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#### 11th Public Exam March 2019 Model Test

11th Standard

Reg.No. :
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Maths

Time : 02:30:00 Hrs
Total Marks : 90
20 x 1 = 20
1. If two sets A and B have 17 elements in common, then the number of elements common to the set A x B and B x A is

(a)

217

(b)

172

(c)

34

(d)

insufficient data

2. The domain of the function $f(x)=\sqrt{ x - 5 }+ \sqrt{6 - x}$is

(a)

[5, )

(b)

(- , 6)

(c)

[5, 6]

(d)

(-5, ≠6)

3. If a and b are the roots of the equation x2-kx+c = 0 then the distance between the points (a, 0) and (b, 0)

(a)

$\sqrt { { 4k }^{ 2 }-c }$

(b)

$\sqrt { { k }^{ 2 }-4c }$

(c)

$\sqrt { 4c-{ k }^{ 2 } }$

(d)

$\sqrt { k-8c }$

4. $\left( 1+cos\frac { \pi }{ 8 } \right) \left( 1+cos\frac { 3\pi }{ 8 } \right) \left( 1+cos\frac { 5\pi }{ 8 } \right) \left( 1+cos\frac { 7\pi }{ 8 } \right)$=

(a)

$\frac { 1 }{ 8 }$

(b)

$\frac { 1 }{ 2 }$

(c)

$\frac { 1 }{ \sqrt { 3 } }$

(d)

$\frac { 1 }{ \sqrt { 2 } }$

5. In any ΔABC, a(b cosC-c Cos B)=

(a)

a2

(b)

b2-c2

(c)

0

(d)

b2+c2

6. In an examination there are three multiple choice questions and each question has 5 choices. Number of ways in which a student can fail to get all answer correct is

(a)

125

(b)

124

(c)

64

(d)

63

7. The number of parallelogram formed if 5 parallel lines intersect with 4 other paralle llines is:

(a)

10

(b)

45

(c)

30

(d)

60

8. $\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+...$is:

(a)

$\frac{e^{-1}}{2}$

(b)

$\frac{e+e^{-1}}{2}$

(c)

$\frac{e-e^{-1}}{2}$

(d)

none of these

9. The length of $\bot$ from the origin to the line $\frac{x}{3}-\frac{y}{4}=1$ is

(a)

$\frac{11}{5}$

(b)

$\frac{5}{12}$

(c)

$\frac{12}{5}$

(d)

$\frac{-5}{12}$

10. The equation x2+kxy+y2-5x-7y+6=0 represents a pair of straight lines then k=

(a)

$\frac{5}{3}$

(b)

$\frac{10}{3}$

(c)

$\frac{3}{2}$

(d)

$\frac{3}{10}$

11. If x1,x2,x3 as well as y1,y2,y3 are in geometric progression with the same common ratio, then the points (x1, y1 ), (x2,y2), (x3,y3 ) are

(a)

vertices of an equilateral triangle

(b)

vertices of a right angled triangle

(c)

vertices of a right angled isosceles triangle

(d)

collinear

12. The negative of a matrix is obtained by multiplying it by

(a)

1

(b)

-1

(c)

I

(d)

AT

13. If $\overrightarrow{a}=\hat{i}+2\hat{j}+2\hat{k},|\overrightarrow{b}|=5$ and the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is ${\pi\over 6},$ then the area of the triangle formed by these two vectors as two sides, is

(a)

$7\over4$

(b)

$15\over4$

(c)

$3\over4$

(d)

$17\over4$

14. If $lim_{x \rightarrow 0}{sin \ px\over tan \ 3x}=4$ , then the value of p is

(a)

6

(b)

9

(c)

12

(d)

4

15. The points of discontinuity of the function $\frac { { x }^{ 2 }+6x+8\quad }{ { x }^{ 2 }-5x+6\quad } is$

(a)

3,2

(b)

3,-2

(c)

-3,2

(d)

-3,-2

16. If g(x)=(x2+2x+3) f(x) and f(0)=5 and $lim_{x \rightarrow 0}{f(x)-5\over x}=4$,then g'(0) is

(a)

20

(b)

14

(c)

18

(d)

12

17. $\int \sqrt{{1-x\over 1+x}}dx$ is

(a)

$\sqrt{1+x^2}+sin^{-1}x+c$

(b)

$sin^{-1}x-\sqrt{1+x^2}+c$

(c)

$log|x+\sqrt{1-x^2}|-\sqrt{1-x^2}+c$

(d)

$\sqrt{1-x^2}+log|x+\sqrt{1-x^2}|+c$

18. $\int { \frac { sin\sqrt { x } }{ x } }$ dx = ________ +c.

(a)

2 cos $\sqrt { x }$

(b)

2 sin $\sqrt { x }$

(c)

-2 sin $\sqrt { x }$

(d)

-2 cos $\sqrt { x }$

19. If X and Y be two events such that P(X/Y)=${1\over2},P(Y/X)={1\over3}$ and $P(X\cap Y)={1\over6}$then P(x$\cup$y)is

(a)

${1\over3}$

(b)

${2\over5}$

(c)

${1\over6}$

(d)

${2\over3}$

20. Two dice are thrown. It is known that the sum of the numbers on the dice was less than 6, the probability of getting a sum 3 is

(a)

$\frac { 1 }{ 18 }$

(b)

$\frac { 5 }{ 18 }$

(c)

$\frac { 1 }{ 5 }$

(d)

$\frac { 2 }{ 5 }$

21. 7 x 2 = 14
22. On the set of natural number let R be the relation defined by aRb if a + b $\le$ 6. Write down the relation by listing all the pairs. Check whether it is transitive

23. A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be 300. If after 100km, the target has an angle of depression of 450, how far is the target from the fighter jet at that instant?

24. A family of 4 brothers and 3 sisters is to be arranged in a row for a photograph, in how many ways can they be seated if
(i) all the sisters sit together
(ii) all the sisters are not together.

25. Find the 5th term in the sequence whose first three terms are 3, 3, 6 and each term after the second is the sum of the two terms preceding it.

26. Find the equation of the line through the point of intersection of the line 5x - 6y = 1 and 3x + 2y + 5 = 0 and cutting off equal intercepts on the coordinate axis.

27. Find the position vector of a point R which divides the line joining points P and Q whose position vectors are $\hat{i}+2\hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively in the ratio 2:1 externally.

28. $If\lim _{ x\rightarrow 2 }{ \frac { { x }^{ n }-{ 2 }^{ n } }{ x-2 } } =80\quad and\quad n\in N,\quad find\quad n.$

29. Find the derivatives of the following functions using first principle.f(x)=-x2+2

30. Evaluate :$\int e^{xlog2}e^x dx$

31. A die is tossed thrice. find the probability of getting an odd number atleast once?

32. 7 x 3 = 21
33. In the set Z of integers, define mRn if m - n is divisible by 7. Prove that R is an equivalence relation.

34. Find x if ${{1}\over{2}}$ log10 $(11+4\sqrt{7})$ = log10 (2+x).

35. Find the general solution of sin$\theta =-{\sqrt{3}\over 2}$.

36. Evaluate the following:
$\sqrt [ 3 ]{ 1003 }$correct to 4 places of decimals

37. The Pamban Sea Bridge is a railway bridge of length about 2065m constructed on the PalkStrait, which connects the Island town of Rameswaram to Mandapam, the main land of India. The Bridge is restricted to a uniform speed of only 12.5 m/s. If a train of length 560 m starts at the entry point of the bridge from Mandapam, then
(i) find an equation of the motion of the train.
(ii) when does the engine touch island.
(iii) when does the last coach cross the entry point of the bridge.
(iv) what is the time taken by a train to cross the bridge.

38. Determine the values of b so that the following matrices are singular:$\begin{bmatrix}b-1 &2 &3 \\3 & 1 & 2 \\ 1 & -2 &4 \end{bmatrix}$

39. Calculate $\lim _{ x\rightarrow0}{|x| }$.

40. Differentiate$\log _{ 7 }{ (\log _{ 7 }{ x } ) }$

41. Integrate the following functions with respect to x: sin 3x

42. The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that both the oil and filter need changing is 0.15.
(i) If the oil had to be changed, what is the probability that a new oil filter is needed?
(ii) If a new oil filter is needed, what is the probability that the oil has to be changed

43. 7 x 5 = 35
44. Discuss the following relations for reflexivity, symmetricity and transitivity :
On the set of natural numbers, the relation R is defined by "xRy if x + 2y = 1".

45. Determine the region in the Plane determined by the inequalities x+y≥4,2x-y>0

46. Solve $\sqrt{3}tan^2\theta+(\sqrt{3}-1)tan\theta-1=0$

47. By the principle of mathematical induction, prove that for n > 1,
$1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = {n(2n-1)(2n+1)\over 3}$

48. Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 42) + (1 + 4 + 42 + 43) + ...

49. A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time shown in the following table.

 Weight, (kg) 2 4 5 8 Length, (cm) 3 4 4.5 6

(i) Draw a graph showing the results.
(ii) Find the equation relating the length of the spring to the weight on it.
(iii) What is the actual length of the spring.
(iv) If the spring has to stretch to 9 cm long, how much weight should be added?
(v) How long will the spring be when 6 kilograms of weight on it?

50. A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x -12y = 3. The equation of its locus is .................

51. If A and B are symmetric matrices of same order, prove that
AB - BA is a skew-symmetric matrix.

52. Let A, Band C represent the angles of a $\triangle$ABC and a, band c represent the lengths of the sides opposite to them, then prove that a2 = b2 + c2 - 2bc cos A (Law of cosines)

53. According to Einstein’s theory of relativity, the mass m of a body moving with velocity v is m =${m_O\over \sqrt{1-{v^2\over c^2}}},$  where m0 is the initial mass and c is the speed of light. What happens to m as $v\rightarrow{c}^-$ Why is a left hand limit necessary?

54. $If\quad f\left( 2 \right) =4\ and f^{ ' }\left( 2 \right) =1, then find\quad \lim _{ x\rightarrow 2 }{ \frac { xf\left( 2 \right) -(2)f\left( x \right) }{ x } }$

55. Integrate the following with respect to x:$(1+x^2)^{-{1\over 2}}$

56. Evaluate $\int { \frac { { e }^{ x }dx }{ { e }^{ 2x }+6{ e }^{ x }+5 } }$

57. The probability of simultaneous occurrence of atleast one of two events A and B is p. if the probability that exactly one A, B occurs is q then prove that P($\bar { A }$) + P($\bar { B }$) = 2-2 p+q.


text:
Question

Calculate the mass percentage composition of the elements in nitric acid. (H = 1, N = 14, O = 16).

Hint: The percentage of the elements or constituents can be calculated by dividing the number of parts of the mass of the element or constituents with the molar mass of the compound multiplied to 100.

