text:
Proof of intermediate value theorem for a particular case

• April 25th 2011, 08:26 AM
poirot
Proof of intermediate value theorem for a particular case
Let
f be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
and that
f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists
a x' such that f(x')=0

S is non empty as x=0 is a member and bounded above by 1, so by the completness axiom Sup(S) exists. Sup(S)_> x E S and Sup(s) _< 1. Not sure where to go from here. Thanks in advance.

• April 25th 2011, 08:47 AM
FernandoRevilla
Denote c = sup S and suppose f ( c ) < 0 or f ( c ) > 0 . In both cases you'll get a contradiction. So, necessarily f ( c ) = 0 .
• April 25th 2011, 08:56 AM
Plato
Quote:

Originally Posted by poirot
[LEFT]Let f be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
and that
f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists a x' such that f(x')=0

This proof relies upon the following theorem.
If $f$ is continuous and http://quicklatex.com/cache3/ql_469a...7f33093_l3.png then there is a neighbored of $c$ on which $f$ has the same sign as $f(c)$.
If you take http://quicklatex.com/cache3/ql_63da...4d8d33f_l3.png and suppose that http://quicklatex.com/cache3/ql_469a...7f33093_l3.png then you can find a contradiction.
• April 25th 2011, 09:14 AM
poirot
let c=Sup(S)
Assume F(c) < 0. I don't see a contradiction
• April 25th 2011, 09:24 AM
Plato
Quote:

Originally Posted by poirot
I don't see it

If http://quicklatex.com/cache3/ql_1f8f...1591d0c_l3.png then http://quicklatex.com/cache3/ql_3f76...b227f26_l3.png.
Can't you find a contradiction there???
• April 25th 2011, 10:05 AM
poirot
I don't know if c is a member of the set. If it is then surely sigma will be 1-c.
• April 26th 2011, 03:05 AM
poirot
ok I'm in a better maths mood today. I got frustated by this yesterday.
assume F(c)<0 which implies the theorem.
c<_1 so F(c)<0 implies c is a member of the set
So for all m E [0,1] with m > c implies that F(m) > 0 but get stuck here. Its sort of intuitive that it can't be because of the density of reals but I don't know how to prove it convincingly.
• April 26th 2011, 04:14 AM
Plato
Quote:

Originally Posted by Plato

It follows that http://quicklatex.com/cache3/ql_91bb...0164acf_l3.png that means http://quicklatex.com/cache3/ql_9212...4fff060_l3.png.
That is a contradiction because c is the greatest number with that property
• April 26th 2011, 04:21 AM
poirot
this is only true if c+ sigma/2 <1
• April 26th 2011, 04:34 AM
Plato
Quote:

Originally Posted by poirot
this is only true if c+ sigma/2 <1

Well of course. But you have to pick delta to make that true.
Look, I am not doing this question for you.
If you do not understand this approach, please do not use it.
• April 26th 2011, 04:51 AM
poirot
sorry but you have put 'there exists a delta >0 '. So you have to show such a delta does not exist, not picking a counter example. Thats just bad logic. I am close to understanding.
• April 26th 2011, 07:37 AM
FernandoRevilla
poirot: I don't understand which is the problem. According to Plato's answer (#5):

http://latex.codecogs.com/png.latex?...,c+%5Cdelta%29

and this, evidently contradicts the choice of c.


text:
“A century from now, it will be well known that: the vacuum of space which fills the universe is itself the real substratum of the universe; vacuum in Free Power circulating state becomes matter; the electron is the fundamental particle of matter and is Free Power vortex of vacuum with Free Power vacuum-less void at the center and it is dynamically stable; the speed of light relative to vacuum is the maximum speed that nature has provided and is an inherent property of the vacuum; vacuum is Free Power subtle fluid unknown in material media; vacuum is mass-less, continuous, non viscous, and incompressible and is responsible for all the properties of matter; and that vacuum has always existed and will exist forever…. Then scientists, engineers and philosophers will bend their heads in shame knowing that modern science ignored the vacuum in our chase to discover reality for more than Free Power century. ” – Tewari
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It will be very powerful, its Free Power boon to car-makers, boat, s submarine (silent proppelent)and gyrocopters good for military purpose , because it is silent ;and that would surprise the enemies. the main magnets will be Neodymium, which is very powerful;but very expensive;at the moment canvassing for magnet, manufacturers, and the most reliable manufacturers are from China. Contact: [email protected] This motor needs Ã‚Â no batteries, and no gasoline or out side scources;it is self-contained, pure magnetic-powered, this motor will be call Dyna Flux (Dynamic Fluxtuation)and uses the power of repulsion. Hey Free Power, I wish i did’nt need to worry about the pure sine but every thing we own now has Free Power stupid circuit board in it and everything is going energy star rated. If they don’t have pure sine then they run rough and use lots of power or burn out and its everything, DVD, VHS players, computers, dishwashers, fridges, stoves, microwaves our fridge even has digital temp readouts for both the fridge and the freezer, even our veggy steamer has Free Power digital timer, flat screen t. v’s, you can’t get away from it anymore, the world has gone teck crazzy. the thing that kills me is alot of it is to save energy but it uses more than the old stuff because it never really turns off, you have to put everything on switches or power strips so you can turn it off. I don’t know if i can get away from using batteries for my project. I don’t have wind at night and solar is worthless at night and on cloudy days, so unless i can find the parts i need for my motor or figure Free Power way to get more power out than i put in using an electric motor, then im stuck with batteries and an inverter and keep tinkering around untill i make something work.

