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# The Improving Mathematics Education in Schools (TIMES) Project
## THE REAL NUMBERS
A guide for teachers - Years 810
Module 28
June 2011
810
YEARS
## The Real Numbers
(Number and Algebra : Module 28)
For teachers of Primary and Secondary Mathematics
510
Cover design, Layout design and Typesetting by Claire Ho
The Improving Mathematics Education in Schools (TIMES)
Project 20092011 was funded by the Australian Government
Department of Education, Employment and Workplace
Relations.
The views expressed here are those of the author and do not
necessarily represent the views of the Australian Government
Department of Education, Employment and Workplace Relations.
The University of Melbourne on behalf of the International
Centre of Excellence for Education in Mathematics (ICEEM),
the education division of the Australian Mathematical Sciences
Institute (AMSI), 2010 (except where otherwise indicated). This
Module 28
## THE REAL NUMBERS
A guide for teachers - Years 810
June 2011
Peter Brown
Michael Evans
David Hunt
Janine McIntosh
810
Bill Pender
Jacqui Ramagge
YEARS
{4}
## A guide for teachers
THE REAL
NUMBERS
ASSUMED KNOWLEDGE
Fluency with integer arithmetic.
Familiarity with fractions and decimals.
Facility with converting fractions to decimals and vice versa.
Familiarity with Pythagoras Theorem.
MOTIVATION
This module differs from most of the other modules, since it is not designed to summarise
in one document the content of material for one given topic or year. Material related
to the real numbers is scattered throughout the modules. We felt, however, that it was
important to have a short module on the real numbers to bring together some of the
important ideas that arise in school mathematics.
From the beginning of their mathematical studies, students are introduced to the whole
numbers. Later, fractions and decimals are introduced, leading to the notion of a rational
number, then the integers and negative fractions and decimals are covered. Finally the
notion of real number is gradually introduced after Pythagoras theorem has been covered.
Throughout this module we will use the term rational numbers for positive and negative
m
fractions including the integers. The rational numbers are numbers of the form n where
m is an in integer and n a non-zero integer. An irrational number is a number which is not
rational.
Real
Numbers
Rationale
Numbers
Integers
Whole
numbers
## The Improving Mathematics Education in Schools (TIMES) Project
The integers and rational numbers arise naturally from the ideas of arithmetic. The real
numbers essentially arise from geometry. Finding the length of the diagonal of a square
leads to square roots of numbers that are not squares. When we draw a circle whose
irrational number .
1
2
2
1
9
2
3
2
## It was as a consequence of Pythagoras theorem that the Greeks discovered irrational
numbers, which shook their understanding of number to its foundations. They also realised
that several of their geometric proofs were no longer valid. The Greek mathematician
Eudoxus considered this problem (see the History section), and mathematicians remained
unsettled by irrational numbers for a very long time. The modern understanding of real
number only began to be developed during the 19th century.
It is not possible to give a rigorous treatment of real numbers at high school. Nonetheless,
we can give a reasonable answer to the student who asks `What is a real number? and
we can also explain why such numbers are important.
CONTENT
RATIONAL NUMBERS
p
A rational number is one that can be expressed in the form q where p is an integer,
and q is a non-zero whole number. Fractions, terminating and recurring decimals are all
examples of rational numbers.
Thus:
5 = 1 , 3 =
3
1 ,
2 3 = 3 , 6.19 =
619
100 ,
20% =
1
5
## are all examples of rational numbers.
In the module, Integers, we showed, in an appendix, how the integers could be
constructed from the whole numbers using ordered pairs. In the first Appendix to
this module we show how the rational numbers can be constructed in a similar way.
3
6
This is needed to cope with the multiple representation of a fraction. For example, 4 = 8 .
{5}
{6}
## A guide for teachers
You will have seen in the module, Decimals and Percentages. That every fraction can be
expressed as either:
## a terminating decimal, for example,
a recurring decimal, for example,
2
7
5
8
= 0.625, or
= 0.285714, and
1
6
## = 0.16 where the dots indicate
repetition.
p
A fraction, q , in reduced form where p and q are whole numbers with no common
factors, will have a terminating decimal representation if and only if q has no prime factors
except 2 and 5.
All fractions can be converted to a decimal by using long division. This involves dividing
the denominator into the numerator. A zero remainder at some stage in the division
will result in a terminating decimal. If the remainder is never zero, then division by a
denominator q will result in a repeating pattern of remainders after at most (q 1) steps
(see the module, Decimals and Percentages).
EXERCISE 1
Use long division to find the decimal expansion of
1
31 .
EXERCISE 2
a Find a rational number midway between the rational numbers
a
b
and
c
d.
b Explain how this can be used to show that there are infinitely many rational numbers
a
c
between b and d .
Decimals to Fractions
The technique for converting decimals to fractions depends on whether the decimal
is terminating, recurring or eventually recurring. A terminating decimal can be easily
converted back to a fraction by using a denominator which is a power of 10.
14
7
50
157
50 .
## The following example demonstrates the method for converting recurring
decimals to fractions.
## The Improving Mathematics Education in Schools (TIMES) Project
EXAMPLE
Convert 0.123 to a fraction.
SOLUTION
Let x = 0.123 = 0.123123123 ... ...
We multiply by 103 as the repeating block has length 3.
Then 1000x = 123.123123 ... ..
= 123 + x
123
## Hence 999x = 123 and so x = 999 =
Thus we obtain 0.123 =
41
333 .
41
333 .
## An eventually recurring decimal can easily be handled by firstly multiplying it by a
sufficiently high power of 10 to make its decimal part purely recurring.
EXAMPLE
Convert 0.69123 to a fraction.
SOLUTION
Let x = 0.69123 = 0.69123123123 ...
Then 100 x = 69.123123 ... .. = 69.123.
Multiply by 1000
69054
11509
11509
## Thus we obtain 0.69123 = 16650
We first prove the following result about fractions:
If a fraction
a
b
a2
b2
## is not a whole number either.
Proof
a
Suppose that b is in reduced form, so that a and b have no common factors except 1.
Then a2 and b2 also have no common factors, because if any prime p were a common
factor of a2 and b2, it would also be a common factor of a and b.
a2
Hence b2 is not a whole number.
{7}
{8}
## A guide for teachers
Since 2 is not a whole number, but its square is the whole number 2, it follows from the
above result that 2 is not a rational.
This caused great upset to the Greek mathematicians, since it introduced a new
sort of number which they called an irrational number. An older word for this is
incommensurable, which meant that it could not be measured as a ratio of two whole
numbers. This discovery caused a dramatic rethink into the nature of number. The validity
of many of their geometric proofs, which assumed that all lengths could be measured as
ratios of whole numbers, was also called into question.
The proof given above can easily be adapted to prove that if a whole number x is not an
n
nth power, then x is not a rational number. Thus we have infinitely many examples of
irrational numbers, such as:
12
5 , 7 , 31
Such numbers are called surds and will be discussed in detail in the module, Surds.
In addition, numbers such as , log10 3, log2 6, sin 22, and so on, are also irrational,
although it is harder to prove this fact for the first and last of these examples. There is no
general method for telling when a number is irrational, and indeed there are numbers
such as + e that arise in mathematics whose status is currently unknown.
EXAMPLE
Prove that log2 5 is irrational.
SOLUTION
As with the proof that 2 is irrational, we begin by supposing the contrary.
Suppose that log2 5 =
p
q,
5 = 2q.
## Raising both sides to the power gives,
5q = 2p.
Now this equation is impossible, since the left hand side is odd, while the right hand side
is even. Thus, log2 5 is irrational.
The Fundamental Theorem of Arithmetic (The module, Prime and Prime Factorisation )
can be used to generalize this result.
## THE REAL NUMBERS
Think about graphing the rational numbers between 0 and 2 on the number line. First we
graph 21 , 1 21 , then the thirds, then the quarters, then the fifths, . As we keep going, the
gaps between the dots get smaller and smaller, and as we graph more and more rational
numbers, the largest gap between successive dots tends to zero.
If we imagine the situation when all the infinitely many rational numbers have been
graphed, there appears to be no gaps at all, and the rational numbers are spread out like
pieces of dust along the number line. Surely every point on the number line has been
accounted for by some rational number?
Not true! There are infinitely many numbers that we have not graphed, all rational
multiples of 2, including:
2,
1
2
2,
1
3
2,
1
4
2,
1
5
2, ...
Of course, there are many more missing numbers, like 3 and log2 3 and 2 . We need
a new definition of numbers that will cover all these irrational objects, which are not
rational numbers, but which we nevertheless want to think of as numbers. The solution is
very simple we make an appeal to geometry and define numbers using the geometrical
idea of points on a line:
Definition
The real numbers are all of the points on the number line.
The set of real numbers consists of both the rational numbers and the irrational numbers.
Constructing real numbers
We have seen in the module Constructions that
every rational number can be plotted on the
3
number line. For example, to plot 5 we first
divide the interval from 0 to 1 into 5 equal
subinterval (this requires the construction of
0
parallels, but that detail is not shown in the diagram).
Then we count along 3 subintervals and place a dot.
Thus rational numbers are indeed special cases of real numbers.
3
5
{9}
{10}
## We also saw how to place the square root of any
whole number on the number line using Pythagoras
theorem. For example, to plot 2 we first construct a
square on the interval from 0 to 1 (the constructions
of the right angles are not shown on the diagram).
Then we draw the diagonal from 0, which has
length 2, and use compasses to place this length
on the number line.
## The real numbers and the rational numbers
We have seen that as we place halves, thirds, quarters, fifths, on the number line, the
maximum gap between successive fractions tends to zero. Thus if is any real number
then rational numbers will be placed increasingly close to . Thus we can use rational
numbers to approximate a real number correct to any required order of accuracy.
2
2
1
For example, the diagrams above show a circle of area enclosed in a square of area
4, and enclosing a square of area 2, which proves that 2 < < 4. Archimedes improved
greatly on this result by using regular polygons with 96 sides, and was able to prove that
1
10
3 7 < < 3 71 .
Modern computer calculations, and the extremely clever algorithms they are based on,
give approximations to up to a trillion decimal places.
But beware! These observations might lead one to believe naively that the rational and
irrational numbers somehow alternate on the number line. Nothing could be further
from the truth. Even though there are infinitely many rational numbers and infinitely many
irrational numbers between 0 and 1, there are vastly more real numbers in that interval
than rational numbers.
This spectacular, but rather vague, claim can be made into a theorem as precise as any
other mathematics, and proven rigorously see the Appendix 2 for the details, which are
an excellent challenge for interested and able students. Intuitively, one should see the real
number line as a continuum, with the points joined up to make a line, whereas the rational
numbers are like disconnected specks of dust scattered along it.
## The real numbers and decimals
We have seen that every rational number can be written as a terminating or recurring
decimal. Conversely, every terminating or recurring decimal can be written as a fraction,
and thus as a rational number. Now suppose that we have a decimal that is neither
terminating nor recurring, such as 1.01001000100001 . where the number of zeroes
increases by 1 each time. This decimal represents a definite point on the number line,
and so is a real number, but it is not a rational number, because it neither terminates nor
recurs. Indeed, any infinite decimal that is neither terminating nor recurring represents
an irrational real number, and two different such infinite decimals represent different real
numbers.
Now suppose that is an irrational real number between 0 and 1. As we add tenths,
hundredths, thousandths, ..to the number line between 0 and 1, the gap between adjacent
dots decreases to zero.
By placing successively between the tenths, between the hundredths, between the
thousandths, we can produce an infinite decimal expansion that represents . The
conclusion of all this is the following theorem.
Every real number can be represented by one, and only one, infinite decimal (excluding
recurring nines). If the expansion terminates or recurs, the number is rational, otherwise
the number is irrational.
Approximating real numbers by decimals
There are many ways of approximating real number by decimals. The two simplest are
described below.
Sequence of truncations
Now consider the infinite decimal expansion of a real number If we truncate this
expansion at 1 place, 2 places, 3 places, , the result is a sequence of rational numbers
(terminating decimals) converging to . For example, the decimal expansion of is known
to nearly 3 trillion places, and begins
= 3.141592653589793238462643383279502884197169399375
{11}
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## A guide for teachers
Sequence of approximations
To approximate the real number to a given number of decimal places, we truncate it one
place more and then round in the usual way. Thus the sequence of approximations to are
## Of course this procedure can be applied just as easily to a recurring decimal, or to an
overlong terminating decimal. This method of approximating real numbers by terminating
decimals of whatever length is required is one reason why decimal expansions are so
useful in science and everywhere that mathematics is applied.
Arithmetic with real numbers using approximations
Having defined real numbers as points on the number line, how are we to define addition,
subtraction, multiplication and division of real numbers? The most obvious approach
is to work with their decimal expansions, and add, subtract, multiply and divide suitable
truncations of these expansions. For example, suppose we want to add, subtract, multiply
and divide and 2 correct to two decimal places.
We obtain correct to two decimal places
## + 2 4.56, 2 1.73, 2 4.44, 2 2.22,
For people such as calculator and computer programmers who are concerned with the
accuracy of such calculations, there are serious problems here about the number of
decimal places necessary in the calculations and the sizes of possible errors, but these
things need not concern us here.