The percentage of any element or constituent in a compound is the number of parts by mass of that element or constituents present in 100 parts by mass of the compound. It can be calculated by the two steps:
Step 1: Calculated the molecular mass of the compound from its formula by adding the atomic masses of the elements present.
Step 2: Calculate the percentage of the elements or the constituents by applying the following relation:
It can be calculated by dividing the number of parts of the mass of the element or constituents with the molar mass of the compound multiplied to 100.
$\text{Percentage of the element or constituents = }\dfrac{\text{no}\text{. of parts by mass of the elements or constituents}}{\text{Mol}\text{. mass of the compound}}\text{ x 100}$
The formula of nitric acid is $HN{{O}_{3}}$
The molar mass of hydrogen is 1, the molar mass of nitrogen is 14, and the molar mass of oxygen is 16.
The molar mass of nitric acid = 1 + 14 + 3(16) = 1 + 14 + 48 = 63.
Mass percentage of hydrogen = $\dfrac{\text{mass of hydrogen}}{\text{molar mass of nitric acid}}\text{ x 100}$
Mass percentage of hydrogen = $\dfrac{1}{63}\text{ x 100 = 1}\text{.586 }\!\!%\!\!\text{ }$
Mass percentage of nitrogen = $\dfrac{\text{mass of nitrogen}}{\text{molar mass of nitric acid}}\text{ x 100}$
Mass percentage of nitrogen = $\dfrac{14}{63}\text{ x 100 = 22}\text{.22 }\!\!%\!\!\text{ }$
Mass percentage of oxygen = $\dfrac{\text{mass of oxygen}}{\text{molar mass of nitric acid}}\text{ x 100}$
Mass percentage of oxygen = $\dfrac{48}{63}\text{ x 100 = 76}\text{.19 }\!\!%\!\!\text{ }$
So, the mass percentage of hydrogen is 1.58%, nitrogen is 22.22%, and oxygen is 76.19%.

Note: In some of the compounds the mass composition of a molecule is calculated like ${{H}_{2}}O$ etc. In some compounds the mass composition of ions is calculated, the mass of ions is the same as that of its neutral atom or molecule.

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3 videos

3 videos

4 videos

3 videos

# Collection Jean Morlet Chair - 2016 - Sem 1 - Prasad - Heiermann

Relative Aspects in Representation Theory, Langlands Functoriality and Automorphic Forms Aspects relatifs en théorie de la représentation, fonctorialité de Langlands et formes automorphes

January-June 2016

Representation theory of real and $p$-adic groups, Automorphic representations, and associated $L$-functions have been among the most important topics of study in Number theory, attracting a lot of attention through many spectacular successes in the recent decades. The proof of the Shimura-Taniyama conjecture, and the Sato-Tate conjecture are two of the highlights besides the Local Langlands conjecture for GL(n) and now through the work of Arthur, the Local Langlands Conjecture for all classical groups. Thus it can be said that, at least for classical groups, representation theory and automorphic forms are reasonably well understood. If one takes this to mean the understanding of $L^{2}(G)$, or of $L^{2}(G(Q) \backslash G(A))$, then the next natural question in this optic is to understand $L^{2}(H \backslash G)$ locally, or $H(Q) \backslash H(A)$ period integrals on $G(Q) \backslash G(A)$ for $H$ a closed subgroup of G. A good framework to understand these questions in 'relative representation theory' is already there, with distinct inputs:

1. The work of Gan-Gross-Prasad, and that of Ichino-Ikeda sets the stage for some of the natural questions and their proposed answers.

2. The work of Sakellaridis and Venkatesh which sets up an ideal stage, that of Spherical subgroups where these questions could be considered.

3. The work of Jacquet and collaborators on Relative Trace formula as a tool to prove global theorems.

Our project hopes to contribute to this theme, using the expertise in Marseille on symmetric varieties.

Appears in collection : Chaire Jean-Morlet

Organizer(s) Prof. Dipendra PRASAD Tata Institute of Fundamental Research School of Mathematics, Mumbai and Prof. Volker HEIERMANN I2M Marseille Aix-Marseille Université
Date(s) 1/1/16 - 6/30/16
• 5 videos
published on October 27, 2021

## Jean-Morlet Chair: Relative trace formula, periods, L-functions and harmonic analysis / Chaire Jean-Morlet : Formule des traces relatives, périodes, fonctions L et analyse harmonique

5 videos
published on May 20, 2016

## Jean-Morlet Chair - Doctoral school: Introduction to relative aspects in representation theory, Langlands functoriality and automorphic forms / Chaire Jean-Morlet - Ecole doctorale : Introduction aux aspects relatifs dans la théorie de la représentation, la fonctorialité de Langlands

published on July 4, 2016


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# Marble game theory

In a game, there exist three piles of marbles, each pile with ,a,b, and c marbles respectively, where a,b,c are natural numbers and all different. At each turn, you can double the number of marbles in one pile by transporting marbles from one other larger pile (relative to the pile that is going to be doubled, before the doubling). The game is won when any two of the piles have an equal number of marbles.

Either show that the game can be won from any starting a,b,c, or prove that this is not the case. (In particular, prove that the game cannot be won from every starting a,b,c.)

An interesting one, but the solution come does not.

Ernst

-

A duplicate of this question, Emptying buckets by moving pebbles around, was asked (and, interestingly, received a lot more upvotes than this one). Before Brian pointed out that it was a duplicate, I came up with a solution different from the shortlist solution. (I treat the problem of emptying one of the piles, which, as discussed in Phira's answer and comments, is equivalent.)

The idea is to successively produce $0$s in the binary representations of two of the numbers, starting with the least significant bit. So assume that the last $k$ bits of $b$ and $c$ are already zero; then if neither $b$ nor $c$ is zero yet, our aim is to make the last $k+1$ bits of two of the numbers zero.

So consider the $(k+1)$-th bits of $b$ and $c$. If they're both $0$, we're done. If they're both $1$, we just need one transfer between $b$ and $c$ to make them both $0$. If one is $0$ and one is $1$, we can put the $1$ in the lesser of the two by transferring from the greater to the lesser until it becomes the lesser.

Without loss of generality, assume $b\lt c$. Now there are two cases. If $a\ge b$, we can get rid of the $1$ by a transfer from $a$ to $b$. Otherwise, $a\lt b\lt c$, and we transfer first from $b$ to $a$ and then from $c$ to $b$, thus going from $a,b,c$ to $2a,b-a,c$ to $2a,2(b-a),c+a-b$. Now the sum of the first two numbers is $2b$, which has $0$s in the last $k+1$ bits, so we can make the last $k+1$ bits of those two numbers $0$ by making transfers between them, each of which gets rid of their last $1$ bits.

If we initially assign $a$, $b$ and $c$ such that $a$ has the fewest final zeros, this strategy seems to be slightly more efficient than the shortlist strategy: Compared to the total of $103505$ transfers minimally required to solve all distinct instances with totals less than $100$, this strategy makes $172865$ transfers while the shortlist strategy makes $190994$.

-
The $2b$ trick is nice. ($2b$ or $\lnot 2b\dots$) –  Brian M. Scott Dec 8 '11 at 21:33
:-) ${}{}{}{}{}$ –  joriki Dec 8 '11 at 21:39

For strictly positive numbers this is essentially problem C3 of the IMO shortlist 1994 where we want to empty an account in the same situation. (Because you have to have two equal amounts before an amount can get zero.)

There are several websites that offer shortlist solutions, for example:

http://mks.mff.cuni.cz/kalva/short/soln/sh94c3.html

I don't know if I am supposed to copy a solution here.

-
Presumably, "natural numbers" here is meant to exclude 0 (pretty common usage). –  Alon Amit Oct 9 '11 at 8:54
For natural number read positive integer. It’s a regrettably common usage. –  Brian M. Scott Oct 9 '11 at 9:02
Positive is also not always advisable as there are languages/countries where 0 is positive and negative. –  Phira Oct 9 '11 at 9:09
@Brian M.Scott Note that the IMO problem has two parts. The first part is about emptying just one of the three accounts which is exactly equivalent to making two accounts the same size. –  Phira Oct 10 '11 at 7:22
You’re right. $\quad$ –  Brian M. Scott Oct 10 '11 at 7:42


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# Synopsis: A Quantum Machine Made of Ions

Experiments with trapped ions could prove that a quantum machine can churn through a calculation faster than a classical one.

In computer science, the renowned “extended Church-Turing thesis” says a Turing machine can perform a computation as efficiently as any physical device.  Proof the thesis is wrong would come from showing a quantum machine solves a problem significantly faster than a classical one—and would shake up the current way in which computer scientists classify the difficulty of a problem. In Physical Review Letters, Chao Shen at the University of Michigan, Ann Arbor, and colleagues propose a way to make such a quantum machine with ions.

Shen et al.’s theorized machine would perform a task, first proposed in 2011, called boson sampling. Roughly, the idea is to start with an “input” of N identical bosons, allow them to make a random walk and then find the probability (the “output”) that they are distributed over a certain set of positions. The time it takes to calculate this probability on a classical machine grows exponentially with N, while an appropriately designed quantum machine could find a solution much more efficiently.

Boson sampling experiments have already been carried out with photons, but these tests involved, at most, four particles; according to theory, ten or more are needed to see a quantum speedup. Shen et al.’s boson sampling machine instead consists of a line of trapped ions, spaced roughly $10$ micrometers apart. The “input” bosons are the quantized vibrations (phonons) in the ion string, which could be initialized by a laser; the “output” is the phonons’ final states, which shift the atoms’ internal energy levels and could therefore be read out with another laser.  The authors argue that, using existing ion trap technology, their machine could handle $20$ to $30$ bosons. – Jessica Thomas

More Features »

### Announcements

More Announcements »

## Subject Areas

Quantum Information

## Previous Synopsis

Nonlinear Dynamics

Soft Matter

## Related Articles

Magnetism

### Viewpoint: Watching a Quantum Magnet Grow in Ultracold Atoms

Two experiments watch an antiferromagnetic phase of matter emerge in ultracold Rydberg atoms, opening up a new platform for quantum simulation. Read More »

Photonics

### Viewpoint: Lost Photons Won’t Derail Quantum Sampling

A photon-based method for demonstrating the advantage of quantum over classical machines can handle photon loss, facilitating experiments. Read More »

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### Synopsis: Security Breach for Qubit Communication

Theorists predict that quantum information can’t be securely “locked” in a vault. Read More »


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# NCERT Class 11-Math's: Chapter –2 Relations and Functions Part 4

Get top class preparation for NCO right from your home: fully solved questions with step-by-step explanation- practice your way to success.

Question 10:

Find the domain of the function given by

Given that is defined

Or

Or

Or

Hence Domain

## Objective Type Questions

Choose the correct answer out of the four given possible answers (M.C.Q.)

Question 11:

The domain of the function defined by is

(A)

(B)

(C)

(D) None of these

The correct answer is (D). Given that

Where

Thus is not defined for any .

Hence f is not defined for any , i.e. Domain of is none of the given options.

Question 12:

If is equal to

(A)

(B)

(C)

(D)

The correct choice is C.

Since

Hence,

Question 13:

Let A and B be any two sets such that then the total number of functions is equal to __________.

Any element of set A, say can be connected with the element of set B in p ways. Hence, there are exactly functions.

Question 14:

Let f and g be two functions given by

then. Domain of is __________

Since Domain of and Domain of , therefore the domain of

## Short Answer Type

Question 1:

Let and . Determine

(i)

(ii)

(iii)

(iv)

(i)

(ii)

(iii)

(iv)

Question 2:

If , where

W is the set of whole numbers.


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# Exponential distribution¶

The exponential distribution is the probability distribution that describes the waiting times between successive events following a Poisson distribution with constant average rate $$\lambda$$. The probability density function for $$x \ge 0$$ is given by

$f(x; \lambda) = \lambda e^{-\lambda x},$

and the cumulative distribution function is

$F(x; \lambda) = 1 - e^{-\lambda x}.$

The expected value and variance are as follows

$\mathrm{E}[X] = \frac{1}{\lambda}. \quad \mathrm{Var}[X] = \frac{1}{\lambda^2}.$

The exponential distribution is often used to test revenue metrics like ARPPU (average revenue per paying user), i.e, ARPPU = Revenue / Paying Users.

class cprior.models.ExponentialModel(name='', shape=0.001, rate=0.001)

Bases: cprior.cdist.gamma.GammaModel

Bayesian model with an exponential likelihood and a gamma prior distribution.