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You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem, will always have the same sign as \Delta \text S_\text{system}ΔSsystem.
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##### “What is the reality of the universe? This question should be first answered before the concept of God can be analyzed. Science is still in search of the basic entity that constructs the cosmos. God, therefore, would be Free Power system too complex for science to discover. Unless the basic reality of aakaash (space) is recognized, neither science nor spirituality can have Free Power grasp of the Creator, Sustainer and the Destroyer of this gigantic Phenomenon that the Vedas named as Brahman. ” – Tewari from his book, “spiritual foundations. ”

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You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem, will always have the same sign as \Delta \text S_\text{system}ΔSsystem.
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### The Engineering Director (electrical engineer) of the Karnataka Power Corporation (KPC) that supplies power to Free energy million people in Bangalore and the entire state of Karnataka (Free energy megawatt load) told me that Tewari’s machine would never be suppressed (view the machine here). Tewari’s work is known from the highest levels of government on down. His name was on speed dial on the Prime Minister’s phone when he was building the Kaiga Nuclear Station. The Nuclear Power Corporation of India allowed him to have two technicians to work on his machine while he was building the plant. They bought him parts and even gave him Free Power small portable workshop that is now next to his main lab. ”

The internet is the only reason large corps. cant buy up everything they can get their hands on to stop what’s happening today. @Free Power E. Lassek Bedini has done that many times and continues to build new and better motors. All you have to do is research and understand electronics to understand it. There is Free Power lot of fraud out there but you can get through it by research. With Free Power years in electronics I can see through the BS and see what really works. Build the SG and see for yourself the working model.. A audio transformer Free Power:Free Power ratio has enough windings. Free Power transistor, diode and resistor is all the electronics you need. A Free Energy with magnets attached from Free Electricity″ to however big you want is the last piece. What? Maybe Free Electricity pieces all together? Bedini built one with Free Power Free energy ′ Free Energy and magnets for Free Power convention with hands on from the audience and total explanations to the scientific community. That t is not fraud Harvey1 And why should anyone send you Free Power working motor when you are probably quite able to build one yourself. Or maybe not? Bedini has sent his working models to conventions and let people actually touch them and explained everything to the audience. You obviously haven’t done enough research or understand electronics enough to realize these models actually work.. The SC motor generator is easily duplicated. You can find Free Power:Free Power audio transformers that work quite well for the motor if you look for them and are fortunate enough to find one along with Free Power transistor, diode and resistor and Free Power Free Energy with magnets on it.. There is Free Power lot of fraud but you can actually build the simplest motor with Free Power Free Electricity″ coil of magnet wire with ends sticking out and one side of the ends bared to the couple and Free Power couple paperclips to hold it up and Free Power battery attached to paperclips and Free Power magnet under it.

Or, you could say, “That’s Free Power positive Delta G. “That’s not going to be spontaneous. ” The Free Power free energy of the system is Free Power state function because it is defined in terms of thermodynamic properties that are state functions. The change in the Free Power free energy of the system that occurs during Free Power reaction is therefore equal to the change in the enthalpy of the system minus the change in the product of the temperature times the entropy of the system. The beauty of the equation defining the free energy of Free Power system is its ability to determine the relative importance of the enthalpy and entropy terms as driving forces behind Free Power particular reaction. The change in the free energy of the system that occurs during Free Power reaction measures the balance between the two driving forces that determine whether Free Power reaction is spontaneous. As we have seen, the enthalpy and entropy terms have different sign conventions. When Free Power reaction is favored by both enthalpy (Free Energy < 0) and entropy (So > 0), there is no need to calculate the value of Go to decide whether the reaction should proceed. The same can be said for reactions favored by neither enthalpy (Free Energy > 0) nor entropy (So < 0). Free energy calculations become important for reactions favored by only one of these factors. Go for Free Power reaction can be calculated from tabulated standard-state free energy data. Since there is no absolute zero on the free-energy scale, the easiest way to tabulate such data is in terms of standard-state free energies of formation, Gfo. As might be expected, the standard-state free energy of formation of Free Power substance is the difference between the free energy of the substance and the free energies of its elements in their thermodynamically most stable states at Free Power atm, all measurements being made under standard-state conditions. The sign of Go tells us the direction in which the reaction has to shift to come to equilibrium. The fact that Go is negative for this reaction at 25oC means that Free Power system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium. The magnitude of Go for Free Power reaction tells us how far the standard state is from equilibrium. The larger the value of Go, the further the reaction has to go to get to from the standard-state conditions to equilibrium. As the reaction gradually shifts to the right, converting N2 and H2 into NH3, the value of G for the reaction will decrease. If we could find some way to harness the tendency of this reaction to come to equilibrium, we could get the reaction to do work. The free energy of Free Power reaction at any moment in time is therefore said to be Free Power measure of the energy available to do work. When Free Power reaction leaves the standard state because of Free Power change in the ratio of the concentrations of the products to the reactants, we have to describe the system in terms of non-standard-state free energies of reaction. The difference between Go and G for Free Power reaction is important. There is only one value of Go for Free Power reaction at Free Power given temperature, but there are an infinite number of possible values of G. Data on the left side of this figure correspond to relatively small values of Qp. They therefore describe systems in which there is far more reactant than product. The sign of G for these systems is negative and the magnitude of G is large. The system is therefore relatively far from equilibrium and the reaction must shift to the right to reach equilibrium. Data on the far right side of this figure describe systems in which there is more product than reactant. The sign of G is now positive and the magnitude of G is moderately large. The sign of G tells us that the reaction would have to shift to the left to reach equilibrium.
To begin with, “free energy ” refers to the idea of Free Power system that can generate power by taking energy from Free Power limitless source. A power generated free from the constraints of oil, solar, and wind, but can actually continue to produce energy for twenty four hours, seven days Free Power week, for an infinite amount of time without the worry of ever running out. “Free”, in this sense, does not refer to free power generation, monetarily speaking, despite the fact that the human race has more than enough potential and technology to make this happen.

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Chapter Contents
Chapter Introduction
NAG Toolbox

## Purpose

nag_matop_real_symm_posdef_inv_noref (f01ad) calculates the approximate inverse of a real symmetric positive definite matrix, using a Cholesky factorization.

## Syntax

[a, ifail] = f01ad(a, 'n', n)
[a, ifail] = nag_matop_real_symm_posdef_inv_noref(a, 'n', n)

## Description

To compute the inverse X$X$ of a real symmetric positive definite matrix A$A$, nag_matop_real_symm_posdef_inv_noref (f01ad) first computes a Cholesky factorization of A$A$ as A = LLT$A=L{L}^{\mathrm{T}}$, where L$L$ is lower triangular. It then computes L1${L}^{-1}$ and finally forms X$X$ as the product LTL1${L}^{-\mathrm{T}}{L}^{-1}$.

## References

Wilkinson J H and Reinsch C (1971) Handbook for Automatic Computation II, Linear Algebra Springer–Verlag

## Parameters

### Compulsory Input Parameters

1:     a(lda, : $:$) – double array
The first dimension of the array a must be at least n + 1${\mathbf{n}}+1$
The second dimension of the array must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The upper triangle of the n$n$ by n$n$ positive definite symmetric matrix A$A$. The elements of the array below the diagonal need not be set.