It is, however, worth commenting on the infinitely recurring 9s that result from attempting
non- truncated calculations such as
0. 3 3 3 ...
0. 3 3 3 ...
0. 6 6 6 ... +
0. 9 9 9 ...
0. 9 9 9 ...
## We know that both answers are 1, because
1
3
2
3
= 1 and
1
3
3=1
In the module, Decimals and Percentages we defined 0.9 = 1. To justify this, we remarked
that if 0.9represents a number x, then 1 x is a non-negative number less than every
positive number, so 1 x = 0. Similarly, any decimal expansion with an infinite string of 9s
can always be replaced by a terminating decimal.
## Arithmetic with real numbers using geometry
Since real numbers have been defined geometrically, we should be able to describe
the operations of arithmetic with real numbers using geometry alone. Addition and
subtraction are straightforward. We can construct the sum a + b of two real numbers
a and b in the usual way. We draw a straight line and on it mark off with a compass the
distances OA = a and AB = b. Then OB = a + b. Similarly for a b we mark of OA = a
and AB = b but this time with AB in the opposite direction to OA. Then OB = a b.
The opposite of a real number can be constructed as its reflection in the origin (using
compasses). Multiplication and division require similarity. We have taken the similarity tests
as axioms of our geometry, and we now can construct multiplication and division of real
numbers on the number line in terms of the ratios and products of lengths introduced
in the theory of similarity. The following exercise shows how to construct the product
1
ab, the quotient a b and the reciprocal a on a number line, where a and b are positive
real numbers. Reflection in the origin then extends these constructions to negative real
numbers.
EXERCISE 3
Let a and b be positive real numbers on the number line as shown, where the points
O, I, A and B represent the numbers 0, 1, a and b respectively.
Construct any other line m through O. Then use compasses to make m into a number
line, by constructing points I and B on m such that
OI = OI = 1 and OB = OB = b.
## Join the interval IA, and construct lines parallel to IA:
B
through B, meeting at P,
through B, meeting m at Q,
through I, meeting m at R.
l
O
## Using similar triangles, show that
OP = ab, OQ =
b
a
and OR =
A
a
B
b
1
a
Thus P represents the product a b, Q represents the quotient b a, and R represents the
1
reciprocal a .
In an earlier exercise you proved that between any two rational numbers there is a rational
number. The following exercise shows that between any two rational numbers there is an
irrational number.
{13}
{14}
## A guide for teachers
EXERCISE 4
Suppose that a and b are any two rational numbers, with a < b.
Let x = a +
1
2 (b
a).
## a Prove that x is irrational.
b Show that a < x < b.
Thus there is an irrational number lying between the two rational numbers a and b.
Show that there are infinitely many irrational numbers lying between two rational
numbers.
It is also possible to show that between any two irrationals there is a rational number.
This is harder and is left to the Links Forward section.
The Real Numbers and Algebra
Numbers other than rational number frequently occur in problems involving
measurement, areas and volumes, where the number often makes an appearance.
Quadratic surds also arise quite naturally when we find heights and sides of triangles.
A quadratic surd involves the square root of a non-square whole number. Irrational
numbers also arise in trigonometry since the sine, cosine and tangent ratios of most
angles are irrational. We usually approximate these numbers using a calculator. Certain
special angles have quadratic surds as their sine, cosine or tangent ratios.
For example, cos 30 =
3
2,
sin 45 =
1
2.
Irrational numbers also arise when we solve equations of degree greater than one. Thus,
3
the real solution of x3 = 5 is x = 5 .
Quadratic surds often arise when we solve quadratic equations using either the method of
completing the square or the quadratic formula.
EXAMPLE
Solve x2 3x 7 = 0.
SOLUTION
x=
3 9 + 28
2
b b2 4ac
,
2a
with a = 1, b = 3, c = 7 gives
3 + 37 3 37
, 2 .
2
## The Improving Mathematics Education in Schools (TIMES) Project
When performing calculations, it is best to leave real numbers in exact form, at least until
the end of a problem. We can then, if required, find approximate values. Converting real
numbers to decimal approximations can lead to cumulative rounding errors.
SEQUENCES
We can approximate all irrational numbers by rational numbers. This is often done by
means of sequences.
Define the sequence a1, a2, by a1 = 1 and for n > 1 = an + 1 =
1
2
an +
2
an
## Substituting in successive values, we obtain
a1 =1 , a2 = 1.5, a3 =1.416666 , a4 = 1.4142156, and so on.
These numbers appear to be getting closer to the decimal 1.41421 which is the
beginning of the decimal expansion for 2. If we assume that this sequence converges
to some real number that is, we assume that the terms of the sequence approach as
grows without bound. As n gets bigger and an + 1 gets closer to an a good idea of what
should be can be found by replacing an and an + 1 in the equation above by and we have
1
2
## which simplifies to, 2 = 2. So = 2 since is positive. Thus we have a sequence of
rational numbers which converge to the real number 2.
Approximations to many real numbers can be obtained in this way.
EXERCISE 5
Define the sequence a1, a2, by a1 = 1 and for n > 1, an + 1 =
1
2
an +
3
an
Find the first five terms of this sequence and, assuming that the sequence converges
show that it converges to 3.
Rationals between irrationals
We have seen that
between any two rational numbers there is a rational number
between any two rational numbers there is an irrational number.
It is also possible to show that
{15}
{16}
## between any two irrational numbers there is a rational number
between any two irrational numbers there is an irrational number.
Here is a proof of the third dot point.
Proof
Suppose we take two irrational numbers, which we may as well assume are positive,
where < .
We imagine that these two numbers are very close to each other, (although the proof
works the same if they are not), so that the number = is small.
n( ) = n
n
Now no matter how small is, we can always multiply it by a whole number n to make
the product greater than 2.
(For example, if = 0.00003145..., then we take n = 60 000 say, then n 2.07 > 2.)
Now look at the numbers na, nb. The distance between these numbers is
nb na = n( ) = n > 2.
Since there is a gap of at least 2 between these numbers, then there is at least one
whole number m, between them. That is,
na < m < n.
Dividing by n we have,
a<m
n < .
Since m and n are whole numbers, we have constructed a rational number between
the two given irrational numbers.
EXERCISE 6
Adapt the above proof to prove There is an irrational number between any two irrational
numbers
EXERCISE 7
Use the method in the proof (and your calculator) to find a rational number between 10
and .
## The Improving Mathematics Education in Schools (TIMES) Project
CONTINUED FRACTIONS
A continued fraction is an expression with either a finite or infinite number of steps
such as 2 + 3 +
1
1
5+ 1
## If we truncate the continued fraction somewhere, it can be simplified to produce a
rational number. For example,
2+ 3+
=2+
1
5+
1
7
1
3+
1
36
7
=2+ 3+
7
36
=2+
36
115
= 266
.
115
The continued fraction given here is usually represented using the notation [2; 3, 5, 7].
It is easy to see that a rational number has a finite continued fraction. Therefore continued
fractions that continue indefinitely represent irrational numbers. Continued fractions
that repeat, such as [1; 2, 2, 2, 2, ...], which we write as [1; 2], represent quadratic surds,
that is, surds of the form a + b, with a, b rational and conversely, every quadratic surd is
represented by an (eventually) repeating continued fraction.
EXERCISE 8
Expand as rational number the first five terms of the continued fraction for [1; 2] (the first
term is just 1), and find their squares as decimals, correct to six decimal places.
Thus the continued fraction [1; 2, 2, 2, 2, ...] appears to represent the number 2.
To show this, we write x = [1; 2, 2, 2, 2, ...] = 1 +
Thus, from x = 1 +
1
1+x
1
2+
1
2 + ...
=1+
1
.
2 + (x 1)
## we obtain x + x2 = 1 + x + 1 which gives x2 = 2.
Since x > 0, x = 2.
We can use the calculator to discover the continued fractions for other irrational
numbers. For example, to find the continued fraction for 3, we enter this into the
calculator, copy down the whole number part, 1, subtract it and push the reciprocal key
( x1 button) giving 1.3660. Again copy the whole number part, 1, subtract it and push
the reciprocal button giving 2.73. Copy 2 and repeat, giving 1.3660 (again). Hence
(assuming the pattern continues), we have the continued fraction 3 = [1; 1, 2, 1, 2, ...].
EXERCISE 9
Using the method outlined above for 2 show that the repeating continued fraction [1; 1,
2, 1, 2, ...] does indeed equal to 3.
Beyond quadratic surds, we do not know a lot about continued fractions. They are rather
mysterious and there are many unsolved problems related to them. The continued
fractions for the numbers and e appear to have very different forms
{17}
{18}
## with no apparent pattern here, while
e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8...]
where the obvious pattern continues forever.
The continued fraction for the golden ratio =
is [1; 1, 1, 1, 1, 1].
1+ 5
2
## Continued fractions give good rational approximations to irrational numbers.
EXERCISE 10
1
The first two terms of the continued fraction for give the usual rational approximation 3 7
Use the first 3 terms and then the first four terms to find better rational approximations to .
COMPLEX NUMBERS
We saw that real numbers arise algebraically when we try to solve equations such as x3 =
5. The real numbers alone are, however, not sufficient to solve all polynomial equations.
Since the square of any real number is positive, it is not possible to solve x2 + 1 = 0 , using
real numbers. To solve this equation, we need to introduce a new kind of number, which
is called i, and has the property that i 2 = 1. We can then construct the set of all complex
numbers. This consists of the set of all numbers of the form a + ib where a and b are real
numbers. It turns out that all polynomial equations with real or complex coefficients can
be solved using the complex numbers.
HISTORY
The ancient Greeks were probably the first to have discovered the existence of irrational
numbers. The led them to realize that many of their geometric proofs were no longer
valid. Here is an example of this.
Consider the simple problem of proving that two triangles with the same height have their
areas in the same ratio as their bases.
A
|ABC|
|BC|
|DE| .
## The Improving Mathematics Education in Schools (TIMES) Project
Here is the Pythagorean proof of the given theorem. It assumes that all lengths
are rational.
Suppose u is a unit of length going n times into BC and m times into DE. Mark off the
points of division on BC and DE.
A
Then triangles BCA and DEA are divided into n and m smaller triangles respectively, all
having the same area. Hence
|ABC|
m
n
|BC|
.
|DE|
## The objection to this proof is that BC and DE could be
the construction breaks down.
|BC|
|DE| could
## be irrational. In this case
Eudoxus developed a new definition and new approaches for comparing rational and
irrational magnitudes.
## THE NUMBERS AND e
The Arabic mathematician Muhammad ibn Musa al-Khwarizmi (9th century), who wrote
the first major treatise on algebra, believed that was irrational but had no proof.
In 1768, over 800 years later, Johann Lambert gave a proof that is irrational. His proof
is clever and not too difficult, but it uses ideas from calculus. A modified proof, which
is accessible to very good students in their final year of high school level mathematics,
is outlined in Question 8 of the NSW Higher School Certificate Extension 2 Paper from
2003, at the following web page: http://www.boardofstudies.nsw.edu.au/hsc_exams/
hsc2003exams/pdf_doc/mathematics_ext2_03.pdf
The number e, which arises in Calculus and elsewhere, is also irrational. Its approximate
value is 2.718281828 correct to 9 decimal places. (Despite the apparent pattern, this is not
a repeating decimal.) The proof that e is irrational is generally attributed to Euler. Once
again, a proof which is accessible to very good students in their final year of high school,
is outlined in Question 8 of the NSW Higher School Certificate Extension 2 Paper from
2001, at the following web page: http://www.boardofstudies.nsw.edu.au/hsc_exams/
hsc2001exams/pdf_doc/mathemat_ext2_01.pdf
The late eighteenth and early nineteenth century saw the beginnings of what we call
analysis which examines the theoretical underpinnings of calculus. The mathematicians
{19}
{20}
## A guide for teachers
Cauchy, Weierstrass and Dedekind were the founders of analysis. In 1872 Richard
Dedekind published a paper in which he showed how to construct the real number
system using a procedure known nowadays as a Dedekind cut. This is a way of
constructing a real number by dividing the rational line into two sets. For example, the
number 2 can be identified with the sets (cut) x rational and x2 < 2 or x 0 and the set x
rational and x2 2 and x > 0. Using these cuts Dedekind was able prove all of the basic
properties of the real numbers. The other standard method of defining real numbers is via
sequences, called Cauchy sequences.
Finally we mention the mysterious number , called Eulers constant, which is defined by
the following limit. It first appeared in a 1735 paper of Euler.
= lim 1 +
n
1
2
1
3
+ ... +
1
n
log (n) .
## It is not known whether is rational or irrational. It is believed to be a very difficult
question to answer, although it is expected that is irrational.
## APPENDIX 1 CONSTRUCTING THE
RATIONAL NUMBERS
While we generally accept the set of natural numbers N = {0, 1, 2, 3, ...} as given, it is
possible to formally construct them using the Peano axioms. One version of these axioms
are listed as follows:
1 Zero belongs to N.
2 If belongs to N, then the successor of a belongs to N. (Intuitively, we think of the
successor of a to be a + 1.)
