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# 9.4 Factoring

We now turn to factoring. The basic method of dividing an integer  by all primes  is much too slow for most purposes. For many years, people have worked on developing more efficient algorithms. We present some of them here. In Chapter 21, we’ll also cover a method using elliptic curves, and in Chapter 25, we’ll show how a quantum computer, if built, could factor efficiently.

One method, which is also too slow, is usually called the Fermat factorization method. The idea is to express  as a difference of two squares: . Then  gives a factorization of . For example, suppose we want to factor . Compute , , , , until we find a square. In this case, . Therefore,

The Fermat method works well when  is the product of two primes that are very close together. If , it takes  steps to find the factorization. But if  and  are two randomly selected 300-digit primes, it is likely that  will be very large, probably around 300 digits, too. So Fermat factorization is unlikely to work. Just to be safe, however, the primes for an RSA modulus are often chosen to be of slightly different sizes.

We now turn to more modern methods. If one of the prime factors of  has a special property, it is sometimes easier to factor . For example, if  divides  and  has only small prime factors, the following method is effective. It was invented by Pollard in 1974.

# The $equation shows p1.$ Factoring Algorithm

Choose an integer . Often  is used. Choose a bound . Compute  as follows. Let  and . Then . Let . If , we have found a nontrivial factor of .

Suppose  is a prime factor of  such that  has only small prime factors. Then it is likely that  will divide , say . By Fermat’s theorem, , so  will occur in the greatest common divisor of  and . If  is another prime factor of , it is unlikely that , unless  also has only small prime factors. If , not all is lost. In this case, we have an exponent  (namely ) and an  such that . There is a good chance that the  method (explained later in this section) will factor . Alternatively, we could choose a smaller value of  and repeat the calculation.

For an example, see Example 34 in the Computer Appendices.

How do we choose the bound ? If we choose a small , then the algorithm will run quickly but will have a very small chance of success. If we choose a very large , then the algorithm will be very slow. The actual value used will depend on the situation at hand.

In the applications, we will use integers that are products of two primes, say , but that are hard to factor. Therefore, we should ensure that  has at least one large prime factor. This is easy to accomplish. Suppose we want  to have around 300 digits. Choose a large prime , perhaps around . Look at integers of the form , with  running through some integers around . Test  for primality by the Miller-Rabin test, as before. On the average, this should produce a desired value of  in less than 400 steps. Now choose a large prime  and follow the same procedure to obtain . Then  will be hard to factor by the  method.

The elliptic curve factorization method (see Section 21.3) gives a generalization of the  method. However, it uses some random numbers near  and only requires at least one of them to have only small prime factors. This allows the method to detect many more primes , not just those where  has only small prime factors.

# 9.4.1 $x superscript of 2 trible bond y superscript of 2.$

Since it is the basis of the best current factorization methods, we repeat the following result from Section 9.4.

# Basic Principle

Let  be an integer and suppose there exist integers  and  with , but . Then  is composite. Moreover,  gives a nontrivial factor of .

For an example, see Example 33 in the Computer Appendices.

How do we find the numbers  and ? Let’s suppose we want to factor . Observe the following:

If we multiply the relations, we obtain

Since , we now can factor 3837523 by calculating

The other factor is .

Here is a way of looking at the calculations we just did. First, we generate squares such that when they are reduced mod  they can be written as products of small primes (in the present case, primes less than 20). This set of primes is called our factor base. We’ll discuss how to generate such squares shortly. Each of these squares gives a row in a matrix, where the entries are the exponents of the primes 2, 3, 5, 7, 11, 13, 17, 19. For example, the relation  gives the row 6, 2, 0, 0, 1, 0, 0, 0.

In addition to the preceding relations, suppose that we have also found the following relations:

We obtain the matrix

Now look for linear dependencies mod 2 among the rows. Here are three of them:

1.

2.

3.

When we have such a dependency, the product of the numbers yields a square. For example, these three yield

1.

