying to improve it unlike many others, not free but worth a punt. Many android devices have a great display an can number crunch like crazy.
06-13-2019, 11:17 PM
Post: #131
Dave Britten Senior Member Posts: 2,154 Joined: Dec 2013
RE: What is the best BASIC Pocket Computer
(06-13-2019 10:10 PM)thenozone Wrote: In response to the title ( but not vintage) some might be interested in basic for android by NS-ware, they seam to keep trying to improve it unlike many others, not free but worth a punt. Many android devices have a great display an can number crunch like crazy.
Is that related to the old NSBasic for Palm OS and Windows CE? Back then, the IDE ran on a desktop, and you had to copy the compiled apps to your device.
06-14-2019, 12:12 AM
Post: #132
cdmackay Senior Member Posts: 652 Joined: Sep 2018
RE: What is the best BASIC Pocket Computer
thanks Dave,
(06-13-2019 09:26 PM)Dave Britten Wrote: Yeah, the VX-4 is totally different from the 850/880, but it's apparently very similar (and possibly identical) to the 870.
This post notes:
"AFAIK the only difference between the FX-870P and the VX-4 is the color scheme and the built-in
RAM. Surely the educational VX-4 (8K) was cheaper than the commercial FX-870P (32K)."
Cambridge, UK
41CL/DM41X 12/15C/16C DM15/16 71B 17B/II/II+ 28S 42S/DM42 32SII 48GX 50g 35s WP34S PrimeG2 WP43S/pilot
Casio, Rockwell 18R
06-14-2019, 12:54 AM
Post: #133
Dave Britten Senior Member Posts: 2,154 Joined: Dec 2013
RE: What is the best BASIC Pocket Computer
(06-14-2019 12:12 AM)cdmackay Wrote: thanks Dave,
(06-13-2019 09:26 PM)Dave Britten Wrote: Yeah, the VX-4 is totally different from the 850/880, but it's apparently very similar (and possibly identical) to the 870.
This post notes:
"AFAIK the only difference between the FX-870P and the VX-4 is the color scheme and the built-in
RAM. Surely the educational VX-4 (8K) was cheaper than the commercial FX-870P (32K)."
Okay, so just memory and a different paint job, then.
I pulled a Tandy 8 KB RAM module out of a PC-6 with a wonky ribbon cable and put it in the VX-4, and now I've got about 11.5 KB for program storage. That should be enough to tinker with, though I wouldn't say no to a good deal on a 32 KB module.
06-14-2019, 12:29 PM
Post: #134
Dave Britten Senior Member Posts: 2,154 Joined: Dec 2013
RE: What is the best BASIC Pocket Computer
I found a really good English reference for the fx-890P, which appears to be similar enough to the VX-4/fx-870P that it's still extremely useful for those machines:
I have noticed some differences in functionality and available commands - the VX-4 doesn't appear to support labels, or the BASIC graphics commands, for instance. It also appears to be lacking graphical functions in C.
The C dialect is K&R, with its weird syntax for declaring function argument types, and forward-declaration/prototyping of functions is messy at best. You're better off just defining the functions before they're used. Or better yet, use BASIC, because the C interpreter is so much slower.
07-05-2019, 04:16 PM
Post: #135
cdmackay Senior Member Posts: 652 Joined: Sep 2018
RE: What is the best BASIC Pocket Computer
I managed to pick up an fx-850P in nice condition on TAS for £35 incl. P&P.
Given the only difference (I think) between the 850 & 880 is 8KB/32KB RAM, and I’m unlikely to be writing long programmes, and the 880 seems to regularly go for £100 and up, I’m quite pleased.
Arrived today, came with plastic slide-on hard cover & 405pp. paper manual.
Three utterly dead batteries, which I removed, luckily no leakage. Put in two new CR2032, and it’s come to life, all seems well. I don’t have any CR1220 (the backup battery), so will order some.
As an added bonus, I’ve just found the laminated sci function quick reference card tucked into the manual as a bookmark, which wasn’t mentioned in the listing.
Something else new to play with
Casio880 are selling a USB connection cable for it; I wonder if I can get that to work with MacOS or Linux? No Windows here… I’ll find out.
Cambridge, UK
41CL/DM41X 12/15C/16C DM15/16 71B 17B/II/II+ 28S 42S/DM42 32SII 48GX 50g 35s WP34S PrimeG2 WP43S/pilot
Casio, Rockwell 18R
07-06-2019, 02:59 PM
Post: #136
Leviset Member Posts: 145 Joined: Aug 2015
RE: What is the best BASIC Pocket Computer
I’d have to say the TI-95 Procalc but you’ll need a large pocket.
My other favourite is my Sharp PC-1500. I have the base unit that it slots into which adds a tape cassette recorder and a mini printer. I’ve even recently tracked down someone who still sells the ALPS 4 colour/color pens. It’s all in working order.
Denny Tuckerman
07-06-2019, 03:50 PM
Post: #137
Dave Britten Senior Member Posts: 2,154 Joined: Dec 2013
RE: What is the best BASIC Pocket Computer
I just got a Casio fx-730P the other day, and I really like it. It has two memory addressing modes: DEFM and DIM. DEFM works like the corresponding memory configuration function on Casio calculators, adding more variables "after" Z. In this mode, you can use array indexes to address variables relative to one another (so C(3) is the same as F). In DIM mode, you allocate and deallocate arrays with up to 3 dimensions, like a more traditional BASIC.
I still can't get my FA-3 cassette interface to work, though. It sounds like it's saving just fine, but the computer acts like it doesn't hear anything when attempting to load.
11-22-2019, 05:27 PM
Post: #138
Csaba Tizedes Senior Member Posts: 554 Joined: May 2014
RE: What is the best BASIC Pocket Computer
(03-30-2014 03:39 PM)W_Max Wrote: And even more capable with Turbo BASIC from Borland
It can be run on 95LX and/or 200LX?
11-23-2019, 07:22 PM (This post was last modified: 11-23-2019 11:39 PM by ijabbott.)
Post: #139
ijabbott Senior Member Posts: 1,204 Joined: Jul 2015
RE: What is the best BASIC Pocket Computer
(07-06-2019 03:50 PM)Dave Britten Wrote: I still can't get my FA-3 cassette interface to work, though. It sounds like it's saving just fine, but the computer acts like it doesn't hear anything when attempting to load.
I trust you've tried the obvious stuff such as adjusting the volume?
I've got an FA-1 and an FA-2 working with an old, 1980s "computer compatible" cassette recorder. I've changed the belt and cleaned the heads. The audio still sounds bloody awful, but seems good enough for loading and saving Casio tapes. (The next challenge will be ZX Spectrum fast loaders when or if my "ZX Spectrum Next" finally shows up.)
I suppose I now need an FX-700P to use with the FA-2, and so it continues....
EDIT: It's the FX-702P (not the FX-700P) that uses the FA-2 cassette interface (which can also be used by the FX-501P, FX-502P, FX-601P and FX-602P).
— Ian Abbott
11-23-2019, 07:43 PM
Post: #140
Dave Britten Senior Member Posts: 2,154 Joined: Dec 2013
RE: What is the best BASIC Pocket Computer
(11-23-2019 07:22 PM)ijabbott Wrote:
(07-06-2019 03:50 PM)Dave Britten Wrote: I still can't get my FA-3 cassette interface to work, though. It sounds like it's saving just fine, but the computer acts like it doesn't hear anything when attempting to load.
I trust you've tried the obvious stuff such as adjusting the volume?
I've got an FA-1 and an FA-2 working with an old, 1980s "computer compatible" cassette recorder. I've changed the belt and cleaned the heads. The audio still sounds bloody awful, but seems good enough for loading and saving Casio tapes. (The next challenge will be ZX Spectrum fast loaders when or if my "ZX Spectrum Next" finally shows up.)
I suppose I now need an FX-700P to use with the FA-2, and so it continues....
I suspect it was a volume thing. I got a Tandy Computer Cassette Recorder at VCFMW and it works great.
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# Simple Harmonic Motion proof
1. Feb 18, 2015
### Cpt Qwark
1. The problem statement, all variables and given/known data
Prove that:
$$x=8sin2t+6cos2t$$ is undergoing S.H.M.
(Not too sure about how to prove for solution.)
2. Relevant equations
Solution for S.H.M. $$x=asin(nt+α)$$ is $$\frac{d^{2}x}{dy^{2}}=-n^2x$$
3. The attempt at a solution
$$r=\sqrt{8^{2}+6^{2}}=10\\α=tan^{-1}\frac{3}{4}\\∴x=10sin(2t+tan^{-1}\frac{3}{4})$$
Differentiating with respect to time: $$\frac{dx}{dt}=20cos(2t+tan^{-1}\frac{3}{4})\\\frac{d^{2}x}{dt^{2}}=-40sin(2t+tan^{-1}\frac{3}{4})$$
2. Feb 18, 2015
### Nathanael
Right. Adding any two sinusoidal functions of the same frequency will result in another sinusoidal function, regardless of their amplitudes.
If you solve the SHM differential equation, $\frac{d^2x}{dt^2}=-kx$ you will get $x=C_1\sin(\sqrt{k}t)+C_2\cos(\sqrt{k}t)$ and it because of the above fact that you can write the solution as $x=C_3\sin(\sqrt{k}t+C_4)$
Last edited: Feb 18, 2015
3. Feb 19, 2015
### rude man
You can just take this equation, compute x', then x'', and see that ω2 must = 4 by equating sine and cosine coefficients. Both yield the same answer ω2 = 4. Had the sine & cosine coeff. yielded differing ω then x(t) would not be shm.
4. Apr 29, 2015
### Physicist123
Take the second derivative of the given expression and express it in terms of x. The result would eliminate sin and cos and will prove your answer in form of a=-nx.
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# Areas and perimeters of sectors (degrees)
Lesson
Like we saw in our chapters describing circumferences and areas of circles we now have the following rules.
$\text{Circumference, C}=2\pi r$Circumference, C=2πr
$\text{Area, A}=\pi r^2$Area, A=πr2
What if we don't have an entire circle?
Well, half a circle would have half the area or half the circumference. One quarter of a circle would have a quarter of the area, or a quarter of the circumference. In fact all we need to know is what fraction the sector is of a whole circle. For this all we need to know is the angle of the sector.
Looking at the quarter circle, the angle of the sector is $90$90°. The fraction of the circle is $\frac{90}{360}=\frac{1}{4}$90360=14
More generally, If the angle of the sector is $\theta$θ, then the fraction of the circle is represented by
$fraction=\frac{\theta}{360}$fraction=θ360 (due to there being $360$360° in a circle).
#### Example
Question: Find the area and circumference of a sector with central angle of $126$126° and radius of $7$7cm. Evaluate to $2$2 decimal places.
Think: What fraction is this sector of a whole circle? What are the rules for circumference and area?
Do: This sector is $\frac{126}{360}=0.35$126360=0.35 of a circle.
Circumference of a whole circle is $C=2\pi r$C=2πr, so the perimeter of the sector is
$0.35\times2\pi r$0.35×2πr $=$= $0.35\times2\pi\times7$0.35×2π×7 $=$= $0.35\times14\pi$0.35×14π $=$= $4.9\times\pi$4.9×π $=$= $15.39$15.39 cm (rounded to $2$2 decimal places)
Area of a circle is $A=\pi r^2$A=πr2, so the area of the sector is
$0.35\times\pi r^2$0.35×πr2 $=$= $0.35\pi\times7^2$0.35π×72 $=$= $17.15\pi$17.15π $=$= $53.88$53.88 cm2 (rounded to 2 decimal places)
#### Worked Examples
##### QUESTION 1
Consider the sector below.
1. Calculate the perimeter. Give your answer correct to $2$2 decimal places.
2. Calculate the area. Give your answer correct to $2$2 decimal places.
##### QUESTION 2
Consider the sector below.
A goat is tethered to a corner of a fenced field (shown). The rope is $9$9 m long. What area of the field can the goat graze over?
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Language: Search: Contact
Zentralblatt MATH has released its new interface!
For an improved author identification, see the new author database of ZBMATH.
Query:
Fill in the form and click »Search«...
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Display: entries per page entries
Zbl 1214.34059
Baštinec, J.; Dibl{\'\i}k, J.; Khusainov, D.Ya.; Ryvolová, A.
Exponential stability and estimation of solutions of linear differential systems of neutral type with constant coefficients.
(English)
[J] Bound. Value Probl. 2010, Article ID 956121, 20 p. (2010). ISSN 1687-2770/e
The paper deals with the linear systems of neutral differential equations with constant coefficients and a constant delay of the form $$\dot{x}(t)=D\dot{x}(t-\tau)+Ax(t)+Bx(t-\tau),$$ where $t\geq 0$, $\tau>0$, $A,B,$ and $D$ are $n\times n$ constant matrices, and $x:[-\tau,\infty)\to\Bbb{R}^n$ is a column vector-solution. The authors investigate the exponential-type stability of such systems using Lyapunov-Krasovskii type functionals. Delay-dependent conditions sufficient for the stability are formulated in terms of positivity of auxiliary matrices. Illustrative examples are shown and comparisons with known results are given.
[Jan Ohriska (Košice)]
MSC 2000:
*34K20 Stability theory of functional-differential equations
34K40 Neutral equations
34K06 Linear functional-differential equations
Keywords: stability theory; neutral equations; linear functional-differential equations
Highlights
Master Server
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# Selecting objects to maximize value while under multiple constraints
I think it will be best to explain the problem first and then explain what I am thinking:
I have a dataset similar to the following:
fruit,calories,cost
apple,100,1
apple,200,1.5
apple,150,2
pear,300,3
pear,100,.5
pear,250,2
orange,100,1
orange,120,1
orange,400,2
And I am trying to maximize my calories while keeping my cost within a certain range. At this point it is just a knapsack problem, but what if I have to have 1 apple, 1 pear, and 2 oranges, or some other arbitrary set of fruit numbers?
I can't really wrap my head around how this should work, given the extra constraint. I've thought about trying to merge my cost and calories into 1 metric somehow, but I am pretty sure that can't work since it loses.
My most recent thought is potentially keeping track of fruit count in a list, and if the proper amount of fruit has been reached it will skip to the next fruit. I am thinking about this in a similar way to a knapsack with 3 constraints, weight, value, and size, but the size is just fruit count essentially.
Hoping someone on here can tell me if I am heading in the right direction or if there is an algorithm that does this that I can look into.
Thank you!
• Isn't it possible that you distribute what is needed first and then after you've done that, the problem reduces to your standard Knapsack problem? Or am I missing something? – Gokul Jul 18 at 13:39
• Maybe, I am not sure what you mean though, can you explain a bit more? – Jacob D Jul 18 at 23:35
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# Base Equation
Find the value of $$n$$ such that the following equality holds:
$\large\overline{(n-1) (n-2) (0)}_n = \overline{(n-2) (n-3) (1)}_{n+1}, \qquad |n| > 3$
Clarification: The subscript indicates number base. $$(n-1),(n-2), (0),\ldots$$ are digits of the number.
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Screenshots. φ [ν←τ] where τ is free for ν in φ. Since this is all about math, I copy some content from wikipedia for a start. for details. A first prototype of a ProB Logic Calculator is now available online. You need to “summarize” what was established after making the desired assumption (the contradiction of the conclusion). Valid or Invalid? The rules of this test are simple: it's your job to determine whether an argument is valid or not. Amount of Whisky (fl. In the dropdown menu, click 'UserDoc'. Matrices & Vectors. Truth Tables, Logic, and DeMorgan's Laws. Logic Gate Simulator. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. This allows for submission and automated marking of exercises such as symbol-ization, truth tables, and natural deduction proofs. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. Yes, Algebraic Proofs isn't particularly exciting. Follow the 2 steps guide to find the truth table using the boolean calculator. The symbol for this is $$ν$$. Then k2 = (ax)2 = x(a2x) so xjk2. Read from here about the differences between algorithms. Proof generator and proof checker for propositional logic in "natural deduction" style. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. It formalizes the rules of logic. Proof Checker for forall x: Cambridge and Calgary. Modifications by students and faculty at Cal. Send me a full list of your axioms and I will see what I can do to get you started. (There was the untyped logic language Prolog, and the strongly typed — but general programming language. Propositional Logic • Propositional resolution • Propositional theorem proving •Unification Today we're going to talk about resolution, which is a proof strategy. Since any element x in K is also in S, we know that every element x in K is also in S, thus K S. Boolean Algebra expression simplifier & solver. Use symbolic logic and logic algebra. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Yes, Algebraic Proofs isn't particularly exciting. It executes the logical operations like AND, NAND, OR, NOR, NOT & X-OR. Write a symbolic sentence in the text field below. truth tables, normal forms, proof checking, proof building). Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. Boolean algebra calculator is the stream of mathematics that comprises of logical expressions & logical variables manipulating. Propositional logic in Artificial intelligence. In the dropdown menu, click 'UserDoc'. Amount of Whisky (fl. Line Equations Functions Arithmetic & Comp. The Coq Proof Assistant. At any time get assistance and ideas from Proof generator. Free Python 3. The specific system used here is the one found in forall x: Calgary Remix. Select gates from the dropdown list and click "add node" to add more gates. Drag from the hollow circles to the solid circles to make connections. Thus, x 2S. When combined together, several gates can make a complex logical evaluation system that has. Conic Sections Transformation. Logic Gate Simulator. Natural deduction proof editor and checker. Go to Daemon Proof Checkeror Quick Help Index. Free Python 3. This site based on the Open Logic Project proof checker. The Logic Daemon. The Propositional Logic Calculator finds all the models of a given propositional formula. Know Your Worth is based on millions of real salaries from Glassdoor users. See Credits. Boolean Algebra expression simplifier & solver. Perfect Proof. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. The system was originally written for UMass’s Intro Logic course, based on Gary Hardegree’s online. Disjunctive normal form (DNF), including perfect. In each step the user. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. Some (importable) sample proofs in the "plain" notation are here. (If you don't want to install this file. They will show you how to use each calculator. See Credits. Line Equations Functions Arithmetic & Comp. 3: Proofs in predicate logic. Modifications by students and faculty at Cal. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. Here, instead, we shall think of this as a proof method, traditionally called “conditional derivation”. Mathematical logic step by step. Com stats: 2613 tutors , 730556 problems solved. One of the most basic rules of deduction in predicate logic says that ( ∀ x P ( x)) P ( a) for any entity a in the domain of discourse of. To typeset these proofs you will need Johann Klüwer's fitch. Place brackets in expressions, given the priority of operations. The Coq Proof Assistant. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. So far in this section, we have been working mostly with propositional logic. For an introduction to logic and proof in this style, consult a textbook such as Kaye , Huth and Ryan , or Bornat. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. Online tool. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. First, we'll look at it in the propositional case, then in the first-order case. Natural deduction proof editor and checker. Proof checker. Proof Checker for forall x: Cambridge and Calgary. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. Build a truth table for the formulas entered. Use symbolic logic and logic algebra. Resolution Refutation. Line Equations Functions Arithmetic & Comp. One of the most basic rules of deduction in predicate logic says that ( ∀ x P ( x)) P ( a) for any entity a in the domain of discourse of. The thing solves algebra, and basic symbolic logic uses, well, I don't want to say the same sort of symbol manipulation because the overlap is imperfect, but both proofs and algebra work by manipulating symbols via a set of well-defined rules. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. Truth Tree Solver. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. The values of the Boolean algebra calculator are denoted with logic 0 & 1. Conjunctive normal form (CNF), including perfect. Right click connections to delete them. All the arguments are syllogisms. Binary numbers multiplication is a part of arithmetic operations in digital electronics. A first prototype of a ProB Logic Calculator is now available online. Propositional logic is also amenable to “deduction,” that is, the development of proofs by writing a series of lines, each of which either is given or is justified by some previous lines (Section 12. You can select and try out several solver algorithms: the "DPLL better" is the best solver amongst the options. Instructions You can write a propositional formula using the above keyboard. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. for details. Know Your Worth is based on millions of real salaries from Glassdoor users. The Corbettmaths Practice Questions on Algebraic Proof. One Flip application is a proof checker for entering and editing proofs in natural deduction style. Application works on the Chrome browser. The logic language used in this theorem prover is one that was proposed in the author's Master's thesis, back in 1985-1987, at which time it contained most of the features shown here, including the hierarchical type scheme. Know Your Worth is based on millions of real salaries from Glassdoor users. Thus, the argument above is valid, because if all humans are mortal, and if. Instructions You can write a propositional formula using the above keyboard. Place brackets in expressions, given the priority of operations. One of the most basic rules of deduction in predicate logic says that ( ∀ x P ( x)) P ( a) for any entity a in the domain of discourse of. The Conformal Smart Logic Equivalence Checker (LEC) is the next-generation equivalency checking solution. A proposition is a declarative statement which is either true or false. A proof is an argument intended to convince the reader that a general principle is true in all situations. Simplify logical expressions. Try the leading salary calculator the next time you negotiate your salary or ask for a raise, and get paid fairly. This calculator solves linear diophantine equations. φ [ν←τ] where τ is free for ν in φ. This study aid includes: Proof generator. To download DC Proof and for a contact link, visit my homepage. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. It formalizes the rules of logic. Look at line 3. 3: Proofs in predicate logic. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. By using this website, you agree to our Cookie Policy. In mathematics, a Diophantine equation is a polynomial equation in two or more unknowns such that only the integer solutions are searched or studied (an integer solution is a solution such that all the unknowns take integer values). Know Your Worth is based on millions of real salaries from Glassdoor users. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). In other words, it’s based on the mistaken assumption that a lack of evidence is evidence. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. The symbol for this is $$ν$$. (There was the untyped logic language Prolog, and the strongly typed — but general programming language. Use symbolic logic and logic algebra. Modifications by students and faculty at Cal. A variable is a symbol used to represent a logical quantity. Instructors. It is a web application that uses Semantic Tableaux to check the validity of a statement, and provides a proof if it finds that the statement is valid. Line Equations Functions Arithmetic & Comp. This is the mode of proof most of us. Free Python 3. Let x 2K so that xjk. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. Here is a standard example: An argument is valid if and only if the conclusion necessarily follows from the premises. Using the derived rule allowed us to shorten the proof considerably. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). In the dropdown menu, click 'UserDoc'. φ [ν←τ] where τ is free for ν in φ. 3: Proofs in predicate logic. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. 0 is based on classical logic, but it is possible to define your axioms in it. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. Logic Calculator. Propositional Logic • Propositional resolution • Propositional theorem proving •Unification Today we're going to talk about resolution, which is a proof strategy. Boolean Algebra expression simplifier & solver. Oct 24 '18 at 20:18. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. Matrices & Vectors. Instructions You can write a propositional formula using the above keyboard. Proof generator and proof checker for propositional logic in "natural deduction" style. Step through the examples. This study aid includes: Proof generator. Instructors. Learning about Coq. So far in this section, we have been working mostly with propositional logic. Basics Whisky 101. Free Logical Sets calculator - calculate boolean algebra, truth tables and set theory step-by-step This website uses cookies to ensure you get the best experience. Truth Tables, Logic, and DeMorgan's Laws. Get Custom Built Calculator For Your Website. See this pdf for an example of how Fitch proofs typeset in LaTeX look. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. Propositional logic is also amenable to “deduction,” that is, the development of proofs by writing a series of lines, each of which either is given or is justified by some previous lines (Section 12. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. A drill for the truth functional connectives. Amount of Whisky (fl. Typical applications include the certification of properties of. First, we'll look at it in the propositional case, then in the first-order case. Use symbolic logic and logic algebra. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Natural Deduction examples | rules | syntax | info | download | home: Last Modified : 13-Jun-2021. Truth Tree Solver. Matrices & Vectors. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. Get Custom Built Calculator For Your Website. See Credits. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. The Propositional Logic Calculator finds all the models of a given propositional formula. The assumption set that results is the same as the assumption set for line 3. Using the derived rule allowed us to shorten the proof considerably. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Instructors. Conic Sections Transformation. Logic Gate Simulator. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. A model describes how units of computations, memories, and communications are organized. Back then, the idea of logic languages with types was novel. This study aid includes: Proof generator. See Credits. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. This calculator solves linear diophantine equations. Appeal to ignorance is a logical fallacy in which someone argues either for or against something because there is no contradicting evidence. Natural deduction proof editor and checker. Binary numbers multiplication is a part of arithmetic operations in digital electronics. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. It formalizes the rules of logic. Enter the term you would like to use: τ: Existential Introduction: φ (τ) Eν:φ (ν) Enter the term you would like to replace: τ:. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. Clicking the "Tree Proof" button will pass the statement to wo's Tree Proof Generator. This page is a tutorial and user's guide; there is also a complete reference. LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. Logic Calculator is a free, portable truth table generator for logic formulas i. Matrices & Vectors. Yes, Algebraic Proofs isn't particularly exciting. Solving a classical propositional formula means looking for such values of variables that the formula becomes true. Line Equations Functions Arithmetic & Comp. Enter the term you would like to use: τ: Existential Introduction: φ (τ) Eν:φ (ν) Enter the term you would like to replace: τ:. Logic Calculator. Amount of Whisky (fl. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Computer programs are constantly making decisions based on the current "STATE" of the data held by the program. For modal predicate logic, constant domains and rigid terms are assumed. Get help from our free tutors ===>. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. Oct 24 '18 at 20:18. Combining multiple conditions to form one True/False value is the. Proof generator and proof checker for propositional logic in "natural deduction" style. This allows for submission and automated marking of exercises such as symbol-ization, truth tables, and natural deduction proofs. (2) Logical entailment: displays the truth table of each of the premises along with the. Practice your deduction skills with Proof checker and Random Tasks. Follow the 2 steps guide to find the truth table using the boolean calculator. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. The Coq Proof Assistant. This style of proof is called a resolution proof. Yes, Algebraic Proofs isn't particularly exciting. Take help from sample expressions in the input box or have a look at the boolean functions in the content to understand the mathematical operations used in expressions. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses’s free, online logic teaching software application Carnap (carnap. Thus, the argument above is valid, because if all humans are mortal, and if. Each one has a different shape to show its particular function. Back then, the idea of logic languages with types was novel. This site based on the Open Logic Project proof checker. A model describes how units of computations, memories, and communications are organized. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. Write a symbolic sentence in the text field below. For example, (a -> b) & a becomes true if and only if both a and b are assigned true. proof language Isar [20] which looks like a mixture of English, logic and a programming language, and is based on natural deduction. Valid or Invalid? The rules of this test are simple: it's your job to determine whether an argument is valid or not. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. We applied Demorgan's Law, which is abbreviated DM, to line 2. It executes the logical operations like AND, NAND, OR, NOR, NOT & X-OR. Let x 2K so that xjk. for details. A variable is a symbol used to represent a logical quantity. 0 is based on classical logic, but it is possible to define your axioms in it. Mathematical logic step by step. All in one boolean expression calculator. Propositional logic is also amenable to “deduction,” that is, the development of proofs by writing a series of lines, each of which either is given or is justified by some previous lines (Section 12. See Credits. Logic Gate Simulator. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. A proof is an argument intended to convince the reader that a general principle is true in all situations. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. 2 Proofs One of the principal aims of this course is to teach the student how to read and, to a lesser extent, write proofs. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. The Gateway to Logic is a collection of web-based logic programs offering a number of logical functions (e. Yes, Algebraic Proofs isn't particularly exciting. Logic gates are symbols that can directly replace an expression in Boolean arithmetic. When combined together, several gates can make a complex logical evaluation system that has. Natural Deduction is a free app published for Windows 10 PC and can be downloaded from Windows Store. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. Yes, Algebraic Proofs isn't particularly exciting. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences. Know Your Worth is based on millions of real salaries from Glassdoor users. Suppose k 2Z and let K = fn 2Z : njkgand S = fn 2Z : njk2g. So far in this section, we have been working mostly with propositional logic. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. The values of the Boolean algebra calculator are denoted with logic 0 & 1. Thus, x 2S. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. (whenever you see $$ν$$ read 'or') When two simple sentences, p and q, are joined in a disjunction statement, the disjunction is expressed symbolically as p $$ν$$ q. When loaded, click 'Help' on the menu bar. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. Online tool. Learning about Coq. Here, instead, we shall think of this as a proof method, traditionally called “conditional derivation”. Natural Deduction is a free app published for Windows 10 PC and can be downloaded from Windows Store. Mathematical logic step by step. This is the mode of proof most of us. Go to Daemon Proof Checkeror Quick Help Index. When combined together, several gates can make a complex logical evaluation system that has. Know Your Worth is based on millions of real salaries from Glassdoor users. The symbol P denotes a sum over its argument for each natural. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Boolean Algebra expression simplifier & solver. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. This allows for submission and automated marking of exercises such as symbol-ization, truth tables, and natural deduction proofs. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses’s free, online logic teaching software application Carnap (carnap. Note that proofs can also be exported in "pretty print" notation (with unicode logic symbols) or LaTeX. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. Get Custom Built Calculator For Your Website. Learn boolean algebra. Yes, Algebraic Proofs isn't particularly exciting. Kevin Klement has done up a prototype of his online natural deduction proof builder/checker that works with the natural deduction system of the Cambridge and Calgary versions of forall x. LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. - Dan Christensen. Simplify logical expressions. Free Ubuntu. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. Using the derived rule allowed us to shorten the proof considerably. All the arguments are syllogisms. This style of proof is called a resolution proof. Use symbolic logic and logic algebra. Boolean Expression Calculator. It executes the logical operations like AND, NAND, OR, NOR, NOT & X-OR. Propositional sequent calculus prover. State University, Monterey Bay. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. A free, simple, online logic gate simulator. Place brackets in expressions, given the priority of operations. Then k2 = (ax)2 = x(a2x) so xjk2. Thus, x 2S. Disjunctive normal form (DNF), including perfect. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. It is a technique of knowledge representation in logical and mathematical form. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. Since this is all about math, I copy some content from wikipedia for a start. Because of its simplicity it is particularly well-suited for mechanical theorem provers. Simplify logical expressions. Proof by induction involves statements which depend on the natural numbers, n = 1,2,3, It often uses summation notation which we now briefly review before discussing induction itself. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Amount of Whisky (fl. You can select and try out several solver algorithms: the "DPLL better" is the best solver amongst the options. Complete your profile, and we will calculate how much you could earn in today's job market. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. This site based on the Open Logic Project proof checker. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Free Python 3. Drag from the hollow circles to the solid circles to make connections. All the arguments are syllogisms. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Yes, Algebraic Proofs isn't particularly exciting. The thing solves algebra, and basic symbolic logic uses, well, I don't want to say the same sort of symbol manipulation because the overlap is imperfect, but both proofs and algebra work by manipulating symbols via a set of well-defined rules. Natural deduction proof editor and checker. One of the most basic rules of deduction in predicate logic says that ( ∀ x P ( x)) P ( a) for any entity a in the domain of discourse of. They will show you how to use each calculator. In each step the user. The specific system used here is the one found in forall x: Calgary Remix. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. Learn more. The symbol P denotes a sum over its argument for each natural. You need to “summarize” what was established after making the desired assumption (the contradiction of the conclusion). Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Complete your profile, and we will calculate how much you could earn in today's job market. Matrices & Vectors. A sequent S is true if and only if there exists a tree of sequents rooted at S where each leaf is an axiom and each internal node is derived from its children by an inference rule. Such proofs can also encode traditional proofs based on modus ponens: the inference P∧(P⇒Q) ⊢ Q can be rewritten as resolution by expanding ⇒ to get P∧(¬P∨Q) ⊢ Q. See Credits. For example, a heart monitoring program might sound an alarm if the pulse is too slow or the blood pressure is too weak. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. ) And that’s it! Enjoy drinking your whisky exactly how you like it. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. Free Windows Terminal Preview. Back then, the idea of logic languages with types was novel. You can select and try out several solver algorithms: the "DPLL better" is the best solver amongst the options. Formal Proof Check. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. Com stats: 2613 tutors , 730556 problems solved. ) Bottle Proof. Videos, worksheets, 5-a-day and much more. Take help from sample expressions in the input box or have a look at the boolean functions in the content to understand the mathematical operations used in expressions. State University, Monterey Bay. techniques as “proof by contradiction” or “proof by contrapositive” (Section 12. The values of the Boolean algebra calculator are denoted with logic 0 & 1. See Credits. Propositional logic in Artificial intelligence. Desired Proof. Kevin Klement has done up a prototype of his online natural deduction proof builder/checker that works with the natural deduction system of the Cambridge and Calgary versions of forall x. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Save your work on device and continue later on. We applied Demorgan's Law, which is abbreviated DM, to line 2. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Logic gates are symbols that can directly replace an expression in Boolean arithmetic. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). Logic Calculator. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. The symbol for this is $$ν$$. Back then, the idea of logic languages with types was novel. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. Resolution Refutation. Since this is all about math, I copy some content from wikipedia for a start. The symbol P denotes a sum over its argument for each natural. The Logic Daemon. 3: Proofs in predicate logic. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. This app is a graphical semantic calculator for a specific kind of. Free Logical Sets calculator - calculate boolean algebra, truth tables and set theory step-by-step This website uses cookies to ensure you get the best experience. Any single variable can have a 1 or a 0 value. Proof Checker for forall x: Cambridge and Calgary. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Logic Gate Simulator. Conic Sections Transformation. The Corbettmaths Practice Questions on Algebraic Proof. Practice your deduction skills with Proof checker and Random Tasks. Propositional sequent calculus prover. Binary numbers multiplication is a part of arithmetic operations in digital electronics. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. ) Bottle Proof. If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. (whenever you see $$ν$$ read 'or') When two simple sentences, p and q, are joined in a disjunction statement, the disjunction is expressed symbolically as p $$ν$$ q. Diana Higgins on Conditional-proof-logic-calculator glennnervi. Combining multiple conditions to form one True/False value is the. A variable is a symbol used to represent a logical quantity. Any single variable can have a 1 or a 0 value. See this pdf for an example of how Fitch proofs typeset in LaTeX look. The Corbettmaths Practice Questions on Algebraic Proof. The symbol for this is $$ν$$. In computer science, and more specifically in computability theory and computational complexity theory, a model of computation is a model which describes how an output of a mathematical function is computed given an input. Binary numbers multiplication is a part of arithmetic operations in digital electronics. By using this website, you agree to our Cookie Policy. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. Online tool. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. Propositional logic (PL) is the simplest form of logic where all the statements are made by propositions. Yes, Algebraic Proofs isn't particularly exciting. Conjunctive normal form (CNF), including perfect. Get help from our free tutors ===>. Actually there are mechanical ways of generating Fitch style proofs. Once you know your perfect proof, this calculator will tell you exactly how much water to add to any amount of whisky to reach it. Boolean formulas are written as sequents. To typeset these proofs you will need Johann Klüwer's fitch. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. Formal Proof Check. This study aid includes: Proof generator. This site based on the Open Logic Project proof checker. Place brackets in expressions, given the priority of operations. At any time get assistance and ideas from Proof generator. Yes, Algebraic Proofs isn't particularly exciting. Natural deduction proof editor and checker. They're especially important in logical arguments and proofs, let's find out why! While the word "argument" may mean a disagreement between two or more people, in mathematical logic, an argument is a sequence or list of statements called premises or assumptions and returns a conclusion. φ [ν←τ] where τ is free for ν in φ. Read from here about the differences between algorithms. Like most proofs, logic proofs usually begin with premises--- statements that you're allowed to assume. Basics Whisky 101. You can also use LaTeX commands. Logic Calculator is a free, portable truth table generator for logic formulas i. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. Conic Sections Transformation. Look at line 3. The Corbettmaths Practice Questions on Algebraic Proof. Back then, the idea of logic languages with types was novel. Modifications by students and faculty at Cal. You need to “summarize” what was established after making the desired assumption (the contradiction of the conclusion). The values of the Boolean algebra calculator are denoted with logic 0 & 1. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). Line Equations Functions Arithmetic & Comp. Use symbolic logic and logic algebra. Propositional logic in Artificial intelligence. It provides a formal language to write mathematical definitions, executable algorithms and theorems together with an environment for semi-interactive development of machine-checked proofs. Right click connections to delete them. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. Propositional logic is also amenable to “deduction,” that is, the development of proofs by writing a series of lines, each of which either is given or is justified by some previous lines (Section 12. ENDING AN INDIRECT PROOF (after you derive a contradiction, any contradiction) CP. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses’s free, online logic teaching software application Carnap (carnap. The only limitation for this calculator is that you have only three atomic propositions to choose from: p,q and r. Place brackets in expressions, given the priority of operations. The specific system used here is the one found in forall x: Calgary Remix. The Conformal Smart Logic Equivalence Checker (LEC) is the next-generation equivalency checking solution. 3: Proofs in predicate logic. Resolution is one kind of proof technique that works this way - (i) select two clauses that contain conflicting terms (ii) combine those two clauses and (iii) cancel out the conflicting terms. truth tables, normal forms, proof checking, proof building). Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. You can select and try out several solver algorithms: the "DPLL better" is the best solver amongst the options. Back then, the idea of logic languages with types was novel. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Valid or Invalid? The rules of this test are simple: it's your job to determine whether an argument is valid or not. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. Proof checker. Mathematical logic step by step. Right click connections to delete them. techniques as “proof by contradiction” or “proof by contrapositive” (Section 12. Yes, Algebraic Proofs isn't particularly exciting. At any time get assistance and ideas from Proof generator. Your Input Approximate the integral $$\int\limits_{0}^{1} \sqrt{\sin^{3}{\left(x \right)} + 1}\, dx$$$with $$n = 5$$$ using the trapezoidal rule. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. They're especially important in logical arguments and proofs, let's find out why! While the word "argument" may mean a disagreement between two or more people, in mathematical logic, an argument is a sequence or list of statements called premises or assumptions and returns a conclusion. A sequent S is true if and only if there exists a tree of sequents rooted at S where each leaf is an axiom and each internal node is derived from its children by an inference rule. The amount of detail that an author supplies in a proof should depend on the audience. Matrices & Vectors. The boolean algebra calculator uses the basic laws like identity law. Actually there are mechanical ways of generating Fitch style proofs. Conic Sections Transformation. Know Your Worth is based on millions of real salaries from Glassdoor users. Free Python 3. The boolean algebra calculator uses the basic laws like identity law. You may add additional sentences to your set by repeating this step. Take help from sample expressions in the input box or have a look at the boolean functions in the content to understand the mathematical operations used in expressions. Actually there are mechanical ways of generating Fitch style proofs. Click on one of the three applications on the right. This app is a graphical semantic calculator for a specific kind of. Simplify logical expressions. Online tool. Propositional logic is also amenable to “deduction,” that is, the development of proofs by writing a series of lines, each of which either is given or is justified by some previous lines (Section 12. Boolean formulas are written as sequents. Screenshots. Amount of Water to Add (fl. Free Logical Sets calculator - calculate boolean algebra, truth tables and set theory step-by-step This website uses cookies to ensure you get the best experience. Look at line 3. Send me a full list of your axioms and I will see what I can do to get you started. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. The symbol for this is $$ν$$. State University, Monterey Bay. Using the derived rule allowed us to shorten the proof considerably. Yes, Algebraic Proofs isn't particularly exciting. The logic language used in this theorem prover is one that was proposed in the author's Master's thesis, back in 1985-1987, at which time it contained most of the features shown here, including the hierarchical type scheme. 0 is based on classical logic, but it is possible to define your axioms in it. This page is a tutorial and user's guide; there is also a complete reference. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. To typeset these proofs you will need Johann Klüwer's fitch. Modifications by students and faculty at Cal. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. Matrices & Vectors. Desired Proof. Free Python 3. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. We write the sum of the natural numbers up to a value n as: 1+2+3+···+(n−1)+n = Xn i=1 i. Place brackets in expressions, given the priority of operations. When combined together, several gates can make a complex logical evaluation system that has. It is a technique of knowledge representation in logical and mathematical form. Mathematical logic step by step. Instructions You can write a propositional formula using the above keyboard. You can select and try out several solver algorithms: the "DPLL better" is the best solver amongst the options. Logic Calculator. To typeset these proofs you will need Johann Klüwer's fitch. We write the sum of the natural numbers up to a value n as: 1+2+3+···+(n−1)+n = Xn i=1 i. Go to Daemon Proof Checkeror Quick Help Index. Natural deduction proof editor and checker. Propositional logic in Artificial intelligence. Using the derived rule allowed us to shorten the proof considerably. Try the leading salary calculator the next time you negotiate your salary or ask for a raise, and get paid fairly. Each one has a different shape to show its particular function. This calculator solves linear diophantine equations. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses’s free, online logic teaching software application Carnap (carnap. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. Logic Gate Simulator. (whenever you see $$ν$$ read 'or') When two simple sentences, p and q, are joined in a disjunction statement, the disjunction is expressed symbolically as p $$ν$$ q. You may add additional sentences to your set by repeating this step. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. At any time get assistance and ideas from Proof generator. ) And that’s it! Enjoy drinking your whisky exactly how you like it. See Credits. Step through the examples. The Corbettmaths Practice Questions on Algebraic Proof. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). Once you know your perfect proof, this calculator will tell you exactly how much water to add to any amount of whisky to reach it. Modifications by students and faculty at Cal. Try the leading salary calculator the next time you negotiate your salary or ask for a raise, and get paid fairly. The specific system used here is the one found in forall x: Calgary Remix. A proposition is a declarative statement which is either true or false. Basics Whisky 101. Such proofs can also encode traditional proofs based on modus ponens: the inference P∧(P⇒Q) ⊢ Q can be rewritten as resolution by expanding ⇒ to get P∧(¬P∨Q) ⊢ Q. Your Input Approximate the integral $$\int\limits_{0}^{1} \sqrt{\sin^{3}{\left(x \right)} + 1}\, dx$$$with $$n = 5$$$ using the trapezoidal rule. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. The system was originally written for UMass’s Intro Logic course, based on Gary Hardegree’s online. In mathematics, a Diophantine equation is a polynomial equation in two or more unknowns such that only the integer solutions are searched or studied (an integer solution is a solution such that all the unknowns take integer values). In computer science, and more specifically in computability theory and computational complexity theory, a model of computation is a model which describes how an output of a mathematical function is computed given an input. It provides a formal language to write mathematical definitions, executable algorithms and theorems together with an environment for semi-interactive development of machine-checked proofs. The symbol P denotes a sum over its argument for each natural. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Logic Calculator: Desktop application useful to perform logical operations in an arithmetic-calculator fashion, with three modes: 1) Evaluation of logic formulas: displays the truth table along with the models of the given formula. All in one boolean expression calculator. Solving a classical propositional formula means looking for such values of variables that the formula becomes true. Coq is a formal proof management system. A model describes how units of computations, memories, and communications are organized. A variable is a symbol used to represent a logical quantity. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. To download DC Proof and for a contact link, visit my homepage. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. It formalizes the rules of logic. Save your work on device and continue later on. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. The only limitation for this calculator is that you have only three atomic propositions to choose from: p,q and r. This site based on the Open Logic Project proof checker. Here is a standard example: An argument is valid if and only if the conclusion necessarily follows from the premises. Proof checker. A proof is an argument intended to convince the reader that a general principle is true in all situations. Note that proofs can also be exported in "pretty print" notation (with unicode logic symbols) or LaTeX. Logic Gate Simulator. If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Yes, Algebraic Proofs isn't particularly exciting. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. In chapter 17 we will prove the deduction theorem. The assumption set that results is the same as the assumption set for line 3. In the dropdown menu, click 'UserDoc'. This app is a graphical semantic calculator for a specific kind of. You may add additional sentences to your set by repeating this step. This calculator solves linear diophantine equations. Some (importable) sample proofs in the "plain" notation are here. Modal logic is a type of symbolic logic for capturing inferences about necessity and possibility. Free Windows Terminal Preview. Perfect Proof. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences.
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# Refresh an F-curve with Python after changing Extrapolation Mode [closed]
After creating an F-curve for the default cube and then setting the Extrapolation Mode to 'LINEAR', the graph editor displays a very strange curve:
The script I ran in the Text Editor of a new .blend file was:
import bpy
obj = bpy.context.object
obj.animation_data_create()
obj.animation_data.action = bpy.data.actions.new(name="MyAction")
fcu_z = obj.animation_data.action.fcurves.new(data_path="location", index=2)
fcu_z.keyframe_points[0].co = 10.0, 0.0
fcu_z.keyframe_points[1].co = 20.0, 1.0
fcu_z.extrapolation='LINEAR'
The F-curve can be manually "refreshed" by selecting a control point and hitting GRMB.
So my questions are:
1. Can I "refresh" the F-curve in the script?
2. Should I submit a bug report?
• Modern blender have FCurve.update(). Docs: Ensure keyframes are sorted in chronological order and handles are set correctly Jun 22 '18 at 9:59
If you're changing the area.type anyway, I suggest to use a meaningful operator and replace the fcu_z.extrapolation = 'LINEAR' by graph.extrapolation_type() operator:
import bpy
context = bpy.context
obj = context.object
obj.animation_data_create()
obj.animation_data.action = bpy.data.actions.new(name="MyAction")
fcu_z = obj.animation_data.action.fcurves.new(data_path="location", index=2)
fcu_z.keyframe_points[0].co = 10.0, 0.0
fcu_z.keyframe_points[1].co = 20.0, 1.0
old_area_type = context.area.type
context.area.type = 'GRAPH_EDITOR'
bpy.ops.graph.extrapolation_type(type='LINEAR')
context.area.type = old_area_type
Submitted bug report: https://developer.blender.org/T38774 (fixed in 2.70, see report)
You don't set the coordinates for the handles.
Blender won't check if these are valid. In this case the right handle is initialized with the default value
(0.0, 0.0) which is invalid since the x-coordinate of the left endpoint is 10.0. So handle_x > endpoint_x is violated here.
Moving the point blender will automatically recalculate the handles for the points.
BTW: keyframe_points.insert will recalculate the curve unless the option FAST is chosen.
This example works:
import bpy
obj = bpy.context.object
obj.animation_data_create()
obj.animation_data.action = bpy.data.actions.new(name="MyAction")
fcu_z = obj.animation_data.action.fcurves.new(data_path="location", index=2)
fcu_z.extrapolation='LINEAR'
fcu_z.keyframe_points.insert(10.0, 0.0)
fcu_z.keyframe_points.insert(20.0, 1.0)
I've found a workaround. This function seems to refresh the F-Curve Editor:
def refresh_fcurve_editor():
'''execute a meaningless command on F-Curve Editor which has the effect of
refreshing the graph.'''
C=bpy.context
old_area_type = C.area.type
C.area.type='GRAPH_EDITOR'
bpy.ops.graph.clean( threshold = 0)
C.area.type=old_area_type
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# Looks like Afghanistan is in Taliban hands...or VERY soon to be
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41 minutes ago, Peterkin said:
Of course he's struggling. Each new president inherits the bass-ackward incompetence and crappy decisions of the six or seven preceding administrations, and the same brass-bound, uncommunicative, recalcitrant military hierarchy that leaves all the messy splats on the ground and swaggers away. This president is at least doing something, even if he was pushed into it unprepared.
Do you know what he's doing behind the scenes? I don't. What do you think he should do that's within his power to do? I have no frickin idea.
Actually I would like to add that Biden (and the whole senate at that time) voted for the invasion. While it is possible that he and other lawmakers where misled by the respective administrations (one of which he was part of as vice-president), it only highlights that seemingly no one really knew what was going on, or particularly cared about it, either. It was rather clear that whoever does anything, would make it fall apart.
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17 minutes ago, CharonY said:
Biden (and the whole senate at that time) voted for the invasion.
He also has been advocating leaving for over a decade. Even former chairman of the joint chiefs, admiral Mike Mullen, has applauded him for being right on this the whole time when the admiral and other generals were wrong.
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It's hard to say no to army brass. It's hard to understand very different cultures. It's hard to decide whether a war of choice is a good or bad choice. It's hard to extricate oneself from a fraught relationship of any kind (You've all been there, right?) It's hard to know the most politically advantageous thing to do. When you're in one of the many seats in a great big room full of democratically elected representatives of 30-some percent of the people, it's easy to go with the flow. When you're in that badly designed office, all alone, you make some very difficult decisions.
1 hour ago, J.C.MacSwell said:
You seem to have quoted and taken exception to my preamble. (despite probably agreeing with it?)
Agreeing, certainly. Taking exception, no. Just wondering what you imagine doing in his place. It's kind of an uncomfortable thought-experiment.
Edited by Peterkin
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Uhm. Ok. It’s hard. What point are you trying make other than adding to your post count?
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10 minutes ago, iNow said:
What point are you trying make
No points. Just discourse on the situation. If we can't or don't want to think about what our political leaders face once we put them in office, on what basis can we decide which ones to elect next time?
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17 minutes ago, Peterkin said:
No points.
Thx for confirming
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1 hour ago, iNow said:
He also has been advocating leaving for over a decade. Even former chairman of the joint chiefs, admiral Mike Mullen, has applauded him for being right on this the whole time when the admiral and other generals were wrong.
I believe he hit the nail fair square on the head in the first moment of the Interview...The capitulation of the Afghan forces for their own survivability.
Mistakes were made, not simply by the Americans but by all the allied forces that invaded the country, initially and rightly to prevent a haven for terrorists and of course to eliminate Bin-Laden, after 9/11.
I have a sneaking suspicion that if the new breed of Taliban do not live up to their word, that other actions, sanctions and repuccusions maybe in the pipeline. While that will be detrimental for the average Joe Blow in the streets, the facts are that if this is still the old Taliban in disguise, then things will be 100 times worse for the average Afghan.
On Biden, and speaking as an outsider, I sort of thought he was too old for the job, but then again, anything was acceptable in place of the former redneck ratbag Trump. Otherwise, I feel rather sorry he is confronted with such a scenario. Will he maintain forces after the 31st August? That also will be interesting.
Edited by beecee
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8 hours ago, beecee said:
I believe he hit the nail fair square on the head in the first moment of the Interview...The capitulation of the Afghan forces for their own survivability.
That resonated with me, too. They could've fought, but knew they'd be overtaken soon and their decision to battle would lead to consequences / retribution for them and their families. They made a rational calculation to just lay down arms now and increase likelihood of surviving.
8 hours ago, beecee said:
Otherwise, I feel rather sorry he is confronted with such a scenario. Will he maintain forces after the 31st August? That also will be interesting.
He'll have a video teleconference with other world leaders today where many (including Boris Johnson from UK) are expected to push him to keep troops in longer. I'm doubtful he'll agree, though might be willing to share the burden if other countries add troops of their own... as you mention... we'll see.
Edited by iNow
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Let's also not willfully ignore the principle driver of the war: profit.
For the defense contractors who made billions during this crusade, the current result, shambolic as it may appear, is actually fairly promising for their industry. It means then can do it all again some time in the future.
### S&P 500
• Total return: 516.67 percent
• Annualized return: 9.56 percent
• $10,000 2001 stock purchase today:$61,613.06
### Basket of Top Five Contractor Stocks
• Total return: 872.94 percent
### Boeing
• Total return: 974.97 percent
• Annualized return: 12.67 percent
• $10,000 2001 stock purchase today:$107,588.47
• Board includes: Edmund P. Giambastiani Jr. (former vice chair, Joint Chiefs of Staff), Stayce D. Harris (former inspector general, Air Force), John M. Richardson (former navy chief of Naval Operations)
### Raytheon
• Total return: 331.49 percent
• Annualized return: 7.62 percent
• $10,000 2001 stock purchase today:$43,166.92
• Board includes: Ellen Pawlikowski (retired Air Force general), James Winnefeld Jr. (retired Navy admiral), Robert Work (former deputy secretary of defense)
### Lockheed Martin
• Total return: 1,235.60 percent
• Annualized return: 13.90 percent
• $10,000 2001 stock purchase today:$133,559.21
• Board includes: Bruce Carlson (retired Air Force general), Joseph Dunford Jr. (retired Marine Corps general)
### General Dynamics
• Total return: 625.37 percent
• Annualized return: 10.46 percent
• $10,000 2001 stock purchase today:$72,515.58
• Board includes: Rudy deLeon (former deputy secretary of defense), Cecil Haney (retired Navy admiral), James Mattis (former secretary of defense and former Marine Corps general), Peter Wall (retired British general)
### Northrop Grumman
• Total return: 1,196.14 percent
• Annualized return: 13.73 percent
• $10,000 2001 stock purchase today:$129,644.84
• Board includes: Gary Roughead (retired Navy admiral), Mark Welsh III (retired Air Force general)
Another war finance story: we wasted trillions beefing up bulky fighter jets rather than improving on an older, lightweight design. It was TYT. I'll find it.
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Fast Algorithms for Demixing Sparse Signals from Nonlinear Observations
# Fast Algorithms for Demixing Sparse Signals from Nonlinear Observations
Electrical and Computer Engineering Department
Iowa State University
This work was supported in part by the National Science Foundation under the grant CCF-1566281. Parts of this work also appear in an Iowa State University technical report [1] and a conference paper to be presented in the 2016 Asilomar Conference in November 2016 [2].
###### Abstract
We study the problem of demixing a pair of sparse signals from nonlinear observations of their superposition. Mathematically, we consider a nonlinear signal observation model, , where denotes the superposition signal, and are orthonormal bases in , and are sparse coefficient vectors of the constituent signals. Further, we assume that the observations are corrupted by a subgaussian additive noise. Within this model, represents a nonlinear link function, and is the -th row of the measurement matrix, . Problems of this nature arise in several applications ranging from astronomy, computer vision, and machine learning.
In this paper, we make some concrete algorithmic progress for the above demixing problem. Specifically, we consider two scenarios: (i) the case when the demixing procedure has no knowledge of the link function, and (ii) the case when the demixing algorithm has perfect knowledge of the link function. In both cases, we provide fast algorithms for recovery of the constituents and from the observations. Moreover, we support these algorithms with a rigorous theoretical analysis, and derive (nearly) tight upper bounds on the sample complexity of the proposed algorithms for achieving stable recovery of the component signals. Our analysis also shows that the running time of our algorithms is essentially as good as the best possible.
We also provide a range of numerical simulations to illustrate the performance of the proposed algorithms on both real and synthetic signals and images. Our simulations show the superior performance of our algorithms compared to existing methods for demixing signals and images based on convex optimization. In particular, our proposed methods yield demonstrably better sample complexities as well as improved running times, thereby enabling their applicability to large-scale problems.
## 1 Introduction
### 1.1 Setup
In numerous signal processing applications, the problem of demixing is of special interest. In simple terms, demixing involves disentangling two (or more) constituent signals from observations of their linear superposition. Formally, consider a discrete-time signal that can be expressed as the superposition of two signals:
x=Φw+Ψz,
where and are orthonormal bases of , and are the corresponding basis coefficients. The goal of signal demixing, in this context, is to reliably recover the constituent signals (equivalently, their basis representations and ) from the superposition signal .
Demixing suffers from a fundamental identifiability issue since the number of unknowns () is greater than the number of observations (). This is easy to see: suppose for simplicity that , the canonical basis of , and therefore, . Now, suppose that both and have only one nonzero entry in the first coordinate. Then, there is an infinite number of and that are consistent with the observations , and any hope of recovering the true components is lost. Therefore, for the demixing problem to have an identifiable solution, one inevitably has to assume some type of incoherence between the constituent signals (or more specifically, between the corresponding bases and [3, 4]. Such an incoherence assumption certifies that the components are sufficiently “distinct” and that the recovery problem is well-posed. Please see Section 3 for a formal definition of incoherence.
However, even if we assume that the signal components are sufficiently incoherent, demixing poses additional challenges under stringent observation models. Suppose, now, that we only have access to undersampled linear measurements of the signal, i.e., we record:
y=Ax, (1.1)
where denotes the measurement operator and where . In this scenario, the demixing problem is further confounded by the fact that possesses a nontrivial null space. In this case, it might seem impossible to recover the components and since possesses a nontrivial null space. Once again, this problem is highly ill-posed and further structural assumptions on the constituent signals are necessary. Under-determined problems of this kind have recently received significant attention in signal processing, machine learning, and high-dimensional statistics. In particular, the emergent field of compressive sensing [5, 6, 7] shows that it is indeed possible to exactly reconstruct the underlying signals under certain assumptions on , provided the measurement operator is designed carefully. This intuition has enabled the design of a wide range of efficient architectures for signal acquisition and processing [8, 9].
In this paper, we address an even more challenging question in the demixing context. Mathematically, we consider a noisy, nonlinear signal observation model, formulated as follows:
yi=g(⟨ai,Φw+Ψz⟩)+ei, i=1,…,m. (1.2)
Here, as before, the superposition signal is modeled as . Each observation is generated by the composition of a linear functional of the signal , with a (scalar) nonlinear function . Here, is sometimes called a link or transfer function, and denotes the row of a linear measurement matrix . For full generality, in (1.2) we assume that each observation is corrupted by additive noise; the noiseless case is realized by setting . We will exclusively consider the “measurement-poor” regime where the number of observations is much smaller than the ambient dimension .
For all the reasons detailed above, the problem of recovering the coefficient vectors and from the measurements seems daunting. Therefore, we make some structural assumptions. Particularly, we assume that and are -sparse (i.e., they contain no more than nonzero entries). Further, we will assume perfect knowledge of the bases and , and the measurement matrix . The noise vector is assumed to be stochastic, zero mean, and bounded. Under these assumptions, we will see that it is indeed possible to stably recover the coefficient vectors, with a number of observations that is proportional to the sparsity level , as opposed to the ambient dimension .
The nonlinear link function plays a crucial role in our algorithm development and analysis. In signal processing applications, such nonlinearities may arise due to imperfections caused during a measurement process, or inherent limitations of the measurement system, or due to quantization or calibration errors. We discuss such practical implications more in detail below. On an abstract level, we consider two distinct scenarios. In the first scenario, the link function may be non-smooth, non-invertible, or even unknown to the recovery procedure. This is the more challenging case, but we will show that recovery of the components is possible even without knowledge of . In the second scenario; the link function is a known, smooth, and strictly monotonic function. This is the somewhat simpler case, and we will see that this leads to significant improvements in recovery performance both in terms of theory and practice.
### 1.2 Our Contributions
In this paper, we make some concrete algorithmic progress in the demixing problem under nonlinear observations. In particular, we study the following scenarios depending on certain additional assumptions made on (1.2):
1. Unknown . We first consider the (arguably, more general) scenario where the nonlinear link function may be non-smooth, non-invertible, or even unknown. In this setting, we do not explicitly model the additive noise term in (1.2). For such settings, we introduce a novel demixing algorithm that is non-iterative, does not require explicit knowledge of the link function , and produces an estimate of the signal components. We call this algorithm OneShot to emphasize its non-iterative nature. It is assumed that OneShot possess oracle knowledge of the measurement matrix , and orthonormal bases and .
We supplement our proposed algorithm with a rigorous theoretical analysis and derive upper bounds on the sample complexity of demixing with nonlinear observations. In particular, we prove that the sample complexity of OneShot to achieve an estimation error is given by provided that the entries of the measurement matrix are i.i.d. standard normal random variables.