Given data samples $$\mathbf{x} = (x_1, \ldots, x_n)$$ from an exponential distribution with parameter $$\lambda$$, the posterior distribution is

$\lambda | \mathbf{x} \sim \mathcal{G}\left(\alpha + n, \beta + \sum_{i=1}^n x_i \right).$

with prior parameters $$\alpha$$ (shape) and $$\beta$$ (rate).

Parameters: name (str (default="")) – Model name. shape (float (default=0.001)) – Prior parameter shape. rate (float (default=0.001)) – Prior parameter rate.
n_samples_

Number of samples.

Type: int
cdf(x)

Cumulative distribution function of the posterior distribution.

Parameters: x (array-like) – Quantiles. cdf – Cumulative distribution function evaluated at x. numpy.ndarray
credible_interval(interval_length)

Credible interval of the posterior distribution.

Parameters: interval_length (float (default=0.9)) – Compute interval_length% credible interval. This is a value in [0, 1]. interval – Lower and upper credible interval limits. tuple
mean()

Mean of the posterior distribution.

Returns: mean float
pdf(x)

Probability density function of the posterior distribution.

Parameters: x (array-like) – Quantiles. pdf – Probability density function evaluated at x. numpy.ndarray
ppf(q)

Percent point function (quantile) of the posterior distribution.

Parameters: x (array-like) – Lower tail probability. ppf – Quantile corresponding to the lower tail probability q. numpy.ndarray
ppmean()

Posterior predictive mean.

If $$X$$ follows an exponential distribution with parameter $$\lambda$$, then the posterior predictive expected value is given by

$\mathrm{E}[X] = \frac{\beta}{\alpha - 1},$

where $$\alpha$$ and $$\beta$$ are the posterior values of the parameters.

Returns: mean float
pppdf(x)

Posterior predictive probability density function.

If $$X$$ follows an exponential distribution with parameter $$\lambda$$, then the posterior predictive probability density function is given by

$f(x; \alpha, \beta) = \frac{\alpha \beta^{\alpha}}{ (\beta + x)^{\alpha + 1}}.$

where $$\alpha$$ and $$\beta$$ are the posterior values of the parameters.

Parameters: x (array-like) – Quantiles. pdf – Probability density function evaluated at x. float
ppvar()

Posterior predictive variance.

If $$X$$ follows a Poisson distribution with parameter $$\lambda$$, then the posterior predictive variance is given by

$\mathrm{Var}[X] = \frac{\alpha \beta^2}{(\alpha - 1)^2 (\alpha - 2)},$

where $$\alpha$$ and $$\beta$$ are the posterior values of the parameters.

Returns: var float
rate_posterior

Posterior parameter beta (rate).

Returns: beta float
rvs(size=1, random_state=None)

Random variates of the posterior distribution.

Parameters: size (int (default=1)) – Number of random variates. random_state (int or None (default=None)) – The seed used by the random number generator. rvs – Random variates of given size. numpy.ndarray or scalar
shape_posterior

Posterior parameter alpha (shape).

Returns: alpha float
std()

Standard deviation of the posterior distribution.

Returns: std float
update(data)

Update posterior parameters with new data.

Parameters: data (array-like, shape = (n_samples)) – Data samples from an exponential distribution.
var()

Variance of the posterior distribution.

Returns: var float
class cprior.models.ExponentialABTest(modelA, modelB, simulations=1000000, random_state=None)

Bases: cprior.cdist.gamma.GammaABTest

Exponential A/B test.

Parameters: modelA (object) – The control model. modelB (object) – The variation model. simulations (int or None (default=1000000)) – Number of Monte Carlo simulations. random_state (int or None (default=None)) – The seed used by the random number generator.
expected_loss(method='exact', variant='A', lift=0)

Compute the expected loss. This is the expected uplift lost by choosing a given variant.

• If variant == "A", $$\mathrm{E}[\max(B - A - lift, 0)]$$
• If variant == "B", $$\mathrm{E}[\max(A - B - lift, 0)]$$
• If variant == "all", both.

If lift is positive value, the computation method must be Monte Carlo sampling.

Parameters: method (str (default="exact")) – The method of computation. Options are “exact” and “MC”. variant (str (default="A")) – The chosen variant. Options are “A”, “B”, “all”. lift (float (default=0.0)) – The amount of uplift. exoected_loss float or tuple of floats
expected_loss_ci(method='MC', variant='A', interval_length=0.9, ci_method='ETI')

Compute credible intervals on the difference distribution of $$Z = B-A$$ and/or $$Z = A-B$$.

• If variant == "A", $$Z = B - A$$
• If variant == "B", $$Z = A - B$$
• If variant == "all", both.
Parameters: method (str (default="MC")) – The method of computation. Options are “asymptotic” and “MC”. variant (str (default="A")) – The chosen variant. Options are “A”, “B”, “all”. interval_length (float (default=0.9)) – Compute interval_length% credible interval. This is a value in [0, 1]. ci_method (str (default="ETI")) – Method to compute credible intervals. Supported methods are Highest Density interval (method="HDI) and Equal-tailed interval (method="ETI"). Currently, method="HDI is only available for method="MC". expected_loss_ci np.ndarray or tuple of np.ndarray
expected_loss_relative(method='exact', variant='A')

Compute expected relative loss for choosing a variant. This can be seen as the negative expected relative improvement or uplift.

• If variant == "A", $$\mathrm{E}[(B - A) / A]$$
• If variant == "B", $$\mathrm{E}[(A - B) / B]$$
• If variant == "all", both.
Parameters: method (str (default="exact")) – The method of computation. Options are “exact” and “MC”. variant (str (default="A")) – The chosen variant. Options are “A”, “B”, “all”. expected_loss_relative float or tuple of floats
expected_loss_relative_ci(method='MC', variant='A', interval_length=0.9, ci_method='ETI')

Compute credible intervals on the relative difference distribution of $$Z = (B-A)/A$$ and/or $$Z = (A-B)/B$$.

• If variant == "A", $$Z = (B-A)/A$$
• If variant == "B", $$Z = (A-B)/B$$
• If variant == "all", both.
Parameters: method (str (default="MC")) – The method of computation. Options are “asymptotic”, “exact” and “MC”. variant (str (default="A")) – The chosen variant. Options are “A”, “B”, “all”. interval_length (float (default=0.9)) – Compute interval_length% credible interval. This is a value in [0, 1]. ci_method (str (default="ETI")) – Method to compute credible intervals. Supported methods are Highest Density interval (method="HDI) and Equal-tailed interval (method="ETI"). Currently, method="HDI is only available for method="MC". expected_loss_relative_ci np.ndarray or tuple of np.ndarray
probability(method='exact', variant='A', lift=0)

Compute the error probability or chance to beat control.

• If variant == "A", $$P[A > B + lift]$$
• If variant == "B", $$P[B > A + lift]$$
• If variant == "all", both.

If lift is positive value, the computation method must be Monte Carlo sampling.

Parameters: method (str (default="exact")) – The method of computation. Options are “exact” and “MC”. variant (str (default="A")) – The chosen variant. Options are “A”, “B”, “all”. lift (float (default=0.0)) – The amount of uplift. probability float or tuple of floats
update_A(data)

Update posterior parameters for variant A with new data samples.

Parameters: data (array-like, shape = (n_samples)) –
update_B(data)

Update posterior parameters for variant B with new data samples.

Parameters: data (array-like, shape = (n_samples)) –
class cprior.models.ExponentialMVTest(models, simulations=1000000, random_state=None, n_jobs=None)

Bases: cprior.cdist.gamma.GammaMVTest

Exponential Multivariate test.

Parameters: models (dict) – The control and variations models. simulations (int or None (default=1000000)) – Number of Monte Carlo simulations. random_state (int or None (default=None)) – The seed used by the random number generator.
expected_loss(method='exact', control='A', variant='B', lift=0)

Compute the expected loss. This is the expected uplift lost by choosing a given variant, i.e., $$\mathrm{E}[\max(control - variant - lift, 0)]$$.

If lift is positive value, the computation method must be Monte Carlo sampling.

Parameters: method (str (default="exact")) – The method of computation. Options are “exact” and “MC”. control (str (default="A")) – The control variant. variant (str (default="B")) – The tested variant. lift (float (default=0.0)) – The amount of uplift. expected_loss float
expected_loss_ci(method='MC', control='A', variant='B', interval_length=0.9, ci_method='ETI')

Compute credible intervals on the difference distribution of $$Z = control-variant$$.

Parameters: method (str (default="MC")) – The method of computation. Options are “asymptotic” and “MC”. control (str (default="A")) – The control variant. variant (str (default="B")) – The tested variant. interval_length (float (default=0.9)) – Compute interval_length% credible interval. This is a value in [0, 1]. ci_method (str (default="ETI")) – Method to compute credible intervals. Supported methods are Highest Density interval (method="HDI) and Equal-tailed interval (method="ETI"). Currently, method="HDI is only available for method="MC". expected_loss_ci np.ndarray or tuple of np.ndarray
expected_loss_relative(method='exact', control='A', variant='B')

Compute expected relative loss for choosing a variant. This can be seen as the negative expected relative improvement or uplift, i.e., $$\mathrm{E}[(control - variant) / variant]$$.

Parameters: method (str (default="exact")) – The method of computation. Options are “exact” and “MC”. control (str (default="A")) – The control variant. variant (str (default="B")) – The tested variant. expected_loss_relative float
expected_loss_relative_ci(method='MC', control='A', variant='B', interval_length=0.9, ci_method='ETI')

Compute credible intervals on the relative difference distribution of $$Z = (control - variant) / variant$$.

Parameters: method (str (default="MC")) – The method of computation. Options are “asymptotic”, “exact” and “MC”. control (str (default="A")) – The control variant. variant (str (default="B")) – The tested variant. interval_length (float (default=0.9)) – Compute interval_length% credible interval. This is a value in [0, 1]. ci_method (str (default="ETI")) – Method to compute credible intervals. Supported methods are Highest Density interval (method="HDI) and Equal-tailed interval (method="ETI"). Currently, method="HDI is only available for method="MC". expected_loss_relative_ci np.ndarray or tuple of np.ndarray
expected_loss_relative_vs_all(method='MLHS', control='A', variant='B', mlhs_samples=1000)

Compute the expected relative loss against all variations. For example, given variants “A”, “B”, “C” and “D”, and choosing variant=”B”, we compute $$\mathrm{E}[(\max(A, C, D) - B) / B]$$.

Parameters: method (str (default="MLHS")) – The method of computation. Options are “MC” (Monte Carlo), “MLHS” (Monte Carlo + Median Latin Hypercube Sampling) and “quad” (numerical integration). variant (str (default="B")) – The chosen variant. mlhs_samples (int (default=1000)) – Number of samples for MLHS method. expected_loss_relative_vs_all float
expected_loss_vs_all(method='quad', variant='B', lift=0, mlhs_samples=1000)

Compute the expected loss against all variations. For example, given variants “A”, “B”, “C” and “D”, and choosing variant=”B”, we compute $$\mathrm{E}[\max(\max(A, C, D) - B, 0)]$$.

If lift is positive value, the computation method must be Monte Carlo sampling.

Parameters: method (str (default="quad")) – The method of computation. Options are “MC” (Monte Carlo), “MLHS” (Monte Carlo + Median Latin Hypercube Sampling) and “quad” (numerical integration). variant (str (default="B")) – The chosen variant. lift (float (default=0.0)) – The amount of uplift. mlhs_samples (int (default=1000)) – Number of samples for MLHS method. expected_loss_vs_all float
probability(method='exact', control='A', variant='B', lift=0)

Compute the error probability or chance to beat control, i.e., $$P[variant > control + lift]$$.