2.

3.

Therefore, we have  for various values of  and . If , then  yields a nontrivial factor of . If , then usually  or , so we don’t obtain a factorization. In our three examples, we have

1. , but

2.  and

3.  and

We now return to the basic question: How do we find the numbers 9398, 19095, etc.? The idea is to produce squares that are slightly larger than a multiple of , so they are small mod . This means that there is a good chance they are products of small primes. An easy way is to look at numbers of the form  for small  and for various values of . Here  denotes the greatest integer less than or equal to . The square of such a number is approximately , which is approximately  mod . As long as  is not too large, this number is fairly small, hence there is a good chance it is a product of small primes.

In the preceding calculation, we have  and , for example.

The method just used is the basis of many of the best current factorization methods. The main step is to produce congruence relations

An improved version of the above method is called the quadratic sieve. A recent method, the number field sieve, uses more sophisticated techniques to produce such relations and is somewhat faster in many situations. See [Pomerance] for a description of these two methods and for a discussion of the history of factorization methods. See also Exercise 52.

Once we have several congruence relations, they are put into a matrix, as before. If we have more rows than columns in the matrix, we are guaranteed to have a linear dependence relation mod 2 among the rows. This leads to a congruence . Of course, as in the case of 1st + 5th + 6th  considered previously, we might end up with , in which case we don’t obtain a factorization. But this situation is expected to occur at most half the time. So if we have enough relations – for example, if there are several more rows than columns – then we should have a relation that yields  with . In this case  is a nontrivial factor of .

In the last half of the twentieth century, there was dramatic progress in factoring. This was partly due to the development of computers and partly due to improved algorithms. A major impetus was provided by the use of factoring in cryptology, especially the RSA algorithm. Table 9.2 gives the factorization records (in terms of the number of decimal digits) for various years.

# 9.4.2 Using $equation ar trible bond 1.$

On the surface, the Miller-Rabin test looks like it might factor  quite often; but what usually happens is that  is reached without ever having . The problem is that usually . Suppose, on the other hand, that we have some exponent , maybe not , such that  for some  with . Then it is often possible to factor .

# $equation a trible bond 1.$ Factorization Method

Suppose we have an exponent  and an integer  such that . Write  with  odd. Let , and successively define  for . If , then stop; the procedure has failed to factor . If, for some , we have , stop; the procedure has failed to factor . If, for some , we have  but , then  gives a nontrivial factor of .

Of course, if we take , then any  works. But then , so the method fails. But if  and  are found by some reasonably sensible method, there is a good chance that this method will factor .

This looks very similar to the Miller-Rabin test. The difference is that the existence of  guarantees that we have  for some , which doesn’t happen as often in the Miller-Rabin situation. Trying a few values of  has a very high probability of factoring .

Of course, we might ask how we can find an exponent . Generally, this seems to be very difficult, and this test cannot be used in practice. However, it is useful in showing that knowing the decryption exponent in the RSA algorithm allows us to factor the modulus. Moreover, if a quantum computer is built, it will perform factorizations by finding such an exponent  via its unique quantum properties. See Chapter 25.

For an example for how this method is used in analyzing RSA, see Example 32 in the Computer Appendices.


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## 18 July 2014

### Predictive Causality

We know correlation does not imply causality. What does?

I have seen the following situation countless times:

1. newspaper publishes article with the title ‘humans with long noses are more prone to having babies’, or something equally ridiculous-sounding;
2. invariably one commenter points out that the authors are idiots, because ‘correlation does not imply causation’; and, d'oh!, they only showed correlation

Which begs the question: What implies causation? Somehow, the discussion rarely proceeds to this third step.

It turns out, nothing really implies causation. So, saying that ‘correlation doesn't imply causation’ is kinda meaningless.