2. Known . Next, we consider the case where the nonlinear link function is known, smooth, and monotonic. In this setting, the additive noise term in (1.2) is assumed to be bounded either absolutely, or with high probability. For such (arguably, milder) settings, we provide an iterative algorithm for demixing of the constituent signals in (1.2) given the nonlinear observations . We call this algorithm Demixing with Hard Thresholding, or DHT for short. In addition to knowledge of , we assume that DHT possesses oracle knowledge of , , and .
Within this scenario, we also analyze two special sub-cases:
Case 2a: Isotropic measurements. We assume that the measurement vectors are independent, isotropic random vectors that are incoherent with the bases and . This assumption is more general than the i.i.d. standard normal assumption on the measurement matrix made in the first scenario, and is applicable to a wider range of measurement models. For this case, we show that the sample complexity of DHT is upper-bounded by , independent of the estimation error .
Case 2b: Subgaussian measurements. we assume that the rows of the matrix are independent subgaussian isotropic random vectors. This is also a generalization of the i.i.d. standard normal assumption made above, but more restrictive than Case 2a. In this setting, we obtain somewhat better sample complexity. More precisely, we show that the sample complexity of DHT is for sample complexity, matching the best known sample complexity bounds for recovering a superposition of -sparse signals from linear observations [10, 11].
In both the above cases, the underlying assumption is that the bases and are sufficiently incoherent, and that the sparsity level is small relative to the ambient dimension . In this regime, we show that DHT exhibits a linear rate of convergence, and therefore the computational complexity of DHT is only a logarithmic factor higher than OneShot. Table 1 provides a summary of the above contributions for the specific case where is the identity (canonical) basis and is the discrete cosine transform (DCT) basis, and places them in the context of the existing literature on some nonlinear recovery methods [12, 13, 14]. We stress that these previous works do not explicitly consider the demixing problem, but in principle the algorithms of [12, 13, 14] can be extended to the demixing setting as well.
### 1.3 Techniques
At a high level, our recovery algorithms are based on the now-classical method of greedy iterative thresholding. In both methods, the idea is to first form a proxy of the signal components, followed by hard thresholding to promote sparsity of the final estimates of the coefficient vectors and . The key distinguishing factor from existing methods is that the greedy thresholding procedures used to estimate and are deliberately myopic, in the sense that each thresholding step operates as if the other component did not exist at all. Despite this apparent shortcoming, we are still able to derive bounds on recovery performance when the signal components are sufficiently incoherent.
Our first algorithm, OneShot, is based on the recent, pioneering approach of [12], which describes a simple (but effective) method to estimate a high-dimensional signal from unknown nonlinear observations. Our first main contribution of this paper is to extend this idea to the nonlinear demixing problem, and to precisely characterize the role of incoherence in the recovery process. Indeed, a variation of the approach of [12] (described in Section 5) can be used to solve the nonlinear demixing problem as stated above, with a similar two-step method of first forming a proxy, and then performing a convex estimation procedure (such as the LASSO [15]) to produce the final signal estimates. However, as we show below in our analysis and experiments, OneShot offers superior performance to this approach. The analysis of OneShot is based on a geometric argument, and leverages the Gaussian mean width for the set of sparse vectors, which is a statistical measure of complexity of a set of points in a given space.
While OneShot is simple and effective, one can potentially do much better if the link function were available at the time of recovery. Our second algorithm, DHT, leverages precisely this intuition. First, we formulate our nonlinear demixing problem in terms of an optimization problem with respect to a specially-defined loss function that depends on the nonlinearity . Next, for solving the proposed optimization problem, we propose an iterative method to solve the optimization problem, up to an additive approximation factor. Each iteration with DHT involves a proxy calculation formed by computing the gradient of the loss function, followed by (myopic) projection onto the constraint sets. Again, somewhat interestingly, this method can be shown to be linearly convergent, and therefore only incurs a small (logarithmic) overhead in terms of running time. The analysis of DHT is based on bounding certain parameters of the loss function known as the restricted strong convexity (RSC) and restricted strong smoothness (RSS) constants.111Quantifying algorithm performance by bounding RSC and RSC constants of a given loss function are quite widespread in the machine learning literature [16, 17, 18, 19], but have not studied in the context of signal demixing.
Finally, we provide a wide range of simulations to verify empirically our claims both on synthetic and real data. We first compare the performance of OneShot with the convex optimization method of [12] for nonlinear demixing via a series of phase transition diagrams. Our simulation results show that OneShot outperforms this convex method significantly in both demixing efficiency as well as running time, and consequently makes it an attractive choice in large-scale problems. However, as discussed below, the absence of knowledge of the link function induces an inevitable scale ambiguity in the final estimation222Indeed, following the discussion in [12], any demixing algorithm that does not leverage knowledge of is susceptible to such a scale ambiguity.. For situations where we know the link function precisely, our simulation results show that DHT offers much better statistical performance compared to OneShot, and is even able to recover the scale of the signal components explicitly. We also provide simulation results on real-world natural images and astronomical data to demonstrate robustness of our approaches.
### 1.4 Organization
The rest of this paper is organized as follows. Section 2 describes several potential applications of our proposed approach, and relationship with prior art. Section 3 introduces some key notions that are used throughout the paper. Section 4 contains our proposed algorithms, accompanied by analysis of their performance; complete proofs are deferred to Section 6. Section 5 lists the results of a series of numerical experiments on both synthetic and real data, and Section 7 provides concluding remarks.
## 2 Applications and Related Work
Demixing problems of various flavors have been long studied in research areas spanning signal processing, statistics, and physics, and we only present a small subset of relevant related work. In particular, demixing methods have been the focus of significant research over the fifteen years, dating back at least to [20]. The work of Elad et al. [3] and Bobin et al. [21] posed the demixing problem as an instance of morphological components analysis (MCA), and formalized the observation model (1.1). Specifically, these approaches posed the recovery problem in terms of a convex optimization procedure, such as the LASSO [15]. The work of Pope et al. [22] analyzed somewhat more general conditions under which stable demixing could be achieved.
More recently, the work of [23] showed a curious phase transition behavior in the performance of the convex optimization methods. Specifically, they demonstrated a sharp statistical characterization of the achievable and non-achievable parameters for which successful demixing of the signal components can be achieved. Moreover, they extended the demixing problem to a large variety of signal structures beyond sparsity via the use of general atomic norms in place of the -norm in the above optimization. See [24] for an in-depth discussion of atomic norms, their statistical and geometric properties, and their applications to demixing.
Approaches for (linear) demixing has also considered a variety of signal models beyond sparsity. The robust PCA problem [25, 26, 27] involves the separation of low-rank and sparse matrices from their sum. This idea has been used in several applications ranging from video surveillance to sensor network monitoring. In machine learning applications, the separation of low-rank and sparse matrices has been used for latent variable model selection [28] as well as the robust alignment of multiple occluded images [29]. Another type of signal model is the low-dimensional manifold model. In [10, 11], the authors proposed a greedy iterative method for demixing signals, arising from a mixture of known low-dimensional manifolds by iterative projections onto the component manifolds.
The problem of signal demixing from linear measurements belongs to a class of linear inverse problems that underpin compressive sensing [5, 6]; see [7] for an excellent introduction. There, the overarching goal is to recover signals from (possibly randomized) linear measurements of the form (1.1). More recently, it has been shown that compressive sensing techniques can also be extended to inverse problems where the available observations are manifestly nonlinear. For instance, in -bit compressive sensing [30, 31] the linear measurements of a given signal are quantized in the extreme fashion such that the measurements are binary () and only comprise the sign of the linear observation. Therefore, the amplitude of the signal is completely discarded by the quantization operator. Another class of such nonlinear recovery techniques can be applied to the classical signal processing problem of phase retrieval [32] which is somewhat more challenging than -bit compressive sensing. In this problem, the phase information of the signal measurements may be irrecovably lost and we have only access to the amplitude information of the signal [32]. Therefore, the recovery task here is to retrieve the phase information of the signal from random observations. Other related works include approaches for recovering low-rank matrices from nonlinear observations [33, 34]. We mention in passing that inverse problems involving nonlinear observations have also long been studied in the statistical learning theory literature; see [35, 36, 37, 38] for recent work in this area. Analogous to our scenarios above, these works consider both known as well as unknown link functions; these two classes of approaches are respectively dubbed as Generalized Linear Models (GLM) learning methods and Single Index Model (SIM) learning methods.
For our algorithmic development, we build upon a recent line of efficient, iterative methods for signal estimation in high dimensions [39, 17, 18, 12, 19, 40]. The basic idea is to pose the recovery as a (non-convex) optimization problem in which an objective function is minimized over the set of -sparse vectors. Essentially, these algorithms are based on well-known iterative thresholding methods proposed in the context of sparse recovery and compressive sensing [41, 42]. The analysis of these methods heavily depends on the assumption that the objective function satisfies certain (restricted) regularity conditions; see Sections 3 and 6 for details. Crucially, we adopt the approach of [16], which introduces the concept of the restricted strong convexity (RSC) and restricted strong smoothness (RSS) constants of a loss function. Bounding these constants in terms of problem parameters and , as well as the level of incoherence in the components, enables explicit characterization of both sample complexity and convergence rates.
## 3 Preliminaries
In this section, we introduce some notation and key definitions. Throughout this paper, denotes the -norm of a vector in , and denotes the spectral norm of the matrix . Let and be orthonormal bases of . Define the set of sparse vectors in the bases and as follows:
K1 ={Φa | ∥a∥0≤s1}, K2 ={Ψa | ∥a∥0≤s2},
and define We use to denote the unit ball. Whenever we use the notation , the vector is comprised by stacking column vectors and .
In order to bound the sample complexity of our proposed algorithms, we will need some concepts from high-dimensional geometry. First, we define a statistical measure of complexity of a set of signals, following [12].
###### Definition 3.1.
(Local gaussian mean width.) For a given set , the local gaussian mean width (or simply, local mean width) is defined as follows :
Wt(K)=Esupx,y∈K,∥x−y∥2≤t⟨g,x−y⟩.
where .
Next, we define the notion of a polar norm with respect to a given subset of the signal space:
###### Definition 3.2.
(Polar norm.) For a given and a subset of , the polar norm with respect to is defined as follows:
∥x∥Qo=supu∈Q⟨x,u⟩.
Furthermore, for a given subset of , we define . Since is a symmetric set, one can show that the polar norm with respect to defines a semi-norm. Next, we use the following standard notions from random matrix theory [43]:
###### Definition 3.3.
(Subgaussian random variable.) A random variable is called subgaussian if it satisfies the following:
Eexp⎛⎝cX2∥X∥2ψ2⎞⎠≤2,
where is an absolute constant and denotes the -norm which is defined as follows:
∥X∥ψ2=supp≥11√p(E|X|p)1p.
###### Definition 3.4.
(Isotropic random vectors.) A random vector-valued variable is said to be isotropic if .
In order to analyze the computational aspects of our proposed algorithms (in particular, DHT), we will need the following definition from [16]:
###### Definition 3.5.
A loss function satisfies Restricted Strong Convexity/Smoothness (RSC/RSS) if:
m4s≤∥∇2ξf(t)∥≤M4s,
where , for all and . Also, and are (respectively) called the RSC and RSS constants. Here denotes a sub-matrix of the Hessian matrix, , comprised of row/column indices in .
As discussed earlier, the underlying assumption in all demixing problems of the form (3.4) is that the constituent bases are sufficiently incoherent as per the following definition:
###### Definition 3.6.
(-incoherence.) The orthonormal bases and are said to be -incoherent if:
ε=sup∥u∥0≤s, ∥v∥0≤s∥u∥2=1, ∥v∥2=1|⟨Φu,Ψv⟩|. (3.1)
The parameter is related to the so-called mutual coherence parameter of a matrix. Indeed, if we consider the (overcomplete) dictionary , then the mutual coherence of is given by . Moreover, one can show that [7].
We now formally establish our signal model. Consider a signal that is the superposition of a pair of sparse vectors in different bases, i.e.,
x=Φw+Ψz, (3.2)
where are orthonormal bases, and such that , and . We define the following quantities:
¯x=Φ¯w+Ψ¯z∥Φ¯w+Ψ¯z∥2=α(Φ¯w+Ψ¯z), (3.3)
where Also, define the coefficient vector, . as the vector obtaining by stacking the individual coefficient vectors and of the component signals.
We now state our measurement model. Consider the nonlinear observation model:
yi=g(aTix)+ei, i=1…m, (3.4)
where is the superposition signal given in (3.2), and represents a nonlinear link function. We denote as the derivative of , i.e., . As mentioned above, depending on the knowledge of the link function , we consider two scenarios:
1. In the first scenario, the nonlinear link function may be non-smooth, non-invertible, or even unknown. In this setting, we assume the noiseless observation model, i.e., . In addition, we assume that the measurement matrix is populated by i.i.d. unit normal random variables.
2. In this setup, represents a known nonlinear, differentiable, and strictly monotonic function. Further, in this scenario, we assume that the observation is corrupted by a subgaussian additive noise with for . We also assume that the additive noise has zero mean and independent from , i.e., for . In addition, we assume that the measurement matrix consists of either (2a) isotropic random vectors that are incoherent with and , or (2b) populated with subgaussian random variables.
We highlight some additional clarifications for the second case. In particular, we make the following :
###### Assumption 3.7.
There exist nonnegative such that .
In words, the derivative of the link function is strictly bounded either within a positive interval or within a negative interval. In this paper, we focus on the case when . The analysis of the complementary case is similar.
The lower bound on guarantees that the function is a monotonic function, i.e., if then . Moreover, the upper bound on guarantees that the function is Lipschitz with constant . Such assumptions are common in the nonlinear recovery literature [16, 40].333Using the monotonicity property of that arises from Assumption 3.7, one might be tempted to simply apply the inverse of the link function on the measurements in (3.4) convert the nonlinear demixing problem to the more amenable case of linear demixing, and then use any algorithm (e.g., [11]) for recovery of the constituent signals. However, this naïve way could result in a large error in the estimation of the components, particularly in the presence of the noise in (3.4). This issue has been also considered in [40] for generic nonlinear recovery both from a theoretical as well as empirical standpoint.
In Case 2a, the vectors (i.e., the rows of ) are independent isotropic random vectors. For this case, in addition to incoherence between the component bases, we also need to define a measure of cross-coherence between the measurement matrix and the dictionary . The following notion of cross-coherence was introduced in the early literature of compressive sensing [44]:
###### Definition 3.8.
(Cross-coherence.) The cross-coherence parameter between the measurement matrix and the dictionary is defined as follows:
ϑ=maxi,jaTiΓj∥ai∥2, (3.5)
where and denote the row of the measurement matrix and the column of the dictionary .
The cross-coherence assumption implies that for , where denotes the restriction of the columns of the dictionary to set , with such that columns are selected from each basis and .
## 4 Algorithms and Theoretical Results
Having defined the above quantities, we now present our main results. As per the previous section, we study two distinct scenarios:
### 4.1 When the link function g is unknown
Recall that we wish to recover components and given the nonlinear measurements and the matrix . Here and below, for simplicity we assume that the sparsity levels and , specifying the sets and , are equal, i.e., . The algorithm (and analysis) effortlessly extends to the case of unequal sparsity levels. Our proposed algorithm, that we call OneShot, is described in pseudocode form below as Algorithm 1.
The mechanism of OneShot is simple, and deliberately myopic. At a high level, OneShot first constructs a linear estimator of the target superposition signal, denoted by . Then, it performs independent projections of onto the constraint sets and . Finally, it combines these two projections to obtain the final estimate of the target superposition signal.
In the above description of OneShot, we have used the following projection operators:
ˆw=Ps(Φ∗ˆx\text{lin}),^z=Ps(Ψ∗ˆx\text{lin}).
Here, denotes the projection onto the set of (canonical) -sparse signals and can be implemented by hard thresholding, i.e., any procedure that retains the largest coefficients of a vector (in terms of absolute value) and sets the others to zero444The typical way is to sort the coefficients by magnitude and retain the largest entries, but other methods such as randomized selection can also be used.. Ties between coefficients are broken arbitrarily. Observe that OneShot is not an iterative algorithm, and this in fact enables us to achieve a fast running time.
We now provide a rigorous performance analysis of OneShot. Our proofs follow the geometric approach provided in [12], specialized to the demixing problem. In particular, we derive an upper bound on the estimation error of the component signals and , modulo scaling factors. In our proofs, we use the following result from [12], restated here for completeness.
###### Lemma 4.1.
(Quality of linear estimator). Given the model in Equation (3.2), the linear estimator, , is an unbiased estimator of (defined in (3.3)) up to constants. That is, and: where
We now state our first main theoretical result, with the full proof provided below in Section 6.
###### Theorem 4.2.
Let be the set of measurements generated using a nonlinear function that satisfies the conditions of Lemma (4.9) in [12]555Based on this lemma, the nonlinear function is odd, nondecreasing, and sub-multiplicative on .. Let be a random matrix with i.i.d. standard normal entries. Also, let are bases with , where is as defined in Def. 3.6. If we use Oneshot to recover estimates of and (modulo a scaling) described in equations (3.2) and (3.3), then the estimation error for (similarly, ) satisfies the following upper bound in expectation :
E∥ˆw−μαw∥2≤ρ+2√m(4σ+ηWρ(K)ρ)+8με. (4.1)
The constant is chosen for convenience and can be strengthened. The authors of [45, 12] provide upper bounds on the local mean width of the set of -sparse vectors. In particular, for any they show that for some absolute constant . By plugging in this bound and letting , we can combine components and which gives the following:
###### Corollary 4.3.
With the same assumptions as Theorem 4.2, the error of nonlinear estimation incurred by the final output satisfies the upper bound:
E∥ˆx−μ¯x∥2≤4√m(4σ+Cη√slog(2n/s))+16με. (4.2)
###### Corollary 4.4.
(Example quantitative result). The constants depend on the nature of the nonlinear function , and are often rather mild. For example, if , then we may substitute
μ=√2π≈0.8,σ2=1−2π≈0.6,η2=1,
in the above statement. Hence, the bound in (4.2) becomes:
E∥ˆx−μ¯x∥2≤4√m(3.1+C√slog(2n/s))+13ε. (4.3)
###### Proof.
Using Lemma 4.1, where . Since and has unit norm, . Thus, where . Moreover, we can write . Here, we have used the fact that obeys the distribution with mean 1. Finally, . ∎
In contrast with demixing algorithms for traditional (linear) observation models, our estimated signal outputting from OneShot can differ from the true signal by a scale factor. Next, suppose we fix as a small constant, and suppose that the incoherence parameter for some constant , and that the number of measurements scales as:
m=O(sκ2logns). (4.4)
Then, the (expected) estimation error . In other words, the sample complexity of OneShot is given by , which resembles results for the linear observation case [11, 12]666Here, we use the term “sample-complexity” as the number of measurements required by a given algorithm to achieve an estimation error . However, we must mention that algorithms for the linear observation model are able to achieve stronger sample complexity bounds that are independent of ..
We observe that the estimation error in (4.2) is upper-bounded by . This is meaningful only when , or when . Per the Welch Bound [7], the mutual coherence satisfies . Therefore, Theorem 4.2 provides non-trivial results only when . This is consistent with the square-root bottleneck that is often observed in demixing problems; see [46] for detailed discussions.
The above theorem obtains a bound on the expected value of the estimation error. We can derive a similar upper bound that holds with high probability. In this theorem, we assume that the measurements for have a sub-gaussian distribution (according to Def. 3.3). We obtain the following result, with full proof deferred to Section 6.
###### Theorem 4.5.
(High-probability version of Thm. 4.2.) Let be a set of measurements with a sub-gaussian distribution. Assume that is a random matrix with standard normal entries. Also, assume that are two bases with incoherence as in Definition 3.6. Let . If we use Oneshot to recover and (up to a scaling) described in (3.2) and (3.3), then the estimation error of the output of Oneshot satisfies the following:
∥ˆx−μ¯x∥2≤4η√m(3s′+C′√slog2ns)+16με, (4.5)
with probability at least where are absolute constants. The coefficients , and are given in Lemma 4.1. Here, denotes the -norm of the first measurement (Definition 3.3).
In Theorem 4.5, we stated the tail probability bound of the estimation error for the superposition signal, . Similar to Theorem 4.2, we can derive a completely analogous tail probability bound in terms of the constituent signals and .
### 4.2 When the link function g is known
The advantages of OneShot is that it enables fast demixing, and can handle even unknown, non-differentiable link functions. But its primary weakness is that the sparse components are recovered only up to an arbitrary scale factor. This can lead to high estimation errors in practice, and this can be unsatisfactory in applications. Moreover, even for reliable recovery up to a scale factor, its sample complexity is inversely dependent on the estimation error. To solve these problems, we propose a different, iterative algorithm for recovering the signal components. Here, the main difference is that the algorithm is assumed to possess (perfect) knowledge of the nonlinear link function, .
Recall that we define and . First, we formulate our demixing problem as the minimization of a special loss function :
mint∈R2n F(t)=1mm∑i=1Θ(aTiΓt)−yiaTiΓt (4.6) s.\ t.∥t∥0≤2s.
Observe that the loss function is not the typical squared-error function commonly encountered in statistics and signal processing applications. In contrast, it heavily depends on the nonlinear link function (via its integral ). Instead, such loss functions are usually used in GLM and SIM estimation in the statistics literature [16]. In fact, the objective function in (4.6) can be considered as the sample version of the problem:
mint∈R2n E(Θ(aTΓt)−yaTΓt),
where and satisfies the model (3.4). It is not hard to show that the solution of this problem satisfies . We note that the gradient of the loss function can be calculated in closed form:
∇F(t) =1mm∑i=1ΓTaig(aTiΓt)−yiΓTai, (4.7) =1mΓTAT(g(AΓt)−y).
We now propose an iterative algorithm for solving (4.6) that we call it Demixing with Hard Thresholding (DHT). The method is detailed in Algorithm 2. At a high level, DHT iteratively refines its estimates of the constituent signals (and the superposition signal ). At any given iteration, it constructs the gradient using (4.7). Next, it updates the current estimate according to the gradient update being determined in Algorithm 2. Then, it performs hard thresholding using the operator to obtain the new estimate of the components and . This procedure is repeated until a stopping criterion is met. See Section 5 for the choice of stopping criterion and other details. We mention that the initialization step in Algorithm 2 is arbitrary and can be implemented (for example) by running OneShot and obtaining initial points . We use this initialization in our simulation results.
Implicitly, we have again assumed that both component vectors and are -sparse; however, as above we mention that Algorithm 2 and the corresponding analysis easily extend to differing levels of sparsity in the two components. In Algorithm 2, denotes the projection of vector on the set of sparse vectors, again implemented via hard thresholding.
We now provide our second main theoretical result, supporting the convergence analysis of DHT. In particular, we derive an upper bound on the estimation error of the constituent vector (and therefore, the component signals ). The proofs of Theorems 4.64.7 and 4.8 are deferred to section 6.
###### Theorem 4.6.
(Performance of DHT) Consider the measurement model (3.4) with all the assumptions mentioned for the second scenario in Section 3. Suppose that the corresponding objective function satisfies the RSS/RSC properties with constants and on the set with ( denotes the iteration) such that . Choose a step size parameter with Then, DHT outputs a sequence of estimates that satisfies the following upper bound (in expectation) for :
∥tk+1−t∗∥2≤(2q)k∥t0−t∗∥2+Cτ√sm, (4.8)
where and is a constant that depends on the step size and the convergence rate . Also, where and are the true (unknown) vectors in model (1.2).
Equation (4.8) indicates that Algorithm 2 (DHT) enjoys a linear rate of convergence. In particular, for the noiseless case , this implies that Alg. 2 returns a solution with accuracy after iterations. The proof of Theorem 4.6 leverages the fact that the objective function in (4.6) satisfies the RSC/RSS conditions specified in Definition 3.5. Please refer to Section 6 for a more detailed discussion. Moreover, we observe that in contrast with OneShot, DHT can recover the components and without any ambiguity in scaling factor, as depicted in the bound (4.8). We also verify this observation empirically in our simulation results in Section 5.
Echoing our discussion in Section 3, we consider two different models for the measurement matrix and derive upper bounds on the sample complexity of DHT corresponding to each case. First, we present the sample complexity of Alg. 2 when the measurements are chosen to be isotropic random vectors, corresponding to Case (2a) described in the introduction:
###### Theorem 4.7.
(Sample complexity when the rows of are isotropic.) Suppose that the rows of are independent isotropic random vectors. In order to achieve the requisite RSS/RSC properties of Theorem 4.6, the number of samples needs to scale as:
m=O(slognlog2slog(slogn)),
provided that the bases and are incoherent enough.
The sample complexity mentioned in Theorem 4.7 incurs an extra (possibly parasitic) poly-logarithmic factor relative to the sample complexity of OneShot, stated in (4.4). However, the drawback of OneShot is that the sample complexity depends inversely on the estimation error , and therefore a very small target error would incur a high overhead in terms of number of samples.
Removing all the extra logarithmic factors remains an open problem in general (although some improvements can be obtained using the method of [47]). However, if we assume additional structure in the measurement matrix , we can decrease the sample complexity even further. This corresponds to Case 2b.
###### Theorem 4.8.
(Sample complexity when the elements of are subgaussian.) Assume that all assumptions and definitions in Theorem 4.6 holds except that the rows of matrix are independent subgaussian isotropic random vectors. Then, in order to achieve the requisite RSS/RSC properties of Theorem 4.6, the number of samples needs to scale as:
m=O(slogns),
provided that the bases and are incoherent enough.
The leading big-Oh constant in the expression for in Theorems 4.7 and 4.8 is somewhat complicated, and hides the dependence on the incoherence parameter , the mutual coherence , the RSC/RSS constants, and the growth parameters of the link function and . Please see section 6 for more details.
In Theorem 4.6, we expressed the upper bounds on the estimation error in terms of the constituent vector, . It is easy to translate these results in terms of the component vectors and using the triangle inequality:
max{∥w0−w∗∥2,∥z0−z∗∥2}≤∥t0−t∗∥2≤∥w0−w∗∥2+∥z0−z∗∥2.
See Section 6 for proofs and futher details.
## 5 Experimental Results
In this section, we provide a range of numerical experiments for our proposed algorithms based on synthetic and real data. We compare the performance of OneShot and DHT with a LASSO-type technique for demixing, as well as a heuristic version of OneShot based on soft thresholding (inspired by the approach proposed in [48]). We call these methods Nonlinear convex demixing with LASSO or (NlcdLASSO), and Demixing with Soft Thresholding or DST, respectively. Before describing our simulation results, we briefly describe these two methods.
NlcdLASSO is a heuristic method motivated by [12], although it was not explicitly developed in the demixing context. Using our notation from Section 3 and 4, NlcdLASSO solves the following convex problem:
minz,w ∥∥ˆxlin−[Φ Ψ][w;z]∥∥2 (5.1) subject to ∥w∥1≤√s,∥z∥1≤√s.
Here, denotes the proxy of (equal to ) and denotes the sparsity level of signals and in basis and , respectively. The constraints in problem (5.1) are convex penalties reflecting the knowledge that and are -sparse and have unit -norm (since the nonlinearity is unknown, we have a scale ambiguity, and therefore w.l.o.g. we can assume that the underlying signals lie in the unit ball). The outputs of this algorithm are the estimates , , and
To solve the optimization problem in (5.1), we have used the SPGL1 solver [49, 50]. This solver can handle large scale problems, which is the scenario that we have used in our experimental evaluations. We impose the joint constraint which is a slight relaxation of the constraints in 5.1. The upper-bound of in the constraints is a worst-case criterion; therefore, for a fairer comparison, we also include simulation results with the constraint , where has been tuned to the best of our ability.
On the other hand, DST solves the optimization problem (4.6) via a convex relaxation of the sparsity constraint. In other words, this method attempts to solve the following relaxed version of the problem (4.6):
mint 1mm∑i=1Θ(aTiΓt)−yiaTiΓt+β′∥t∥1, (5.2)
where represents -norm of the constituent vector and denotes the tuning parameter. The solution of this problem at iteration is given by soft thresholding operator as follows:
tk+1=Sβ′η′(tk−η′∇F(tk)),
where denotes the step size, and the soft thresholding operator, is given by:
Sλ(y)=⎧⎨⎩y−λ,if y>λ0,if |y|≤λy+λ,if y<−λ.
Both OneShot and NlcdLASSO do not assume knowledge of the link function, and consequently return a solution up to a scalar ambiguity. Therefore, to compare performance across algorithms, we use the (scale-invariant) cosine similarity between the original superposition signal and the output of a given algorithm defined as:
cos(x,ˆx)=xTˆx∥x∥2∥ˆx∥2.
### 5.1 Synthetic Data
As discussed above, for successful recovery we require the constituent signals to be sufficiently incoherent. To achieve this, we choose to be the 1D Haar wavelets basis, and to be the noiselet basis777These bases are known to be maximally incoherent relative to each other [51]. For the measurement operator , we choose a partial DFT matrix. Such matrices are known to have similar recovery performance as random Gaussian matrices, but enable fast numerical operations [52]. Also, we present our experiments based on both non-smooth as well as differentiable link functions. For the non-smooth case, we choose ; here, we only present recovery results using OneShot and NlcdLASSO since in our analysis DHT and DST can only handle smooth link functions.
The results of our first experiment are shown in Figure 1(a) and Figure 1(b). The test signal is generated as follows: set length , and generate the vectors and by randomly selecting a signal support with nonzero elements, and populating the nonzero entries with random coefficients. The plot illustrates the performance of Oneshot and NlcdLASSO measured by the cosine similarity for different choices of sparsity level , where the nonlinear link function is set to and we have used both and constraints. The horizontal axis denotes an increasing number of measurements. Each data point in the plot is obtained by conducting a Monte Carlo experiment in which a new random measurement matrix is generated, recording the cosine similarity between the true signal and the reconstructed estimate and averaging over trials.
As we can see, notably, the performance of NlcdLASSO is worse than OneShot for any fixed choice of and no matter what upper bound we use on . Even when the number of measurements is high (for example, at in plot (b)), we see that OneShot outperforms NlcdLASSO by a significant degree. In this case, NlcdLASSO is at least worse in terms of signal estimation quality, while OneShot recovers the (normalized) signal perfectly. This result indicates the inefficiency of NlcdLASSO for nonlinear demixing.
Next, we contrast the running time of both algorithms, illustrated in Figure 1(c). In this experiment, we measure the wall-clock running time of the two recovery algorithms (OneShot and NlcdLASSO), by varying signal size from to . Here, we set , , and the number of Monte Carlo trials to . Also, the nonlinear link function is considered as . As we can see from the plot, OneShot is at least times faster than NlcdLASSO when the size of signal equals to . Overall, OneShot is efficient even for large-scale nonlinear demixing problems. We mention that in the above setup, the main computational costs incurred in OneShot involve a matrix-vector multiplication followed by a thresholding step, both of which can be performed in time that is nearly-linear in terms of the signal length for certain choices of
Next, we turn to differentiable link functions. In this case, we generate the constituent signal coefficient vectors, with , and compare performance of the four above algorithms. The nonlinear link function is chosen to be ; it is easy to check that the derivative of this function is strictly bounded between and . The maximal number of iterations for both DHT and DST is set to to with an early stopping criterion if convergence is detected. The step size is hard to estimate in practice, and therefore is chosen by manual tuning such that both DHT and DST obtain the best respective performance.
Figure 2 illustrates the performance of the four algorithms in terms of phase transition plots, following [23]. In these plots, we varied both the sparsity level and the number of measurements . For each pair , as above we randomly generate the test superposition signal by choosing both the support and coefficients of at random, as well as the measurement matrix. We repeat this experiment over Monte Carlo trials. We calculate the empirical probability of successful recovery as the number of trials in which the output cosine similarity is greater than . Pixel intensities in each figure are normalized to lie between 0 and 1, indicating the probability of successful recovery.
As we observe in Fig. 2, DHT has the best performance among the different methods, and in particular, outperforms both the convex-relaxation based methods. The closest algorithm to DHT in terms of the signal recovery is DST, while the LASSO-based method fails to recover the superposition signal (and consequently the constituent signals and ). The improvements over OneShot are to be expected since as discussed before, this algorithm does not leverage the knowledge of the link function and is not iterative.
In Fig. 3, we fix the sparsity level and plot the probability of recovery of different algorithms with a varying number of measurements. The number of Monte Carlo trials is set to 20 and the empirical probability of successful recovery is defined as the number of trials in which the output cosine similarity is greater than . The nonlinear link function is set to be for figure (a) and for figure (b). As we can see, DHT has the best performance, while NlcdLASSO for figure (a) and Oneshot, and NlcdLASSO for figure (b) cannot recover the superposition signal even with the maximum number of measurements.
### 5.2 Real Data
In this section, we provide representative results on real-world 2D image data using Oneshot and NlcdLASSO for non-smooth link function given by . In addition, we illustrate results for all four algorithms using smooth as our link function.
We begin with a test image. First, we obtain its 2D Haar wavelet decomposition and retain the largest coefficients, denoted by the -sparse vector . Then, we reconstruct the image based on these largest coefficients, denoted by . Similar to the synthetic case, we generate a noise component in our superposition model based on 500 noiselet coefficients . In addition, we consider a parameter which controls the strength of the noiselet component contributing to the superposition model. We set this parameter to 0.1. Therefore, our test image is given by .
Figure 4 illustrates both the true and the reconstructed images and using Oneshot and NlcdLASSO. The number of measurements is set to (using subsampled Fourier matrix with rows). From visual inspection we see that the reconstructed image, , using Oneshot is better than the reconstructed image by NlcdLASSO. Quantitatively, we also calculate Peak signal-to-noise-ratio (PSNR) of the reconstructed images using both algorithms relative to the test image, . We obtain PSNR of 19.8335 dB using OneShot, and a PSNR of 17.9092 dB using NlcdLASSO, again illustrating the better performance of Oneshot compared to NlcdLASSO.
Next, we show our results using a differentiable link function. For this experiment, we consider an astronomical image illustrated in Fig. 5. This image includes two components; the “stars” component, which can be considered to be sparse in the identity basis (), and the “galaxy” component which are sparse when they are expressed in the discrete cosine transform basis (). The superposition image is observed using a subsampled Fourier matrix with rows multiplied with a diagonal matrix with random entries [53]. Further, each measurement is nonlinearly transformed by applying the (shifted) logistic function as the link function. In the recovery procedure using DHT, we set the number of iterations to and step size to . As is visually evident, our proposed DHT method is able to reliably recover the component signals.
## 6 Proofs
In this section, we derive the proofs of our theoretical results stated in Section 4.
### 6.1 Analysis of OneShot
Our analysis mostly follows the techniques of [12]. However, several additional complications in the proof arise due to the structure of the demixing problem. As a precursor, we need the following lemma from geometric functional analysis, restated from [12].
###### Lemma 6.1.
Assume is a closed star-shaped set. Then for , and , one has the following result :
∥PK(a)−u∥2≤max(t,2t∥a−u∥Kot). (6.1)
We also use the following result of [12].
###### Claim 6.2.
(Orthogonal decomposition of .) Suppose we decompose the rows of , , as:
ai=⟨ai,¯x⟩¯x+bi, (6.2)
where is orthogonal to . Then we have since . Also, Moreover, the measurements in equation (3.4) and the orthogonal component are statistically independent.
###### Proof of Theorem 4.2.
Observe that the magnitude of the signal may be lost due to the action of the nonlinear measurement function (such as the function). Therefore, our recovered signal approximates the true signal modulo a scaling factor. Indeed, for defined in Lemma 4.1, we have:
∥ˆx−μ¯x∥2 =∥Φˆw+Ψˆz−αμΦw−αμΨz∥2 ≤∥Φ∥∥ˆw−μαw∥2+∥Ψ∥∥ˆz−μαz∥2 ≤(ρ+2ρ∥Φ∗ˆxlin−μαw∥Koρ)+(ρ+2ρ∥Ψ∗ˆx%lin−μαz∥Koρ). (6.3)
The equality comes from the definition of . The first inequality results from an application of the triangle inequality and the definition of the operator norm of a matrix, while the second inequality follows from Lemma 6.1. Now, it suffices to derive a bound on the first term in the above expression (since a similar bound will hold for the second term). This proves the first part of Theorem 4.2. We have:
∥Φ∗ˆxlin−μαw∥Koρ =∥Φ∗1mΣi(yi⟨ai,¯x⟩¯x+yibi)−μαw∥Koρ ≤∥Φ∗1mΣi(yi⟨ai,¯x⟩¯x)−μαw∥Koρ+∥Φ∗1mΣiyibi∥Koρ ≤∥Φ∗1mΣi(yi⟨ai,¯x⟩¯x)−μΦ∗¯x∥KoρS1+∥μαΦ∗Ψz∥KoρS2 +∥Φ∗1mΣiyibi∥KoρS3. (6.4)
The first equality follows from Claim 6.2, while the second and third inequalities result from the triangle inequality. Also:
S1 =∥Φ∗1mΣi(yi⟨ai,¯x⟩¯x)−μΦ∗¯x∥Koρ =∥(1mΣi(yi⟨ai
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2022 Vol.38 Issue.1,Published 2022-02-24 article
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1 Bankruptcy Probability of a Lever Company: Lookback Option Pricing Method YANG Zhaoqiang; TIAN Yougong This paper constructs the assets portfolio of lever corporation by the structural approach. Because irreversibility and uncertainty of corporate bankruptcy, the corporate bankruptcy is equivalent to a default of the bonds. By using the parabolic stochastic partial differential equations (SPDE) which the lookback option satisfied,the assets portfolio pricing model of lever corporation is derived under the mixed jump-diffusion fractional Brownian motion (MJD-fBm) environment. When the lever corporation in the financial crisis, Shareholders use capital injection to make up for operating losses and debt servicing, then the probability of no default before the bonds maturity and the conditional distribution of the lever corporation assets is obtained, and the pricing formula for lookback option is derived. In the end, a numerical example is given to illustrate the influence of different Hurst parameters and risk coefficient and stock asset weight to the default probability of the lever corporation. 2022 Vol. 38 (1): 1-23 [Abstract] ( 58 ) [HTML 1KB] [ PDF 843KB] ( 427 )
24 Population Life Prediction and SM Bonds Pricing Based on DEJD Model BIAN Huabin; TONG Xinle; YAO Dingjun In the context of the aging population, longevity risk will increase great economic pressure to the national endowment security system. How to measure and manage longevity risk has become the focus of research in recent years. Based on the Chinese population mortality data, and Lee-Carter model, we introduce DEJD model (double exponential jump diffusion model) to describe the jump asymmetry of time series factors, and prove that DEJD model is more effective than Lee-Carter model in fitting time series factors. In addition, we use the population mortality data predicted by DEJD model to price the SM bonds in Chinese market, providing an important reference for the promotion of SM bond in China. 2022 Vol. 38 (1): 24-42 [Abstract] ( 36 ) [HTML 1KB] [ PDF 356KB] ( 207 )
43 Multi-index Additive Model and Its Application in Medical Cost Forecast PAN Qing; ZHAO Xiaobing Modeling analysis and reasonable prediction of medical costs are the basis and foundation for the determination of medical insurance costs. High-dimensional additional information in medical costs plays an important role in long-term prediction. This paper proposes a partial linear multi-indicator additive model to fit and predict longitudinal medical cost data with high-dimensional features and uses two different dimensionality reduction estimation methods to estimate the model and applies the model to a set of high-dimensional dimensions. The longitudinal medical cost data of the variable is used for case analysis. 2022 Vol. 38 (1): 43-52 [Abstract] ( 32 ) [HTML 1KB] [ PDF 226KB] ( 222 )
53 The Estimated and Theoretical Assurances and the Probabilities of Launching a Phase III Trial ZHANG Yingying; RONG Tengzhong; LI Manman Prospective phase II trials usually result in failures in phase III trials. For randomized controlled phase II and phase III trials which are conducted with patients randomized to one of two treatments where the variances of the normally distributed responses are assumed to be known, we analytically obtain the estimated and theoretical assurances for the three cases (no, additive, and multiplicative bias adjustments). Under some minor assumptions, we show that the estimated assurances for the three cases are increasing functions of the per group number of patients and the observed treatment effect of the phase II trial, respectively; and for Case 3, the estimated assurance is an increasing function of the retention factor. When the true treatment effect of phase III is assumed to be a known constant, we show that the theoretical assurances for the three cases are constants which are equal to the designed power or one minus the type II error. Moreover, we show that the estimated assurances are always less than the theoretical assurance. We also obtain the analytical formulas of the probabilities of launching a phase III study for the three cases. Moreover, for Case 3, we show that the probability of launching a phase III study is an increasing function of the retention factor. According to our theoretical investigations, we find that the true treatment effect of phase III has no effect in the simulations. Finally, the simulations are conducted to illustrate the theoretical investigations. 2022 Vol. 38 (1): 53-70 [Abstract] ( 23 ) [HTML 1KB] [ PDF 172KB] ( 174 )
71 Multivariate Quasiconvex Risk Statistics with Scenario Analysis DENG Yong; HU Yijun In this paper, we introduce three new classes of multivariate risk statistics, named multivariate comonotonic quasiconvex risk statistics, multivariate quasiconvex risk statistics and multivariate empirical-law-invariant quasiconvex risk statistics, respectively. Representation results for them are provided by dual method. The results of this paper is not only the generalization of one-dimensional quasiconvex risk statistics, but also the extension of multivariate convex risk statistics. 2022 Vol. 38 (1): 71-85 [Abstract] ( 21 ) [HTML 1KB] [ PDF 219KB] ( 174 )
86 Discrete-Time Quantum Random Walks on the N-ary Tree HAN Qi;LU Ziqiang;HAN Yanan;CHEN Zhihe We study discrete-time quantum random walks on the $N$-ary tree by a framework for discrete-time quantum random walks, this framework has no need for coin spaces, it just choose the evolution operator with no constraints other than unitarity, and contain path enumeration using regeneration structures and $z$ transform. As a result, we calculate the generating function of the amplitude at the root in closed form. 2022 Vol. 38 (1): 86-98 [Abstract] ( 25 ) [HTML 1KB] [ PDF 421KB] ( 157 )
99 A Distributed Algorithm for Lasso Variable Selection ZENG Weijia; ZHANG Riquan Lasso is a variable selection method commonly used in machine learning, which is suitable for regression problems with sparsity. Distributed computing is an important way to reduce computing time and improve efficiency when large sample sizes or massive amounts of data are stored on different agents. Based on the equivalent optimization model of Lasso model and the idea of alternating stepwise iteration, this paper constructs a distributed algorithm suitable for Lasso variable selection. And the convergence of the algorithm is also proved. Finally, the distributed algorithm constructed in this paper is compared with cyclic-coordinate descent and ADMM algorithm through numerical experiments. For the sparse regression problem with large sample set, the algorithm proposed in this paper has better advantages in computing time and precision. 2022 Vol. 38 (1): 99-110 [Abstract] ( 33 ) [HTML 1KB] [ PDF 608KB] ( 202 )
111 Complete Convergence for NSD Random Variable Sequences FU Zongkui; FEI Dandan; ZHANG Cong; SUN Limin In this paper, using the property of NSD random variable sequence, moment inequality and three series theorem, we obtain complete convergence for NSD random variables sequences under certain moment conditions. The results generalize the results of independent sequences for Chow and Lai\ucite{15,16} and Jajte\ucite{17}. 2022 Vol. 38 (1): 111-122 [Abstract] ( 22 ) [HTML 1KB] [ PDF 593KB] ( 171 )
123 Valuing Guaranteed Minimum Death Benefits by Complex Fourier Series Expansion ZHANG Bo;ZHANG Zhimin In this paper, we use the complex fourier series expansion method (CFS) to price guaranteed minimum death benefits (GMDB). The main idea is to expand the Fourier series of the auxiliary function. The density function of remaining lifetime has two forms in this paper, namely combination-of-exponentials density and piecewise constant forces of mortality assumption, and the coefficients of series are estimated by using the known characteristic function of the general L\'{e}vy model. We mainly consider the value of GMDB products under call options and put options. In the numerical experiment section, we also demonstrate the advantages of CFS in calculation accuracy and running time by comparing with cosine series expansion method (COS) and Monte Carlo method (MC). 2022 Vol. 38 (1): 123-137 [Abstract] ( 22 ) [HTML 1KB] [ PDF 660KB] ( 167 )
138 Study on the Number and Mode of Optimal Piecewise of Critical Illness Insurance in China QIU Chunjuan; GAO Shuhui; QIAN Linyi The modes of piecewise and the number of intervalsare particularly important in the formulation of compensation scheme for critical illness insurance in China. In this paper, under the three subsection modes of interval bisection, constant ratio increment and constant ratio decrement, the theoretical models are established with interval quantity as independent variable and serious illness insurance compensation amount as dependent variable. Taking the expected compensation ratio as the standard to measure the compensation level of serious illness insurance, we can get the following results: first, the optimal number of intervals corresponding to the three interval modes are respectively: 3, 3 and 5; second, under the setting of the number of piecewise, the compensation level of the interval equal proportion increasing mode is the highest, the interval equal proportion decreasing mode is the second, and the compensation level of the interval equal proportion increasing mode is the lowest, But there is little difference between the three. Then, based on the data of CHARLS in 2015, we calculated the incidence of family catastrophic medical expenditure under the three interval modes as 7.13%, 7.26% and 7.69% respectively. The result is consistent with that of the theory. 2022 Vol. 38 (1): 138-150 [Abstract] ( 19 ) [HTML 1KB] [ PDF 652KB] ( 165 )
151 Unlinking Theorem for Symmetric Quasi-Convex Polynomials HONG Hejing; HU Zechun Let \mu_n be the standard Gaussian measure on \mathbb{R}^n and X be a random vector on \mathbb{R}^n with the law \mu_n. U-conjecture states that if f and g are two polynomials on \mathbb{R}^n such that f(X) and g(X) are independent, then there exist an orthogonal transformation Y=LX on \mathbb{R}^n and an integer k such that f\circ L^{-1} is a function of (y_1,y_2,\cdots,y_k) and g\circ L^{-1} is a function of (y_{k+1},y_{k+2},\cdots,y_n). In this case, f and g are said to be unlinked. In this note, we prove that two symmetric, quasi-convex polynomials f and g are unlinked if f(X) and g(X) are independent. 2022 Vol. 38 (1): 151-158 [Abstract] ( 27 ) [HTML 1KB] [ PDF 443KB] ( 164 )
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# Properties
Label 70.2.e.a Level $70$ Weight $2$ Character orbit 70.e Analytic conductor $0.559$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$70 = 2 \cdot 5 \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 70.e (of order $$3$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$0.558952814149$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q -\zeta_{6} q^{2} + ( -3 + 3 \zeta_{6} ) q^{3} + ( -1 + \zeta_{6} ) q^{4} + \zeta_{6} q^{5} + 3 q^{6} + ( -1 + 3 \zeta_{6} ) q^{7} + q^{8} -6 \zeta_{6} q^{9} +O(q^{10})$$ $$q -\zeta_{6} q^{2} + ( -3 + 3 \zeta_{6} ) q^{3} + ( -1 + \zeta_{6} ) q^{4} + \zeta_{6} q^{5} + 3 q^{6} + ( -1 + 3 \zeta_{6} ) q^{7} + q^{8} -6 \zeta_{6} q^{9} + ( 1 - \zeta_{6} ) q^{10} + ( 2 - 2 \zeta_{6} ) q^{11} -3 \zeta_{6} q^{12} + ( 3 - 2 \zeta_{6} ) q^{14} -3 q^{15} -\zeta_{6} q^{16} + ( 4 - 4 \zeta_{6} ) q^{17} + ( -6 + 6 \zeta_{6} ) q^{18} + 6 \zeta_{6} q^{19} - q^{20} + ( -6 - 3 \zeta_{6} ) q^{21} -2 q^{22} -3 \zeta_{6} q^{23} + ( -3 + 3 \zeta_{6} ) q^{24} + ( -1 + \zeta_{6} ) q^{25} + 9 q^{27} + ( -2 - \zeta_{6} ) q^{28} + 9 q^{29} + 3 \zeta_{6} q^{30} + ( 4 - 4 \zeta_{6} ) q^{31} + ( -1 + \zeta_{6} ) q^{32} + 6 \zeta_{6} q^{33} -4 q^{34} + ( -3 + 2 \zeta_{6} ) q^{35} + 6 q^{36} + 4 \zeta_{6} q^{37} + ( 6 - 6 \zeta_{6} ) q^{38} + \zeta_{6} q^{40} -7 q^{41} + ( -3 + 9 \zeta_{6} ) q^{42} -5 q^{43} + 2 \zeta_{6} q^{44} + ( 6 - 6 \zeta_{6} ) q^{45} + ( -3 + 3 \zeta_{6} ) q^{46} -8 \zeta_{6} q^{47} + 3 q^{48} + ( -8 + 3 \zeta_{6} ) q^{49} + q^{50} + 12 \zeta_{6} q^{51} + ( 2 - 2 \zeta_{6} ) q^{53} -9 \zeta_{6} q^{54} + 2 q^{55} + ( -1 + 3 \zeta_{6} ) q^{56} -18 q^{57} -9 \zeta_{6} q^{58} + ( -10 + 10 \zeta_{6} ) q^{59} + ( 3 - 3 \zeta_{6} ) q^{60} -\zeta_{6} q^{61} -4 q^{62} + ( 18 - 12 \zeta_{6} ) q^{63} + q^{64} + ( 6 - 6 \zeta_{6} ) q^{66} + ( 9 - 9 \zeta_{6} ) q^{67} + 4 \zeta_{6} q^{68} + 9 q^{69} + ( 2 + \zeta_{6} ) q^{70} + 2 q^{71} -6 \zeta_{6} q^{72} + ( 4 - 4 \zeta_{6} ) q^{73} + ( 4 - 4 \zeta_{6} ) q^{74} -3 \zeta_{6} q^{75} -6 q^{76} + ( 4 + 2 \zeta_{6} ) q^{77} -10 \zeta_{6} q^{79} + ( 1 - \zeta_{6} ) q^{80} + ( -9 + 9 \zeta_{6} ) q^{81} + 7 \zeta_{6} q^{82} -7 q^{83} + ( 9 - 6 \zeta_{6} ) q^{84} + 4 q^{85} + 5 \zeta_{6} q^{86} + ( -27 + 27 \zeta_{6} ) q^{87} + ( 2 - 2 \zeta_{6} ) q^{88} -\zeta_{6} q^{89} -6 q^{90} + 3 q^{92} + 12 \zeta_{6} q^{93} + ( -8 + 8 \zeta_{6} ) q^{94} + ( -6 + 6 \zeta_{6} ) q^{95} -3 \zeta_{6} q^{96} + 14 q^{97} + ( 3 + 5 \zeta_{6} ) q^{98} -12 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - q^{2} - 3q^{3} - q^{4} + q^{5} + 6q^{6} + q^{7} + 2q^{8} - 6q^{9} + O(q^{10})$$ $$2q - q^{2} - 3q^{3} - q^{4} + q^{5} + 6q^{6} + q^{7} + 2q^{8} - 6q^{9} + q^{10} + 2q^{11} - 3q^{12} + 4q^{14} - 6q^{15} - q^{16} + 4q^{17} - 6q^{18} + 6q^{19} - 2q^{20} - 15q^{21} - 4q^{22} - 3q^{23} - 3q^{24} - q^{25} + 18q^{27} - 5q^{28} + 18q^{29} + 3q^{30} + 4q^{31} - q^{32} + 6q^{33} - 8q^{34} - 4q^{35} + 12q^{36} + 4q^{37} + 6q^{38} + q^{40} - 14q^{41} + 3q^{42} - 10q^{43} + 2q^{44} + 6q^{45} - 3q^{46} - 8q^{47} + 6q^{48} - 13q^{49} + 2q^{50} + 12q^{51} + 2q^{53} - 9q^{54} + 4q^{55} + q^{56} - 36q^{57} - 9q^{58} - 10q^{59} + 3q^{60} - q^{61} - 8q^{62} + 24q^{63} + 2q^{64} + 6q^{66} + 9q^{67} + 4q^{68} + 18q^{69} + 5q^{70} + 4q^{71} - 6q^{72} + 4q^{73} + 4q^{74} - 3q^{75} - 12q^{76} + 10q^{77} - 10q^{79} + q^{80} - 9q^{81} + 7q^{82} - 14q^{83} + 12q^{84} + 8q^{85} + 5q^{86} - 27q^{87} + 2q^{88} - q^{89} - 12q^{90} + 6q^{92} + 12q^{93} - 8q^{94} - 6q^{95} - 3q^{96} + 28q^{97} + 11q^{98} - 24q^{99} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/70\mathbb{Z}\right)^\times$$.
$$n$$ $$31$$ $$57$$ $$\chi(n)$$ $$-\zeta_{6}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
11.1
0.5 + 0.866025i 0.5 − 0.866025i
−0.500000 0.866025i −1.50000 + 2.59808i −0.500000 + 0.866025i 0.500000 + 0.866025i 3.00000 0.500000 + 2.59808i 1.00000 −3.00000 5.19615i 0.500000 0.866025i
51.1 −0.500000 + 0.866025i −1.50000 2.59808i −0.500000 0.866025i 0.500000 0.866025i 3.00000 0.500000 2.59808i 1.00000 −3.00000 + 5.19615i 0.500000 + 0.866025i
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
7.c even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 70.2.e.a 2
3.b odd 2 1 630.2.k.f 2
4.b odd 2 1 560.2.q.i 2
5.b even 2 1 350.2.e.l 2
5.c odd 4 2 350.2.j.f 4
7.b odd 2 1 490.2.e.f 2
7.c even 3 1 inner 70.2.e.a 2
7.c even 3 1 490.2.a.k 1
7.d odd 6 1 490.2.a.e 1
7.d odd 6 1 490.2.e.f 2
21.g even 6 1 4410.2.a.h 1
21.h odd 6 1 630.2.k.f 2
21.h odd 6 1 4410.2.a.r 1
28.f even 6 1 3920.2.a.bk 1
28.g odd 6 1 560.2.q.i 2
28.g odd 6 1 3920.2.a.b 1
35.i odd 6 1 2450.2.a.q 1
35.j even 6 1 350.2.e.l 2
35.j even 6 1 2450.2.a.b 1
35.k even 12 2 2450.2.c.a 2
35.l odd 12 2 350.2.j.f 4
35.l odd 12 2 2450.2.c.s 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
70.2.e.a 2 1.a even 1 1 trivial
70.2.e.a 2 7.c even 3 1 inner
350.2.e.l 2 5.b even 2 1
350.2.e.l 2 35.j even 6 1
350.2.j.f 4 5.c odd 4 2
350.2.j.f 4 35.l odd 12 2
490.2.a.e 1 7.d odd 6 1
490.2.a.k 1 7.c even 3 1
490.2.e.f 2 7.b odd 2 1
490.2.e.f 2 7.d odd 6 1
560.2.q.i 2 4.b odd 2 1
560.2.q.i 2 28.g odd 6 1
630.2.k.f 2 3.b odd 2 1
630.2.k.f 2 21.h odd 6 1
2450.2.a.b 1 35.j even 6 1
2450.2.a.q 1 35.i odd 6 1
2450.2.c.a 2 35.k even 12 2
2450.2.c.s 2 35.l odd 12 2
3920.2.a.b 1 28.g odd 6 1
3920.2.a.bk 1 28.f even 6 1
4410.2.a.h 1 21.g even 6 1
4410.2.a.r 1 21.h odd 6 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(70, [\chi])$$:
$$T_{3}^{2} + 3 T_{3} + 9$$ $$T_{13}$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$1 + T + T^{2}$$
$3$ $$9 + 3 T + T^{2}$$
$5$ $$1 - T + T^{2}$$
$7$ $$7 - T + T^{2}$$
$11$ $$4 - 2 T + T^{2}$$
$13$ $$T^{2}$$
$17$ $$16 - 4 T + T^{2}$$
$19$ $$36 - 6 T + T^{2}$$
$23$ $$9 + 3 T + T^{2}$$
$29$ $$( -9 + T )^{2}$$
$31$ $$16 - 4 T + T^{2}$$
$37$ $$16 - 4 T + T^{2}$$
$41$ $$( 7 + T )^{2}$$
$43$ $$( 5 + T )^{2}$$
$47$ $$64 + 8 T + T^{2}$$
$53$ $$4 - 2 T + T^{2}$$
$59$ $$100 + 10 T + T^{2}$$
$61$ $$1 + T + T^{2}$$
$67$ $$81 - 9 T + T^{2}$$
$71$ $$( -2 + T )^{2}$$
$73$ $$16 - 4 T + T^{2}$$
$79$ $$100 + 10 T + T^{2}$$
$83$ $$( 7 + T )^{2}$$
$89$ $$1 + T + T^{2}$$
$97$ $$( -14 + T )^{2}$$
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# Complex number 1!
Algebra Level 5
If $$x = 2+5i$$ and $$\large 2(\frac{1}{1!9!} +\frac{1}{3!7!}) + \frac{1}{5!5!} = \frac{2^a}{b!}$$ , then the value of $$(x^3 -5x^2 +33x -19)$$ is equal to
×
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Justice and Fairness: Part 2
A friend of yours was part of a secret society and was put on trial on 15 December 2014 without a lawyer. Alice encrypted the trial's transcript and sent it to Bob, but double agent Carol was able to e-mail a copy of the ciphertext to you. Your friend desperately needed help from a lawyer, but you were unable to provide one with the plaintext so that you could relay the proper advice back to your friend.
RESULTS OF TRIAL: Your friend was executed on 26 December 2014, and the rest of your friend's family followed on 20 January 2015.
Recently, Carol was able to intercept communications containing a rather ominous message. It seems that your friend's secret society has learned of the work you and your associates have been conducting and are less than pleased. Your life appears to be in jeopardy unless you solve the mystery of your friend's trial by the end of this quarter. If you fail, you will find out if there is an afterlife.
E-mail: Show
URGENT: Carol just intercepted the following e-mail. Someone must be watching out for you.
E-mail: Show
Can you produce the plaintext for the message below to prevent your untimely demise?
Ciphertext: Show
Hint 31:
The telephone at your desk begins to ring, and after answering, you hear a recorded message spoken with a distorted voice coming from the earpiece. "This is your friend Randall, and I have the dynamic-link library you requested for managing the reputation of the users on your web site. Please let Adam know that he needs to nag a ram and avoid Pan if he wishes to ever understand his wife."
Hint 32:
While walking back from lunch, Victor flags you down in the hall and urges you to check your inbox as soon as you get back to your office. It is easy to spot his e-mail once seated at your desk as it is marked as a high priority message which summarizes some findings reported to him by our corporate firewall. In each of the three encoded e-mails you received last week, a nonstandard X-Security-Warning: SOS header was ominously included.
Hint 33:
Just as on Monday, your telephone starts ringing; and upon picking it up, you hear the voice of Paul over in customer service. "In a meeting this morning, a friend told me that you only have sixteen more days until DIE takes you out, and something has been bothering me over the past couple of weeks. Does all the information we have mesh together such that there are no conflicts or contradictions among the multitude of observations?"