If lift is positive value, the computation method must be Monte Carlo sampling.

Parameters: method (str (default="exact")) – The method of computation. Options are “exact” and “MC”. control (str (default="A")) – The control variant. variant (str (default="B")) – The tested variant. lift (float (default=0.0)) – The amount of uplift. probability float
probability_vs_all(method='quad', variant='B', lift=0, mlhs_samples=1000)

Compute the error probability or chance to beat all variations. For example, given variants “A”, “B”, “C” and “D”, and choosing variant=”B”, we compute $$P[B > \max(A, C, D) + lift]$$.

If lift is positive value, the computation method must be Monte Carlo sampling.

Parameters: method (str (default="quad")) – The method of computation. Options are “MC” (Monte Carlo), “MLHS” (Monte Carlo + Median Latin Hypercube Sampling) and “quad” (numerical integration). variant (str (default="B")) – The chosen variant. lift (float (default=0.0)) – The amount of uplift. mlhs_samples (int (default=1000)) – Number of samples for MLHS method. probability_vs_all float
update(data, variant)

Update posterior parameters for a given variant with new data samples.

Parameters: data (array-like, shape = (n_samples)) – variant (str) –


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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Problem Solving with Linear Graphs ( Read ) | Algebra | CK-12 Foundation

# Problem Solving with Linear Graphs

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Problem Solving with Linear Graphs

What if you borrowed $2000 from your parents for a used car? At the end of the summer you pay them back$750 that you earned from your summer job. Throughout the school year, you will pay them another $125 per month from your part-time job. How many months will it take you to pay them back? After completing this Concept, you'll be able to solve problems like this one using graphs. ### Watch This ### Guidance In this chapter, we’ve been solving problems where quantities are linearly related to each other. In this section, we’ll look at a few examples of linear relationships that occur in real-world problems, and see how we can solve them using graphs. Remember back to our Problem Solving Plan: 1. Understand the Problem 2. Devise a Plan—Translate 3. Carry Out the Plan—Solve 4. Look—Check and Interpret #### Example A A cell phone company is offering its costumers the following deal: You can buy a new cell phone for$60 and pay a monthly flat rate of $40 per month for unlimited calls. How much money will this deal cost you after 9 months? Solution Let’s follow the problem solving plan. Step 1: The phone costs$60; the calling plan costs $40 per month. Let $x =$ number of months. Let $y =$ total cost. Step 2: We can solve this problem by making a graph that shows the number of months on the horizontal axis and the cost on the vertical axis. Since you pay$60 when you get the phone, the $y-$ intercept is (0, 60).

You pay $40 for each month, so the cost rises by$40 for 1 month, so the slope is 40.

We can graph this line using the slope-intercept method.

Step 3: The question was “How much will this deal cost after 9 months?” We can now read the answer from the graph. We draw a vertical line from 9 months until it meets the graph, and then draw a horizontal line until it meets the vertical axis.

We see that after 9 months you pay approximately $420. Step 4: To check if this is correct, let’s think of the deal again. Originally, you pay$60 and then $40 a month for 9 months. $\text{Phone} & = \60\\\text{Calling plan} & = \40 \times 9 = \360\\\text{Total cost} & = \420.$ The answer checks out. #### Example B A stretched spring has a length of 12 inches when a weight of 2 lbs is attached to the spring. The same spring has a length of 18 inches when a weight of 5 lbs is attached to the spring. What is the length of the spring when no weights are attached? Solution Step 1: We know: the length of the spring = 12 inches when weight = 2 lbs the length of the spring = 18 inches when weight = 5 lbs We want: the length of the spring when weight = 0 lbs Let $x =$ the weight attached to the spring. Let $y =$ the length of the spring. Step 2: We can solve this problem by making a graph that shows the weight on the horizontal axis and the length of the spring on the vertical axis. We have two points we can graph: When the weight is 2 lbs, the length of the spring is 12 inches. This gives point (2, 12). When the weight is 5 lbs, the length of the spring is 18 inches. This gives point (5, 18). Graphing those two points and connecting them gives us our line. Step 3: The question was: “What is the length of the spring when no weights are attached? We can answer this question by reading the graph we just made. When there is no weight on the spring, the $x-$ value equals zero, so we are just looking for the $y-$ intercept of the graph. On the graph, the $y-$ intercept appears to be approximately 8 inches. Step 4: To check if this correct, let’s think of the problem again. You can see that the length of the spring goes up by 6 inches when the weight is increased by 3 lbs, so the slope of the line is $\frac{6 \ inches}{3 \ lbs} = 2 \ inches/lb$ . To find the length of the spring when there is no weight attached, we can look at the spring when there are 2 lbs attached. For each pound we take off, the spring will shorten by 2 inches. If we take off 2 lbs, the spring will be shorter by 4 inches. So, the length of the spring with no weights is 12 inches - 4 inches = 8 inches. The answer checks out. #### Example C Christine took 1 hour to read 22 pages of Harry Potter. She has 100 pages left to read in order to finish the book. How much time should she expect to spend reading in order to finish the book? Solution Step 1: We know - Christine takes 1 hour to read 22 pages. We want - How much time it takes to read 100 pages. Let $x =$ the time expressed in hours. Let $y =$ the number of pages. Step 2: We can solve this problem by making a graph that shows the number of hours spent reading on the horizontal axis and the number of pages on the vertical axis. We have two points we can graph: Christine takes 1 hour to read 22 pages. This gives point (1, 22). A second point is not given, but we know that Christine would take 0 hours to read 0 pages. This gives point (0, 0). Graphing those two points and connecting them gives us our line. Step 3: The question was: “How much time should Christine expect to spend reading 100 pages?” We can find the answer from reading the graph - we draw a horizontal line from 100 pages until it meets the graph and then we draw the vertical until it meets the horizontal axis. We see that it takes approximately 4.5 hours to read the remaining 100 pages. Step 4: To check if this correct, let’s think of the problem again. We know that Christine reads 22 pages per hour - this is the slope of the line or the rate at which she is reading. To find how many hours it takes her to read 100 pages, we divide the number of pages by the rate. In this case, $\frac{100 \ pages}{22 \ pages/hour} = 4.54 \ hours$ . This is very close to the answer we got from reading the graph. The answer checks out. Watch this video for help with the Examples above. ### Vocabulary The four steps of the problem solving plan when using graphs are: 1. Understand the Problem 2. Devise a Plan—Translate: Make a graph. 3. Carry Out the Plan—Solve: Use the graph to answer the question asked. 4. Look—Check and Interpret ### Guided Practice Aatif wants to buy a surfboard that costs$249. He was given a birthday present of $50 and he has a summer job that pays him$6.50 per hour. To be able to buy the surfboard, how many hours does he need to work?

Solution

Step 1: We know - The surfboard costs $249. Aatif has$50.

His job pays $6.50 per hour. We want - How many hours Aatif needs to work to buy the surfboard. Let $x =$ the time expressed in hours Let $y =$ Aatif’s earnings Step 2: We can solve this problem by making a graph that shows the number of hours spent working on the horizontal axis and Aatif’s earnings on the vertical axis. Aatif has$50 at the beginning. This is the $y-$ intercept: (0, 50).

He earns $6.50 per hour. This is the slope of the line. We can graph this line using the slope-intercept method. We graph the $y-$ intercept of (0, 50), and we know that for each unit in the horizontal direction, the line rises by 6.5 units in the vertical direction. Here is the line that describes this situation. Step 3: The question was: “How many hours does Aatif need to work to buy the surfboard?” We find the answer from reading the graph - since the surfboard costs$249, we draw a horizontal line from $249 on the vertical axis until it meets the graph and then we draw a vertical line downwards until it meets the horizontal axis. We see that it takes approximately 31 hours to earn the money. Step 4: To check if this correct, let’s think of the problem again. We know that Aatif has$50 and needs $249 to buy the surfboard. So, he needs to earn $\249 - \50 = \199$ from his job. His job pays$6.50 per hour. To find how many hours he need to work we divide: $\frac{\199}{\6.50/hour} = 30.6 \ hours$ . This is very close to the answer we got from reading the graph.

### Explore More

Solve the following problems by making a graph and reading it.

1. A gym is offering a deal to new members. Customers can sign up by paying a registration fee of $200 and a monthly fee of$39.
1. How much will this membership cost a member by the end of the year?
2. The old membership rate was $49 a month with a registration fee of$100. How much more would a year’s membership cost at that rate?
3. Bonus: For what number of months would the two membership rates be the same?
2. A candle is burning at a linear rate. The candle measures five inches two minutes after it was lit. It measures three inches eight minutes after it was lit.
1. What was the original length of the candle?
2. How long will it take to burn down to a half-inch stub?
3. Six half-inch stubs of candle can be melted together to make a new candle measuring $2\frac{5}{6}$ inches (a little wax gets lost in the process). How many stubs would it take to make three candles the size of the original candle?
3. A dipped candle is made by taking a wick and dipping it repeatedly in melted wax. The candle gets a little bit thicker with each added layer of wax. After it has been dipped three times, the candle is 6.5 mm thick. After it has been dipped six times, it is 11 mm thick.
1. How thick is the wick before the wax is added?
2. How many times does the wick need to be dipped to create a candle 2 cm thick?
4. Tali is trying to find the thickness of a page of his telephone book. In order to do this, he takes a measurement and finds out that 55 pages measures $\frac{1}{8} \ inch$ . What is the thickness of one page of the phone book?
5. Bobby and Petra are running a lemonade stand and they charge 45 cents for each glass of lemonade. In order to break even they must make $25. 1. How many glasses of lemonade must they sell to break even? 2. When they’ve sold$18 worth of lemonade, they realize that they only have enough lemons left to make 10 more glasses. To break even now, they’ll need to sell those last 10 glasses at a higher price. What does the new price need to be?
6. Dale is making cookies using a recipe that calls for 2.5 cups of flour for two dozen cookies. How many cups of flour does he need to make five dozen cookies?
7. To buy a car, Jason makes a down payment of $1500 and pays$350 per month in installments.
1. How much money has Jason paid at the end of one year?
2. If the total cost of the car is $8500, how long will it take Jason to finish paying it off? 3. The resale value of the car decreases by$100 each month from the original purchase price. If Jason sells the car as soon as he finishes paying it off, how much will he get for it?
8. Anne transplants a rose seedling in her garden. She wants to track the growth of the rose so she measures its height every week. On the third week, she finds that the rose is 10 inches tall and on the eleventh week she finds that the rose is 14 inches tall. Assuming the rose grows linearly with time, what was the height of the rose when Anne planted it?
9. Ravi hangs from a giant spring whose length is 5 m. When his child Nimi hangs from the spring its length is 2 m. Ravi weighs 160 lbs and Nimi weighs 40 lbs. Write the equation for this problem in slope-intercept form. What should we expect the length of the spring to be when his wife Amardeep, who weighs 140 lbs, hangs from it?
10. Nadia is placing different weights on a spring and measuring the length of the stretched spring. She finds that for a 100 gram weight the length of the stretched spring is 20 cm and for a 300 gram weight the length of the stretched spring is 25 cm.
1. What is the unstretched length of the spring?
2. If the spring can only stretch to twice its unstretched length before it breaks, how much weight can it hold?
11. Andrew is a submarine commander. He decides to surface his submarine to periscope depth. It takes him 20 minutes to get from a depth of 400 feet to a depth of 50 feet.
1. What was the submarine’s depth five minutes after it started surfacing?
2. How much longer would it take at that rate to get all the way to the surface?
12. Kiersta’s phone has completely run out of battery power when she puts it on the charger. Ten minutes later, when the phone is 40% recharged, Kiersta’s friend Danielle calls and Kiersta takes the phone off the charger to talk to her. When she hangs up 45 minutes later, her phone has 10% of its charge left. Then she gets another call from her friend Kwan.
1. How long can she spend talking to Kwan before the battery runs out again?
2. If she puts the phone back on the charger afterward, how long will it take to recharge completely?
13. Marji is painting a 75-foot fence. She starts applying the first coat of paint at 2 PM, and by 2:10 she has painted 30 feet of the fence. At 2:15, her husband, who paints about $\frac{2}{3}$ as fast as she does, comes to join her.
1. How much of the fence has Marji painted when her husband joins in?
2. When will they have painted the whole fence?
3. How long will it take them to apply the second coat of paint if they work together the whole time?