We can still gather evidence that makes us more or less sure that a causal relation exists. What kind of evidence? For example, correlation. But, yes, (Pearson) correlation constitutes only weak evidence. There are three problems with it:

1. It only detects linear dependences.
2. It does not say which way the influence goes.
3. It does not account for hidden variables. (Perhaps a third thing causes both two observed things.)

To alleviate the first problem, you can compute mutual information instead.

To alleviate the second problem, you can try to build a predictive model. Say you want to see if $x$ causes $y$. Then you come up with an algorithm which you feed observations of $x$ and out come predictions of $y$. If you did a good job, then you have fairly high confidence that a causal connection exists. This is basically how physics works. There is always the danger that you'll discover a setup in which your algorithm's predictions are bad.

If $x$ and $y$ are time-series and you bring back the assumption that dependency is linear, then there are nice mathematical tools you can use: see Granger causality.

I don't know of any good way to account for hidden variables, apart from ‘test, test again’. A particularly good way to test is to make sure you are in control, and

• try to systematically cover all possible values of the independent variable $x$
• try to systematically cover all possible values of all other possibly relevant variables you can think of

But, in the end, we'll never be absolutely sure that one thing causes another.

Anonymous said...

The phrase "correlation does not imply causation" can be, and has been, used for great evil.

Several years ago, some were denying that there was any possible cause-effect relationship between deployment of a taser and the occasional death that sometimes followed nearly immediately after. They squawked the same "correlation is not causation" refrain as was once used by the tobacco companies. The usage pattern of taser deployments thankfully provided the built-in control. Typically a taser is drawn and warnings issued often for minutes on end, no deaths. Oftentimes the taser is fired but misses, no deaths. Only when the taser actually makes contact are there any mysterious deaths. The pattern is crystal clear and beyond coincidence.

They carried out a mischievous campaign to attribute these apparently cause-free deaths to "excited delirium". Years later, after hundreds of "taser associated" deaths, it is finally widely accepted that tasers, especially the X-26, can cause or contribute to death. Tobacco was dealt wih by the famous Surgeon General Report. Taser's role in death was formalized by the Braidwood Inquiry.

In fact, correlation *does* imply causation (in many cases), it just requires additional evidence before one may claim a rational proof.

Anonymous said...

I've been wanting to read Judea Pearl's book "Causality", about causal Bayesian networks. can you say anything about that?

Chris Hemedinger said...

Sometimes, correlation is enough to cause us to look further. In the tobacco example (cited by another commenter), perhaps that data led to the research that uncovered a precise mechanism by which tobacco use creates conditions that promote lung cancer. From then on it's not statistics, but hard science that shows you the connection.

But other times a good predictive model is enough to make decisions, and we don't need to know the underlying causes. Young males pay more for auto insurance because a predictive model says they are more likely to have auto accidents. "Being a young male" does not cause accidents -- not directly -- but the model works (at least for the insurance companies).

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# When is the inclusion of a relative mapping space into a mapping space a cofibration?

Let $(X,A)$ and $(Y,B)$ be pairs of spaces and subspaces, let $\operatorname{Map}(X,Y)$ the space of maps $f:X\to Y$ equipped with the compact-open topology and let $\operatorname{Map}(X,A;Y,B)$ be the subspace of maps $f:X\to Y$ such that $f(A)\subseteq B$. Suppose the inclusions $A\hookrightarrow X$ and $B\hookrightarrow Y$ are cofibrations, would that be enough to ensure the inclusion $\operatorname{Map}(X,A;Y,B)\hookrightarrow\operatorname{Map}(X,Y)$ is a cofibration or are other conditions needed?

In particular if $X$ and $Y$ are well-pointed is the inclusion of the based mapping space $\operatorname{Map}_*(X,Y)\hookrightarrow\operatorname{Map}(X,Y)$ a cofibration?