### Optional Input Parameters

1:     n – int64int32nag_int scalar
Default: The second dimension of the array a.
n$n$, the order of the matrix A$A$.
Constraint: n0${\mathbf{n}}\ge 0$.

lda

### Output Parameters

1:     a(lda, : $:$) – double array
The first dimension of the array a will be n + 1${\mathbf{n}}+1$
The second dimension of the array will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
ldan + 1$\mathit{lda}\ge {\mathbf{n}}+1$.
The lower triangle of the inverse matrix X$X$ is stored in the elements of the array below the diagonal, in rows 2$2$ to n + 1$n+1$; xij${x}_{ij}$ is stored in a(i + 1,j)${\mathbf{a}}\left(i+1,j\right)$ for ij$i\ge j$. The upper triangle of the original matrix is unchanged.
2:     ifail – int64int32nag_int scalar
${\mathrm{ifail}}={\mathbf{0}}$ unless the function detects an error (see [Error Indicators and Warnings]).

## Error Indicators and Warnings

Errors or warnings detected by the function:
ifail = 1${\mathbf{ifail}}=1$
The matrix A$A$ is not positive definite, possibly due to rounding errors.
ifail = 2${\mathbf{ifail}}=2$
 On entry, n < 0${\mathbf{n}}<0$, or lda < n + 1$\mathit{lda}<{\mathbf{n}}+1$.

## Accuracy

The accuracy of the computed inverse depends on the conditioning of the original matrix. For a detailed error analysis see page 39 of Wilkinson and Reinsch (1971).

The time taken by nag_matop_real_symm_posdef_inv_noref (f01ad) is approximately proportional to n3${n}^{3}$. nag_matop_real_symm_posdef_inv_noref (f01ad) calls functions nag_lapack_dpotrf (f07fd) and nag_lapack_dpotri (f07fj) from LAPACK.

## Example

```function nag_matop_real_symm_posdef_inv_noref_example
a = [5, 7, 6, 5;
7, 10, 8, 7;
6, 8, 10, 9;
5, 7, 9, 10;
0, 0, 0, 0];
[aOut, ifail] = nag_matop_real_symm_posdef_inv_noref(a)
```
```

aOut =

5.0000    7.0000    6.0000    5.0000
68.0000   10.0000    8.0000    7.0000
-41.0000   25.0000   10.0000    9.0000
-17.0000   10.0000    5.0000   10.0000
10.0000   -6.0000   -3.0000    2.0000

ifail =

0

```
```function f01ad_example
a = [5, 7, 6, 5;
7, 10, 8, 7;
6, 8, 10, 9;
5, 7, 9, 10;
0, 0, 0, 0];
```
```

aOut =

5.0000    7.0000    6.0000    5.0000
68.0000   10.0000    8.0000    7.0000
-41.0000   25.0000   10.0000    9.0000
-17.0000   10.0000    5.0000   10.0000
10.0000   -6.0000   -3.0000    2.0000

ifail =

0

```


text:
# What is the 99% confidence interval?

I'm having some trouble with this problem. The way I'd go about solving it is:

$56.2 +- 2.576 * \frac{53.6}{\sqrt5}$

However this does not give me the correct answer which is:

$56.2 +- 15.07$.

Can some one help me out?

EDIT:

I worked the problem out with the guidance from below:

56 +- 4.604 * (7.3212/sqrt(5))

• You have to use square root of 53.6 – user83346 Oct 19 '17 at 4:23
• 99% CI's are in many cases meaningless, you could as well use 100% interval: $(-\infty, \infty)$, in both cases it would be very wide and would not provide you much information. – Tim Oct 19 '17 at 7:54

The key to this problem is that they say you are working with $s^2$, which is an estimate of the population variance. You have applied a z-distribution to the problem, which would be correct if you had the actual population variance ($\sigma^2$) . However, when you are working with an estimate of the population variance you must apply a student's t-distribution to the problem.

The form of the t-distribution equation is similar to the z-distribution equation except that you replace the "z" with a "t", and the population standard deviation with the sample standard deviation (s):

$t = \frac{xbar - \mu}{\frac{s}{\sqrt{n}}}$

where $s$ is the square root of the sample variance, $n$ is the sample size, and $t$ is a value to be calculated from a t-table. You will need two things to find the t-value to plug into your equation:

1. Confidence interval - in this case you are asked for 99% confidence, meaning that your total acceptable Type I error rate ($\alpha$) is 1-.99 = .01. That error rate must be split over both ends of the t-distribution, thus the heading of the t - table column you seek should read .005.
2. Degrees of freedom ($n$). In this case, your Degrees of Freedom will be 5-1 = 4

Go to a t-table with a t($\alpha$/2=.005, DOF=4), and you will have enough information to find your answer.

• So just to clarify: If you are given an estimate you ALWAYS have to use the t-statistic? Also, when do you seek 0.01/2 or 0.005 vs 0.01? Because some questions I simply use the value of 2.576. Is that only without degrees of freedom? – Koalafications Oct 19 '17 at 2:38
• Yes, if the sample is 'small'. The t-distribution converges on the normal distribution as the number of samples grows. By the time you have 50 samples from an infinite population, the difference is usually irrelevant. – eSurfsnake Oct 19 '17 at 5:28


text:
Rational Numbers Class 7 Extra Questions Maths Chapter 9

# Rational Numbers Class 7 Extra Questions Maths Chapter 9

## Rational Numbers Class 7 Extra Questions Maths Chapter 9

Extra Questions for Class 7 Maths Chapter 9 Rational Numbers

### Rational Numbers Class 7 Extra Questions Very Short Answer Type

Question 1.
Find three rational numbers equivalent to each of the following rational numbers.
(i) $$\frac { -2 }{ 5 }$$
(ii) $$\frac { 3 }{ 7 }$$
Solution:

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+91

Verify OTP Code (required)

Question 2.
Reduce the following rational numbers in standard form.
(i) $$\frac { 35 }{ -15 }$$
(ii) $$\frac { -36 }{ -216 }$$
Solution:

Question 3.
Represent $$\frac { 3 }{ 2 }$$ and $$\frac { -3 }{ 4 }$$ on number lines.
Solution:

Question 4.
Which of the following rational numbers is greater?
(i) $$\frac { 3 }{ 4 }$$, $$\frac { 1 }{ 2 }$$
(ii) $$\frac { -3 }{ 2 }$$, $$\frac { -3 }{ 4 }$$
Solution:

Question 5.
Find the sum of

Solution:

Question 6.
Subtract:

Solution:

Question 7.
Find the product:

Solution:

### Rational Numbers Class 7 Extra Questions Short Answer Type

Question 8.
If the product of two rational numbers is $$\frac { -9 }{ 16 }$$ and one of them is $$\frac { -4 }{ 15 }$$, find the other number.
Solution:
Let the required rational number be x.