### Texas Instruments Resources

In the CK-12 Texas Instruments Algebra I FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9614 .


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# Experimentally based description of harp plucking

1 IJLRDA-LAM - Lutheries - Acoustique - Musique
DALEMBERT - Institut Jean Le Rond d'Alembert
Abstract : This paper describes an experimental study of string plucking for the classical harp. Its goal is to characterize the playing parameters that play the most important roles in expressivity, and in the way harp players recognize each other, even on isolated notes-what we call the acoustical signature'' of each player. We have designed a specific experimental setup using a high-speed camera that tracks some markers on the fingers and on the string. This provides accurate three-dimensional positioning of the finger and of the string throughout the plucking action, in different musical contexts. From measurements of ten harp players, combined with measurements of the soundboard vibrations, we extract a set of parameters that finely control the initial conditions of the string's free oscillations. Results indicate that these initial conditions are typically a complex mix of displacement and velocity, with additional rotation. Although remarkably reproducible by a single player-and the more so for professional players-we observe that some of these control parameters vary significantly from one player to another. [DOI: 10.1121/1.3651246]
Document type :
Journal articles
Complete list of metadatas

https://hal.archives-ouvertes.fr/hal-01461750
Contributor : Gestionnaire Hal-Upmc <>
Submitted on : Wednesday, February 8, 2017 - 2:13:29 PM
Last modification on : Thursday, July 4, 2019 - 11:00:07 AM

### Citation

Delphine Chadefaux, Jean-Loic Le Carrou, Benoît Fabre, Laurent Daudet. Experimentally based description of harp plucking. Journal of the Acoustical Society of America, Acoustical Society of America, 2012, 131 (1, 2), pp.844-855. ⟨10.1121/1.3651246⟩. ⟨hal-01461750⟩

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# “Understanding” $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$

I have heard people say that a major goal of number theory is to understand the absolute Galois group of the rational numbers $G = \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. What do people mean when they say this? The Kronecker-Weber theorem gives a good idea of what the abelianization of $G$ looks like. But in one of Richard Taylor's MSRI talks, Taylor said that he's never heard of anyone proposing a similar direct description of $G$ and that to understand $G$ one studies the representations of $G$.

I know that there is a strong interest in showing the Langlands reciprocity conjecture [Edit: What I had in mind in writing this is evidently Clozel's conjecture, not the Langlands reciprocity conjecture - see Kevin Buzzard's post below] - that $L$-functions attached to $\ell$-adic Galois representations coincide with $L$-functions attached to certain automorphic representations. And I've heard people refer to the Tannakian philosophy which I understand as (roughly speaking) asserting that $G$ is determined by all of its finite dimensional representations.

Here is a representation of $G$ understood not to be a representation of $G$ as an abstract group but as a group together with a labeling of some of the conjugacy classes of G by rational primes (the Frobenius elements)?

When people talk about "understanding $G$" do they mean proving [Edit: Clozel's conjecture] (in view of the Tannakian philosophy)? If not, what do they mean? If so, this conceptualization seems quite abstract to me. Is this what people mean when they say "understand $G$"? Can [Edit: Clozel's conjecture] be used to give more tangible statements about $G$?

Something that I have in mind as I write this is the inverse Galois problem (does every finite group occur as a Galois group of a normal extension of $\mathbb{Q}$?) and Gross' conjecture (mostly proven by now) that for each prime $p$ there exists a nonsolvable extension of $\mathbb{Q}$ ramified only at $p$. But I am open to and interested in other senses and respects in which one might "understand" $G$.

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Thanks to the respondees so far. I'm hoping for a more complete answer and so decided to offer a bounty. –  Jonah Sinick Oct 31 '09 at 5:05
Also, sorry for the nitpick, but I changed slightly the TeXing in the title -- feel free to revert. + Here's a quote from A.N.Parshin: "Langlands program sounds nice, but then I go and ask people -- if you go through with it, would you be able to get the full structure of absolute Galois group? And they say no, it won't be still completely known. And [studying the absolute Galois group] is, I think, more interesting. –  Ilya Nikokoshev Oct 31 '09 at 10:57
Ilya, there are some interesting things in your post and I thank you for writing it, the respect in which I did not feel satisfied by your response is that it didn't (directly) address any of the questions that I asked above (about what people mean when they talk about understanding G, and whether the Langlands reciprocity conjecture has more tangible/explicit interpretations concerning the structure of G such as offering the possibility of a solution to the inverse Galois problem) –  Jonah Sinick Oct 31 '09 at 18:22
The conjecture that Kevin attributes to Clozel, I would in fact attribute to Langlands, and call "the Langlands reciprocity conjecture", as you did originally. (Here "reciprocity" is used in distinction to "functoriality", which is roughly what Kevin is referring to when he writes about Langlands's conjectures.) –  Emerton Feb 12 '10 at 6:56
Dear @Hassan Jolany, you have edited over 10 old threads in the past hour. Please never edit more than three old threads in any period of 24 hours. Please consult the relevant meta thread. Also, some people at MathOverflow are not terribly fond of having posts edited simply for the purpose of adding latex/mathjax: see the meta thread meta.mathoverflow.net/questions/478/… for more information. –  Ricardo Andrade May 9 '14 at 20:07

What would it mean to understand this Galois group? You could mean several things.

You could mean trying to give the group in terms of some smallish generators and relations. This would be nice, and help to answer questions like the inverse Galois problem that Greg Muller mentioned, and having a certain family of "generating" Galois automorphisms would allow you to study questions about e.g. the representation theory in quite explicit terms. However, the Galois group is an uncountable profinite group, and so to give any short description in terms of generators and relations leads you into subtle issues about which topology you want to impose.

You could also ask for a coherent system of names for all Galois automorphisms, so that you can distinguish them and talk about them on an individual basis. One system of names comes from the dessins d'enfant that Ilya mentioned: associated to a Galois automorphism we have some associated data.

• We have its image under the cyclotomic character, which tells us how it acts on roots of unity. By the Kronecker-Weber theorem this tells us about the abelianization of the Galois group.
• We also have an element in the free profinite group on two generators, which (roughly speaking) tells us something about how abysmally acting on the coefficients of a power series fails to commute with analytic continuation.

These two names satisfy some relations, called the $2$-, $3$-, and $5$-cycle relation, which are conjectured to generate all relations (at least the last time I checked), but it is difficult to know whether they actually do so. If they do, then the Galois group is the so-called Grothendieck-Teichmüller group.

The problem with this perspective is that the names aren't very explicit (and we don't expect them to be: we may need the axiom of choice to show they exist, and there are only two Galois automorphisms of $\mathbb{C}$ that are measurable functions!) and it seems to be a difficult problem to determine whether the Grothendieck-Teichmuller group really is the whole thing. (Or it was the last time I checked.)

However, the cyclotomic character is a nice, and fairly canonical, name associated for Galois automorphisms. We could try to generalize this: there are Kummer characters telling us what a Galois automorphism does to the system of real positive roots of a positive rational number number (these determine a compatible system of roots of unity, or equivalent an element of the Tate module of the roots of unity). This points out one of the main difficulties, though: we had to make choices of roots of unity to act on, and if Galois theory taught us nothing else it is that different choices of roots of an irreducible polynomial should be viewed as indistinguishable. Different choices differ by conjugation in the Galois group.

This brings us to the point JSE was making: if we take the "symmetry" point of view seriously, we should only be interested in conjugacy-invariant information about the Galois group. Assigning names to elements or giving a presentation doesn't really mesh with the core philosophy.

So this brings us to how many people here have mentioned understanding the Galois group: you understand it by how it manifests, in terms of its representations (as permutations, or on dessins, or by representations, or by its cohomology), because this is how it's most useful. Then you can study arithmetic problems by applying knowledge about this. If I have two genus $0$ curves over $\mathbb{Q}$, what information distinguishes them? If I have two lifts of the same complex elliptic curve to $\mathbb{Q}$, are they the same? How can I get information about a reduction of an abelian variety mod $p$ in terms of the Galois action on its torsion points? Et cetera.

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The 2, 3 and 5-cycle relations of GT translate the action on the moduli space curbes of genus 0 with n marked points. I don't think it was seriously conjectured that these generate all the relations. Grothendieck's original conjecture is that you have to consider all the M_{g,n} for 3g-3+n < 3 to get a full set of relations. Lochack, Nakamura and Schneps have defined a subgroup of GT (containing G_Q) that acts on the fundamental groups of all the M_{g,n}'s. It is probably strictly smaller than GT but that hasn't been proved. –  AFK Nov 25 '09 at 0:19
What is the most current status on the 2-, 3- and 5-cycle relations? –  Dr Shello Dec 15 '10 at 9:45

$\newcommand{\bb}{\mathbb}$ This is a response to Charles' remark to JSE's answer, why doesn't $\bar{\bb Q}$ come with a standard algebraic closure inside the complex numbers $\bb C$?

First, if one considers an abstract extension $K/\bb Q$, then $K$ has $d = [K:\bb Q]$ embeddings into the complex numbers, which can not, a priori, be distinguished in any way. (e.g. maps from $K = \bb Q[x]/(x^3-2)$ to $\bb C$ require a "choice" of $2^{1/3}$ in $\bb C$.

Of course, this doesn't answer Charles' question, which, I imagine, is more along the following lines. Why doesn't one simply start with the complex numbers, and then consider the set of algebraic numbers inside $\bb C$? The resulting field is clearly isomorphic to $\bar{\bb Q}$, and, moreover, comes with a canonical embedding into $\bb C$.

The problem arises when one wants to define Frobenius elements. Defining such elements amounts to giving a choice of embedding from $\bar{\bb Q}$ into $\bar{\bb Q}_p$. So there is a choice to be made for every $p$! Thinking of $\bar{\bb Q}$ inside $\bb C$ fixes this choice for "$p=0$" only.

To make this completely explicit, consider the splitting field $K$ of $x^3 - 2$. In the "fields live inside $\bb C$" optic, $K$ is the field $\bb Q(2^{1/3},\sqrt{-3})$ where $2^{1/3}$ is real and the imaginary part of $\sqrt{-3}$ is positive. Clearly $\mathrm{Gal}(K/\bb Q) = S_3$, where we can think of $(123)$ as sending $2^{1/3}\mapsto e^{2\pi i/3} 2^{1/3}$. If $p = 7$, then $\mathrm{Frob}_p$ has order $3$ in $S_3$. We can ask, does $\mathrm{Frob}_p = (123)$ or $(132)$? We find that $$\mathrm{Frob}_p = (123) \quad\mbox{if } \sqrt{-3}\equiv 2\mod 7$$ $$\mathrm{Frob}_p = (132) \quad\mbox{if } \sqrt{-3}\equiv 5\mod 7$$ and knowing that the imaginary part of $\sqrt{-3}$ is positive does not allow us to determine $\mathrm{Frob}_p$ without choosing an embedding of $K$ into $\bar{\bb Q}_7$.