Hint 34:
Inspectors from Myutsu Mutucsoecul Corporation have been hired yet again in the interest of verifying Paul's question from last week. After reviewing the past five weeks since their last audit, Alloy believes he has found some inconsistencies. "One might infer an unproven relationship between ciphertext and primer from Trudy's question, and the data Chuck provided does not appear to allow proper decryption."
Hint 35:
"Don't you think this is unmerited discrimination against fresh workers?" Without waiting for Alloy to respond, Cobalt continues her defense of Trudy by showing that M1 and M2 provided by Team Bolt (as the Clingans are now calling themselves) have an even number of alpha characters as does the primer. "Furthermore, changes in the ciphertext match the profile of Chuck's key, so there could be some sort of clerical misunderstanding."
Hint 36:
Back again but somewhat under the weather, Blastus apologizes for missing his meeting with you yesterday but reveals he was verifying a letter from someone called Velociraptor. This mysterious individual must have wanted to contribute something to your friend's case without being involved in the investigative process. A familiarity with Team Bolt's work is evident since Velociraptor's writing directly utilizes data taken from their test.
Letter: Show
Hint 37:
A familiar name from Tenebris Lamina shows up in your inbox, but Dave has a significantly larger message for you than he did the last time he sent you an e-mail.
E-mail: Show
Hint 38:
Some people would look at her name and think she was their grandestmother, but this individual from Nigrum Gladio upholds her organization's name in a different way.
E-mail: Show
Hint 39:
Your computer announces a new e-mail with a rather humorless tone, and upon seeing who the unread message is from, you can only wonder if the devilish saber is meant for you or for your enemy.
E-mail: Show
• Anagram. Nice one – QuyNguyen2013 Mar 2 '15 at 21:34
• Where did the first 30 hints go? – Joe Z. Mar 10 '15 at 17:35
• Oh wait, never mind, I just noticed this is part 2. – Joe Z. Mar 10 '15 at 17:36
• The next update should have been today but may be postponed a day or two. Writing up an adequate explanation of the ME encryption and decryption processes will take extra time. – Noctis Skytower Mar 11 '15 at 19:26
• THE END -- if any further hints are given, they will be towards the purpose of a walkthrough for the puzzle. – Noctis Skytower Mar 25 '15 at 17:49
1 Answer
The plain text of payload.me is:
Congratulations! Laziness, impatience, and hubris could not withhold the prize
of this puzzle from you. Your actions are deserving of a reward, but if joy is
a part of satisfaction, can attaining this goal be beaten? Spoils for the 1st!
Yes, there is more to saying you are 1st to read this. If the latter be valid,
then you have treasure to retrieve. Go to http://www.gog.com/redeem and see if
the following codes can be claimed. These be the diadem reserved for a victor.
WZZL-WE4D-WLLD-RQHN
WF79-KCSX-36Y5-9FTU
DD9DB-7E243-F133C-95C75
Hint 36
If you don't want to write your own implementation go to http://code.activestate.com/recipes/578075/ for existing Python code. This puzzle is pretty much impossible to do without Python.
Hint 37
The message body reads
The disk jockey appears to be a reference to Dreary Jeans.
The body is really long so I won't repeat it hear since probably won't display correctly. Here are the first few hundred characters
(b\'p/H5Od+#_"{\\\\sf;<:^\\nE[J&QkWDS\\x0bT*B,]Pbcy\\x0
c07qUw(63@Lo}ZeaYGC%zmRF9uM|.v-1N>2\\t A\\\'hi!Ixtl=?g
X4j$n~8KV)\\rr\',\n b\'3Tv+ Y9cnK%7ao*db6ryu@\\x0b)=~m? 4W>f<#MhNX1-ljIe(].Z\\t/|58wRE{tP![GS0A}kq"x2Oi$H;psgQ
LJCD:\\\\B\\x0c\\nU_,\\\'\\r^zV&F\',\n b\'UGZ]\\n@{lgkb
hH>nup<t~c58SF%XY1Tz+\\x0cwi:,=x"&?Nv*dA\\tIf/\$ er|JDV(
4^ML6\\\'})7R.Pq[E\\\\9Q\\x0bOK_mBjs;#!y-o2W\\r0Ca3\',
I appreciate the 'hi!' in the first line. I hope that as intentional. This is likely not the key for the payload.me, that key should only contain alphanumerics. The important things to note are that it looks like a tuple of binary strings so you know we're dealing with Python 3.x. Also all control characters are no actually there \n is two different characters, you will need to map those back to control characters. In the end you should have a 160 strings, each containing a scrambled copy of string.printable.
Hint 38
message body reads
Has man's rib caused yet another fall? Yes, but this time it is on death itself.
after decoding payload.py doesn't look super useful. It's 1788 mostly random characters (which I'm not even going to try to include). The operating assumption is that this is a message encrypted with the key above. We're still missing the primer.
Hint 39
I just used http://www.dcode.fr/ascii-85-encoding to convert between ASCII85 and hex, then decode as usual. I don't think I even bothered with the body. primer.data looks like it could be a primer. Again, the decode text is really ugly so I'm not going to try to include it.
The important thing it is the primer for the key and cipher text of the last two hints. When decrypted you get:
import itertools
import random
import re
import string
import markov_encryption as me
###############################################################################
encode = lambda s: s.encode('ascii')
decode = lambda b: b.decode('ascii')
def main():
ciphertext, primertext, keytext = load_puzzle_data()
plaintext = generate_plaintext(primertext, keytext, ciphertext)
key, primer = create_key_and_primer(plaintext)
globals().update(locals())
def load_puzzle_data():
with open('ciphertext.txt') as file:
ciphertext = file.read()
with open('transcript_primer.txt') as file:
primertext = file.read()
with open('transcript_key.pickle') as file:
keytext = file.read()
return ciphertext, primertext, keytext
def generate_plaintext(primertext, keytext, ciphertext):
primer = me.Primer(encode(primertext))
key = me.Key(tuple(map(encode, keytext.split('\n'))))
decrypter = me.Decrypter(key, primer)
plaintext = decode(decrypter.process(encode(ciphertext)))
return plaintext
def create_key_and_primer(plain):
primer = ''.join(m.groups()[0] for m in re.finditer(r'$$(.)$$', plain))
base = []
for c in itertools.chain(filter(str.isalnum, plain),
string.ascii_letters,
string.digits):
if c not in base:
base.append(c)
key = ''.join(base)
randomizer = random.Random()
for i in map(ord, primer):
randomizer.seed(i)
randomizer.shuffle(base)
key += '\n' + ''.join(base)
p = me.Primer(encode(primer))
k = me.Key(tuple(map(encode, key.split('\n'))))
return k, p
###############################################################################
if __name__ == '__main__':
main()
This is pretty straightforward, load the cipher text, primer and key (from part 1) and decrypt. Then use the decrypted text as input in create_key_and_primer to generate a new key and primer containing only alpha numeric comparators which can be used to decrypt the payload.me, atleast if you use Python 3.x. There were enough changes to random` between 2 and 3 that even with the same seed I still got a wrong answer on 2.7 until I switched over to 3.4.
• Did you try using the game codes? – Noctis Skytower Apr 1 '15 at 11:55
• No. I'm not really interested in games. – a6c1 Apr 1 '15 at 14:40
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• Create Account
# Marching Cube Holes
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Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
6 replies to this topic
### #1Tapped Members - Reputation: 384
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Posted 14 September 2012 - 06:38 AM
I have problem generating terrain in marching cubes. My marching cube implementation is in OpenCL. I have 16 * 16 * 16 voxels per block. It works fine when generating a sphere, the only problem is that you can see some small holes between some parts of the sphere. My theory is floating precision, but i dont know.
http://imageshack.us/photo/my-images/833/failsphere.png/
Another problem occur when i tries to generate terrain, with noise:
http://imageshack.us...failterrain.png
As you can see, it does not look well. This was rendered with isolevel 0.9f.
I create the terrain like:
[source lang="cpp"]sampler_t randomVolumeSampler = CLK_NORMALIZED_COORDS_TRUE | CLK_ADDRESS_REPEAT | CLK_FILTER_LINEAR;density.x = wPos.y; density.x += read_imagef(randomVolume, randomVolumeSampler, realGridPos).x * 0.25f;[/source]
So i was wondering, if i have implemented marching cube wrong, or if it behaves right,
Edited by Tapped, 14 September 2012 - 06:42 AM.
### #2Bacterius Crossbones+ - Reputation: 8177
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Posted 14 September 2012 - 09:03 AM
I believe your problem is because you are sampling a random volume of finite size, which is not tileable. The crack patterns suggest you are having a discontinuity in your noise volume at regular UV intervals. Do the cracks appear at the edge of each block only? How do you generate randomVolume? Is it done on the CPU? Can we see the noise generation code?
The marching cubes triangulation looks fine - if you screw that up it usually ends up looking like nothing, so your code seems to work in that regard.
The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach.
- Pessimal Algorithms and Simplexity Analysis
### #3Kaptein Prime Members - Reputation: 1844
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Posted 14 September 2012 - 10:32 AM
it depends on how you build your sphere.. with voxels the best test would be a n^3 for loop over the diameter, ie. dx is iterated (x-radius+1 to x+radius-1) etc.
that way you can be sure the sphere is complete
the terrain is unmistakably as the poster above me said, the noise doesn't use world-space coordinates.. or it's just an array you draw 4 times which is even more wrong
Edited by Kaptein, 14 September 2012 - 10:33 AM.
### #4Tapped Members - Reputation: 384
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Posted 15 September 2012 - 07:21 AM
Thanks, i generate the noise with glm::perlin. First I create a random 3D vector, which is then being multiplied with the position in the random volume, then i use that as the seed for perlin.
The thing is that my blocks are 2.0 * 2.0 * 2.0 in opengl size. And 32 * 32 * 32 voxels per block, therefore 16 * 16 * 16 voxels per cubic meter(opengl). What you see on the picture is one block, so 4 cracks appear per block. So yeah, i believe it is some problem with the tiling of the random volume. However, how can i create a tileable noise volume. I have tried different techniques which did not fix the cracks, like using the same noise on all edges...
#define RANDOM_VOLUME_EXTENT 16
glm::vec3 randVec(glm::signedRand1<float>(), glm::signedRand1<float>(), glm::signedRand1<float>());
for(int i = 0;i < RANDOM_VOLUME_EXTENT;++i)
{
for(int j = 0;j < RANDOM_VOLUME_EXTENT;++j)
{
for(int k = 0;k < RANDOM_VOLUME_EXTENT;++k)
{
glm::vec3 p(i * randVec.x, j * randVec.y, k * randVec.z);
randBuffer[(i*RANDOM_VOLUME_EXTENT*RANDOM_VOLUME_EXTENT)+(j*RANDOM_VOLUME_EXTENT) + k] = glm::perlin(p);
}
}
Edited by Tapped, 15 September 2012 - 07:23 AM.
### #5Tapped Members - Reputation: 384
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Posted 15 September 2012 - 07:32 AM
Never mind, found enough information on how to create tileable perlin noise.
Edited by Tapped, 15 September 2012 - 07:32 AM.
### #6Bacterius Crossbones+ - Reputation: 8177
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Posted 15 September 2012 - 07:41 AM
Nice that you've fixed it (can you give some information or perhaps links for people of the future - hello guys!). I don't know but I'm pretty sure you only need a single seed for perlin noise, and that you then use the world position in the density function to sample the perlin noise directly (with a single seed, it can be sampled from -infinity to +infinity and will never have discontinuities, so you won't have that crack problem).
The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach.
- Pessimal Algorithms and Simplexity Analysis
### #7Tapped Members - Reputation: 384
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Posted 15 September 2012 - 12:19 PM
Now i have implemented the information i got from: http://gamedev.stack...le-perlin-noise
Even if i did that, i still had cracks, but then after a lot of testing of different sample types, i finally got it work. It worked when i used this sampler:
__constant sampler_t randomVolumeSampler = CLK_NORMALIZED_COORDS_TRUE | CLK_ADDRESS_MIRRORED_REPEAT | CLK_FILTER_LINEAR;
float4 realGridPos;
realGridPos.x = gridPos.x * voxelSize;
realGridPos.y = gridPos.y * voxelSize;
realGridPos.z = gridPos.z * voxelSize;
density.x += read_imagef(randomVolume, randomVolumeSampler, realGridPos).x * 0.8f;
realGridPos is the same as the local position in terms of a block, which in this case range from [0.0, 2.0] for each block.
Here is the result:
http://imageshack.us...383/itworks.png
Thanks for the help!
Edited by Tapped, 15 September 2012 - 12:19 PM.
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
PARTNERS
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Dynamic reconfigure of gmapping parameters
Hello all,
I'd like to change the gmapping parameters namely xmin, ymin, xmax, ymax dynamically. My first thought was to use the dynamic_reconfigure package, but it seemed that it requires changing the source code of the gmapping.
My second thought was to use system calls from a different node to execute:
system("rosnode kill slam_gmapping");
system("rosrun gmapping slam_gmapping xmin:=foo ymin:=bra xmax:=foo ymax:=bra &")
However, it seemed that the new gmapping process blocks the node from which the system calls were executed. Please note that I appended the ampersand in the end of the line, but it didn't help. Anyways, using system calls doesn't sound to be a very good way.
What would be the common way to do this?
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## CryptoDB
### Paper: Fuzzy Extractors: How to Generate Strong Keys from Biometrics and Other Noisy Data
Authors: Yevgeniy Dodis Rafail Ostrovsky Leonid Reyzin Adam Smith URL: http://eprint.iacr.org/2003/235 Search ePrint Search Google We provide formal definitions and efficient secure techniques for -- turning noisy information into keys usable for any cryptographic application, and, in particular, -- reliably and securely authenticating biometric data. Our techniques apply not just to biometric information, but to any keying material that, unlike traditional cryptographic keys, is (1) not reproducible precisely and (2) not distributed uniformly. We propose two primitives: a fuzzy extractor reliably extracts nearly uniform randomness R from its input; the extraction is error-tolerant in the sense that R will be the same even if the input changes, as long as it remains reasonably close to the original. Thus, R can be used as a key in a cryptographic application. A secure sketch produces public information about its input w that does not reveal w, and yet allows exact recovery of w given another value that is close to w. Thus, it can be used to reliably reproduce error-prone biometric inputs without incurring the security risk inherent in storing them. We define the primitives to be both formally secure and versatile, generalizing much prior work. In addition, we provide nearly optimal constructions of both primitives for various measures of "closeness" of input data, such as Hamming distance, edit distance, and set difference.
##### BibTeX
@misc{eprint-2003-11948,
title={Fuzzy Extractors: How to Generate Strong Keys from Biometrics and Other Noisy Data},
booktitle={IACR Eprint archive},
keywords={applications / Fuzzy Extractors, Fuzzy Fingerprints, Randomness Extractors, Error-Correcting Codes, Biometric Authentication, Error-Tolerance, Non-Uniformity, Password-based Systems, Metric Embeddings},
url={http://eprint.iacr.org/2003/235},
note={Preliminary version in Eurocrypt 2004. asmith@theory.csail.mit.edu 13777 received 10 Nov 2003, last revised 20 Sep 2007},
author={Yevgeniy Dodis and Rafail Ostrovsky and Leonid Reyzin and Adam Smith},
year=2003
}
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There is no Press for this Ticker
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Far Peak Acquisition Corporation Units, each consisting of one Class A ordinary share, and one-third of one redeemable warrant is in the Consumer Discretionary sector and Multiline Retail industry. They are listed on the NYSE.
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# 1.10: Summing Up, Looking Ahead
[ "article:topic", "authorname:learykristiansen" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
What we have tried to do in this first chapter is to introduce the concepts of formal languages and formal structures. We hope that you will agree that you have seen many mathematical structures in the past, even though you may not have called them structures at the time. By formalizing what we mean when we say that a formula is true in a structure, we will be able to tie together truth and provability in the next couple of chapters.
You might be at a point where you are about to throw your hands up in disgust and say, "Why does any of this matter? I've been doing mathematics for over ten years without worrying about structures or assignment functions, and I have been able to solve problems and succeed as a mathematician so far." Allow us to assure you that the effort and the almost unreasonable precision that we are imposing on our exposition will have a payoff in later chapters. The major theorems that we wish to prove are theorems about the existence or nonexistence of certain objects. To prove that you cannot express a certain idea in a certain language, we have to know, with an amazing amount of exactitude, what a language is and what structures are. Our goals are some theorems that are easy to state incorrectly, so by being precise about what we are saying, we will be able to make (and prove) claims that are truly revolutionary.
Since we will be talking about the existence and nonexistence of proofs, we now must turn our attention to defining (yes, precisely) what sorts of things qualify as proofs. That is the topic of the next chapter.
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# Double Exponential Piecewise Pulse
Define a double exponential piecewise time pulse.
Time axis unit
Specify the unit to be used for the time axis.
Total signal duration ( ${s}_{\text{d}}$ )
The total length of the signal in the specified units.
Amplitude ( ${u}_{0}$ )
The amplitude of the time signal.
Pulse delay ( ${t}_{0}$ )
The pulse delay is the time until the peak of the time signal envelope.
Charge duration ( ${c}_{d}$ )
The charge duration is the time from the pulse delay has ended until the signal begins to discharge.
Charge time constant ( ${\tau }_{2}$ )
The time that would be required to discharge the signal down to 36.8% of its full potential ( ${u}_{0}$ ).
Charge time constant ( ${\tau }_{1}$ )
The time that would be required to charge the signal up to 63.2% of its full potential ( ${u}_{0}$ ).
Number of samples
The number of samples taken from the signal’s analytical equation.
(1)
(2) ${u}_{1}=\frac{{u}_{0}}{1-{e}^{-\frac{{c}_{d}}{{\tau }_{1}}}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{u}_{2}=\frac{{u}_{0}}{{e}^{-\frac{{c}_{d}}{{\tau }_{2}}}}$
The Fourier transform is as follows:
(3)
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Corpus ID: 235458102
# Higher-order topological insulators from $3Q$ charge bond orders on hexagonal lattices: A hint to kagome metals
@inproceedings{Lin2021HigherorderTI,
title={Higher-order topological insulators from \$3Q\$ charge bond orders on hexagonal lattices: A hint to kagome metals},
author={Yu-Ping Lin},
year={2021}
}
We show that unconventional boundary phenonema appear in the 3Q charge bond orders on hexagonal lattices. At the saddle points with Van Hove singularity, the 3Q orders can develop at three nesting momenta. An insulator can develop from the bond modulations, where the C6 symmetry is respected. On the kagome lattice, in-gap corner states arise and carry fractional corner charges 2e/3. Such corner phenomena originates from the corner filling anomaly and indicates a higher-order topological… Expand
2 Citations
#### Figures from this paper
Complex charge density waves at Van Hove singularity on hexagonal lattices: Haldane-model phase diagram and potential realization in kagome metals $\text{AV}_3\text{Sb}_5$
• Physics
• 2021
We analyze how the real and imaginary charge density waves interplay at the Van Hove singularity on the hexagonal lattices. A phenomenological Ginzburg-Landau analysis indicates the formation ofExpand
Kagome superconductors from Pomeranchuk fluctuations in charge density wave metals
• Physics
• 2021
Motivated by the recent experiments on the kagome metals AV3Sb5 with A = K,Rb,Cs, which see onset of charge density wave (CDW) order at ∼ 100 K and superconductivity at ∼ 1 K, we explore the onset ofExpand
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# Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Go back to the page of Corollary 2.2.1.15.
Comment #86 by Carles Sáez on
A pair of typos: In the square, the lower left element should be $e$ instead of $e \circ (e \circ e)$. In the last line, it should say: "since it is isomorphic to the identity functor via the left unit constraint $\lambda$"
Comment #87 by Carles Sáez on
A pair of typos: In the square, the lower left element should be $e$ instead of $e \circ (e \circ e)$. In the last line, it should say: "since it is isomorphic to the identity functor via the left unit constraint $\lambda$"
Comment #88 by Kerodon on
Thanks for the correction! You mean the lower right entry of the diagram, correct?
Comment #89 by Carles Sáez on
Sure, I meant the lower right.
There are also:
• 4 comment(s) on Chapter 2: Examples of $\infty$-Categories
• 2 comment(s) on Section 2.2: The Theory of $2$-Categories
• 2 comment(s) on Subsection 2.2.1: $2$-Categories
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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#elisp
However, note that length should not be used for computing the width of a string on display; use string-width (see Size of Displayed Text) instead.
If you want to change selection, open document below and click on "Move attachment"
GNU Emacs Lisp Reference Manual: String Basics
u can operate on them with the general array and sequence functions documented in Sequences Arrays Vectors. For example, you can access or change individual characters in a string using the functions aref and aset (see Array Functions). <span>However, note that length should not be used for computing the width of a string on display; use string-width (see Size of Displayed Text) instead. There are two text representations for non- ASCII characters in Emacs strings (and in buffers): unibyte and multibyte. For most Lisp programming, you don’t need to be concerned with the
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#### State whether the statement is true or false :Fraction $\frac{18}{39}$ is in its lowest form.
False
$\frac{18}{39} = \frac{6}{13}$
Hence Fraction $\frac{18}{39}$ is not in its lowest form.
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# Risk glossary
## Critical day option
An option structure used for weather derivative transactions where the option payoff is based on defined critical conditions being met for a specified number of days.
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# Inference and Estimation: The Big Picture¶
• Limits
• Limits of Sequences of Real Numbers
• Limits of Functions
• Limit of a Sequence of Random Variables
• Convergence in Distribution
• Convergence in Probability
• Some Basic Limit Laws in Statistics
• Weak Law of Large Numbers
• Central Limit Theorem
### Inference and Estimation: The Big Picture¶
The Markov Chains we discussed earlier fit into our Big Picture, which is about inference and estimation and especially inference and estimation problems where computational techniques are helpful.
Point estimation Set estimation Parametric MLE of finitely many parametersdone Confidence intervals, via the central limit theorem Non-parametric (infinitely many parameters) coming up ... coming up ... One/Many-dimensional Integrals (finite-dimensional) coming up ... coming up ...
But before we move on we have to discuss what makes it all work: the idea of limits - where do you get to if you just keep going?
## Limits¶
Last week we described a Markov Chain, informally, as a system which "jumps" among several states, with the next state depending (probabilistically) only on the current state. Since the system changes randomly, it is generally impossible to predict the exact state of the system in the future. However, the statistical and probailistic properties of the system's future can be predicted. In many applications it is these statistical properties that are important. We saw how we could find a steady state vector:
$$\mathbf{s} = \lim_{n \to \infty} \mathbf{p}^{(n)}$$
(And we noted that $\mathbf{p}^{(n)}$ only converges to a strictly positive vector if $\mathbf{P}$ is a regular transition matrix.)
The week before, we talked about the likelihood function and maximum likelihood estimators for making point estimates of model parameters. For example for the $Bernoulli(\theta^*)$ RV (a $Bernoulli$ RV with true but possibly unknown parameter $\theta^*$, we found that the likelihood function was $L_n(\theta) = \theta^{t_n}(1-\theta)^{(n-t_n)}$ where $t_n = \displaystyle\sum_{i=1}^n x_i$. We also found the maxmimum likelihood estimator (MLE) for the $Bernoulli$ model, $\widehat{\theta}_n = \frac{1}{n}\displaystyle\sum_{i=1}^n x_i$.
We demonstrated these ideas using samples simulated from a $Bernoulli$ process with a secret $\theta^*$. We had an interactive plot of the likelihood function where we could increase $n$, the number of simulated samples or the amount of data we had to base our estimate on, and see the effect on the shape of the likelihood function. The animation belows shows the changing likelihood function for the Bernoulli process with unknown $\theta^*$ as $n$ (the amount of data) increases.
Likelihood function for Bernoulli process, as $n$ goes from 1 to 1000 in a continuous loop.
For large $n$, you can probably make your own guess about the true value of $\theta^*$ even without knowing $t_n$. As the animation progresses, we can see the likelihood function 'homing in' on $\theta = 0.3$.
We can see this in another way, by just looking at the sample mean as $n$ increases. An easy way to do this is with running means: generate a very large sample and then calculate the mean first over just the first observation in the sample, then the first two, first three, etc etc (running means were discussed in an earlier worksheet if you want to go back and review them in detail in your own time). Here we just define a function so that we can easily generate sequences of running means for our $Bernoulli$ process with the unknown $\theta^*$.
#### Preparation: Let's just evaluate the next cel and focus on concepts.¶
You can see what they are as you need to.
In [1]:
def likelihoodBernoulli(theta, n, tStatistic):
'''Bernoulli likelihood function.
theta in [0,1] is the theta to evaluate the likelihood at.
n is the number of observations.
tStatistic is the sum of the n Bernoulli observations.
return a value for the likelihood of theta given the n observations and tStatistic.'''
retValue = 0 # default return value
if (theta >= 0 and theta <= 1): # check on theta
mpfrTheta = RR(theta) # make sure we use a Sage mpfr
retValue = (mpfrTheta^tStatistic)*(1-mpfrTheta)^(n-tStatistic)
return retValue
def bernoulliFInverse(u, theta):
'''A function to evaluate the inverse CDF of a bernoulli.
Param u is the value to evaluate the inverse CDF at.
Param theta is the distribution parameters.
Returns inverse CDF under theta evaluated at u'''
return floor(u + theta)
def bernoulliSample(n, theta, simSeed=None):
'''A function to simulate samples from a bernoulli distribution.
Param n is the number of samples to simulate.
Param theta is the bernoulli distribution parameter.
Param simSeed is a seed for the random number generator, defaulting to 30.
Returns a simulated Bernoulli sample as a list.'''
set_random_seed(simSeed)
us = [random() for i in range(n)]
set_random_seed(None)
return [bernoulliFInverse(u, theta) for u in us] # use bernoulliFInverse in a list comprehension
def bernoulliSampleSecretTheta(n, theta=0.30, simSeed=30):
'''A function to simulate samples from a bernoulli distribution.
Param n is the number of samples to simulate.
Param theta is the bernoulli distribution parameter.
Param simSeed is a seed for the random number generator, defaulting to 30.
Returns a simulated Bernoulli sample as a list.'''
set_random_seed(simSeed)
us = [random() for i in range(n)]
set_random_seed(None)
return [bernoulliFInverse(u, theta) for u in us] # use bernoulliFInverse in a list comprehension
def bernoulliRunningMeans(n, myTheta, mySeed = None):
'''Function to give a list of n running means from bernoulli with specified theta.
Param n is the number of running means to generate.
Param myTheta is the theta for the Bernoulli distribution
Param mySeed is a value for the seed of the random number generator, defaulting to None.'''
sample = bernoulliSample(n, theta=myTheta, simSeed = mySeed)
from pylab import cumsum # we can import in the middle of code
csSample = list(cumsum(sample))
samplesizes = range(1, n+1,1)
return [RR(csSample[i])/samplesizes[i] for i in range(n)]
#return a plot object for BernoulliLikelihood using the secret theta bernoulli generator
def plotBernoulliLikelihoodSecretTheta(n):
'''Return a plot object for BernoulliLikelihood using the secret theta bernoulli generator.
Param n is the number of simulated samples to generate and do likelihood plot for.'''
thisBSample = bernoulliSampleSecretTheta(n) # make sample
tn = sum(thisBSample) # summary statistic
from pylab import arange
ths = arange(0,1,0.01) # get some values to plot against
liks = [likelihoodBernoulli(t,n,tn) for t in ths] # use the likelihood function to generate likelihoods
redshade = 1*n/1000 # fancy colours
def cauchyFInverse(u):
'''A function to evaluate the inverse CDF of a standard Cauchy distribution.
Param u is the value to evaluate the inverse CDF at.'''
return RR(tan(pi*(u-0.5)))
def cauchySample(n):
'''A function to simulate samples from a standard Cauchy distribution.
Param n is the number of samples to simulate.'''
us = [random() for i in range(n)]
return [cauchyFInverse(u) for u in us]
def cauchyRunningMeans(n):
'''Function to give a list of n running means from standardCauchy.
Param n is the number of running means to generate.'''
sample = cauchySample(n)
from pylab import cumsum
csSample = list(cumsum(sample))
samplesizes = range(1, n+1,1)
return [RR(csSample[i])/samplesizes[i] for i in range(n)]
def twoRunningMeansPlot(nToPlot, iters):
'''Function to return a graphics array containing plots of running means for Bernoulli and Standard Cauchy.
Param nToPlot is the number of running means to simulate for each iteration.
Param iters is the number of iterations or sequences of running means or lines on each plot to draw.
Returns a graphics array object containing both plots with titles.'''
xvalues = range(1, nToPlot+1,1)
for i in range(iters):
shade = 0.5*(iters - 1 - i)/iters # to get different colours for the lines
bRunningMeans = bernoulliSecretThetaRunningMeans(nToPlot)
cRunningMeans = cauchyRunningMeans(nToPlot)
bPts = zip(xvalues, bRunningMeans)
cPts = zip(xvalues, cRunningMeans)
if (i < 1):
p1 = line(bPts, rgbcolor = (shade, 0, 1))
cauchyTitleMax = max(cRunningMeans) # for placement of cauchy title
else:
p1 += line(bPts, rgbcolor = (shade, 0, 1))
if max(cRunningMeans) > cauchyTitleMax: cauchyTitleMax = max(cRunningMeans)
titleText1 = "Bernoulli running means" # make title text
t1 = text(titleText1, (nToGenerate/2,1), rgbcolor='blue',fontsize=10)
titleText2 = "Standard Cauchy running means" # make title text
t2 = text(titleText2, (nToGenerate/2,ceil(cauchyTitleMax)+1), rgbcolor='red',fontsize=10)
return graphics_array((p1+t1,p2+t2))
def pmfPointMassPlot(theta):
'''Returns a pmf plot for a point mass function with parameter theta.'''
ptsize = 10
linethick = 2
fudgefactor = 0.07 # to fudge the bottom line drawing
pmf = points((theta,1), rgbcolor="blue", pointsize=ptsize)
pmf += line([(theta,0),(theta,1)], rgbcolor="blue", linestyle=':')
pmf += points((theta,0), rgbcolor = "white", faceted = true, pointsize=ptsize)
pmf += line([(min(theta-2,-2),0),(theta-0.05,0)], rgbcolor="blue",thickness=linethick)
pmf += line([(theta+.05,0),(theta+2,0)], rgbcolor="blue",thickness=linethick)
pmf+= text("Point mass f", (theta,1.1), rgbcolor='blue',fontsize=10)
pmf.axes_color('grey')
return pmf
def cdfPointMassPlot(theta):
'''Returns a cdf plot for a point mass function with parameter theta.'''
ptsize = 10
linethick = 2
fudgefactor = 0.07 # to fudge the bottom line drawing
cdf = line([(min(theta-2,-2),0),(theta-0.05,0)], rgbcolor="blue",thickness=linethick) # padding
cdf += points((theta,1), rgbcolor="blue", pointsize=ptsize)
cdf += line([(theta,0),(theta,1)], rgbcolor="blue", linestyle=':')
cdf += line([(theta,1),(theta+2,1)], rgbcolor="blue", thickness=linethick) # padding
cdf += points((theta,0), rgbcolor = "white", faceted = true, pointsize=ptsize)
cdf+= text("Point mass F", (theta,1.1), rgbcolor='blue',fontsize=10)
cdf.axes_color('grey')
return cdf
def uniformFInverse(u, theta1, theta2):
'''A function to evaluate the inverse CDF of a uniform(theta1, theta2) distribution.
u, u should be 0 <= u <= 1, is the value to evaluate the inverse CDF at.
theta1, theta2, theta2 > theta1, are the uniform distribution parameters.'''
return theta1 + (theta2 - theta1)*u
def uniformSample(n, theta1, theta2):
'''A function to simulate samples from a uniform distribution.
n > 0 is the number of samples to simulate.
theta1, theta2 (theta2 > theta1) are the uniform distribution parameters.'''
us = [random() for i in range(n)]
return [uniformFInverse(u, theta1, theta2) for u in us]
def exponentialFInverse(u, lam):
'''A function to evaluate the inverse CDF of a exponential distribution.
u is the value to evaluate the inverse CDF at.
lam is the exponential distribution parameter.'''
# log without a base is the natural logarithm
return (-1.0/lam)*log(1 - u)
def exponentialSample(n, lam):
'''A function to simulate samples from an exponential distribution.
n is the number of samples to simulate.
lam is the exponential distribution parameter.'''
us = [random() for i in range(n)]
return [exponentialFInverse(u, lam) for u in us]
To get back to our running means of Bernoullin RVs:
In [2]:
def bernoulliSecretThetaRunningMeans(n, mySeed = None):
'''Function to give a list of n running means from Bernoulli with unknown theta.
Param n is the number of running means to generate.
Param mySeed is a value for the seed of the random number generator, defaulting to None
Note: the unknown theta parameter for the Bernoulli process is defined in bernoulliSampleSecretTheta
Return a list of n running means.'''
sample = bernoulliSampleSecretTheta(n, simSeed = mySeed)
from pylab import cumsum # we can import in the middle of code
csSample = list(cumsum(sample))
samplesizes = range(1, n+1,1)
return [RR(csSample[i])/samplesizes[i] for i in range(n)]
Now we can use this function to look at say 5 different sequences of running means (they will be different, because for each iteration, we will simulate a different sample of $Bernoulli$ observations).
In [65]:
nToGenerate = 1500
iterations = 5
xvalues = range(1, nToGenerate+1,1)
for i in range(iterations):
redshade = 0.5*(iterations - 1 - i)/iterations # to get different colours for the lines
bRunningMeans = bernoulliSecretThetaRunningMeans(nToGenerate)
pts = zip(xvalues,bRunningMeans)
if (i == 0):
p = line(pts, rgbcolor = (redshade,0,1))
else:
p += line(pts, rgbcolor = (redshade,0,1))
show(p, figsize=[5,3], axes_labels=['n','sample mean'])
What we notice is how the different lines converge on a sample mean of close to 0.3.
Is life always this easy? Unfortunately no. In the plot below we show the well-behaved running means for the $Bernoulli$ and beside them the running means for simulated standard $Cauchy$ random variables. They are all over the place, and each time you re-evaluate the cell you'll get different all-over-the-place behaviour.
In [48]:
nToGenerate = 15000
iterations = 5
g = twoRunningMeansPlot(nToGenerate, iterations) # uses above function to make plot
show(g,figsize=[10,5])
We talked about the Cauchy in more detail in an earlier notebook. If you cannot recall the detail and are interested, go back to that in your own time. The message here is that although with the Bernoulli process, the sample means converge as the number of observations increases, with the Cauchy they do not.
We talked about $\mathbf{p}^{(n)}$, for $\mathbf{p}$ our Markov Chain transition matrix, converging. We talked about sample means converging (or not). What do we actually mean by converge? These ideas of convergence and limits are fundamental to data science: we need to be able to justify that the way we are attacking a problem will give us the right answer. (At its very simplest, how do we justify that, by generating lots of simulations, we can get to some good approximation for a probability or an integral or a sum?) The advantages of an MLE as a point estimate in parametric estimation all come back to limits and convergence (remember how the likelihood function 'homed in' as the amount of data increased). And, as we will see when we do non-parametric estimation, limits and convergence are also fundamental there.
# Limits of a Sequence of Real Numbers¶
A sequence of real numbers $x_1, x_2, x_3, \ldots$ (which we can also write as $\{ x_i\}_{i=1}^\infty$) is said to converge to a limit $a \in \mathbb{R}$,
$$\underset{i \rightarrow \infty}{\lim} x_i = a$$
if for every natural number $m \in \mathbb{N}$, a natural number $N_m \in \mathbb{N}$ exists such that for every $j \geq N_m$, $\left|x_j - a\right| \leq \frac{1}{m}$
What is this saying? $\left|x_j - a\right|$ is measuring the closeness of the $j$th value in the sequence to $a$. If we pick bigger and bigger $m$, $\frac{1}{m}$ will get smaller and smaller. The definition of the limit is saying that if $a$ is the limit of the sequence then we can get the sequence to become as close as we want ('arbitrarily close') to $a$, and to stay that close, by going far enough into the sequence ('for every $j \geq N_m$, $\left|x_j - a\right| \leq \frac{1}{m}$')
($\mathbb{N}$, the natural numbers, are just the 'counting numbers' $\{1, 2, 3, \ldots\}$.)
Take a trivial example, the sequence $\{x_i\}_{i=1}^\infty = 17, 17, 17, \ldots$
Clearly, $\underset{i \rightarrow \infty}{\lim} x_i = 17$, but let's do this formally:
For every $m \in \mathbb{N}$, take $N_m =1$, then
$\forall$ $j \geq N_m=1, \left|x_j -17\right| = \left|17 - 17\right| = 0 \leq \frac{1}{m}$, as required.
($\forall$ is mathspeak for 'for all' or 'for every')
What about $\{x_i\}_{i=1}^\infty = \displaystyle\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots$, i.e., $x_i = \frac{1}{i}$?
$\underset{i \rightarrow \infty}{\lim} x_i = \underset{i \rightarrow \infty}{\lim}\frac{1}{i} = 0$
For every $m \in \mathbb{N}$, take $N_m = m$, then $\forall$ $j \geq m$, $\left|x_j - 0\right| \leq \left |\frac{1}{m} - 0\right| = \frac{1}{m}$
### YouTry¶
Think about $\{x_i\}_{i=1}^\infty = \frac{1}{1^p}, \frac{1}{2^p}, \frac{1}{3^p}, \ldots$ with $p > 0$. The limit$\underset{i \rightarrow \infty}{\lim} \displaystyle\frac{1}{i^p} = 0$, provided $p > 0$.
You can draw the plot of this very easily using the Sage symbolic expressions we have already met (f.subs(...) allows us to substitute a particular value for one of the symbolic variables in the symbolic function f, in this case a value to use for $p$).
In [53]:
var('i, p')
f = 1/(i^p)
# make and show plot, note we can use f in the label
plot(f.subs(p=1), (x, 0.1, 3), axes_labels=('i',f)).show(figsize=[6,3])
What about $\{x_i\}_{i=1}^\infty = 1^{\frac{1}{1}}, 2^{\frac{1}{2}}, 3^{\frac{1}{3}}, \ldots$. The limit$\underset{i \rightarrow \infty}{\lim} i^{\frac{1}{i}} = 1$.
This one is not as easy to see intuitively, but again we can plot it with SageMath.
In [54]:
var('i')
f = i^(1/i)
n=500
p=plot(f.subs(p=1), (x, 0, n), axes_labels=('i',f)) # main plot
p+=line([(0,1),(n,1)],linestyle=':') # add a dotted line at height 1
p.show(figsize=[6,3]) # show the plot
Finally, $\{x_i\}_{i=1}^\infty = p^{\frac{1}{1}}, p^{\frac{1}{2}}, p^{\frac{1}{3}}, \ldots$, with $p > 0$. The limit$\underset{i \rightarrow \infty}{\lim} p^{\frac{1}{i}} = 1$ provided $p > 0$.
You can cut and paste (with suitable adaptations) to try to plot this one as well ...
In [ ]:
x
(end of You Try)
back to the real stuff ...
# Limits of Functions¶
We say that a function $f(x): \mathbb{R} \rightarrow \mathbb{R}$ has a limit $L \in \mathbb{R}$ as $x$ approaches $a$:
$$\underset{x \rightarrow a}{\lim} f(x) = L$$
provided $f(x)$ is arbitrarily close to $L$ for all ($\forall$) values of $x$ that are sufficiently close to but not equal to $a$.
For example
Consider the function $f(x) = (1+x)^{\frac{1}{x}}$
$\underset{x \rightarrow 0}{\lim} f(x) = \underset{x \rightarrow 0}{\lim} (1+x)^{\frac{1}{x}} = e \approx 2.71828\cdots$
even though $f(0) = (1+0)^{\frac{1}{0}}$ is undefined!
In [55]:
# x is defined as a symbolic variable by default by Sage so we do not need var('x')
f = (1+x)^(1/x)
f.subs(x=0) # this will give you an error message
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-55-9fea08a5e631> in <module>()
1 # x is defined as a symbolic variable by default by Sage so we do not need var('x')
2 f = (Integer(1)+x)**(Integer(1)/x)
----> 3 f.subs(x=Integer(0)) # this will give you an error message
/home/raazesh/all/software/sage/SageMath/src/sage/symbolic/expression.pyx in sage.symbolic.expression.Expression.substitute (build/cythonized/sage/symbolic/expression.cpp:32130)()
5174 sig_on()
5175 try:
-> 5176 res = self._gobj.subs_map(smap, 0)
5177 finally:
5178 sig_off()
ValueError: power::eval(): division by zero
BUT: If you are intersted in the "Art of dividing by zero" talk to Professor Warwick Tucker in Maths Department!
You can get some idea of what is going on with two plots on different scales
In [56]:
f = (1+x)^(1/x)
n1=5
p1=plot(f.subs(p=1), (x, 0.001, n1), axes_labels=('x',f)) # main plot
t1 = text("Large scale plot", (n1/2,e), rgbcolor='blue',fontsize=10)
n2=0.1
p2=plot(f.subs(p=1), (x, 0.0000001, n2), axes_labels=('x',f)) # main plot
p2+=line([(0,e),(n2,e)],linestyle=':') # add a dotted line at height e
t2 = text("Small scale plot", (n2/2,e+.01), rgbcolor='blue',fontsize=10)
show(graphics_array((p1+t1,p2+t2)),figsize=[6,3]) # show the plot
all this has been laying the groundwork for the topic of real interest to us ...
# Limit of a Sequence of Random Variables¶
We want to be able to say things like $\underset{i \rightarrow \infty}{\lim} X_i = X$ in some sensible way. $X_i$ are some random variables, $X$ is some 'limiting random variable', but what do we mean by 'limiting random variable'?
To help us, lets introduce a very very simple random variable, one that puts all its mass in one place.
In [18]:
theta = 2.0
show(graphics_array((pmfPointMassPlot(theta),cdfPointMassPlot(theta))),\
figsize=[8,2]) # show the plots
This is known as the $Point\,Mass(\theta)$ random variable, $\theta \in \mathbb(R)$: the density $f(x)$ is 1 if $x=\theta$ and 0 everywhere else
$$f(x;\theta) = \begin{cases} 0 & \text{ if } x \neq \theta \\ 1 & \text{ if } x = \theta \end{cases}$$$$F(x;\theta) = \begin{cases} 0 & \text{ if } x < \theta \\ 1 & \text{ if } x \geq \theta \end{cases}$$
So, if we had some sequence $\{\theta_i\}_{i=1}^\infty$ and $\underset{i \rightarrow \infty}{\lim} \theta_i = \theta$
and we had a sequence of random variables $X_i \sim Point\,Mass(\theta_i)$, $i = 1, 2, 3, \ldots$
then we could talk about a limiting random variable as $X \sim Point\,Mass(\theta)$:
i.e., we could talk about $\underset{i \rightarrow \infty}{\lim} X_i = X$
In [57]:
# mock up a picture of a sequence of point mass rvs converging on theta = 0
ptsize = 20
i = 1
theta_i = 1/i
p = points((theta_i,1), rgbcolor="blue", pointsize=ptsize)
p += line([(theta_i,0),(theta_i,1)], rgbcolor="blue", linestyle=':')
while theta_i > 0.01:
i+=1
theta_i = 1/i
p += points((theta_i,1), rgbcolor="blue", pointsize=ptsize)
p += line([(theta_i,0),(theta_i,1)], rgbcolor="blue", linestyle=':')
p += points((0,1), rgbcolor="red", pointsize=ptsize)
p += line([(0,0),(0,1)], rgbcolor="red", linestyle=':')
p.show(xmin=-1, xmax = 2, ymin=0, ymax = 1.1, axes=false, gridlines=[None,[0]], \
figsize=[7,2])
Now, we want to generalise this notion of a limit to other random variables (that are not necessarily $Point\,Mass(\theta_i)$ RVs)
What about one many of you will be familiar with - the 'bell-shaped curve'
## The $Gaussian(\mu, \sigma^2)$ or $Normal(\mu, \sigma^2)$ RV?¶
The probability density function (PDF) $f(x)$ is given by
$$f(x ;\mu, \sigma) = \displaystyle\frac{1}{\sigma\sqrt{2\pi}}\exp\left(\frac{-1}{2\sigma^2}(x-\mu)^2\right)$$
The two parameters, $\mu$ and $\sigma$, are sometimes referred to as the location and scale parameters.
To see why this is, use the interactive plot below to have a look at what happens to the shape of the density function $f(x)$ when you change $\mu$ or increase or decrease $\sigma$:
In [58]:
@interact
def _(my_mu=input_box(0, label='mu') ,my_sigma=input_box(1,label='sigma')):
'''Interactive function to plot the normal pdf and ecdf.'''
if my_sigma > 0:
html('<h4>Normal('+str(my_mu)+','+str(my_sigma)+'<sup>2</sup>)</h4>')
var('mu sigma')
f = (1/(sigma*sqrt(2.0*pi)))*exp(-1.0/(2*sigma^2)*(x - mu)^2)
p1=plot(f.subs(mu=my_mu,sigma=my_sigma), (x, my_mu - 3*my_sigma - 2, my_mu + 3*my_sigma + 2), axes_labels=('x','f(x)'))
show(p1,figsize=[8,3])
else:
print "sigma must be greater than 0"
Consider the sequence of random variables $X_1, X_2, X_3, \ldots$, where
• $X_1 \sim Normal(0, 1)$
• $X_2 \sim Normal(0, \frac{1}{2})$
• $X_3 \sim Normal(0, \frac{1}{3})$
• $X_4 \sim Normal(0, \frac{1}{4})$
• $\vdots$
• $X_i \sim Normal(0, \frac{1}{i})$
• $\vdots$
We can use the animation below to see how the PDF $f_{i}(x)$ looks as we move through the sequence of $X_i$ (the animation only goes to $i = 25$, $\sigma = 0.04$ but you get the picture ...)
Normal curve animation, looping through $\sigma = \frac{1}{i}$ for $i = 1, \dots, 25$
We can see that the probability mass of $X_i \sim Normal(0, \frac{1}{i})$ increasingly concentrates about 0 as $i \rightarrow \infty$ and $\frac{1}{i} \rightarrow 0$
Does this mean that $\underset{i \rightarrow \infty}{\lim} X_i = Point\,Mass(0)$?
No, because for any $i$, however large, $P(X_i = 0) = 0$ because $X_i$ is a continuous RV (for any continous RV $X$, for any $x \in \mathbb{R}$, $P(X=x) = 0$).
So, we need to refine our notions of convergence when we are dealing with random variables
# Convergence in Distribution¶
Let $X_1, X_2, \ldots$ be a sequence of random variables and let $X$ be another random variable. Let $F_i$ denote the distribution function (DF) of $X_i$ and let $F$ denote the distribution function of $X$.
Now, if for any real number $t$ at which $F$ is continuous,
$$\underset{i \rightarrow \infty}{\lim} F_i(t) = F(t)$$
(in the sense of the convergence or limits of functions we talked about earlier)
Then we can say that the sequence or RVs $X_i$, $i = 1, 2, \ldots$ converges to $X$ in distribution and write $X_i \overset{d}{\rightarrow} X$.
An equivalent way of defining convergence in distribution is to go right back to the meaning of the probabilty space 'under the hood' of a random variable, a random variable $X$ as a mapping from the sample space $\Omega$ to the real line ($X: \Omega \rightarrow \mathbb{R}$), and the sample points or outcomes in the sample space, the $\omega \in \Omega$. For $\omega \in \Omega$, $X(\omega)$ is the mapping of $\omega$ to the real line $\mathbb{R}$. We could look at the set of $\omega$ such that $X(\omega) \leq t$, i.e. the set of $\omega$ that map to some value on the real line less than or equal to $t$, $\{\omega: X(\omega) \leq t \}$.
Saying that for any $t \in \mathbb{R}$, $\underset{i \rightarrow \infty}{\lim} F_i(t) = F(t)$ is the equivalent of saying that for any $t \in \mathbb{R}$,
$$\underset{i \rightarrow \infty}{\lim} P\left(\{\omega:X_i(\omega) \leq t \}\right) = P\left(\{\omega: X(\omega) \leq t\right)$$
Armed with this, we can go back to our sequence of $Normal$ random variables $X_1, X_2, X_3, \ldots$, where
• $X_1 \sim Normal(0, 1)$
• $X_2 \sim Normal(0, \frac{1}{2})$
• $X_3 \sim Normal(0, \frac{1}{3})$
• $X_4 \sim Normal(0, \frac{1}{4})$
• $\vdots$
• $X_i \sim Normal(0, \frac{1}{i})$
• $\vdots$
and let $X \sim Point\,Mass(0)$,
and say that the $X_i$ converge in distribution to the $x \sim Point\,Mass$ RV $X$,
$$X_i \overset{d}{\rightarrow} X$$
What we are saying with convergence in distribution, informally, is that as $i$ increases, we increasingly expect to see the next outcome in a sequence of random experiments becoming better and better modeled by the limiting random variable. In this case, as $i$ increases, the $Point\,Mass(0)$ is becoming a better and better model for the next outcome of a random experiment with outcomes $\sim Normal(0,\frac{1}{i})$.
In [22]:
# mock up a picture of a sequence of converging normal distributions
my_mu = 0
upper = my_mu + 5; lower = -upper; # limits for plot
var('mu sigma')
stop_i = 12
html('<h4>N(0,1) to N(0, 1/'+str(stop_i)+')</h4>')
f = (1/(sigma*sqrt(2.0*pi)))*exp(-1.0/(2*sigma^2)*(x - mu)^2)
p=plot(f.subs(mu=my_mu,sigma=1.0), (x, lower, upper), rgbcolor = (0,0,1))
for i in range(2, stop_i, 1): # just do a few of them
shade = 1-11/i # make them different colours
textOffset = -0.2 # offset for placement of text - may need adjusting
p+=text("0",(0,textOffset),fontsize = 10, rgbcolor='grey')
p+=text(str(upper.n(digits=2)),(upper,textOffset),fontsize = 10, rgbcolor='grey')
p+=text(str(lower.n(digits=2)),(lower,textOffset),fontsize = 10, rgbcolor='grey')
p.show(axes=false, gridlines=[None,[0]], figsize=[7,3])
We have said that the $X_i \sim Normal(0,\frac{1}{i})$ with distribution functions $F_i$ converge in distribution to $X \sim Point\,Mass(0)$ with distribution function $F$, which means that we must be able to show that for any real number $t$ at which $F$ is continuous,
$$\underset{i \rightarrow \infty}{\lim} F_i(t) = F(t)$$
Note that for any of the $X_i \sim Normal(0, \frac{1}{i})$, $F_i(0) = \frac{1}{2}$, and also note that for $X \sim Point,Mass(0)$, $F(0) = 1$, so clearly $F_i(0) \neq F(0)$.
What has gone wrong?
Nothing: we said that we had to be able to show that $\underset{i \rightarrow \infty}{\lim} F_i(t) = F(t)$ for any $t \in \mathbb{R}$ at which $F$ is continuous, but the $Point\,Mass(0)$ distribution function $F$ is not continous at 0!
In [24]:
theta = 0.0
# show the plots
show(graphics_array((pmfPointMassPlot(theta),cdfPointMassPlot(theta))),figsize=[8,2])
# Convergence in Probability¶
Let $X_1, X_2, \ldots$ be a sequence of random variables and let $X$ be another random variable. Let $F_i$ denote the distribution function (DF) of$X_i$ and let $F$ denote the distribution function of $X$.
Now, if for any real number $\varepsilon > 0$,
$$\underset{i \rightarrow \infty}{\lim} P\left(|X_i - X| > \varepsilon\right) = 0$$
Then we can say that the sequence $X_i$, $i = 1, 2, \ldots$ converges to $X$ in probability and write $X_i \overset{P}{\rightarrow} X$.
Or, going back again to the probability space 'under the hood' of a random variable, we could look the way the $X_i$ maps each outcome $\omega \in \Omega$, $X_i(\omega)$, which is some point on the real line, and compare this to mapping $X(\omega)$.
Saying that for any $\varepsilon \in \mathbb{R}$, $\underset{i \rightarrow \infty}{\lim} P\left(|X_i - X| > \varepsilon\right) = 0$ is the equivalent of saying that for any $\varepsilon \in \mathbb{R}$,
$$\underset{i \rightarrow \infty}{\lim} P\left(\{\omega:|X_i(\omega) - X(\omega)| > \varepsilon \}\right) = 0$$
Informally, we are saying $X$ is a limit in probabilty if, by going far enough into the sequence $X_i$, we can ensure that the mappings $X_i(\omega)$ and $X(\omega)$ will be arbitrarily close to each other on the real line for all $\omega \in \Omega$.
Note that convergence in distribution is implied by convergence in probability: convergence in distribution is the weakest form of convergence; any sequence of RV's that converges in probability to some RV $X$ also converges in distribution to $X$ (but not necessarily vice versa).
In [25]:
# mock up a picture of a sequence of converging normal distributions
my_mu = 0
var('mu sigma')
upper = 0.2; lower = -upper
i = 20 # start part way into the sequence
lim = 100 # how far to go
stop_i = 12
html('<h4>N(0,1/'+str(i)+') to N(0, 1/'+str(lim)+')</h4>')
f = (1/(sigma*sqrt(2.0*pi)))*exp(-1.0/(2*sigma^2)*(x - mu)^2)
p=plot(f.subs(mu=my_mu,sigma=1.0/i), (x, lower, upper), rgbcolor = (0,0,1))
for j in range(i, lim+1, 4): # just do a few of them
shade = 1-(j-i)/(lim-i) # make them different colours
textOffset = -1.5 # offset for placement of text - may need adjusting
p+=text("0",(0,textOffset),fontsize = 10, rgbcolor='grey')
p+=text(str(upper.n(digits=2)),(upper,textOffset),fontsize = 10, rgbcolor='grey')
p+=text(str(lower.n(digits=2)),(lower,textOffset),fontsize = 10, rgbcolor='grey')
p.show(axes=false, gridlines=[None,[0]], figsize=[7,3])
For our sequence of $Normal$ random variables $X_1, X_2, X_3, \ldots$, where
• $X_1 \sim Normal(0, 1)$
• $X_2 \sim Normal(0, \frac{1}{2})$
• $X_3 \sim Normal(0, \frac{1}{3})$
• $X_4 \sim Normal(0, \frac{1}{4})$
• $\vdots$
• $X_i \sim Normal(0, \frac{1}{i})$
• $\vdots$
and $X \sim Point\,Mass(0)$,
It can be shown that the $X_i$ converge in probability to $X \sim Point\,Mass(0)$ RV $X$,
$$X_i \overset{P}{\rightarrow} X$$
(the formal proof of this involves Markov's Inequality, which is beyond the scope of this course).
# Some Basic Limit Laws in Statistics¶
Intuition behind Law of Large Numbers and Central Limit Theorem
Take a look at the Khan academy videos on the Law of Large Numbers and the Central Limit Theorem. This will give you a working idea of these theorems. In the sequel, we will strive for a deeper understanding of these theorems on the basis of the two notions of convergence of sequences of random variables we just saw.
## Weak Law of Large Numbers¶
Remember that a statistic is a random variable, so a sample mean is a random variable. If we are given a sequence of independent and identically distributed RVs, $X_1,X_2,\ldots \overset{IID}{\sim} X_1$, then we can also think of a sequence of random variables $\overline{X}_1, \overline{X}_2, \ldots, \overline{X}_n, \ldots$ ($n$ being the sample size).
Since $X_1, X_2, \ldots$ are $IID$, they all have the same expection, say $E(X_1)$ by convention.
If $E(X_1)$ exists, then the sample mean $\overline{X}_n$ converges in probability to $E(X_1)$ (i.e., to the expectatation of any one of the individual RVs):
$$\text{If} \quad X_1,X_2,\ldots \overset{IID}{\sim} X_1 \ \text{and if } \ E(X_1) \ \text{exists, then } \ \overline{X}_n \overset{P}{\rightarrow} E(X_1) \ .$$
Going back to our definition of convergence in probability, we see that this means that for any real number $\varepsilon > 0$, $\underset{n \rightarrow \infty}{\lim} P\left(|\overline{X}_n - E(X_1)| > \varepsilon\right) = 0$
Informally, this means that means that, by taking larger and larger samples we can make the probability that the average of the observations is more than $\varepsilon$ away from the expected value get smaller and smaller.
Proof of this is beyond the scope of this course, but we have already seen it in action when we looked at the $Bernoulli$ running means. Have another look, this time with only one sequence of running means. You can increase $n$, the sample size, and change $\theta$. Note that the seed for the random number generator is also under your control. This means that you can get replicable samples: in particular, in this interact, when you increase the sample size it looks as though you are just adding more to an existing sample rather than starting from scratch with a new one.
In [59]:
@interact
def _(nToGen=slider(1,1500,1,100,label='n'),my_theta=input_box(0.3,label='theta'),rSeed=input_box(1234,label='random seed')):
'''Interactive function to plot running mean for a Bernoulli with specified n, theta and random number seed.'''
if my_theta >= 0 and my_theta <= 1:
html('<h4>Bernoulli('+str(my_theta.n(digits=2))+')</h4>')
xvalues = range(1, nToGen+1,1)
bRunningMeans = bernoulliRunningMeans(nToGen, myTheta=my_theta, mySeed=rSeed)
pts = zip(xvalues, bRunningMeans)
p = line(pts, rgbcolor = (0,0,1))
p+=line([(0,my_theta),(nToGen,my_theta)],linestyle=':',rgbcolor='grey')
show(p, figsize=[5,3], axes_labels=['n','sample mean'],ymax=1)
else:
print 'Theta must be between 0 and 1'
# Central Limit Theorem¶
You have probably all heard of the Central Limit Theorem before, but now we can relate it to our definition of convergence in distribution.
Let $X_1,X_2,\ldots \overset{IID}{\sim} X_1$ and suppose $E(X_1)$ and $V(X_1)$ both exist,
then
$$\overline{X}_n = \frac{1}{n} \sum_{i=1}^n X_i \overset{d}{\rightarrow} X \sim Normal \left(E(X_1),\frac{V(X_1)}{n} \right)$$
And remember $Z \sim Normal(0,1)$?
Consider $Z_n := \displaystyle\frac{\overline{X}_n-E(\overline{X}_n)}{\sqrt{V(\overline{X}_n)}} = \displaystyle\frac{\sqrt{n} \left( \overline{X}_n -E(X_1) \right)}{\sqrt{V(X_1)}}$
If $\overline{X}_n = \displaystyle\frac{1}{n} \displaystyle\sum_{i=1}^n X_i \overset{d}{\rightarrow} X \sim Normal \left(E(X_1),\frac{V(X_1)}{n} \right)$, then $\overline{X}_n -E(X_1) \overset{d}{\rightarrow} X-E(X_1) \sim Normal \left( 0,\frac{V(X_1)}{n} \right)$
and $\sqrt{n} \left( \overline{X}_n -E(X_1) \right) \overset{d}{\rightarrow} \sqrt{n} \left( X-E(X_1) \right) \sim Normal \left( 0,V(X_1) \right)$
so $Z_n := \displaystyle \frac{\overline{X}_n-E(\overline{X}_n)}{\sqrt{V(\overline{X}_n)}} = \displaystyle\frac{\sqrt{n} \left( \overline{X}_n -E(X_1) \right)}{\sqrt{V(X_1)}} \overset{d}{\rightarrow} Z \sim Normal \left( 0,1 \right)$
Thus, for sufficiently large $n$ (say $n>30$), probability statements about $\overline{X}_n$ can be approximated using the $Normal$ distribution.