3 Zero is not the successor of a number in N.
4 Two numbers in N whose successors are equal are themselves equal.
5 If a set S of numbers contains zero, and also contains the successor of every number
in S, then S contains N. (This is the principle of induction.)
The natural numbers can be constructed from these axioms. Then, the operation of
the addition can be defined and shown to have all the standard properties, such as
commutativity, and associativity. Similarly the other operations can be defined in turn.
In the appendix to the module, The Integers, we showed how to construct the integers
from the whole numbers using ordered pairs and equivalence classes. Here also, addition
and multiplication were defined and all the usual properties developed.
## CONSTRUCTION OF THE RATIONALS
In the module on Fractions, we developed the rational numbers intuitively in a way that
would be appropriate for classroom use.
## The Improving Mathematics Education in Schools (TIMES) Project
In this appendix, we present a more formal construction for the rationals using ordered
pairs and the idea of an equivalence class. This is very similar to the construction of the
integers we undertook in the appendix to the module on the Integers.
While we hope that teachers find this an interesting approach, the material in this
appendix is not meant for the classroom.
ORDERED PAIRS
The starting point is to take the set of ordered pairs (a, b) of integers, with b 0.
a
Intuitively, we will think of this ordered pair as representing the rational number b . Thus (7,
7
4) will be thought of as representing the rational number 4 , while (6, 2) will represent the
6
number 2 = 3, and (4, 6) will represent the number 2.
3
You will realize that (2, 6) and (1, 3) both represent the rational number 3 . Hence we
will say that two ordered pairs (a, b) and (c, d) are equivalent if and only if ad = bc. (This
a
c
corresponds to b = d but the definition only refer to integers.).
We need to find a way of defining the addition of two ordered pairs that will model the
a
c
rule for addition of fractions. We want to say b + d = bd , by only referring to the
integers and so we define the addition of two ordered pairs as follows:
1
3
5
2
17
6.
## Find (1, 5) + (13, 6) using the above definition.
Notice that since we are using ordered pairs of integers, all the usual rules of arithmetic,
(the commutative, associative and distributive rules) for integers automatically hold. So we
can see, for example, that the commutative law holds for addition of ordered pairs, since
(a, b) + (c, d) = (ad + bc, bd) = (cb + da, db) = (c, d) + (a, b).
EXERCISE 11
Check, by expanding out [(a, b) + (c, d)] + (e, f), that the associative law also holds for the
MULTIPLICATION
The multiplication rule is easier. We want to obtain
of the ordered pairs by
a
b
c
d
ac
1
3
5
2
5
6.
{21}
{22}
## A guide for teachers
EXERCISE 12
Find the sum and product of the following ordered pairs and interpret these in terms of
rational numbers.
a (8, 3), (7, 3)
## c (5, 11), (2, 4)
EXERCISE 13
Check that the commutative law for multiplication holds.
EXERCISE 14
Check, by expanding out [(a, b).(c, d)].(e, f), that the associative law for multiplication of
ordered pairs holds.
EXERCISE 15
Check, by expanding out (a, b).[(c, d)] + (e, f)], that the distributive law holds.
We again stress that all these results are proven assuming only the rules of arithmetic for
the integers.
The alert reader will realize that we have skipped over a rather important question. Since
various ordered pairs are equivalent, how do we know that the addition and multiplication
of equivalent ordered pairs will always produce equivalent ordered pairs? For example,
## (3, 6) + (6, 9) = (63, 54).
Now (3,6) is equivalent to (1, 2) and (6, 9) is equivalent to (2, 3), and
Similarly,
## (3, 6).(6, 9) = (18, 54)
and
(1, 2).(2, 3) = (2, 6), but (18, 54) is equivalent to (2, 6).
In mathematical language, we say that the definitions of addition and multiplication are
well-defined.
EXERCISE 16
Suppose (a1, b1) is equivalent to (a2, b2) and (c1, d1) is equivalent to (c2, d2). Show that
(a1, b1) + (c1, d1) is equivalent to (a2, b2 ) + (c2, d2). This shows that addition is well-defined.
Try doing the same for multiplication.
GRAPHICAL REPRESENTATION
We saw earlier how to tell algebraically when two ordered pairs are equivalent. We can
now view this graphically. For example, the ordered pairs
(2, 4), (1, 2), (1, 2), (2, 4), (3, 6), (4, 8), (5, 10) ..
are all equivalent, and in the diagram we see that they all lie on the dotted line shown.
y = 2x
y
8
(4, 8)
(3, 6)
(2, 4)
(1, 2)
0
1
4 3 2 1
(1, 2)
(2, 4)
(3, 6)
4
6
The set of all ordered pairs that are equivalent to each other is called an equivalence
class. Thus {.(2, 4), (1, 2), (1, 2), (2, 4), (3, 6), (4, 8), (5, 10) ..} is an equivalence class.
The equivalence class can be denoted by [(1, 2)]
EXERCISE 17
Write down the equivalence class for the ordered pair (3, 5).
We can choose from each equivalence class, any one ordered pair as a representative for
that class. Hence (1, 2) (or indeed (4, 8)) is a representative of the equivalence class [(1, 2)].
{24}
## CONSTRUCTING THE RATIONALS
We now see how to recover the rational numbers from these equivalence classes. We will
identify these equivalence classes with the rational numbers as follows:
[(1, 2)]
1
2,
[(1, 3)]
1
3,
[(2, 3)]
2
3,
[(3, 4)]
3
4,
[(1, 4)]
[(4, 5)]
1
4,
[(1, 5)]
4
5,
1
5,
[(1, 6)]
1
6,
and so on.
## Thus we identify an equivalence class of ordered pairs with a rational number.
We can similarly construct negative rationals by:
[(1, 2)]
2 , [(1, 3)]
3 , [(1, 4)]
4 , and so on.
EXERCISE 18
Show [(a, b)] = [(a, b)].
Ordering of the Rationals
We conclude this appendix by showing how the ordered pairs can be used to define the
ordering relations < and > on the rationals.
We will say that [(a, b)] < [(c, d)] if ad < bc.
For example, [(2, 3)] < [(4, 5)] since 10 < 12.
EXERCISE 19
By interpreting [(a, b)] as
a
b,
## explain why the above definition makes sense.
EXERCISE 20
Show that the relation < is well-defined.
Similarly, we say that [(a, b)] > [(c, d)] if ad > bc.
For example, [(3, 7)] > [(2, 9)], since 27 > 14.
Henceforth in this appendix we will only refer to a representative ordered pair from an
equivalence class.
You will recall the rule from algebra, that if x < y then x > y. We can show this holds for
rationals using the ordered pair construction.
Now
(a, b) = (a, b)
> (c, d)
= (c, d)
So
## (a, b) > (c, d).
EXERCISE 21
Arrange the following ordered pairs from smallest to largest using the definition
of < above.
(3, 4), (9, 5), (2, 7), (3, 6), (12, 2).
## APPENDIX 2 THE INFINITIES OF THE RATIONAL NUMBERS
AND THE REAL NUMBERS
When discussing the way that the rational numbers and the real numbers are arranged on
the number line, we promised to make precise mathematical sense of the rather vague
statement, Even though there are infinitely many rational numbers and infinitely many
irrational numbers between 0 and 1, there are vastly more real numbers in that interval
than rational numbers.
The following theory about the sizes of infinite sets was developed by the German (and
Russian- born) mathematician Georg Cantor in the late nineteenth century. Cantors
theory became notorious, and was the occasion of many vicious personal attacks. It was
one of the few pieces of pure mathematics ever to be attacked by the Catholic Church
on the grounds that only theologians should be discussing infinity and Kronecker,
another famous German mathematician, attacked Cantor as a scientific charlatan and a
corrupter of youth.
We shall call two sets equivalent if there is a one-to-one correspondence between them.
If there is a one-to-one correspondence between two finite sets, then they have the same
number of elements, and conversely if two finite sets have the same number of elements,
then they can be put into one-to-one correspondence. Thus for finite sets, equivalent
simple means have the same number of elements.
For infinite sets, however, things quickly become counter-intuitive, because a set can be
equivalent to a proper subset of itself. Here are some examples:
The set of whole numbers is equivalent to the set of even whole numbers. This is easily
proven, even though the conclusion is so strange:
0
6
...
10
12
{25}
{26}
## A guide for teachers
he important point of this proof is that every whole number occurs exactly once in the
T
top line, and every even whole number occurs exactly once in the bottom line.
The set of whole numbers is equivalent to the set of integers. Again, this is easily proven:
0
6
...
The important point is the same as before every whole number occurs exactly once in
the top line, and every integer occurs exactly once in the bottom line.
The set of whole numbers is equivalent to the set of rational numbers.
This is quite a surprising result, because there are infinitely many rational numbers just
between 0 and 1. To prove this, it is necessary to show how to write the rational numbers
down in a list so that they can subsequently be paired with the whole numbers in order:
0
...
1
2
2
2
3
2
4
2
5
2
6
2
7
2
8
2
...
1
3
2
3
3
3
4
3
5
3
6
3
7
3
8
3
...
1
4
2
4
3
4
4
4
5
4
6
4
7
4
8
4
...
1
5
2
5
3
5
4
5
5
5
6
5
7
5
8
5
...
1
6
2
6
3
6
4
6
5
6
6
6
7
6
8
6
...
1
7
2
7
3
7
4
7
5
7
6
7
7
7
8
7
...
1
8
2
8
3
8
4
8
5
8
6
8
7
8
8
8
...
...
...
...
...
...
...
...
...
In the diagram above, every positive rational number has been written down exactly once,
because the fractions that cancel have been crossed out. We can now write down all the
rational numbers in a list, proceeding in the direction of the arrows, and interleaving the
negative rational numbers. The list begins:
0
10
11
12
13
14
...
1
2
21
1
3
31
1
4
41
2
3
23
## The Improving Mathematics Education in Schools (TIMES) Project
{27}
As always, the important point is that every whole number occurs exactly once in the top
line, and every rational number occurs exactly once in the bottom line.
We have now proven that the sets of even whole numbers, of whole numbers, of integers,
and of rational numbers are all equivalent, even though each is embedded as a proper
subset of the succeeding set. Once these pieces of set-theory trickery have been done, the
definition of equivalent sets may seem to be vacuous, because it now looks as if there could
always be some similar trick to demonstrate that any two infinite sets are equivalent.
Thus the next result and its proof is usually a complete surprise.
Theorem
The set of real numbers is not equivalent to the set of whole numbers.
Proof
We prove this result by contradiction. Suppose, then, that the set of real numbers
is equivalent to the set of whole numbers. Then there would be a one-to-one
correspondence between the set of whole numbers and the set of real numbers,
and we could arrange all the real numbers in a list. If we represent each real number by
its decimal expansion (excluding repeating 9s), the list would look something like this:
3.641559072 ...
167.115902726 ...
28.256115294 ...
0.008711369 ...
1.125266991 ...
900.050411818 ...
478.515242229 ...
529.257188231 ...
55.000001251 ...
23.152556863 ...
We shall now produce a contradiction by finding a real number that is not on the list
on the right. Define by defining the nth digit an in its decimal expansion as follows:
If the nth digit of the real number paired with n is 1, let an = 2. Otherwise let an = 1.
We have underlined the nth digit of each number in the list, so given our list, the
decimal expansion of would begin = 1.211122112.
{28}
## A guide for teachers
The real number is a well-defined real number, but it is not on the list, because it
differs at its nth digit from the real number paired with n. This proves that the pairing is
not a one-to-one correspondence at all, giving the required contradiction. Hence the
set of real numbers is not equivalent to the set of whole numbers.
We are now justified in saying, rather more loosely, that the infinity of real numbers is
vastly greater than the infinity of rational numbers. As we remarked before, one should see
the real number line as a continuum, with the points joined up to make a line, whereas the
rational numbers are like disconnected specks of dust scattered along it.
## ALGEBRAIC AND TRANSCENDENTAL NUMBERS
If a real number satisfies an equation P(x) = 0 where P(x) is a polynomial with integer
coefficients we say that is an algebraic number. If a real number satisfies no such
equation it is said to be a transcendental number. Clearly every rational number is
3
algebraic. Other examples of algebraic numbers are 2 and 7 . The transcendental
numbers include and e.
It can be shown that as the algebraic numbers are countable. We have seen that this is
not true for the real numbers. Hence there are many more transcendental numbers than
algebraic numbers. See the book Numbers: Rational and Irrational, given in the references
REFERENCES
What is Mathematics, Richard Courant and Herbert Robbins, Oxford University Press, (1941)
Numbers: Rational and Irrational, Ivan Niven, New Mathematical Library, The Mathematical
Association of America, (1961)
Continued fractions: C D Olds, New Mathematical Library, The Mathematical Association
of America, (1963)
EXERCISE 1
0.0322580645161209.
EXERCISE 2
a
1 a
2 b
a
b
2bd
## The Improving Mathematics Education in Schools (TIMES) Project
EXERCISE 4
a x = a +
1
2
(b a)
Assume x is rational
Then
Therefore x is irrational
ba
xa
## = 2 is rational which is a contradiction
1
2 (b
b x a =
a) since b > a.