Question 9.
Arrange the following rational numbers in ascending order.

Solution:

Question 10.
Insert five rational numbers between:

Solution:

### Rational Numbers Class 7 Extra Questions Long Answer Type

Question 11.
Evaluate the following:

Solution:

Question 12.
Subtract the sum of $$\frac { -5 }{ 6 }$$ and -1$$\frac { 3 }{ 5 }$$ from the sum 2$$\frac { 2 }{ 3 }$$ and -6$$\frac { 2 }{ 5 }$$.
Solution:

Question 13.

Solution:
We have

Question 14.
Divide the sum of -2$$\frac { 15 }{ 17 }$$ and 3$$\frac { 5 }{ 34 }$$ by their difference.
Solution:

Question 15.
During a festival sale, the cost of an object is ₹ 870 on which 20% is off. The same object is available at other shops for ₹ 975 with a discount of 6$$\frac { 2 }{ 3 }$$ %. Which is a better deal and by how much?
Solution:
The cost of the object = ₹ 870
Discount = 20% of ₹ 870 = $$\frac { 20 }{ 100 }$$ × 870 = ₹ 174
Selling price = ₹ 870 – ₹ 174 = ₹ 696
The same object is available at other shop = ₹ 975

Selling price = ₹ 975 – ₹ 65 = ₹ 910
Since ₹ 910 > ₹ 696
Hence, deal at first shop is better and by ₹ 910 – ₹ 696 = ₹ 214

### Rational Numbers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 16.
Simplify:
21.5 ÷ 5 – $$\frac { 1 }{ 5 }$$ of (20.5 – 5.5) + 0.5 × 8.5
Solution:
Using BODMAS rule, we have
21.5 ÷ 5 – $$\frac { 1 }{ 5 }$$ of (20.5 – 5.5) + 0.5 × 8.5
= 21.5 ÷ 5 – $$\frac { 1 }{ 5 }$$ of 15 + 0.5 × 8.5
= 21.5 × $$\frac { 1 }{ 5 }$$ – $$\frac { 1 }{ 5 }$$ × 15 + 0.5 × 8.5
= 4.3 – 3 + 4.25
= 4.3 + 4.25 – 3
= 8.55 – 3
= 5.55

Question 17.
Simplify:

Solution:
Using BODMAS rule, we have
2.3 – [1.89 – {3.6 – (2.7 – 0.77)}] = 2.3 – [1.89 – {3.6 – 1.93}] = 2.3 – [1.89 – 1.67] = 2.3 – 0.22
= 2.08

Question 18.

Solution:
Using BODMAS rule, we have

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text:
# Revision history [back]

### Re-distorting a set of points after camera calibration

I am working on a project in Python to calibrate a small thermal camera sensor (FLIR Lepton). Because of limited resolution initial distortion removal is not very exact. By using an iterative method I should be able to refine this calibration (for those of you with access to scientific articles, see this link). This requires me to take the following steps:

1. Use a set of images of a calibration pattern to estimate the initial distortion
2. Undistort the images
3. Apply a perspective correction to the undistorted images
4. Re-estimate the calibration point positions
5. Remap these refined calibration points back to the original images
6. Use the refined points to re-estimate the distortion
7. Repeat until the RMS-error converges

I am stuck at step four. Below you see the commands I used to remove the camera distortion from the original image using the camera matrices and the distortion matrix.

mapx,mapy = cv2.initUndistortRectifyMap(mtx,dist,None,newcameramtx,(w,h),5)
dst = cv2.remap(img,mapx,mapy,cv2.INTER_LINEAR)

So far I am not able to figure out how to reverse these commands to remap the new point positions to the original image. So far I am able to do (roughly in order of the steps above):

I have looked online and found the same question a bunch of times, with several examples in C++, which I cannot fully comprehend and modify for my purposes. I have tried the solution suggested by this post but this has not yielded the desired results, see last image above. There is my code of that solution

def distortBackPoints(x, y, cameraMatrix, dist):

fx = cameraMatrix[0,0]
fy = cameraMatrix[1,1]
cx = cameraMatrix[0,2]
cy = cameraMatrix[1,2]
k1 = dist[0][0] * -1
k2 = dist[0][1] * -1
k3 = dist[0][4] * -1
p1 = dist[0][2] * -1
p2 = dist[0][3] * -1
x = (x - cx) / fx
y = (y - cy) / fy

r2 = x*x + y*y

xDistort = x * (1 + k1 * r2 + k2 * r2 * r2 + k3 * r2 * r2 * r2)
yDistort = y * (1 + k1 * r2 + k2 * r2 * r2 + k3 * r2 * r2 * r2)

xDistort = xDistort + (2 * p1 * x * y + p2 * (r2 + 2 * x * x))
yDistort = yDistort + (p1 * (r2 + 2 * y * y) + 2 * p2 * x * y)

xDistort = xDistort * fx + cx;
yDistort = yDistort * fy + cy;

return xDistort, yDistort

Then using this command to call the function

corners2 = []
for point in corners:
x, y = distortBackPoints(point[0][0], point[0][1], newcameramtx, dist)
corners2.append([x,y])

I am new to OpenCV and computer vision, so my knowledge about the algebra of these solutions is limited. Any hands on examples or correction to my current code would be greatly appreciated.

Kind regards,

Bart


text:
# Thread: Prove the following equality

1. ## Prove the following equality

if you did not see the image please inlarge ctrl+

any idea that help in proving this equality is appretiated.

Thanks

2. Substitute the MacLaurin series for $\displaystyle e^z, e^{-\frac{1}{2}z}$ and $\displaystyle \cos{\frac{z\sqrt{3}}{2}}$ into RHS and simplify...


text:
Biot-Savart Law: Magnetic Fields on an equilateral triangle

1. Apr 14, 2008

alovesong

1. The problem statement, all variables and given/known data

Two long straight wires sit at the lower corners of an equilateral triangle and carry current I. Find the magnitude and direction of the B field at the top vertex of the triangle for the case where:

a) the current in both lower wires flows out of the page
b) the current in the left wire flows out of the page, the current of the right wire flows into the page.