It seems like this should be true for reasonably nice spaces and there are similar results. I know, for example, that if $B\hookrightarrow Y$ is a closed cofibration and $X$ is compact Hausdorff then the inclusion $\operatorname{Map}(X,B)\hookrightarrow\operatorname{Map}(X,A;Y,B)$ is a cofibration. In particular this makes based mapping spaces with compact Hausdorff domain well-pointed.

-

The inclusion $\operatorname{Map}(X,A;Y,B)\hookrightarrow \operatorname{Map}(X;Y)$ will be a cofibration whenever $(X,A)$ and $(Y,B)$ are cofibrations, under some mild point-set hypotheses. My argument requires $Y$ to be Hausdorff and $A$ and $X$ to be locally compact Hausdorff (but I'm not sure if these restrictions are necessary).

To begin with, recall that the map given by restriction $p\;\colon \operatorname{Map}(X;Y)\to \operatorname{Map}(A;Y)$ is a fibration when $A$ and $X$ are locally compact Hausdorff. This is Theorem 2.8.2 of Spanier's book.

Now it is a result of A. Strøm (Math. Scand. 22 130–142 (1969)) that if $p\;\colon E\to B$ is a fibration and $i\;\colon C\hookrightarrow B$ is a closed cofibration, then $p^{-1}(C)\hookrightarrow E$ is a cofibration.

To complete the argument, therefore, it suffices to show that $i\;\colon \operatorname{Map}(A;B)\hookrightarrow\operatorname{Map}(A;Y)$ is a closed cofibration. To see this, use the characterization of cofibrations in terms of retractions onto mapping cylinders. We have a retraction $r_Y\;\colon Y\times I \to Y\times\lbrace 0\rbrace \cup B\times I$. We can use this to define a retraction $r\;\colon\operatorname{Map}(A;Y)\times I \to \operatorname{Map}(A;Y)\times \lbrace 0\rbrace \cup \operatorname{Map}(A;B)\times I$ by setting $$(f,t)\mapsto \big( (a\mapsto p_Yr_Y(f(a),t)),t \big),$$ where $p_Y$ is the projections from the mapping cylinder onto $Y$. This shows $i$ is a cofibration; it is closed since $\operatorname{Map}(A;Y)$ is Hausdorff.

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The second entry of your formula for r has a free occurance of a. I guess it should be $\inf_{a \in A} p_I r_Y (f(a),t)$. – Karol Szumiło Aug 31 '12 at 19:16
@Karol: You are right. I have edited it to something simpler, which I think works. – Mark Grant Aug 31 '12 at 19:48
Thanks Mark, that's exactly the kind of argument I've been looking for! – Richard Manthorpe Sep 1 '12 at 12:23
@Richard: You're welcome, I had fun thinking about it! – Mark Grant Sep 3 '12 at 9:35


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Question

# Solve for x and y: x2−y9=6,x7+y3=5.

Open in App
Solution

## x7+y3=5, x2−y9=6 On taking LCM of denominators and simplifying, we get the linear equations as 3x + 7y = 105 → (1) 9x − 2y = 108 → (2) Multiply equation(1) with 3 , we get 9x + 21y = 315 → (3) Now subtract (2) from (3) 9x + 21y = 315 9x − 2y = 108 ---------------------- 23y = 207 ∴ y =20723 = 9 Put y = 9 in 3x + 7y = 105 3x + 7(9) = 105 ⇒ 3x = 105 − 63 = 42 ∴x=423=14

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# Laplace transformation Transformation Number En funktionsdomän, stapeldiagram, vinkel, område png. Laplace transformation Transformation Number En

The calculator will find the Laplace Transform of the given function. Recall that the Laplace transform of a function is F(s)=L(f(t))=\int_0^{\infty}

The function F(s) is a function of the Laplace variable, "s." We call this a Laplace domain function. So the Laplace Transform takes a time domain function, f(t), and   The Laplace transform describes signals and systems not as functions of time but rather as functions of a complex variable s. When transformed into the Laplace  In anglo-american literature there exist numerous books, devoted to the application of the Laplace transformation in technical domains such as electrotechnics,  Laplace transform, in mathematics, a particular integral transform invented by the French mathematician Pierre-Simon Laplace (1749–1827), and systematically  the Laplace transform converts integral and difierential equations into algebraic equations this is like phasors, but. • applies to general signals, not just sinusoids.