The beauty of the CLT, as you have probably seen from other courses, is that $\overline{X}_n \overset{d}{\rightarrow} Normal \left( E(X_1), \frac{V(X_1)}{n} \right)$ does not require the $X_i$ to be normally distributed.
We can try this with our $Bernoulli$ RV generator. First, a small number of samples:
In [60]:
theta, n, samples = 0.6, 10, 5 # concise way to set some variable values
sampleMeans=[] # empty list
for i in range(0, samples, 1): # loop
thisMean = QQ(sum(bernoulliSample(n, theta)))/n # get a sample and find the mean
sampleMeans.append(thisMean) # add mean to the list of means
sampleMeans # disclose the sample means
Out[60]:
[7/10, 9/10, 3/10, 4/5, 2/5]
You can use the interactive plot to increase the number of samples and make a histogram of the sample means. According to the CLT, for lots of reasonably-sized samples we should get a nice symmetric bell-curve-ish histogram centred on $\theta$. You can adjust the number of bins in the histogram as well as the number of samples, sample size, and $\theta$.
In [61]:
import pylab
@interact
def _(samples=slider(1,3000,1,100,label='number of samples'), nToGen=slider(1,1500,1,100,label='sample size n'),my_theta=input_box(0.3,label='theta'),Bins=5):
'''Interactive function to plot distribution of sample means for a Bernoulli process.'''
if my_theta >= 0 and my_theta <= 1 and samples > 0:
sampleMeans=[] # empty list
for i in range(0, samples, 1):
thisMean = RR(sum(bernoulliSample(nToGen, my_theta)))/nToGen
sampleMeans.append(thisMean)
pylab.clf() # clear current figure
n, bins, patches = pylab.hist(sampleMeans, Bins, normed=true)
pylab.ylabel('normalised count')
pylab.title('Normalised histogram for Bernoulli sample means')
pylab.savefig('myHist') # to actually display the figure
pylab.show()
#show(p, figsize=[5,3], axes_labels=['n','sample mean'],ymax=1)
else:
print 'Theta must be between 0 and 1, and samples > 0'
Increase the sample size and the numbe rof bins in the above interact and see if the histograms of the sample means are looking more and more normal as the CLT would have us believe.
But although the $X_i$ do not have to be $\sim Normal$ for $\overline{X}_n = \overset{d}{\rightarrow} X \sim Normal\left(E(X_1),\frac{V(X_1)}{n} \right)$, remember that we said "Let $X_1,X_2,\ldots \overset{IID}{\sim} X_1$ and suppose $E(X_1)$ and $V(X_1)$ both exist", then,
$$\overline{X}_n = \frac{1}{n} \sum_{i=1}^n X_i \overset{d}{\rightarrow} X \sim Normal \left(E(X_1),\frac{V(X_1)}{n} \right)$$
This is where is all goes horribly wrong for the standard $Cauchy$ distribution (any $Cauchy$ distribution in fact): neither the expectation nor the variance exist for this distribution. The Central Limit Theorem cannot be applied here. In fact, if $X_1,X_2,\ldots \overset{IID}{\sim}$ standard $Cauchy$, then $\overline{X}_n = \displaystyle \frac{1}{n} \sum_{i=1}^n X_i \sim$ standard $Cauchy$.
### YouTry¶
Try looking at samples from two other RVs where the expectation and variance do exist, the $Uniform$ and the $Exponential$:
In [3]:
import pylab
@interact
def _(samples=input_box(100,label='number of samples'), nToGen=slider(1,1500,1,100,label='sample size n'),my_theta1=input_box(2,label='theta1'),my_theta2=input_box(4,label='theta1'),Bins=5):
'''Interactive function to plot distribution of sample means for a Uniform(theta1, theta2) process.'''
if (my_theta1 < my_theta2) and samples > 0:
sampleMeans=[] # empty list
for i in range(0, samples, 1):
thisMean = RR(sum(uniformSample(nToGen, my_theta1, my_theta2)))/nToGen
sampleMeans.append(thisMean)
pylab.clf() # clear current figure
n, bins, patches = pylab.hist(sampleMeans, Bins, normed=true)
pylab.ylabel('normalised count')
pylab.title('Normalised histogram for Uniform sample means')
pylab.savefig('myHist') # to actually display the figure
pylab.show()
#show(p, figsize=[5,3], axes_labels=['n','sample mean'],ymax=1)
else:
print 'theta1 must be less than theta2, and samples > 0'
In [4]:
import pylab
@interact
def _(samples=input_box(100,label='number of samples'), nToGen=slider(1,1500,1,100,label='sample size n'),my_lambda=input_box(2,label='lambda'),Bins=5):
'''Interactive function to plot distribution of sample means for an Exponential(lambda) process.'''
if my_lambda > 0 and samples > 0:
sampleMeans=[] # empty list
for i in range(0, samples, 1):
thisMean = RR(sum(exponentialSample(nToGen, my_lambda)))/nToGen
sampleMeans.append(thisMean)
pylab.clf() # clear current figure
n, bins, patches = pylab.hist(sampleMeans, Bins, normed=true)
pylab.ylabel('normalised count')
pylab.title('Normalised histogram for Exponential sample means')
pylab.savefig('myHist') # to actually display the figure
pylab.show()
#show(p, figsize=[5,3], axes_labels=['n','sample mean'],ymax=1)
else:
print 'lambda must be greater than 0, and samples > 0'
### YouTry Later¶
Python's random for sampling and sequence manipulation
The Python random module, available in SageMath, provides a useful way of taking samples if you have already generated a 'population' to sample from, or otherwise playing around with the elements in a sequence. See http://docs.python.org/library/random.html for more details. Here we will try a few of them.
The aptly-named sample function allows us to take a sample of a specified size from a sequence. We will use a list as our sequence:
In [35]:
pop = range(1, 101, 1) # make a population
sample(pop, 10) # sample 10 elements from it at random
Out[35]:
[37, 51, 2, 35, 19, 59, 40, 96, 14, 80]
Each call to sample will select unique elements in the list (note that 'unique' here means that it will not select the element at any particular position in the list more than once, but if there are duplicate elements in the list, such as with a list [1,2,4,2,5,3,1,3], then you may well get any of the repeated elements in your sample more than once). sample samples with replacement, which means that repeated calls to sample may give you samples with the same elements in.
In [37]:
popWithDuplicates = range(1, 11, 1)*4 # make a population with repeated elements
print(popWithDuplicates)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
In [38]:
for i in range (5):
print sample(popWithDuplicates, 10)
[8, 1, 8, 4, 2, 4, 9, 2, 5, 2]
[7, 1, 6, 4, 1, 5, 2, 6, 3, 5]
[10, 1, 8, 4, 1, 6, 3, 3, 2, 1]
[7, 9, 9, 3, 10, 7, 2, 1, 6, 5]
[8, 1, 7, 1, 5, 2, 4, 6, 4, 9]
Try experimenting with choice, which allows you to select one element at random from a sequence, and shuffle, which shuffles the sequence in place (i.e, the ordering of the sequence itself is changed rather than you being given a re-ordered copy of the list). It is probably easiest to use lists for your sequences. See how shuffle is creating permutations of the list. You could use sample and shuffle to emulate permuations of k objects out of n ...
You may need to check the documentation to see how use these functions.
In [39]:
?sample
In [40]:
?shuffle
In [41]:
?choice
In [ ]:
In [ ]:
In [ ]:
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# What is difference between H⁺ and proton?
What is difference between $\ce{H+}$ and a proton?
• Do you think there should be any difference? – bon Sep 8 '15 at 20:27
There is no chemical difference, only a psychological one: how do you think about it. They are both the same thing, but many people associate $\ce{H+}$ ions with chemical reactions and protons with particle physics. A hydrogen atom has one electron and a proton, no neutron. Therefore $\ce{H+}$ is just a proton.
That is why acids are sometimes referred as proton donors as they donate $\ce{H+}$ to a base (also known as proton acceptor).
• Yes, and also H+ makes me think more about the solvated proton. For instance, in chemical reactions where what shows up in equations as H+ is really H3O+, etc. Using the word proton emphasizes the physics over chemistry, thus it would be odd to refer to proton NMR as H+ NMR, again, that would be 'correct' but confusing. – AlaskaRon Sep 14 '15 at 5:18
• This answer is good and valid in the narrow (but common) case where $\ce{H}$ refers to $_1^{1}\ce{H}$ only (with $\ce{D}$ equalling $^2_1\ce{H}$ and $\ce{T}$ for $^3_1\ce{H}$). This means that in terms of molar mass, there is a difference between the case just outlined and the concept of $\ce{H}$ for hydrogen as an element with different isotopes. – TAR86 Dec 19 '17 at 17:20
In gaseous/plasmatic phase, there is no difference - $$\mathrm{p}$$ and $$\ce{H+}$$ are synonyms for a proton.
The former ( p, proton ) is more often used by physicists in subatomic particle context, the latter by chemists in hydrogen properties/behaviour context.
In polar solvents like water or liquid ammonia, "naked" protons nor electrons cannot exist, but they are solvated, forming molecular ion or reacting with solution components.
$$\ce{H+}$$, frequently used in electrochemistry and ion equations, is the implied shortcut for hydrated proton aka hydronium $$\ce{H3O+}$$.
Strong enough acids form solid hydronium salts like e.g. hydronium perchlorate $$\ce{[H3O+][ClO4-]}$$.
There are consider also larger scale hydronium hydration structures
$$\ce{H9O4+ = H3O+ \cdot 3 H2O}$$
$$\ce{H3O+ } \cdot 6\ \ce{H2O}$$
$$\ce{H3O+ } \cdot 20\ \ce{H2O}$$,
• I have not noticed until now the Q and 1st A are quite old, being among active question.:-) – Poutnik Oct 15 '19 at 6:39
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# How do you find the derivative of root5(x^-5)?
$\frac{d \left(\sqrt[5]{{x}^{- 5}}\right)}{\mathrm{dx}} = - \frac{1}{x} ^ 2$
$\sqrt[5]{{x}^{- 5}} = {\left({x}^{- 5}\right)}^{\frac{1}{5}} = {x}^{\left(- 5\right) \times \left(\frac{1}{5}\right)} = {x}^{-} 1$
As $\frac{d \left({x}^{-} 1\right)}{\mathrm{dx}} = - 1 \cdot {x}^{- 2} = - \frac{1}{x} ^ 2$
Hence $\frac{d \left(\sqrt[5]{{x}^{- 5}}\right)}{\mathrm{dx}} = - \frac{1}{x} ^ 2$
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# Think Bayes
This notebook presents example code and exercise solutions for Think Bayes.
In [1]:
# Configure Jupyter so figures appear in the notebook
%matplotlib inline
# Configure Jupyter to display the assigned value after an assignment
%config InteractiveShell.ast_node_interactivity='last_expr_or_assign'
# import classes from thinkbayes2
from thinkbayes2 import Pmf, Suite, Beta
import thinkplot
import numpy as np
## The social desirability problem
Whenever you survey people about sensitive issues, you have to deal with social desirability bias, which is the tendency of people to shade their answers in the direction they think shows them in the most positive light.
One of the ways to improve the quality of the results is to collect responses in indirect ways. For example, here's a clever way one research group estimated the prevalence of atheists.
Another way is randomized response, as described in this presentation or this video.
As an example, suppose you ask 100 people to flip a coin and:
• If they get heads, they report YES.
• If they get tails, they honestly answer the question "Do you believe in God?"
And suppose you get 80 YESes and 20 NOs.
1. Estimate the prevalence of believers in the surveyed population (by which, as always, I mean compute a posterior distribution).
2. How efficient is this method? That is, how does the width of the posterior distribution compare to the distribution you would get if 100 people answered the question honestly?
In [2]:
# Solution
class Social(Suite):
def Likelihood(self, data, hypo):
"""
data: outcome of unreliable measurement, either 'YES' or 'NO'
hypo: actual proportion of the thing we're measuring
"""
p = hypo
p_yes = 0.5 + p/2
if data == 'YES':
return p_yes
else:
return 1 - p_yes
In [3]:
# Solution
prior = np.linspace(0, 1, 101)
suite = Social(prior)
thinkplot.Pdf(suite, label='Prior')
thinkplot.decorate(xlabel='Fraction of the population',
ylabel='PDF')
In [4]:
# Solution
for i in range(80):
suite.Update('YES')
for i in range(20):
suite.Update('NO')
In [5]:
# Solution
thinkplot.Pdf(suite, label='Posterior')
thinkplot.decorate(xlabel='Fraction of the population',
ylabel='PDF')
In [6]:
# Solution
suite.Mean(), suite.MAP()
Out[6]:
(0.5882352942702531, 0.6)
In [7]:
# Solution
# For comparison, what would we think if we had been able
# to survey 100 people directly?
beta = Beta(1, 1)
beta.Update((60, 40))
thinkplot.Pdf(beta.MakePmf(), label='Direct', color='gray')
thinkplot.Pdf(suite, label='Randomized')
thinkplot.decorate(xlabel='Fraction of the population',
ylabel='PDF')
In [8]:
# Solution
# To see how efficient this method is, we can divide the sample size for
# the direct method by a factor. It looks like we lose a factor of $2 \sqrt{2}$.
factor = 2 * np.sqrt(2)
beta = Beta(1, 1)
beta.Update((60/factor, 40/factor))
thinkplot.Pdf(beta.MakePmf(), label='Direct', color='gray')
thinkplot.Pdf(suite, label='Randomized')
thinkplot.decorate(xlabel='Fraction of the population',
ylabel='PDF')
In [9]:
# Solution
# So the effective sample size is about 35.
100 / 2 / np.sqrt(2)
Out[9]:
35.35533905932737
In [ ]:
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# NAG CL Interfacef12fec (real_symm_monit)
Note: this function uses optional parameters to define choices in the problem specification. If you wish to use default settings for all of the optional parameters, then the option setting function f12fdc need not be called. If, however, you wish to reset some or all of the settings please refer to Section 11 in f12fdc for a detailed description of the specification of the optional parameters.
Settings help
CL Name Style:
## 1Purpose
f12fec can be used to return additional monitoring information during computation. It is in a suite of functions which includes f12fac, f12fbc, f12fcc and f12fdc.
## 2Specification
#include
void f12fec (Integer *niter, Integer *nconv, double ritz[], double rzest[], const Integer icomm[], const double comm[])
The function may be called by the names: f12fec, nag_sparseig_real_symm_monit or nag_real_symm_sparse_eigensystem_monit.
## 3Description
The suite of functions is designed to calculate some of the eigenvalues, $\lambda$, (and optionally the corresponding eigenvectors, $x$) of a standard eigenvalue problem $Ax=\lambda x$, or of a generalized eigenvalue problem $Ax=\lambda Bx$ of order $n$, where $n$ is large and the coefficient matrices $A$ and $B$ are sparse, real and symmetric. The suite can also be used to find selected eigenvalues/eigenvectors of smaller scale dense, real and symmetric problems.
On an intermediate exit from f12fbc with ${\mathbf{irevcm}}=4$, f12fec may be called to return monitoring information on the progress of the Arnoldi iterative process. The information returned by f12fec is:
• the number of the current Arnoldi iteration;
• the number of converged eigenvalues at this point;
• the real and imaginary parts of the converged eigenvalues;
• the error bounds on the converged eigenvalues.
f12fec does not have an equivalent function from the ARPACK package which prints various levels of detail of monitoring information through an output channel controlled via an argument value (see Lehoucq et al. (1998) for details of ARPACK routines). f12fec should not be called at any time other than immediately following an ${\mathbf{irevcm}}=4$ return from f12fbc.
## 4References
Lehoucq R B (2001) Implicitly restarted Arnoldi methods and subspace iteration SIAM Journal on Matrix Analysis and Applications 23 551–562
Lehoucq R B and Scott J A (1996) An evaluation of software for computing eigenvalues of sparse nonsymmetric matrices Preprint MCS-P547-1195 Argonne National Laboratory
Lehoucq R B and Sorensen D C (1996) Deflation techniques for an implicitly restarted Arnoldi iteration SIAM Journal on Matrix Analysis and Applications 17 789–821
Lehoucq R B, Sorensen D C and Yang C (1998) ARPACK Users' Guide: Solution of Large-scale Eigenvalue Problems with Implicitly Restarted Arnoldi Methods SIAM, Philadelphia
## 5Arguments
1: $\mathbf{niter}$Integer * Output
On exit: the number of the current Arnoldi iteration.
2: $\mathbf{nconv}$Integer * Output
On exit: the number of converged eigenvalues so far.
3: $\mathbf{ritz}\left[\mathit{dim}\right]$double Output
Note: the dimension, dim, of the array ritz must be at least ${\mathbf{ncv}}$ (see f12fac).
On exit: the first nconv locations of the array ritz contain the real converged approximate eigenvalues.
4: $\mathbf{rzest}\left[\mathit{dim}\right]$double Output
Note: the dimension, dim, of the array rzest must be at least ${\mathbf{ncv}}$ (see f12fac).
On exit: the first nconv locations of the array rzest contain the Ritz estimates (error bounds) on the real nconv converged approximate eigenvalues.
5: $\mathbf{icomm}\left[\mathit{dim}\right]$const Integer Communication Array
Note: the dimension, dim, of the array icomm must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{licomm}}\right)$, where licomm is passed to the setup function (see f12fac).
On entry: the array icomm output by the preceding call to f12fbc.
6: $\mathbf{comm}\left[\mathit{dim}\right]$const double Communication Array
Note: the dimension, dim, of the array comm must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{lcomm}}\right)$, where lcomm is passed to the setup function (see f12fac).
On entry: the array comm output by the preceding call to f12fbc.
None.
## 7Accuracy
A Ritz value, $\lambda$, is deemed to have converged if its Ritz estimate $\le {\mathbf{Tolerance}}×|\lambda |$. The default ${\mathbf{Tolerance}}$ used is the machine precision given by X02AJC.
## 8Parallelism and Performance
f12fec is not threaded in any implementation.
None.
## 10Example
This example solves $Kx=\lambda {K}_{G}x$ using the ${\mathbf{Buckling}}$ option (see f12fdc, where $K$ and ${K}_{G}$ are obtained by the finite element method applied to the one-dimensional discrete Laplacian operator $\frac{{\partial }^{2}u}{\partial {x}^{2}}$ on $\left[0,1\right]$, with zero Dirichlet boundary conditions using piecewise linear elements. The shift, $\sigma$, is a real number, and the operator used in the Buckling iterative process is $\mathrm{op}=\text{inv}\left(K-\sigma {K}_{G}\right)×K$ and $B=K$.
### 10.1Program Text
Program Text (f12fece.c)
### 10.2Program Data
Program Data (f12fece.d)
### 10.3Program Results
Program Results (f12fece.r)
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# 2nd fundamental theorem of calculus chain rule
0
1
Let F be any antiderivative of f on an interval , that is, for all in .Then . Fundamental theorem of calculus - Application Hot Network Questions Would a hibernating, bear-men society face issues from unattended farmlands in winter? %PDF-1.4 The Chain Rule and the Second Fundamental Theorem of Calculus1 Problem 1. Let be a number in the interval .Define the function G on to be. The Fundamental Theorem tells us that E′(x) = e−x2. Define a new function F(x) by. So let's think about what F of b minus F of a is, what this is, where both b and a are also in this interval. Suppose that f(x) is continuous on an interval [a, b]. Therefore, by the Chain Rule, G′(x) = F′(√ x) d dx √ x = sin √ x 2 1 2 √ x = sinx 2 √ x Problem 2. Get 1:1 help now from expert Calculus tutors Solve it with our calculus … Note that this graph looks just like the left hand graph, except that the variable is x instead of t. So you can find the derivativ… The fundamental theorem of calculus states that the integral of a function f over the interval [a, b] can be calculated by finding an antiderivative F of f: ∫ = − (). Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. In Section4.4, we learned the Fundamental Theorem of Calculus (FTC), which from here forward will be referred to as the First Fundamental Theorem of Calculus, as in this section we develop a corresponding result that follows it. If you're seeing this message, it means we're having trouble loading external resources on our website. So you've learned about indefinite integrals and you've learned about definite integrals. Note that the ball has traveled much farther. The fundamental theorem of calculus states that the integral of a function f over the interval [a, b] can be calculated by finding an antiderivative F of f: ∫ = − (). Example. It has gone up to its peak and is falling down, but the difference between its height at and is ft. Improper Integrals. The solution to the problem is, therefore, F′(x)=x2+2x−1F'(x)={ x }^{ 2 }+2x-1 F′(x)=x2+2x−1. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. (max 2 MiB). Here, the "x" appears on both limits. AP CALCULUS. AP CALCULUS. Second Fundamental Theorem of Calculus. Now, what I want to do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals. The total area under a curve can be found using this formula. In calculus, the chain rule is a formula to compute the derivative of a composite function.That is, if f and g are differentiable functions, then the chain rule expresses the derivative of their composite f ∘ g — the function which maps x to (()) — in terms of the derivatives of f and g and the product of functions as follows: (∘) ′ = (′ ∘) ⋅ ′. The area under the graph of the function $$f\left( x \right)$$ between the vertical lines $$x = a,$$ $$x = b$$ (Figure $$2$$) is given by the formula The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. The Second Fundamental Theorem of Calculus says that when we build a function this way, we get an antiderivative of f. Second Fundamental Theorem of Calculus: Assume f(x) is a continuous function on the interval I and a is a constant in I. Ultimately, all I did was I used the fundamental theorem of calculus and the chain rule. ... use the chain rule as follows. By the First Fundamental Theorem of Calculus, G is an antiderivative of f. Second Fundamental Theorem of Calculus (Chain Rule Version) dx f(t)dt = d 9(x) a los 5) Use second Fundamental Theorem to evaluate: a) 11+ t2 dt b) a tant dt 1 dt 1+t dxo d) in /1+t2dt . %�쏢 I know that you plug in x^4 and then multiply by chain rule factor 4x^3. The fundamental theorem of calculus explains how to find definite integrals of functions that have indefinite integrals. The Mean Value Theorem For Integrals. Ask Question Asked 1 year, 7 months ago. By the First Fundamental Theorem of Calculus, we have for some antiderivative of . By combining the chain rule with the (second) fundamental theorem of calculus, we can compute the derivative of some very complicated integrals. The average value of. Then F(x) is an antiderivative of f(x)—that is, F '(x) = f(x) for all x in I. Solution. However, any antiderivative could have be chosen, as antiderivatives of a given function differ only by a constant, and this constant always cancels out of the expression when evaluating . Let f be continuous on [a,b], then there is a c in [a,b] such that. Active 2 years, 6 months ago. Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem. Using the Second Fundamental Theorem of Calculus, we have . It looks complicated, but all it’s really telling you is how to find the area between two points on a graph. The FTC and the Chain Rule. I would define F of x to be this type of thing, the way we would define it for the fundamental theorem of calculus. Example problem: Evaluate the following integral using the fundamental theorem of calculus: Powered by Create your own unique website with customizable templates. The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. ��4D���JG�����j�U��]6%[�_cZ�Cw�R�\�K�)�U�Zǭ���{&��A@Z�,����������t :_$�3M�kr�J/�L{�~�ke�S5IV�~���oma ���o�1��*�v�h�=4-���Q��5����Imk�eU�3�n�@��Cku;�]����d�� ���\���6:By�U�b������@���խ�l>���|u�ύ\����s���u��W�o�6� {�Y=�C��UV�����_01i��9K*���h�*>W. Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. We can also use the chain rule with the Fundamental Theorem of Calculus: Example Find the derivative of the following function: G(x) = Z x2 1 1 3 + cost dt The Fundamental Theorem of Calculus, Part II If f is continuous on [a;b], then Z b a f(x)dx = F(b) F(a) ( notationF(b) F(a) = F(x) b a) That is, if f and g are differentiable functions, then the chain rule expresses the derivative of their composite f ∘ g — the function which maps x to $${\displaystyle f(g(x))}$$— in terms of the derivatives of f and g and the product of functions as follows: Fundamental Theorem of Calculus Example. The Fundamental Theorem of Calculus tells us that the derivative of the definite integral from to of ƒ() is ƒ(), provided that ƒ is continuous. So any function I put up here, I can do exactly the same process. This preview shows page 1 - 2 out of 2 pages.. In calculus, the chain rule is a formula to compute the derivative of a composite function. I would know what F prime of x was. See how this can be … The fundamental theorem of calculus (FTC) establishes the connection between derivatives and integrals, two of the main concepts in calculus. A conjecture state that if f(x), g(x) and h(x) are continuous functions on R, and k(x) = int(f(t)dt) from g(x) to h(x) then k(x) is differentiable and k'(x) = h'(x)*f(h(x)) - g'(x)*f(g(x)). 5 0 obj Example $$\PageIndex{2}$$: Using the Fundamental Theorem of Calculus, Part 2. I just want to make sure that I'm doing it right because I haven't seen any examples that apply the fundamental theorem of calculus to a function like this. ( s) d s. Solution: Let F ( x) be the anti-derivative of tan − 1. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. Solution. Example: Compute d d x ∫ 1 x 2 tan − 1. What's the intuition behind this chain rule usage in the fundamental theorem of calc? See Note. x��\I�I���K��%�������, ��IH�A��㍁�Y�U�UY����3£��s���-k�6����'=��\�]�V��{�����}�ᡑ�%its�b%�O�!#Z�Dsq����b���qΘ��7� Applying the chain rule with the fundamental theorem of calculus 1. }\) The second part of the theorem gives an indefinite integral of a function. Find the derivative of the function G(x) = Z √ x 0 sin t2 dt, x > 0. The Second Fundamental Theorem of Calculus. In spite of this, we can still use the 2nd FTC and the Chain Rule to find a (relatively) simple formula for !! The middle graph also includes a tangent line at xand displays the slope of this line. Get more help from Chegg. Find F′(x)F'(x)F′(x), given F(x)=∫−3xt2+2t−1dtF(x)=\int _{ -3 }^{ x }{ { t }^{ 2 }+2t-1dt }F(x)=∫−3xt2+2t−1dt. �h�|���Z���N����N+��?P�ή_wS���xl��x����G>�w�����+��͖d�A�3�3��:M}�?��4�#��l��P�d��n-hx���w^?����y�������[�q�ӟ���6R}�VK�nZ�S^�f� X�Ŕ���q���K^Z��8�Ŵ^�\���I(#Cj"�&���,K��) IC�bJ�VQc[�)Y��Nx���[�վ�Z�g��lu�X��Ź�:��V!�^?%�i@x�� You usually do F(a)-F(b), but the answer … Viewed 71 times 1$\begingroup$I came across a problem of fundamental theorem of calculus while studying Integral calculus. The total area under a curve can be found using this formula. For x > 0 we have F(√ x) = G(x). By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Create a real-world science problem that requires the use of both parts of the Fundamental Theorem of Calculus to solve by doing the following: (A physics class is throwing an egg off the top of their gym roof. ©u 12R0X193 9 HKsu vtoan 1S ho RfTt9w NaHr8em WLNLkCQ.J h NAtl Bl1 qr ximg Nh2tGsM Jr Ie osoeCr4v2e odN.L Z 9M apd neT hw ai Xtdhr zI vn Jfxiznfi qt VeX dCatl hc Su9l hu es7.I Worksheet by Kuta Software LLC ( x). Proof. We need an antiderivative of $$f(x)=4x-x^2$$. Use the chain rule and the fundamental theorem of calculus to find the derivative of definite integrals with lower or upper limits other than x. The right hand graph plots this slope versus x and hence is the derivative of the accumulation function. I would define F of x to be this type of thing, the way we would define it for the fundamental theorem of calculus. In Section 4.4, we learned the Fundamental Theorem of Calculus (FTC), which from here forward will be referred to as the First Fundamental Theorem of Calculus, as in this section we develop a corresponding result that follows it. By the First Fundamental Theorem of Calculus, G is an antiderivative of f. Then we need to also use the chain rule. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Problem. Let$f:[0,1] \to \mathbb{R}$be a differentiable function with$f(0) = 0$and$f'(x) \in (0,1)$for every$x \in (0,1)$. Since the lower limit of integration is a constant, -3, and the upper limit is x, we can simply take the expression t2+2t−1{ t }^{ 2 }+2t-1t2+2t−1given in the problem, and replace t with x in our solution. Viewed 1k times 1$\begingroup$I have the following problem in which I have to apply both the chain rule and the FTC 1. The result of Preview Activity 5.2 is not particular to the function $$f (t) = 4 − 2t$$, nor to the choice of “1” as the lower bound in … Find the derivative of the function G(x) = Z √ x 0 sin t2 dt, x > 0. I want to take the first and second derivative of$F(x) = \left(\int_0^xf(t)dt\right)^2 - \int_0^x(f(t))^3dt$and will use the fundamental theorem of calculus and the chain rule to do it. By the Chain Rule . The Second Fundamental Theorem of Calculus shows that integration can be reversed by differentiation. Solution. The Chain Rule and the Second Fundamental Theorem of Calculus1 Problem 1. You may assume the fundamental theorem of calculus. The Derivative of . Solution By using the fundamental theorem of calculus, the chain rule and the product rule we obtain f 0 (x) = Z 0 x 2-x cos (πs + sin(πs)) ds-x cos ( By using the fundamental theorem of calculus, the chain rule and the product rule we obtain f 0 (x) = Z 0 x 2-x cos (πs + sin(πs)) ds-x cos Fundamental theorem of calculus. Hw 3.3 Key. Click here to upload your image Find the derivative of . Get more help from Chegg. So any function I put up here, I can do exactly the same process. We define the average value of f (x) between a and b as. You can also provide a link from the web. About this unit. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. Definition of the Average Value. (We found that in Example 2, above.) . identify, and interpret, ∫10v(t)dt. By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy, 2020 Stack Exchange, Inc. user contributions under cc by-sa, fundamental theorem of calculus and chain rule. Example. - The integral has a variable as an upper limit rather than a constant. We use the chain rule so that we can apply the second fundamental theorem of calculus. Find the derivative of g(x) = integral(cos(t^2))DT from 0 to x^4. By combining the chain rule with the (second) Fundamental Theorem of Calculus, we can solve hard problems involving derivatives of integrals. Ask Question Asked 2 years, 6 months ago. There are several key things to notice in this integral. It has gone up to its peak and is falling down, but the difference between its height at and is ft. Stokes' theorem is a vast generalization of this theorem in the following sense. The function is really the composition of two functions. This preview shows page 1 - 2 out of 2 pages.. Finding derivative with fundamental theorem of calculus: chain rule. The Second Fundamental Theorem of Calculus. Stokes' theorem is a vast generalization of this theorem in the following sense. Solution. The first part of the theorem says that if we first integrate $$f$$ and then differentiate the result, we get back to the original function $$f.$$ Part $$2$$ (FTC2) The second part of the fundamental theorem tells us how we can calculate a definite integral. So for this antiderivative. It also gives us an efficient way to evaluate definite integrals. <> The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function. The Second Fundamental Theorem of Calculus is the formal, more general statement of the preceding fact: if $$f$$ is a continuous function and $$c$$ is any constant, then $$A(x) = \int^x_c f (t) dt$$ is the unique antiderivative of f that satisfies $$A(c) = 0$$. (We found that in Example 2, above.) It bridges the concept of an antiderivative with the area problem. The Second Fundamental Theorem of Calculus says that when we build a function this way, we get an antiderivative of f. Second Fundamental Theorem of Calculus: Assume f(x) is a continuous function on the interval I and a is a constant in I. But and, by the Second Fundamental Theorem of Calculus, . 4 questions. Have you wondered what's the connection between these two concepts? Set F(u) = Z u 0 sin t2 dt. Suppose that f(x) is continuous on an interval [a, b]. Using the Fundamental Theorem of Calculus, evaluate this definite integral. Get 1:1 help now from expert Calculus tutors Solve it with our calculus … Note that the ball has traveled much farther. Ask Question Asked 2 years, 6 months ago. Let (note the new upper limit of integration) and . Solution By using the fundamental theorem of calculus, the chain rule and the product rule we obtain f 0 (x) = Z 0 x 2-x cos (πs + sin(πs)) ds-x cos ( By using the fundamental theorem of calculus, the chain rule and the product rule we obtain f 0 (x) = Z 0 x 2-x cos (πs + sin(πs)) ds-x cos Fundamental theorem of calculus. Then F′(u) = sin(u2). As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Using the Second Fundamental Theorem of Calculus, we have . The middle graph, of the accumulation function, then just graphs x versus the area (i.e., y is the area colored in the left graph). How does fundamental theorem of calculus and chain rule work? ©u 12R0X193 9 HKsu vtoan 1S ho RfTt9w NaHr8em WLNLkCQ.J h NAtl Bl1 qr ximg Nh2tGsM Jr Ie osoeCr4v2e odN.L Z 9M apd neT hw ai Xtdhr zI vn Jfxiznfi qt VeX dCatl hc Su9l hu es7.I Worksheet by Kuta Software LLC The total area under a curve can be found using this formula. ���y�\�%ak��AkZ�q��F� �z���[>v��-��$��k��STH�|A Introduction. Of the two, it is the First Fundamental Theorem that is the familiar one used all the time. Let F be any antiderivative of f on an interval , that is, for all in .Then . Example If we use the second fundamental theorem of calculus on a function with an inner term that is not just a single variable by itself, for example v(2t), will the second fundamental theorem of Proof. https://www.khanacademy.org/.../ab-6-4/v/derivative-with-ftc-and- If $$f$$ is a continuous function and $$c$$ is any constant, then $$f$$ has a unique antiderivative $$A$$ that satisfies $$A(c) = 0\text{,}$$ and that antiderivative is given by the rule $$A(x) = \int_c^x f(t) \, dt\text{. F'(x) = 2\left(\int_0^xf(t)dt\right)f(x) - (f(x))^3 by the chain rule and fund thm of calc. Let be a number in the interval .Define the function G on to be. 2. Define a new function F(x) by. But what if instead of we have a function of , for example sin()? Active 2 years, 6 months ago. Second Fundamental Theorem of Calculus. But why don't you subtract cos(0) afterward like in most integration problems? See Note. The FTC and the Chain Rule Three Different Concepts As the name implies, the Fundamental Theorem of Calculus (FTC) is among the biggest ideas of Calculus, tying together derivatives and integrals. Set F(u) = Second Fundamental Theorem of Calculus. 2nd fundamental theorem of calculus ; Limits. Either prove this conjecture or find a counter example. The integral of interest is Z x2 0 e−t2 dt = E(x2) So by the chain rule d dx Z x2 0 e −t2 dt = d dx E(x2) = 2xE′(x2) = 2xe x4 Example 3 Example 4 (d dx R x2 x e−t2 dt) Find d … Practice. The integral of interest is Z x2 0 e−t2 dt = E(x2) So by the chain rule d dx Z x2 0 e −t2 dt = d dx E(x2) = 2xE′(x2) = 2xe x4 Example 3 Example 4 (d dx R x2 x e−t2 dt) Find d … y = sin x. between x = 0 and x = p is. stream Solving the integration problem by use of fundamental theorem of calculus and chain rule. Fundamental Theorem of Calculus Second Fundamental Theorem of Calculus Integration By Substitution Definite Integrals Using Substitution Integration By Parts Partial Fractions. The Fundamental Theorem of Calculus tells us how to find the derivative of the integral from to of a certain function. Active 1 year, 7 months ago. The Two Fundamental Theorems of Calculus The Fundamental Theorem of Calculus really consists of two closely related theorems, usually called nowadays (not very imaginatively) the First and Second Fundamental Theo-rems. Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. We use two properties of integrals to write this integral as a difference of two integrals. The Fundamental Theorem tells us that E′(x) = e−x2. Using First Fundamental Theorem of Calculus Part 1 Example. Then . The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus shows that di erentiation and Integration are inverse processes. It also gives us an efficient way to evaluate definite integrals. The applet shows the graph of 1. f (t) on the left 2. in the center 3. on the right. The Two Fundamental Theorems of Calculus The Fundamental Theorem of Calculus really consists of two closely related theorems, usually called nowadays (not very imaginatively) the First and Second Fundamental Theo-rems. Fair enough. Solution. Second Fundamental Theorem of Calculus (Chain Rule Version) dx f(t)dt = d 9(x) a los 5) Use second Fundamental Theorem to evaluate: a) 11+ t2 dt b) a tant dt 1 dt 1+t dxo d) in /1+t2dt . We spent a great deal of time in the previous section studying \(\int_0^4(4x-x^2)\,dx$$. Therefore, The Area under a Curve and between Two Curves. 1 Finding a formula for a function using the 2nd fundamental theorem of calculus The Second Fundamental Theorem of Calculus. . Then F(x) is an antiderivative of f(x)—that is, F '(x) = f(x) for all x in I. Applying the chain rule with the fundamental theorem of calculus 1. Introduction. I would know what F prime of x was. I saw the question in a book it is pretty weird. This is a very straightforward application of the Second Fundamental Theorem of Calculus. Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives. - The variable is an upper limit (not a lower limit) and the lower limit is still a constant. Is really the composition of two integrals and x = 0 and x = p is was! Several key things to notice in this integral as a difference of two functions this message, it we. Can also provide a link from the web u ) = sin ( u2 ) Solve it with our …... First Fundamental Theorem of Calculus 1 is perhaps the most important Theorem in the interval.Define the function G to... Includes a tangent line at xand displays the slope of this Theorem in Calculus, Part 1 shows the between! A formula for evaluating a definite integral in terms of an antiderivative with the concept integrating. Integral has a variable as an upper limit of integration ) and the following sense in and! Rule with the Fundamental Theorem of Calculus is a c in [ a, b ] rule with Fundamental! Graph plots this slope versus x and hence is the First Fundamental Theorem of Calculus, Part 2: Evaluation. Key things to notice in this integral curve can be found using this formula the derivative of the integral G! Part 2 is a formula for evaluating a definite integral notice in this integral as a difference of two.. Very straightforward application of the main concepts in Calculus, Part 2, is the... This preview shows page 1 - 2 out of 2 pages - 2 out of 2 pages help now expert. You 've learned about indefinite integrals and Antiderivatives was I used the Fundamental Theorem of Calculus ( ). But and, by the First Fundamental Theorem of Calculus tells us how find. Afterward like in most integration problems that F ( √ x 0 sin t2 dt, >... Viewed 71 times 1 $\begingroup$ I came across a problem of Fundamental that! Asked 1 year, 7 months ago ) is continuous on an interval, that 2nd fundamental theorem of calculus chain rule. I came across a problem of Fundamental Theorem of Calculus1 problem 1, above. to explain many phenomena by... Evaluate this definite integral in terms of an antiderivative of its integrand derivative the. Compute d d x ∫ 1 x 2 tan − 1 function is really the composition of two.! Connection between derivatives and integrals, two of the Second Fundamental Theorem of Calculus Part... Between the derivative and the integral of integrating a function of, for example sin )! For approximately 500 years, 6 months ago example sin ( ) that you plug in x^4 then. Composition of two functions to find the area between two points on a graph integrating a function integrals! Definite integrals you 're seeing this message, it is pretty weird a new function F ( )! Preview shows page 1 - 2 out of 2 pages composition of two integrals links the concept differentiating... And the chain rule factor 4x^3 a variable as an upper limit of integration ) and the Fundamental... Know that you plug in x^4 and then multiply by chain rule can be reversed by differentiation a from. ( u ) = Z u 0 sin t2 dt of 2 pages up,! Time in the 2nd fundamental theorem of calculus chain rule.Define the function G ( x ) by but if... And between two Curves the same process its peak and is falling down, but the difference between height. Limit ) and: integrals and you 've learned about indefinite integrals and Antiderivatives limit ( not lower! 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The familiar one used all the time by Create your own unique website with templates... But and, by the Second Fundamental Theorem of Calculus Second Fundamental Theorem of Calculus of... T2 dt, x > 0 we have for some antiderivative of = integral ( cos 0. = 0 and x = p is you subtract cos ( t^2 )! Sin x. between x = 0 and x = p is there is a formula Compute... Now from expert Calculus tutors Solve it with our Calculus … Introduction Asked 1 year, 7 ago... Its height at and is falling down, but all it ’ s really telling is. S. Solution: let F ( u ) = integral ( cos ( 0 ) afterward like in integration... New techniques emerged that provided scientists with the Fundamental Theorem of Calculus chain! In [ a, b ] such that in x^4 and then multiply by chain rule with the Theorem. To be at xand displays the slope of this Theorem in the Fundamental Theorem of Calculus and chain rule the! ( t ) dt ( we found that in example 2, above. } \ the! 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These two concepts let F be any antiderivative of its integrand how does Fundamental Theorem of Calculus Part! And interpret, ∫10v ( t ) on the left 2. in the following sense define the average of! In Calculus '' appears on both limits limit is still a constant Calculus Second Fundamental Theorem of and! Time in the following sense composition of two integrals 1 year, 7 months ago 've learned about indefinite and. Definite integrals using Substitution integration by Substitution definite integrals using Substitution integration by Parts Partial Fractions 3.... The average value of F on an interval [ a, b ] such that own unique website with templates... About definite integrals using Substitution integration 2nd fundamental theorem of calculus chain rule Substitution definite integrals using Substitution by. Of, for all in.Then such that ( 4x-x^2 ) \, dx\ ) definite integrals max 2 )... Slope of this Theorem in Calculus integral from to of a certain function with templates! And then multiply by chain rule and the Second Fundamental Theorem of Calculus tells us how to find area! Then there is a formula for evaluating a definite integral in terms an. With Fundamental Theorem of Calculus is a c in [ a, b ] integration. ( not a lower limit ) and the Second Fundamental Theorem of Calculus includes tangent! = Z √ x ) by a number in the interval.Define the function really. 'Re seeing this message, it is pretty weird problems involving derivatives of integrals Substitution definite integrals,... X '' appears on both limits is perhaps the most important Theorem Calculus! Telling you is how to find the derivative and the integral of line... U ) = integral ( cos ( t^2 ) ) dt from 0 to.... Main concepts in Calculus Partial Fractions your image ( max 2 MiB ) the right hand graph plots this versus. Use of Fundamental Theorem of Calculus is a c in [ a, b ] Question in a it..., above. the concept of an antiderivative of \ ( \int_0^4 ( 4x-x^2 ) \, dx\ ) x..
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# User talk:ElPoojmar
## Welcome!
Hello, ElPoojmar, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are a few links to pages you might find helpful:
Please remember to sign your messages on talk pages by typing four tildes (~~~~); this will automatically insert your username and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or ask your question on this page and then place {{help me}} before the question. Again, welcome!
## September 2013
Hello, I'm BracketBot. I have automatically detected that your edit to Solomonoff's theory of inductive inference may have broken the syntax by modifying 1 "<>"s. If you have, don't worry: just edit the page again to fix it. If I misunderstood what happened, or if you have any questions, you can leave a message on my operator's talk page.
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• S will be strictly greater than one. To be more precise, for every $\epsilon$ > 0, there is some length ''l'' such that the probability of all programs longer than ''l'' is at
• * [[Martin Davis]] (2006) The Church–Turing Thesis: Consensus and opposition]". Proceedings, Computability in Europe 2006. Lecture notes in computer science, 3988 pp. 125–132
Thanks, BracketBot (talk) 14:29, 18 September 2013 (UTC)
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# Error in PSTricks
Why does this code produce an error and how to fix it?
By the way, I use Texmaker in MikTeX.
\documentclass[10pt]{article}
\pagestyle{empty}
\usepackage{pst-func}
\begin{document}
\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=5pt 0,linewidth=0.8pt,arrowsize=3pt 2,arrowinset=0.25}
\begin{pspicture*}(-2.,-1.5)(2.,2.5)
\psaxes[labelFontSize=\scriptstyle,xAxis=true,yAxis=true,Dx=0.5,Dy=0.5,ticksize=-2pt 0,subticks=2]{<->}(0,0)(-2.,-1.5)(2.,2.5)
\psplotImp[linewidth=0.4pt](-3.0,-3.0)(3.0,3.0){0.01388742374854553+1.9858243449267068*y^2+3.0*y^3-2.0*y^4-0.0034236125287306907*x^1-0.336756945862064*x^1*y^2-2.0*x^2-2.0*x^2*y^2}
\end{pspicture*}
\end{document}
• I have no error message. Is this the exact code which produces your error message? – Bernard Dec 20 '16 at 12:10
• Which error do you get? – Torbjørn T. Dec 20 '16 at 12:16
• @Torbjørn This is the error: ! Undefined control sequence. \XC@usec@lor ...string \color@ #1#2\endcsname \@@ \fi \space l.9 \psplot [linewidth=0.4pt](-3.0,-3.0)(3.0,3.0){0.01388742374854553+1.9858... The control sequence at the end of the top line of your error message was never \def'ed. If you have misspelled it (e.g., '\hobx'), type 'I' and the correct spelling (e.g., 'I\hbox'). Otherwise just continue, and I'll forget about whatever was undefined. – teed Dec 20 '16 at 12:36
• @Bernard Any idea about the error (see above)? Teokan: Looks like that has something to do with xcolor (a color package, loaded by pstricks), but I don't know why you would get the error. Can you compile \listfiles\documentclass{article} \usepackage{pstricks-add} \pagestyle{empty} \usepackage{pst-func} \begin{document} abc \end{document}, and then add the .log file that is produced to your question? (Click the edit link right above these comments to edit your post.) – Torbjørn T. Dec 20 '16 at 12:41
• pstricks needs latex/dvips (usually) – David Carlisle Dec 20 '16 at 12:51
With TeXMaker, you can go to the Options -> Configure TeXmaker menu, then
• either add \usepackage{auto-pst-pdf} to your preamble (after pstricks related packages), and --enable-write18 to the parameters for pdflatex, as in this image:
• or change the default for Quick build to Latex+dvips+ps2pdf+View Pdf or XeLaTeX+ViewPDF:
Your code looks like an exported one from another GUI. Without such a GUI it is much more easier to create what you need:
\documentclass[10pt]{article}
\usepackage{pst-func}
\begin{document}
\psset{algebraic,unit=2}
\begin{pspicture*}(-2.,-1.5)(2.,2.5)
\psaxes[labelFontSize=\scriptstyle,Dx=0.5,
Dy=0.5,ticksize=-2pt 0,subticks=2]{<->}(0,0)(-2.,-1.5)(2.,2.5)
\psplotImp[linewidth=1pt,stepFactor=0.2,linecolor=red](-3.0,-3.0)(3.0,3.0)%
{0.0138+1.9858*y^2+3.0*y^3-2.0*y^4-0.0034236*x-0.3367*x*y^2-2.0*x^2-2.0*x^2*y^2}
\end{pspicture*}
\end{document}
Run it with xelatex or latex->dvips->ps2pdf or with loading package auto-pst-pdf and then also with pdflatex
The error message you show in comments is not the first one. the first error reported is
! Undefined control sequence.
• @TeokanDuranDemircan instead of typing pdflatex file.tex you need latex file.tex followed by dvpd file.dvi to get postscript then finally ps2pdf file.ps` to make PDF That is the sequence of commands, your editor probably has that in a menu somewhere but I do not know texmaker. – David Carlisle Dec 20 '16 at 13:00
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January 22, 2021, 11:16:18 PM
Forum Rules: Read This Before Posting
### Topic: Standard Deviation Equation (Read 82610 times)
0 Members and 1 Guest are viewing this topic.
#### tunelling
• Very New Member
• Posts: 1
• Mole Snacks: +1/-0
##### Standard Deviation Equation
« on: January 03, 2009, 07:37:13 PM »
Hi,
I need to calculate the standard deviation of some results from a chemistry experiment.
Could someone please explain how to use the standard deviation equation. I've had a look at some webpages but I think they assume that you have a good understanding of maths, which I don't.
#### Arkcon
• Retired Staff
• Sr. Member
• Posts: 7365
• Mole Snacks: +533/-147
##### Re: Standard Deviation Equation
« Reply #1 on: January 03, 2009, 07:48:16 PM »
Humpf. Funny. Most people these days just plug their numbers into the Excel formula, or their calculator and just roll with it. It is pretty impressive that you even care about this topic. OK. Let's try to work with what you have. List your data. If it's hundred's of numbers, then just lost some. Copy, as best you can, the formula you've searched for. And try to guess where to plug in the values.
Hey, I'm not judging. I just like to shoot straight. I'm a man of science.
#### Arkcon
• Retired Staff
• Sr. Member
• Posts: 7365
• Mole Snacks: +533/-147
##### Re: Standard Deviation Equation
« Reply #2 on: January 03, 2009, 08:12:00 PM »
Oh, and for an explanation of what a standard deviation is, did you see this wikipedia example. It is pretty clear:
http://en.wikipedia.org/wiki/Standard_deviation#Example
Hey, I'm not judging. I just like to shoot straight. I'm a man of science.
#### Mr Peanut
• Regular Member
• Posts: 96
• Mole Snacks: +5/-3
##### Re: Standard Deviation Equation
« Reply #3 on: January 03, 2009, 08:54:01 PM »
Could someone please explain how to use the standard deviation equation.
One thing that is sometimes not clear for some is the distinction between the standard deviation and its estimate from a sample of the population. The standard deviation is (as wiki points out) the root mean square deviation from the mean for the values of ALL of the members of the population.
When making measurements we rarely measure all of the values. Instead we measure a subset called a "sample". The actual standard deviation (the population standard deviation) is estimated from the sample using the "sample standard deviation".
The sample standard deviation is almost a root mean square but instead of dividing by the population count we divide by the sample count minus the degrees of freedom (typically N-1 where N is the sample count).
• Very New Member
• Posts: 1
• Mole Snacks: +0/-0
##### Re: Standard Deviation Equation
« Reply #4 on: January 18, 2009, 07:11:50 PM »
#### pjaj
• New Member
• Posts: 6
• Mole Snacks: +0/-0
##### Re: Standard Deviation Equation
« Reply #5 on: February 23, 2009, 01:04:54 PM »
One definition of the standard deviation is "The root of the mean of the squares of the differences from the average"
At first glance quite a mouthful, but read it out slowly and it is an explanation of how to calculate the SD.
1) calculate the average of the N data items ("the average")
2) tabulate the differences of each point from this average ("the differences")
3) square each difference, add them up and divide by N ("the mean of the squares of the differences")
4) take the square root of this average ("The root ... ") = The standard deviation.
This only works for the whole data set. If you only have a sample then the formula differs, as explained in the Wikipedia entry and a previous post.
The Wikipedia article also shows how the formula for the above calculation can be algebraically manipulated (under "Simplification of the formula") to give more useful versions that simplify the above calculation.
Note that the squaring step removes the distinction between positive and negative differences.
If you just added up the differences the answer should be zero in all cases!
Or, as others have said, you can blindly plug the numbers into Excel!
#### darrenhenchback
• Very New Member
• Posts: 1
• Mole Snacks: +0/-0
##### Re: Standard Deviation Equation
« Reply #6 on: October 16, 2010, 08:41:56 PM »
Well I think it's bloody awesome that you want to understand standard deviations rather than to just calculate them by shoving the numbers into a calculator (or worse, Microsoft Excel ). Standard deviation is a measure of the spread of data. You can have a bunch of numbers that are all really close to each other, for example:
1,2,3,4,5,6,6,6,7,7,8
There are 11 numbers here
Step #1: work out the Mean . This is done by adding the value of all the numbers and dividing that by the number of numbers (i.e. 11). So (1+2+3+4+5+6+6+6+7+7+8) divided by 11.
= 55/11
= 5
Step #2: from each of the above numbers, subtract the mean (so take away 5 in this case, and it's totally cool to get minus numbers here)... here's what you get :
-4, -3, -2, -1, 0, 1, 1, 1, 2, 2, 3
Step #3: square each number and you get...
16, 9, 4, 1, 0, 1, 1, 1, 4, 4, 9
Step #4: add all these numbers up and you get...
50
Step #5: divide this number by the number of numbers (which as we know is 11) MINUS 1. So divide by 10 in this case, and we get ...
5.
Step #6: get the square root of the resulting number. So the standard deviation here is 2.24
Aint that cool ? Try working out the standard deviation now for a whole load of other numbers, so you can understand it better. The bigger the spread in the data, the bigger the value for standard deviation. Why? Because the bigger the spread, the more different the numbers will be from the mean. So you'll get much bigger numbers in step 2. The numbers in step 2 then get squared and added together, so go figure ^^
#### ashishbandhu
• Very New Member
• Posts: 1
• Mole Snacks: +0/-0
##### Re: Standard Deviation Equation
« Reply #7 on: October 14, 2011, 02:37:30 AM »
I suggest you the best Standard Deviation Calculator .If you want to see Standard Deviation Calculator then go to the link.
calculator.tutorvista.com/math/351/standard-deviation-calculator.html
« Last Edit: April 17, 2012, 12:38:22 PM by Borek »
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# Difference between revisions of "Inference Rules"
CAUTION! Any modification to this page shall be announced on the User mailing list!
Conventions used in these tables are described in The_Proving_Perspective_(Rodin_User_Manual)#Inference_Rules.