2. Relevant equations

dB=(μ_o Idlsinθ)/(4πr^2 ) for the magnetic field at a point P in space

3. The attempt at a solution

First, I don't really know what effect being physically connected to the current-bearing wires has on P. Assuming it's negligible, though, then working with just one wire

r= length of triangle side from P to current-bearing wire
R= distance directly from P to the wire?

B= (u_o*I/4π) integral[dlsinθ/r^2]

dl= R(csc^2θdθ = r^2dθ/R

B= (u_o*I/4πR) [int from θ=0 to π] sinθdθ = -(u_o*I/4πR)cosθ evaluated from 0 to π

This is based mostly around an example for another problem, but I think most of it applies... However I am confused on the integral boundaries (if it's an equilateral triangle, shouldn't θ be fixed at 60?) and am unsure how how incorporate the second wire. Help, please!

2. Apr 14, 2008

Nabeshin

I'm not sure if you have a correct picture of what's going on here in your head, because P is not connected to the current-bearing wires. Imagine an equilateral triangle. Now, the two lower VERTICES have wires passing through them perpendicular to the plane of the page with directions given in a) and b). This should help to simplify the problem into a point experiencing a force due to two magnetic fields from two wires.

3. Apr 14, 2008

alovesong

Okay, yeah, you're right - for some reason I thought that the "equilateral triangle" was physical, when it's not... but I'm still confused about what to do with the two separate wires.

4. Apr 14, 2008

Nabeshin

Well you're going to need to know how to calculate the magnetic field due to the wires, which it seems you need to do via biot-savart.
The biot-savart equation, in its differential form, is actually this:
$$dB=\frac{\mu_{o}IdLx\hat{r}}{4 \pi r^{2}}$$
Where $$\hat{r}$$ is the unit vector in the direction of the point. Now, think about what the theta from the cross product represents, and you should be able to develop the general form for magnetic field a distance r away from a charge carrying wire. Hint: What is constant and what is changing?


text:
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x values are : 3,5,6,8 y values are : 14,19,21.5,26.5 ( in my book these values are in a table )    I tried to find the slope of (3,14) (5,19) (6,21.5) ( 8,26.5). The slopes weren't constant so... more
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#### equation for a linear function (14,0) perpendicular to the given of g(x) = 2x + 5

Need help making the equation

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text:
# Let X be an irreducible, normal variety over an algebraically closed field of characteristic z

Let $X$ be an irreducible, normal variety over an algebraically closed field of characteristic zero. Let $x,y\in X$ be two points such that $f\left(x\right)=f\left(y\right)$ for every $f\in K\left(X\right)$ which is defined at $x$ and at $y$. Can I conclude that $x=y$?
I feel the answer should be affirmative. In fact, the statement can be reduced to the following: Given two effective prime divisors ${D}_{x}$ and ${D}_{y}$ on $X$, there exists a rational function $f\in K\left(X\right)$ with ${v}_{{D}_{x}}\left(f\right)\ne 0$ and ${v}_{{D}_{y}}\left(f\right)=0$.
If this is true, then assuming $x\ne y$ we could find a divisor ${D}_{x}$ containing $x$ but not $y$ and a divisor ${D}_{y}$ containing $y$ but not $x$, so a function $f$ as above would yield a contradiction. However, I just can't prove the statement, even though I also think it should be true.
You can still ask an expert for help

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Ronald Hickman
As mentioned by Cantlog there is true if $X$ is integral and separated. Here is a counterexample showing this is not true in the integral non-separated case.
Take $X$ to be the affine line with doubled origin, $x$ the north pole and $y$ the south pole. Let $f\left(t\right)=p\left(t\right)/q\left(t\right)\in k\left(t\right)$ be any rational function on $X$. We see

but $x\ne y$.


text:
# Isomorphic encryption or homomorphic encryption?

Many encryption functions are said to be homomorphic:

http://en.wikipedia.org/wiki/Homomorphic_encryption

As encryption functions are invertible, they can be considered one-to-one and onto on properly defined domains and ranges.

So, my basic question is why we don't use the term "isomorphic encryption" rather than "homomorphic encryption"?

First, consider the definition of group isomorphism:

Given two groups $(G, \otimes)$ and $(H, \odot)$, a group isomorphism from $(G, \otimes)$ to $(H, \odot)$ is a bijective function $f : G \to H$ such that for all $u$ and $v$ in $G$ it holds that $f(u \otimes v) = f(u) \odot f(v)$.

Now, consider a homomorphic encryption such as ElGamal cryptosystem: It takes a message from a cyclic group $G$, and outputs a pair $(c_1, c_2) \in G^2$. That is, $\mathcal{E} \colon G \to G^2$.

Notice that under this definition, $\mathcal{E}$ is not a bijection from the message space $G$ to the ciphertext space $G^2$. However, the decryption of ElGamal is unique.

Homomorphic Encryption does not necessarily mean that the encryption function is a group homomorphism.

Let $E: M \rightarrow C$ be the encryption function where $M$ and $C$ are the plaintext and ciphertext space respectively. Now $E$ is said to be homomorphic if $E(m_1 * m_2)=E(m_1)\cdot E(m_2)$, where $*$ and $\cdot$ are binray operations on $M$ and $C$ respectively. However, $M$ and $C$ are not necessarily possessing an algebraic structure like groups, rings etc. Thus, even though the encryption function $E$ is invertible, the question of isomorphism does not arise.

• A set endowed with an operation possesses an algebraic structure, even if the structure is not a group. We are sure that the set M is closed under *, otherwise, the "multiplication of two valid plaintexts will not result in a valid plaintext. So, in the very least, we have an algebraic structure called Magma: en.wikipedia.org/wiki/Magma_(algebra). Jun 26 '15 at 2:14


text:
# Why is matrix not a different field altogether?

According to Wikipedia-

In mathematics, a field is a set on which addition, subtraction, multiplication, and division are defined, and behave as when they are applied to rational and real numbers.

Now, when we work with matrices we define addition,subtraction,multiplication operations (not division) on matrices.