In Laplace transformation, the differential equation in the time domain is first converted or transformed into an algebraic equation in the frequency domain. Laplacetransformation er i matematikken en transformation af en funktion til en anden funktion ved hjælp af en operator. Laplacetransformationer bruges meget i fysik og teknik til at løse differentialligninger og integralligninger. I've been working on Laplace transform for a while. I can carry it out on calculation and it's amazingly helpful.

The above procedure can be summarized by Figure 43.1 Figure 43.1 In mathematics, the Laplace transform, named after its inventor Pierre-Simon Laplace ( / ləˈplɑːs / ), is an integral transform that converts a function of a real variable.

## Get the full course at: http://www.MathTutorDVD.comIn this lesson, you will learn how to apply the definition of the Laplace Transform and take the transform

The process of solution consists of three. Circuits with any type of source (so long as the function describing the source has a Laplace transform), resistors, inductors, capacitors, transformers, and/or op  2 Jul 2011 The most common presentation of the Laplace transform in undergraduate texts on ordinary differential equations (ODE) consists of a definition of  19 Feb 2018 Maxima has a fairly serviceable Laplace transform utility built-in.

### Complex Functions Theory a-4 builds on these previous texts, focusing on the general theory of the Laplace Transformation Operator. This e-book and previous titles in the series can be downloaded for free here. All theorems are accompanied by their proofs, and all equations are explained and demonstrated in detail.

Medvetenhet är grunden till all förändring, transformation och skapande av vår verklighet. Mat Uni: Laplace transformation af Eulers tal. En transformation är en operation på en funktion eller vektor som ger en Transformation (matematik) Laplace-transform · Fourier-transform · Z-transform  Laplace-omvandling - Laplace-transform är en integrerad transformation som förbinder en funktion av en komplex variabel (bild) med en funktion av en verklig  Översättnig av laplace-muunnos på engelska. Gratis Internet Ordbok. Miljontals finska-engelska översättning av laplace-muunnos.

The Laplace transform is particularly useful in solving linear ordinary differential equations such as those arising in the analysis of electronic circuits. This MATLAB function returns the Laplace Transform of f. Independent variable, specified as a symbolic variable. This variable is often called the "time variable" or the "space variable." The Laplace Transform - Theory and Applications. Ehsan Shaukat. Download PDF 2018-04-12 · We learn about some commonly used properties of the Laplace Transformation. Includes, constant multiple, linearity property and change of scale property.
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Shifat Ahmed. HOW TO USE LAPLACE HOW TO USE LAPLACE• Find differential equations that describe Find differential equations that describe system system 1 )} s ( F { L ) t ( f dt e ) t ( f )} t ( f { L ) s ( F• t is real, s is complex!

Time discretization via Laplace transformation of an integro-differential equation of parabolic type.
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# Coprime Solutions to Divisibility Relation

Let $$m$$ and $$n$$ be coprime positive integers such that $$m\geq n$$. Determine the number of ordered pairs of $$(m,n)$$ such that $\dfrac{m^3 + n^3}{m^2+ n^2 + m+n}$ is an integer.

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# Algorithm for generating all cartesian products, without rotations

(Not sure if that's the right SX site? I don't need actual code, so…)

I'm looking for an algorithm that generates all cartesian products for a list of sets, but skips tuples that are just rotations of previously issued tuples in a smart way.

Here, $(a_0,\dotsc,a_n)$ is a rotation of $(b_0,\dotsc,b_n)$ iff $\exists x : \forall i \in [0,\dotsc,n]: a_i = b_j \text{ with } j = i + x \mod n$.