Therefore, x > a.
b x = b (a +
= (b a)
1
2
(b a))
21
2
>0
## Therefore a < x < b.
EXERCISE 5
a1 = 1, a2 = 2, a3 = 1.25, a4 = 1.73214285, a5 = 1.73205081
a=
1
2
a+
3
a
Solving for a
2a = a +
3
a
2a2 = a2 + 3
a2 = 3
a= 3
EXERCISE 7
3.15
EXERCISE 8
t1 = 1, t12 = 1
3
t2 = 2 , t12 = 1.25
7
t3 = 5 , t12 = 1.96
17
t5 =
41
2
29 , t5
= 1.9988
{29}
{30}
## A guide for teachers
EXERCISE 9
[1;1,2,1,2,]
x = 1+
=1+
1
1+
1
2 + (x 1)
x+1
x+2
x2 = 3
x= 3
EXERCISE 10
= [3; 7, 15, 1, 292, 1
First five terms
3,
## 22 333 365 103993
7 , 106 , 113 , 33102
EXERCISE 11
[(a, b) + (c, d)] + (e, f) = (ad + bc, bd) + (e, f)
= (fad + fbc + ebd, bdf)
= (b(fc + ed) + adf, bdf)
= (a, b) + (cf + de, df)
= (a, b) + [(c, d) + (e, f)]
The associative law for addition holds.
EXERCISE 12
a (1,3),
1
3
; (56, 9),
b (55, 18),
55
18
c (42, 44),
21
22 ;
56
9
; (28, 18),
(10, 44),
14
9
5
22
EXERCISE 13
(a, b)(c, d) = (ac, bd) = (ca, db) = (c, d)(a, b)
## The Improving Mathematics Education in Schools (TIMES) Project
EXERCISE 14
[(a, b)(c, d)](e, f) = (ac, bd)(e, f)
= ((ac)e, (bd)f)
= (a(ce), b(df))
= (a, b)(ce, df)
= (a, b)[(c, d)(e, f)]
EXAMPLE 15
(a, b)[(c, d) + (e, f)] = (a, b)(cf + de, df)
= (acbf + bdae, b2fd)
= (ac, bd) + (ae, bf)
= (a, b)(c),d) + (a,b)(e, f)
(using equivalence)
EXAMPLE 16
(a1, b1) is equivalent to (a2, b2)
Hence a1 b2 = a2 b1
(c1, d1) is equivalent to (c2, b2)
Hence c1d2 = c2 d1
(a1, b1) + (c1, d1) = (a1d1 + b1c1, b1d1)
= (a1d1c2b2 + b1c1 c2b2, b1d1b2d2)
=(a2b1d1d2 +b2b1d1c2, b1d1b2d2)
=( a2d2 + b2c2, b2d2)
= (a2, b2) + (c2, d2)
For multiplication
(a1, b1) (c1, d1) = (a1c1, b1d1)
(a2, b2) (c2, d2) = (a2c2, b2d2)
Also, a1c1 b2d2 = b1a2d1c1
Multiplication is well defined.
{31}
{32}
## A guide for teachers
EXAMPLE 17
{(3, 5), (6,10), (9, 15), (12, 20)}
EXAMPLE 18
[(a, b)] + [(a. b)] = [(ab ab, b2)] = [(0, 1)]
Therefore the additive inverse of [(a, b)] is [(a. b)]
Uniqueness can be shown.
EXAMPLE 19
a
b
<
c
d
## if and only if ad < bc
EXAMPLE 20
If (a1, b1) < (c1, d1) if and only if a1d1 < b1c1
(a1, b1) is equivalent to (a2, b2)
Hence a1 b2 = a2 b1
(c1, d1) is equivalent to (c2, b2)
Hence c1 d2 = c2 d1
a1 d1 < b1 c1
a1 d1 d2 < b1 c1 d2
a1 d1 d2 < b1 c2 d1
a1 d2 < b1 c2
a2 d2 b1 < b1 c2 b2
a2 d2 < c2 b2
{33}
## The aim of the International Centre of Excellence for
Education in Mathematics (ICE-EM) is to strengthen
education in the mathematical sciences at all levelsfrom school to advanced research and contemporary
applications in industry and commerce.
ICE-EM is the education division of the Australian
Mathematical Sciences Institute, a consortium of
27 university mathematics departments, CSIRO
Mathematical and Information Sciences, the Australian
Bureau of Statistics, the Australian Mathematical Society
and the Australian Mathematics Trust.
## The ICE-EM modules are part of The Improving
Mathematics Education in Schools (TIMES) Project.
The modules are organised under the strand
titles of the Australian Curriculum:
## Number and Algebra
Measurement and Geometry
Statistics and Probability
The modules are written for teachers. Each module
contains a discussion of a component of the
mathematics curriculum up to the end of Year 10.
www.amsi.org.au | 12,679 | 47,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-35 | latest | en | 0.882154 |
https://theunconditionalguru.in/tag/there/ | 1,653,329,402,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00441.warc.gz | 628,777,930 | 20,044 | # Tag: there
## How many lone pairs of electrons are on the S atom in sulphur tetra floride (SF4) ?
In case of SF4 there are 6 electrons in the outermost shell of Sulphur, out of these 6 electrons, 4 are utilized in bonding with fluorine and thus two unshared electrons are there. Which leads to the lone pair as shown below. Contribute to make this site more informative.
## How many atoms are there in 1 gram of carbon?
For calculation of number of atoms we have to calculate the number of moles present in 1gram carbon. Multiply the number of moles with Avogadro number. Step I Number of moles in 1 gram of carbon = given mass / molar mass Molar mass (molecular or atomic mass expressed in grams) = 12g So the | 184 | 718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-21 | longest | en | 0.888393 |
https://rdrr.io/cran/mbbefd/man/distr-genbeta.html | 1,506,313,840,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690318.83/warc/CC-MAIN-20170925040422-20170925060422-00146.warc.gz | 710,596,677 | 15,057 | # distr-genbeta: The generalized Beta of the first kind Distribution In mbbefd: Maxwell Boltzmann Bose Einstein Fermi Dirac Distribution and Destruction Rate Modelling
## Description
Density, distribution function, quantile function and random generation for the GB1 distribution with parameters `shape0`, `shape1` and `shape2`.
## Usage
```1 2 3 4 5 6``` ```dgbeta(x, shape0, shape1, shape2, log = FALSE) pgbeta(q, shape0, shape1, shape2, lower.tail = TRUE, log.p = FALSE) qgbeta(p, shape0, shape1, shape2, lower.tail = TRUE, log.p = FALSE) rgbeta(n, shape0, shape1, shape2) ecgbeta(x, shape0, shape1, shape2) mgbeta(order, shape0, shape1, shape2) ```
## Arguments
`x, q` vector of quantiles. `p` vector of probabilities. `n` number of observations. If `length(n) > 1`, the length is taken to be the number required. `shape0, shape1, shape2` positive parameters of the GB1 distribution. `log, log.p` logical; if TRUE, probabilities p are given as log(p). `lower.tail` logical; if TRUE (default), probabilities are P[X ≤ x], otherwise, P[X > x]. `order` order of the raw moment.
## Details
The GB1 distribution with parameters `shape0` = g, `shape1` = a and `shape2` = b has density
Γ(a+b)/(Γ(a)Γ(b))x^(a/g-1)(1-x^{1/g})^(b-1)/g
for a,b,g > 0 and 0 ≤ x ≤ 1 where the boundary values at x=0 or x=1 are defined as by continuity (as limits).
## Value
`dgbeta` gives the density, `pgbeta` the distribution function, `qgbeta` the quantile function, and `rgbeta` generates random deviates.
## References
Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole.
Abramowitz, M. and Stegun, I. A. (1972) Handbook of Mathematical Functions. New York: Dover. Chapter 6: Gamma and Related Functions.
Johnson, N. L., Kotz, S. and Balakrishnan, N. (1995) Continuous Univariate Distributions, volume 2, especially chapter 25. Wiley, New York.
`1` ```#TODO ``` | 579 | 1,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-39 | longest | en | 0.690748 |
http://www.instructables.com/id/Wedging-Table-for-Clay-Studio/ | 1,498,618,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322275.28/warc/CC-MAIN-20170628014207-20170628034207-00539.warc.gz | 570,862,957 | 17,515 | This is an instructable on how to make a free standing wedging table for your own studio. I am writing this instructable in hopes of claiming one of the awesome sets of glazes or even a rad T-shirt! This is my very first instructable!!! :)
Basic knowledge of tool safety should be used, if you don't know how to use the tool, please seek assistance.
This table is roughly 31 inches tall and 2 ft squared top. This was the largest size I felt comfortable making for my garage studio.
Materials Needed:
4x4 lumber: Total Length = 4 x height of table(-table top width-caster height if applicable).
30 inches tall - 1 inch (i forgot to do this,so im going to lie on my math)= 30 inches x 4=120 inches or 10 feet
2 x 6 lumber(or similar 2 x 4): Perimeter of Table Top ADD 6 SIDES. (you will need cross supports)
24 inches x 6 sides= 120 inches or 10 ft.
2 inch screws
1/2 inch plywood or better
*Canvas (*if you choose to cover it, some don't because of the issues with dust accumulation)
*Upholstery Nails
Other Tools:
Kreg --not necessary but easier to make joints -kinda
Drill
Wood Glue
Clamps
*Upholstery Nails
Table Saw
Miter Saw
Ruler
*Box Cutter
Pencil
Safety Gear
## Step 1: Set the Miter for the Table Top Frame, Glue, and Screw
Set your miter for the table top frame.
You can also make a frame however you like.
Cut the 2 x 6's to create a frame
I used a stop block to better measure my cuts.
The inside measurement is different from the outside measurement.
Set the Kreg tool to the appropriate depth depending on your wood, and use the correct guide screws.
Glue and screw the frame together. Use clamps if you have them!
## Step 2: Cut the Legs for the Table
Reset the miter to 0 degrees.
Cut the 4 x 4's to the height you determined.
## Step 3: Attach Legs to Frame
Screw the legs onto the frame.
It's tricky getting it level, don't give up!
## Step 4: Measure Inside Frame and Make Supports
Measure the distance between the frames at two different places.
Cut those using 2 x 6's or other wood
Kreg the supports and put into place
## Step 5: Cut the Table Top From Wood and Screw On
I had some beautiful birch plywood left over from another project, so I chose this wood. It would also be acceptable to have this wood as the final wood and not cover it in canvas.
I cut my wood to 2 ft x 2 ft. And I placed screws every 3 -4 inches. I wanted as little banging as possible when I slam the clay to wedge it. ( I didn't take a picture of this process.)
## Step 6: Wrap in Canvas and Hammer Down
You can optionally wrap your wedging table in canvas just make sure to pull it tight. Hammer it down using fabric tacks, then cut.
## Step 7: Yay! a Wedging Table I Don't Feel Like I'm Going to Break!
Finished project. I am going to add a second canvas top if i switch to red clay. I also plan on adding a longer vertical piece i can attach a wire cutter to so that I can slam wedge more efficiently.
Thanks for looking at my first Instructable. I hope you enjoyed it . If so, please vote for me in the Clay Contest.
Happy Muddin'
<p>Nice, solid, wedging table. Got my vote...</p>
Thank you so much!
Thank you, I can wedge and store things underneath. I wanted to be able to throw as much clay as I could carry and not feel like I was going to break it. | 839 | 3,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-26 | longest | en | 0.858572 |
https://mathoverflow.net/questions/13933/dropping-three-bodies | 1,624,161,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00219.warc.gz | 352,150,411 | 29,921 | # Dropping three bodies
Consider the usual three-body problem with Newtonian $1/r^2$ force between masses. Let the three masses start off at rest, and not collinear. Then they will become collinear a finite time later by a theorem I proved some time ago. (See the papers "Infinitely Many Syzygies" and "The zero angular momentum three-body problem: all but one solution has syzygies" available on my web site or the arXivs.) Let $t_c$ denote the first such time.
Write $r_{ij} (t)$ for the distance between mass $i$ and mass $j$ at time $t$.
Question 1. For general masses $m_i >0$, is it true that the "moment of inertia" $I = m_1 m_2 r_{12}^2 + m_2 m_3 r_{23}^2 + m_1 m_3 r_{13}^2$ monotonically decreases over the interval $(0, t_c)$?
Question 2. If the masses are all equal and if the initial side-lengths satsify $0 < r_{12}(0) < r_{23} (0)< r_{13} (0)$
is it true that these inequalities remain in force: $0 < r_{12} (t) < r_{23} (t) < r_{13}(t)$ for $0 < t < t_c$? In other words: if the triangle starts off as scalene (not isosceles, and having nonzero area) does it remain scalene up to collinearity?
Motivation: The space of collinear triangles, consisting of triangles of zero area, acts like a global Poincare section for the zero-angular momentum, negative energy three-body problem. To obtain some understanding of the return map from this space to itself the "brake orbits"-- those solutions for which all velocities vanish at some instant -- seem to play an organizing role. Answering either questions would yield useful information about brake orbits.