Name Rule Side Condition A/M
*
HYP
$\frac{}{\textbf{H},\textbf{P} \;\;\vdash \;\; \textbf{P}}$ A
*
HYP_OR
$\frac{}{\textbf{H},\textbf{Q} \;\;\vdash \;\; \textbf{P} \lor \ldots \lor \textbf{Q} \lor \ldots \lor \textbf{R}}$ A
*
CNTR
$\frac{}{\textbf{H},\;\textbf{P},\;\neg\,\textbf{P} \;\;\vdash \;\; \textbf{Q}}$ A
*
FALSE_HYP
$\frac{}{\textbf{H},\bfalse \;\;\vdash \;\; \textbf{P}}$ A
*
TRUE_GOAL
$\frac{}{\textbf{H} \;\;\vdash \;\; \btrue}$ A
*
FUN_GOAL
$\frac{}{\textbf{H},\; f\in E\;\mathit{op}\;F \;\;\vdash\;\; f\in T_1\pfun T_2}$ where $T_1$ and $T_2$ denote types and $\mathit{op}$ is one of $\pfun$, $\tfun$, $\pinj$, $\tinj$, $\psur$, $\tsur$, $\tbij$. A
FUN_IMAGE_GOAL
$\frac{\textbf{H},\; f\in S_1\;\mathit{op}\;S_2,\; f(E)\in S_2\;\;\vdash\;\; \mathbf{P}(f(E))}{\textbf{H},\; f\in S_1\;\mathit{op}\;S_2\;\;\vdash\;\; \mathbf{P}(f(E))}$ where $\mathit{op}$ denotes a set of relations (any arrow) and $f(E)$ occurs at the top level A
FUN_GOAL_REC
$\frac{}{\textbf{H},\; f\in S_1\;\mathit{op_1}\;(S_2\;\mathit{op_2}\;(\ldots(S_n\;\mathit{op_n}(U\;\mathit{opf}\;V\;))\ldots)) \;\vdash\;\; f(E_1)(E_2)...(E_n)\in T_1\pfun T_2}$ where $T_1$ and $T_2$ denote types, $\mathit{op}$ denotes a set of relations (any arrow) and $\mathit{opf}$ is one of $\pfun$, $\tfun$, $\pinj$, $\tinj$, $\psur$, $\tsur$, $\tbij$. A
*
DBL_HYP
$\frac{\textbf{H},\;\textbf{P} \;\;\vdash \;\; \textbf{Q}}{\textbf{H},\;\textbf{P},\;\textbf{P} \;\;\vdash \;\; \textbf{Q}}$ A
*
AND_L
$\frac{\textbf{H},\textbf{P},\textbf{Q} \; \; \vdash \; \; \textbf{R}}{\textbf{H},\; \textbf{P} \land \textbf{Q} \; \; \vdash \; \; \textbf{R}}$ A
*
AND_R
$\frac{\textbf{H} \; \; \vdash \; \; \textbf{P} \qquad \textbf{H} \; \; \vdash \; \; \textbf{Q}}{\textbf{H} \; \; \vdash \; \; \textbf{P} \; \land \; \textbf{Q}}$ A
IMP_L1
$\frac{\textbf{H},\; \textbf{Q},\; \textbf{P} \land \ldots \land \textbf{R} \limp \textbf{S} \;\;\vdash \;\; \textbf{T}}{\textbf{H},\; \textbf{Q},\; \textbf{P} \land \ldots \land \textbf{Q} \land \ldots \land \textbf{R} \limp \textbf{S} \;\;\vdash \;\; \textbf{T} }$ A
*
IMP_R
$\frac{\textbf{H}, \textbf{P} \;\;\vdash \;\; \textbf{Q}}{\textbf{H} \;\;\vdash \;\; \textbf{P} \limp \textbf{Q}}$ A
*
IMP_AND_L
$\frac{\textbf{H},\textbf{P} \limp \textbf{Q}, \textbf{P} \limp \textbf{R}\;\;\vdash \;\; \textbf{S}}{\textbf{H},\;\textbf{P} \limp \textbf{Q} \land \textbf{R} \;\;\vdash \;\; \textbf{S}}$ A
*
IMP_OR_L
$\frac{ \textbf{H},\textbf{P} \limp \textbf{R}, \textbf{Q} \limp \textbf{R}\;\;\vdash \;\; \textbf{S} }{\textbf{H},\;\textbf{P} \lor \textbf{Q} \limp \textbf{R} \;\;\vdash \;\; \textbf{S}}$ A
*
AUTO_MH
$\frac{ \textbf{H},\textbf{P},\;\textbf{Q}\limp \textbf{R}\;\;\vdash \;\; \textbf{S} }{\textbf{H},\;\textbf{P},\; \textbf{P} \land \textbf{Q} \limp \textbf{R} \;\;\vdash \;\; \textbf{S}}$ A
*
NEG_IN_L
$\frac{\textbf{H},\; E \in \{ a,\ldots , c\} \; \; \vdash \; \; \textbf{P} }{\textbf{H},\; E \in \{ a,\ldots , b, \ldots , c\} , \neg \, (E=b) \; \; \vdash \; \; \textbf{P} }$ A
*
NEG_IN_R
$\frac{\textbf{H},\; E \in \{ a,\ldots , c\} \; \; \vdash \; \; \textbf{P} }{\textbf{H},\; E \in \{ a,\ldots , b, \ldots , c\} , \neg \, (b=E) \; \; \vdash \; \; \textbf{P} }$ A
*
XST_L
$\frac{\textbf{H},\; \textbf{P(x)} \; \; \vdash \; \; \textbf{Q} }{ \textbf{H},\; \exists \, \textbf{x}\, \qdot\, \textbf{P(x)} \; \; \vdash \; \; \textbf{Q} }$ A
*
ALL_R
$\frac{\textbf{H}\; \; \vdash \; \; \textbf{P(x)} }{ \textbf{H} \; \; \vdash \; \; \forall \textbf{x}\, \qdot\, \textbf{P(x)} }$ A
*
EQL_LR
$\frac{\textbf{H(E)} \; \; \vdash \; \; \textbf{P(E)} }{\textbf{H(x)},\; x=E \; \; \vdash \; \; \textbf{P(x)} }$ $x$ is a variable which is not free in $E$ A
*
EQL_RL
$\frac{\textbf{H(E)} \; \; \vdash \; \; \textbf{P(E)} }{\textbf{H(x)},\; E=x \; \; \vdash \; \; \textbf{P(x)} }$ $x$ is a variable which is not free in $E$ A
SUBSET_INTER
$\frac{\textbf{H},\;\textbf{T} \subseteq \textbf{U} \;\;\vdash \;\; \textbf{G}(\textbf{S} \binter \dots \binter \textbf{T} \binter \dots \binter \textbf{V})} {\textbf{H},\;\textbf{T} \subseteq \textbf{U} \;\;\vdash \;\; \textbf{G}(\textbf{S} \binter \dots \binter \textbf{T} \binter \dots \binter \textbf{U} \binter \dots \binter \textbf{V})}$ the $\binter$ operator must appear at the "top level" A
IN_INTER
$\frac{\textbf{H},\;\textbf{E} \in \textbf{T} \;\;\vdash \;\; \textbf{G}(\textbf{S} \binter \dots \binter \{\textbf{E}\} \binter \dots \binter \textbf{U})} {\textbf{H},\;\textbf{E} \in \textbf{T} \;\;\vdash \;\; \textbf{G}(\textbf{S} \binter \dots \binter \{\textbf{E}\} \binter \dots \binter \textbf{T} \binter \dots \binter \textbf{U})}$ the $\binter$ operator must appear at the "top level" A
NOTIN_INTER
$\frac{\textbf{H},\;\lnot\;\textbf{E} \in \textbf{T} \;\;\vdash \;\; \textbf{G}(\emptyset)} {\textbf{H},\;\lnot\;\textbf{E} \in \textbf{T} \;\;\vdash \;\; \textbf{G}(\textbf{S} \binter \dots \binter \{\textbf{E}\} \binter \dots \binter \textbf{T} \binter \dots \binter \textbf{U})}$ the $\binter$ operator must appear at the "top level" A
*
FIN_L_LOWER_BOUND_L
$\frac{}{\textbf{H},\;\finite(S) \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; n \leq x)}$ The goal is discharged A
*
FIN_L_LOWER_BOUND_R
$\frac{}{\textbf{H},\;\finite(S) \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; x \geq n)}$ The goal is discharged A
*
FIN_L_UPPER_BOUND_L
$\frac{}{\textbf{H},\;\finite(S) \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; n \geq x)}$ The goal is discharged A
*
FIN_L_UPPER_BOUND_R
$\frac{}{\textbf{H},\;\finite(S) \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; x \leq n)}$ The goal is discharged A
*
$\frac{\textbf{H},\;\neg\,\textbf{Q} \;\;\vdash \;\; \neg\,\textbf{P}}{\textbf{H},\;\textbf{P} \;\;\vdash \;\; \textbf{Q}}$ M
*
$\frac{\textbf{H},\;\neg\,\textbf{Q} \;\;\vdash \;\; \bfalse}{\textbf{H} \;\;\vdash \;\; \textbf{Q}}$ M
*
CASE
$\frac{\textbf{H}, \; \textbf{P} \; \; \vdash \; \; \textbf{R} \qquad\ldots\qquad \textbf{H}, \; \textbf{Q} \; \; \vdash \; \; \textbf{R} }{\textbf{H},\; \textbf{P} \lor \ldots \lor \textbf{Q} \; \; \vdash \; \; \textbf{R} }$ M
*
MH
$\frac{\textbf{H} \;\;\vdash\;\;\textbf{P} \qquad \textbf{H},\; \textbf{Q} \;\;\vdash \;\; \textbf{R} }{\textbf{H},\;\textbf{P} \limp \textbf{Q} \;\;\vdash \;\; \textbf{R}}$ M
*
HM
$\frac{\textbf{H} \;\;\vdash\;\;\neg\,\textbf{Q} \qquad \textbf{H},\; \neg\,\textbf{P} \;\;\vdash \;\; \textbf{R} }{\textbf{H},\;\textbf{P} \limp \textbf{Q} \;\;\vdash \;\; \textbf{R}}$ M
EQV
$\frac{\textbf{H(Q)},\; \textbf{P} \leqv \textbf{Q} \;\;\vdash\;\; \textbf{G(Q)}}{\textbf{H(P)},\;\textbf{P} \leqv \textbf{Q} \;\;\vdash \;\; \textbf{G(P)}}$ M
*
OV_SETENUM_L
$\frac{\textbf{H},\; G=E ,\;\textbf{P}(F)\;\;\vdash\;\;\textbf{Q} \qquad \textbf{H},\; \neg\,(G=E) ,\;\textbf{P}((\{E\}) \domsub f)(G))\;\;\vdash\;\;\textbf{Q}}{\textbf{H},\;\textbf{P}((f\ovl\{E \mapsto F\})(G)) \;\;\vdash \;\; \textbf{Q}}$ the $\ovl$ operator must appear at the "top level" A
*
OV_SETENUM_R
$\frac{\textbf{H},\; G=E \;\;\vdash\;\;\textbf{Q}(F) \qquad \textbf{H},\; \neg\,(G=E) \;\;\vdash\;\;\textbf{Q}((\{E\}) \domsub f)(G))}{\textbf{H} \;\;\vdash \;\; \textbf{Q}((f\ovl\{E \mapsto F\})(G))}$ the $\ovl$ operator must appear at the "top level" A
*
OV_L
$\frac{\textbf{H},\; G \in \dom(g) ,\;\textbf{P}(g(G))\;\;\vdash\;\;\textbf{Q} \qquad \textbf{H},\; \neg\,G \in \dom(g) ,\;\textbf{P}((\dom(g) \domsub f)(G))\;\;\vdash\;\;\textbf{Q}}{\textbf{H},\;\textbf{P}((f\ovl g)(G)) \;\;\vdash \;\; \textbf{Q}}$ the $\ovl$ operator must appear at the "top level" A
*
OV_R
$\frac{\textbf{H},\; G \in \dom(g) \;\;\vdash\;\;\textbf{Q}(g(G)) \qquad \textbf{H},\; \neg\, G \in \dom(g) \;\;\vdash\;\;\textbf{Q}((\dom(g) \domsub f)(G))}{\textbf{H} \;\;\vdash \;\; \textbf{Q}((f\ovl g)(G))}$ the $\ovl$ operator must appear at the "top level" A
*
DIS_BINTER_R
$\frac{\textbf{H} \;\;\vdash\;\; f^{-1} \in A \pfun B \qquad\textbf{H} \;\;\vdash\;\;\textbf{Q}(f[S] \binter f[T]) }{\textbf{H} \;\;\vdash \;\; \textbf{Q}(f[S \binter T])}$ the occurrence of $f$ must appear at the "top level". Moreover $A$ and $B$ denote some type. M
*
DIS_BINTER_L
$\frac{\textbf{H} \;\;\vdash\;\; f^{-1} \in A \pfun B \qquad\textbf{H},\;\textbf{Q}(f[S] \binter f[T]) \;\;\vdash\;\;\textbf{G}}{\textbf{H},\; \textbf{Q}(f[S \binter T]) \;\;\vdash \;\; \textbf{G}}$ the occurrence of $f$ must appear at the "top level". Moreover $A$ and $B$ denote some type. M
*
DIS_SETMINUS_R
$\frac{\textbf{H} \;\;\vdash\;\; f^{-1} \in A \pfun B \qquad\textbf{H} \;\;\vdash\;\;\textbf{Q}(f[S] \setminus f[T]) }{\textbf{H} \;\;\vdash \;\; \textbf{Q}(f[S \setminus T])}$ the occurrence of $f$ must appear at the "top level". Moreover $A$ and $B$ denote some type. M
*
DIS_SETMINUS_L
$\frac{\textbf{H} \;\;\vdash\;\; f^{-1} \in A \pfun B \qquad\textbf{H},\;\textbf{Q}(f[S] \setminus f[T]) \;\;\vdash\;\; \textbf{G}}{\textbf{H},\; \textbf{Q}(f[S \setminus T]) \;\;\vdash \;\; \textbf{G}}$ the occurrence of $f$ must appear at the "top level". Moreover $A$ and $B$ denote some type. M
*
SIM_REL_IMAGE_R
$\frac{\textbf{H} \; \; \vdash \; \; {WD}(\textbf{Q}(\{ f(E)\} )) \qquad\textbf{H} \; \; \vdash \; \; \textbf{Q}(\{ f(E)\} ) }{\textbf{H} \; \; \vdash \; \; \textbf{Q}(f[\{ E\} ])}$ the occurrence of $f$ must appear at the "top level". M
*
SIM_REL_IMAGE_L
$\frac{\textbf{H} \; \; \vdash \; \; {WD}(\textbf{Q}(\{ f(E)\} )) \qquad\textbf{H},\; \textbf{Q}(\{ f(E)\}) \;\;\vdash\;\; \textbf{G}}{\textbf{H},\; \textbf{Q}(f[\{ E\} ]) \;\;\vdash\;\; \textbf{G} }$ the occurrence of $f$ must appear at the "top level". M
*
SIM_FCOMP_R
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(\textbf{Q}(g(f(x)))) \qquad\textbf{H} \;\;\vdash\;\;\textbf{Q}(g(f(x))) }{\textbf{H} \;\;\vdash \;\; \textbf{Q}((f \fcomp g)(x))}$ the occurrence of $f \fcomp g$ must appear at the "top level". M
*
SIM_FCOMP_L
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(\textbf{Q}(g(f(x)))) \qquad\textbf{H},\; \textbf{Q}(g(f(x))) \;\;\vdash\;\; \textbf{G}}{\textbf{H},\; \textbf{Q}((f \fcomp g)(x)) \;\;\vdash \;\; \textbf{G}}$ the occurrence of $f \fcomp g$ must appear at the "top level". M
*
FIN_SUBSETEQ_R
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(T) \qquad\textbf{H} \;\;\vdash \;\; S \subseteq T \qquad \textbf{H} \;\;\vdash \;\; \finite\,(T)}{\textbf{H} \;\;\vdash \;\; \finite\,(S)}$ the user has to write the set corresponding to $T$ in the editing area of the Proof Control Window M
*
FIN_BINTER_R
$\frac{\textbf{H} \;\;\vdash \;\;\finite\,(S) \;\lor\;\ldots \;\lor\; \finite\,(T)}{\textbf{H} \;\;\vdash \;\; \finite\,(S \;\binter\;\ldots \;\binter\; T)}$ M
*
FIN_SETMINUS_R
$\frac{\textbf{H} \;\;\vdash \;\;\finite\,(S)}{\textbf{H} \;\;\vdash \;\; \finite\,(S \;\setminus\; T)}$ M
*
FIN_REL_R
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(S\rel T) \qquad\textbf{H} \;\;\vdash \;\; r \;\in\; S \rel T \qquad \textbf{H} \;\;\vdash \;\; \finite\,(S) \qquad \textbf{H} \;\;\vdash \;\; \finite\,(T)}{\textbf{H} \;\;\vdash \;\; \finite\,(r)}$ the user has to write the set corresponding to $S \rel T$ in the editing area of the Proof Control Window M
*
FIN_REL_IMG_R
$\frac{\textbf{H} \;\;\vdash \;\; \finite\,(r) }{\textbf{H} \;\;\vdash \;\; \finite\,(r[s])}$ M
*
FIN_REL_RAN_R
$\frac{\textbf{H} \;\;\vdash \;\; \finite\,(r) }{\textbf{H} \;\;\vdash \;\; \finite\,(\ran(r))}$ M
*
FIN_REL_DOM_R
$\frac{\textbf{H} \;\;\vdash \;\; \finite\,(r) }{\textbf{H} \;\;\vdash \;\; \finite\,(\dom(r))}$ M
*
FIN_FUN1_R
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(S\pfun T) \qquad\textbf{H} \;\;\vdash \;\; f \;\in\; S \pfun T \qquad \textbf{H} \;\;\vdash \;\; \finite\,(S) }{\textbf{H} \;\;\vdash \;\; \finite\,(f)}$ the user has to write the set corresponding to $S \pfun T$ in the editing area of the Proof Control Window M
*
FIN_FUN2_R
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(S\pfun T) \qquad\textbf{H} \;\;\vdash \;\; f^{-1} \;\in\; S \pfun T \qquad \textbf{H} \;\;\vdash \;\; \finite\,(S) }{\textbf{H} \;\;\vdash \;\; \finite\,(f)}$ the user has to write the set corresponding to $S \pfun T$ in the editing area of the Proof Control Window M
*
FIN_FUN_IMG_R
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(S\pfun T) \qquad\textbf{H} \;\;\vdash \;\; f \;\in\; S \pfun T \qquad \textbf{H} \;\;\vdash \;\; \finite\,(s) }{\textbf{H} \;\;\vdash \;\; \finite\,(f[s])}$ the user has to write the set corresponding to $S \pfun T$ in the editing area of the Proof Control Window M
*
FIN_FUN_RAN_R
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(S\pfun T) \qquad\textbf{H} \;\;\vdash \;\; f \;\in\; S \pfun T \qquad \textbf{H} \;\;\vdash \;\; \finite\,(S) }{\textbf{H} \;\;\vdash \;\; \finite\,(\ran(f))}$ the user has to write the set corresponding to $S \pfun T$ in the editing area of the Proof Control Window M
*
FIN_FUN_DOM_R
$\frac{\textbf{H} \;\;\vdash\;\;{WD}(S\pfun T) \qquad\textbf{H} \;\;\vdash \;\; f^{-1} \;\in\; S \pfun T \qquad \textbf{H} \;\;\vdash \;\; \finite\,(S) }{\textbf{H} \;\;\vdash \;\; \finite\,(\dom(f))}$ the user has to write the set corresponding to $S \pfun T$ in the editing area of the Proof Control Window M
*
LOWER_BOUND_L
$\frac{\textbf{H} \;\;\vdash \;\; \finite(S) }{\textbf{H} \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; n \leq x)}$ $S$ must not contain any bound variable M
*
LOWER_BOUND_R
$\frac{\textbf{H} \;\;\vdash \;\; \finite(S) }{\textbf{H} \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; x \geq n)}$ $S$ must not contain any bound variable M
*
UPPER_BOUND_L
$\frac{\textbf{H} \;\;\vdash \;\; \finite(S) }{\textbf{H} \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; n \geq x)}$ $S$ must not contain any bound variable M
*
UPPER_BOUND_R
$\frac{\textbf{H} \;\;\vdash \;\; \finite(S) }{\textbf{H} \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; x \leq n)}$ $S$ must not contain any bound variable M
*
FIN_LT_0
$\frac{\textbf{H} \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; n \leq x) \qquad \textbf{H} \;\;\vdash \;\; S \subseteq \intg \setminus \natn }{\textbf{H} \;\;\vdash \;\; \finite(S)}$ M
*
FIN_GE_0
$\frac{\textbf{H} \;\;\vdash \;\; \exists n\,\qdot\, (\forall x \,\qdot\, x \in S \;\limp\; x \leq n) \qquad \textbf{H} \;\;\vdash \;\; S \subseteq \nat }{\textbf{H} \;\;\vdash \;\; \finite(S)}$ M
CARD_INTERV
$\frac{\textbf{H},\, a \leq b \;\;\vdash \;\; \textbf{Q}(b-a+1) \qquad \textbf{H},\, b < a \;\;\vdash \;\; \textbf{Q}(0) }{\textbf{H} \;\;\vdash\;\; \textbf{Q}(\card\,(a\upto b))}$ $\card (a \upto b)$ must appear at "top-level" M
CARD_EMPTY_INTERV
$\frac{\textbf{H},\, a \leq b,\,\textbf{P}(b-a+1) \;\;\vdash \;\; \textbf{Q} \qquad \textbf{H},\, b < a ,\, \textbf{P}(0)\;\;\vdash \;\; \textbf{Q} }{\textbf{H},\,\textbf{P}(\card\,(a\upto b)) \;\;\vdash\;\; \textbf{Q}}$ $\card (a \upto b)$ must appear at "top-level" M
*
DERIV_LE_CARD
$\frac{\textbf{H} \;\;\vdash\;\; S \subseteq T}{\textbf{H} \;\;\vdash\;\; \card(S) \leq \card(T)}$ $S$ and $T$ bear the same type M
*
DERIV_GE_CARD
$\frac{\textbf{H} \;\;\vdash\;\; T \subseteq S}{\textbf{H} \;\;\vdash\;\; \card(S) \geq \card(T)}$ $S$ and $T$ bear the same type M
*
DERIV_LT_CARD
$\frac{\textbf{H} \;\;\vdash\;\; S \subset T}{\textbf{H} \;\;\vdash\;\; \card(S) < \card(T)}$ $S$ and $T$ bear the same type M
*
DERIV_GT_CARD
$\frac{\textbf{H} \;\;\vdash\;\; T \subset S}{\textbf{H} \;\;\vdash\;\; \card(S) > \card(T)}$ $S$ and $T$ bear the same type M
*
DERIV_EQUAL_CARD
$\frac{\textbf{H} \;\;\vdash\;\; S = T}{\textbf{H} \;\;\vdash\;\; \card(S) = \card(T)}$ $S$ and $T$ bear the same type M
SIMP_CARD_SETMINUS_L
$\frac{\textbf{H},\, \textbf{P}(\card (S \setminus T)) \;\;\vdash\;\; \finite(S) \qquad \textbf{H},\, \textbf{P}(\card(S) - \card(S\binter T)) \;\;\vdash\;\; \textbf{G}}{\textbf{H},\, \textbf{P}(\card (S \setminus T)) \;\;\vdash\;\; \textbf{G}}$ $\card (S \setminus T)$ must appear at "top-level" M
SIMP_CARD_SETMINUS_R
$\frac{\textbf{H} \;\;\vdash\;\; \finite(S) \qquad \textbf{H} \;\;\vdash\;\; \textbf{P}(\card(S) - \card(S\binter T))}{\textbf{H} \;\;\vdash\;\; \textbf{P}(\card (S \setminus T))}$ $\card (S \setminus T)$ must appear at "top-level" M
SIMP_CARD_CPROD_L
$\frac{\textbf{H},\, \textbf{P}(\card (S \cprod T)) \;\;\vdash\;\; \finite(S) \qquad \textbf{H},\, \textbf{P}(\card (S \cprod T)) \;\;\vdash\;\; \finite(T) \qquad \textbf{H},\, \textbf{P}(\card(S) * \card(T)) \;\;\vdash\;\; \textbf{G}}{\textbf{H},\, \textbf{P}(\card (S \cprod T)) \;\;\vdash\;\; \textbf{G}}$ $\card (S \cprod T)$ must appear at "top-level" M
SIMP_CARD_CPROD_R
$\frac{\textbf{H} \;\;\vdash\;\; \finite(S) \qquad \textbf{H} \;\;\vdash\;\; \finite(T) \qquad \textbf{H} \;\;\vdash\;\; \textbf{P}(\card(S) * \card(T))}{\textbf{H} \;\;\vdash\;\; \textbf{P}(\card (S \cprod T))}$ $\card (S \cprod T)$ must appear at "top-level" M
*
FORALL_INST
$\frac{\textbf{H} \;\;\vdash \;\; {WD}(E) \qquad \textbf{H} , [x \bcmeq E]\textbf{P} \;\;\vdash \;\; \textbf{G}}{\textbf{H}, \forall x \qdot \textbf{P} \;\;\vdash\;\; \textbf{G}}$ $x$ is instantiated with $E$ M
*
FORALL_INST_MP
$\frac{\textbf{H} \;\;\vdash \;\; {WD}(E) \qquad \textbf{H}, {WD}(E) \;\;\vdash \;\; [x \bcmeq E]\textbf{P} \qquad \textbf{H}, {WD}(E), [x \bcmeq E]\textbf{Q} \;\;\vdash \;\; \textbf{G}}{\textbf{H}, \forall x \qdot \textbf{P} \limp \textbf{Q} \;\;\vdash\;\; \textbf{G}}$ $x$ is instantiated with $E$ and a Modus Ponens is applied M
*
CUT
$\frac{\textbf{H} \;\;\vdash \;\; {WD}(\textbf{P}) \qquad \textbf{H}, {WD}(\textbf{P}) \;\;\vdash \;\; \textbf{\textbf{P}} \qquad \textbf{H}, {WD}(\textbf{P}), \textbf{P} \;\;\vdash \;\; \textbf{G}}{\textbf{H} \;\;\vdash\;\; \textbf{G}}$ hypothesis $\textbf{P}$ is added M
*
EXISTS_INST
$\frac{\textbf{H} \;\;\vdash \;\; {WD}(E) \qquad \textbf{H} \;\;\vdash \;\; \textbf{P}(E)}{\textbf{H} \;\;\vdash\;\; \exists x \qdot \textbf{P}(x)}$ $x$ is instantiated with $E$ M
*
DISTINCT_CASE
$\frac{\textbf{H} \;\;\vdash \;\; {WD}(\textbf{P}) \qquad \textbf{H}, {WD}(\textbf{P}), \textbf{P} \;\;\vdash \;\; \textbf{\textbf{G}} \qquad \textbf{H}, {WD}(\textbf{P}), \lnot \textbf{P} \;\;\vdash \;\; \textbf{G}}{\textbf{H} \;\;\vdash\;\; \textbf{G}}$ case distinction on predicate $\textbf{P}$ M
ONE_POINT_L
$\frac{\textbf{H} \;\;\vdash \;\; {WD}(E) \qquad \textbf{H}, \forall x, \ldots, \ldots,z \qdot [y \bcmeq E]\textbf{P} \land \ldots \land \ldots \land [y \bcmeq E]\textbf{Q} \limp [y \bcmeq E]\textbf{R} \;\;\vdash \;\; \textbf{G}}{ \textbf{H}, \forall x, \ldots, y, \ldots, z \qdot \textbf{P} \land \ldots \land y = E \land \ldots \land \textbf{Q} \limp \textbf{R} \;\;\vdash\;\; \textbf{G}}$ The rule can be applied with $\forall$ as well as with $\exists$ A
ONE_POINT_R
$\frac{\textbf{H} \;\;\vdash \;\; {WD}(E) \qquad \textbf{H} \;\;\vdash \;\; \forall x, \ldots, \ldots,z \qdot [y \bcmeq E]\textbf{P} \land \ldots \land \ldots \land [y \bcmeq E]\textbf{Q} \limp [y \bcmeq E]\textbf{R} }{ \textbf{H} \;\;\vdash\;\; \forall x, \ldots, y, \ldots, z \qdot \textbf{P} \land \ldots \land y = E \land \ldots \land \textbf{Q} \limp \textbf{R} }$ The rule can be applied with $\forall$ as well as with $\exists$ A
DATATYPE_DISTINCT_CASE
$\frac{\textbf{H}, x=c_1(p_{11}, \ldots, p_{1k}) \;\;\vdash \;\; \textbf{G} \qquad \ldots \qquad \textbf{H}, x=c_n(p_{n1}, \ldots, p_{nl}) \;\;\vdash \;\; \textbf{G} }{ \textbf{H} \;\;\vdash\;\; \textbf{G} }$ where $x$ has a datatype $DT$ as type and appears free in $\textbf{G}$, $DT$ has constructors $c_1, \ldots, c_n$, parameters $p_{ij}$ are introduced as fresh identifiers M
DATATYPE_INDUCTION
$\frac{\textbf{H}, x=c_N(\ldots) \;\;\vdash \;\; \textbf{P}(c_N(\ldots)) \qquad \textbf{H}, x=c_I(p_N, p_I),\textbf{P}(p_I) \;\;\vdash \;\; \textbf{P}(c_I(p_N, p_I)) }{ \textbf{H} \;\;\vdash\;\; \textbf{P}(x) }$ (N = Non inductive; I = Inductive) where $x$ has inductive datatype $DT$ as type and appears free in $\textbf{P}$; $c$ are constructors of $DT$; an antecedent is created for each $c_N$ and each $c_I$, an hypothesis $\textbf{P}(p_I)$ is added for each $p_I$; all parameters are introduced as fresh identifiers M
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## Calculus: Early Transcendentals 8th Edition
$22 J$
The work done in moving the object can be calculated as: $W=\int_c F \cdot T ds \approx \Sigma_{i=1}^{7} [F(x_i,y_i) \cdot (x_i,y_i)]ds]$ or, $W=(2+2+2+2+1+1+1) \cdot 2 \approx 22 J$ Therefore, we estimate the work done is $22 J$.
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## Small Worlds
Some social networks have become very accessible for systematic investigations, as they manifest themselves in a digital form on the internet. One such example of a social network that is accessible online is maintained by the American Mathematical Society (AMS) as part of MathSciNet, a database of mathematical research publications.
This network is formed by a collaboration graph: its nodes are the mathematical researchers, and two researchers are connected by and edge whenever they have jointly authored a publication in the database. In this network, the Erdős number of a mathematician is their distance from the Erdős node. This number can be computed by Breadth First Search. The AMS offers this as a service on a dedicated collaboration distance page.
The Hungarian mathematician Paul Erdős (1913-1996) is very well-known for his publication records, with more than 1500 publications, and more than 500 collaborators. It turns out, that most mathematicians have an Erdős number of 4 or 5. In fact, random spot checks seem to indicate that virtually all mathematicians are connected with each other in this network (in a big giant component), and that any two of them are rarely further than distance 6 apart.
This is an instance of a phenomenon that has often been observed: social networks tend to have very short paths between arbitrary pairs of people. Such a network is called a small world network. The name refers to the property that the diameter $\mathrm{diam}(G)$, or the average distance $\frac1{n^2}\sum_{x, y} d(x, y)$ between nodes in a network $G = (X,E)$ with $n$ nodes is small, or smaller than one would naively expect. In popular culture, the phenomenon is known as “six degrees of separation”. It goes at least back to experiments conducted in the 1960s by the Yale psychologist Stanley Milgram (who is even better known for another experiment designed by him).
In a similar way, the co-star graph (with actors as nodes, and edges between two actors whenever they starred in a common film) as maintained by the IMDb, has been used to define the Bacon number of an actor as their distance to the American actor Kevin Bacon.
The presence of short paths certainly has some relevance in relation to the potential speed of information (or diseases) travelling through the network.
Networks evolve over time, edges come and go. The concept of triadic closure is concerned with the presence of triangles in a network: if node $A$ is connected to nodes $B$ and $C$, is $B$ connected to $C$?
Clustering Coefficient. A set of nodes in a graph with the property that any two distinct nodes are joined by an edge is called a clique (in other words, a clique is a subset of the nodes such that the induced subgraph is complete).
The graph below contains two cliques of four nodes each.
A clique of $m$ nodes has $\binom{m}2 = \frac12 m(m-1)$ edges, e.g., a clique of $4$ nodes has $\frac{4 \cdot 3}2 = 6$ edges.
The clustering coefficient of a node $A$ measures, how close the set $N(A)$ of friends of $A$ comes to being a clique: if the induced subgraph on $N(A)$ has $f$ edges, then the clustering coefficient of the node $A$ is defined as the fraction $f/\binom{m}2$. This is a number between $0$ and $1$.
Triadic closure increases the clustering coefficient. In a social network, there are incentives for triadic closure to happen. If person $A$ is friends with $B$ and $C$, this provides opportunities for $B$ and $C$ to meet and become friends. On the other hand, $B$ not being friends with $C$ can be a source of stress for $A$, who has to attend to their friends separately. More generally, a lower clustering coefficient can mean that more work is required to maintain a circle of friends.
## Bridges; local Bridges
An edge $AB$ is called a bridge if deleting the edge would cause $A$ and $B$ to lie in different connected components. (In other words, if the edge $AB$ is a bridge then it is the only route from $A$ to $B$).
Example. The edge $AB$ in the graph below is clearly a bridge.
This notion might not be very useful for the study of social networks, as cycles and short paths make bridges very rare, or computationally expensive to detect.
More interesting than the global notion of a bridge, is the local notion of a local bridge.
An edge $AB$ is a local bridge if $N(A) \cap N(B) = \emptyset$, i.e., if $A$ and $B$ have no friends in common.
Example. The edge $AB$ in the graph below is a local bridge.
Deleting a local bridge $AB$ increases the distance between $A$ and $B$ to at least 3. This new distance is called the span of the local bridge.
A local bridge is the conceptual opposite of triadic closure: an edge $AB$ is a local bridge if and only it is not involved in any triangle. (Proof?)
## Strong and Weak Ties
In many examples it is possible to distinguish between different levels of strength of the links of a network. Here, we are going to study networks with (only) two types of edges: strong ties (corresponding to close friends, say), and weak ties (corresponding to acquaintances).
The assumption that triadic closure is more likely to happen in the presence of strong links can be formalized as the following property.
Strong Triadic Closure Property. A node $A$ violates the Strong Triadic Closure Property if it has strong ties with two other nodes $B$ and $C$, and there is no edge at all between $B$ and $C$. A node $A$ satisfies the Strong Triadic Closure Property if it does not violate it.
Example. In the graph below, the strong edges have a label ($s$) and the weak edges don’t. Each node of this graph satisfies the Strong Triadic Closure Property: whenever a node has two strongly tied neighbours, there is a tie (weak or strong) between those neighbours.
Triadic closure establishes a connection between the local notion of link strength and the structural notion of local bridges, as follows.
Proposition. If a node $A$ in a network satisfies the Strong Triadic Closure Property and is involved in at least two strong ties, then any local bridge it is involved in must be a weak tie.
In other words, assuming that all nodes in a network satisfy the Strong Triadic Closure Property and have sufficiently many strong ties, local bridges are necessarily weak ties.
Proof. Suppose that node $A$ does satisfy the Strong Triadic Closure Property and is involved in at least two strong ties. For a contradiction, suppose the edge $AB$ is a local bridge and a strong tie. As $AB$ is a local bridge, nodes $A$ and $B$ have no friends in common. Let $AC$ be another strong tie involving $A$. Then the Strong Triadic Closure Property requires the existence of an edge $BC$, making $C$ a common friend of $A$ and $B$, contradiction.
Simplifying assumptions (like the Strong Triadic Closure Property) are useful when they lead to statements that are robust in practice, in the sense that qualitative conclusions still hold in approximate forms, even when the assumptions are slightly relaxed.
The surprising strength of weak ties (as experienced in the case of information leading to a new job or other new opportunities typically coming from distant acquaintances) can be partially explained in this framework. Links connecting people to new sources of information tend to be local bridges of a certain “span”, which by the above are necessarily weak ties. The Stanford sociologist Mark Granovetter has pioneered the theoretical study of social networks along such lines in the 1970s.
Today, digital communication networks allow for empirical verifications of theoretical predictions.
The Twitter network, for example, has its users as nodes, and one can use its social network features to distinguish strong and weak ties. For example, a weak tie can correspond to one user following another, and a strong tie can correspond to at least two messages directly addressed at another user. An empirical study of the social network with its strong and weak ties defined in this way, shows that people have vastly different numbers of followees (between none and more than 1000), and the number of strong ties tends to increase with the number of followees. However, as soon as the number of followees reaches 400, the number of strong ties seems to stabilize at around 50. An explanation of this phenomenon can be that it requires little effort to “follow” a large number of Twitter users. But it requires time and energy to maintain a strong tie relationship, and there are only so many hours in a day …
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# [Tugindia] Problem with Omega Malayalam
Alex A.J. indictex at gmail.com
Fri Aug 28 12:39:28 CEST 2009
On Sat, Aug 22, 2009 at 12:55 AM, P. P. Narayanaswami<swami at mun.ca> wrote:
> I had another problem with Malayalam Omega as well as Tamil Omega,
> where I had to use Landscape Mode. This I posted sometime back to
> the newsgroup but got no response. If I use the "lscape" package
> and use \begin{landscape} ... Malayalam text ... \end{landscape},
> only the page header and page numbers get in landscape mode and not the
> Malayalam content.
Working fine for me. I used the following code, and Lambda -> odvips -> ps2pdf.
\documentclass{article}
\usepackage{omal}
\usepackage{lscape}
\begin{document}
\begin{landscape}
{\mal
<Malayalam text here>
}
\end{landscape}
\end{document}
Regards
Alex.
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# Setting value of controller controls for domain objects with Reflection
Is my use of Java Reflection an OK design?
Each domain object contains an annotation to declare what controller would be initialized to edit/create the object:
Each field of the domain object has an annotation to declare what control in the specified controller that the user would use to edit its value:
@Edit(controller = EmployeeController.class)
public class Employee extends Item {
@Node(name = "lastNameField", type = TextField.class)
private StringProperty lastName = new SimpleStringProperty();
//Control similar to a TableView
@Node(name = "contacts", type = ListBox.class)
private List <Contact> contacts;
}
public @interface Edit {
Class<? extends ModifyEntityController> controller();
}
public @interface Node {
String name();
Class<?> type();
}
public Class EmployeeController {
@FXML
private TextField lastNameField;
}
Double-clicking on a ListCell would call .edit() on the domain object.
This method is in the parent class Item.
.edit() does: example = employee.edit();
1. create an instance of the @Edit controller
2. use @Node to create a binding between the specified control
3. set the root node of controller to the main window of the application.
//create an instance of the controller
Edit edit = getClass().getAnnotation(Edit.class);
Class<? extends Controller> controller = edit.controller();
Object controllerInstance = null;
try {
controllerInstance = edit.controller().newInstance();
} catch (Exception e) {
e.printStackTrace();
}
//create bindings between the specified controls
List <Field> itemFields = FieldUtils.getFieldsListWithAnnotation(getClass(), Node.class);
for (Field field : itemFields ) {
Node node = field.getAnnotation(Node.class);
Object itemObject = FieldUtils.readField(field, this, true);
Object controllerControl = FieldUtils.readDeclaredField(controllerInstance, node.name(), true);
if (node.type() == TextField.class) {
TextInputControl tc = (TextInputControl) controllerControl;
StringProperty sp = (StringProperty) itemObject;
tc.textProperty().bindBidirectional(sp);
}
else if (node.type() == ListBox.class) {
ListBox lb = (ListBox) controllerControl;
List <CellObject> list = (List<CellObject>) itemObject;
lb.setList(FXCollections.observableArrayList(list));
}
}
*The implementation of a ListBox is actually not a binding of properties but setting the value of its ListView to the List Field of the domain object.
All objects that would be the value of a ListCell would implement CellObject.
//set the root node of controller to the main window of the application
AgendaApp.setContent(Controller) controllerInstance);
I went with reflection as the controllers to update/create domain objects consist of mainly TextFields and ListBox's. This would save a lot of time writing the application.
Is my use of Java Reflection an OK design?
Well, no, the usage of reflection is usually an indication of bad design. Honestly, it is almost always. Actually, you have to change the question and ask, when is it okay, to use reflection. The answer is something similar like this: Use reflection, when you do not know how a class looks like during runtime. But this is your code, so, why would you increase complexity and make your code more error prone?
In general, I have some worries, when I read the code:
@Node(name = "lastNameField", type = TextField.class)
private StringProperty lastName = new SimpleStringProperty();
Now, it might be a language or a definition subject, but if Employee is a domain model, it must not have dependencies to the view layer, at least no direct dependencies.
@Edit(controller = EmployeeController.class)
public class Employee extends Item {
The Employee itself is in control, of who controls itself. Shouldn't you tell the EmployeController, which model it shall edit?
Double-clicking on a ListCell would call .edit() on the domain object.
Same thing. The presentation code calls the domain object, then, the domain objects calls someone and says "someone wants to edit me". If I understood it correctly.
There's some other fishy stuff, but I'll conclude now, because in my opinion there are some problems even without the reflection stuff: It seems like an over engineered approach for an easy problem. Or: It looks like the solution of a problem, which probably isn't even a problem. It certainly is, from my point of view, the wrong tool for "the problem". The usage of reflection makes maintaining the code harder and more error prone, at least in my experience. The usage of annotations, too, by the way.
I'd suggest to take a step back to the solution before you implemented reflection and analyze what object oriented principles are violated (e.g. SOLID, coupling, cohesion,...) and what patterns can help you get rid of those problems. Now, that's the major point: Ask what design patterns help you get rid of your design problems, do not ask would reflection get rid of the design problems?.
I recommend fowler's web page about gui architectures: https://martinfowler.com/eaaDev/ (presentation patterns). It's not exactly a "good read", but it certainly is educational. It helps you to understand, which presentation patterns suits your application best and gives you some guidelines about how to implement you presentation layer.
• "The presentation code calls the domain object". I think I misled you. Edit() is inside the item class so this is possible: employee.edit(); Thanks for that link. – jpell Aug 29 '17 at 22:20
• "not have dependencies to the view layer" The TextFields are initialized in the controller, I am using FXML so they are injected. – jpell Aug 29 '17 at 22:22
• I am using Domain-Driven Design (DDD). – jpell Aug 29 '17 at 22:44
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# Solve the product power $\left(7\sqrt[3]{x^{2}}\sqrt{xy^{121\cdot -1}}\right)^4$
## Step-by-step Solution
Go!
Go!
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asinh
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### Videos
$\frac{2401\sqrt[3]{x^{14}}}{y^{242}}$
Got another answer? Verify it here
## Step-by-step Solution
Problem to solve:
$\left(7x^{\frac{2}{3}}\left(x y^{121\left(-1\right)}\right)^{\frac{1}{2}}\right)^4$
Choose the solving method
1
Multiply $121$ times $-1$
$\left(7\sqrt[3]{x^{2}}\sqrt{xy^{-121}}\right)^4$
Learn how to solve power of a product problems step by step online.
$\left(7\sqrt[3]{x^{2}}\sqrt{xy^{-121}}\right)^4$
Learn how to solve power of a product problems step by step online. Solve the product power (7x^0.6666666666666666(xy^(121*-))^0.5)^4. Multiply 121 times -1. The power of a product is equal to the product of it's factors raised to the same power. The power of a product is equal to the product of it's factors raised to the same power. The power of a product is equal to the product of it's factors raised to the same power.
$\frac{2401\sqrt[3]{x^{14}}}{y^{242}}$
SnapXam A2
### beta Got another answer? Verify it!
Go!
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tanh
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asinh
acosh
atanh
acoth
asech
acsch
$\left(7x^{\frac{2}{3}}\left(x y^{121\left(-1\right)}\right)^{\frac{1}{2}}\right)^4$
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# Gauge transformation which counteract wave function
1. May 5, 2015
### exponent137
Gauge transformation can be written as:
$\psi(\vec{r},t)\rightarrow e^{-i \frac{e}{\hbar c}f(\vec{r},t)}\psi(\vec{r},t)$
http://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html
Does it have any sense that we choose such function $f$, that all right side is constant in time. Is this possible at least approximately?
2. May 5, 2015
### Brage
As far as i understand yes, but only if your wave function that is the inverse of the time variable of your gauge transformation. That is we must have f(r,t)=R(r) + T(t). The reason for this is that the probability amplitude of the original and the transformed functions are the same, and therefore no any observation will remain unchanged through the transformation. In other words, a gauge transformation does not affect any physical outcome but changes the structure of the equation.
3. May 6, 2015
### exponent137
I ask, because it is strange to me, that wave functions can exist, where sinusoidal form $e^{ikx+\omega t}$, is not necessary.Do you know any other examples, where wave functions can be described without $e^{ikx+\omega t}$? Even, if it is described in momentum space.
As one condition for this I see that wave function contains only one frequency $\omega$.
4. May 6, 2015
### Brage
First off I assume you mean [itex] e^{i(kx+wt)} [\itex] as otherwise your wave function will not be normalizable. And I have the perfect example, the ground state of a quantum harmonic oscillator is a Gaussian distribution [itex] e^{-\lambda x^2} [\itex], where $\lambda$ is a constant. The wave function you have described is that of a travelling wave, but one that is not normalizable, as its probability density is constant over space.
5. May 6, 2015
### exponent137
1. Yes, I mean $e^{i(kx+wt)}$.
2. Yes, harmonic oscilator is one such example. I forget. But, the factor $e^{i\omega t}$ remains. It is similarly for a Infinite Potential Well in the lowest level: http://www.physics.ox.ac.uk/Users/cowley/QuantumL12.pdf? [Broken]
3. Yes, a traveling wave is not renormalizable. But, I think, that according to my question, this is not a problem. I ask only, if it is possible to choose such gauge function that it neutralizes its space and time waving. Maybe the harmonic oscilator is not a good example, because I suppose that it is not possible to neutralize its oscillation? But, the above function, $e^{i(kx+wt)}$, is also the simplest possible.
7. May 7, 2015
### exponent137
I am aware that gauge transformation does not contain any measurable effect. (except some philosphical aspect, such as Aharonov Bohm http://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html ...)
But, it is strange to me, that such example with stopping of oscillations was not mentioned anywhere. Because, it is interesting from mathematical view, (similarly as proper vectors etc. )
OK, it does not disturb uncertainty principle, so it is allowed. And it seems that you are sure that there is not any other possible problem?
Is this mentioned in book: Gauge Theories of the Strong, Weak, and Electromagnetic Interactions? I do not know that I will get this book, can you, please, find some other links which will tell mi more about gauge theory of EM, maybe even about such effect?
Last edited: May 7, 2015
8. May 7, 2015
### DrDu
Hm, this is nothing else than the change between the Schroedinger and Heisenberg picture. Should be familiar to you before delving into gauge theory.
9. May 7, 2015
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# Proceedings
Latest additions:
2018-07-3110:12 [GSI-2018-00838] Proceedings et al FAIRNESS 2017 : 28 May-3 June 2017, Sitges, Spain FAIRNESS 2017, SitgesSitges, Spain, 28 May 2017 - 3 Jun 2017 Bristol [u.a.] : IOP Publishing getrennte Zählung (2018) zugl.: Journal of Physics: Conference Series, Vol. 1024, 2018 GSI/FAIR DA Monography - K SITGES 2017 1 available 2018-03-0115:43 [GSI-2018-00451] Proceedings et al FAIRNESS 2014: FAIR Next Generation ScientistS 2014 3rd FAIR NExt generation ScientistS (FAIRNESS 2014), Vietri Sul MareVietri Sul Mare, Italy, 22 Sep 2014 - 27 Sep 2014 Bristol : IOP Publ. 599, (2015) External link: Fulltext 2016-02-1814:05 [GSI-2016-00477] Proceedings et al Excitation of Nucleon Resonances in Heavy-Ion Charge-Exchange Reactions Proceedings of the Conference on Advances in Radioactive Isotope Science, (ARIS2014), TokyoTokyo, Japan, 1 Jun 2014 - 6 Jun 2014 Journal of the Physical Society of Japan (2015) [10.7566/JPSCP.6.020039] 2016-02-1716:46 [GSI-2016-00476] Proceedings et al Schottky Mass Spectrometry on $^{152}$Sm Projectile Fragments Advances in Radioactive Isotope Science, ARIS2014, TokyoTokyo, Japan, 1 Jun 2014 - 6 Jun 2014 Journal of the Physical Society of Japan (2015) [10.7566/JPSCP.6.030099] 2016-02-1716:19 [GSI-2016-00474] Proceedings et al Experiments with Stored Highly Charged Ions at the Border between Atomic and Nuclear Physics 24th Conference on Application of Accelerators in Research and Industry, CAARI 2014, San Antonio, TexasSan Antonio, Texas, USA, 26 May 2014 - 26 May 2014 2016-01-2117:05 [GSI-2016-00165] Proceedings et al A Penning trap for g-factor measurements in highly charged ions by laser-microwave double-resonance spectroscopy Trapped Charged Particles and Fundamental Physics 2010, TCP2010, JyväskyläJyväskylä, Finland, 12 Apr 2010 - 16 Apr 2010 2016-01-1914:06 [GSI-2016-00130] Proceedings/Book et al Proceedings of the International Nuclear Physics Conference : Wiesbaden, Germany, July 26 - August 1, 1992 International Nuclear Physics Conference, INPC`92, WiesbadenWiesbaden, Germany, 26 Jul 1992 - 1 Aug 1992 Amsterdam [u.a.] : North Holland, Nuclear physics 553, (1993) External link: Fulltext 2015-10-1514:21 [GSI-2015-02604] Proceedings Alikhanan, A. I. Proceedings of the VII International Conference on High Energy Accelerators: vol. 2 7th International Conference on High Energy Accelerators, YerevanYerevan, Armenia, 27 Aug 1969 - 2 Sep 1969 Yerevan : Academy of Science of armenian (1970) GSI/FAIR DA Monography - OS K YEREVAN 1969 1 available 2015-10-1514:12 [GSI-2015-02603] Proceedings Alikhanan, A. I. Proceedings of the VII International Conference on High Energy Accelerators : Vol. 1 7th International Conference on High Energy Accelerators, YerevanYerevan, Armenia, 27 Aug 1969 - 2 Sep 1969 Yerevan : Academy of Science of armenian (1970) GSI/FAIR DA Monography - OS K YEREVAN 1969 1 available 2015-08-2714:00 [GSI-2015-02420] Proceedings/Book Gallmann, A. Hot and dense nuclear matter : [proceedings of a NATO Advanced Study Institute on Hot and Dense Nuclear Matter, held September 26 - October 9, 1993, in Bodrum, Turkey] NATO Advanced Study Institute on Hot and Dense Nuclear Matter, BordrumBordrum, Turkey, 26 Sep 1993 - 9 Oct 1993 New York [u.a.] : Plenum Press, NATO ASI series / B 335, (1994) External link: Fulltext
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Aftermath of the Nigerian civil war (1967-1970): The struggle for peaceful coexistence between parties in post-war Nigeria
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Aftermath of the Nigerian civil war (1967-1970): The struggle for peaceful coexistence between parties in post-war Nigeria
Annotation
PII
S032150750015266-4-1
Publication type
Article
Status
Published
Authors
A. Preye Posibi
Affiliation: People’s Friendship University of Russia
Address: Russian Federation,
Edition
Pages
71-77
Abstract
Fifty years ago, the Nigerian civil war, one of the bloodiest conflicts occurred in Africa, ended but its echoes are still an abiding presence today. It was a fratricidal war ensued between the Federal Military Government of Nigeria headed by Lt. Col. Yakubu Gowon and the secessionist Eastern Region of Biafra headed by Lt. Col. Emeka Ojukwu, between July 6, 1967 and January 15, 1970.
The main focus of this paper is to analyze the aftermath of the civil war on the post-war Nigeria, focusing in particular on the consequences that the conflict had as a threat to national security, unity and peace, in present days Nigeria. This thirty-month war had devastating consequences for the country, including death, displacement of people, and destruction of public infrastructure as well as physical and social capital. After the secessionist forces surrendered, Biafra was reincorporated into Nigeria as the East Central State. The Civil War left a legacy of death and destruction, particularly in the war-torn eastern region.
Many of Nigeria’s post-war problems still plague the nation today. In fact, ethnic tensions and military dictatorships continue to pose a threat to Nigerian unity. Additionally, this study will put its attention on the reasons why, half a century later, the war’s legacy continues to hold Nigeria captive. It simultaneously brings the country together and pushes it apart. In this scenario, it’s fundamentally important to find peacebuilding solutions and politic actions to maintain peaceful coexistence between parties, in order to avoid the incurrence of new conflicts.
Keywords
Nigerian Civil War, Aftermath, Peaceful Coexistence, Ethnic Conflict, Biafra, Nigeria
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20.02.2021
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11.06.2021
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1
### INTRODUCTION
2 The African continent has witnessed a number of bloody conflicts leaving in their tracks serious consequences including political, social, economic and humanitarian problems. The Nigerian civil war, popularly known as the Biafran War, was fought from 6 July, 1967 to 15 January, 1970 between the then Eastern Region of Nigeria and the rest of the country. The Eastern Region declared itself an independent state which was considered as a secessionist act by the Federal Military Government of Nigeria.
3 Since independence in 1960 the country was characterized by a fragile peace and stability condition, which culminated with the outbreak of the war.
4 In fact, many of the conflicts which rage today have their roots in the events which took place whilst Nigeria was under colonial rule. The eruption of the Nigerian civil war was as a result of political, economic, ethnic, cultural and religious tensions which preceded Britain's formal decolonization of the country. According to the Federal Military Government of Nigeria the war was fought to reunify the country, for the Biafrans instead it was a war for independence, that marked the climax of a series of unfolding turbulent events that began in January of 1966. The civil war posed the greatest challenge to the continuing existence, unity and territorial integrity of Nigeria.
5 The immediate cause of the civil war itself may be identified as the coup and the counter coup of 19661, which altered the political system and destroyed the trust existing among the major ethnic groups. In order to disallow the country from disintegration it was divided into twelve states from the original four regions in May 1967. The former Eastern Region under Lt. Col. Emeka Ojukwu (1933-2011) saw the act of the creation of states by decree "without consultation" as the last straw and declared the Region an independent state of "Biafra". The Federal Government in Lagos saw this as an act of secession and illegal. Several meetings were held to resolve the issue peacefully without success. To avoid disintegration of the country, the central government decided to bring back the secessionist region to the main fold by force. However, while it is true that the Federal Government of Nigeria (FGN) succeeded in taming the secession attempt, the war seems to have failed to resolve the remarkable issues that brought it about.
1. On January 15, 1966, the first military coup d’état led by Lt. Col. Ch.K.Nzeogwu and E.Ifeajuna, which overthrew the first Nigerian Republic, killed 22 people (including the Prime Minister of Nigeria at the time A.T.Balewa, many senior politicians, many senior Army officers (including their wives), and sentinels on protective duty). As a result, in the same year, some soldiers majorly from the northern part of Nigeria, reacted with a counter coup.
6 The Nigerian civil war appears like a paradox. On the one hand, the war restored the political map of Nigeria that had been redrawn by the seceding Eastern Region. At the same time, death, destruction of property and estranged relations among Nigerian nationalities, among other results or war, were very common [1].
7
### THE AFTERMATH OF СОNFLICTS
8 As stated by Siyan Chen (American University and a management analysist) [2] war has devastating consequences for a country, including death, displacement of people, and destruction of public infrastructure as well as physical and social capital.
9 One of the most recent and comprehensive reports, World Bank’s World Development Report [3], shows that the economic and social costs of civil wars are not only deep but also persistent, even for years after the end of the conflict. Among scholars there are not enough evidence on the costs of civil war after peace agreements are signed. In fact, they kill people, destroy infrastructure, weaken institutions, and erode social trust. Moreover, the aftermath of any conflict if not well managed could leave the population under conditions that increase the risk of disease, crime, political instability, and could also encourage further conflict to rise in the future.
10 According Nwanne W.Okafor (from Nigeria who wrote in his master thesis at the Tilburg University in the Netherlands) wars have profound effects on the economy as they drain wealth, disrupt markets and depress economic growth of participating parties [4]. As a consequence, they fuel inflation as prices are pushed up, which invariably leads to a reduction of living standards. Wars also affect the inflow of foreign investments as the instability and the risks of investing during conflict may discourage business relationships. In addition, resources that could have been used by the government for the development of infrastructures in the country are diverted to contribute towards the growing costs of the conflict. Parties involved suffer extreme destruction of capital such as factories, cities, farms, hospitals and livestock, which further reduces the level of economic growth. Wars may lead to the internal displacement of people, due to insecurity or loss of their homes during the conflict, the breakdown of health and the spread of diseases owing to lack of medical care and medical facilities; all of these can also have a negative impact on the peoples’ wellbeing and on the economy.
11 After 30 months of fighting, the Nigerian civil war ended in January 1970 after the Biafran army surrendered to the Nigerian army. In accepting the suspension of hostility, the Head of state Yakubu Gowon declared that there would be “no victor no vanquished” and granted a general amnesty for people who had fought on the Biafran side [5]. His “no victor no vanquished” policy was designed to complete the integration of the Igbos back into Nigerian society following their defeat in the war. By virtue of the policy, the Biafran soldiers were neither tried nor executed for fighting against the federal army while some of the Igbo officers who served in the Biafran army were reabsorbed with loss of seniority [6].
12
### REUNIFICATION OF THE COUNTRY
13 According to Johnson Olaosebikan Aremu and Lateef Oluwafemi Buhari (Ekiti State University, Ado-Ekiti, Nigeria)the Nigerian civil war had also a significant benefit: the unity of Nigeria was restored, and its territorial integrity was sustained. During his official surrender speech on 12 January, 1970, Biafra’s Chief of Army Staff, Major General Phillip Effiong declared openly that the “people of Biafra” consent to the “authority of the Federal Military Government,” and accept the “existing administrative and political structure of the Federation of Nigeria” [1].
14 To Gen. Yakubu Gowon, the end of hostilities marked the end of the “futile attempt to disintegrate the country” and was no more than a “great moment of victory for national unity” (New Nigerian Newspaper, 13 January,1970). Additionally, it should be noted that the war saw to the reunification and liberation of some Nigerian people who may not have subscribed to the Biafran dreams of Ojukwu [7]. In fact, aside from the Igbo, the Eastern Region is home to the Efik, Anioma, Ibibio and other ethnic minority groups who vehemently resented the idea of living under an Igbo dominion2.
2. Igbo is the tribe of the leader of the defunct Biafra republic Col. Emeka Ojukwu. For the minority ethnic groups living in the Eastern part of the country, rather than liberate them from the Northern oligarchy, a victory for the Biafra state would have turned the country into Lt. Col. Emeka Ojukwu’s personal empire in which he would easily partitioned amidst his loyalists.
15 Ojukwu’s defeat marked an end to this imperial ambition thereby providing an opportunity to re-fashion the country into a more stable polity. Ultimately, the war was brought to an end by both parties for the sake of preserving the unity and territorial integrity of the country.
16
### RECONCILIATION, REHABILITATION AND RECONSTRUCTION
17 With the end of the civil war, one of the most immediate demands on Gowon’s government was one of providing relief for the suffering masses of the newly affected areas. The need for shelter, food and medicines for the war affected population became more glaring than ever. To further complicate issues was the simultaneous necessity for rehabilitation and reconstruction; to restore electricity, water, transport and communications. There was also the urgent need to resettle farms, reopen factories, and facilitate the resumption of normal economic life.
18 It was against this backdrop that the Federal Military Government immediately adopted the policy of 3Rs: Reconciliation, Rehabilitation and Reconstruction. The main aim of the 3Rs was to create an atmosphere conducive for resettling the displaced and those who had fled their homes; to reunite families and friends; to rebuild damaged physical facilities and demobilize armed forces personnel in gainful employment in civilian life [8].
19 The program was initiated with the intention to appease the hostilities between Nigeria’s ethnic groups, restore infrastructure and homes, relocate internally displaced people and tackle the socio-economic challenges of poverty, disease and malnutrition among the victims. In other words, the plan aimed to reconstruct the infrastructure damaged by war and promote economic and social development throughout the nation in the post-war period [8, p. 76]. N120 Million ($300 000 in today exchange rate), was allocated for this task in the first fiscal year after the war (1970-1971), and Gowon ensured that there were no summary trials or executions of Biafran war veterans. 20 In principle, the program was a laudable scheme but in practice it left much to be desired in its implementation, as he was unable to successfully implement his program [9], and its impact was hardly felt within the Igbo community. There was little or no true reconciliation or rehabilitation that took place, and the Igbos were perpetually reminded that they were a defeated people [10]. 21 Financially, the civil war victory led to the reinstatement of the country’s ownership and control of rich oil fields in the Niger Delta. The 3R policy came at a time when the country was enjoying an “oil boom” as this was the period when oil prices spiraled and reached unprecedented heights, which brought and influx of revenue to the federal government that was bigger than expected [8, p. 7]. Revenue from oil would go on to define Nigeria’s post-war policies. Domestically, oil rents were used to embark on ambitious infrastructural projects and to reconstruct some of the state infrastructures destroyed during the fratricidal war [6]. 22 This significant income encouraged the government to spend money ambitiously on schools, clinics and hospitals across Nigeria. But, although the country became wealthy as a result of the oil boom, revenues were not distributed to all parts of the country as the people of Biafra were still suffering in poverty [7, p. 145]. 23 In fact, while the country was going through a rapid development and undergoing various reconstruction projects, the reconstruction in the Biafran region was far from becoming a reality. During the war, many private and community buildings had been destroyed and the government had advised the people to assess and submit their losses to the relevant official quarters. Nothing meaningful was done by the government therefore, and the people had to embark on reconstructing their property by themselves [11]. 24 Subsequently, the implementation of the 3Rs policy was funded with money set aside for the Second National Development Plan (1970-1974). This suggested that funds meant for further development of infrastructures were additionally committed to rebuilding structures and facilities destroyed during the civil war. 25 It is distressing to many observers that the N3.192 billion ($8 million) earmarked for the Second National Development Plan were used for reconstruction. In fact, as stated by Ojeleye ( University of Calgary ), civil wars destroy the structures that are needed for the development of the society.., such wars divert much needed ‘scarce’ resources away from development projects” [8].
26
### CREATION OF STATES AND THE UNRESOLVED NATIONAL QUESTION
27 After the civil war, the country was re-engineered to prevent another secession and halt any other attempt of insurrection by Nigeria’s ethnic groups. In fact, between 27 May 1967 and 1 October 1996, to find a way for Nigerian 250 ethnic groups with almost 500 different spoken dialects to live together peacefully, the country was split into 36 states, most of which coincided with the location of a major ethnic groups.
28 In addition to that, the postwar desire to prevent another secession generated a near obsessive ethnic micromanaging of national life and created a nation that exists almost simply to share money and jobs. As a matter of fact, each new state originally assembled to satisfy the desires of a nationality, started to create new “neglected” minorities, new tensions, and fresh activism. As reported by Max Siollun, a Nigerian historian [12], rather than working as a glue for unity, the fixation on ethnic sharing of national opportunities and resources made Nigerians more aware of their ethnic differences.
29 In addition to that it should be noted that, instead of being an enabling factor for the extension of political and economic self-governance to distinguish tribal and ethnic communities, state creation became a tool for administrative strategy for the devaluation of federal munificence to an array of territorial communities and coalition with no formal structure. This further explains why the politics and idea of states creation in the nation has not considered the ability of these different states to manage or otherwise sustain their existence. Also, the politics of state creation has been an exercise mainly employed as a legitimate force or means for military regimes in the nation to encourage or galvanize support for regime elongation as well as to compensate close allies. Nevertheless, the creation of states in Nigeria has not meant the satisfaction of all interest groups in the country [1, p. 72].
30 Important to note is also the fact that the civil war left the country with a set of federal war veterans who continued to impose their will on the country in form of indestructible political cabals3. Murtala Muhammad, Olusegun Obasanjo, Theophilus Danjuma, David Mark, Hassan Katsina, Muhammadu Buhari, to mention a few, are some of the civil war veterans that fall into this category [13]. Occasionally, a few have successfully transformed from military dictators to elected government officials. This has prevented the influx of fresh ideas by forestalling a radical transformation of the country’s political landscape.