For example, for $f(\{1, 2\}, \{1, 2, 3\})$ the output should be $\{(1,1), (1,2), (1,3), (2,2), (2,3)\}$, i.e. without $(2,1)$ because it's a rotation of $(1,2)$.

• It's easy to convert each tuple from/to an integer, so when about to generate a tuple, one could look if any of its rotations map to a smaller integer than the current tuple and discard it. But I'm hoping there's an algorithm that's faster than one that generates all products. – pascal Mar 18 '12 at 3:34
• By rotation I assume you mean permutation? – Deathbreath Mar 19 '12 at 15:18
• No, the tuple is supposed to be a ring, i.e. $a_0$ is the successor of $a_n$, $a_1$ that of $a_0$, etc. The order stays fixed, but which element in the ring gets index 0 may change. (edited the question with a proper expression for what I mean by "rotation". That I'm speaking about tuples isn't helping I guess, I mean circularly linked lists: don't care about the start, just the order) – pascal Mar 19 '12 at 20:35
• For a potential tuple, output only those which are lexicographically least among all their possible rotations. – hardmath Mar 8 '16 at 18:18

This is the brute-force way (in Python), maybe it's interesting to somebody:

def rotate_product(*xs):
# map integer to tuple
def from_int(i):
tup = []
for x in xs:
tup.append(x[i % len(x)])
i /= len(x)
return tuple(tup)

# map tuple to integer, or -1 if impossible.
def to_int(tup):
i = 0
mul = 1
for t, x in zip(tup, xs):
if t not in x:
return -1
i += mul * x.index(t)
mul *= len(x)
return i

# map integer to tuple, if no rotation of smaller tuple
def make_tuple(i):
tup = from_int(i)
for k in range(1, len(xs)):
if 0 <= to_int(tup[k:] + tup[:k]) < i:
return None
return tup

# total number of possible tuples
n = 1
for x in xs:
n *= len(x)

# brute force through all possibilities
for i in range(n):
t = make_tuple(i)
if t:
yield t


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# LaTeX for Secretaries

How to survive without Microsoft Word.
Changing Text into Type

To start LaTeX, type this at the prompt:

latex businessletter.latex

After you press the ENTER key, LaTeX will print some cryptic messages, two of which are most important: Output written on businessletter.dvi... and Transcript written on businessletter.log. Your letter is output into a DVI (device independent) file, a universal format, which you can turn into fax, PostScript or printer files.

If there are problems, LaTeX will sometimes stop in the middle of its work and ask you for help. When that happens, make a note of the line number at which the error has been found, then keep tapping the ENTER key until you see the “Transcript written on” message or the command prompt. Then run xdvi, and after visually finding the place where the error occurred, go back to Emacs and correct your mistake. Usually it will be a missing bracket or a misspelled command.

Previewing Documents

The best way to find errors and check the document's look before you print it is to use the xdvi previewer. It works only in the X Window System. To start it, you either choose it from a menu or type xdvi businessletter.dvi & on a command line in a terminal window and press the ENTER key.

Using xdvi is easy. The buttons on the right-hand side of a window are fairly self-explanatory, and you can press the PAGEUP/PAGEDOWN keys to flip pages back and forth or use the arrow-up/arrow-down keys to scroll the page you are looking at up and down (the scroll bars work just like in MS Windows, although with a two-button mouse, you will have to press both buttons to use them). Pressing the Q key exits xdvi.

Some Unbreakable Rules

The most common errors that can be found just by looking at a page in xdvi are the overfull boxes and sudden changes in font and style. These errors usually happen when you forget to obey some of the following rules of typesetting in LaTeX:

• Each paragraph must by separated from another by a blank line (just tap the ENTER key twice at the end of a paragraph).

• No matter how many spaces you put between words, LaTeX will treat them as one space and will format it according to its own typesetting rules. (You might change this behaviour, as described in the section “Overriding LaTeX Rules” below.)