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Thanks, this is a point that I have long wondered about and your remark clarifies the matter completely. –  Jonah Sinick Nov 1 '09 at 18:44
Thanks! So what I take from this is that the only canonical constructions are the completions $\mathbb{Q}_p$ (with $\mathbb{Q}_\infty = \mathbb{R}$). So if $X$ is the algebraic space associated to $\mathbb{Q}$, then for each $p$ we get a covering space $X_p \to X$, associated to the field of algebraic numbers in $Q_p$. If $X$ had a distinguished base point, we could name the $X_p$'s using subgroups of $\pi_1X$. Since we can't, all we know is the conjugacy class of $\pi_1X_p$ inside $\pi_1X$. For rational primes $p$, this is the congugacy class of the subgroup "generated" by $Frob_p$. –  Charles Rezk Nov 1 '09 at 20:12
I hate the fact that you can't even use html characters in comments. –  Charles Rezk Nov 1 '09 at 20:13
Charles, yes, that's correct, although literally to have a Frobenius at $p$ one should work with extensions unramified at $p$, which amounts to working with $\pi_1(\mathbb{Z}[1/N])$ for some set $N$, and taking $p$ not dividing $N$. (For $p \mid n$, one has to work with the still complicated group $Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$, rather than $Gal(\bar{F}_p/F_p)$, which is topologically cyclic and generated by Frobenius. –  user631 Nov 1 '09 at 23:29
Very very nice!!! –  Filippo Alberto Edoardo Nov 6 '12 at 23:06

Taylor said that he's never heard of anyone proposing a similar direct description of $G$ and that to understand $G$ one studies the representations of $G$.

I remember Mazur telling me this when I was a grad student. He made this point in the following way. You shouldn't really think of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ as a group which has elements, but as a "group up to conjugacy" - thus, the aspects of Galois groups that really make sense to think about are the conjugacy-invariant things: conjugacy classes (like Frobenii) and representations.

To unpack this a bit more: a Galois group is a fundamental group. But to talk about a fundamental group (as opposed to a groupoid) you need to choose a basepoint. To talk about an absolute Galois group you also need to choose a basepoint, which is to say an algebraic closure $\bar{\mathbb{Q}}/\mathbb{Q}$. (So just as one should talk about $\pi_1(X,*)$ rather than $\pi_1(X)$, one should talk about $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ rather than $\mathrm{Gal}(\mathbb{Q})$.) But a basepoint you can just draw with a pencil. A Galois closure of $\mathbb{Q}$ is not so easy.

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I like this answer very much. But I wonder: doesn't Q come with a standard algebraic closure? Namely, the algebraic closure of Q inside C. Why doesn't this serve as a good "canonical basepoint" for Qbar/Q? (Okay, canonical up to Gal(C/R) ...) –  Charles Rezk Oct 31 '09 at 16:46
Seconding Charles' question! (Which I think also covers mine, namely why Gal(Qbar/Q) is deep and magical but the absolute Galois group of F_p isn't so much.) –  Harrison Brown Oct 31 '09 at 17:24
What people mean by "canonical" would be: for an equation x^2 + 1 = 0 we have chosen canonically a solution x. The fact that you're in C doesn't help --- there are still 2 solutions, you need a choice. –  Ilya Nikokoshev Nov 3 '09 at 9:33
" why Gal(Qbar/Q) is deep and magical but the absolute Galois group of F_p isn't so much." --- it's much larger, to start with and contains all Gal F_p in some sense. –  Ilya Nikokoshev Nov 3 '09 at 9:34
I feel like this is true of most groups. E.g. the symmetric group - you need to choose an initial ordering of the letters you're permuting. –  David Corwin Apr 29 '13 at 13:16

I am pretty sure that when different number theorists say "one of the main goals of number theory is to understand Gal(Q-bar/Q)" they may well mean different things.

One example of what someone might mean, which has been touched upon above, but which I'd like to stress (because it's what I mean whenever I start seminars with this generic-sounding statement!) is Clozel's conjectures in his Ann Arbor paper.

Langlands would conjecture that for any automorphic representation of GL_n over a number field K, there should be an associated n-dimensional complex representation of what people might now call the Langlands group of K (Langlands really conjectured something a little more precise: he conjectured that a certain category that he could not quite define, but whose objects he understood, should have the structure of a Tannakian category; I've tried to make this statement "concrete" in the above).

One of the problems with Langlands' conjecture was simply that it's not the sort of thing that can be checked by example, because no-one really knows an intrinsic construction of the Langlands group of K.

However, there are theorems in the literature about constructing Galois representations from automorphic representations, for example class field theory (which tells us what the abelianisation of this Langlands group should look like: ((adeles of K)^* / K_^* ) ) and Deligne-et-al's theorem attaching p-adic Galois representations to holomorphic cusp forms for GL_2 over Q.

Hang on a minute though---Langlands was conjecturing the existence of complex Galois representations of some group we don't have a definition of---how does Deligne's theorem fit into this? Here we have a p-adic representation (which wouldn't be continuous if you chose an isomorphism Q_p-bar = C and considered the induced complex representation) of a group that at least we have a definition of (Gal(Q-bar/Q) in Deligne's case). So in fact Deligne isn't proving a special instance of Langlands' conjecture, he's proving something else.

Clozel formulated precisely what that "something else" should be: he conjectured that to any algebraic automorphic representation of GL_n(adeles of K) there should be an attached p-adic Galois representation. I think that nowadays one would also conjecture that conversely given a geometric p-adic Galois representation (i.e. a representation Gal(K-bar/K)-->GL_n(Q_p-bar) which looks reasonable in some way that can be made precise) it should arise in this way. People like Emerton and Kisin are well on the way to proving this for K=Q and n=2.

So "proving Clozel's conjecture" would be one way of making precise "understanding Gal(Q-bar/Q)". What this would boil down to in this case would be understanding all the representations of Gal(Q-bar/Q) which were (a) taking values in Aut(V) with V a f.d. vector space over the p-adics, (b) unramified outside a finite set of primes (c) de Rham at p, and one would be understanding them in the sense that one would be parametrising the set of unramified conjugacy classes Frob_ell in terms of something totally different (automorphic representations, which are analytic gadgets).

The book of Harris and Taylor is to a large extent the state of the art nowadays (but the theory is currently moving so fast) [EDIT: see Toby Gee's comment below]. However the base field K is essentially always either totally real or CM in this book, so in some sense the general case of Clozel's conjecture is still wide wide open.

Note finally that in some sense what Langlands conjectures is still open even for modular elliptic curves. If E is a modular elliptic curve over Q then the Galois representation naturally associated to E is its Tate module. But the representation of the Langlands group attached to E would be a representation to GL2(C), whose image would land in U2(C), the trace of Frob_p would be a sub p/sqrt(p), and (in the non-CM case) the conjugacy classes of the Frob_p's would be "evenly distributed in U(2)"---a statement which turns into the Sato-Tate conjecture when you unravel it. Note that Taylor et al proved the Sato-Tate conjecture but they certainly did not prove it by constructing this representation! Indeed it's in some sense an ill-defined question to construct this representation, because we don't know what the Langlands group is.

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A much better answer than mine, +1. –  Ilya Nikokoshev Nov 2 '09 at 11:34
A pedantic note: Harris-Taylor isn't quite state of the art (though I guess it is to a "large extent"): in particular the hypotheses there on the existence of a discrete series place have been completely removed by Shin and Chenevier-Harris, amongst others. –  TSG Nov 2 '09 at 12:50
Thanks Toby. When I wrote the comment about Harris-Taylor, what I really meant was "I know things have happened since then but don't quite know who to attribute things to so I won't go there". Passing remark: I now don't quite understand the canonical thing to do. Am I supposed to edit my post to reflect your comments, which will then make your comment look ridiculous? I think I'll just leave it. –  Kevin Buzzard Nov 2 '09 at 14:42
Kevin - Thanks for your interesting and informative post. I added an edit to my post (under the heading edit to avoid taking your response out of context just as you wish to avoid taking Toby's comment out of context) to account for your correction. –  Jonah Sinick Nov 3 '09 at 4:21
I should perhaps clarify something: amongst the algebraic automorphic representations are those "of Artin type" (I think there must be a better name for them, but it escapes me)---I am talking about the ones which should correspond to finite image Galois representations. For these representations, Clozel's conjecture and Langlands' conjecture should "basically coincide" and should both predict finite image Galois reps to GL(n,complexes). Even proving this statement (attaching reps to pi's of Artin type on GL(n)) would be an extraordinary achievement. –  Kevin Buzzard Nov 3 '09 at 16:20

Another aspect of "what it might mean to understand Gal(Q)" -- Shafarevich conjectured that the absolute Galois group of Q^{ab} is a free profinite group. That's pretty explicit!

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does "free profinite group" mean "profinite completion of a free group"? –  Ben Webster Oct 27 '09 at 14:12
From one of Harbater's papers on this: "A profinite group Π is free on a generating set S that converges to 1 if every map S → G to a profinite group G that converges to 1 uniquely extends to a group homomorphism Π → G." Here "converges to 1" refers to the profinite topology. –  JSE Oct 27 '09 at 14:19
A good reference for this is Ribes-Zaleski "Profinite groups" page 91-94. One should note that any finite set is "converging to 1" and that in the abelian case a free profinite group (free proabelian group!) on a set converging to 1 is just a product of \Z^hat's. –  Lars Oct 27 '09 at 20:51
Another quite good reference is "Field Arithmetic" of Fried-Jarden. It also discusses the connection between arithmetic/geometric properties of fields and group theoretic properties of the absolute Galois groups. E.g., Ax' Theorem: if a field K is PAC ("pseudo-algebraically closed") which just means that any geometrically irreducible variety has a rational point, then its absolute Galois group is projective, and its counter part due to Lubotzky-v.d.Dries any projective profinite group can be realized as the absolute Galois group of a PAC field. –  Lior Bary-Soroker Nov 25 '09 at 2:46
Is there any serious reason to believe (or disbelieve) this conjecture? –  David Hansen Jun 4 '11 at 3:37

There are already 8 good answers to this old question, but surprisingly there is something which I believe is true and important and that none of the answers says explicitly: Understanding Gal$(\overline{\mathbb Q}/\mathbb Q)$ as a group is not a major goal of number theory, and actually is not even a problem that really belongs to number theory. This is not to say this problem is not interesting. But it is a problem that belongs to algebra, field theory if you like, not to number theory. For example, the conjecture of Shafarevich mentioned by JSE (that's a natural subgroup of tat group is free profinite), while interesting, is not really number theory.

What number theory is concerned about is understanding Gal$(\overline{\mathbb Q}/\mathbb Q)$, together with its family of subgroups $D_p$ (for $p$ a prime or infinity), the decomposition groups well defined up to conjugacy, and isomorphic to Gal$(\overline{\mathbb Q}_p/\mathbb Q_p)$ for $p$ finite, and to $\mathbb Z/2$ when $p=\infty$. I think that this what most people mean, but sometimes forget to say, when they say that understanding is a major goal of Number Theory. Indeed, if we "understand" sufficiently well Gal$(\overline{\mathbb Q}/\mathbb Q)$ with its decomposition subgroup, we understand in principle a lot of finite Galois groups which can be obtained with elementary operations from the above, and that can be interpreted say as various class groups of number fields. In turn those groups governs a lot of classical problems in number theory, like classifying quadratic forms over $\mathbb Z$ (a problem considered since Legendre), understanding which prime can be represented by a give quadratic forms, understanding all kind of Selmer groups, etc.