So,going by the above definition, does a matrix qualify to be a field as we define the mentioned operations on them?

I have two doubts-

• The definition mentions "...a field is a set..." . I know a set is any well defined collection of numbers. A matrix is also a collection of numbers but not well defined. Is this a reason why we cannot call a matrix a set?
• Division of matrices is not defined. Is this also a reason that eliminates the possibilities of a matrix to be called a set?

Thanks for any help!

1. A matrix is definitely not a set of its entries, but all matrices of a specified size over a specified field is a set. This is what we call "the set of matrices".
2. Indeed, this set does not form a field, by two reasons:
1. Matrix multiplication is not commutative: in general, $AB \neq BA$.
2. In a field, all non-zero elements are invertible. This is not the case for matrices: a matrix $A$ is invertible only if $\det A$ is invertible (is not zero, if the scalars form a field), and there do exist non-zero matrices with zero determinant.

So, the structure that the set of matrices forms is a non-commutative ring.

• Does invertible mean division in $\mathbb R$? – tatan Mar 21 '18 at 14:47
• @tatan Invertible means $\exists \frac{1}{x}$ in the ring of scalars. For example, in $\mathbb{R}$ all non-zero elements are invertible. However, in $\mathbb Z$ only $\pm 1$ are invertible, and a matrix with integer coefficients is invertible (with the inverse also having integer coefficients) iff its determinant is $\pm 1$. – lisyarus Mar 21 '18 at 15:17

As you mentioned, the set of all matrices is not a field.

There are many problems with the matrices which keep it far from being a field.

Even if we consider the set of all invertible $n\times n$ matrices along with the $n\times n$ matrix of $0$ we still do not have a field.

The big problem is that even if $AB$ and $BA$ are defined, we do not have $AB=BA$ for all $A$ and $B$.


text:
Proof. Let $U_ d \to S$ be the scheme representing the functor $F_ d$ defined above. Let $\mathcal{L}_ d$, $\psi ^ d : \pi _ d^*\mathcal{A}^{(d)} \to \bigoplus _{n \geq 0} \mathcal{L}_ d^{\otimes n}$ be the universal object. If $d | d'$, then we may consider the quadruple $(d', \pi _ d, \mathcal{L}_ d^{\otimes d'/d}, \psi ^ d|_{\mathcal{A}^{(d')}})$ which determines a canonical morphism $U_ d \to U_{d'}$ over $S$. By construction this morphism corresponds to the transformation of functors $F_ d \to F_{d'}$ defined above.

For every affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ setting $A = \Gamma (V, \mathcal{A})$ we have a canonical identification of the base change $U_{d, V}$ with the corresponding open subscheme of $\text{Proj}(A)$, see Lemma 27.16.2. Moreover, the morphisms $U_{d, V} \to U_{d', V}$ constructed above correspond to the inclusions of opens in $\text{Proj}(A)$. Thus we conclude that $U_ d \to U_{d'}$ is an open immersion.

This allows us to construct $X$ by glueing the schemes $U_ d$ along the open immersions $U_ d \to U_{d'}$. Technically, it is convenient to choose a sequence $d_1 | d_2 | d_3 | \ldots$ such that every positive integer divides one of the $d_ i$ and to simply take $X = \bigcup U_{d_ i}$ using the open immersions above. It is then a simple matter to prove that $X$ represents the functor $F$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# The probability that 25 people do not have the same birthday

I figure that it would be something like this

$$\frac{365}{365} \times \frac{364}{365} \times \cdots \times \frac{340}{365}$$

But that is a lot of fractions to enter into a calculator, even so I think this is the right answer but I was wondering if there is a shortcut or something? I have a casio CFX-9850GB.

-
How accurate do you want the answer? –  Aryabhata Jul 14 '11 at 7:54
The book says 43,1%. –  Algific Jul 14 '11 at 8:02
You may (or, may not) be interested to know that the exact answer is 4314778256349470854170763414024795540030075259130696695808 / 10004116148447957520459906484225353834116619892120361328125. –  Chris Taylor Jul 14 '11 at 8:21
And they said quantum electrodynamics gives very precise results! That's nothing compared to elementary biology. –  André Nicolas Jul 14 '11 at 12:45
But I think the assumption of equidistribution of birthdates does not really hold; there is plenty of data to this effect.Specifically, the ref. cites a collection of n=480,040 data points, where the Chi-squared for (Actual # of births for a given day) -$\frac{1}{365}$480,040 failed at a 95% confidence. –  gary Jul 14 '11 at 21:29

Yes, you could write the numerator as $365! / (365 - 25)!$ where $n!$ is the factorial function, and the denominator as $365^{25}$, so your probability is

$$365! / ((365-25)! \times 365^{25})$$

You might try entering this into Wolfram Alpha to get a result.

Also, the expression you entered is incorrect. You have a total of 26 terms in your product, whereas you only want 25. Don't worry, this kind of off-by-one error is common even among experienced mathematicians and computer scientists. I make it all the time!

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I only count 25. :( How? I just assumed 365-25=340 would be correct. Thanks btw. Perfect. I was thinking about factorial. –  Algific Jul 14 '11 at 8:13
For the same reason that 10...15 has 6 numbers, not 5: 10, 11, 12, 13, 14, 15. You started counting from zero instead of from one. –  Chris Taylor Jul 14 '11 at 8:19
Spot on, thanks again! –  Algific Jul 14 '11 at 8:25
But if counting starts on 0 shouldn't you subtract 24 instead? Or would you just remove the term that is equal to 1? –  Algific Jul 14 '11 at 8:44
On the subtraction issue, consider if you only had one term. Then you would calculate $365-365$ and think that you had 0 terms, which you obviously don't. –  gereeter Jul 14 '11 at 14:06

I think that it is

$$\frac{365}{365}\times \frac{364}{365}\times \cdots \times \frac{341}{365},$$

because $365-341+1=25$.

Call $P(n)$ the probability that $n$ people do not have the same birthday. It is given by $$\begin{eqnarray*} P(n) &=&\frac{365}{365}\times \frac{364}{365}\times \cdots \times \frac{% 365-n+1}{365} \\ &=&\frac{\prod_{k=2}^{n}\left( 366-k\right) }{365^{n-1}}=\frac{364!}{% 365^{n-1}\left( 365-n\right) !} \\ &=&\frac{\Gamma (365)}{365^{n-1}\Gamma (366-n)}. \end{eqnarray*}$$

Notation: The $\prod$ symbol stands for a product

$$\prod_{k=2}^{25}\left( 336-k\right) =\left( 336-2\right) \left( 336-3\right) \left( 336-4\right)\times \cdots \times\left( 336-25\right) .$$

and $\Gamma (x)$ is the gamma function with the property $\Gamma (n+1)=n!$.