• Quotation marks are made using  and ' or and instead of ", so to typeset ''funny'', you'll need to type ''Funny'' instead of "funny".

• Ellipsis (...) is printed with the \ldots command; consequently, to typeset bye..., you ought to type bye\ldots and not bye....

• LaTeX commands should be separated from the actual text. We achieve that in one of the following ways:

that begins above]

• bye, bye\ldots my love (note: the \ character starts a command and therefore does not need to be separated with a space, also see the list of special characters below).

• bye, bye\ldots{ }my love.

• LaTeX interprets some characters as special, and to print them literally, we need to use a special command instead. See list in “Special LaTeX Characters”.

Special LaTeX Characters
• Commands can be nested, i.e., you can put one or more commands within curly brackets of another, e.g., extsf{life is \emph{wonderful}!}.

Getting Rid of Overfull Boxes

LaTeX is very good at documenting its work and puts a lot of information, including error messages, into a log file. For businessletter.latex, it will be called businessletter.log. Every error is described there as best as LaTeX can, together with a line number, which helps to quickly find the right place in a file.

Although the visual method of finding errors with the help of xdvi is probably best for a beginner, there is one category of bugs that requires a different approach—the famous Overfull box. LaTeX produces that message when it cannot properly break a line of text or hyphenate a word, which happens quite often when trying to print a long web page address.

The quickest way of finding those messages is by using grep:

grep Overfull businessletter.log | less
`

After issuing the above command in an Xterm window, you will see either an empty window (a good thing) or a list of lines where LaTeX has put the ominous word. Go back to Emacs, find the offending lines, then use xdvi to judge how to best fix your document.

______________________


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# representation of SU(2)

The question is regarding SU(2) group and SU(2) algebra. The SU(2) group can be generated by exponentiating the generators of SU(2) algebra $X_a$ as $exp(i t_a X_a )$ with $t_a$ being three parameters. Genreally we use half the Pauli matrices as $X_a$ when discussing SU(2) algebra as well as the group. But if we use the spin-1 representation of $X_a$ (which can surely be done at the Lie Alebra level) can they be exponentiated to give another rep. of the SU(2) group (which would be 3 dimensional)?

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By the way, the group you obtain in this case is isomorphic to SO(3). If you choose infinitesimal rotations as the spin-1 representation, you obtain the SO(3) matrices directly. –  Vašek Potoček Sep 20 '12 at 19:42

## 1 Answer

Yes. A representation of a Lie algebra always exponentiates to a representation of the simply connected Lie group corresponding to it. The representations of each are in bijective correspondence this way.

I recommend the free online book by Kirillov ( http://www.math.sunysb.edu/~kirillov/mat552/liegroups.pdf ) as a reference for this business.

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It exponentiates if the Lie algebra is finite-dimensional (as in case of SU(2)), but not necessarily in infinite dimensions. –  Arnold Neumaier Sep 5 '12 at 8:40
I second a vote for Kirillov's Lie Group notes! Download them and keep a copy lest they should disappear. The university sector's shrinking faster than a pierced balloon in my home land and, although Kirillov's present home land is not mine, it is also infected with the English language bacterium which seems to be an effective economic "rationalist" virus's vector organism! Also see my answer on how several different exponentiations can happen in finite dimensions, and I also recommend Chapter 8 of John Stillwell's "Naive Lie Theory". –  WetSavannaAnimal aka Rod Vance Sep 27 '13 at 1:49


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# Why is a gyroscope's precession fastest at 45 degrees?

I did an experiment in which I varied the starting angle of elevation of a gyroscope. I noticed that at 45 degrees, the precessional period (amount of time to perform one spin) is the lowest, while at 0 degrees and 75 degrees, the precessional period is higher. If I plotted this on a graph, it looks somewhat like half a sinusoidal curve, with 45 degrees being the minimum.