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One aspect of understanding the absolute Galois group I've heard of is the 'Inverse Galois Problem'. This simply askes, Is every finite group the Galois group of some extension of Q? Its known for solvable groups, but it is unsolved (and apparently very hard) for general finite groups.

The inverse Galois problem is equivalent to asking whether any finite group can be the quotient of Gal(Q^bar, Q). In this way, its a more concrete question about the absolute Galois group than Langlands reciprocity.

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Khare-Larsen-Savin have used functoriality (ie the Langlands program) to help solve the inverse Galois problem, see for example link text. They show that some finite groups of Lie type occur as Galois groups.

As for the Langlands program in general, solving it may not give an answer to "What is Gal(Qbar/Q)?", however it can certainly help. For example, the use of automorphic forms in Iwasawa theory has been quite fruitful such as Ribet's proof of the converse to Herbrand's theorem, the proof of the Main conjecture by Mazur-Wiles and Wiles, etc. For example, Eisenstein series, their congruences with cusp forms and the associated Galois representations are used in Ribet's paper (Inv. Math. 34 (1976)) to construct unramified p-extensions. It can be easier to hide an automorphic form up your sleeve.

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Perhaps some articles in this book and the articles here help, and the discussion on dessin d'enfants (more) on this site.

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The book link has changed. This is the correct link: library.msri.org/books/Book41/index.html –  arsmath Dec 20 '10 at 19:21
Thanks for the corrected link! –  Thomas Riepe Dec 21 '10 at 5:49

I'm far from expect in this topic, but here's my attempt.

First, and that's something quite straightforward, people want to study Gal Q (this is how I will denote it; this common shortcut is defined as Gal F := Gal \bar F/F) because we like Galois groups. For simple fields we know their Galois groups and we know how tremendously important they have been, from solving the equations (Abel et al) to doing group cohomology in class field theory.

As algebraic geometry matured, people learned that one of the deep reasons for importance of Galois group is that it's the fundamental group pi_1(Spec F) (depending on the definition, sometimes this is true only after a profinite completion, I'll omit this fine point further below). This allows us to use the whole apparatus of topology as well as the geometric intuition.

Representations of Gal F thus have a natural geometric meaning as some bundles (local systems) on the coverings of Spec F. This alone would make their study pretty important. It's very reasonable to study a space in terms of geometric objects that live on it.

The whole Galois group is very complicated. Fortunately, since the Galois groups of F_q and Q_p (p-adic numbers) are known, we know lots of factorgroups of it. This is the standard topic of algebraic number theory courses.

Moreover, if we restrict ourselves to the *Abelian part of Galois group, its structure has been completely established by the class field theory. In other words, all one-dimensional representations of Gal Q are known. That begs a natural question about higher-dimensional reps — indeed this goes on by the name of Langlands program.

The Langlands program is such a huge topic that I don't feel able even starting to talk about it. It might be a good idea to post questions here if you'll feel brave enough to learn it :)

One more topic in the discussion of the structure of Gal F is about the specific (conjugacy classes) of operators that live there, called Frobenius operators. The amazing thing about them is that they behave, formally, in a way similar to the knot operators in the fundamental group of a threefold without knots. This image is reinforced by the computations of the dimension of Spec Z that give an answer of 3, providing fruitful connections to the theory of real 3-manifolds, knots and their L-functions.

Finally, here's the direction that was explored by many people and was probably made famous through Grothendieck's work. Consider a variety Z := P^1-(0, 1, \infty), the projective plane without three points. It is really interesting as it makes sense over any field. Now suppose we consider it as a scheme over Q and ask for its fundamental group. Then one would have an exact sequence, if I'm not mistaken,

0  -->  pi_1(Z/\bar Q)  -->  pi_1(Z/Q)  -->  Gal Q --> 0

which implies, by standard reasoning, that Gal Q acts on pi_1(Z/\bar Q), which is a very straightforward group — it's simply a topological fundamental group of a plane without two points, so it's freely generated by two loops. Grothendieck called it a cartographic group and related it to dessin's d'enfants. The standard references are very well-written in T.'s answers and there are hopefully more questions on this topic coming to MathOverflow ;)

As you see, this is a very interesting direction that we should continue to explore. One more thing I like about it is that it's also very suitable to MathOverflow format, since there are both many specific questions as well as opportunities to write reviews and to produce new results by collaboration.

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Conc. primes as knots and Spec Z as 3-manifold: Fits that to the Poincare conjecture? Topologists view 3-manifolds as Kirby-equivalence classes of framed links. How would that be with Spec Z? Then, topologists have things like virtual 3-manifolds, has that analogies in arithmetics? –  Thomas Riepe Nov 3 '09 at 14:01
I'm not a specialist in this topic, and it's a different from the original question here. Do you mind posting a separate question about this? –  Ilya Nikokoshev Nov 3 '09 at 14:59


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Tuesday, 27 November 2012

Absolute Cell References With Exercise 5.5

Previous post I discussed relative cell referencing with Exercise 5.4 Here I will talk about absolute cell referencing.

Relative cell referencing is very useful in many cases by adjusting the cell references when a formula is copied from but one location to another. Example can be seen here.

There might be times that you need to use one cell in each formula. To prevent Excel to copy all cell references we use Absolute cell referencing. In Excel these formulas are constructed by placing dollar  (\$) signs in the cell reference.

Check this simple example video.

Here I made a simple currency converter. From US Dollars to Malaysian Ringgit. Let's say 1 dolar = 3.02 Ringgit. I simply put this information in cell D1 and I write different amounts of dollars in column C. Ringgit currency is calculated in column D. Just pay attention how the Absolute cell referencing is used in formulas. When we copy the formula we can see that cell D1 never changes in copied formulas.


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## Precalculus: Mathematics for Calculus, 7th Edition

By law 1: a$^{m}$ $\times$ a$^{n}$= a$^{n+m}$ So we have: 5$^{1/3}$ $\times$ 5$^{2/3}$= 5$^{\frac{1}{3}+ \frac{2}{3}}$ = 5$^{3/3}$ = 5.


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# Problems on Understanding Matrices | Matrix Problems and Solutions

Various tricks and tips are included in the Matrices Worksheet for easy solving of matrix questions. Check problems on matrices along with the solutions. Find the step-by-step procedure to know how to solve matrix problems. Refer to all the solutions present in the below sections to prepare for the exam.

Scroll down the page to get the matrices word problems with solutions pdf. Assess your knowledge level by taking the help of the matrices examples available. All the matrices-related articles are given in our 10th Grade Math concepts. Quickly go through the entire article and gain high knowledge of matrix problems.

### Matrices Questions with Solutions

Problems 1: Multiply the given matrix by 3. The matrix is,
$$A = \left[ \begin{matrix} 1 & 2\cr 3 & 4\cr \end{matrix} \right]$$ What is the final value of Matrix?

Solution:
As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 1 & 2\cr 3 & 4\cr \end{matrix} \right]$$
Now, multiply given matrix by 3.
After, multiplication the matrix is $$A = \left[ \begin{matrix} 1 *3 & 2*3\cr 3*3 & 4*3\cr \end{matrix} \right]$$
Therefore, the final matrix value of A is $$A = \left[ \begin{matrix} 3 & 6\cr 9 & 12\cr \end{matrix} \right]$$

Problem 2: Matrix A is of 4×1 order. Matrix B is of 1×3 order. Both the products of AB and BA are defined or not?

Solution:
In the given question,
The Matrix A is of order 4×1.
The Matrix B is of order 1×3.
The product AB will be defined as the number of columns of A is the same as the number of rows of the B.
By using the multiplication of matrices rule, the product matrix hence obtained is of order 4×3.
Next, the product BA is not defined as the number of columns of B is not equal to the number of rows of A.
So, we can say that the matrix multiplication is not commutative, AB is not necessarily equal to BA and sometimes one of the products may also be not defined.
Hence, the product AB is defined but not BA.

Problem 3: Write the elements of the sum matrix C = A+B is explicitly by addition of matrices A and B of dimension 1 × 2 whose elements are given as a11 = 1, a12 = 4 and b11 = -1, and b12 = -8.

Solution:
Given the a and b values.
Now, we need to find the value of C using the addition method.
The addition of matrices is defined as when the matrices have the same order and we will add the corresponding elements of the two matrices A and B.
Now, adding the corresponding elements, we have
C11= a11 + b11  = 1 + (-1) = 0
Next, the value is C12= a12 + b12 = 4 + (-8) = -4
Therefore, the value of C11 is 0, and the value of C12 is -4.

Problem 4: The matrix $$A = \left[ \begin{matrix} 11 & 2\cr 3 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 7 & 8\cr 6 & 5\cr \end{matrix} \right]$$. What is the added value of Matrices A and B?

Solution:
As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 11 & 2\cr 3 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 7 & 8\cr 6 & 5\cr \end{matrix} \right]$$
Now, we will find the addition value of A and B.
Here, both the matrices have the same order of 2 × 2. Now, we can add the elements of the first matrix with the respective elements of the second matrix.
After, addition the matrices is $$A +B= \left[ \begin{matrix} 11+7 & 2+8\cr 3+6 & 4+5\cr \end{matrix} \right]$$
Therefore, the addition matrices value of A +B is $$\left[ \begin{matrix} 18 & 10\cr 9 & 9\cr \end{matrix} \right]$$

Problem 5: Determine the element of the second row and third column of the matrix X + Y using the addition of matrices definition. If x23 = -19 is the element of X and the y23 = 30 is the element of Y.

Solution:
As given in the question, X23 = -19 is the element of X and Y23= 30 is the element of Y.
Now, we need to add X23and Y23 to find the element of the second row and third column of the matrix X + Y.
So, the finding value is, X23+ Y23
= -19 + 30 = 11.
Therefore, the element in the second row and third column of A + B is 11.

Problem 6:  Find the transpose of a column matrix $$A = \left[ \begin{matrix} 3 \cr 1 \cr 7\cr \end{matrix} \right]$$

Solution:
The given matrix is $$A = \left[ \begin{matrix} 3 \cr 17 \cr \end{matrix} \right]$$
To find the transpose of the given column matrix, the elements of the column matrix must be written as row elements.
A=[3 17]
Therefore, the transpose of the given column matrix is A=[3 17]

Problem 7: What is the example of a null matrix having two rows and three columns.

Solution:
In the given question, a null matrix has two rows and three columns.
Actually, the null matrix will have all the elements are zero.
So, all the elements of two rows and three columns in a null matrix are zero.
$$A = \left[ \begin{matrix} 0 & 0 & 0\cr 0 & 0 & 0\cr \end{matrix} \right]$$
Therefore, the answer is$$A = \left[ \begin{matrix} 0 & 0 & 0\cr 0 & 0 & 0\cr \end{matrix} \right]$$

Problem 8: Write the elements of the matrix C = A – B explicitly. If A = [4 10 18] and B = [2 18 24] using matrix subtraction formula?

Solution:
The given matrices are A = [4 10 18] and B = [2 18 24].
Now, we will find the value of the C.
The dimensions of the matrices A and B are the same, that is, 1 × 3. We can perform subtraction of matrices as two matrices have the same order.
First, subtract the elements of the first matrix with the respective elements of the second matrix. i.e.,
A – B = [4-2 10-18 18-24]
= [2 -8 -6]
Hence, the elements of C = A – B are C11 = 2, C12 = -8, C13 = -6.