Aside: I suspect that if the answers to either question is yes for the standard $1/r^2$ force, then it is also yes for any attractive "power law" $1/r^a$ force between masses, any $a > 0$.
added, Sept 20, 2010. The bounty is for an answer to either question 1 or 2.
I've made partial progress toward 2 using variational methods (direct method of the calculus of variations). I can prove that if a syzygy is chosen anywhere in a neighborhood of binary collision (so $r_{12}(t_c) = \delta$, small, $r_{23} (t_) = r_{13}(t_c) + \delta$) then there exists a brake orbit solution arc ending in this syzygy and satisfying the inequality of question 2. The proof suggests, but does not prove, that the result holds locally near isosceles, meaning for brake initial conditions in a neighborhood of isosceles brake initial conditions ( so
$r_{13} (0) = r_{12} (0) + \epsilon$). If I had uniqueness [modulo rotation and reflection] of brake orbits with specified syzygy endpoints, then my proof would yield a proof of this local version of the alleged theorem.
Unfortunately, my proof does not exclude the possibility of more than one orbit ending in the chosen syzygy, one of which violates the inequality.
• I just wanted to chime in to say that this is perhaps the most "gangster" question title on MO. – Harrison Brown Feb 3 '10 at 6:40
• I think the title of the post might be a bit misleading since it could be taken to imply a fourth body. In other words, one could read that as dropping three bodies in a single gravitational field which is a different problem than the one you are proposing (which is a fascinating problem, by the way). – Ian Durham Feb 3 '10 at 12:32
• Can you answer these questions for infinitesimal perturbations of an isosceles configuration? If so, perhaps it would help to restrict the cases on the boundary of those which satisfy those conditions, e.g., ones which become isosceles or singular at the moment of collinearity. – Douglas Zare Feb 3 '10 at 16:18
• @Harrison: As long as Stringer isn't dropping bodies, he's not our problem. – Sheikraisinrollbank Sep 16 '10 at 10:56
• "they will become collinear a finite time later": You hardly need this, but I think this is a remarkable theorem! – Joseph O'Rourke Sep 20 '10 at 23:00
I believe your first proof is trivial since any 3 points may always be placed on a plane and since their gravity will attract along the plane they will remain on it in perpetuity. Then any 2 points say $a$ and $b$ can be placed on a line and their centre of gravity plotted on the line.
Since the centre of gravity of $a$ and $b$ is on this line, this line must accelerate across the plane towards $c$ until $c$ crosses the line. For $c$ never to reached $ab$ would require the entire system to be generating a state of its own perpetual acceleration opposing the direction from $c$ to its nearest point on $ab$, and creating a perpetual angular acceleration around the centre of gravity of $a$ and $b$; a contradiction since there is no fourth mass with respect to which the system $abc$ could be accelerating.
Moving on to the monotonic decrease in momentum, if we choose $c$ such that $c$ is the body which passes between the other two then holding the line $ab$ as a reference point, $a$ and $b$ must move towards each other along $ab$ so $I_{ab}$ monotonically decreases, and $c$ moves towards some point between $a$ and $b$ until it passes between them so again $I_{ca}+I_{cb}$ must decrease monotonically.
Re the final part, the triangles will remain scalene until $t_c$ but will not necessarily do so in perpetuity. This is because it will be the body which is closest to the other two which passes between them. If we draw a scalene triangle, exactly one of the vertices connects the shortest two sides. This body (say $a$) will move between the other two and since it is closer to both of the other two than they are to each other, it will approach both of them faster than they close the distance between each other.
Furthermore due to the inverse square law it will approach the body nearest to it faster than it approaches the other, and therefore there is no means by which the ordering of their distances can change before $a$ reaches the line between $b$ and $c$, which is what would be required to create an isosceles or equilateral triangle from a scalene one.
• Thanks Robert for the feedback. My first proof'' result (about existence of collinear instants) is known to be false when the total energy is positive: simply send each mass outward faster than its escape velocity'. But you do not use negativity of energy in your argument. Can you think of a way that negativity of energy might enter into your thinking on this situation? – Richard Montgomery May 16 '17 at 16:56
• @RichardMontgomery if I interpret your meaning correctly, one or more masses are moving outwards faster than escape velocity would negate my argument that "this line ($ab$) must accelerate across the plane towards $c$ until $c$ crosses the line. In stating that, I assumed the bodies start out stationary. Although they would still accelerate together they would do so having started out on divergent paths, and the relative acceleration of $ab$ towards $c$ would be asymptotic towards some permanently divergent path; insufficient to halt their divergence. Is that what you meant? – samerivertwice May 16 '17 at 17:12
• @RichardMontgomery re-reading your question and the matters you're touching upon, perhaps this is the answer you are looking for. If you were to define in some way the divergent paths which the bodies move asymptotically towards, I believe this would encode some of the energy information similarly to the ways the collinear triangles do for negative energy states. – samerivertwice May 16 '17 at 17:30
• I will have time to think about what you wrote more carefully next week. Your first proof' is heuristically good. The problem with turning it into a rigorous proof is that the line connecting any two of the bodies keeps moving. How do you know it is not moving `away' from the third body faster than the third body is approaching it? That is true, for at least one decomposition ab; c, but since by fixing a line you are working in a non-inertial reference plane, I do not think you've proved it , even when starting all three bodies at rest. – Richard Montgomery May 18 '17 at 3:54
• @RichardMontgomery A good point! That would require it to have attained escape velocity by drawing energy off the other two. I'm not an expert and you almost certainly know the answer to this, but I think it would have to encircle them both first to do that, also completing the proof. – samerivertwice May 18 '17 at 10:03 | 2,024 | 8,271 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-25 | longest | en | 0.898775 |
https://liteconvert.com/units/k%E2%84%A7kilomho-electrical-conductance/ | 1,696,425,115,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00155.warc.gz | 413,473,799 | 5,415 | Unit k℧ kilomho Electrical Conductance The unit k℧ (kilomho) in the field of electrical conductance is equivalent to one thousand mhos. Electrical conductance is a measure of a material's ability to allow the flow of electric current. A mho is the reciprocal of the ohm (1/ohm), which is the SI unit for electrical resistance. Therefore, one kilomho (1 k℧) is equal to 1,000 mhos or 1,000 siemens (S), where siemens is the standard unit for electrical conductance in the International System of Units (SI). In summary, 1 k℧ represents 1,000 times the conductance of a hypothetical ideal conductor with zero resistance. | 157 | 618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.847626 |
https://boyslife.org/hobbies-projects/funstuff/1483/optical-illusions/comment-page-22/?replytocom=29637 | 1,597,277,921,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00055.warc.gz | 234,435,625 | 19,533 | Your eyes can play tricks on you. Pictures that confuse your eyes and brain, tricking them into seeing something differently, are called optical illusions. See if you can figure out these optical illusions.
## WHICH OBJECT IS TALLER?
None: They’re all the same size.
## WHICH LINE IS LONGER?
Neither: They’re both the same.
## IS THIS GRAY HAZE SHRINKING?
Stare at the black dot. After a while the gray haze will appear to shrink.
## DO THESE COLORED LINES BEND?
No, they’re perfectly straight—but try telling your eyes that!
## ARE THESE WHEELS SPINNING?
Stare at the center. Now move your head back and forth toward and away from the page. The circles will appear to spin.
## HOW MANY BLACK DOTS CAN YOU COUNT?
Look closely and you will see them.
## DO YOU SEE TWO FACES OR A VASE?
If you see one, close your eyes for a moment, then look for the other.
## ARE THESE LINES STRAIGHT OR CROOKED?
Yep, you guessed it. The horizontal lines look crooked, but they are perfectly straight.
## ARE YOU SEEING RED, WHITE & BLUE?
Stare at the center of this flag for one minute. Then look at an empty white sheet and you’ll see a red, white and blue flag.
HOW THIS FLAG WORKS
Your eyes see color as measures of red or green, blue or yellow and bright or dark. When you look at a green object for a long time, your eyes get tired and start seeing red. When you look at yellow, after a while you’ll start to see blue. And darkness turns into brightness. The result: Even this wacky flag can be good, old red, white and blue—after a while!
#### 10 Comments on 9 Optical Illusions to Confuse Your Brain
1. THEY ARE CRAZY AWESOME
2. COOL
3. sndbfjirhsjkvbfjkd1 // April 7, 2008 at 11:36 am // Reply
got any more
I RULE
4. I noticed them all
5. The pictures that the people made were the best op art pictures i have ever seen!!!!!!!
6. I thought the flag & wheel ones were awsome!
7. the wheel didn’t work for me
8. those were all cool but I liked the flage one the best.
9. My eyes hurt now
10. These are SO cool! | 525 | 2,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-34 | latest | en | 0.869184 |
https://mathoverflow.net/questions/447924/bounding-minimal-absolute-value-of-a-point-on-a-complex-algebraic-variety | 1,696,166,810,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00221.warc.gz | 419,911,494 | 27,577 | # Bounding minimal absolute value of a point on a complex algebraic variety
Given a system of complex polynomial equations in $$n$$ variables, giving rise to an affine variety $$V \subseteq \mathbb C^n$$, is there a bound $$b \in \mathbb R$$ such that if $$V(\mathbb C) \neq \emptyset$$ then there is a point $$a \in V(\mathbb C)$$ such that $$||a|| < b$$?
I am looking for some bound $$b$$ in terms of $$n$$, the degrees of the polynomials and the absolute values of the coefficients. Or indeed is anything known in this sort of direction?
• A bound for any hypersurface follows from a bound for a one variable polynomial, which I think is not difficult. May 31 at 12:08
• Consider, for example, two almost parallel lines in $\mathbb{C}^2$. Clearly, it's impossible to bound the distance to the point of intersection in terms of absolute values of coefficients. May 31 at 13:18
• @OlegEroshkin In that case, the corresponding algebraic variety (in complex affine space) is empty. The OP does stipulate that the algebraic variety is nonempty. May 31 at 13:24
• @JasonStarr Wait, I wrote "almost parallel", like $az+bw=c$ and $az+(b+\epsilon)w=d$. The distance to the intersection point depends on how small is $\epsilon$ May 31 at 13:44
• @JasonStarr Already for several linear equations one needs to bound from below the lowest singular value of the matrix of coefficients. I don't know the analog for non-linear algebraic equations. May 31 at 13:55
No (for one interpretation of the question). There does not exist a continuous function $$b$$ from the set of systems of polynomial equations to $$\mathbb R^{>0} \cup \{\infty\}$$ that takes a finite value on all systems with a solution, and such that if there is a solution there exists one of norm at most the value of $$b$$.
One considers the vanishing locus of the polynomials $$xy, y(1-y), ty+tx+y-1$$. This has the solution $$(1/t,0)$$ for $$t\neq 0$$ and $$(0,1)$$ for $$t=0$$, and these are the only solutions.
So $$b$$ would have to go to $$\infty$$ as $$t\to 0$$ but then it would have to take the value $$\infty$$ at $$0$$ which contradicts the assumption.
• Of course there is a locally closed stratification of the Chow variety of projective space that "flattens" the intersection of the universal cycle with the hyperplane at infinity. Over each locally closed stratum of this flattening stratification, there is, indeed, a continuous bound $b$. May 31 at 15:32
• Thank you. My question is evidently too naive and general to be true. In the counterexample, you have a fixed variety $V$ given by the first two equations $xy=0$ and $y(1-y) = 0$, intersecting with a family of curves dependent on the parameter t. $V$ is not irreducible, and the discontinuity comes from that. If we replace $V$ by its component $y=0$, we get the solution $(1/t,0)$ and so taking $b=1/t$ works. I am actually interested in this situation where $V$ is fixed and positive dimensional, intersected with a varying family of varieties (but with unbounded degree, not an algebraic family). Jun 2 at 7:28
• @JonathanKirby I don't think your distinction is meaningful. We can take $V$ in $\mathbb A^4$ defined by the equations $xy=z, y(1-y)=w$ and then intersect with the family with equations $ty+tx+y-1 = w=z=0$. Then $V$ is irreducible, smooth, and otherwise nice and the varying family is a smooth family of affine subspaces. Jun 2 at 10:13 | 904 | 3,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-40 | latest | en | 0.906385 |
https://socratic.org/questions/how-do-you-simplify-5b-3-4-and-write-it-using-only-positive-exponents | 1,581,975,550,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00121.warc.gz | 587,482,897 | 6,158 | # How do you simplify (5b^-3)^4 and write it using only positive exponents?
Aug 30, 2016
$= \frac{625}{b} ^ 12$
#### Explanation:
${\left(5 {b}^{-} 3\right)}^{4}$
$= {\left(\frac{5}{b} ^ 3\right)}^{4}$
$= \frac{625}{b} ^ 12$
Aug 30, 2016
${5}^{4} / {b}^{12} = \frac{625}{b} ^ 12$
#### Explanation:
There are 3 laws of indices being applied here,
A power outside applies to all the factors inside: ${\left(x y\right)}^{m} = {x}^{m} {y}^{m}$
Raising to a power, multiply the indices: ${\left({x}^{m}\right)}^{n} = {x}^{m n}$
Negative index becomes positive in the denominator and v.v.