For $n=25$, we have $$\begin{eqnarray*} P(25) &=&\frac{\prod_{k=2}^{25}\left( 366-k\right) }{365^{25-1}}=\frac{364!}{% 365^{25-1}\left( 365-25\right) !}\approx 0.431\,3 \\ &=&\frac{\Gamma (365)}{365^{25-1}\Gamma (366-25)}\approx 0.431\,3 \end{eqnarray*}$$

Here is a plot of $1-P(x)$ (número de pessoas = number of people $=x$)

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I started writting this answer before Chris Taylor had posted his answer. I used this post of mine problemasteoremas.wordpress.com/2009/07/15/… –  Américo Tavares Jul 14 '11 at 8:55
What does the II symbol and the cut of T stand for? –  Algific Jul 14 '11 at 9:07
It stands for product. It works in the same manner as the summation notation : you "multiply" things from $k = 2$ up to $k = n$ and write $\prod_{k=2}^n x_k$. –  Patrick Da Silva Jul 14 '11 at 9:10
nevermind got it.. its to cancel out all the remaining terms from the numerator.. –  Tyler Hilton Jul 14 '11 at 15:26
Got it, beautiful solution by the way. –  Tyler Hilton Jul 15 '11 at 20:19

I seem to remember a surprisingly good approximation of your probability $P_{25}$ is obtained using $1-x\le\exp(-x)$ for $x=k/365$ from $k=1$ to $k=25$. Hence $$P_{25}<\exp(-x_1-\cdots-x_{25})=\exp(-25\cdot26/(2\cdot365))<41.05\%.$$ Likewise, if one needs to go from $k=1$ to $k=24$, one gets $$P_{24}<\exp(-24\cdot25/(2\cdot365))<43.96\%.$$ One can even get not-so-crude lower bounds starting from the fact that on a small interval $(0,z)$, the derivative of $u:x\mapsto1/(1-x)$ is $u'(x)=1/(1-x)^2$ hence $u'\le cu$ with $c=1/(1-z)$, which yields $u(x)\le u(0)\exp(cx)$, hence $1-x\ge \exp(-cx)$. One gets $$P_{24}>\exp(-24\cdot25/(2\cdot365\cdot(1-x_{24})))=\exp(-24\cdot25/(2\cdot341))>41.48\%,$$ and $$P_{25}>\exp(-25\cdot26/(2\cdot340))>38.44\%.$$ To sum it up, for every $1\le n<N$, $$\exp\left(-\frac{n(n+1)}{2(N-n)}\right)<\prod_{k=1}^n\left(1-\frac{k}N\right)<\exp\left(-\frac{n(n+1)}{2N}\right).$$ In the Birthday paradox context ($N=365$), the funny thing is that the upper bound for $n=22$ is $<50\%$ and the lower bound for $n=21$ is $>50\%$ hence these crude approximations are enough to be sure that $P_{22}<50\%<P_{21}$ and that the so-called magic number is $22+1=23$.

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I got $P\approx 0.4313$. –  Américo Tavares Jul 14 '11 at 8:51
@Américo: Which one, $P_{24}$ or $P_{25}$? –  Did Jul 14 '11 at 9:02
$P_25$ (I called it $P(25)$). So that's right $0.43213>0.3844$. –  Américo Tavares Jul 14 '11 at 9:07
@Américo: Your $P(25)$ is my $P_{24}$ so the relevant inqualities are $41.48\% <P_{24}<43.96\%$, which are compatible with your $P(25)=43.13\%$. We happy. –  Did Jul 14 '11 at 9:33
Thanks! +1 for your higher level answer. –  Américo Tavares Jul 14 '11 at 9:36

That's almost the right probability -- there should be 25 factors, so you don't need the $340/365$ at the end. The last factor is $341/365$. You're right that it's going to be quite a bit of work to compute it exactly. And what's really interesting is the numerical value -- is it less than one-half? Less than ten percent? There are lots of approximate methods for figuring out answers to problems like this, and I'll outline one here.

You can rewrite the probability you're looking for in terms of factorials as

$$P = {365! \over 340! 365^{25}}$$

and this is useful because there is a very useful approximation for $n!$, called Stirling's formula:

$$n! \approx \sqrt{2 \pi n} (n/e)^n$$

Now, if you were to plug in $340$ or $365$ to Stirling's formula you would get two very large numbers, which a standard calculator cannot handle; but we know $P$ should be reasonably sized. The trick here is to take logarithms. Stirling's formula can be rewritten as

$$\log n! \approx {1 \over 2} \log {2 \pi} + \left( n + {1 \over 2} \right) \log n - n$$

and the logarithm of your original probability is

$$\log P = \log 365! - \log 340! - 25 \log 365.$$

From Stirling's formula $\log 365! \approx 1792.331$ and $\log 340! \approx 1645.675$; this gives

$$\log P = 1792.331 - 1645.675 - 147.497 = -0.841$$

and so $P \approx e^{-0.841} = 0.432$.

(You might worry if there's an error in Stirling's approximation. It turns out that there is, and that the estimated value of $\log n!$ will always be lower than its true value by about $1/(12n)$. For example $\log 8! = 10.6046$ but Stirling gives ${1 \over 2} \log (2\pi) + 8.5 \log 8 - 8 = 10.5942$; these differ by $0.0104$ or about $1/96$. But the errors in using Stirling to approximate $\log 365!$ and $\log 340!$ are very small and in the same direction; they nearly cancel out.)

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Nope, no shortcut. The first is 1, and all denominators are the same (so don't keep dividing by 365). And in the grand scheme of things, be thankful that this number of calculations is relatively small. It amounts to only 2 dozen operations, you know?

Keep us posted on your result. This is known as the "Birthday Paradox." And (thanks to Chris Taylor for giving me the exact number) the 'magic number' is 23, meaning that with more than 23 people this becomes (without ruining the question) surprisingly likely.