Problem 9: Subtract X and Y where
$$X = \left[ \begin{matrix} 3 & 8\cr 2 & -5\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} -4 & -3\cr 2 & 1\cr \end{matrix} \right]$$

Solution:
Given matrices are $$X = \left[ \begin{matrix} 3 & 8\cr 2 & -5\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} -4 & -3\cr 2 & 1\cr \end{matrix} \right]$$
Both the matrices have the same order of 2 × 2. Now, we can subtract the elements of the first matrix with the respective elements of the second matrix.
So, the matrices X-Y is $$= \left[ \begin{matrix} 3 & 8\cr 2 & -5\cr \end{matrix} \right]$$ – $$\left[ \begin{matrix} -4 & -3\cr 2 & 1\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3 +4 & 8+3 \cr 2-2 & -5-11\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 7 & 11\cr 0 & -6\cr \end{matrix} \right]$$
Thus, the subtraction of X and Y is $$\left[ \begin{matrix} 7 & 11\cr 0 & -6\cr \end{matrix} \right]$$

Problem 10: The matrix is,
$$A = \left[ \begin{matrix} 1 & 0\cr 2 & 4\cr \end{matrix} \right]$$  and $$B = \left[ \begin{matrix} 6 & 8\cr 4 & 3\cr \end{matrix} \right]$$. What is the final product value of two matrices?

Solution:
As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 1 & 0\cr 2 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 6 & 8\cr 4 & 3\cr \end{matrix} \right]$$
Now, we will find the final product value of two matrices.
After, multiplication of two matrix is  $$\left[ \begin{matrix} (1*6) + (0*4) & (1*8) + (0*3) \cr (2*6) + (4*4) & (2*8) + (4*3) \cr \end{matrix}\right]$$
$$\left[ \begin{matrix} 6+0 & 8+0\cr 12+16 & 16+12\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 6 & 8\cr 28 & 28\cr \end{matrix} \right]$$
Therefore, the final product value of the two matrix is $$\left[ \begin{matrix} 6 & 8\cr 28 & 28\cr \end{matrix} \right]$$

Problem 11: Find 3B where
$$B= \left[ \begin{matrix} 4 & 5\cr 11 & 10\cr \end{matrix} \right]$$

Solution:
The given matrix is $$B = \left[ \begin{matrix} 4 & 5\cr 11 & 10\cr \end{matrix} \right]$$
To find the value of 3B, we need to multiply the value of 3 by all the elements of matrix B.
i.e., 3B= $$\left[ \begin{matrix} 3*4 & 3*5\cr 3*11 & 3*10\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 12 & 15\cr 33 & 30\cr \end{matrix} \right]$$
Therefore, the required matrix is 3B $$=\left[ \begin{matrix} 12 & 15\cr 33 & 30\cr \end{matrix} \right]$$

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# Convergence of a sequence with practical example and more complex question about the workings of a sequence

I am really trying to plug in a point to gain a practical understanding.

For $$N \in \mathbb{N}$$

$$x_n = 3(1 + \frac{1}{n}) \to 3$$ as $$n \to \infty$$

$$\vert 3 + \frac{3}{n} -3 \vert \lt \epsilon$$

$$\vert \frac{3}{n} \vert \lt \epsilon$$

since $$n$$ is always positive

$$\frac{3}{n} \lt \epsilon$$

There exists an $$N$$ such that $$\frac{3}{\epsilon} \lt N$$

Now I want pick a random number say $$\epsilon = .002$$

Then in order for $$x_n$$ to be closer to $$a$$ then $$\epsilon$$ we solve for $$N$$

$$\frac{3}{.002} \leq N$$

which is $$1,500 \leq N$$

lets choose $$1,600$$

$$3 + \frac{3}{1,600} = 3.001875$$

$$\vert 3.001875 - 3 \vert = .001875 \lt .002$$

Is this correct? Should $$N$$ be strictly greater than the fraction or is it acceptable to have have $$N \geq \frac{3}{\epsilon}$$ Also these seem like special cases where the algebra just happens to work out perfectly, what are you supposed to do if it doesn't? How do you even tell if it converges? If you can't isolate n from the sequence like in this example then is that a pretty good indicator that it may not converge or that you may not be able to find $$N \in \mathbb{N} S.T.......$$?

A big question of mine is what is the difference between lets say $$1 + 2x_n$$ and $$2-\frac{1}{n}$$ They have very different techniques in the book.

Yes, there is no mistake in your example, and actually it has sense. I read it as « in order to be close to 3, $$N$$ has to be large », especifically, larget than $$1500$$ if you want the error be smaller than $$0.002$$.

For your second question « what if you cannot isolate $$n$$? ». No, it is not an indicator of not convergence of the sequence. In my opinion, a good indicator might be to try to recognize other well known limits inside the limit your currently studying. So, if you only recognize well known comvergent sequences, it is much easier to figure out if the new sequence converges or not. If you recognize not convergent sequence, then you have to be careful because can be some cancelations which can make the new sequence a convergent one. Since you only have two possibilities (it converges or not), if you don’t recognize any well-known sequence and cannot prove the convergence either, you can try to find a counterexample to prove that the sequence does not converge, or to apply some theorem (like L’hôpital rule, super useful theorem to find limits of sequences) in order to prove the (not) converge.

Finally, for your last question, there is no difference between your two example. Note that the only interesting part of both sequence is the factor $$\dfrac{c}{n}$$, where $$c$$ is some constant depending on if you are in the first example or in the second one. Note that in both cases, independently pf what $$c$$ is, the limit $$\dfrac{c}{n}\to0$$ when $$n$$ goes to infinity. So, why they have so different techniques? Probably just because pedagogical purposes. I guess they wantsd to show two different methods to solve the same kind of problems, but I insist, both problem arr essentially the same (it only changes the constant $$c$$).

Quick check of the definition of the limit of a sequence: The sequence $$(x_n)$$ converges to $$x$$ if for all $$\epsilon > 0$$ there exists a non-negative integer $$N$$ such that if $$n \geq N$$ then $$|x_n - x| < \epsilon$$.

What you have there looks correct to me.

I tend to obtain $$N$$ so that the inequality for $$\epsilon$$ is satisfied strictly. I do this to match the definition exactly so that I know that I can use any results based on that definition.

If I have understood you correctly then you have raised two important points:

• Existence of the limit vs actually calculating it. Sometimes the value $$x$$ may not be apparent (follow the link above for examples).

• Existence of the limit vs calculating the rate of convergence. For the definition to work, we only need to know that an $$N$$ exists for any given $$\epsilon$$ in principle. We don’t necessarily need to solve for it explicitly, but an explicit solution is a very good way of establishing existence. Explicit calculation of $$N$$ for $$\epsilon$$ does tell us how big $$N$$ needs to be to achieve $$\epsilon$$ — the “rate of convergence”.

In general it can indeed be hard to tell whether an arbitrary sequence converges and to calculate the limit if it does. But there are increasingly sophisticated theorems that can give results (for example the Squeeze Theorem).


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# Bone¶

Attributes:

class Leap.Bone

The Bone class represents a bone of a finger.

All fingers contain 4 bones that make up the anatomy of the finger. Get valid Bone objects from a Finger object.

Bones are ordered from base to tip, indexed from 0 to 3. Additionally, the bone’s Type enum may be used to index a specific bone anatomically.

The thumb does not have a base metacarpal bone and therefore contains a valid, zero length bone at that location.

Note that Bone objects can be invalid, which means that they do not contain valid tracking data and do not correspond to a physical Bone. Invalid Bone objects can be the result of asking for a Bone object from an invalid Finger object. A Bone object created from the Bone constructor is also invalid. Test for validity with the Bone.is_valid() attribute.

New in version 2.0.

classmethod Bone([Bone])

Constructs a Bone object.

Get valid Bone objects from a Finger object.

bone = Leap.Bone() #invalid

Parameters: Bone (Bone) – An object representing a Bone. If no Bone parameter is supplied, or the object does not represent a Bone, an invalid Bone object is returned. Bone

New in version 2.0.

basis
Type: Matrix

The orthonormal basis vectors for this Bone as a Matrix.

Basis vectors specify the orientation of a bone.

• x_basis. Perpendicular to the longitudinal axis of the bone; exits the sides of the finger.
• y_basis or up vector. Perpendicular to the longitudinal axis of the bone; exits the top and bottom of the finger. More positive in the upward direction.
• z_basis. Aligned with the longitudinal axis of the bone. More positive toward the base of the finger.
basis = bone.basis
x_basis = basis.x_basis
y_basis = basis.y_basis
z_basis = basis.z_basis
origin = basis.origin

The bases provided for the right hand use the right-hand rule; those for the left hand use the left-hand rule. Thus, the positive direction of the x-basis is to the right for the right hand and to the left for the left hand. You can change from right-hand to left-hand rule by multiplying the basis vectors by -1.

You can use the basis vectors for such purposes as measuring complex finger poses and skeletal animation.

Note that converting the basis vectors directly into a quaternion representation is not mathematically valid. If you use quaternions, create them from the derived rotation matrix not directly from the bases.

New in version 2.0.

center
Type: Vector

The midpoint of the bone.

middle = bone.center

New in version 2.0.

direction
Type: Vector

The normalized direction of the bone from base to tip.

direction = bone.direction

This is the negative of the bone’s z-basis.

New in version 2.0.

is_valid
Type: boolean

Reports whether this Bone object contains valid data.

bone = finger.bone(Bone.TYPE_PROXIMAL)
if(bone.is_valid):
# ... Use the bone data

New in version 2.0.

length
Type: float

The length of the bone in millimeters.

length = bone.length

New in version 2.0.

next_joint
Type: Vector

The position of the end of the bone closest to the finger tip.

bone_end = bone.next_joint

New in version 2.0.

prev_joint
Type: Vector

The position of the end of the bone closest to the wrist.

bone_start = bone.prev_joint

New in version 2.0.

type
Type: integer

The name of this Bone.

type = bone.type

The anatomical type of this Bone:

• 0 = TYPE_METACARPAL – the bone inside the hand connecting the wrist and finger
• 1 = TYPE_PROXIMAL – the first exposed bone of the finger, closest to the hand.
• 2 = TYPE_INTERMEDIATE – the middle bone of the finger
• 3 = TYPE_DISTAL – the bone at the tip of the finger

New in version 2.0.

width
Type: float

The average width of the finger along the bone (including the fleshy sheathe).

width = bone.width

New in version 2.0.

invalid
Type: Bone

An invalid Bone object.

bone = Bone.invalid

You can use this invalid object in comparisons testing whether a given Bone instance is valid or invalid. (You can also use the Bone.is_valid attribute.)

New in version 2.0.

TYPE_DISTAL
Type: integer

The phalanx bone at the end for the finger (most distant from the body). The distal phalanx has index 3.

distal_bone = finger.bone(Bone.TYPE_DISTAL)

New in version 2.0.

TYPE_INTERMEDIATE
Type: integer

The bone in between the distal and proximal phanlanges. The intermediate phalanx has index 2.

intermediate_bone = finger.bone(Bone.TYPE_INTERMEDIATE)

New in version 2.0.

TYPE_PROXIMAL
Type: integer

The finger bone at the base of the finger (in closest proximity to the body). The proximal phalanx has index 1.

proximal_bone = finger.bone(Bone.TYPE_PROXIMAL)

New in version 2.0.

TYPE_METACARPAL
Type: integer

The bone connecting the wrist area to the base of the proximal finger bone. The metacarpal has index 0.

metacarpal_bone = finger.bone(Bone.TYPE_METACARPAL)

Note: A real thumb has one less bone than the other fingers; therefore the Leap Motion assigns the thumb model a valid, zero-length bone at the metacarpal index. While this differs from the normal anatomical naming system, which dispenses with the intermediate phalanx instead, it does mean that each finger, including the thumb, has the same bone at the same index.

New in version 2.0.