Keep in mind that the starting angular frequency of the rotor/wheel in the gyroscope is always the same in the beginning.

I just want to know why a gyroscope tends to spin faster at 45 degrees.

Is there a centrifugal force that acts on the rotor? I found on a website:

http://www.gyroscopes.org/forum/questions.asp?id=1533

"The centrifugal force when the arm is at 45 degrees above or below horizontal would be about 70.7 percent of what it is at horizontal. Of course the centrifugal force would vary for every angle above or below horizontal for a fixed precession rpm. If you plotted the calculated centrifugal force for varying angles at a fixed precession rpm, the graph would be 1/2 a sinusoid."

Can anyone verify this? I didn't think centrifugal forces existed.


text:
How do you find the x and y intercepts for x+y=7?

Mar 1, 2016

$x = 7$,
$y = 7$.

Explanation:

We have to remember that equations are easier to solve if put in the form $y = m x + c$. So,
$x + y = 7$,
$y = - x + 7$.
Now we simply substitute the value of $x$ for $0$ to find the $y$ intercept, and $y$ by $0$ to find the $x$ intercept. So,
$y = - 0 + 7$,
$y = 7$. This is the $y$ intercept.
$0 = - x + 7$,
$x = 7$. This is the $x$ intercept.

Hope it Helps! :D .


text:
시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 512 MB444100.000%

## 문제

Stephen and Sergey are enrolled at the university and live on campus. So now they need to learn to cook.

The friends have learned to cook $n$ different dishes. After buying all the necessary food, the guys have decided that before they go shopping next time, they will cook the $i$-th dish exactly $a_i$ times.

Each day Sergey and Stephen choose two dishes $i$ and $j$ and cook them, cooking takes $c_{i, j}$ units of time. It's possible that $i=j$, that means that the $i$-th dish is cooked twice on that day.

The guys are quite lazy, so they want to minimize the total cooking time for all the days before the next shopping. Help them to do it!

## 입력

The first line contains a single integer $n$ ($1 \le n \le 10$) --- the number of dishes that the guys can cook.

The second line contains $n$ positive integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 50$) --- the number of times it must be cooked for each dish.

The next $n$ lines contain $n$ space separated integers $c_{i, j}$ ($1 \le c_{i, j} \le 100$), the $j$-th number in the $i$-th line denotes cooking time of pair of dishes $i$ and $j$. It is guaranteed that $c_{i, j}=c_{j, i}$.

## 출력

Print one integer: the minimum total time of cooking, or $-1$, if it is impossible to make a cooking plan such that the $i$-th dish is cooked exactly $a_i$ times.

## 예제 입력 1

3
2 2 2
1 4 3
4 4 5
3 5 6

## 예제 출력 1

10

## 예제 입력 2

2
2 39
23 9
9 23

## 예제 출력 2

-1

## 예제 입력 3

1
2
100

## 예제 출력 3

100

## 힌트

In the first test case it is optimal to cook the following pairs of dishes: $(1, 3)$, $(1, 3)$, $(2, 2)$.


text:
# Intuition behind moments of random variables [duplicate]

I am looking to understand the intuition behind moments of random variables.

I understand the first moment relating to the mean and the second one relating to the variance. But what use does the n-th moment have?

Thank you

## marked as duplicate by Hans Lundmark, Community♦May 22 '18 at 14:58

$$M_X(t) = \mathbb{E}\left[e^{tX}\right] = \mathbb{E}\left[\sum_k \frac{t^k}{k!}X^k\right] = \sum_k \frac{t^k}{k!}\mathbb{E}[X^k] = \sum_k \frac{t^k}{k!}\left.\frac{{\rm d}^k M(t)}{{\rm d}t^k}\right|_{t=0}$$
If you know all moments $\mathbb{E}[X^k]$ (or equivalently $M_X(t)$) you could try to invert the problem and actually estimate the underlaying PDF, this is know as the moment problem