${x}^{-} m = \frac{1}{x} ^ m$ or $\frac{1}{x} ^ - n = {x}^{n}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
${\left(5 {b}^{-} 3\right)}^{4} = {5}^{4} {b}^{-} 12$
$= {5}^{4} / {b}^{12}$
This can also be written as $\frac{625}{b} ^ 12$ | 332 | 843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-10 | longest | en | 0.614266 |
https://resources.arcgis.com/en/help/arcobjects-net/componentHelp/002m/002m000003q2000000.htm | 1,620,718,000,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991904.6/warc/CC-MAIN-20210511060441-20210511090441-00228.warc.gz | 491,845,794 | 3,644 | ArcObjects Library Reference (Geometry)
# ISpatialReferenceResolution.ConstructFromHorizon Method
Defines the XY resolution and domain extent of this spatial reference based on the extent of its horizon. Low precision SRs will have minimum resolution of 1/10mm in current units.
```[Visual Basic .NET]
Public Sub ConstructFromHorizon ( _
)```
```[C#]
public void ConstructFromHorizon (
);```
```[C++]
HRESULT ConstructFromHorizon(
void
);```
#### Product Availability
Available with ArcGIS Engine, ArcGIS Desktop, and ArcGIS Server.
#### Remarks
ConstructFromHorizon defines a domain extent and resolution sufficient to cover the horizon of a given coordinate system for high precision spatial references. For low-precision spatial references the domain extent is centered on the horizon center. It does not define an extent for the Z- or M-domains; for these use methods available on ISpatialReference.
Projected Coordinate Systems
For a high precision ProjectedCoordinateSystem (PCS), ConstructFromHorizon defines the domain extent to be a square completely covering, and slightly larger than, the horizon extent of the PCS (which is an arbitrary rectangle). The scale factor (1/precision) is chosen to fit this domain.
For a low precision PCS, the center of the domain extent is aligned with the center of the horizon extent and expanded to achieve a target resolution of 1mm.
The example below lists high and low precision extents for WGS 1984 UTM Zone 11N.
Name Low Precision ProjectedCoordinateSystem High Precision ProjectedCoordinateSystem
MinX -573,741.824 -5,120,900.0
MinY -1,073,741.824 -9,998,100.0
MaxX 1,573,741.823 14,875,300.0
MaxY 1,073,741.823 9,998,100.0
Resolution 0.001 2.2 x 10^-9
Geographic Coordinate Systems
For a high precision GeographicCoordinateSystem (GCS), ConstructFromHorizon produces the square domain (-400, -400, 400, 400) (expressed in the units of the SR).
For a low precision GCS the upper right hand corner is adjusted to achieve a default resolution of 1/500 of an arc-second.
Name Low Precision GeographicCoordinateSystem High Precision GeographicCoordinateSystem
MinX -400 -400
MinY -400 -400
MaxX 793.046469444444 400
MaxY 793.046469444444 400
Resolution 5.55555555555556E-07 (1/500 arc-second) 8.8 x 10^-14
Unknown Coordinate Systems
For an UnknownCoordinateSystem (UCS), the "horizon" is defined to be a square that produces a resolution of 1 millimeter for a low precision UCS or 1/10 mm for a high precision UCS.
The example below lists high and low precision extents for an UnknownCoordinateSystem.
Name Low Precision UnknownCoordinateSystem High Precision UnknownCoordinateSystem
MinX -1,073,741.8245 -450,359,962,737.05
MinY -1,073,741.8245 -450,359,962,737.05
MaxX 1,073,741.8225 450,359,962,737.049
MaxY 1,073,741.8225 450,359,962,737.049
Resolution 0.001 0.0001
ISpatialReferenceResolution Interface
#### .NET Snippets
Make Spatial Reference | Create High Precision Spatial Reference
#### .NET Samples
Create a custom raster type from the ground up for DMCII data (Code Files: DMCIIRasterType) | Creating a toolbar of globe tools (Code Files: SpatialReference)
#### .NET Related Topics
Constructing a high- or low-precision spatial reference | Creating feature classes | How to import or export a spatial reference | Working with GeometryEnvironment | 857 | 3,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | latest | en | 0.64344 |
https://gonzoecon.com/2021/11/covid-and-kids/ | 1,713,105,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00459.warc.gz | 260,692,996 | 24,750 | # COVID and Kids
[Updated November 3, 2021 to reflect comments from Phil Kerpen (@kerpen). I have appended Phil’s Twitter thread at the end of this article. I have also added a thread by Emma Woodhouse (@EWoodHouse7) debunking the 744 total deaths reported by CDC. Finally, there is a new version of my Excel workbook that lets you plug in any figure you want for the percentage of the CDC estimate you’d like to assume. The Excel workbook uses 65% as an example. There is a third workbook that uses NCHS numbers as cited by Emma Woodhouse.]
Phil Kerpen Emma Woodhouse
I’ll note in passing that Emma Woodhouse may well be a nom de Twit. Regardless, she’s very good with data.
As noted, Phil Kerpen offers a lengthy critique of the CDC data. Here’s a summary tweet:
(click for larger image)
That put me on to this table from Emma Woodhouse:
(click for larger image)
With that as background, let’s look at the data.
## How Many Kids Have Died From COVID?
How many kids have died from COVID? Using two different datasets, I calculated deaths from all causes as listed by the CDC in 2019. COVID deaths are the latest available from CDC. The CDC is also responsible for the odd age ranges. Bottom line on COVID and kids ages 1 through 17: 774 total deaths out of a total population of 72,187,014 or 0.00107%. Conclusion: spend more on protecting kids from accidents, homicides, and suicides. Masking kids in schools — or anywhere for that matter — has costs that far outweigh any conceivable benefits.
(click for larger image)
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My total above is 9 more than Emma’s CDC total. Given the uncertainty about these figures, that difference is negligible.
I’ll go through the various assumptions and calculations I made to get the numbers at the end of this article. The main issue is that the age brackets used by CDC for deaths by causes and ages are different from the COVID age brackets. There are also a few other inconsistencies. But first let’s see the real dangers kids face. (The graphs are difficult to read. Click here to download the Excel workbook with the data and clear graphs.)
## Causes of Death by Age Group
Let’s start with the babies, ages 1 to 4.
(click for larger image)
Accidents are the leading cause of death for each of the four age groups. For the 1-4 age group, the ubiquitous “all other causes” is second, followed by congenital malformations, malignant neoplasms, and homicide. Yes, you read that correctly. Very young kids have a greater chance of being murdered than of dying from COVID. One item missing from this list is suicide. Sadly, that makes its first appearance among the 5 to 11 group.
The percentages are important. The vertical axis is percentage of population in that age group. Of the 19,422,515 kids in that age group, 1,149 died from accidents (0.00592%). COVID deaths total 244, 0.00126%.
Next up, the group 5 to 11.
(click for larger image)
Accidents once again are the leading cause (1,025 or 0.00364% of the 28,162,647 in this group). All other causes (883, 0.00314%) is next followed by malignant neoplasms (533, 0.00189%), homicide (231, 0.00082%), and suicide (214, 0.00076%). COVID is in seventh place with 171 (0.00061%).
Before going on, note one thing. These percentages are tiny, miniscule, very small. The chances of a kid dying in the U.S. today are incredibly small. And the chances of dying from COVID are tiny relative to the leading causes of death.
Moving on to 12 to 15,
(click for larger image)
Accidents (1,174 and 0.00711%) once again are the leading cause of death. Sadly, suicides have moved up to second (762 and 0.00462%). In order we have all other causes (712 and 0.00431%), homicide (490 and 0.00297%), malignant neoplasms (360 and 0.00218%) and COVID (198 and 0.00120%).
The final group is odd, ages 16 and 17. I suspect CDC wanted to make sure the age groups only included kids in schools up to 12th grade. That has the detrimental effect of a much smaller age group than the first three (8 million compared to 16.5 million in the 12-15 group). For whatever reason, here are the statistics.
(click for larger image)
Accidents are once again the number one cause of death (1,415 and 0.00502%). Suicide is next (884 and 0.00314%) followed by homicide (751 and 0.00267%). All other causes (528 and 0.00188%) and malignant neoplasms (236 and 0.00084%) are just ahead of COVID (161 and 0.00057%).
## The Kerpen – Woodhouse Alternatives
Phil Kerpen says the actual number of COVID deaths is 35% less than the CDC numbers. Here’s what that looks like. (Click here to download the Excel workbook that lets you pick any percentage you want. For Phil’s numbers, the percentage is 100% – 35% = 65%.
(click for larger image)
(click for larger image) (click for larger image) (click for larger image) (click for larger image)
And here are the Woodhouse numbers:
(click for larger image)
(click for larger image) (click for larger image) (click for larger image) (click for larger image)
As promised, here are the links to the two workbooks. For the 65% of CDC figures, click here. For the Woodhouse – NCHS numbers, click here.
## Conclusion
Spending more on child safety and suicide prevention would seem to be a better use of public funds than worrying about masking kids in schools (or anywhere else for that matter).
## Methodology
Translating the mortality from all causes age brackets to match the COVID data took some effort. I first tried various interpolation methods (cubic spline, Bissel) to see if I could get meaningful data for each individual age. That was not successful (to put it mildly). Instead, I just used linear interpolation. Thus for the COVID age group was 16 to 17 and the all mortality group 15 to 19, I multiplied the all mortality deaths by 0.4 (two years out of five). Similar procedures were used for the other three groups. These calculations do not appear in the Excel tables. To be sure the tables were sorted correctly, I replaced the calculations with the actual values. (This is pretty easy in Excel. Copy the data, then paste it into the same area using Paste Special – Values.)
Also, the all mortality data is from 2019. This is before COVID. I wanted to make sure COVID deaths were not double-counted in the data. Naturally, COVID deaths are from January, 2020 through September, 2021.
I urge anyone who has a better methodology for any of this to download the Excel workbook and fix it. When you’re done, please e-mail me the results so I can learn better techniques.
## Appendix: Phil Kerpen and Emma Woodhouse Threads
Share if you feel like it
## About Tony Lima
Retired after teaching economics at California State Univ., East Bay (Hayward, CA). Ph.D., economics, Stanford. Also taught MBA finance at the California University of Management and Technology. Occasionally take on a consulting project if it's interesting. Other interests include wine and technology. | 1,696 | 6,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-18 | latest | en | 0.928979 |
https://www.teachoo.com/8458/2098/Ex-6.1--5/category/Ex-6.1/ | 1,686,344,233,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00232.warc.gz | 1,133,574,075 | 33,667 | Ex 5.1
Chapter 5 Class 8 Squares and Square Roots
Serial order wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 5.1, 5 Observe the following pattern and supply the missing numbers. 〖11〗^2 = 1 2 1 〖 101〗^2 = 1 0 2 0 1 〖 10101〗^2 = 102030201 〖1010101〗^2 = ........................... 〖"................" 〗^2= 10203040504030201 1012 = 10201 Number of 1s in 101 is = 2 So we write numbers upto 2 & then decrease 1012 = Now, Adding 1 zero between the Digits 1012 = 101012 = 102030201 Number of 1s in 101 is = 3 So we write numbers upto 3 & then decrease 101012 = Now, Adding 1 zero between the Digits 101012 = 10101012 = ………….. Number of 1s in 101 is = 4 So we write numbers upto 4 & then decrease 10101012 = Now, Adding 1 zero between the Digits 10101012 = 〖"................" 〗^2= 10203040504030201 Since number on left is till 5 Number of 1s on left number = 5 and we will add one zero between it The number will be = 10203040504030201 | 342 | 987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-23 | longest | en | 0.74829 |
https://discuss.leetcode.com/topic/99643/simple-iterative-ruby-solution-w-explanation-accepted | 1,513,574,453,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948608836.84/warc/CC-MAIN-20171218044514-20171218070514-00042.warc.gz | 585,510,333 | 8,695 | # Simple Iterative Ruby solution w/ explanation (accepted)
• I broke up the problem into 2 functions. This is merely an application of the Cartesian Product generalized to n sets where n >= 2. Here is a great resource http://ndp.jct.ac.il/tutorials/discrete/node28.html
Using cartesian product for the input of "23" we get ["a", "b", "c"] x ["d", "e", "f"] = ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]
Say we want "234" then we would have the following
["a", "b", "c"] x ( ["d", "e", "f"] x ["g", "h", "i"] ).
Therefore we only need to write a function that returns the cartesian product of two arrays. We then can generalize to n > 2 arrays by using the reduce function. The code follows
``````def letter_combinations(digits)
dict = {
"1" => [],
"2" => %w(a b c),
"3" => %w(d e f),
"4" => %w(g h i),
"5" => %w(j k l),
"6" => %w(m n o),
"7" => %w(p q r s),
"8" => %w(t u v),
"9" => %w(w x y z)
}
digits.split("").reverse.map do |digit|
dict[digit]
end.reduce([]) do |result, arr|
cartesian_product(arr, result)
end
end
def cartesian_product(arrA, arrB)
return arrB if arrA.length == 0
return arrA if arrB.length == 0
result = []
for i in (0..arrA.length - 1)
for j in (0..arrB.length - 1)
result += [arrA[i] + arrB[j]]
end
end
result
end
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 441 | 1,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-51 | latest | en | 0.724135 |
https://mathlake.com/Important-4-Marks-Questions-for-CBSE-11-Maths | 1,716,023,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00762.warc.gz | 345,343,394 | 4,908 | # Important 4 Marks Questions for CBSE 11 Maths
Important 4 marks questions for CBSE Class 11 Maths subject are provided here for students. The questions are prepared as per the latest exam pattern and will help students to score good marks. The Central Board of Secondary education is responsible for conducting an examination of the schools affiliated to the Central Board.