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Actually, to prevent overflow, it would make more sense to divide by 365 at each step... –  Zhen Lin Jul 14 '11 at 7:56
I didn't mean to do them all at once, but instead to do them at say, 3 or 4 at a time. –  mixedmath Jul 14 '11 at 7:57
Fwiw the 'magic number' is 23. –  Chris Taylor Jul 14 '11 at 7:57
Ah! Thank you - I'm glad you remembered. Even lower than I had thought, and I knew the paradox! I'm suddenly in grade school again, stunned at little things like multiplication. –  mixedmath Jul 14 '11 at 8:01
3 6 5 OPTN F6 ► F3 PROB F2 nPr 2 5 EXE

÷ 3 6 5 ^ 2 5 EXE


text:
zbMATH — the first resource for mathematics

The $$\pi$$-$$\pi$$-theorem for manifold pairs with boundaries. (English. Russian original) Zbl 1141.57012
Math. Notes 81, No. 3, 356-364 (2007); translation from Mat. Zametki 81, No. 3, 405-416 (2007).
The $$\pi$$-$$\pi$$ theorem of [C. T. C. Wall, Surgery on Compact Manifolds, second edition. Mathematical Surveys and Monographs. 69. (Providence), RI: American Mathematical Society (AMS). (1999; Zbl 0935.57003)] states that in higher dimensions a normal map of a manifold with boundary to a simple Poincaré pair $$(X,\partial X)$$ is normally bordant to a simple homotopy equivalence if the map $$\pi_1(\partial X) \to \pi_1(X)$$ is an isomorphism. Wall also defined the surgery obstruction groups $$LP_*$$ for normal maps to manifold pairs and splitting obstruction groups $$LS_*$$. From this point of view, the authors generalized the $$\pi$$-$$\pi$$ theorem in the following sense: Let $$(f, \partial f): (M, \partial M) \to (X, \partial X)$$ be a topological normal map to a pair of manifolds with boundary $$(Y, \partial Y) \subset (X, \partial X)$$ with topological normal block fibration $$\xi_Y=\xi_{Y\subset X}$$ and $$\dim Y \geq 6$$. If the inclusion $$\partial X \to X$$ induces an isomorphism between the push-out squares of fundamental groups of $$(X; Y, X-Y; S(\xi_Y))$$ and $$(\partial X; \partial Y, \partial X -\partial Y; S(\xi_{\partial Y}))$$, then
(1) $$(f,\partial f)$$ is normally bordant to an $$s$$-triangulation of the pair of manifolds.
(2) if $$(f, \partial f)$$ is a simple homotopy equivalence of the pairs, then it is concordant to an $$s$$-triangulation of the pair of manifolds.
(3) if the restricted map $$(f,\partial f) :(f^{-1}(Y), f^{-1}(\partial Y)) \to (Y, \partial Y)$$ is a simple homotopy equivalence of pairs, then $$(f, \partial f)$$ is normally bordant to an $$s$$-triangulation of the pair of manifolds.
Where an $$s$$-triangulation of a pair of manifolds is in the sense of [A. A. Ranicki, Exact Sequences in the Algebraic Theory of Surgery. Mathematical Notes, 26. Princeton, New Jersey: Princeton University Press; University of Tokyo Press. (1981; Zbl 0471.57012)]. The results can be carried over to $$Diff$$- and $$PL$$-categories. The authors applied the obtained results to surgery on filtered manifolds.
Reviewer: Yang Su (Beijing)
MSC:
 57R65 Surgery and handlebodies 57R67 Surgery obstructions, Wall groups
Full Text:
References:
 [1] C. T. C. Wall, Surgery on Compact Manifolds, Ed. by A. A. Ranicki, 2nd ed. (Amer. Math. Soc., Providence, RI, 1999). · Zbl 0935.57003 [2] A. A. Ranicki, Exact Sequences in the Algebraic Theory of Surgery (Mathem. Notes, Princeton, 1981). · Zbl 0471.57012 [3] W. Browder and F. Quinn, ”A surgery theory for G-manifolds and stratified spaces,” in Manifolds (Univ. Tokyo Press, 1975), pp. 27–36. · Zbl 0343.57017 [4] Yu. V. Muranov, D. Repovš, and R. Jimenez, ”Surgery spectral sequence and manifolds with filtration,” Trudy Moskov. Mat. Obshch. 67, 294–325 (2006) [Trans. Moscow Math. Soc.]. [5] S. Weinberger, The Topological Classification of Stratified Spaces (Univ. Chicago Press, Chicago-London, 1994). · Zbl 0826.57001 [6] S. Lopez de Medrano, Involutions on Manifolds (Springer-Verlag, Berlin-Heidelberg-New York, 1971). · Zbl 0199.58603 [7] M. M. Cohen, A Course in Simple-Homotopy Theory (Springer-Verlag, New York, 1973). · Zbl 0261.57009 [8] Yu. V. Muranov, D. Repovš, and F. Spaggiari, ”Surgery on triples of manifolds,” Mat. Sb. 8, 139–160 (2003) [Russian Acad. Sci. Sb. Math. 194 (8), 1251–1271 (2003)]. · Zbl 1067.57032 · doi:10.4213/sm764
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# sample.prior

0th

Percentile

##### Sample the prior distribution

Sample quadruples $$(\beta, \sigma^2, \phi, \tau^2_{rel})$$ from the prior distribution of parameters specifying a Gaussian random field. Typically the prior is specified in the same manner as when calling krige.bayes.

Keywords
distribution, spatial
##### Usage
sample.prior(n, kb.obj=NULL, prior=prior.control())
##### Arguments
n

number of samples

kb.obj

on object with an output of krige.bayes.

prior

an call to prior.control. Unnecessary if kb.obj is provided.

##### Value

A $$p+3 \times 4$$ data-frame with a sample of the prior distribution of model parameters, where $$p$$ is the length of the mean parameter $$\beta$$.

##### References

Further information on the package geoR can be found at: http://www.leg.ufpr.br/geoR.

krige.bayes and sample.posterior.

• sample.prior
##### Examples
# NOT RUN {
sample.prior(50,
prior=prior.control(beta.prior = "normal", beta = .5, beta.var.std=0.1,
sigmasq.prior="sc", sigmasq=1.2, df.sigmasq= 2,
phi.prior="rec", phi.discrete = seq(0,2, l=21)))
# }

Documentation reproduced from package geoR, version 1.8-1, License: GPL (>= 2)

### Community examples

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