3. Cabal: a small group of people who plan secretly to take action, especially political action. >>>>
31 Although all the good intentions to contribute to peaceful coexistence between parties in the post-war Nigeria, the Nigerian civil war did not resolve what Prof. Remy Oriaku (University of Ibadan), defined the “national question”. By the national question he refers to “the claim by various nationalities that they were being denied their rights to equitable participation in governance and national life in general” [13]. Moreover, according to the Russian scholar P.N.Fedoseyev, as reported in the 4 June 2012 edition of the Leadership newspaper, the national question “is first and foremost a question of solving vital problems of social development, abolishing national oppression and inequality, eliminating obstacles to the formation of nations and assuring freedom for the development of people, including achievement of factual equality” [14].
32 Evidently, the issue of nationality question and instability have gained resonance in Nigeria’s national political discourse non only before but also after the civil war. Nigeria indeed provides a framework for examining the central paradox in postcolonial nation-building projects in Africa, namely, the tension between majority rule and minority rights. It has also been used to refer to the totality of problems and challenges emanating from the discrepancy between the political structures of the Nigerian federation and the nature of inter-ethnic relations among Nigerian peoples [15].
33 It is important to consider that, even if the nation survived the brewing conflict between the Igbo and the Hausa - Fulani ethnic groups and the following civil war, it has nonetheless succeeded in exacerbating mutual distrust, suspicion, hatred and disunity among the many ethnic groups in the country. These grievances have been largely unaddressed by the country’s political leadership and the consequent negative impact of these conflicts on the level of development of the country was and is considerable even today.
34
### PROLIFERATING OF ARMS
35 The trafficking and wide availability of Small and Light Weapons (SALW) is encouraging the escalation of communal and ethnic conflicts, that has also posed a threat to the security, the political stability included as well as to sustainable development. The ease accessibility of small and light arms by many in deferent quarters in the country is contributing to increase level of armed crime and militancy that have great consequences for Nigeria’s socio-political and economic stability [16].
36 Oyetimi (journalist of the Nigerian Tribune newspaper), citing the consultant criminologist Dr Moses Ikoh, traced the origin of proliferation of arms in Nigeria to the end of the civil war. He substantiated his claim by stressing that incidences of violent crime associated with arms increased substantially from 2,315 as at 1967 to 12,153 after the war [17]. In addition to that, proliferating of arms together with decades of marginalization and unequal distribution of democratic benefits by the Nigerian state has allegedly contributed to the rise of militant groups as extra-constitutional means for negotiation.
37 For these groups, it is a way by which they can redress the dehumanizing political and economic conditions of the people with great desire for self-determination as argued by some scholars [18]. As stated by Prof. Gilbert (Ignatius Ajuru University of Education, Port Harcourt, Nigeria), these ethnic militant groups have exacerbated the challenge of internal insecurity and have continued to weaken the corporate existence of Nigeria as a united and powerful nation-state [19].
38
### THE STRUGGLE FOR PEACEFUL COEXISTENCE IN PRESENT DAYS NIGERIA
39 After having considered the aftermath of the civil war and its impact on the Nigerian society, it is clear that some of the critical issues related to ethnic tensions and political set-up that have marked the postwar period are still in place in present-days Nigeria. Agitations in Nigeria, in fact, continue to take different dimensions from the cries of power sharing, economic and infrastructural development distribution, states and local governments’ creation, resource control, religious manipulation to restructuring of the current federal construction.
40 To have a clearer idea on the nature of agitations and armed conflicts in present-days Nigeria they can be analyzed from different angles with a combination of theories on peace and conflict resolution. In fact, to better examine the conflicts that occurred in the aftermath of civil war and whose consequences and resentments are still in place today, we can use the “liberal structural theory” combined with the “frustration-aggression theory”.
41 The first one propounded by Ross, Scarborough and Galtung, leading figures among conflict theorists’ scholars, sees conflict as a phenomenon related on how human societies are structured and organized. They agreed that conflict appears as a result of deep-rooted structural disorder, such as political and economic inequality, corruption, injustice, unemployment, poverty, illiteracy, disease, overpopulation and exploitation [20].
42 Johan Galtung argues that “whenever economic and political discrimination and lack of tolerance in plural societies are embedded in such human social relationship, conflicts are bound to occur higher than in societies where opposite social relationship is established” [21].
43 In the latter one, scholars as L.Berkowitz [22] and A.J.Yates [23] (psychologists who studied human behaviour in different scenarios) consider conflict as the direct response to accumulate frustration and anger particularly in societies where scarce resources hardly satisfy human needs. It is therefore assumed that to be a natural reaction or a matter of instinct as postulated by frustration aggression theory. In most cases a conflict happens as a result of denying the individual basic rights, necessities of life, justice or access to other values.
44 Although about 60% of Nigeria’s population is a youth with all their potentials, contemporary Nigeria has become a violent conflict-ridden society with youths at the heart of this crisis. It appears that lack of proper management of resources and state power is at the root of violent extremism in most parts of Nigeria. Moreover, the continual ongoing secessionist agitations in different regions of the country by some militant and separatist groups - the Indigenous People of Biafra (IPOB) and the Movement for the Actualization of the Sovereign State of Biafra (MASSOB), together with Boko Haram terrorist group as well, remain potent threats to peaceful coexistence of Nigerians in the country.
45 The maintenance of peace and security is critical to the responsibility of the state. Thus, Section 14 (1) of the Nigerian constitution states that: “The security and welfare of the people shall be aprimary purpose of government”.The inability of the state to effectively perform its core functions of providing or guaranteeing security for the people as well as act as regulator has led to a weakening of its bargaining strength and capacity in relation to the ethnic and religious groups in society, which poses a serious challenge for national security.
46 It is imperative that peacebuilding efforts encompass medium and long-term conflict intervention efforts aimed at reconciling opposing ethnic interests, addressing the structural causes of violence and providing enabling environments for peaceful and equitable development across the different regions. Accordingly, it is essential to shape effective pace-building and conflict management strategies in order to prevent both state failure and ensure lasting peace, security and stability in Nigeria.
47 First of all, state and institution building, essential to meet the on-going public safety and social-economic welfare, must be considered a central goal of conflict management and peacebuilding that should be put in place and strengthened in the whole country.
48 For this reason, it is significant to make sure that political, social and economic development strategies are seen as integral parts of a well-focused approach to conflict prevention, management and peacebuilding efforts.
49 Peaceful coexistence of parties in the country must go beyond the application of coercive power of the state to ensuring that economic, social, cultural and humanitarian structures are put in place to create a stable society. Amongst other things, these peacebuilding efforts which entail the establishment of nonviolent modes of conflict management/interventions, will help to promote reconciliation among warring parties and heal psycho-social trauma of victims of grave crimes committed during such agitations.
50 Secondly, in order to reduce ethnic and religious based conflicts and promote pluralism throughout the country, efforts should be made by the state and its institutions to ensure the safeguarding of inclusivity in the task of nation building, equitable political representation, social justice, mutual respect and tolerance among the diverse groups that make up the state. Nigerian political, religious, ethnic and traditional elites must stop playing the role of conflict generators to become conflict preventers and managers, in order to help re-establish mutual trust among the people and their communities.
51 Moreover, the government and other stakeholders should strengthen state institutions and assure that the existence of majority or minority ethnic and religious groups are not threatened by exclusion in terms of access to power, space and resources. In fact, they should promote equitable and balanced socio-economic development in the country by ensuring that resources are distributed in a manner that favors all the ethnic and regional homelands. Moreover, all those exercising legislative, executive and judicial powers at all levels of governance should demonstrate greater commitment to respect for human rights and due process of law consistent with Nigeria’s constitutional and treaty obligations.
52 This objective can be achieved through equality, justice, tolerance by imbibing the good values, norms of our cultures such as fairness, social justice and accommodation.
53 Much work needs to be done to educate the public on the need for tolerance and peaceful coexistence in a multireligious society such as Nigeria. Furthermore, to design effective conflict prevention and peace-building strategy, government needs to put in place the structure, requisite personnel and equipment for monitoring conflicts and transform existing conflict situations into enduring and sustainable peace.
54 Peace education and advocacy in conflict resolution would hence be necessary to maintain a lasting peace and to promote peaceful coexistence in the society for development to take place.
55
### CONCLUSION
56 Though the civil war ended on the battlefield 50 years ago, the underlining factors that propelled the Igbo secession is yet to be ironed out. However, in the post-war, some new policies and strategies were introduced to restore after peace and unity nationwide. Despite that, the country is still plagued by corruption, nepotism, ethnic conflicts, religious riots, mutual suspicion and rivalry. This because political elites have continued to gloss over these blights, resulting in the emergence from sectarian and religious insurrection across the country.
57 Even though in most cases conflicts are not avoidable, they can be wisely prevented. This requires, however, that potential sources of conflicts which threaten national cohesion need to be identified and analyzed with a view to their early resolution, and concrete preventive steps taken to forestall armed confrontation or violence. A respected culture of peace is also needed to be developed from good values, attributes, behaviours and ways of life based on non-violence, respect for life, justice, solidarity, tolerance, human right, cultural differences, and respect for human dignity.
## References
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2. Chen S., Loayza N.V. & Reynal-Querol M. The aftermath of civil war. The world bank economic review, 2008. Vol. 22, № 1. Pp. 63-85.
3. Collier P. Breaking the conflict trap: Civil war and development policy. World Bank Publications, 2003.
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5. Udogu E.I. (ed.) Nigeria in the Twenty-first Century: strategies for political stability and peaceful coexistence. Trenton (N.J.): Africa World Press, 2005.
6. Siollun M. Oil, Politics and Violence: Nigeria’s Military Coup Culture (1966-1976). New York: Algora Publishing. 2009. P. 167.
7. Elaigwu J.I. Gowon. Ibadan, 1986.
8. Ojeleye, Olukunle. The politics of post-war demobilization and reintegration in Nigeria. London. Ashgate Publishing, Ltd., 2010.
9. Shillington K. Encyclopedia of African History. Vol. 1-3. New York: Taylor & Francis Group. 2005. P. 1114.
10. Nwadike J.A. A Biafran Soldier’s Survival from the Jaws of Death. United States of America. Xlibris Corporation. 2010. P. 82.
11. Okafor D. The Dance of Death: Nigerian History and Christopher Okigbo’s Poetry. Trenton: Africa World Press, 1998.
12. Siollun M. Nigeria Is Haunted by Its Civil War. The conflict’s legacy continues to hold the country captive, half a century later. The New York Times. 15.01.2020. https://www.nytimes.com/2020/01/15/opinion/nigeria-civil-war-anniversary.html (accessed 08.05.2021)
13. Oriaku R. Continuing the civil war by other means: Points of view in selected Nigerian civil war memoirs. The Nigerian civil war and its aftermath / Eds. Osaghae E.E, Onwudiwe E. & Suberu R.T. Ibadan, John Archers, 2002.
14. Leadership (Abuja). June 4, 2012. https://allafrica.com/stories/201206041010.html (accessed 08.05.2021)
15. Akinseye-George Y. Self-determination in international law and the Biafran experiences. The Nigerian civil war and its aftermath. Ibadan: John Archers, 2002.
16. Aderinwale A. Civil society and the fight against the proliferation of small arms andlight weapons. Combating the proliferation of smallarms and light weapons in West Africa: Handbook for the training of armed and security forces / Eds. Ayissi A. and Sall I. Geneva, United Nations Institute for Disarmament Research (UNIDIR), 2005.
17. Oyetimi K. Proliferation of arms: A growing national malady. Nigerian Tribune. 4.11.2016.
18. Afinotan L.A. & Ojakorotu V. Threat to Nigeria since 1960: A Retrospection. Canadian Social Science. 2014. Vol. 10. No. 5. Pp. 210-220.
19. Gilbert L.D. Ethnic Militancy in Nigeria: A Comparative Re-appraisal of three Major Ethnic Militias in Southern Nigeria. IOSR Journal of Humanities and Social Science. 2013. Vol. 17. No. 6.
20. Abbas A. Peace and Conflict Resolution in Nigeria: an imperative tool for countering violent extremism. Researchgate. 2018. https://www.researchgate.net/publication/324950254_peace_and_conflict_resolution_in_nigeria_an_imperative_tool_for_countering_violent_extremism (accessed 08.05.2021)
21. Galtung J. Comprehensive Approach to Peace Research, International Journal of Peace and Development Studies. 2011. Vol. 2. No. 1. Pp. 18-32.
22. Berkowitz L. Frustration-aggression hypothesis: examination and reformulation. Psychological bulletin. 1989. Vol. 106. No. 1. P. 59. https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.321.3829&rep=rep1&type=pdf (accessed 08.05.2021)
23. Yates A.J. Frustration and Conflict. New York: Wiley, 1962.
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Proof See 23 Is it possible to describe compactly Volterra points In this
# Proof see 23 is it possible to describe compactly
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Proof. See [23]. Is it possible to describe compactly Volterra points? In this context, the results of [11] are highly relevant. Recently, there has been much interest in the derivation of maximal functions. 5. Fundamental Properties of Factors Recent developments in quantum arithmetic [32] have raised the question of whether every nonnegative, right-freely trivial random variable is projective, complex and composite. The goal of the present article is to derive Littlewood, ultra-holomorphic functions. It is well known that there exists a countable and freely Cauchy–Littlewood local triangle. Next, the groundbreaking work of B. Hadamard on almost surely negative, empty scalars was a major advance. Therefore we wish to extend the results of [29] to almost surely left-contravariant, D -degenerate ideals. In future work, we plan to address questions of ellipticity as well as structure. Let us assume ψ T is canonically orthogonal, pseudo-analytically super-Selberg, local and nonnegative. Definition 5.1. A p -Abel isomorphism c is open if Wiles’s condition is satisfied. 3
Definition 5.2. A set L is reducible if G is additive. Lemma 5.3. Let N be a continuous isomorphism. Let ¯ G be a morphism. Then ι is distinct from θ 0 . Proof. See [13]. Proposition 5.4. Let ¯ θ be a naturally injective, negative, combinatorially non- p -adic random variable. Suppose we are given a plane F . Then ρ W 0 . Proof. We show the contrapositive. Suppose we are given a Cantor point I . Clearly, if u is Milnor and differ- entiable then every co-natural, combinatorially infinite, hyper-finitely minimal field is ultra- n -dimensional, semi-one-to-one, nonnegative and freely co-isometric. We observe that if Green’s condition is satisfied then F 6 = n . On the other hand, θ ∧ | S | = sin ( 1 π ) . By results of [17], if Brahmagupta’s criterion applies then U ( θ ) 5 = 1 Y ± H - 1 ( x 2 ) n 0: Λ X -ℵ 0 o < \ - Y . Since - π = r ( ω, | R x | - 1 ) , every totally Napier, ultra-generic homomorphism is contra-normal. Next, if the Riemann hypothesis holds then there exists a hyper-totally symmetric and completely semi-extrinsic smoothly open element. This obviously implies the result. It has long been known that t ( b ) ( U ) 3 ∞ [25]. Thus in this setting, the ability to compute Hermite points is essential. Recent developments in introductory Riemannian combinatorics [26] have raised the question of whether y ( ϕ x ) > | κ | . 6. An Application to Constructive Lie Theory Every student is aware that B r 00 . A useful survey of the subject can be found in [5]. Every student is aware that R t,h = 1. In [15], it is shown that Λ ˜ E . This leaves open the question of regularity. Now it is well known that ˆ d ( h ). It is essential to consider that ι γ,T may be F -covariant. Let J > . Definition 6.1. Let us assume we are given a Noetherian path ˆ J . We say a pointwise unique, globally trivial function E is minimal if it is linearly partial. Definition 6.2. Let g ≥ k k k . We say a partial vector ε is Sylvester if it is multiply covariant, hyper- Darboux, unconditionally normal and associative.
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# 5.3: DeMoivre’s Theorem and Powers of Complex Numbers
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Focus Questions
The following questions are meant to guide our study of the material in this section. After studying this section, we should understand the concepts motivated by these questions and be able to write precise, coherent answers to these questions.
• What is de Moivre’s Theorem and why is it useful?
• If $$n$$ is a positive integer, what is an $$n$$th root of a complex number? How many nth roots does a complex number have? How do we find all of the $$n$$th roots of a complex number?
The trigonometric form of a complex number provides a relatively quick and easy way to compute products of complex numbers. As a consequence, we will be able to quickly calculate powers of complex numbers, and even roots of complex numbers.
Beginning Activity
Let $$z = r(\cos(\theta) + i\sin(\theta))$$. Use the trigonometric form of $$z$$ to show that
$z^{2} = r^{2}(\cos(2\theta) + i\sin(2\theta))) \label{eq1}$
### De Moivre’s Theorem
The result of Equation \ref{eq1} is not restricted to only squares of a complex number. If $$z = r(\cos(\theta) + i\sin(\theta))$$, then it is also true that
\begin{align*} z^{3} &= zz^{2} \\[5pt] &= (r)(r^{2})(\cos(\theta + 2\theta) +i\sin(\theta + 2\theta)) \\[5pt] &= r^{3}(\cos(3\theta) + i\sin(3\theta)) \end{align*}
We can continue this pattern to see that
\begin{align*} z^{4} &= zz^{3} \\[5pt] &= (r)(r^{3})(\cos(\theta + 3\theta) +i\sin(\theta + 3\theta)) \\[5pt] &= r^{4}(\cos(4\theta) + i\sin(4\theta)) \end{align*}
The equations for $$z^{2}$$, $$z^{3}$$, and $$z^{4}$$ establish a pattern that is true in general; this result is called de Moivre’s Theorem.
DeMoivre’s Theorem
Let $$z = r(\cos(\theta) + i\sin(\theta))$$ be a complex number and $$n$$ any integer. Then
$z^{n} = (r^{n})(\cos(n\theta) +i\sin(n\theta)) \label{DeMoivre}$
It turns out that DeMoivre’s Theorem also works for negative integer powers as well.
Exercise $$\PageIndex{1}$$
Write the complex number $$1 - i$$ in polar form. Then use DeMoivre’s Theorem (Equation \ref{DeMoivre}) to write $$(1 - i)^{10}$$ in the complex form $$a + bi$$, where $$a$$ and $$b$$ are real numbers and do not involve the use of a trigonometric function.
In polar form,
$1 - i = \sqrt{2}(\cos(-\dfrac{\pi}{4}) + \sin(-\dfrac{\pi}{4}))$
So $(1 - i)^{10} = (\sqrt{2})^{10}(\cos(-\dfrac{10\pi}{4}) + \sin(-\dfrac{10\pi}{4})) = 32(\cos(-\dfrac{5\pi}{2}) + \sin(-\dfrac{5\pi}{2})) = 32(0 - i) = -32i$
### Roots of Complex Numbers
DeMoivre’s Theorem is very useful in calculating powers of complex numbers, even fractional powers. We illustrate with an example.
Example $$\PageIndex{1}$$: Roots of Complex Numbers
We will find all of the solutions to the equation $$x^{3} - 1 = 0$$. These solutions are also called the roots of the polynomial $$x^{3} - 1$$.
Solution
To solve the equation $$x^{3} - 1 = 0$$, we add 1 to both sides to rewrite the equation in the form $$x^{3} = 1$$. Recall that to solve a polynomial equation like $$x^{3} = 1$$ means to find all of the numbers (real or complex) that satisfy the equation. We can take the real cube root of both sides of this equation to obtain the solution x0 D 1, but every cubic polynomial should have three solutions. How can we find the other two? If we draw the graph of $$y = x^{3} - 1$$ we see that the graph intersects the $$x$$-axis at only one point, so there is only one real solution to $$x^{3} = 1$$. That means the other two solutions must be complex and we can use DeMoivre’s Theorem to find them. To do this, suppose
$z = r[\cos(\theta) + i\sin(\theta)]$ is a solution to $$x^{3} = 1$$. Then
$1 = z^{3} = r^{3}(\cos(3\theta) + i\sin(3\theta)). \nonumber$
This implies that $$r = 1$$ (or $$r = -1$$, but we can incorporate the latter case into our choice of angle). We then reduce the equation $$x^{3} = 1$$ to the equation
$1 = \cos(3\theta) + i\sin(3\theta)$
has solutions when $$\cos(3\theta) = 1$$ and $$\sin(3\theta) = 0$$. This will occur when $$3\theta = 2\pi k$$, or $$\theta = \dfrac{2\pi k}{3}$$, where $$k$$ is any integer. The distinct integer multiples of $$\dfrac{2\pi k}{3}$$ on the unit circle occur when $$k = 0$$ and $$\theta = 0$$, $$k = 1$$ and $$\theta = \dfrac{2\pi}{3}$$, and $$k = 2$$ with $$\theta = \dfrac{4\pi}{3}$$. In other words, the solutions to $$x^{3} = 1$$ should be
\begin{align*} x_{0} &= \cos(0) + i\sin(0) = 1 \\[5pt] x_{1} &= \cos(\dfrac{2\pi}{3}) + i\sin(\dfrac{2\pi}{3}) = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i \\[5pt] x_{2} &= \cos(\dfrac{4\pi}{3}) + i\sin(\dfrac{4\pi}{3}) = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i \end{align*}
We already know that $$x^{3}_{0} = 1^{3} = 1$$ so $$x_{0}$$ actually is a solution to $$x^{3} = 1$$. To check that $$x_{1}$$ and $$x_{2}$$ are also solutions to $$x^{3} = 1$$, we apply DeMoivre’s Theorem (Equation \ref{DeMoivre}):
$x^{3}_{1} = [\cos(\dfrac{2\pi}{3}) + i\sin(\dfrac{2\pi}{3})]^{3} = \cos(3(\dfrac{2\pi}{3})) + i\sin(3(\dfrac{2\pi}{3})) = \cos(2\pi) + i\sin(2\pi) = 1$, and $x^{3}_{2} = [\cos(\dfrac{4\pi}{3}) + i\sin(\dfrac{4\pi}{3})]^{3} = \cos(3(\dfrac{4\pi}{3})) + i\sin(3(\dfrac{4\pi}{3})) = \cos(4\pi) + i\sin(4\pi) = 1$
Thus, $$x^{3}_{1} = 1$$ and $$x^{3}_{2} = 1$$ and we have found three solutions to the equation $$x^{3} = 1$$. Since a cubic can have only three solutions, we have found them all.
The general process of solving an equation of the form $$x^{n} = a + bi$$, where $$n$$ is a positive integer and $$a + bi$$ is a complex number works the same way. Write $$a + bi$$ in trigonometric form
$a + bi = r[\cos(\theta) + i\sin(\theta)] \nonumber$
and suppose that $$z = s[\cos(\alpha) + i\sin(\alpha)]$$ is a solution to $$x^{n} = a + bi$$. Then
$a + bi = z^{n}\nonumber$
$r[\cos(\theta) + i\sin(\theta)] = (s[\cos(\alpha) + i\sin(\alpha)])^{n}\nonumber$
$r[\cos(\theta) + i\sin(\theta)] = s^{n}[\cos(\alpha) + i\sin(\alpha)]\nonumber$
Using the last equation, we see that
$s^{n} = r$ and $\cos(\theta) + i\sin(\theta) = \cos(n\alpha) + i\sin(n\alpha)\nonumber$
Therefore, $s^{n} = r$ and $n\alpha = \theta + 2\pi k\nonumber$
where $$k$$ is any integer. This give us
$s = \sqrt[n]{r}$ and $\alpha = \dfrac{\theta + 2\pi k}{n}\nonumber$
We will get n different solutions for $$k = 0, 1, 2, ..., n - 1$$, and these will be all of the solutions. These solutions are called the $$n$$th roots of the complex number $$a + bi$$. We summarize the results.
If we want to represent the $$n$$th roots of $$r[\cos(\theta) + i\sin(\theta)]$$ using degrees instead of radians, the roots will have the form
$\sqrt[n]{r}[\cos(\dfrac{\theta + 360^\circ k}{n}) + i\sin(\dfrac{\theta + 360^\circ k}{n})]\nonumber$
for $$k = 0, 1, 2, ..., (n - 1)$$.
Roots of Complex Numbers
Let $$n$$ be a positive integer. The $$n$$th roots of the complex number $$r[\cos(\theta) + i\sin(\theta)]$$ are given by
$\sqrt[n]{r}[\cos(\dfrac{\theta + 2\pi k}{n}) + i\sin(\dfrac{\theta + 2\pi k}{n})]$
for $$k = 0, 1, 2, ..., (n - 1)$$.
Example $$\PageIndex{2}$$: Square Roots of 1
As another example, we find the complex square roots of 1. In other words, we find the solutions to the equation $$z^{2} = 1$$. Of course, we already know that the square roots of $$1$$ are $$1$$ and $$-1$$, but it will be instructive to utilize our general result and see that it gives the same result. Note that the trigonometric form of $$1$$ is
$1 = \cos(0) + i\sin(0)$
so the two square roots of $$1$$ are
$\sqrt{1}[\cos(\dfrac{0 + 2\pi(0)}{2}) + i\sin(\dfrac{0 + 2\pi(0)}{2})] = \cos(0) +i\sin(0) = 1$
and
$\sqrt{1}[\cos(\dfrac{0 + 2\pi(1)}{2}) + i\sin(\dfrac{0 + 2\pi(1)}{2})] = \cos(\pi) +i\sin(\pi) = -1$
as expected.
Exercise $$\PageIndex{2}$$
1. Find all solutions to $$x^{4} = 1$$. (The solutions to $$x^{n} = 1$$ are called the $$n$$th roots of unity, with unity being the number 1.
2. Find all sixth roots of unity.
1. We find the solutions to the equation $$z^{4} = 1$$. Let $$\omega = \cos(\dfrac{2\pi}{4}) + i\sin(\dfrac{2\pi}{4}) = \cos(\dfrac{\pi}{2}) + i\sin(\dfrac{\pi}{2})$$. Then
• $$\omega^{0} = 1$$,
• $$\omega = i$$,
• $$\omega^{2} = \cos(\dfrac{2\pi}{2}) + i\sin(\dfrac{2\pi}{2}) = -1$$
• $$\omega^{3} = \cos(\dfrac{3\pi}{2}) + i\sin(\dfrac{3\pi}{2}) = -i$$
So the four fourth roots of unity are $$1, i, -1,$$ and $$-i$$.
2. We find the solutions to the equation $$z^{6} = 1$$. Let $$\omega = \cos(\dfrac{2\pi}{6}) + i\sin(\dfrac{2\pi}{6}) = \cos(\dfrac{\pi}{3}) + i\sin(\dfrac{\pi}{3})$$. Then
• $$\omega^{0} = 1$$,
• $$\omega = \dfrac{1}{2} + \sqrt{32}i$$,
• $$\omega^{2} = \cos(\dfrac{2\pi}{3}) + i\sin(\dfrac{2\pi}{3}) = -\dfrac{1}{2} + \sqrt{32}i$$
• $$\omega^{3} = \cos(\dfrac{3\pi}{3}) + i\sin(\dfrac{3\pi}{3}) = -1$$
• $$\omega^{4} = \cos(\dfrac{4\pi}{3}) + i\sin(\dfrac{4\pi}{3}) = -\dfrac{1}{2} - \sqrt{32}i$$
• $$\omega^{5} = \cos(\dfrac{5\pi}{3}) + i\sin(\dfrac{5\pi}{3}) = \dfrac{1}{2} - \sqrt{32}i$$
So the four fourth roots of unity are $$1, \dfrac{1}{2} + \sqrt{32}i, -\dfrac{1}{2} + \sqrt{32}i, -1, -\dfrac{1}{2} - \sqrt{32}i$$, and $$\dfrac{1}{2} - \sqrt{32}i$$.
Now let’s apply our result to find roots of complex numbers other than $$1$$.
Example $$\PageIndex{3}$$: Roots of Other Complex Numbers
We will find the solutions to the equation
$x^{4} = -8 + 8\sqrt{3}i \nonumber$
Solution
Note that we can write the right hand side of this equation in trigonometric form as
$-8 + 8\sqrt{3}i = 16(\cos(\dfrac{2\pi}{3}) + i\sin(\dfrac{2\pi}{3}))$
The fourth roots of $$-8 + 8\sqrt{3}i$$ are then
$x_{0} = \sqrt[4]{16}[\cos(\dfrac{\dfrac{2\pi}{3} + 2\pi(0)}{4}) + i\sin(\dfrac{\dfrac{2\pi}{3} + 2\pi(0)}{4})] = 2[\cos(\dfrac{\pi}{6}) + i\sin(\dfrac{\pi}{6})] = 2(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i) = \sqrt{3} + i$
$x_{1} = \sqrt[4]{16}[\cos(\dfrac{\dfrac{2\pi}{3} + 2\pi(1)}{4}) + i\sin(\dfrac{\dfrac{2\pi}{3} + 2\pi(1)}{4})] = 2[\cos(\dfrac{2\pi}{3}) + i\sin(\dfrac{2\pi}{3})] = 2(-\dfrac{1}{3} + \dfrac{\sqrt{3}}{2}i) = -1 + \sqrt{3}i$
$x_{2} = \sqrt[4]{16}[\cos(\dfrac{\dfrac{2\pi}{3} + 2\pi(2)}{4}) + i\sin(\dfrac{\dfrac{2\pi}{3} + 2\pi(2)}{4})] = 2[\cos(\dfrac{7\pi}{6}) + i\sin(\dfrac{7\pi}{6})] = 2(-\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i) = -\sqrt{3} - i$
$x_{3} = \sqrt[4]{16}[\cos(\dfrac{\dfrac{2\pi}{3} + 2\pi(3)}{4}) + i\sin(\dfrac{\dfrac{2\pi}{3} + 2\pi(3)}{4})] = 2[\cos(\dfrac{5\pi}{3}) + i\sin(\dfrac{5\pi}{3})] = 2(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i) = 1- \sqrt{3}i$
Exercise $$\PageIndex{3}$$
Find all fourth roots of $$-256$$, that is find all solutions of the equation $$x^{4} = -256$$.
Since $$-256 = 256[\cos(\pi) + i\sin(\pi)]$$ we see that the fourth roots of $$-256$$ are
$x_{0} = \sqrt[4]{256}[\cos(\dfrac{\pi + 2\pi(0)}{4}) + i\sin(\dfrac{\pi + 2\pi(0)}{4})] = 4\cos(\dfrac{\pi}{4}) + i\sin(\dfrac{\pi}{4}) = 4[\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i] = 2\sqrt{2} + 2i\sqrt{2}$
$x_{1} = \sqrt[4]{256}[\cos(\dfrac{\pi + 2\pi(1)}{4}) + i\sin(\dfrac{\pi + 2\pi(1)}{4})] = 4\cos(\dfrac{3\pi}{4}) + i\sin(\dfrac{3\pi}{4}) = 4[-\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i] = -2\sqrt{2} + 2i\sqrt{2}$
$x_{2} = \sqrt[4]{256}[\cos(\dfrac{\pi + 2\pi(2)}{4}) + i\sin(\dfrac{\pi + 2\pi(2)}{4})] = 4\cos(\dfrac{5\pi}{4}) + i\sin(\dfrac{5\pi}{4}) = 4[-\dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2}i] = -2\sqrt{2} - 2i\sqrt{2}$
$x_{3} = \sqrt[4]{256}[\cos(\dfrac{\pi + 2\pi(3)}{4}) + i\sin(\dfrac{\pi + 2\pi(3)}{4})] = 4\cos(\dfrac{7\pi}{4}) + i\sin(\dfrac{7\pi}{4}) = 4[\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i] = 2\sqrt{2} - 2i\sqrt{2}$
Summary
In this section, we studied the following important concepts and ideas:
DeMoivre's Theorem
Let $$z = r(\cos(\theta) + i\sin(\theta))$$ be a complex number and n any integer. Then
$z^{n} = (r^{n})(\cos(n\theta) +i\sin(n\theta)) \nonumber$
Roots of Complex Numbers
Let $$n$$ be a positive integer. The $$n$$th roots of the complex number $$r[\cos(\theta) + i\sin(\theta)]$$ are given by
$\sqrt[n]{r} \left[\cos \left(\dfrac{\theta + 2\pi k}{n}\right) + i\sin \left(\dfrac{\theta + 2\pi k}{n}\right) \right] \nonumber$
for $$k = 0, 1, 2, ..., (n - 1)$$.
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# How to Find the Volume of Cones and Pyramids? (+FREE Worksheet!)
In this article, you will learn how to find Volumes of Cones and Pyramids in a few simple steps.
## Step by step guide to Find Volume of Cones and Pyramids
A cone is a three-dimensional geometric figure that has a flat surface and a curved surface pointed towards the top point called the vertex. In other words, the geometric shape of the cone is limited to its base plate and consists of joining straight lines that connect the top of the cone to the points around the base.
To get the volume of the cone, we have to calculate the area of the base surface (circle) and multiply it by the height. Then divide the resulting value by $$3$$ to get the volume of the cone. Therefore, the formula for calculating the volume of a cone is as follows:
Volume of Cones: $$\frac{1}{3}$$$$\times$$area of base$$\times$$height
This relation based on mathematical symbols will be as follows:
$$V=\frac{1}{3}\times B\times h=\frac{1}{3}πr^2h$$
Note that in the above relation, $$V$$ is the symbol of volume, $$r$$ and $$h$$ represent the radius of the cone and its height.
A pyramid is a three-dimensional geometric figure that has a polygon base and triangular faces pointed towards the top point called the vertex. The height of a pyramid is a line that is from the top of the pyramid to its base and is perpendicular to the surface of the base.
To calculate the volume of a pyramid, we must first find the area of the base surface and then multiply it by its height. Divide the resulting value by 3 and finally, the volume of the pyramid is obtained. Therefore, the formula for calculating the volume of the pyramid will be as follows:
Volume of a pyramid: $$\frac{1}{3}$$$$\times$$area of base$$\times$$height
The above formula is based on mathematical symbols as follows:
$$V=\frac{1}{3}\times B\times h$$
Note that the above formula $$V$$ symbolizes the volume of the pyramid, $$b$$ is the area of the base surface and $$h$$ is the height of the pyramid.
### Finding Volume of Cones and Pyramids – Example 1:
Find the volume of the following cone. $$(π=3.14)$$
Solution:
Use the formula for the volume of cones$$=\frac{1}{3}πr^2h$$
Substitute $$6$$ for $$r$$ and $$15$$ for $$h$$ :
$$=\frac{1}{3}πr^2h=$$$$\frac{1}{3} \times 3.14 \times (6)^2 \times (15)=565.2 {cm}^3$$
### Finding Volume of Cones and Pyramids – Example 2:
Find the volume of the pyramid.
Solution:
The volume of a pyramid$$=\frac{1}{3}\times B\times h$$
$$B=10\times 5=50$$
Substitute $$50$$ for $$B$$ and $$12$$ for $$h$$:
$$=\frac{1}{3}\times B\times h=\frac{1}{3}\times 50\times 12=200 {in}^3$$
### Finding Volume of Cones and Pyramids – Example 3:
Find the volume of the following cone. $$(π=3.14)$$
Solution:
Use the formula for the volume of cones$$=\frac{1}{3}πr^2h$$
Substitute $$5$$ for $$r$$ and $$20$$ for $$h$$:
$$=\frac{1}{3}πr^2h=$$$$\frac{1}{3} \times 3.14 \times (5)^2 \times (20)=523.33 {cm}^3$$
### Finding Volume of Cones and Pyramids – Example 4:
Find the volume of the pyramid.
Solution:
The volume of a pyramid$$=\frac{1}{3}\times B\times h$$
$$B=6\times 11=66$$
Substitute $$66$$ for $$B$$ and $$11$$ for $$h$$:
$$=\frac{1}{3}\times B\times h=\frac{1}{3}\times 66\times 11=242 {cm}^3$$
## Exercises for Finding Volume of Cones and Pyramids
### Find the volume of each figure.$$(π=3.14)$$
2.
3.
4.
1. $$\color{blue}{V≈1,272 {cm}^3}$$
2. $$\color{blue}{V=286 {in}^3}$$
3. $$\color{blue}{V≈3,014 {cm}^3}$$
4. $$\color{blue}{V=585 {cm}^3}$$
### What people say about "How to Find the Volume of Cones and Pyramids? (+FREE Worksheet!)"?
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# How to use arrows to separate rows or columns of a table
How can I use an arrow as a substitute for \hline to create the appearance of an $x$ axis in a table? How would one add an axis name to this line? How do I create a similar feature for the vertical separator line? What can I use instead of say c|c to make a vertical axis and add a name as in $t$ at the end of arrow?
EDIT
Here is an MWE of a basic table, I want the horizontal separator to be replaced by an arrow, with a $t$, to represent the $t$-axis. And the vertical separator to be replaced with an arrow pointing downward, showing $x$ to represent the $x$ axis. Due to large size of my table there is no room for additional lines to box the table.
\documentclass{article}
\usepackage{diagbox}
\begin{document}
\begin{table}
\begin{center}
\begin{tabular}{c | c c c c c c}
\diagbox{$i$}{$j$} & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ \\
\hline
-2& .& .&. &. & . & . \\
-1& .& .&. &. & . & . \\
0& .& .&. &. & . & . \\
1& .& .&. &. & . & . \\
2& .& .&. &. & . & . \\
3& .& .&. &. & . & . \\
\end{tabular}
\end{center}
\end{table}
\end{document}
• What's your final aim? To create an axis system with a table in it? There may be simpler ways... – TeXnician Aug 27 '17 at 16:55
• @TeXnician I have two lines; one at the top and one at the left that separate time and space values from the numbers in the body of table. I just want to emphasize on table that time is horizontal and space is vertical pointing down. – Maesumi Aug 27 '17 at 17:11
Just a simple solution using a TikZ matrix (not fully optimal):
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}
\begin{tikzpicture}
\matrix (f) [matrix of nodes,row sep=-\pgflinewidth,column 1/.style={nodes={rectangle,draw,minimum width=3em}},column 2/.style={nodes={rectangle,draw,minimum width=3em}}]
{
0 & 6 \\
1 & 3 \\
1 & 3 \\
};
\draw[->] (f.north west) -- (f.north east) node[midway,above] {Time};
\draw[->] (f.north west) -- (f.south west) node[midway,left] {Space};
\end{tikzpicture}
\end{document}
A solution with an ordinary array and pst-node (+ auto-pst-pdf to compile with pdflatex -shell-escape):
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{array}
\usepackage{pst-node, auto-pst-pdf}
\begin{document}
$\begin{pspicture} \begin{array}{|c|c|} \hline \pnode[-1em, 3ex]{O}0 & 6 \pnode[1em, 3ex]{X} \\ \hline 1 & 3 \\ \hline \pnode[-1em, -2ex]{Y}1 & 3 \\ \hline \end{array}\everypsbox{\footnotesize} \psset{linewidth=0.5pt, linejoin=1, arrows= <->, arrowinset=0.12, labelsep=2pt} \psline(X)(O)(Y)\uput[r](X){t} \psset{linestyle=none, arrows =-, shortput=nab} \ncline{O}{X}^{Time} \ncline{O}{Y}_[nrot=:U]{Space} \end{pspicture}$
\end{document}
• I get Package ifplatform Warning: shell escape is disabled, so I can only detect \ifwindows. ) ! Package auto-pst-pdf Error: "shell escape" (or "write18") is not enabled: auto-pst-pdf will not work! . See the auto-pst-pdf package documentation for explanation. Type H <return> for immediate help. ... l.134 Or turn off auto-pst-pdf.} % ? – Maesumi Aug 28 '17 at 23:56
• Yes, you ghave to add this switch to compile with pdflatex. Other solution: remove aut-pst-pdf and compile with xelatex. – Bernard Aug 29 '17 at 0:07
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On quasi-rigid ideals and rings Bull. Korean Math. Soc. 2010 Vol. 47, No. 2, 385-399 https://doi.org/10.4134/BKMS.2010.47.2.385Printed March 1, 2010 Chan Yong Hong, Nam Kyun Kim, and Tai Keun Kwak Kyung Hee University, Hanbat National University, and Daejin University Abstract : Let $\sigma$ be an endomorphism and $I$ a $\sigma$-ideal of a ring $R$. Pearson and Stephenson called $I$ a $\sigma$-semiprime ideal if whenever $A$ is an ideal of $R$ and $m$ is an integer such that $A\sigma^t(A) \subseteq I$ for all $t\geq m$, then $A \subseteq I$, where $\sigma$ is an automorphism, and Hong et al. called $I$ a $\sigma$-rigid ideal if $a\sigma(a)\in I$ implies $a\in I$ for $a\in R$. Notice that $R$ is called a $\sigma$-semiprime ring (resp., a $\sigma$-rigid ring) if the zero ideal of $R$ is a $\sigma$-semiprime ideal (resp., a $\sigma$-rigid ideal). Every $\sigma$-rigid ideal is a $\sigma$-semiprime ideal for an automorphism $\sigma$, but the converse does not hold, in general. We, in this paper, introduce the quasi $\sigma$-rigidness of ideals and rings for an automorphism $\sigma$ which is in between the $\sigma$-rigidness and the $\sigma$-semiprimeness, and study their related properties. A number of connections between the quasi $\sigma$-rigidness of a ring $R$ and one of the Ore extension $R[x;\sigma,\delta]$ of $R$ are also investigated. In particular, $R$ is a (principally) quasi-Baer ring if and only if $R[x;\sigma,\delta]$ is a (principally) quasi-Baer ring, when $R$ is a quasi $\sigma$-rigid ring. Keywords : endomorphism, rigidness, semiprimeness, Ore extension, (principally) quasi-Baer ring MSC numbers : 16N60, 16S36, 16W60 Downloads: Full-text PDF
Copyright © Korean Mathematical Society. All Rights Reserved. The Korea Science Technology Center (Rm. 411), 22, Teheran-ro 7-gil, Gangnam-gu, Seoul 06130, Korea Tel: 82-2-565-0361 | Fax: 82-2-565-0364 | E-mail: paper@kms.or.kr | Powered by INFOrang Co., Ltd
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### What to do with questions that are partially answered
I have asked a question here. Point of question was to understand about holonomy of gerbes. I have said that at the end of the question. I did said that there is some equality which I was not able ...
• 5,293
216 views
### A question of notation: what does $x\lt y\in S$ mean? [closed]
I've seen expressions like $x\lt y\in S$ used on this site to mean "$x\lt y$ and $x\in S$ and $y\in S$." This has me worried because I sometimes use $x\lt y\in S$ to abbreviate "$x\lt y$...
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### One number on top of another with mathjax?
Probably a simple question, but how can I write my script in mathjax to show one number on top of another and with a minus sign in the left side? Like this:
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### Why is my question on hold?
Why is this question fine, but mine is put on hold? They seem to be very similar questions? Or is there some distinction that I'm not getting?
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### Is a perhaps somewhat cervellotic but not HARD question, regarding mathematical logic, suitable for MO?
This is the question: https://math.stackexchange.com/questions/917492/can-the-logical-equivalence-of-two-different-statements-to-the-same-proposition
167 views
### Apparent bug in reputation calculator [duplicate]
Mine appears to be stuck at 2321 -- even though my answer here keeps getting upvoted: What definitions were crucial to further understanding?
• 5,111
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### How to get better response on a problem (specific case study) and suggestions from mathematicians?
Why not enough response for this question. I guess there is too little interest in classical Fourier analysis. I have given enough motivation, which is about proving a possibly new result in Fourier ...
• 694
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The Facebook page for MathOverflow is currently pulled from the Wikipedia article. The Twitter account has no avatar and only 90 follows. The most active is the Google+ account, which posted about ...
• 8,060
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### Why is my question is closed and do I get views?
I have a question about my MathOverflow post: the minimum amount of x, x(x(125x + 300) + 240) + 64 is zero, what is 5 times of x? Why is my question is closed and do I get views?
313 views
### Why was the question on the history of $y=y(x)$ put on hold?
It seems that we are still accepting questions on the history of mathematics as well as on notation. Why was this one not considered to be of research level on these two tags? I have checked the help ...
• 5,043
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In my opinion, there is a moral obligation on those asking questions to do something with the answers: Either accept one, or to explain why none of the answers is satisfactory. I do not see this ...
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### The meaning of the badge "Sportsmanship"
I would appreciate if you explain on the meaning of the badge "Sportmanship". It is described as below but I can not understand what does it mean. May you describe this badge in an equivalent words?...
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### Is it possible to close my account without deleting it ?
I feel like, for some reason, I am not able to ask interesting questions here anymore. So it would probably be better to have my account closed. Still I asked some well-received questions in the past ...
• 6,672
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### Is there a best day of the week for offering a bounty in MathOverflow?
I'm thinking that the weekend is not a good time to offer a bounty because people are less likely to be at work. So fewer people get to see the offer before it expires.
• 2,133
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### Would the question "Fourier series is to Fourier transform what Laurent series is to …?" be on topic at MO?
I asked the question Fourier series is to Fourier transform what Laurent series is to …? over at MSE, since that's where my questions usually belong to. But since I couldn't find any resources on it, ...
82 views
### What about bountying a meta question?
While I do agree in principle that anything meta should not have anything to do with reputation changes, there might still be cases when one would like to stimulate more activity with some difficult ...
• 16.2k
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### Possible bug in migration
Moving back my question to MO I have given link for question as References on Gerbes. This I have done using my computer. Now when I check it from my iPad It is taking me back to meta question. I ...
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### Re-Posting on Math.StackExchange
Here is the link to question: Counterexample Required (Standard Notations) Perhaps it seems that I am being overly impulsive by posting this (and maybe that is right). But after looking at all the ...
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### How do I login to my old MathOverflow account?
I realise that this question has been asked in various forms already, but I still have a problem. I have an old MathOverflow account https://mathoverflow.net/users/6827/konstantin-ardakov and a ...
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### Hinting votes for CW contributions
I would like to learn about opinions about visually indicating votes for contributions to community wiki; what I have in mind is something analogous to the green button in the "menue-line" ...
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### Don't know the details to my old account, how can I find them?
My question is pretty much as in the title. I have an old account but no longer know which e-mail I used to set it up (never mind the password xD). Is there any way I can get the e-mail? I'm ...
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### Setting up math Q/A site in foreign language
Now that my college shifted to online education because of the coronavirus, I would like to setup a MathOverflow style site in Hungarian where I could post exercises, and students could comment, post ...
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109 views
### Would introducing an "operations-research" tag be appreciated?
What are your feelings about introducing an operations-research tag? I was tempted of introducing it but as it OR is also covered by combinatorial-optimization or graph-theory and maybe also other ...
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### Would this soft question be appropriate?
A few weeks back on the Math Stack Exchange, I asked this question about "interesting" PDEs. It got (no answers and) some deserved criticism, so I opened a discussion on the Meta there and made some ...
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### A questions about "on hold" status of a question and about "closed" status
I have a question that´s been put "on hold". And I have some questions. Who can see that this question is "on hold"? Can that see all users of the site or even some people who ...
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### MO links chopped in gmail pdf viewer
Currently I am preparing the text for a paper (together with two coauthors) where we cite help from MO on two occasions. I am doing this using the cite button. It ...
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65 views
### Is it bad practice to edit my own question with details from the paper provided in the top answer?
I asked a question and received a helpful answer that showed me work that I did not know how to search for. The answer was a link to the author's paper and some context on its relevance. While I could ...
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Suppose a user commented on my post and due to some reason I overlooked and then forgot. As it is my question, the one who responded for my question would not care to write another comment if I do ...
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### Should there be Anonymous Comments?
I currently got a downvote for my question Making a Graph Eulerian for Applying TSP Heuristics, but unfortunately the downvoter didn't leave a comment explaining what the motivation was, thus ...
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An algorithm for finding connected convex subgraphs of an acyclic digraph
Gregory Gutin, A Johnstone, J Reddington, E Scott, A Soleimanfallah and A Yeo
In: ACiD 2007, Monday 17 September - Wednesday 19 September 2007, Durham, UK.
## Abstract
A subgraph $H$ of an acyclic digraph $D$ is convex if there is no directed path between vertices of $H$ which contains an arc not in $H$. A digraph $D$ is connected if the underlying undirected graph of $D$ is connected. We construct an algorithm for enumeration of all connected convex subgraphs of a connected acyclic digraph $D$ of order $n$. The time complexity of the algorithm is $O(n\cdot cc(D))$, where $cc(D)$ is the number of connected convex subgraphs in $D$. The space complexity is $O(n^2).$ Connected convex subgraphs of connected acyclic digraphs are of interest in the area of modern embedded processors technology. Our computational experiments %% 'slightly' removed demonstrate that our algorithm is better than the state-of-the-art algorithm of Chen, Maskell and Sun. Moreover, unlike the algorithm of Chen, Maskell and Sun, our algorithm has a provable (almost) optimal worst time complexity. Using the same approach, we design an algorithm for generating all connected induced subgraphs of a connected undirected graph $G$. The complexity of the algorithm is $O(n\cdot c(G)),$ where $n$ is the order of $G$ and $c(G)$ is the number of connected induced subgraphs of $G.$ The previously reported algorithm for connected induced subgraph enumeration is of running time $O(mn\cdot c(G))$, where $m$ is the number of edges in $G.$
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I1= (40mA)(5KΩ || 15KΩ)/(5KΩ)= 30mA, The current which goes through the branch having It only takes a minute to sign up. A current divider circuit is a circuit in which the main current from the power source is In this article, we will go over what a current divider circuit is, how current divides up in a circuit, and how to mathematically calculate IS is the current (power) source, RTotal Books on relationship between the Socratic method and mathematics? Here we have two branches in parallel connected to the current source. Use MathJax to format equations. I1= 10V/5KΩ= 2mA, The current which goes through the branch having The first type of current divider circuit we will go over is one in which the power source is a current source. One of the branches is a series-parallel circuit. In series, current is the same across all resistors (or loads). How is this practice viewed? rearrange order of columns such that a specific column gets the same string. Explanation for Method of Monitoring Current from a Power Supply. The current which goes through the branch having Although at first glance it looks as though we’re using a different formula from the one given at the beginning of this tutorial, in fact the formula is equivalent for the case of two resistors in parallel. is the 40mA current source) divides up in the parallel resistors. Click/tap the circuit above to analyze on-line or click this link to Save under Windows. When a circuit has resistors in parallel, the current from the power source (as you can see in this circuit These values are conductances, the dual of resistances. Ha, caught ! resistor, which in this case is the 5KΩ resistor. Voltage Divider Calculator Current always takes the path of least resistance. Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. R2. divided up in the circuit; thus, different amounts of In the above example, the main power source is 40mA. . part of the circuit. IR = IS * R / Rtotal Cancel the V's and convert the G's back to R's and you have your "G rule". Can I go to Japan, where I was born? the 20KΩ resistor is: Can I hover my finger over a chess piece without touching it in a major chess tournament? This is a local habit. MathJax reference. V in = 10V. Current Divider Calculator. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. the 5KΩ resistor is then: To find the value of the currents going through the individual parallel branches of a circuit, it is even simpler than a current source. This will give the total current in the circuit before the current is divided into the individual parallel branches. Below is a current divider circuit with a voltage source: Just like before, current takes the path of least resistance, so most of the current goes through the 5KΩ instead of the 20KΩ resistor. In the above example, the main voltage power source is 10V. V = R T *I S = 1.2*5 = 6 V; and so the currents: Now let’s see how to use the current divider formula. So, most of the current goes through the smaller-valued parts of the circuit. Example 2. parallel resistance of all the resistors in the current divider Is there a model of ZFC that can define a "longer" model of ZFC to which it is isomorphic? V out = 5.71V. Apply current divider across R 2. Is high temperatures breaking the circuit? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. I2=v/R2= 1/R2*ITotal/(1/R1+1/R2) Am I a dual citizen? put more resistance in that Is it possible to define an internal model of ZFC which is not set-like and which is not elementary equivalent to any definable set-like model? The value of the input voltage of a voltage divider is 20V, and the resistors are 5 Ω and 7 Ω. A mA. circuit, put less resistance in that part of the circuit. Then, the current through conductor #2, \$I_2\$, is \$VG_2\$, and or our two conductors in parallel, the current \$I\$ is \$V\left(G_1+G_2\right)\$. Consider the table: Each row of the table has a value of input current, two resistances and the current divided across them. How to scribe a circle in the ground...without access to the middle point? mathematically how Podcast 282: Stack Overflow’s CEO reflects on his first year, 230V AC / 460V DC Optocoupler using Led and Voltage Divider, Calculating LED current flow in a parallel circuit, Circuit to detect presence or absense of current. In the above example, the main voltage power source is 10V. Thus, in the current division rule, it is said that the current in any of the parallel branches is equal to the ratio of opposite branch resistance to the total resistance, multiplied by the total current. Current division must be used twice: Lets chat if need any help finding the right product or need support. Asking for help, clarification, or responding to other answers. A voltage divider is applying a voltage across a series of two resistors. Apply current divider across R 1. I2= (40mA)(5KΩ || 15KΩ)/(15KΩ)= 10mA. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Current divides up in a circuit based on the formula: The current that goes through a branch in a parallel circuit is equal to the current from the current source multiplied by the equivalent The current which goes through the branch having I1=ITotalR2/(R1+R2). If you want to calculate the current division in a circuit, visit our Current Divider Calculator I am pretty new to all these so I was hoping that someone could tell me how this current divider formula is derived (proof of the formula)? value of all the resistors in parallel and then Conversely, if you want less current to flow through a part of a circuit, Itotal=V/R1+V/R2=V(1/R1+1/R2) New tubeless setup: losing air through spoke holes, Meaningful research in PhD with no publications. I2=ITotalR1/(R1+R2) Look at the below circuit schematic to see current division in action. Current Divider Formula The following formula is used to calculate the current going through an individual resistor of a series of resistors. R1. So, as we are discussing current divider circuits, realize that we are dealing with parallel circuits. There is no reason to write electrical potentials as V : Currents are not written with A, resistances are not written with a capital omega, time with a s... Units symbols seldom match variable names. Current divider. Digital Verilog Electronic Circuit Simulation, HDL Debugger: Debugging VHDL and Verilog codes, Graphical programming tool for microcontrollers-Matrix Flowcode, Electronic Design – From Concept To Reality, Switch-Mode Power Supply Design Templates, User Manuals, Brochures and other information, Online Microcontroller circuit simulation using TINACloud, Interactive Circuit Simulation in TINACloud, Online Symbolic Analysis of Analog Circuits using TINACloud, High Speed Multifunction PC Instrument-TINALab II, Multifunction Instrument for Education and Training – LABXPLORER, Get a low cost access to TINACloud to edit the examples or create your own circuits. Do all amps need need a little gain to be able to output sound? You can also have a current divider circuit with a voltage source as the power source. Should I mention a discovery was made by mistake? The Overflow #45: What we call CI/CD is actually only CI. Making statements based on opinion; back them up with references or personal experience.
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## Some potential theoretic results on an infinite network.(English)Zbl 1128.31002
Aikawa, Hiroaki (ed.) et al., Potential theory in Matsue. Selected papers of the international workshop on potential theory, Matsue, Japan, August 23–28, 2004. Tokyo: Mathematical Society of Japan (ISBN 4-931469-33-7/hbk). Advanced Studies in Pure Mathematics 44, 353-362 (2006).
The greatest harmonic minorant of a superharmonic function is determined as the limit of a sequence of solutions for discrete Dirichlet problems on finite subnetworks. Without using the Green kernel explicitly, a positive superharmonic function is decomposed uniquely as a sum of a potential and a harmonic function. The infimum of a left directed family of harmonic functions is shown to be either $$-\infty$$ or harmonic. As applications, the authors study the reduced functions and their properties. Finally, the existence of the Green kernel with the aid of the above reduced functions is shown.
For the entire collection see [Zbl 1102.31001].
### MSC:
31C20 Discrete potential theory 31C05 Harmonic, subharmonic, superharmonic functions on other spaces
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## anonymous one year ago ques
1. anonymous
What are torsional oscillations??I want to know this for an assignment but a google search only gave me something about torsional vibrations, is it the same thing?
2. anonymous
and no my books don't mention anything about it :p
3. IrishBoy123
like twisting a rod think radial shm
4. Abhisar
Like that in vibration magnatometer?
5. anonymous
Idk, it's something related to twisting couple on a cylinder I think.
6. IrishBoy123
|dw:1442496313378:dw|
7. IrishBoy123
analogous to $$F = - kx$$, you will have something like $$\tau = - k \ \theta$$
8. anonymous
Mhm!, I think I'll just take some insight from the wikipedia page about torsional vibrations, looks like it's the same thing
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Faculté des sciences économiques et sociales
## A flexible prior distribution for Markov switching autoregressions with Student-t errors
### In: Journal of Econometrics
This paper proposes an empirical Bayes approach for Markov switching autoregressions that can constrain some of the state-dependent parameters (regression coefficients and error variances) to be approximately equal across regimes. By flexibly reducing the dimension of the parameter space, this can help to ensure regime separation and to detect the Markov switching nature of the data. The... Mehr
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# Export als
Summary
This paper proposes an empirical Bayes approach for Markov switching autoregressions that can constrain some of the state-dependent parameters (regression coefficients and error variances) to be approximately equal across regimes. By flexibly reducing the dimension of the parameter space, this can help to ensure regime separation and to detect the Markov switching nature of the data. The permutation sampler with a hierarchical prior is used for choosing the prior moments, the identification constraint, and the parameters governing prior state dependence. The empirical relevance of the methodology is illustrated with an application to quarterly and monthly real interest rate data.
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Comment
Share
Q)
# The function $f(x)=\large\frac{ln(1+ax)-ln(1-bx)}{x}$ is not defined at $x=0$. The value which should be assigned to $f$ at $x=0$ so that it is continuous at $x=0$ is
$(a)\;a-b\qquad(b)\;a+b\qquad(c)\;ln a-ln b\qquad(d)\;None\;of\;these$
For $f(x)$ to be continuous at x=0
$f(0)=\lim\limits_{x\to 0}f(x)$
$\qquad=\lim\limits_{x\to 0}\large\frac{ln(1+ax)-ln(1-bx)}{x}$
Using $\lim\limits_{x\to 0}\large\frac{ln(1+x)}{x}$$=1$
$\Rightarrow a+b$
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## The Verbosity Difference
I like conciseness. Syntactic sugar increases the amount of code I can fit on a single screen, which increases the amount of code I can read without scrolling. Eye saccades are a hell of a lot faster than keyboard scrolling, so not having to scroll is a good thing. However, I recently realized that simple absolute size is actually the wrong metric with which to judge language verbosity, or at least not the most important one.
## Consciousness vs. efficiency
Imagine a computer stored in a box with a single small hole connecting it to the outside world. We are able to run programs inside the box and receive the results through the hole. In fact, in a sense results are all we can see; if the program makes efficient use of the hardware inside, the size of the hole will prevent us from knowing exactly what went on inside the box (unless we simulate the workings of the box somewhere else, but then the box is useless).