Students usually find Maths as one of the most difficult papers. The major reason is due to lack of confidence and less practice. It is suggested that practicing Maths on daily basis will help students to develop to their full potential and also help them to solve the question with high accuracy.
## Class 11 Maths Important 4 Marks Questions
We at BYJU’S provide the student with 4 marks Maths important questions for class 11th. Students can practice 4 marks wise questions to be aware of the questions that can be framed in their final examination.
Question 1- In triangle ABC, prove that
$$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{a^{2}+ b^{2}+ c^{2}}{2abc}$$
Question 2- How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letter out of word DAUGHTER?
Question 3- Find the coordinates of the orthocentre of the triangle whose vertices are (-1,3), (2,-1) and (0,0).
Question 4- Find the domain and range of the function $$f(x) = \frac{1}{\sqrt{x – \left | x \right |}}$$.
Question 5- Find the variance for the following data-
Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2
Question 6- Using principle of mathematical induction, prove that
$$\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + …….. + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}, \forall n \in N$$.
Question 7- Express in the form of a+ib,
$$\left ( \frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + i\sqrt{2})(\sqrt{3} – i\sqrt{2})} \right )$$
Question 8- Prove that: $$\cot (15/2)^{\circ} = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$$
Question 9- Four cards are drawn at random from a pack of 52 playing cards. Find the probability of getting one card from each suit.
Question 10- If a is A.M. and b and c be two G.M.s between any two positive numbers, then prove that $$b^{3} + c^{3} = 2abc$$.
Question 11- In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Question 12- Represent the complex number $$1 + \sqrt{3}i$$ in polar form.
Question 13- On an evening a man planned a party of their friends on his 25th marriage anniversary. When all of his friends have arrived, he introduced all to each other and everybody shakes hand with everybody else. Find the total person in a room, if total shake hands are 66.
Question 14- A committee of two members is selected from two men and two women. What is the probability that the committee will have one man.
Question 15- A teacher teaches their students with such a spirit that they must know what he knows? After teaching with same spirit, he decided to check the ability of the students through a test. He has given a question, if a,b are roots of $$x^{2} – 3x + p = 0$$, c and d are roots of $$x^{2} – 12x + 1 = 0$$, where a,b,c,d are in G.P. Evaluate the ratio q+p to q-p.
Question 16- Find the equation of set of points ‘P’ such that its distance from the points (3,4,-5) and (-2,1,4) are equal.
Question 17- A horse is tied to a pole by a rope. If the horse moves along a circular path keeping the rope tight and describe 176m when it has traced out $$72^{\circ}$$ at the pole, find the length of the rope.
Question 18- Evaluate the value of $$\tan \left ( \frac{\pi}{8} \right )$$
Question 19- If $$\left ( 1 + 1/i – i\right )^{m} = 1$$, then find the least integral value of m.
Question 20- Evaluate-
(i) $$\lim \limits_{x \to 0} \left [ \frac{(x+1)^{5}-1}{x} \right ]$$
(ii) $$\lim \limits_{x \to \pi} \left [ \frac{\sin (\pi + x)}{\pi( \pi -x)} \right ]$$ | 1,189 | 4,040 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-22 | latest | en | 0.870226 |
https://physics.stackexchange.com/questions/220968/how-does-curved-spacetime-cause-motion-revisited | 1,618,960,088,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039491784.79/warc/CC-MAIN-20210420214346-20210421004346-00089.warc.gz | 545,046,949 | 38,697 | # How does curved spacetime cause motion revisited
There was a previous question titled "Why would spacetime curvature cause gravity?" asked March 10, 2014.
The answer given was essentially that since the time component of an object in curved space slows that automatically has to generate a motion in space so the total spacetime velocity is still c.
So far so good but there seems to be a problem. Suppose another object is sitting on the surface of the earth. Its clock is also slowed due to curved space but there is NO motion in space.
1. So how can the object still have a total spacetime velocity of c if its coordinate time is slowed but it is not moving in space?
2. It doesn’t seem like the answer to the original question works if the slowing of coordinate time has to produce motion in space because there isn’t any motion in space in this case but there is still slowing of coordinate time. Is this correct?
3. In light of this how does curved spacetime cause motion? Is there some other explanation that works or am I missing something?
Thanks, Edgar
• Note that in (1), the object's world line is accelerated, i.e., an accelerometer attached to the object reads non-zero acceleration. If the surface of the Earth were not there, the object would free-fall (accelerometer reads zero) 'downward'. Indeed, relative to a free-falling observer, the object is accelerating. – Alfred Centauri Nov 28 '15 at 2:38
• Think of it this way. Objects in space follow geodesic paths. By virtue of the fact that spacetime is curved, objects must follow a geodesic path which results in motion. – Horus Nov 28 '15 at 12:27
Consider the example of the Schwarzschild metric. If we take an observer who is at fixed $r$, $\theta$ and $\phi$ the metric simplifies to:
$$d\tau^2 = \left( 1 - \frac{2M}{r}\right) dt^2$$
the four-velocity is $dx^\alpha/d\tau$, so in this case we find:
$$\mathbf{u} = \left( \frac{1}{\sqrt{1-2M/r}}, 0, 0, 0 \right)$$
The norm of the four-velocity is given by:
$$u = \sqrt{g_{\alpha\beta}x^\alpha x^\beta}$$
So in this case we find:
$$u = u^0 \sqrt{g_{00}} = \frac{1}{\sqrt{1-2M/r}} \sqrt{1-2M/r} = 1$$
So the norm of the four velocity is indeed $c$ even in this example of a curved spacetime.
When moving from special relativity to general relativity it's easy to forget that computing the norm of a four-vector requires the metric. This is true in special relativity as well of course, but since in SR the entries in the metric all have modulus unity it's easy to overlook it.
Time duration does not cause motion. If you had a spherical shell of matter then the inside has time dilation compared to distant outside, but there is no motion caused in the inside.
On the other hand your example is also totally wrong. An inertially moving object is in freefall and the non inertial earth based coordinate system would say it is falling down.
In fact, an object sitting on the surface of the earth is accelerated upwards in every inertial frame because of the pressure of the ground below it. It sinks into the earth until the pressure is enough to accelerate it upwards at 1 g.
The word General in General Relativity refers to the fact that there are not global inertial frames. | 792 | 3,222 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-17 | latest | en | 0.93487 |
https://www.education.com/resources/adding-and-subtracting-fractions/CCSS-Math-Content-5-NF/?page=2 | 1,627,600,705,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153897.89/warc/CC-MAIN-20210729203133-20210729233133-00250.warc.gz | 763,542,275 | 25,719 | # Search Our Content Library
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Do some pumpkin patch math with your child this fall. See if you can find the tricky pattern that weaves through this pumpkin patch math worksheet.
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Use these vocabulary cards with the EL Support Lesson: Compare Common Denominators Methods. | 728 | 3,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-31 | latest | en | 0.799533 |
https://math.stackexchange.com/questions/1569031/pivot-positions-and-reduced-row-echelon-form | 1,713,085,778,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00460.warc.gz | 360,930,306 | 35,124 | # Pivot positions and reduced row echelon form
I have the definition of reduced row echelon form (the relevant part) as The leading entry in each non-zero row is 1and each leading $1$ is the only non $0$ entry in its column.
I then have the definition of a pivot position as a location in a matrix that corresponds to a leading $1$ in the RRE form.
So, Suppose matrix A is $11\times 9$. Is it possible that matrix $A$ has a pivot position in every row?
This is no right? Because by definition you can't have more than one pivot position per column.
So this brings me to the point:
My practice test says:
Is $Ax = b$ consistent for a particular given $b$?
The way I would do this is simply solve the augmented matrix and make sure it doesn't have any impossible equations, such as $0=4$.
However, the way the practice says to do it is to:
To answer this question, we look at if the rightmost column of the augmented matrix $[A\quad b]$ is a pivot column.
But this doesn't make any sense to me, because a pivot column is only going to have 1 non 0 variable in it, and the point of an augmented matrix is to solve the system of equations for a column (vector) of multiple variables.
I hope this makes sense, let me know if I need to clarify anything, thanks.
The pivot column in the hint can refer to a column that has a leading entry. You don't need to transform a matrix $A$ to its reduced row echelon form to see whether it has solutions. A row echelon form is enough.
Even if you transform it to its reduced row echelon form, if the last column is a pivot column, the system has no solution. You can still get all zeros for the other entries except the last one. In this case, the last equation is just $0=1$.
Here is an example for you:
$$\left[ \begin{array}{cc|c} 1&2&3\\ 1&1&0\\ 1&3&1 \end{array} \right]$$
The row echelon form of the above augmented matrix is: $$\left[ \begin{array}{cc|c} 1&2&3\\ 0&-1&-3\\ 0&0&-5 \end{array} \right]$$ You can already see that the last column is a pivot column because it has a leading entry.
If you prefer, you can find its reduced row echelon form: $$\left[ \begin{array}{cc|c} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array} \right]$$
• Hmm, ok. Lets say the question was How to do you find the pivot columns of augmented matrix [A b] Would you reduce this to RRE? In your example, it seems like there is a pivot in every column, but you can't tell that until you get it into RRE, if its just in RE then it looks like there is a pivot in the first 2, but not the last. You can only find pivot positions in RRE form right?
– Luke
Dec 10, 2015 at 16:18
• @BNSlug: You can find the pivot positions in RE form. In my example, the second matrix is in RE form. You can see that $1, -1, -5$ are all leading entries in their own rows. Dec 10, 2015 at 17:58 | 773 | 2,797 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-18 | latest | en | 0.939057 |
https://www.teachoo.com/8827/2834/Ex-12.3--9/category/Unitary-Method/ | 1,660,889,472,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573623.4/warc/CC-MAIN-20220819035957-20220819065957-00670.warc.gz | 905,705,479 | 29,761 | Unitary Method
Chapter 12 Class 6 Ratio And Proportion
Concept wise
Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
### Transcript
Ex 12.3, 9 A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km? Diesel required to cover 594 km = 108 litres Diesel required to cover 1 km = (108/594) litres Diesel required to cover 1650 km = 108/594 × 1650 litres = 54/297 × 1650 = 18/99 × 1650 = 6/33 × 1650 = 2/11 × 1650 = 2 × 150 = 300 lites ∴ 300 litres of diesel is required for covering a distance of 1650 km | 192 | 628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-33 | latest | en | 0.84081 |
https://au.mathworks.com/matlabcentral/cody/problems/64-the-goldbach-conjecture-part-2/solutions/2020427 | 1,606,600,628,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00598.warc.gz | 192,161,245 | 16,964 | Cody
# Problem 64. The Goldbach Conjecture, Part 2
Solution 2020427
Submitted on 14 Nov 2019 by Miles Walker
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 6; c_correct = 1; assert(isequal(goldbach2(n),c_correct))
2 Pass
n = 10; c_correct = 2; assert(isequal(goldbach2(n),c_correct))
3 Pass
n = 50; c_correct = 4; assert(isequal(goldbach2(n),c_correct))
4 Pass
n = 480; c_correct = 29; assert(isequal(goldbach2(n),c_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 207 | 690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-50 | latest | en | 0.699189 |
https://www.shaalaa.com/question-bank-solutions/evaluate-the-following-1x-x-dx-methods-of-integration-integration-by-substitution_155554 | 1,702,000,190,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100710.22/warc/CC-MAIN-20231208013411-20231208043411-00145.warc.gz | 1,074,133,978 | 9,408 | # Evaluate the following. ∫1x+x dx - Mathematics and Statistics
Sum
Evaluate the following.
int 1/(sqrt"x" + "x") dx
#### Solution
Let I = int 1/(sqrt"x" + "x") dx
= int 1/(sqrt"x" (1 + sqrt"x"))dx
Put 1 + sqrt"x" = "t"
∴ 1/(2sqrt"x") "dx" = "dt"
∴ 1/sqrt"x"dx = 2 dt
∴ I = int (2 * "dt")/"t"
= 2 int 1/"t" * "dt"
= 2 log | t | + c
∴ I = 2 log |1 + sqrt"x"| + c
Is there an error in this question or solution?
Chapter 5: Integration - Exercise 5.2 [Page 123]
Share | 192 | 479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-50 | latest | en | 0.699318 |
https://msgroups.net/excel.misc/how-to-arrange-list-by-first-or-last-name/87775 | 1,639,002,117,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00003.warc.gz | 456,501,507 | 11,508 | #### how to arrange list by first or last name
```I am trying to make a list with first, middle, last name , street address,
and phone numbers. I like to know how you can review the list by any
combination of first name, last name, city, state, and or phone number?