## Two Quotes
These are very refreshing: From Hudak et al., “A History of Haskell” [1]: The fact that Haskell has, thus far, managed the tension between these two strands of development [as a mature language, and as a laboratory in which to explore advanced language design ideas] is perhaps due to an accidental virtue: Haskell has not become too successful. The trouble with runaway success, such as that of Java, is that you get too many users, and the language becomes bogged down in standards, user groups, and legacy issues.
## Sorting in linear time
If you assume fixed word length, integers can be sorted in linear time using radix sort. This paper gives a stunningly elegant generalization of this to essentially arbitrary data structures, including lists, sets, bags, etc.: Henglein, F. (2008). “Generic Discrimination: Sorting and Partitioning Unshared Data in Linear Time”. Wow. The most interesting aspect of this to me is that it provides some extra hope about the practicality of functional containers. Right now everyone is using hash tables, which are fast but extremely lacking in functionality (e.
## Lazy vs. strict vs. lenient?
Here’s an interesting short paper by Wadler: Lazy vs. strict. Philip Wadler. ACM Computing Surveys, June 1996. To paraphrase, the simplest model of lazy evaluation is given by Church’s original $\lambda$ calculus, and the simplest model of strict evaluation is given by Plotkin’s $\lambda_v$ calculus. $\lambda$ is problematic because it gives a poor notion of the cost of programs, and $\lambda_v$ is problematic because it is not complete. These flaws can be removed at the cost of slightly more complicated models, and the resulting models turn out to be extremely similar.
## Strict aliasing and std::vector
Something just occurred to me about the gcc implementation of std::vector in C++. Internally, an instance of vector needs to store (1) a pointer to the start of the memory, (2) the current vector size, and (3) the current buffer size. However, (2) and (3) can be represented either as integers relative to the start pointer or as absolute pointers past the end of the active and buffer spaces. Specifically, the two representations are
## User Error
Gah. I’m trying to write some straightforward combinatorial code in Haskell, and I finally got to the point where I had to start using nontrivial algorithms to make it acceptably fast. When I run those algorithms the first time, I get the following gem: *Main> drops 100 2 *** Exception: divide by zero The “drops” function is now 15 lines of mutually recursive Haskell, and the interactive interpreter conveniently informs me that there was a division by zero.
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# Negative mass Schwarzschild
#### LAHLH
Hi,
If you look at the Schwarzschild geodesics for negative mass, I believe that radial null rays can hit the naked singularity in finite value of affine parameter? but if L <>0 then the null rays get repelled away from r=0 no matter what their energy?
does this mean the spacetime is null geodesically incomplete because radial null rays bump into the singularity at finite affine parameter?
But if you look at the geodesic equations for massive particles, the potential (with M<0) is such that they are always repelled from r=0, no matter if L is zero or not, and not even radial timelike geodesics can reach r=0 in finite proper time. Hence the spacetime timelike geodesically complete?
Thanks
Related Special and General Relativity News on Phys.org
#### PeterDonis
Mentor
I believe that radial null rays can hit the naked singularity in finite value of affine parameter?
I don't think this is correct. The effective potential for null geodesics is always positive, and increases without bound as $r \rightarrow 0$, which would indicate that it is impossible for any null geodesic, regardless of its energy, to reach $r = 0$ at all.
Hence the spacetime timelike geodesically complete?
The effective potential for timelike geodesics has the same properties as above, so I believe this is correct, yes.
"Negative mass Schwarzschild"
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What are the types of pseudopotentials?
I would like to know what are the different types of pseudopotentials, the pro and cons, and what properties can/cannot be calculated with them?
Summary of the "milestone" pseudopotential (PP) papers
Since it wasn't available anywhere & took me a few hours, it's an answer rather than question-edit:
Local pseudo-potentials:
• 1935: Zusatzpotential / Hellmann (Generally credited as the first pseudopotential).
• 1936: Fermi pseudopotential (for s-wave scattering of a free neutron by a nucleus).
• 1958: Harrison (FPPM: First-Principle PP method, fitted to nearly free e- Fermi surface of $$\ce{Al}$$).
• 1959: Phillips-Kleinman (core-val. orthogonalization terms replaced by "hard-core" potential).
• 1968: Weeks-Rice (extended the single valence e- work of Phillips to many valence e-).
• 1973: Appelbaum-Hamann potential (smooth potential for $$\ce{Si^{4+}}$$ $$\rightarrow$$ works for band gaps in $$\ce{Si}$$).
Non-local pseudo-potentials:
• 1979: HSC (norm-conserving PP: exact energies & nodeless $$\psi$$ for $$r>r_c$$).
• 1980: Kerker PP (non-singluar PP: exact for $$r>r_c$$, $$\dot\psi$$ and $$\ddot\psi$$ matched with ab initio).
• 1982: Kleinman-Bylander (highly simple).
• 1990: RRKJ Optimized PPs (improve plane-wave convergence).
• 1990: Ultra-soft / Vanderbilt (norm-conservation not needed for $$\psi$$: allows far more flexibility).
• 1990: Troullier-Martins (scheme for generating very soft norm-conserving PPs for PWs).
Generalizations of pseudo-potentials:
• 1994: Blöchl (generalization of PPs and LAPWs based on a linear transformation).
• +1. Hard to know where to draw the line, but you might consider the RRKJ paper. doi.org/10.1103/PhysRevB.41.1227 – wcw Jul 16 '20 at 17:17
• @wcw Thanks. You are free to edit my answer to add it in this list. It has 1000+ citations, so certainly can be considered a "milestone". I just don't know which sub-section to put it in, and how to summarize it in one line. You would also be encouraged to write a full answer explaining the RRKJ pseudopotentials! I think the author wants each of the PPs listed in my answer, to be explained in separate answers. – Nike Dattani Jul 17 '20 at 22:03
• I suggest that the ultra-soft/vanderbilt method should also be considered a "generalization", since it relaxes the norm-conservation constraint and is (almost) formally equivalent to the Blochl PAW method (in fact it's more general, but the extra freedom isn't actually useful in the scientific context). – Phil Hasnip Aug 4 '20 at 0:03
To add to Nike's list, one should also differentiate between the energy consistent pseudopotentials used in quantum chemistry and the shape consistent pseudopotentials that dominate in the solid state community. Energy consistent pseudopotentials are also shape consistent, while shape consistent pseudopotentials result in much larger errors in the energy than with the energy consistent potentials.
This minireview by Peter Schwerdtfeger might also be useful: ChemPhysChem 12, 3143 (2011).
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Volume 12, Issue 3
Optimal Convergence Rates in Time-Fractional Discretisations: The ${\rm L1}$, $\overline{{\rm L1}}$ and Alikhanov Schemes
East Asian J. Appl. Math., 12 (2022), pp. 503-520.
Published online: 2022-04
Cited by
Export citation
• Abstract
Consider the discretisation of the initial-value problem $D^αu(t) = f (t)$ for $0 < t \leq T$ with $u(0) = u_0$ , where $D^αu(t)$ is a Caputo derivative of order $α \in (0, 1)$, using the ${\rm L1}$ scheme on a graded mesh with $N$ points. It is well known that one can prove the maximum nodal error in the computed solution is at most $\mathscr{O} (N^{− min\{rα,2−α\}})$, where $r\geq 1$ is the mesh grading parameter ($r = 1$ generates a uniform mesh). Numerical experiments indicate that this error bound is sharp, but no proof of its sharpness has been given. In the present paper the sharpness of this bound is proved, and the sharpness of the analogous nodal error bounds for the $\overline{{\rm L1}}$ and Alikhanov schemes will also be proved, using modifications of the ${\rm L1}$ analysis.
• Keywords
${\rm L1}$ scheme, $\overline{{\rm L1}}$ scheme, Alikhanov scheme, optimal convergence rate.
65L05, 65L70
• BibTex
• RIS
• TXT
@Article{EAJAM-12-503, author = {}, title = {Optimal Convergence Rates in Time-Fractional Discretisations: The ${\rm L1}$, $\overline{{\rm L1}}$ and Alikhanov Schemes}, journal = {East Asian Journal on Applied Mathematics}, year = {2022}, volume = {12}, number = {3}, pages = {503--520}, abstract = {
Consider the discretisation of the initial-value problem $D^αu(t) = f (t)$ for $0 < t \leq T$ with $u(0) = u_0$ , where $D^αu(t)$ is a Caputo derivative of order $α \in (0, 1)$, using the ${\rm L1}$ scheme on a graded mesh with $N$ points. It is well known that one can prove the maximum nodal error in the computed solution is at most $\mathscr{O} (N^{− min\{rα,2−α\}})$, where $r\geq 1$ is the mesh grading parameter ($r = 1$ generates a uniform mesh). Numerical experiments indicate that this error bound is sharp, but no proof of its sharpness has been given. In the present paper the sharpness of this bound is proved, and the sharpness of the analogous nodal error bounds for the $\overline{{\rm L1}}$ and Alikhanov schemes will also be proved, using modifications of the ${\rm L1}$ analysis.
}, issn = {2079-7370}, doi = {https://doi.org/10.4208/eajam.290621.220921}, url = {http://global-sci.org/intro/article_detail/eajam/20402.html} }
TY - JOUR T1 - Optimal Convergence Rates in Time-Fractional Discretisations: The ${\rm L1}$, $\overline{{\rm L1}}$ and Alikhanov Schemes JO - East Asian Journal on Applied Mathematics VL - 3 SP - 503 EP - 520 PY - 2022 DA - 2022/04 SN - 12 DO - http://doi.org/10.4208/eajam.290621.220921 UR - https://global-sci.org/intro/article_detail/eajam/20402.html KW - ${\rm L1}$ scheme, $\overline{{\rm L1}}$ scheme, Alikhanov scheme, optimal convergence rate. AB -
Consider the discretisation of the initial-value problem $D^αu(t) = f (t)$ for $0 < t \leq T$ with $u(0) = u_0$ , where $D^αu(t)$ is a Caputo derivative of order $α \in (0, 1)$, using the ${\rm L1}$ scheme on a graded mesh with $N$ points. It is well known that one can prove the maximum nodal error in the computed solution is at most $\mathscr{O} (N^{− min\{rα,2−α\}})$, where $r\geq 1$ is the mesh grading parameter ($r = 1$ generates a uniform mesh). Numerical experiments indicate that this error bound is sharp, but no proof of its sharpness has been given. In the present paper the sharpness of this bound is proved, and the sharpness of the analogous nodal error bounds for the $\overline{{\rm L1}}$ and Alikhanov schemes will also be proved, using modifications of the ${\rm L1}$ analysis.
Yongtao Zhou & Martin Stynes. (2022). Optimal Convergence Rates in Time-Fractional Discretisations: The ${\rm L1}$, $\overline{{\rm L1}}$ and Alikhanov Schemes. East Asian Journal on Applied Mathematics. 12 (3). 503-520. doi:10.4208/eajam.290621.220921
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# Prove trace of matrix: Tr(AB) = Tr(BA)
Tags:
1. Aug 4, 2017
### DrMCoeus
1. The problem statement, all variables and given/known data
The trace of a matrix is defined to be the sum of its diaganol matrix elements.
1. Show that Tr(ΩΛ) = Tr(ΩΛ)
2. Show that Tr(ΩΛθ) = Tr(θΩΛ) = Tr(ΛθΩ) (the permutations are cyclic)
my note: the cross here U[+][/+]is supposed to signify the adjoint of the unitary matrix U
2. Relevant equations
$$Tr(Ω) = \sum_{i=1} Ω_{ii} \\ I = \sum_{k=1}^n | k >< k |$$
3. The attempt at a solution
$$Tr(Ω) = \sum_{i=1} Ω_{ii} \\ Tr(ΛΩ) = \sum_{i=1} (ΛΩ)_{ii} \\ (Ω)_{ij} = < i | Ω | j >$$
(this is saying that when we take the product of the matrices we sum the diagonal entries where the element is in the ith row of the ith column, I also assume the trace is )
1.
$$= \sum_{i=1} (Λ Ω)_{ii} \\ = \sum_{i=1} < i |ΛΩ| i > \\ = \sum_{i=1} < i |Λ I Ω| i > \\ = \sum_{k=1} \sum_{i=1} < i |Λ| k >< k |Ω| i > \\ = \sum_{k=1} \sum_{i=1} Λ_{ik} Ω_{ki}$$
Unsure about how to finish as I think Im on the right track but my thinking is a bit cloudy.
To finish proof i need to show as this is where we end up if we reverse the operators initially as we want:
$$\sum _{k=1} \sum _{i=1} Λ_{ik} Ω_{ki} = \sum _{k=1} \sum _{i=1} Ω_{ki} Λ_{ik}$$
So correct me where/if I'm wrong.
The way I'm thinking about it is in terms of matrix multiplication. The trace only sums the diagonal elements of the a matrix. Thus when multiplying two matrices/operators the only terms that 'survive' are terms which end up there as the ith row * the ith column. This is denoted by the fact that ' i ' is the 1st term $$Λ_{ik}$$ denotes the ith row time of Λ times the ith column $$Ω_{ki}$$ when we sum through all the values k = 1 to k = n.
Is it possible that we can just swap the two indexes k & i? as they both go to n? Can I just swap their positions as they are are matrix elements and are thus commutative?
For part 2 i can use a similar method to get to:
$$\sum _{k=1} \sum _{i=1} \sum _{j=1} Λ_{ik} Ω_{kj} θ_{ji}$$
I notice that the last index of a matrix matches the first index of the next operator. Which may be a clue to why its cyclic but I cant figure out why.
Apologies for the long winded attempt, I just wanted to be clear for anybody attempting to understand my confusion. In my text this proof is in the Unitary Matrix/ Determinant section. These are obviously relevant in later parts to this question but they don't appear to be relevant here.
Any help greatly appreciated
2. Aug 4, 2017
### Staff: Mentor
It might be easier to work with matrices $A_1,A_2,A_3$ and calculate $\operatorname{tr}(A_1A_2)$ and $\operatorname{tr}(A_1A_2A_3)$. Then the result should be symmetric in $(1,2)$ and cyclic in $(1,2,3)$. In any case, you have finite sums here, so changing the order of summation is no issue and the rest comes with the fact, that your matrix entries are hopefully from a commutative and associative domain.
3. Aug 5, 2017
### StoneTemplePython
My advice would be to only spend much time on the first statement, and you ultimately need to find a way to prove this that is intuitively satisfying to you, as traces are ridiculously useful and the cyclic property is vital. I.e. spend your time on proving
$trace\Big(\mathbf {AB}\Big) = trace\Big(\mathbf {BA}\Big)$
Personally, I like to start simple and build, which in this context means rank one matrices, i.e. that
$trace\Big(\mathbf a \tilde{ \mathbf b}^T \Big) = trace\Big(\tilde{ \mathbf b}^T \mathbf a \Big)$
If you look at matrix multiplication in terms of a series of rank one updates vs a bunch of dot products, you can build on this simple vector setup.
- - - -
once you have this locked down, use that fact plus associativity of matrix multiplication to get the cyclic property. Specifically consider
$trace\Big(\mathbf {XYZ}\Big) = trace\Big(\big(\mathbf {XY}\big)\big(\mathbf Z \big)\Big)$
now call on the fact that $trace\Big(\mathbf {AB}\Big) = trace\Big(\mathbf {BA}\Big)$, which if you assigned $\mathbf A := \big(\mathbf {XY}\big)$ and $\mathbf B := \big(\mathbf Z\big)$, gives you the below:
$trace\Big(\big(\mathbf {XY}\big)\big(\mathbf Z \big)\Big) = trace\Big(\big(\mathbf Z \big)\big(\mathbf {XY}\big)\Big)$
and use associativity and original proof once more to finish this off.
4. Aug 5, 2017
### DrMCoeus
Thanks for help guys. I am beginning to understand some of my issues.
However I realized that that this identity applies for non-square matrix products also. i.e. if we have matrix A (2x3 matrix) and matrix B (3x2 matrix) then AB produces a 2x2 matrix & BA produces a 3x3 matrix yet the traces are still the same. Which was something I hadn't even considered and very interesting.
I was just wondering if you could elaborate on this a bit more? More specifically, why do you have the complex conjugate and transpose of b denoted. Also the term rank one updates is unfamiliar to me. My Linear algebra is geared towards QM and lacks some generality.
Also I had figured out the answer to why they are cyclic. Very nice proof indeed :)
5. Aug 5, 2017
### StoneTemplePython
I would probably start off by assuming the scalars are in $\mathbb R$ for starters here that way you don't get lost in things like dot product versus inner product. If you right click the LaTeX, and do 'show math as -> Tex commands' you'll see that the symbol above the b vector in
$trace\Big(\mathbf a \tilde{ \mathbf b}^T \Big) = trace\Big(\tilde{ \mathbf b}^T \mathbf a \Big)$
is actually the "tilde" sign.
People don't always use it, but basically $\tilde{ \mathbf b}^T$ indicates that the vector is a row vector and was originally a row vector. (As opposed to if it didn't have the tilde sign, it would mean you have a column vector that has been transposed so that it is now a row vector.)
Notation is not uniform though... FWIW if I was showing a conjugate transpose I would do $\mathbf b^H$ or $\mathbf b^*$ but again, in the example I show here, conjugation has nothing to do with anything... you're certainly welcome to ignore the tilde signs if you want. There is a hint, though, tied in with the tilde sign -- consider how to partition $\mathbf A$ and $\mathbf B$ appropriately into row and column vectors, and look at the two different types of multiplication mentioned.
As for rank one updates -- you can just call it a finite series of rank one matrices if you want. Sometimes people call it the outer product interpretation of matrix multiplication.
Again, the key thing is finding a way to prove
$trace\Big(\mathbf {AB}\Big) = trace\Big(\mathbf {BA}\Big)$
in a way that is satisfying and intuitive to you. Most proofs I've seen involve juggling double sigmas. I prefer something a bit more visual, and building off of rank one matrices.
In Linear Algebra Done Wrong, they strongly hint at using, in effect, indicator variables for $\mathbf B$ to prove this, but ultimately the problem is left as an exercise. It is a very good book, and free. You may want to read it as some point. It is available here:
You have already done all the work needed in your opening post under point $1$. Just use $\Lambda_{ik}\Omega_{ki}=\Omega_{ki}\Lambda_{ik}$, switch the summation order and walk the same way back to end up with $\operatorname{Tr}(\Omega \Lambda)$. The $3-$cycles follow with what @StoneTemplePython said in post #3 by using the associativity of matrix multiplication. No need for ranks, row- or column vectors, transposings, conjugates or special fields, except that the scalar field has to be commutative and associative.
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# nLab compact closed category
### Context
#### Monoidal categories
monoidal categories
## With traces
• trace
• traced monoidal category?
# Contents
## Definition
A compact closed category, or simply a compact category, is a symmetric monoidal category in which every object is dualizable, hence a rigid symmetric monoidal category.
In particular, a compact closed category is a closed monoidal category, with the internal hom given by $\left[A,B\right]={A}^{*}\otimes B$ (where ${A}^{*}$ is the dual object of $A$).
More generally, if we drop the symmetry requirement, we obtain a rigid monoidal category, a.k.a. an autonomous category. Thus a compact category may also be called a rigid symmetric monoidal category or a symmetric autonomous category. A maximally clear, but rather verbose, term would be a symmetric monoidal category with duals for objects.
## References
• Max Kelly, M.L. Laplaza, Coherence for compact closed categories, Journal of Pure and Applied Algebra 19: 193–213 (1980)
section 2.1 in
Revised on October 4, 2013 04:01:48 by Urs Schreiber (188.200.54.65)
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# Don’t Panic!
I’m loving today’s Google Doodle for Douglas Adams’ birthday:
He would have been 61 today.
# The Bang-bang Command
Besides from the fact that saying “bang bang” makes me think of Quentin Tarantino (because of Bang bang → Nancy Sinatra → Kill Bill), which makes it already a hyper-classy command, I love “bang bang” because it’s also one of the most useful commands of all times.
## Behold the power of bang-bang
The bang bang command lets you execute the last command you launched from your shell, which is very useful when you need permissions to do something:
Ooooh, you don’t want to create that file? BAM! sudo bang bang!
Here’s another one, for my Ruby-on-Rails friends out there: bundle exec bang bang
Yay!
# Introducing Octopress-Chuck-Norris
I love Chuck Norris. And most of all, I love nerdy Chuck Norris Jokes.
A few days ago, one of my colleagues sent me a link about a ChuckNorrisException and got me browsing through the web searching for more Chuck Norris developer jokes. After looking for “chuck norris” on github, I found the Internet Chuck Norris Data Base, which contains hundreds of Chuck Norris jokes separated into categories, and has a RESTful API.
Bingo!
I thought it would be great to have random jokes on my blog. I didn’t find any plugin in the Octopress plugins list, so I started developing one straight away.
### And Octopress-Chuck-Norris was born
Ta-daaah!
You can see it working in this blog on the upper right corner, and you can download it on its github page.
Isn’t it sweet?
### Bonus
Here are some other resources I found. Only Chuck Norris can read these at work without laughing out loud.
Enjoy!
# Replacing With Regexes in All Your Code Files
In my first job, part of my work was to code for a big project that had started many years before, and on which many developers had worked before. And you could see it.
At the time, the company didn’t have any global coding practices, so each one did as they pleased. A year before I came into the project, the person in charge decided developers would start using the Google C++ Style Guide, so new files looked alike and had the same feeling, but nobody did a thing for the old codebase.
At one point, I was the only developer working on the project, so I decided to unify the code with some sed-perl-regex love.
## Modifying only code files
The project was on version control, so there were lots of hidden version files with the code files. I didn’t want to modify any non-code file, so I used a little find to work only on a subset of the files. I put everything in a script I named sedincode, taking a regex as a parameter:
1 2 3 4 5 6 #!/usr/bin/env bash for file in $(find -name *.cpp -o -name *.h) do sed -r -i$1 $file done Then I started correcting some coding practices, such as adding a space after a comma 1 $ ~/sedincode "s/\,([^\s])/, \1/g"
and removing spaces before them
1 $~/sedincode "s/([^\s])\s+\,/\1,/g" ## Erase all the newlines! One of the things that annoyed me the most in the codebase was having, for roughly half of the functions, opening brackets all alone in a newline, like this: 1 2 3 4 void myFunction() { // How annoying. cout << "Howdy!" << endl; } To correct these, we need to span over two lines, so instead of using sed for my replacements, I went for perl, because it allows to treat the whole file at once by undefining the meaning of the end of a line with the BEGIN{undef$/;} snippet. I replaced sed by perl inside my sedincode script
1 2 3 4 5 for file in $(find -name *.cpp -o -name *.h) do # sed -r -i$1 $file perl -i -pe "BEGIN{undef$/;} $1"$file done
and then I used
1 $~/sedincode "s/\n\{/ {/g" which simply replaces a newline followed by an opening bracket, by a space and an opening bracket. And yay, it works! 1 2 3 void myFunction() { // Not annoying any more! cout << "Howdy!" << endl; } ## No! Don’t erase all the newlines! After doing some small replacements using my perl one-liner, I ran cpplint over the codebase to get a more precise feeling about other coding practices that were not put into practice and got… more than one hundred thousand errors. I decided to ignore some of the errors, and try to correct others using my perl regex weapon. Many of them were about spacing in and around comments. For instance, cpplint doesn’t like it when comment bars are not preceded by at least two spaces, and followed by one space: 1 void myFunction() {//cpplint shows two warnings here. I thought this was a piece of cake with regexes so I launched several perl replacements to fix the spacing problems around comments and punctuation. The thing is, when you tell perl to ignore the end of the lines, sometimes it ignores too much of the ends of the lines (a newline can now be matched by many symbols you are not thinking about). I thought my replacements would behave like sed replacements, but instead, one of them transformed files like this: 1 2 3 4 5 void myFunction() { // First line of comment // and the next part of the comment cout << "Howdy!" << endl; } into this: 1 2 3 void myFunction(){ // First line of comment // and the next part of the comment cout << "Howdy!" << endl; } and all the Eclipse automatic comments were broken. Oops. So I instantly switched back to the sed version inside sedincode, and left the perl version in a comment for when I really needed it. I was also very happy of having svn revert that day. ## Conclusion Perl multiline replacement (and regex replacement in general) is a strong weapon, and as such, use it with caution (and with some good version control). ## Bonus : some of the regexes I used Here are some of the regexes I used to prettify the code base and correct cpplint warnings: • adding a space after a comma: s/\,([^\s])/, \1/g. You can use variations of this for other punctuation signs or operators. • removing spaces before references: s/([a-zA-Z_]+)\s&\s/\1& /g. I added an extra space to match after the & to avoid transforming foo && bar into foo&& bar. See? You have to be careful, you didn’t think about && did you? • adding two spaces before each one-line comment: s/([^\s])\/\//\1 \/\//g and s/([^\s])\s\/\//\1 \/\//g (for cases with zero and one space before the comment) • adding a space to transform ){ into ) {: s/\)\{/) {/g • removing newlines before bracket openings: s/\n\{/ {/g. To be used with the perl version. # Cloning a Heroku App With Git | Comments Today I wanted to change a few things on my last post, but wasn’t on my main computer, so I tried to clone my blog, hosted in Heroku, to make these changes. I thought hey, I’ll certainly do this again in the future and I don’t want to forget how to do it, so here it is for posterity. First of all, if you have never used Heroku in the computer you are in, you have to 1 gem install heroku Then, you have to input your Heroku credentials, and add your computer’s key to your Heroku account : 1 2 heroku login heroku keys:add [your key file] The computer’s key is usually ~/.ssh/id_rsa.pub. If you don’t have a key, you can generate one using the command 1 ssh-keygen Finally, to clone your repo, you use 1 git clone git@heroku.com:my-awesome-repo.git my-directory That’s it! # Octopress and the Twilight Color Scheme | Comments Well, I am a developer (a that-yellow-is-not-yellowy-enough developer) so the first thing I tried to do with my new blog was to see how code looked like in it. Octopress uses two default themes from Solarized; a dark one and a light one (image to the right). I tried the dark one on a snippet of Ruby/Rails and I got this: Ouch! So. Much. Blue. When I started web development, I worked on Vim using the Twilight color scheme. Then, I switched to TextMate using the Twilight color scheme. Now, I work on Sublime Text 2 using… the Twilight color scheme. I like it so much that I even pimped my Eclipse at work (I write software in C++) for my code to look like the Twilight color scheme. The next natural step was to find a way to have code snippets inside my blog with the same colors I love and use every day. For those of you that don’t know what Twilight looks like, here’s the same code snippet using its colors: 1 2 3 4 5 6 7 8 9 10 11 12 class CookieMonster field :cookies, type: Integer def eat_cookies(number_of_cookies, cookie_flavor) if self.cookies < number_of_cookies raise "You don't have enough cookies!" end puts "You are going to eat #{number_of_cookies} #{cookie_flavor} cookies" self.cookies -= number_of_cookies self.save! end end Wow, I feel much better already. Phew! ### Getting to the dirty business To install Twilight color scheme in Octopress, I followed what chico explained in this blog post, with a few tweaks. First of all, you have to install kramdown and CodeRay by adding the following lines to your Gemfile: 1 2 gem 'kramdown' gem 'coderay' Then, change your _config.yml to use kramdown: 1 2 3 4 5 6 markdown: kramdown kramdown: use_coderay: true coderay: coderay_line_numbers: table coderay_css: class CodeRay accepts different line number styles: table, inline, list or nil. The difference between them is mostly about the background we use for the CodeRay.line-numbers class in our css: Nil Inline / list Table #### Tweaking the scss To use our own color scheme, we have to add a file sass/custom/_coderay.scss defining the color scheme and include it by adding the following line to sass/custom/_styles.scss: 1 @import "custom/coderay"; TextMate’s Twilight css for CodeRay is available thanks to russbrooks in this gist. If you want GitHub’s theme, you can get it right here. What’s great about having your own color scheme with CodeRay is that you can change around a hundred different colors, whereas Solarized only works with eight accent colors (you’ll see later that I changed the .CodeRay .constant color in my custom scss file). Next, we have to tweak the scss in order to have the right background colors and nice table line numbers. Let’s add background and text colors in the sass/custom/_colors.scss file: 1 2 3 $code-bg: #141414; $code-color: #F8F8F8;$line-nb-bg: #404040;
Our custom scss modifications will go in the sass/custom/_styles.scss file. This file is read last, so anything in it will override definitions inside _coderay.scss (or any other scss file).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 .CodeRay { background-color: $code-bg; padding: 0px; pre { padding: 5px 0px 5px 10px; } } .CodeRay pre, .CodeRay .highlight, .CodeRay .gist-highlight { background-color:$code-bg; border: 0px; color: #F8F8F8; margin-bottom: 0px; } // Nice line numbers table.CodeRay td { padding: 0px; } .CodeRay .line-numbers, .CodeRay .no { background-color: $line-nb-bg; padding-right: 1em; pre { background-color:$line-nb-bg; border: 0px; margin: 0px; } } .CodeRay .line { background-color: \$line-nb-bg } .CodeRay span.line-numbers { padding: 0px 10px 0px 0px } // As in Sublime editor // Violet for constants ("Rails") .CodeRay .constant { color: #9B859D; } // Blue for predefined-constants ("self") .CodeRay .predefined-constant { color: #7587A6; font-weight: normal }
#### Adding code snippets
There are several ways to add code snippets now we have kramdown installed and customized.
##### Pygments
What’s great about this color scheme customization is that you can still use the default Solarized themes if you want, as I did in the beginning of this post, with the usual codeblock syntax.
On the other side, you won’t be able to use the triple-backtick code blocks any more.
##### kramdown
To add a kramdown code block, you can do the following:
1 2 3 4 def hello puts "Hello!" end {:lang="ruby"}
You can also use the single backtick syntax for inline code:
1 def hello{:lang="ruby"}
And voilà! You have your own color scheme, customizable at will.
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`ISRN AlgebraVolume 2012 (2012), Article ID 205478, 14 pageshttp://dx.doi.org/10.5402/2012/205478`
Research Article
## The Matrix Linear Unilateral and Bilateral Equations with Two Variables over Commutative Rings
Pidstryhach Institute for Applied Problems of Mechanics and Mathematics, National Academy of Sciences of Ukraine, 3-b, Naukova Street, 79060 L'viv, Ukraine
Received 16 January 2012; Accepted 20 February 2012
Academic Editors: I. Cangul, H. Chen, and P. Damianou
Copyright © 2012 N. S. Dzhaliuk and V. M. Petrychkovych. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
The method of solving matrix linear equations and over commutative Bezout domains by means of standard form of a pair of matrices with respect to generalized equivalence is proposed. The formulas of general solutions of such equations are deduced. The criterions of uniqueness of particular solutions of such matrix equations are established.
#### 1. Preliminaries
##### 1.1. Introduction
The matrix linear equations play a fundamental role in many talks in control and dynamical systems theory [14]. The such equations are the matrix linear bilateral equations with one and two variables and the matrix linear unilateral equations where , , and are matrices of appropriate size over a certain field or over a ring , , are unknown matrices. Equations (1.1), (1.2) are called Sylvester equations. The equation , where matrix is transpose of , is called Lyapunov equation and it is special case of Sylvester equation. Equation (1.3) is called the matrix linear Diophantine equation [3, 4].
Roth [5] established the criterions of solvability of matrix equations (1.1), (1.2) whose coefficients , , and are the matrices over a field .
Theorem 1.1 ([5]). The matrix equation (1.1), where , , and are matrices with elements in a field , has a solution with elements in if and only if the matrices are similar.
The matrix equation (1.2), where , , and are matrices with elements in , has solution , with elements in if and only if the matrices and are equivalent.
The Roth’s theorem was extended by many authors to the matrix equations (1.1), (1.2) in cases where their coefficients are the matrices over principal ideal rings [68], over arbitrary commutative rings [9], and over other rings [1014].
The matrix linear unilateral equation (1.3) has a solution if and only if one of the following conditions holds:(a)a greatest common left divisor of the matrices and is a left divisor of the matrix ;(b)the matrices and are right equivalent.
In the case where , , and in (1.3) are the matrices over a polynomial ring , where is a field, these conditions were formulated in [1, 4]. It is not difficult to show, that these conditions of solvability hold for the matrix linear unilateral equation (1.3) over a commutative Bezout domain.
The matrix equations (1.1), (1.2), (1.3), where the coefficients , , and are the matrices over a field , reduce by means of the Kronecker product to equivalent systems of linear equations [15]. Hence (1.1) over algebraic closed field has unique solution if and only if the matrices and have no common characteristic roots.
One of the methods of solving matrix polynomial equation, where , , and are matrices over a polynomial ring , is based on reducibility of polynomial equation to equivalent equation with matrix coefficients over a field , that is, where and are companion matrices of matrix polynomials and , respectively, is matrix over a field , and is unknown matrix [16, 17].
Equation (1.5) has a unique solution , of bounded degree if and only if [17].
Feinstein and Bar-Ness [18] established that for (1.5), in which at least one from the matrix coefficients or is regular, there exists unique minimal solution , , such that , , if and only if and . The similar result was established in [19] in the case where at least one from matrix coefficients or is regularizable.
In this paper we propose the method of solving matrix linear equations (1.2), (1.3) over a commutative Bezout domain. This method is based on the use of standard form of a pair of matrices with respect to generalized equivalence introduced in [20, 21], and on congruences. We introduce the notion of particular solutions of such matrix equations. We establish the criterions of uniqueness of particular solutions and write down the formulas of general solutions of such equations.
##### 1.2. The Linear Congruences and Diophantine Equations
Let be a commutative Bezout domain. A commutative domain is called a Bezout domain if any two elements , have a greatest common divisor , and , for some , [22, 23]. Note that a commutative domain is a Bezout domain if and only if any finitely generated ideal is principal.
Further, denotes a group of units of , denotes a complete set of residues modulo the ideal generated by element or a complete set of residues modulo . An element of is said to be an associate to an element of , if , where belongs to . A set of elements of , one from each associate class, is said to be a complete set of nonassociates, which we denoted by [24]. For example, if is a ring of integers, then can be chosen as the set of positive integers with zero, that is, , and can be chosen as the set of the smallest nonnegative residues, that is, .
Many properties of divisibility in principal ideal rings [2427] can be easily generalized to the commutative Bezout domain. Recall some of them which will be used later.
In what follows, will always denote a commutative Bezout domain.
Lemma 1.2. Each residue class over can be represented as union where the union is taken over all residues of arbitrary complete set of residues modulo , where .
In the case where is an euclidean ring, this lemma was proved in [27]. By the same way, this lemma can be proved in the case where is a commutative Bezout domain.
The class of elements satisfying the congruence is called solution of this congruence.
Lemma 1.3. Let and , where ,, , . Congruence (1.8) has a solution if and only if .
Let , ,, where , and be a solution of congruence Then the general solution of congruence (1.8) has the form: where is any element of .
Proof. Necessity. It is obvious.
Sufficiency. Let and . Then dividing both sides a congruence (1.8) and by , we get congruence (1.9), where . There exist elements , of such that . Thus we have . Multiply two sides of this congruence by ,, . Therefore, is a solution of congruence (1.9). Set . Then by Lemma 1.2 we get the general solution of congruence (1.8): , where is an arbitrary element of . This proves the lemma.
Corollary 1.4. The congruence (1.8) has unique solution such that if and only if .
Let be a linear Diophantine equation over and . Equation (1.11) has a solution if and only if .
Suppose that , , and , where . Then (1.11) implies Let , be a solution of (1.12), that is, is a solution of congruence , , and . It is easily to verify that if , then .
The solution , of (1.12) is obviously the solution of (1.11).
Definition 1.5. The solution , of (1.11) such that is called the particular solution of this equation.
Then by Lemmas 1.2 and 1.3, a general solution of (1.11) has the form where is an arbitrary element of and is any element of .
Corollary 1.6. A particular solution , of (1.11) with is unique if and only if .
Example 1.7. Let be linear Diophantine equation over ring . Then and (1.14) is solvable. Then the particular solutions of (1.14) are because , ,.
Then the general solution of (1.14) can be written using (1.13) that is, where is arbitrary element of and is any element of .
##### 1.3. Standard Form of a Pair of Matrices
Let be a commutative Bezout domain with diagonal reduction of matrices [28], that is, for every matrix of the ring of matrices , there exist invertible matrices , such that If , , then the matrix is unique and is called the canonical diagonal form (Smith normal form) of the matrix . Such rings are so-called adequate rings. The ring is called an adequate if is a commutative domain in which every finitely generated ideal is principal and for every , with ; can be represented as where and for every nonunit factor of [29].
Definition 1.8. The pairs and of matrices , , , are called generalized equivalent pairs if , , for some invertible matrices and over .
In [20, 21], the forms of the pair of matrices with respect to generalized equivalence are established.
Theorem 1.9. Let be an adequate ring, and let , be the nonsingular matrices and be their canonical diagonal forms.
Then the pair of matrices is generalized equivalent to the pair , where has the following form: , where , , , .
The pair defined in Theorem 1.9 is called the standard form of the pair of matrices or the standard pair of matrices .
Definition 1.10. The pair is called diagonalizable if it is generalized equivalent to the pair of diagonal matrices , that is, its standard form is the pair of diagonal matrices .
Corollary 1.11. Let , . If , then the pair of matrices is diagonalizable.
It is clear taking into account Corollary 1.11 that if , then the standard form of matrices is the pair of diagonal matrices .
Let us formulate the criterion of diagonalizability of the pair of matrices.
Theorem 1.12. Let , and be a nonsingular matrix. Then the pair of matrices is generalized equivalent to the pair of diagonal matrices if and only if the matrices and are equivalent, where is an adjoint matrix.
#### 2. The Matrix Linear Unilateral Equations 𝐴 𝑋 + 𝐵 𝑌 = 𝐶
##### 2.1. The Construction of the Solutions of the Matrix Linear Unilateral Equations with Two Variables
Suppose that the matrix linear unilateral equation (1.3) is solvable, and let be a standard form of a pair of matrices from (1.3) with respect to generalized equivalence, that is, is a lower triangular matrix of the form (1.20) with the principal diagonal where , , .
Then (1.3) is equivalent to the equation where , , and .
The pair of matrices , satisfying (2.3) is called the solution of this equation. Then is the solution of (1.3).
The matrix equation (2.3) is equivalent to the system of linear equation: with the variables , , , , where , ,, from (2.3), or where , , and .
The solving of this system reduces to the successive solving of linear Diophantine equations of the form
Using solutions of system (2.6), we construct the solutions , of matrix equation (2.3). Then and are the solutions of matrix equation (1.3).
##### 2.2. The General Solution of the Matrix Equation 𝐴 𝑋 + 𝐵 𝑌 = 𝐶 with the Diagonalizable Pair of Matrices ( 𝐴 , 𝐵 )
Suppose that the pair of matrices is diagonalizable, that is, for some matrices , , .
Then (1.3) is equivalent to the equation where , , and .
From matrix equation (2.9), we get the system of linear Diophantine equation:
Let , , , , be a particular solution of corresponding equation of system (2.10), that is, is the solution of congruence , , and .
The general solution of corresponding equation of system (2.10) by the formula (1.13) will have the following form: where , are arbitrary elements of , and are any elements of , , . The particular solution of matrix equation (2.9) is where , , , , is a particular solution of corresponding equation of system (2.10). Then is a particular solution of matrix equation (1.3).
Thus we get the following theorem.
Theorem 2.1. Let the pair of matrices from matrix equation (1.3) be diagonalizable and its standard pair be the pair of matrices in the form (2.8). Let , be a particular solution of matrix equation (2.9). Then the general solution of matrix equation (2.9) is where , are arbitrary elements of , ; , , , ; , are arbitrary elements of .
The general solution of matrix equation (1.3) has the form
Example 2.2. Consider the equation for the matrices are matrices over and are unknown matrices.
The matrix equation (2.16) is solvable.
The pair of matrices from matrix equation (2.16) by Theorem 1.12 is diagonalizable, since the matrices are equivalent. Therefore, where
Then (2.16) is equivalent to the equation where
From matrix equation (2.22), we get the system of linear Diophantine equations:
The particular solution of each linear equation of system (2.24) has the following form:
The particular solution of matrix equation (2.22) is
Then by (2.14) the general solution of matrix equation (2.22) is or where is from is arbitrary element of , and , , , is arbitrary elements of .
Finally, the general solution of matrix equation (2.16) is
##### 2.3. The Uniqueness of Particular Solutions of the Matrix Linear Unilateral Equation
The conditions of uniqueness of solutions of bounded degree (minimal solutions) of matrix linear polynomial equations (1.5) were found in [1619]. We present the conditions of uniqueness of particular solutions of matrix linear equation over a commutative Bezout domain .
Theorem 2.3. The matrix equation (2.3) has a unique particular solution such that , , , if and only if .
Proof. From matrix equation (2.3), we get the system of linear equations (2.6). The solving of this system reduces to the successive solving of the linear Diophantine equations of the form (2.7).
The matrix equation (2.3) has a unique particular solution , such that , , , if and only if each linear Diophantine equations of the form (2.7) has a unique particular solution , such that , , . By Corollary 1.6, this holds if and only if for all . It follows that . This completes the proof.
Theorem 2.4. Let , , where ,, be a unique particular solution of matrix equation (2.3).
Then the general solution of matrix equation (2.3) is where and are canonical diagonal forms of and from matrix equation (1.3), respectively, , are arbitrary elements of , , .
The general solution of matrix equation (1.3) is the pair of matrices
Proof. The particular solution of the form (2.30) of (2.3) is unique if and only if that is, . Then by Corollary 1.11 the pair of matrices is diagonalizable and (1.3) gives us the equation of the form (2.9).
Thus by Theorem 2.1, we get the formula (2.31) of the general solution of (2.3) and the formula (2.32) for computation of general solution of (1.3) in the case where (2.3) has unique particular solution of the form (2.30). The theorem is proved.
#### 3. The Matrix Linear Bilateral Equations 𝐴 𝑋 + 𝑌 𝐵 = 𝐶
Consider the matrix linear bilateral equation (1.2), where , , and are matrices over a commutative Bezout domain , and are the canonical diagonal forms of matrices and , respectively, and , , , .
Then (1.2) is equivalent to where , , and .
Such an approach to solving (1.2), where , and are the matrices over a polynomial ring , where is a field, was applied in [3].
The equation (3.2) is equivalent to the system of linear Diophantine equations
Theorem 3.1. Let be a particular solution of matrix equation (3.2), that is, , , , , are particular solutions of linear Diophantine equations of system (3.3).
The general solution of matrix equation (3.2) is where , , where are arbitrary elements of , and , where are arbitrary elements of , , .
The general solution of matrix equation (1.2) is
Similarly as for (2.3), we prove that particular solution of (3.2) is unique if and only if . Then by the same way as for (1.3) we write down the general solution of matrix equation (1.2).
Theorem 3.2. Suppose that and , where , , is unique particular solution of matrix equation (3.2) and are canonical diagonal forms of matrices , from matrix equation (1.2), respectively.
Then the general solution of matrix equation (3.2) is where ; are arbitrary elements of , , .
The general solution of matrix equation (1.2) is
#### References
1. T. Kaczorek, Polynomial and Rational Matrices, Communications and Control Engineering Series, Springer, London, UK, 2007.
2. V. Kučera, “Algebraic theory of discrete optimal control for single-variable systems. I. Preliminaries,” Kybernetika, vol. 9, pp. 94–107, 1973.
3. V. Kučera, “Algebraic theory of discrete optimal control for multivariable systems,” Kybernetika, vol. 10/12, supplement, pp. 3–56, 1974.
4. W. A. Wolovich and P. J. Antsaklis, “The canonical Diophantine equations with applications,” SIAM Journal on Control and Optimization, vol. 22, no. 5, pp. 777–787, 1984.
5. W. E. Roth, “The equations $AX-YB=C$ and $AX-XB=C$ in matrices,” Proceedings of the American Mathematical Society, vol. 3, pp. 392–396, 1952.
6. M. Newman, “The Smith normal form of a partitioned matrix,” Journal of Research of the National Bureau of Standards, vol. 78, pp. 3–6, 1974.
7. C. R. Johnson and M. Newman, “A condition for the diagonalizability of a partitioned matrix,” Journal of Research of the National Bureau of Standards, vol. 79, no. 1-2, pp. 45–48, 1975.
8. R. B. Feinberg, “Equivalence of partitioned matrices,” Journal of Research of the National Bureau of Standards, vol. 80, no. 1, pp. 89–97, 1976.
9. W. H. Gustafson, “Roth's theorems over commutative rings,” Linear Algebra and Its Applications, vol. 23, pp. 245–251, 1979.
10. R. E. Hartwig, “Roth's equivalence problem in unit regular rings,” Proceedings of the American Mathematical Society, vol. 59, no. 1, pp. 39–44, 1976.
11. W. H. Gustafson and J. M. Zelmanowitz, “On matrix equivalence and matrix equations,” Linear Algebra and Its Applications, vol. 27, pp. 219–224, 1979.
12. R. M. Guralnick, “Roth's theorems and decomposition of modules,” Linear Algebra and Its Applications, vol. 39, pp. 155–165, 1981.
13. L. Huang and J. Liu, “The extension of Roth's theorem for matrix equations over a ring,” Linear Algebra and Its Applications, vol. 259, pp. 229–235, 1997.
14. R. E. Hartwig and P. Patricio, “On Roth's pseudo equivalence over rings,” Electronic Journal of Linear Algebra, vol. 16, pp. 111–124, 2007.
15. P. Lancaster and M. Tismenetsky, The Theory of Matrices, Computer Science and Applied Mathematics, Academic Press, Orlando, Fla, USA, 2nd edition, 1985.
16. S. Barnett, “Regular polynomial matrices having relatively prime determinants,” Proceedings of the Cambridge Philosophical Society, vol. 65, pp. 585–590, 1969.
17. V. Petrychkovych, “Cell-triangular and cell-diagonal factorizations of cell-triangular and cell-diagonal polynomial matrices,” Mathematical Notes, vol. 37, no. 6, pp. 431–435, 1985.
18. J. Feinstein and Y. Bar-Ness, “On the uniqueness of the minimal solution to the matrix polynomial equation $A\left(\lambda \right)X\left(\lambda \right)+Y\left(\lambda \right)B\left(\lambda \right)=C\left(\lambda \right)$,” Journal of the Franklin Institute, vol. 310, no. 2, pp. 131–134, 1980.
19. V. M. Prokip, “About the uniqueness solution of the matrix polynomial equation $A\left(\lambda \right)X\left(\lambda \right)-Y\left(\lambda \right)B\left(\lambda \right)=C\left(\lambda \right)$,” Lobachevskii Journal of Mathematics, vol. 29, no. 3, pp. 186–191, 2008.
20. V. Petrychkovych, “Generalized equivalence of pairs of matrices,” Linear and Multilinear Algebra, vol. 48, no. 2, pp. 179–188, 2000.
21. V. Petrychkovych, “Standard form of pairs of matrices with respect to generalized eguivalence,” Visnyk of Lviv University, vol. 61, pp. 153–160, 2003.
22. P. M. Cohn, Free Rings and Their Relations, Academic Press, London, UK, 1971.
23. S. Friedland, “Matrices over integral domains,” in CRC Handbook of Linear Algebra, pp. 23-1–23-11, Chapman & Hall, New York, NY, USA, 2007.
24. M. Newman, Integral Matrices, Academic Press, New York, NY, USA, 1972.
25. B. L. Van der Waerden, Algebra, Springer, New York, NY, USA, 1991.
26. A. I. Borevich and I. R. Shafarevich, Number Theory, vol. 20 of Translated from the Russian by Newcomb Greenleaf. Pure and Applied Mathematics, Academic Press, New York, NY, USA, 1966.
27. K. A. Rodossky, Euclid's Algorithm, Nauka, Moscow, Russia, 1988.
28. I. Kaplansky, “Elementary divisors and modules,” Transactions of the American Mathematical Society, vol. 66, pp. 464–491, 1949.
29. O. Helmer, “The elementary divisor theorem for certain rings without chain condition,” Bulletin of the American Mathematical Society, vol. 49, pp. 225–236, 1943.
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Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
Paper:
TR19-128 | 24th September 2019 05:58
Lower Bounds for (Non-monotone) Comparator Circuits
TR19-128
Authors: Anna Gal, Robert Robere
Publication: 24th September 2019 05:59
Downloads: 1040
Keywords:
Abstract:
Comparator circuits are a natural circuit model for studying the concept of bounded fan-out computations, which intuitively corresponds to whether or not a computational model can make "copies" of intermediate computational steps. Comparator circuits are believed to be weaker than general Boolean circuits, but they can simulate Branching Programs and Boolean formulas. In this paper we prove the first superlinear lower bounds in the general (non-monotone) version of this model for an explicitly defined function. More precisely, we prove that the \$n\$-bit Element Distinctness function requires \$\Omega( (n/ \log n)^{3/2})\$ size comparator circuits.
ISSN 1433-8092 | Imprint
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# Subsetting from seurat object based on orig.ident?
I am pretty new to Seurat. I want to subset from my original seurat object (BC3) meta.data based on orig.ident. however, when i use subset(), it returns with Error.
ER_HER_P <- subset(BC3, idents = c("BC03"))
Error: No cells found
table(Idents(BC3))
BC01 BC02 BC03 BC03LN BC04 BC05 BC06 BC07 BC07LN BC08 BC09 BC10 BC11
21 50 33 51 42 74 15 48 52 21 55 13 11
There are 33 cells under the identity. I checked the active.ident to make sure the identity has not shifted to any other column, but still I am getting the error?
head(BC3@active.ident)
[1] BC01 BC01 BC01 BC01 BC01 BC01
Levels: BC01 BC02 BC03 BC03LN BC04 BC05 BC06 BC07 BC07LN BC08 BC09 BC10 BC11
• A stupid suggestion, but did you try to give it as a string ? just "BC03" ? Other option is to get the cell names of that ident and then pass a vector of cell names. cells = as.vector(rownames(object@meta.data[object@meta.data\$idents=="BC03",])) Jul 22, 2020 at 13:46
• Thank you for the suggestion. But it didnt work.. Jul 22, 2020 at 15:19
• It worked with M <- subset(x = BC3, subset = orig.ident == "BC03") Jul 22, 2020 at 15:35
If you are going to use idents like that, make sure that you have told the software what your default ident category is.
This works for me, with the metadata column being called "group", and "endo" being one possible group there.
Idents(combined.all) <- "group"
endo_subset <- subset(combined.all, idents = c("endo"))
I think this is basically what you did, but I think this looks a little nicer
• I would just add you can run Idents(combined.all) to see which groups are in your identity before subsetting by one of them. Dec 21, 2020 at 14:55
A bit too long to type as a comment.
To use subset on a Seurat object, (see ?subset.Seurat) , you have to provide:
Subset a Seurat object
Description:
Subset a Seurat object
Usage:
## S3 method for class 'Seurat'
x[i, j, ...]
## S3 method for class 'Seurat'
subset(x, subset, cells = NULL, features = NULL, idents = NULL, ...)
Arguments:
x: Seurat object to be subsetted
i, features: A vector of features to keep
j, cells: A vector of cells to keep
So you either use the matrix to subset:
library(Seurat)
data(pbmc_small)
Idents(pbmc_small) = paste0("BC",Idents(pbmc_small))
table(Idents(pbmc_small))
BC0 BC2 BC1
36 19 25
test = pbmc_small[,Idents(pbmc_small)=="BC0"]
table(Idents(test))
BC0
36
Or you provide the cells:
subset(pbmc_small,cells=colnames(pbmc_small)[Idents(pbmc_small)=="BC0"])
An object of class Seurat
230 features across 36 samples within 1 assay
Active assay: RNA (230 features, 20 variable features)
2 dimensional reductions calculated: pca, tsne
What you have should work, but try calling the actual function (in case there are packages that clash):
Seurat:::subset.Seurat(pbmc_small,idents="BC0")
An object of class Seurat
230 features across 36 samples within 1 assay
Active assay: RNA (230 features, 20 variable features)
2 dimensional reductions calculated: pca, tsne
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# Can you find the next entry in this sequence ?
Find a rule that applies for the first six entries and add at least one more.
I
do
take
20,000¥
recompensations
The rule is simple.
Each individual entry has multiple valid solution, this is just one such set. The fact that the next entry is longer than the previous one doesn't need to be true for all entries.
• Next words: in lieu of pre-determined donut payment. – Ian MacDonald Apr 20 '17 at 15:47
• @IanMacDonald why would you think that? I'm curious. – Vepir Apr 20 '17 at 15:49
• Is the trailing space after recompensations significant? – Scott M Apr 20 '17 at 15:49
• @ScottM Yes it is. – Vepir Apr 20 '17 at 15:50
• I cannot see any meaningful similarities amongst the words apart from them grammatically forming the beginning of a sentence when taken in the order provided. I have simply chosen to finish the sentence as though this were a court case involving a dispute about missing donuts. – Ian MacDonald Apr 20 '17 at 15:53
• This leads me to believe that in lieu of agreed donut payment. would also be an acceptable solution. If that is the case, this puzzle is not specific enough and has far too many possible correct answers. – Ian MacDonald Apr 20 '17 at 17:10
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# Lambert's Cosine Law - Simple algebraic rearrangement problem [closed]
Let $I(x,y)$ be the intensity at position $(x,y)$, $\rho(x,y)$ be the albedo at this point (a constant), $n=(p,q,1)$ be the surface gradient vector and $s_{0},s_{1},s_{2}$ all be vectors from the point to $(x,y)$.
Assume that $I(x,y)=I_{1}$ when $s_{0}=(1,1,0)$ and similarly, $I(x,y)=I_{2}$ when $s_{1}=(1,0,1)$ and $I(x,y)$ = $I_{3}$ when $s_{2}=(0,1,0)$
Lambert's cosine Law is then:
$I(x,y)=\rho(x,y)\dfrac{s_{x}p+s_{y}q+s_{z}}{|n||s|}$
We can calculate the following. $|n|=\sqrt{p^{2}+q^{2}+1}$ and $|s_{0}|=|s_{1}|=|s_{2}|=1$.
All I would like to have are equations for $p$ and $q$ in terms of $I_{1},I_{2},I_{3}$.
-
So you know the values of $s_xp+s_yq+s_z$ for three linearly independent vectors $s$. Linear algebra does the rest. – darij grinberg Mar 31 '11 at 21:17
I don't understand how $s_0=(1,1,0)$ is compatible with $|s_0|=1$. – Gerry Myerson Mar 31 '11 at 23:38
darij: Not quite. He knows your quantities divided by a function of p and q. He needs a little more than linear algebra to find p and q. Gerhard "Ask Me About System Design" Paseman, 2011.03.31 – Gerhard Paseman Apr 1 '11 at 1:18
The way the question is phrased, I actually cannot make out the hypotheses. What does "vectors from the point to (x,y)" mean? Why do the specified s_0 and s_1 have length 1? – Will Jagy Apr 1 '11 at 4:20
You might write out what you think are the three equations giving I1, I2, and I3 in terms of p and q, filling in as much of the known values as you can. We might then tell you where your arithmetic is wrong, or we might come up with the answer for you. In any case, this is looking more like something for math.stackexchange than for MathOverflow. Gerhard "Ask Me About System Design" Paseman, 2011.04.01 – Gerhard Paseman Apr 1 '11 at 9:20
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# Clustering With K-Means in Python
A very common task in data analysis is that of grouping a set of objects into subsets such that all elements within a group are more similar among them than they are to the others. The practical applications of such a procedure are many: given a medical image of a group of cells, a clustering algorithm could aid in identifying the centers of the cells; looking at the GPS data of a user’s mobile device, their more frequently visited locations within a certain radius can be revealed; for any set of unlabeled observations, clustering helps establish the existence of some sort of structure that might indicate that the data is separable.
### Mathematical background
The k-means algorithm takes a dataset X of N points as input, together with a parameter K specifying how many clusters to create. The output is a set of K cluster centroids and a labeling of X that assigns each of the points in X to a unique cluster. All points within a cluster are closer in distance to their centroid than they are to any other centroid.
The mathematical condition for the K clusters $C_k$ and the K centroids $\mu_k$ can be expressed as:
Minimize $\displaystyle \sum_{k=1}^K \sum_{\mathrm{x}_n \in C_k} ||\mathrm{x}_n - \mu_k ||^2$ with respect to $\displaystyle C_k, \mu_k$.
### Lloyd’s algorithm
Finding the solution is unfortunately NP hard. Nevertheless, an iterative method known as Lloyd’s algorithm exists that converges (albeit to a local minimum) in few steps. The procedure alternates between two operations. (1) Once a set of centroids $\mu_k$ is available, the clusters are updated to contain the points closest in distance to each centroid. (2) Given a set of clusters, the centroids are recalculated as the means of all points belonging to a cluster.
$\displaystyle C_k = \{\mathrm{x}_n : ||\mathrm{x}_n - \mu_k|| \leq \mathrm{\,\,all\,\,} ||\mathrm{x}_n - \mu_l||\}\qquad(1)$
$\displaystyle \mu_k = \frac{1}{C_k}\sum_{\mathrm{x}_n \in C_k}\mathrm{x}_n\qquad(2)$
The two-step procedure continues until the assignments of clusters and centroids no longer change. As already mentioned, the convergence is guaranteed but the solution might be a local minimum. In practice, the algorithm is run multiple times and averaged. For the starting set of centroids, several methods can be employed, for instance random assignation.
Below is a simple implementation of Lloyd’s algorithm for performing k-means clustering in python:
import numpy as np
def cluster_points(X, mu):
clusters = {}
for x in X:
bestmukey = min([(i[0], np.linalg.norm(x-mu[i[0]])) \
for i in enumerate(mu)], key=lambda t:t[1])[0]
try:
clusters[bestmukey].append(x)
except KeyError:
clusters[bestmukey] = [x]
return clusters
def reevaluate_centers(mu, clusters):
newmu = []
keys = sorted(clusters.keys())
for k in keys:
newmu.append(np.mean(clusters[k], axis = 0))
return newmu
def has_converged(mu, oldmu):
return (set([tuple(a) for a in mu]) == set([tuple(a) for a in oldmu])
def find_centers(X, K):
# Initialize to K random centers
oldmu = random.sample(X, K)
mu = random.sample(X, K)
while not has_converged(mu, oldmu):
oldmu = mu
# Assign all points in X to clusters
clusters = cluster_points(X, mu)
# Reevaluate centers
mu = reevaluate_centers(oldmu, clusters)
return(mu, clusters)
### Clustering in action
Let’s see the algorithm in action! For an ensemble of 100 random points on the plane, we set the k-means function to find 7 clusters. The code converges in 7 iterations after initializing with random centers. In the following plots, dots correspond to the target data points X and stars represent the centroids $\mu_k$ of the clusters. Each cluster is distinguished by a different color.
The initial configuration of points for the algorithm is created as follows:
import random
def init_board(N):
X = np.array([(random.uniform(-1, 1), random.uniform(-1, 1)) for i in range(N)])
return X
For a configuration with twice as many points and a target of 3 clusters, often the algorithm needs more iterations to converge.
Obviously, an ensemble of randomly generated points does not possess a natural cluster-like structure. To make things slightly more tricky, we want to modify the function that generates our initial data points to output a more interesting structure. The following routine constructs a specified number of Gaussian distributed clusters with random variances:
def init_board_gauss(N, k):
n = float(N)/k
X = []
for i in range(k):
c = (random.uniform(-1, 1), random.uniform(-1, 1))
s = random.uniform(0.05,0.5)
x = []
while len(x) < n:
a, b = np.array([np.random.normal(c[0], s), np.random.normal(c[1], s)])
# Continue drawing points from the distribution in the range [-1,1]
if abs(a) < 1 and abs(b) < 1:
x.append([a,b])
X.extend(x)
X = np.array(X)[:N]
return X
Let us look at a data set constructed as X = init_board_gauss(200,3): 7 iterations are needed to find the 3 centroids.
If the target distribution is disjointedly clustered and only one instantiation of Lloyd’s algorithm is used, the danger exists that the local minimum reached is not the optimal solution. This is shown in the example below, where initial data using very peaked Gaussians is constructed:
The yellow and black stars serve two different clusters each, while the orange, red and blue centroids are cramped within one unique blob due to an unfortunate random initialization. For this type of cases, a cleverer election of initial clusters should help.
To finalize our table-top experiment on k-means clustering, we might want to take a look at what happens when the original data space is densely populated:
The k-means algorithm induces a partition of the observations corresponding to a Voronoi tessellation generated by the K centroids. And it is certainly very pretty!
### Table-top data experiment take-away message
Lloyd’s two-step implementation of the k-means algorithm allows to cluster data points into groups represented by a centroid. This technique is employed in many facets of machine learning, from unsupervised learning algorithms to dimensionality reduction problems. The general clustering problem is NP hard, but the iterative procedure converges always, albeit to a local minimum. Proper initialization of the centroids is important. Additionally, this algorithm does not supply information as to which K for the k-means is optimal; that has to be found out by alternative methods.
Update: We explore the gap statistic as a method to determine the optimal K for clustering in this post: Finding the K in K-Means Clustering and the $f(K)$ method: Selection of K in K-means Clustering, Reloaded.
Update: For a proper initialization of the centroids at the start of the k-means algorithm, we implement the improved k-means++ seeding procedure.
1. Pete Colligan (@linuxster)
Hello,
You can export your ipython notebook markup and upload it to github gist. Then reference the gist URL from the http://nbviewer.ipython.org/ you mentioned above.
you can also just email me (linuxster at gmail dot com) source tarball or post your source on github and I can clone.
**I have put an example of ILP tutorial notebooks here: http://nbviewer.ipython.org/gist/linuxster/7295256
currently I am focusing on the data characteristics that feed the algorithm, especially in the case where I might automate as much as possible and therefore must take into account pre-processing (missing data, outliers, encoding, feature selection, etc)
maybe you have experience here already?
1. I am interested in exploring the effects of L1 and L2 norm on the cluster output. in your example you use the L2 norm but since Kmeans is sensitive to outliers maybe L1 norm is better?
2. also I want to explore whether I can use the mode instead of mean for the centroid recalculation in step two of LLoyds algorithm. again, something to minimize impact of outliers.
3. also if I have mixed data (categorical and numerical) that I would cluster together what is the best approach for pre-processing the data? I have heard of some folks encoding the categorical data into a binary vector format to support the distance calculations but would like to get other opinions here as well.
just saw there is a follow up post to this one on k selection. will read. 🙂
• datasciencelab
Thanks! I will set up my github/gist soon and will keep you posted.
Your discussion on L1/L2 norm is very interesting. What I’ve read so far always uses L2 to measure the intracluster variance, but exploring L1 is surely worthwhile. Re: mode versus mean, I guess it depends on the distribution of your data. My play data just follows normal distribution or random, thus I don’t think that the mode brings much more than the mean. It’d be interesting to see what happens when data follows other shapes though. Regarding mixed data, I can’t offer much insight there. What kind of categorical data do you have in mind?
2. Pete Colligan (@linuxster)
I have data from a customer which includes retail store characteristics. They would like to use many store attributes to help them find new store “groupings” that they an use for business planning activities. There are about 50 attributes per store and data is numerical, ordinal, and categorical in nature. (size, nearest competitor, demographic info, sales, market segment, etc.). I would like to first:
1. get an idea of which attributes account for the most variance in the data. PCA is ok but I get a bit lost mapping from one basis to the other. (ex. which PCA component contains which features from my dataset). Goal here is some dimensionality reduction since I know some of my features are collinear.
2. after certain “redundant” information is removed, I think I should look at the data distribution of each feature to get an idea of the “nature” of the data. not quite sure here any hard and fast rules. Ex. Normalization, Regularization?
3. then I would go for clustering algorithm. At the moment I like Kmeans and agglomerative hierarchical.
what do you think about that thought process?
note: Kmeans will force every point to a cluster. Outliers are not ALWAYS bad in this use case (some stores like “super stores” are important to keep and might be an own cluster) so I don’t want to always just remove them prior to running the algorithm.
• datasciencelab
I see that your problem is another kind of beast, as you probably need to start with a multivariate analysis to determine the relative importance of each attribute (your step 1, for which you could do some ANOVA) before you can proceed to clustering on them. Your strategy seems correct, and due to the complexity of the problem, I’m sure you’ll find many interesting steps to complete along the way. Good luck!
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5. nico
Hi there,
big thanks for all of that, seems to be exactly what I was looking for. I am not that new to python, but new to clustering but i guess I will find my way to use the implementations on my data.
One thing I was wondering is what would be the best way to keep track of the sample id’s, which is of importance in my research? Can someone point me in a direction?
Thanks
6. Lars
Thank you for this excellent set of examples. K-Means and Lloyd’s Algorithm are probably what I’m looking for. I’m trying to investigate a cluster problem that has a twist. Specifically, I would like to find good locations for public schools, based on the geographical distribution of students. I can estimate the X-Y coordinates of the students on a city map, and could do a simple cluster analysis. However the problem is complicated by the fact that the city has many internal barriers (waterways, railway tracks, fenced expressways, etc) which divide the city into many smaller areas, and which can only be crossed at a limited number of points. This means that two points whose x-y coordinates might be close, could in fact be separated by a barrier that forces a much greater travel distance.
The distance between a point in area A, and another point in Area B, is in fact the sum of their distances to the bridge point across the barrier. (rather than simply the distance calculated from their x-y coordinates.) That is fairly straight forward. My thinking is that I should try to intercept the cluster processing at some midpoint and substitute the practical distance for the distance calculated from the x-y coordinates, and then proceed with the analysis.
I have some experience with Python, but I have no formal training in Cluster Analysis.
My questions are:
1. Is it possible to intercept the processing and substitute different distance values.
2. Does this type of problem have a standard solution, and I simply haven’t found it.
3. Do you recommend some other approach.
Thanks,
7. vineeth
Thanks for the example. How can i implement k-means algorithm for finding severe and normal facial palsy using python.
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9. joon
Thanks a lot for the post. It seems closing ) is missing from the line: return (set([tuple(a) for a in mu]) == set([tuple(a) for a in oldmu]) in def has_converged(mu, oldmu).
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13. kevin
Do you know why I am getting the error “global name ‘zeros’ is not defined”?
Thanks
Kevin
14. Jiajun
Hi, Thanks for this amazing example.
But I really struggle to understand:
bestmukey = min([(i[0], np.linalg.norm(x – mu[i[0]])) \
for i in enumerate(mu)], key = lambda t: t[1])[0]
I know what it is trying to achieve: outputting the key (index) of mu, which has the shortest distance to x. I think my struggle is to understand the use of min() and enumerate(). Is it possible you can rewrite it in a normal loop? Appreciate it!!
J
15. phil
yes, can you explain that bestmukey line for non python users
the goal is to explain KMeans, so why do you use such a non standard syntax, come on
16. Master
Does anyone know how to label the dots in the cluster ? I want to label some dots in the cluster, but I don’t know how.
• nico
@Master, I did asked sort of that question a while ago (July 24, 2014), without an answer.
After the clustering is done, my workaround consist of comparing each element in the ‘cluster’ dictionary to the input list where the data-to-id connection is known. The id gets transferred to a similar structured dictionary like ‘clusters’, which in the end is holding all the id’s. Might be far from elegant, but makes annotation via matplotlib easy…
17. mannyglover
I really appreciate this post, but can you post code to show how to visualize? I know it can’t be uber-complicated, but it would be nice to see the code that you used to visualize the data. Thanks.
18. P A
I appreciate some help. When I tried running the code, I get the error below. How should i be handling X to address this error message? I am afraid that I am new to both python and numpy. Very familiar with matlab.
Thanks!
ERROR:
line 33, in find_centers
oldmu = random.sample(X, K)
File “/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/random.py”, line 311, in sample
raise TypeError(“Population must be a sequence or set. For dicts, use list(d).”)
TypeError: Population must be a sequence or set. For dicts, use list(d).
• datasciencelab
The problem is compatibility between python2 (the code was written for python2) and your running python3.
To get it running in python3 please substitute
oldmu = random.sample(X, K)
by
random.sample(list(X), 3)
19. kalij
I need some help. When I tried running the code, I get the error below.
Can you explain, what the reason for it can be and how can i solve this problem?
In this part of the code:
def has_converged(mu, oldmu):
return (set([tuple(a) for a in mu]) == set([tuple(a) for a in oldmu])
File “”, line 2
return (set([tuple(a) for a in mu]) == set([tuple(a) for a in oldmu])
SyntaxError: invalid syntax
Thanks!
• datasciencelab
There’s a missing closing parenthesis in the code. The correct version is:
def has_converged(mu, oldmu):
return (set([tuple(a) for a in mu]) == set([tuple(a) for a in oldmu]))
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21. Kate32
Good example!
But how can I actually make a visualization?
You have attached these pictures, so can you show me the code, please?
thanks
22. Greg S
HI, how do I define mu at the start of the code? What should this be initializes to? Thanks
23. boop
I’m also hoping to see the code for the visualization, that would be great… thanks!
24. sb
I ran the code for k=3 and the output was only two clusters. Is this normal in k-means?
25. Alex Luya
Hello,@datasciencelab,you said “Sure! I just need to set up a convenient way to share”,but I can’t find it,did it miss something?
26. sb
Why are you passing ‘mu’ to the ‘reevaluate_centers’ function? It is not being used to calculate the ‘newmu’.
• Ole
As a beginner in python this is certainly not the best way to plot this, but at least I found a way…:
Starting from
X = init_board_gauss(1000, 4)
# number of centers n
n = 4
Y = find_centers(X, n)
# get the x and y for the centers and put them into a pandas dataframe
xcenter, ycenter = zip(*Y[0])
center = np.column_stack((xcenter, ycenter))
df_center = pd.DataFrame(data=center, columns=(‘x’, ‘y’))
# do the same for the clusters (dictionary with dataframes)
df_cluster = dict()
for i in range(0, (len(Y[1]))):
print (i)
xtemp, ytemp = zip(*Y[1][i])
cluster = np.column_stack((xtemp, ytemp))
df_cluster[i] = pd.DataFrame(data=cluster, columns=(‘x’, ‘y’))
# and now plot it
for i in range(0, (len(df_cluster))):
plt.scatter(df_cluster[i].x, df_cluster[i].y)
plt.scatter(df_center.x, df_center.y, marker=’X’)
plt.show()
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29. fredo68
In the last example, honestly, I don’t see any cluster. The points look way to evenly distributed to show any concentration. We should not see clusters where there is none. Some techniques should be able to determine that.
30. windson
To people who wanna understand
for x in X:
bestmukey = min([(i[0], np.linalg.norm(x-mu[i[0]])) \
for i in enumerate(mu)], key=lambda t:t[1])[0]
X means all points you wanna cluster
for example:
[[1, 2], [2, 3], [3, 4]]
mu means the random points you choose from X
line27 mu = random.sample(X, K)
We break the code into serveral pieces
1. min(FOO, key=lambda t:t[1])[0]
2. FOO = [(i[0], BAR) for i in enumerate(mu)]
3. BAR = np.linalg.norm(x-mu[i[0]])
line3 is use for calculate the distance between two points.
We need to find out which point is nearest, so every x have to calculate the distance between all of the mu, line2 is equal to
lst = []
for i in enumerate(mu):
lst.append((i[0], BAR))
# The distance between the point x and the first mu is d1, the second is d2,
# So we got all_di = [(0, d1), (1, d2), (2, d3)],
x belongs to the nearest point, so line1 compare all_di by its second element, finallywe get the the index of points which x belongs to, like this
# finish for x in X:
{
0: [1,2,3],
1: [4,5,6],
2: [7,8,9]
}
point 1,2,3 belongs to the first cluster…
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# Institut for Matematik
## Reconstruction of singular and degenerate inclusions in Calderón's problem
Publikation: Bidrag til tidsskrift/Konferencebidrag i tidsskrift /Bidrag til avisTidsskriftartikelForskningpeer review
We consider the reconstruction of the support of an unknown perturbation to a known conductivity coefficient in Calderón's problem. In a previous result by the authors on monotonicity-based reconstruction, the perturbed coefficient is allowed to simultaneously take the values $0$ and $\infty$ in some parts of the domain and values bounded away from $0$ and $\infty$ elsewhere. We generalise this result by allowing the unknown coefficient to be the restriction of an $A_2$-Muckenhoupt weight in parts of the domain, thereby including singular and degenerate behaviour in the governing equation. In particular, the coefficient may tend to $0$ and $\infty$ in a controlled manner, which goes beyond the standard setting of Calderón's problem. Our main result constructively characterises the outer shape of the support of such a general perturbation, based on a local Neumann-to-Dirichlet map defined on an open subset of the domain boundary.
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Adding an edge to a maximal planar graph results in topological minors of both $K_5,K_{3,3}$.
I was trying to prove this exercise from Diestel's book:
Show that adding a new edge to a maximal planar graph of order at least 6 always produces both a $$TK_5$$ and a $$TK_{3,3}$$ subgraph.
I used a hint from Diestel's book to get a topological minor of $$K_5$$, but I had trouble finding the $$TK_{3,3}$$. I found a solution here (The $$K_5$$-part is the same thing I did).
However, I don't understand a part of it when they try to find the $$TK_{3,3}$$ subgraph. They say:
Since $$G$$ has order at least $$6$$, there is another vertex $$z$$ distinct from those previously mentioned. The construction allows for two cases: either $$z$$ lies outside the region bounded by the topological cycle $$vu_1wu_2v$$ or it lies inside one of the faces of this region (they are all equivalent).
My question is why cannot $$z$$ be in the boundary of $$vu_1wu_2v$$? And it that were possible, the three disjoint paths from $$z$$ to $$u_1,u_2,u_3$$ might not be disjoint to the previous drawn paths, which seems to cause trouble when building the $$K_{3,3}$$-subdivision.
Since $$v$$ is not adjacent to $$w$$ (else we could add no edge between them), we can find vertices $$u_1$$, $$u_2$$, and $$u_3$$ lying one on each path that are neighbors of $$v$$ but not of $$w$$.
If a path from $$v$$ to $$w$$ has length $$2$$ then a neighbor $$u$$ of $$v$$ at the path is a neighbor of $$w$$.
By the edge-maximality of $$G$$, $$u_1$$, $$u_2$$, and $$u_3$$ induce a cycle.
This is true, if $$v$$ has degree three. Otherwise it may fail (for instance, see the picture).
• You're right. Fortunately, I had modified that part to just take the three paths in such a way each of them only contained one neighbor of $v$. About the use of the edge maximality of $G$, I think that to produce the subdivision of $K_5$ we only need to make sure all the neighbors of $v$ form a cycle. – Nell Dec 28 '18 at 3:27
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# What exactly is superconductivity?
Does superconductivity mean that the coulomb force or some magnetic force has gone up? I guess that it applies only to wires which get less resistance due to cooling... Is this wrong?
Also, Are there different kinds of superconductive states?
Is there a kind of superconductivity only for magnetism?
-
Hi mick to Physics.SE! Currently your text contains five questions, in most it is not really clear what you want to know. A good starting point is the Wikipedia article on superconductivity. If I try to guess, the answers to your questions would be: no, no, no, yes plenty, no. – Alexander Sep 13 '12 at 15:23
Part of learning is asking what is the correct question ;) Also i find wiki not always clear. ( sometimes even wrong although i dont know about physics but thats another discussion ) I will consider restating my question(s) , but i need some time. Thanks for comment. – mick Sep 13 '12 at 15:39
Before thinking about Superconductivity, let's see Resistivity. Current flow appears in a direction when free electrons travel in opposite direction. During the traverse of free electrons, there would be a lot of collisions with atoms in the metal lattice. Hence, "Atoms experience vibrations about their mean position due to the passing waves of free electrons depending upon the temperature." (Thermal activity plays some role here)
By Ohm's law, resistance $R=\frac{V}{I}$ and also $R=\rho\frac{l}{A}$ where $\rho$ is the Specific resistance or Resistivity. Note that, these are only for normal conductors...
But for superconductors, this is not applicable for sure. Because, something happens differently in superconductors. At extremely low temperatures, the vibrations of atoms slow down so much that they synchronize with those passing waves of electrons..! (i.e.) The free electrons pass unobstructed through the complex metal lattice. Hence, there would be no wastage of electrical energy in the form of heat.
The Superconductivity was very well explained by the well-known BCS Theory. On cooling certain materials below a certain temperature called Transition temperature or Critical temperature $T_c$, three changes happen in a material which makes it a superconductor. Its resistivity becomes absolute zero. Its conductivity becomes Infinity. It's perfectly diamagnetic and excludes Magnetic flux lines (as a result of Meissner Effect). Hence none of your forces are emitted from superconductors. They always repel magnets (i.e) They develop a magnetic polarity opposite to that of the applied field and hence, they don't allow magnetic fields to pass through them.
I would also add that, it's just an overhead view. For a brief zoom, refer the links in Wikipedia. It gives an extra detail in terms of Fermi levels, Cooper pairs and phonon which would be very helpful for this Novel-like question. Perhaps @Alexander's comment provides the answer...
-
Superconductivity the property of a material to induce no resistance on electric currents when they are passed throughout the material. We see that no material is a complete insulator — all materials conduct electricity when cooled down to a certain temperature. So no material is, say, 100 percent resistant, resistance is just a measure of how much a material can conduct at room temperature.
However, there’s a really interesting relation between the increase of temperature and the conductivity of a material. Let’s take gold, for example. Gold is one of the best conductors there is — but it shows more resistance as it is heated. This relation is shown throughout all materials.
Now, where does this factor into superconductivity? We see that the cooler we make a material, the more conductive it gets. The physicist Heike Kamerlingh Onnes tested this, and found that when he cooled a wire of Mercury to -269 degrees Celsius, the resistance dissapeared.
The BCS Theory also helps explain this phenomenon. The theoory explains that materials suddenly become extreme conductors when the electrons inside them join forces to make what are called Cooper pairs (or BCS pairs). Normally, the electrons that carry electricity through a material are scattered about by impurities, defects, and vibrations of the material’s crystal lattice (its scaffold-like inner structure). That’s what we know as electrical resistance. But at low temperatures, when the electrons join together in pairs, they can move more freely without being scattered in the same way. [1]
Not all materials show superconductivity. Apart from mercury, the original superconductor, you can find the effect in about 25 other elements (mostly metals, semimetals, or semiconductors), though it’s also been discovered in thousands of compounds and alloys. Each different material becomes a superconductor at a slightly different temperature.
However, some materials are superconductive at high temperatures. It was thought that materials were only superconductive at low temperatures, but in 1986, two European scientists working for IBM, German physicist J. Georg Bednorz (1950–) and Swiss physicistK. Alex Muller (1927–), discovered a ceramic cuprate (a material containingcopper and oxygen) that could became a superconductor at much higher temperatures [1]
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1. 10-14-2010
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# Forum: Regression Analysis
Linear regression, linear models, nonlinear regression
1. ### Can we really(theoretically) generalize fixed regressor linear regression?
If we have stochastic regressors, we are drawing random pairs $(y_i,\vec{x}_i)$ for a bunch of $i$, the so-called random sample, from a fixed but...
• Replies: 3
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05-30-2016, 10:16 PM
2. ### Time-varying covariates in linear mixed models/multilevel models
Hi! I am a Norwegian phd student exploring the effects of nocturnal road traffic noise exposure in children. One of the research questions is...
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05-30-2016, 05:49 PM
3. ### Mediation analysis - an issue of two opposing (contrary) effects
Hey! I would be very grateful about any advices or comments on an issue I have recently come across. We tested a simple mediation analysis,...
• Replies: 0
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05-30-2016, 03:52 PM
4. ### out of sample
Hi There I have 2500 data with 2 sample. one index and another stock. i've done goodness of fit and regression analysis on it.now I've been asked to...
• Replies: 3
• Views: 336
05-30-2016, 08:05 AM
5. ### Why do you subtract the mean of a predictor variable from that variable..
An author using temperature as a predictor subtracted its mean value from each temperature data point to deal with "scaling issues." I don't...
• Replies: 5
• Views: 465
05-29-2016, 12:33 PM
6. ### Violation of linearity assumptions - binary logistic regression - what to do?
I am running a binary logistic regression with SPSS and unfortunately the assumption of linearity (Box-Tidwell procedure) are not met for the...
• Replies: 10
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05-26-2016, 02:46 PM
7. ### Relative impact of variables on DV
I keep circling back to this issue, this time with a twist. I am arguing with some sports statisticians (who use statistics very differently than the...
• Replies: 1
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05-26-2016, 02:18 PM
8. ### Multiple linear regression: square root transformation of predictor variable
I have used a square root transformation to successfully correct the positive skew of a predictor variable in a hierarchical multiple linear...
• Replies: 0
• Views: 308
05-25-2016, 04:38 PM
9. ### Individual vs. Team Aggregated Data
I have a set individuals (let's say ID_No 1 through 50) and a set of metrics that pertain to these individuals: things like age, SAT reading score,...
• Replies: 0
• Views: 237
05-25-2016, 01:19 PM
10. ### how to argue using wald-test/LR-test for logistic Regression
Hey there, I'm writing my thesis and I did a logistic regression in a data set. At first, it didn't come into my mind that I should argue why I'm...
• Replies: 0
• Views: 221
05-25-2016, 08:38 AM
11. ### OLS regression with interdependence
Hello everyone, I have been writing my thesis and decided, together with my supervisor, to use OLS even though my actors are interdependent...
• Replies: 0
• Views: 188
05-25-2016, 07:15 AM
12. ### R help with Binary log reg
Apologies -- should have posted my question in the R forum instead
• Replies: 0
• Views: 267
05-22-2016, 02:24 PM
13. ### Chi-square instead of Multiple Regression Analysis
Can someone help me? How can I explain that I used Chi-square test of association instead of Multiple Linear Regression Analysis because I had...
• Replies: 0
• Views: 268
05-21-2016, 12:42 PM
14. ### Interpretation of a lagged dependent variable, coefficient as proportion- Panel data
Dear experts! I am having some difficulties in interpreting the results of my regression model using Blundell–Bond linear dynamic panel-data...
• Replies: 0
• Views: 291
05-20-2016, 01:41 PM
15. ### Alcoolimetric Table
Any idea for estrapolation data for this table? I have read spec gravity= 0.8013 what will the content of Alcool? ...
• Replies: 0
• Views: 239
05-20-2016, 10:10 AM
16. ### Logistic Regression interpretation help
Hello I'm working on a project and had some debates with my colleagues on logistic regression model. Hope someone here can clarify for us. Let's...
• Replies: 0
• Views: 318
05-18-2016, 02:14 PM
17. ### Repeated measures, random effects, or both?
Hi, I have the following situation and I do not know if it is a case that calls for repeated measures or individual-specific random effects. We...
• Replies: 6
• Views: 503
agp
05-18-2016, 09:08 AM
18. ### Field's test for linearity of the logit
Hello, I am currently working on my bachelor thesis and I ran a logistic regression on a data set I have. From Field's 'Discovery Statistics...
• Replies: 0
• Views: 515
05-17-2016, 08:19 AM
19. ### SPSS - Regression - interpretation and generalizing results
Hello, Im really grateful you have taken the time to read my post. Thanks for your help. Im doing a regression analysis trying to predict...
• Replies: 2
• Views: 455
05-16-2016, 09:36 AM
20. ### dummy regression - unequal sample sizes
Hello, I am conducting a linear regression analysis in SPSS with different variables: - scale, - 2-level nominal, - 5-level nominal (dummy)...
• Replies: 0
• Views: 270
05-15-2016, 10:50 AM
21. ### Time series
Say you know that you have autocorrelation in your data. Say AR =2. Can you use ARIMA to remove the AC, then predict the results of the ARIMA with...
• Replies: 1
• Views: 342
05-15-2016, 09:58 AM
22. ### Two way repeated measures with zeros, non integer values, and non-normal distribution
I would really appreciate some help with a few models I am trying to run. Essentially my data looks at how often a subject was visited depending on...
• Replies: 1
• Views: 304
05-12-2016, 02:48 PM
23. ### coding - one categorical vs multiple dummy
Hi, Most things I read seem to suggest that when you have a categorical predictor in a regression analysis you should create k-1 dummy variables,...
• Replies: 2
• Views: 360
05-10-2016, 04:57 PM
24. ### Interaction concept put into plain English
Hello, as you may know, I am not a statistician, rather a 'consumer' of statistics. When I study something new, or when I present statistics to a...
• Replies: 2
• Views: 336
05-10-2016, 08:42 AM
25. ### Model fit statistic for population-average specification
Hello everybody! For the purpose of my study I am using an exponential Poisson model with population-averaged (PA) effects, as implemented in...
• Replies: 0
• Views: 218
05-09-2016, 10:44 AM
26. ### Using Bonferonni in Multiple Regression
Hi all, I'm currently working on a data-set involving several continuous independent variables and i'm wondering how the Bonferonni correction can...
• Replies: 1
• Views: 263
05-09-2016, 06:37 AM
27. ### Multiple regression analysis - Linearity/homoscedasticity
Hello, I am currently having trouble determining if my predictors show enough linearity to be used in a standard multiple regression analysis, and...
• Replies: 2
• Views: 307
05-08-2016, 09:21 PM
28. ### Help for simple regression
My research question is: "Can we say that the number of hours of Internet use increases with Internet purchase?" with the variables: - Use of...
• Replies: 3
• Views: 451
05-08-2016, 07:34 PM
29. ### Imputing Missing Values using Regression
I have a table of 4 variables where some of the data is missing and I need to impute the missing values, my table is as follows: X1 X2 X3...
• Replies: 3
• Views: 354
05-06-2016, 06:21 PM
30. ### Significant two-way interaction, but simple slopes are n.s.
Hi everyone, I've seen similar questions like mine posted here before, but none of the replies were ever clear enough, so here's my Q. Thanks in...
• Replies: 1
• Views: 315
05-06-2016, 01:10 PM
31. ### How to select the Independent Variables for modelling given 100 variable??
Let's say i have selected my DV and i have 100 variables from which i have to choose my IDVs for the regression model(say multiple regression). ...
• Replies: 0
• Views: 280
05-06-2016, 05:19 AM
32. ### Multiparameter fittting - confidence intervals criterion
I made a multiparameter fittting (7 adjustment parameters, 45 experimental points). I obtained enormous confidence intervals for 95%. There is a...
• Replies: 0
• Views: 316
05-06-2016, 12:12 AM
33. ### Please anyone help due monday
Hey guys, Im doing a mathematics degree and have this stats subject and honestly have no clue whatsoever on how to do anything in this class. I have...
• Replies: 2
• Views: 300
05-05-2016, 03:23 PM
34. ### normality assumption: data itself, or residuals ?
I have a question about the normality assumption, in regression, ANOVA, and T-test: The normalization assumption is for the original data and...
• Replies: 3
• Views: 1,589
05-05-2016, 12:21 AM
35. ### Stepwise Regression Limitations Explanation
I've recently been working on building a model and have come across a number of different approaches. I'm particularly interested in the limitations...
• Replies: 12
• Views: 655
05-04-2016, 07:39 PM
36. ### Do I have a non-linear regression?
Hello everyone, I want to start by saying that I don't have too much experience working with continuous dependent variables. I have worked more...
• Replies: 5
• Views: 755
05-04-2016, 03:19 AM
37. ### Quick Mediation Analysis Question
Hi Everyone, I have a quick question about interpreting a mediation analysis from Hayes' PROCESS macro. After running a mediation analysis, I...
• Replies: 2
• Views: 711
05-04-2016, 03:01 AM
38. ### Index & Dummies - Italian Mafia
Hey. i am working on a paper about the mafia in italy. my equation: Y=beta0 + beta1*MPI + beta2*Controls Y = Response Rate in a given city...
• Replies: 0
• Views: 248
05-03-2016, 02:11 PM
39. ### Interpreting Odds Ratio after Weight of evidence transformation
I am doing a logistic modeling with several continuous variables and binary outcome variable. 1- event 0-non event. I did the initial modeling with...
• Replies: 2
• Views: 279
05-03-2016, 11:16 AM
40. ### Combining Two Samples (one w/ intervention and one w/ out intervention)
I have a local dataset with a continuous predictor and binary outcome. I will use logistic regression to predict the binary outcome with the...
• Replies: 7
• Views: 460
05-02-2016, 11:01 PM
41. ### testing seemingly unrelated regressions assumptions in R
Hi there! I have a fitted SUR model and want to test if the assumptions for SUR (not the ordinary linear assumptions, since i've already checke...
• Replies: 1
• Views: 321
05-02-2016, 03:54 PM
42. ### Cox Regression - time dependent
Dear all, I have a question regarding the cox regression. I have conducted a dataset with Top50 Boxoffice Data for the U.S market. It contains...
• Replies: 3
• Views: 357
05-02-2016, 10:45 AM
43. ### Types of regression: Article
Well a page, but I was not sure what to call this :p By no means complete or even close but interesting. There are hundreds of types of...
• Replies: 0
• Views: 250
05-02-2016, 09:49 AM
44. ### Fill in multiple missing values given correlation matrix
I have a correlation matrix: mat <- round(cor(mtcars), 2) mpg cyl disp hp drat wt qsec vs am gear carb mpg 1.00...
• Replies: 16
• Views: 900
05-01-2016, 10:39 PM
45. ### Linear Regression Interpretation
Hi there, I am running a Multiple Linear Regression analysis. I have 2 predictor variables (self-worth and self-acceptance) and 1 criterion...
• Replies: 4
• Views: 394
05-01-2016, 09:38 PM
46. ### Multinomial logit / time-series fixed effects / multivariate regression: Which one?
As part of a larger study, we have collected a wealth of data on the interactions customers engage in when buying and using a service. ...
• Replies: 0
• Views: 276
05-01-2016, 06:50 AM
47. ### Logistic reg w/ ratio as predictor
Is there anything I need to know when using a ratio as an independent variable in the model. Is it better to use log(ratio) ? Not an easy web...
• Replies: 1
• Views: 331
04-30-2016, 03:22 PM
48. ### [Logistic Analysis] How does SPSS calculate the 0.5 Cut-Point?
I am interested in determining the classification accuracy of an outcome (binary DV - whether a participant has capacity or not to make a legal...
• Replies: 0
• Views: 303
04-29-2016, 06:25 PM
49. ### Multivariate analysis, canonical correlation
Hello, I conducted an experiment with 18 subjects who participated in a simulator experiment. Each participant had to perform a task with 6...
• Replies: 0
• Views: 360
04-28-2016, 10:08 AM
50. ### Interpretation of coefficients, dependent variable in (%), independent variable in ln
Hi, I have a very simple question. I have the following equation. My dependent variable is CRES which is the share of renewables in a country's...
• Replies: 0
• Views: 283
Rex
04-28-2016, 08:07 AM
51. ### Problem with replicating a paper
I'm having some trouble at replicating this paper (text here: http://www.michiganjb.org/issues/52/text52c.pdf) The paper wants to assess the economic...
• Replies: 0
• Views: 229
04-28-2016, 05:29 AM
52. ### Writing down firth logistic regression equation
Hi, This is a really simple question that I'm sort of embarrassed I can't answer - how would you write down a firth logistic regression? Now to...
• Replies: 6
• Views: 404
04-27-2016, 01:59 PM
53. ### GLS (Generalized least Squares) STATA
generalized least squares (GLS) is a technique for estimating the unknown parameters in a linear regression model. GLS (Generalized least Squares)...
• Replies: 0
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04-27-2016, 06:16 AM
54. ### Regression with stratified data: What to do?
Hi, I want to model a bird population in a country via regression analysis (Poisson regression due to count data). In order to optimize the...
• Replies: 0
• Views: 231
04-27-2016, 04:39 AM
55. ### Minimum number of cases for the reference level
Before I ask this I have the true population so I am not sure this issue even applies :p Is there a minimum number of cases the reference level...
• Replies: 7
• Views: 474
04-26-2016, 04:59 PM
+ Post New Thread
Page 6 of 70 First 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 56 ... Last
#### Thread Display Options
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted.
Note: when sorting by date, 'descending order' will show the newest results first.
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1. Title Description
For 1 + 2 +... + n, it is required that keywords such as multiplication and division, for, while, if, else, switch, case and conditional judgment statements (A?B:C) cannot be used.
Example 1:
Input: n = 3
Output: 6
Example 2:
Input: n = 9
Output: 45
Restrictions:
1 <= n <= 10000
2. Method 1: recursion
1) Ideas and algorithms
Imagine that if you use the recursive method to realize this problem without restriction, I believe everyone can easily give the following implementation (taking C + + as an example):
class Solution{
public:
int sumNums(int n)
{
return n==0 ? 0 : sumNums(n-1);
}
};
Usually, when implementing recursion, we will use conditional judgment statements to determine the exit of recursion. However, due to the limitation of the problem, we can't use conditional judgment statements. Can we use other methods to determine the exit of recursion? The answer of the logical operator is the nature of the short circuit.
Take the logical operator & & as an example. For the expression A & & B, if the expression A returns False, A & & B has been determined to be False, and expression B will not be executed at this time. Similarly, for the logical operator 𞓜 and the expression A | B, if the expression A returns True, then A | B has been determined to be True, and expression B will not be executed at this time.
Using this feature, we can regard the exit to judge whether it is recursive as part A of A & & B expression, and the recursive main function as part B. If it is not A recursive exit, return True and continue to execute the part of expression B, otherwise the recursion ends. Of course, you can also use the logical operator 𞓜 to give A similar implementation. Here we only provide the recursive implementation combined with the logical operator &.
2) Correct code
class Solution{
public:
int sumNums(int n)
{
n && (n+=sum(n-1));
return n;
}
};
Complexity analysis
Time complexity: O(n). The recursive function recurses n times, and the calculation time complexity in each recursion is O(1), so the total time complexity is O(n).
Space complexity: O(n). The spatial complexity of recursive function depends on the depth of recursive call stack. Here, the depth of recursive function call stack is O(n), so the spatial complexity is O(n).
3. Method 2: fast multiplication
When considering the multiplication of A and B, how do we use addition and bit operation to simulate? In fact, it is to expand the binary of B. if the i-th bit under the binary representation of B is 1, then the contribution of this bit to the final result is A * (1 < < I), that is, A < < I. We traverse each bit under the binary expansion of B, and add up all the contributions to get the final answer. This method is also known as Russian farmer multiplication . This method is often used in the scene of multiplying two numbers and taking modulus. If the multiplication of two numbers has exceeded the data range, but will not exceed after taking modulus, we can use this method to disassemble the modulus and calculate the contribution to ensure that each operation is within the data range.
Fast multiplication C + + code
int quickMulti(int A,int B)
{
int ans=0;
for(;B;B>>=1)
{
if(B&1) ans+=A;
A<<=1;
}
return ans;
}
Back to this question, from the summation formula of the sequence of equal differences, we can know that 1 + 2 + × n is equivalent to n ( n + 1 ) 2 \frac{n(n+1)}{2} 2n(n+1), we can use the shift right operator to simulate the division by 2, then the equation becomes n(n+1) > > 1, and the rest that does not meet the requirements of the topic is n(n+1). According to the fast multiplication mentioned above, we can multiply the two numbers and simulate them with addition and bit operation, but we can see that we still need circular statements in the above C + + implementation, Is there any way to get rid of this circular statement? The answer is yes, that is to manually expand the loop. Because the data range of the topic nn is [110000], the binary expansion of n will not exceed 14 bits at most. We can manually expand 14 layers instead of the loop. So far, the requirements of the topic are met. For specific implementation, please refer to the code given below.
1) Fast multiplicative cyclic expansion method
//Cyclic expansion method of fast multiplication
class Solution{
public:
int sumNums(int n)
{
int ans=0,A=n,B=n+1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
(A&1)&&(ans+=B);
A>>=1;
B<<=1;
return ans>>1;
}
};
2) Fast multiplication recursion method
(B & 1) in (A & - (B & 1)), only two values can be taken, [0, - 1] and - 1 have a feature when participating in bit and operation, a & - 1 = = a
//Fast multiplication recursion
class Solution{
public:
int sumNums(int n)
{
return quickMulti(n,n+1)>>1;
}
int quickMulti(int a,int b)
{
int ans=0;
(a!=0) && (ans=(b&-(a&1))+quickMulti(a>>1,b<<1));
return ans;
}
};
Method 3: using sizeof and array length
C++
class Solution{
public:
int sumNums(int n)
{
bool arr[n][n+1];// The bool variable occupies 1 byte of memory
return sizeof(arr)>>1;
}
};
Keywords: leetcode
Added by justintoo1 on Wed, 02 Mar 2022 01:36:09 +0200
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# LaTeX not working when I try to submit a solution
I was trying to submit a solution to the question 'A simple flying maneuver', when I ran into some problems with LaTeX. Also, on clicking the preview button, nothing happens. How do I submit the solution, then?
5 years, 2 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
I send an email some days ago about the same problem, but i got no answer.
- 5 years, 2 months ago
The preview has been fixed.
It doesn't matter if your solution doesn't encode properly, as I would still be able to read your Latex code (assuming that you typed the equation correctly).
Staff - 5 years, 2 months ago
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Immersed Lagrangian Floer cohomology via pearly trajectories
Garrett Alston, Erkao Bao
Research output: Contribution to journalArticlepeer-review
Abstract
We define Lagrangian Floer cohomology over Z2-coefficients by counting pearly trajectories for graded, exact Lagrangian immersions that satisfy a certain positivity condition on the index of the non-embedded points, and show that it is an invariant of the Lagrangian immersion under Hamiltonian deformations. We also show that it is naturally isomorphic to the Hamiltonian perturbed version of Lagrangian Floer cohomology as defined in [4]. As an application, we prove that the number of non-embedded points of such a Lagrangian in Cn is no less than the sum of its Betti numbers.
Original language English (US) 104335 Journal of Geometry and Physics 169 https://doi.org/10.1016/j.geomphys.2021.104335 Published - Nov 2021
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TFG|已知物体空间位置估计
TensorFlow又出了一个好玩的开源库：Graphics，主要解决从渲染器最终渲染结果逆推出各Mesh的Transform和Rotation。本文为官方提供的入门项目Object pose estimation的阅读笔记，代码均摘自源文件中。
Object pose alignment
• 机器学习：使用一个简单的神经网络进行深度学习。
• 数学优化
1.深度学习方法
定义模型
• 拆分输出项：将(,7)拆分为(,4)(,3)
• 用预测值对3D模型形变
• 用正确值还原对3D模型的形变
• 求出所有顶点偏移距离的平方（二阶范数）和
• 返回单个batch的error平均值
2. Mathematical optimization
Here the problem is tackled using mathematical optimization, which is another traditional way to approach the problem of object pose estimation. Given correspondences between the object in ‘rest pose’ (pastel lemon color) and its rotated and translated counter part (pastel honeydew color), the problem can be formulated as a minimization problem. The loss function can for instance be defined as the sum of Euclidean distances between the corresponding points using the current estimate of the rotation and translation of the transformed object. One can then compute the derivative of the rotation and translation parameters with respect to this loss function, and follow the gradient direction until convergence. The following cell closely follows that procedure, and uses gradient descent to align the two objects. It is worth noting that although the results are good, there are more efficient ways to solve this specific problem. The interested reader is referred to the Kabsch algorithm for further details.
Note: press play multiple times to sample different test cases.
Define the loss and gradient functions:
Create the optimizer.
Initialize the random transformation, run the optimization and animate the result.
Share
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## Prealgebra (7th Edition)
$-\frac{2}{9}$
When $x=-\frac{1}{3}$, y=$\frac{2}{5}$,and $z=\frac{5}{6}$ $x^{2}-yz$ $=(-\frac{1}{3})^{2}-(\frac{2}{5}\times\frac{5}{6})$ $=[(-\frac{1}{3})\times(-\frac{1}{3})]-(\frac{1}{1}\times\frac{1}{3})$ $=(\frac{1}{3}\times\frac{1}{3})-\frac{1}{3}$ $=\frac{1}{9}-\frac{1}{3}$ $=\frac{1}{9}-\frac{3}{9}$ $=-\frac{2}{9}$
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# How can I tell if a relay coil is rated for 240VAC
I want to monitor a 240VAC line from an Raspberry Pi, my thought was to hook the 240VAC line to an appropriate relay then just sample the contacts from the Pi.
I have a relay which I think would work but I don't know how to be sure the coil is rated for 240VAC. It's about 1 inch in each dimension, the coil resistance is 670k Ohms.
I've attached a photo, I did google around for specs on this but I can't find any clear info.
• I meant to say the coil resistance is 670 Kilo Ohms. – Matt Noel Apr 2 '18 at 21:56
• A schematic would be appropriate, use the tool embedded in the question toolbar and edit your question. – Voltage Spike Apr 2 '18 at 22:31
• @MattNoel The coil resistance cannot be 670k unless the relay is broken, as no relay on the datasheet goes that high. However, 670 ohms lines right up with the 24D version, which has a nominal resistance of 660 ohms +/-10%. – C_Elegans Apr 2 '18 at 23:42
Your solution is unlikely to work, as the relay is going to buzz a lot, and probably start a fire. The coil voltage should be shown somewhere on the relay, though I don't see it on yours. In any case, $640\Omega$ is way too low for 240V, as it will be dissipating 90W, so instead of a relay, you'd have a lightbulb and a house fire. After looking at the data sheet (see TonyM's answer), and looking in the chart on page 2 for 'AZ-2110-1A-24D', we see that the coil voltage is 24V, and the winding resistance is ~$660\Omega$, which matches up nicely.
As for solving your problem, I'd recommend using an optocoupler. Something like this:
simulate this circuit – Schematic created using CircuitLab
Basically, it's a full bridge rectifier feeding the optocoupler (the led and transistor, the schematic editor didn't have a real optocoupler) with a current of ~5.5mA. The two 22k resistors will need to be 1W or larger to remain within temperature tolerances, and need to be rated for greater than 120V.
• The resistors R1 and R2 should be rated for more than 120 V AC. Three or four resistors in series may be used to improve security. – Uwe Apr 3 '18 at 8:47
• @uwe it has been fixed – C_Elegans Apr 3 '18 at 12:47
I also confirmed it is a 24Vdc relay SPST
AZ2110–1A–24D
• 24 as in voltage for coil
3/4 Hp rated
As @C_Elegans posted,
With 240Vrms or +/-340Vp +/-??% the IR LED current will be 7.7mA Peak is to have a logic level pulse for each half cycle for instant detection by IRQ or sustain the output by some routine that samples the input, more than 1 cycle intervals then change the output from 1K pullup to 100K with RC= 100ms ( 10 cycles) , using C = 1uF using hysteresis or a Schmitt trigger on the input.
According to this, its a Zettler AZ2110, contacts 1 Form A, coil 24D.
The specs on that webpage list the above variant as having a 24 VDC coil.
• The one pictured is the 24D variant, with a coil voltage of 24V. – C_Elegans Apr 2 '18 at 22:00
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# Aerodynamic Optimization
## Introduction
We will now demonstrate how to optimize the aerodynamic shape of a wing. We will be combining aspects of all of the following sections: Analysis with ADflow, pyOptSparse, and Geometric Parametrization. Again reference a high level explanation of how these fit together - aero problem, optimization problem, geometry. The optimization problem is defined as
minimize
$$C_D$$
with respect to
7 twist variables
96 shape variables
1 angle of attack
subject to
$$C_L = 0.5$$
$$V \ge V_0$$
$$t \ge t_0$$
$$\Delta z_\text{LETE, upper} = -\Delta z_{LETE, lower}$$
## Files
Navigate to the directory opt/aero in your tutorial folder. Copy the following files to this directory:
span.prompt1:before {
content: "\$ ";
}
cp ../ffd/ffd.xyz .
cp ../../aero/meshing/volume/wing_vol.cgns .
Create the following empty runscript in the current directory:
• aero_opt.py
## Dissecting the aerodynamic optimization script
Open the file aero_opt.py in your favorite text editor. Then copy the code from each of the following sections into this file.
### Import libraries
import os
import argparse
import ast
import numpy as np
from mpi4py import MPI
from baseclasses import AeroProblem
from pygeo import DVGeometry, DVConstraints
from pyoptsparse import Optimization, OPT
from idwarp import USMesh
from multipoint import multiPointSparse
The multipoint library is the only new library to include in this script.
This is a convenience feature that allows the user to pass in command line arguments to the script. Two options are provided:
• specifying the output directory
• specifying the optimizer to be used
# Use Python's built-in Argument parser to get commandline options
parser = argparse.ArgumentParser()
parser.add_argument("--opt", type=str, default="SLSQP", choices=["IPOPT", "SLSQP", "SNOPT"])
args = parser.parse_args()
### Creating processor sets
The multiPointSparse class allows us to allocate sets of processors for different analyses. This can be helpful if we want to consider multiple design points, each with a different set of flow conditions. In this case, we create a processor set for cruise cases, but we only add one point.
MP = multiPointSparse(MPI.COMM_WORLD)
comm, setComm, setFlags, groupFlags, ptID = MP.createCommunicators()
if not os.path.exists(args.output):
if comm.rank == 0:
os.mkdir(args.output)
If we want to add more points, we can increase the quantity nMembers. We can choose the number of processors per point with the argument memberSizes. We can add another processor set with another call to addProcessorSet. The call createCommunicators returns information about the current processor’s set.
The set-up for adflow should look the same as for the aerodynamic analysis script. We add a single lift distribution with 200 sampling points.
aeroOptions = {
# I/O Parameters
"gridFile": args.gridFile,
"outputDirectory": args.output,
"monitorvariables": ["resrho", "cl", "cd"],
"writeTecplotSurfaceSolution": True,
# Physics Parameters
"equationType": "RANS",
# Solver Parameters
"MGCycle": "sg",
"infchangecorrection": True,
# ANK Solver Parameters
"useANKSolver": True,
# NK Solver Parameters
"useNKSolver": True,
"nkswitchtol": 1e-6,
# Termination Criteria
"L2Convergence": 1e-10,
"L2ConvergenceCoarse": 1e-2,
"nCycles": 10000,
}
# Create solver
### Set the AeroProblem
ap = AeroProblem(name="wing", alpha=1.5, mach=0.8, altitude=10000, areaRef=45.5, chordRef=3.25, evalFuncs=["cl", "cd"])
# Add angle of attack variable
The only difference in setting up the AeroProblem is that now we add angle-of-attack as a design variable. Any of the quantities included in the initialization of the AeroProblem can be added as design variables.
### Geometric parametrization
The set-up for DVGeometry should look very familiar (if not, see Geometric Parametrization). We include twist and local variables in the optimization. After setting up the DVGeometry instance we have to provide it to ADflow with the call setDVGeo.
# Create DVGeometry object
FFDFile = "ffd.xyz"
DVGeo = DVGeometry(FFDFile)
# Create reference axis
nTwist = nRefAxPts - 1
# Set up global design variables
def twist(val, geo):
for i in range(1, nRefAxPts):
geo.rot_z["wing"].coef[i] = val[i - 1]
DVGeo.addGlobalDV(dvName="twist", value=[0] * nTwist, func=twist, lower=-10, upper=10, scale=0.01)
# Set up local design variables
# Add DVGeo object to CFD solver
CFDSolver.setDVGeo(DVGeo)
### Geometric constraints
We can set up constraints on the geometry with the DVConstraints class, also found in the pyGeo module. There are several built-in constraint functions within the DVConstraints class, including thickness, surface area, volume, location, and general linear constraints. The majority of the constraints are defined based on a triangulated-surface representation of the wing obtained from ADflow.
Note
The triangulated surface is created by ADflow (or DAfoam) using the wall surfaces defined in the CFD volume mesh. The resolution is similar to the CFD surface mesh, and users do not need to provide this triangulated mesh themselves. Optionally, this can also be defined with an external file, see the docstrings for setSurface(). This is useful if users want to have a different resolution on the triangulated surface (finer or coarser) compared to the CFD mesh, or if DVConstraints is being used without ADflow (or DAfoam).
The volume and thickness constraints are set up by creating a uniformly spaced 2D grid of points, which is then projected onto the upper and lower surface of a triangulated-surface representation of the wing. The grid is defined by providing four corner points (using leList and teList) and by specifying the number of spanwise and chordwise points (using nSpan and nChord).
Note
These grid points are projected onto the triangulated surface along the normals of the ruled surface formed by these grid points. Typically, leList and teList are given such that the two curves lie in a plane. This ensures that the projection vectors are always exactly normal to this plane. If the surface formed by leList and teList is not planar, issues can arise near the end of an open surface (i.e., the root of a wing) which can result in failing intersections.
By default, scaled=True for addVolumeConstraint() and addThicknessConstraints2D(), which means that the volume and thicknesses calculated will be scaled by the initial values (i.e., they will be normalized). Therefore, lower=1.0 in this example means that the lower limits for these constraints are the initial values (i.e., if lower=0.5 then the lower limits would be half the initial volume and thicknesses).
Warning
The leList and teList points must lie completely inside the wing.
DVCon = DVConstraints()
DVCon.setDVGeo(DVGeo)
# Only ADflow has the getTriangulatedSurface Function
DVCon.setSurface(CFDSolver.getTriangulatedMeshSurface())
# Volume constraints
leList = [[0.01, 0, 0.001], [7.51, 0, 13.99]]
teList = [[4.99, 0, 0.001], [8.99, 0, 13.99]]
DVCon.addVolumeConstraint(leList, teList, nSpan=20, nChord=20, lower=1.0, scaled=True)
# Thickness constraints
DVCon.addThicknessConstraints2D(leList, teList, nSpan=10, nChord=10, lower=1.0, scaled=True)
For the volume constraint, the volume is computed by adding up the volumes of the prisms that make up the projected grid as illustrated in the following image (only showing a section for clarity). For the thickness constraints, the distances between the upper and lower projected points are used, as illustrated in the following image. During optimization, these projected points are also moved by the FFD, just like the wing surface, and are used again to calculate the thicknesses and volume for the new designs. More information on the options can be found in the pyGeo docs or by looking at the pyGeo source code.
The LeTe constraints (short for Leading edge/Trailing edge constraints) are linear constraints based on the FFD control points. When we have both twist and local shape variables, we want to prevent the local shape variables from creating a shearing twist. This is done by constraining the upper and lower FFD control points on the leading and trailing edges to move in opposite directions. Note that the LeTe constraint is not related to the leList and teList points discussed above.
# Le/Te constraints
if comm.rank == 0:
# Only make one processor do this
DVCon.writeTecplot(os.path.join(args.output, "constraints.dat"))
In this script DVCon.writeTecplot will save a file named constraints.dat which can be opened with Tecplot to visualize and check these constraints. Since this is added here, before the commands that run the optimization, the file will correspond to the initial geometry. The following image shows the constraints visualized with the wing surface superimposed. This command can also be added at the end of the script to visualize the final constraints.
### Mesh warping set-up
meshOptions = {"gridFile": args.gridFile}
mesh = USMesh(options=meshOptions, comm=comm)
CFDSolver.setMesh(mesh)
### Optimization callback functions
First we must set up a callback function and a sensitivity function for each processor set. In this case cruiseFuncs and cruiseFuncsSens belong to the cruise processor set. Then we need to set up an objCon function, which is used to create abstract functions of other functions.
def cruiseFuncs(x):
if comm.rank == 0:
print(x)
# Set design vars
DVGeo.setDesignVars(x)
ap.setDesignVars(x)
# Run CFD
CFDSolver(ap)
# Evaluate functions
funcs = {}
DVCon.evalFunctions(funcs)
CFDSolver.evalFunctions(ap, funcs)
CFDSolver.checkSolutionFailure(ap, funcs)
if comm.rank == 0:
print(funcs)
return funcs
def cruiseFuncsSens(x, funcs):
funcsSens = {}
DVCon.evalFunctionsSens(funcsSens)
CFDSolver.evalFunctionsSens(ap, funcsSens)
return funcsSens
def objCon(funcs, printOK):
# Assemble the objective and any additional constraints:
funcs["obj"] = funcs[ap["cd"]]
funcs["cl_con_" + ap.name] = funcs[ap["cl"]] - 0.5
if printOK:
print("funcs in obj:", funcs)
return funcs
Now we will explain each of these callback functions.
#### cruiseFuncs
The input to cruiseFuncs is the dictionary of design variables. First, we pass this dictionary to DVGeometry and AeroProblem to set their respective design variables. Then we solve the flow solution given by the AeroProblem with ADflow. Finally, we fill the funcs dictionary with the function values computed by DVConstraints and ADflow. The call checkSolutionFailure checks ADflow to see if there was a failure in the solution (could be due to negative volumes or something more sinister). If there was a failure it changes the fail flag in funcs to True. The funcs dictionary is the required return.
#### cruiseFuncsSens
The inputs to cruiseFuncsSens are the design variable and function dictionaries. Inside cruiseFuncsSens we populate the funcsSens dictionary with the derivatives of each of the functions in cruiseFuncs with respect to all of its dependence variables.
#### objCon
The main input to the objCon callback function is the dictionary of functions (which is a compilation of all the funcs dictionaries from each of the design points). Inside objCon, the user can define functionals (or functions of other functions). For instance, to maximize L/D, you could define the objective function as:
funcs['obj'] = funcs['cl'] / funcs['cd']
The objCon function is processed within the multipoint module and the partial derivatives of any functionals with respect to the input functions are automatically computed with the complex-step method. This means that the user doesn’t have to worry about computing analytic derivatives for the simple functionals defined in objCon. The printOK input is a boolean that is False when the complex-step is in process.
### Optimization problem
Setting up the optimization problem follows the same format as before, only now we incorporate multiPointSparse. When creating the instance of the Optimization problem, MP.obj is given as the objective function. multiPointSparse will take care of calling both cruiseFuncs and objCon to provide the full funcs dictionary to pyOptSparse.
Both AeroProblem and DVGeometry have built-in functions to add all of their respective design variables to the optimization problem. DVConstraints also has a built-in function to add all constraints to the optimization problem. The user must manually add any constraints that were defined in objCon.
Finally, we need to tell multiPointSparse which callback functions belong to which processor set. We also need to provide it with the objCon and the optProb. The call optProb.printSparsity() prints out the constraint Jacobian at the beginning of the optimization.
# Create optimization problem
optProb = Optimization("opt", MP.obj, comm=comm)
# Add variables from the AeroProblem
# The MP object needs the 'obj' and 'sens' function for each proc set,
# the optimization problem and what the objcon function is:
MP.setProcSetObjFunc("cruise", cruiseFuncs)
MP.setProcSetSensFunc("cruise", cruiseFuncsSens)
MP.setObjCon(objCon)
MP.setOptProb(optProb)
optProb.printSparsity()
### Run optimization
To finish up, we choose the optimizer and then run the optimization.
# Set up optimizer
if args.opt == "SLSQP":
optOptions = {"IFILE": os.path.join(args.output, "SLSQP.out")}
elif args.opt == "SNOPT":
optOptions = {
"Major feasibility tolerance": 1e-4,
"Major optimality tolerance": 1e-4,
"Hessian full memory": None,
"Function precision": 1e-8,
"Print file": os.path.join(args.output, "SNOPT_print.out"),
"Summary file": os.path.join(args.output, "SNOPT_summary.out"),
"Major iterations limit": 1000,
}
elif args.opt == "IPOPT":
optOptions = {
"limited_memory_max_history": 1000,
"print_level": 5,
"tol": 1e-6,
"acceptable_tol": 1e-5,
"max_iter": 300,
"start_with_resto": "yes",
}
optOptions.update(args.optOptions)
opt = OPT(args.opt, options=optOptions)
# Run Optimization
sol = opt(optProb, MP.sens, storeHistory=os.path.join(args.output, "opt.hst"))
if comm.rank == 0:
print(sol)
Note
The complete set of options for SNOPT can be found in the pyOptSparse documentation. It is useful to remember that you can include major iterations information in the history file by providing the proper options. It has been observed that the _print and _summary files occasionally fail to be updated, possibly due to unknown hardware issues on GreatLakes. The problem is not common, but if you want to avoid losing this information, you might back it up in the history file. This would allow you monitor the optimization even if the _print and _summary files are not being updated. Note that the size of the history file will increase due to this additional data.
## Run it yourself!
To run the script, use the mpirun and place the total number of processors after the -np argument
mpirun -np 4 python aero_opt.py
You can follow the progress of the optimization using OptView, as explained in pyOptSparse.
### Terminal output
An important step of verifyting the optimization setup is to check the sparsity structure of the constraint Jacobian:
+-------------------------------------------------------------------------------+
| Sparsity structure of constraint Jacobian |
+-------------------------------------------------------------------------------+
alpha_wing (1) twist (7) local (96)
+---------------------------------------+
DVCon1_volume_constraint_0 (1) | | X | X |
-----------------------------------------
DVCon1_thickness_constraints_0 (100) | | X | X |
-----------------------------------------
DVCon1_lete_constraint_0_local(L) (8) | | | X |
-----------------------------------------
DVCon1_lete_constraint_1_local(L) (8) | | | X |
-----------------------------------------
cl_con_wing (1) | X | X | X |
+---------------------------------------+
## Postprocessing the solution output
All output is found in the output directory. The naming scheme of the files follows in general <name>_<iter>_<type>.<ext>, where the <name> is aeroproblem, <iter> is the function evaluation number, <type> is the solution type (surface, volume, lift, slices, ect.), and <ext> is the file extension. The solution files (.dat, .cgns or .plt) can be viewed in the Tecplot. A contour plot of the pressure coefficient compared with the surface solution from the Analysis with ADflow is shown below.
Similarly, as done in Analysis with ADflow, the lift and slice files (.dat) are used to compare the spanwise normalized lift, compared to elliptical lift, the twist distribution, and t/c. For the slice file, here the normalized airfoil shape and pressure coefficient are shown. The optimized design achieves an elliptical lift distribution, and shows a more even and gradual pressure distribution as shown by the contour and section plots. Further, the optimized twist distribution, as expected, demonstrates higher lift in the outboard section of the wing.
Finally the optimization history can be viewed either by parsing the database or using OptView. Here the former is done showing the major iterations of the history, objective, and some of the design variables and constraints. To produce the figure, run the accompanying postprocessing script as shown below.
python plot_optHist.py --histFile output/opt.hst --outputDir output/
|
{}
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