```
0
Utf
5/27/2010 7:59:01 PM
excel.misc 78881 articles. 5 followers.
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[PageSpeed] 47
```hi
can you give examples of how your list is layed out.
i am thinking a helper column with an extraction formula might work.
regards
FSt1
"Andy" wrote:
> I am trying to make a list with first, middle, last name , street address,
> and phone numbers. I like to know how you can review the list by any
> combination of first name, last name, city, state, and or phone number?
```
0
Utf
5/27/2010 10:02:50 PM
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# Python Set
## Operations on sets
with other sets
``````# Intersection
{1, 2, 3, 4, 5}.intersection({3, 4, 5, 6}) # {3, 4, 5}
{1, 2, 3, 4, 5} & {3, 4, 5, 6} # {3, 4, 5}
# Union
{1, 2, 3, 4, 5}.union({3, 4, 5, 6}) # {1, 2, 3, 4, 5, 6}
{1, 2, 3, 4, 5} | {3, 4, 5, 6} # {1, 2, 3, 4, 5, 6}
# Difference
{1, 2, 3, 4}.difference({2, 3, 5}) # {1, 4}
{1, 2, 3, 4} - {2, 3, 5} # {1, 4}
# Symmetric difference with
{1, 2, 3, 4}.symmetric_difference({2, 3, 5}) # {1, 4, 5}
{1, 2, 3, 4} ^ {2, 3, 5} # {1, 4, 5}
# Superset check
{1, 2}.issuperset({1, 2, 3}) # False
{1, 2} >= {1, 2, 3} # False
# Subset check
{1, 2}.issubset({1, 2, 3}) # True
{1, 2} <= {1, 2, 3} # True
# Disjoint check
{1, 2}.isdisjoint({3, 4}) # True
{1, 2}.isdisjoint({1, 4}) # False``````
with single elements
``````# Existence check
2 in {1,2,3} # True
4 in {1,2,3} # False
4 not in {1,2,3} # True
s = {1,2,3}
s.remove(2) # s == {1,4}
s.remove(2) # KeyError!``````
Set operations return new sets, but have the corresponding in-place versions:
For example:
``````s = {1, 2}
s.update({3, 4}) # s == {1, 2, 3, 4}
``````
## Get the unique elements of a list
Let’s say you’ve got a list of restaurants — maybe you read it from a file. You care about the unique restaurants in the list. The best way to get the unique elements from a list is to turn it into a set:
``````restaurants = ["McDonald's", "Burger King", "McDonald's", "Chicken Chicken"]
unique_restaurants = set(restaurants)
print(unique_restaurants)
# prints {'Chicken Chicken', "McDonald's", 'Burger King'}``````
Note that the set is not in the same order as the original list; that is because sets are unordered, just like dicts.
This can easily be transformed back into a List with Python’s built in list function, giving another list that is the same list as the original but without duplicates:
``````list(unique_restaurants)
# ['Chicken Chicken', "McDonald's", 'Burger King']``````
It’s also common to see this as one line:
``````# Removes all duplicates and returns another list
list(set(restaurants))
``````
Now any operations that could be performed on the original list can be done again.
## Set of Sets
``{{1,2}, {3,4}}``
``TypeError: unhashable type: 'set'``
``{frozenset({1, 2}), frozenset({3, 4})}``
## Set Operations using Methods and Builtins
We define two sets a and b
``````>>> a = {1, 2, 2, 3, 4}
>>> b = {3, 3, 4, 4, 5}
``````
NOTE: {1} creates a set of one element, but {} creates an empty dict. The correct way to create an
empty set is set().
Intersection
a.intersection(b) returns a new set with elements present in both a and b
``````>>> a.intersection(b)
{3, 4}``````
Union
a.union(b) returns a new set with elements present in either a and b
``````>>> a.union(b)
{1, 2, 3, 4, 5}``````
Difference
a.difference(b) returns a new set with elements present in a but not in b
``````>>> a.difference(b)
{1, 2}
>>> b.difference(a)
{5}``````
Symmetric Difference
a.symmetric_difference(b) returns a new set with elements present in either a or b but not in both
``````>>> a.symmetric_difference(b)
{1, 2, 5}
>>> b.symmetric_difference(a)
{1, 2, 5}``````
NOTE: a.symmetric_difference(b) == b.symmetric_difference(a)
Subset and superset
c.issubset(a) tests whether each element of c is in a.
a.issuperset(c) tests whether each element of c is in a.
``````>>> c = {1, 2}
>>> c.issubset(a)
True
>>> a.issuperset(c)
True``````
The latter operations have equivalent operators as shown below:
Sets a and d are disjoint if no element in a is also in d and vice versa.
``````>>> d = {5, 6}
>>> a.isdisjoint(b) # {2, 3, 4} are in both sets
False
>>> a.isdisjoint(d)
True
# This is an equivalent check, but less efficient
>>> len(a & d) == 0
True
# This is even less efficient
>>> a & d == set()
True``````
Testing membership
The builtin in keyword searches for occurances
``````>>> 1 in a
True
>>> 6 in a
False``````
Length
The builtin len() function returns the number of elements in the set
``````>>> len(a)
4
>>> len(b)
3``````
## Sets versus multisets
Sets are unordered collections of distinct elements. But sometimes we want to work with unordered collections of elements that are not necessarily distinct and keep track of the elements’ multiplicities.
Consider this example:
``````>>> setA = {'a','b','b','c'}
>>> setA
set(['a', 'c', 'b'])``````
By saving the strings ‘a’, ‘b’, ‘b’, ‘c’ into a set data structure we’ve lost the information on the fact that ‘b’
occurs twice. Of course saving the elements to a list would retain this information
``````>>> listA = ['a','b','b','c']
>>> listA
['a', 'b', 'b', 'c']
``````
but a list data structure introduces an extra unneeded ordering that will slow down our computations.
For implementing multisets Python provides the Counter class from the collections module (starting from version 2.7)
``````Python 2.x Version ≥ 2.7
>>> from collections import Counter
>>> counterA = Counter(['a','b','b','c'])
>>> counterA
Counter({'b': 2, 'a': 1, 'c': 1})
``````
Counter is a dictionary where where elements are stored as dictionary keys and their counts are stored as
dictionary values. And as all dictionaries, it is an unordered collection. | 1,759 | 5,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-38 | latest | en | 0.683505 |
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Struggling with Assignments? | 1,285 | 7,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-26 | latest | en | 0.92908 |
http://www.physicsforums.com/showthread.php?s=373c94f9d077e1d47bfbd7641c88a5b2&p=4254404 | 1,369,039,089,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698646863/warc/CC-MAIN-20130516100406-00009-ip-10-60-113-184.ec2.internal.warc.gz | 649,497,002 | 8,405 | ## Definition of Fourier transform
Hi All,
Usually the fourier transform is defined as the one in the Wiki page here (http://en.wikipedia.org/wiki/Fourier_transform), see the definition.
My question is can I define fourier transform as $\int$f(x)e$^{2\pi ix \varsigma}$dx instead, i.e., with the minus sign removed, as the forward fourier transform? The backward one is the one with the minus sign. So the definition is the opposite to the definition on the wiki page.
Can I define this? Will the so-transformed frequency domain still bear the physical meanings as we usually talk about?
Thanks in advance. Any comment will help.
Jo
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Recognitions: Homework Help Welcome to PF, jollage! Yep. You can do that. Fourier transforms are defined haphazardly as you may already have noticed. Changing the sign or the constants does not change the way it operates, nor the physical meaning.
Recognitions: Science Advisor Just be aware that the result you get might differ from one found the other way.
## Definition of Fourier transform
OK, thank you for confirming this. This is great. I guess I could move on with this definition.
Recognitions: Gold Member Science Advisor It just means that what you call a positive frequency, everyone else calls a negative frequency, and vice-versa. If you are dealing with real-valued functions only (i.e. not complex), it won't make much difference, because in that case the negative-frequency spectrum is just a mirror image of the positive-frequency spectrum.
Mentor
Quote by jollage Hi All, Usually the fourier transform is defined as the one in the Wiki page here (http://en.wikipedia.org/wiki/Fourier_transform), see the definition. My question is can I define fourier transform as $\int$f(x)e$^{2\pi ix \varsigma}$dx instead, i.e., with the minus sign removed, as the forward fourier transform? The backward one is the one with the minus sign. So the definition is the opposite to the definition on the wiki page.
See equations 15 and 16 here:
http://mathworld.wolfram.com/FourierTransform.html
To get a "general" Fourier transform there are two free parameters that you can set. Different groups use different choices of those free parameters as their "standard", but it is all just a matter of convention. | 534 | 2,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2013-20 | latest | en | 0.906344 |
https://www.eevblog.com/forum/chat/hex-confusion-what-value-is-attained-when-i-decrement-0x00/msg117438/ | 1,579,465,929,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00038.warc.gz | 856,560,377 | 10,089 | ### Author Topic: Hex confusion: what value is attained when I decrement 0x00? (Read 2926 times)
0 Members and 1 Guest are viewing this topic.
#### iamwhoiam
• Contributor
• Banned!
• Posts: 37
##### Hex confusion: what value is attained when I decrement 0x00?
« on: June 03, 2012, 04:07:58 pm »
Hi guys. Being used to decimal, naturally, I am confused by this; the part highlighted in red:
LED_PAUSE
MOVLW 0xFF
MOVWF LED_LOOP2
LED_PAUSE2
CLRF LED_LOOP
DECFSZ LED_LOOP,F
GOTO \$-1
DECFSZ LED_LOOP2,F
GOTO LED_PAUSE2
END
Seeing as "LED_LOOP" has been issued CLRF, and hence contains 0x00, what value does decrementing 0x00 result in; 0xFF?
Hex n00b!
Thanks.
« Last Edit: June 03, 2012, 04:11:06 pm by iamwhoiam »
#### MikeK
• Frequent Contributor
• Posts: 577
• Country:
##### Re: Hex confusion: what value is attained when I decrement 0x00?
« Reply #1 on: June 03, 2012, 04:14:49 pm »
The opposite of what happens when you increment 0xFF.
#### AntiProtonBoy
• Frequent Contributor
• Posts: 770
• Country:
##### Re: Hex confusion: what value is attained when I decrement 0x00?
« Reply #2 on: June 03, 2012, 04:42:06 pm »
Google says 0x00 -1 is -0x1... or is that just Google fail?
Not necessarily a fail, it depends how the byte is interpreted.
If you interpret 11111111 as an unsigned binary, then its hex equivalent is 0xFF, or 255 decimal.
If you interpret 11111111 as a two's complement signed binary, then hex is -0x01, or -1 decimal.
That being said, hex constant values are generally interpreted as a raw collection of bits with the sign absent, thus I'd find the -0x01 notation unusual.
#### caroper
• Regular Contributor
• Posts: 193
• Country:
##### Re: Hex confusion: what value is attained when I decrement 0x00?
« Reply #3 on: June 03, 2012, 08:26:35 pm »
Thanks to binary arithmetic we have finally solved Shakespeare's famous dilemma.
2B OR NOT 2B = -1
Cheers
Chris
#### Mechatrommer
• Super Contributor
• Posts: 9318
• Country:
• reassessing directives...
##### Re: Hex confusion: what value is attained when I decrement 0x00?
« Reply #4 on: June 04, 2012, 06:44:42 am »
Quote from: OP
what value is attained when I decrement 0x00?
The opposite of what happens when you increment 0xFF.
(its been a while). what value will depend on how many bits per byte/register we have and architecture of the "flipflop" (is it the right term?). deducted it to 8bit PIC8 (from your example).. my answer is 0xFF with 1 in carry (or zero) bit. incrementing 0xFF will result in 0x00 with 1 in carry/zero bit.
« Last Edit: June 04, 2012, 06:47:02 am by Mechatrommer »
if something can select, how cant it be intelligent? if something is intelligent, how cant it exist?
Smf | 804 | 2,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-05 | latest | en | 0.765908 |
https://www.quizzes.cc/metric/percentof.php?percent=27&of=8470 | 1,585,829,111,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00077.warc.gz | 1,094,550,989 | 3,123 | #### What is 27 percent of 8,470?
How much is 27 percent of 8470? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 27% of 8470 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 27% of 8,470 = 2286.9
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating twenty-seven of eight thousand, four hundred and seventy How to calculate 27% of 8470? Simply divide the percent by 100 and multiply by the number. For example, 27 /100 x 8470 = 2286.9 or 0.27 x 8470 = 2286.9
#### How much is 27 percent of the following numbers?
27 percent of 8470.01 = 228690.27 27 percent of 8470.02 = 228690.54 27 percent of 8470.03 = 228690.81 27 percent of 8470.04 = 228691.08 27 percent of 8470.05 = 228691.35 27 percent of 8470.06 = 228691.62 27 percent of 8470.07 = 228691.89 27 percent of 8470.08 = 228692.16 27 percent of 8470.09 = 228692.43 27 percent of 8470.1 = 228692.7 27 percent of 8470.11 = 228692.97 27 percent of 8470.12 = 228693.24 27 percent of 8470.13 = 228693.51 27 percent of 8470.14 = 228693.78 27 percent of 8470.15 = 228694.05 27 percent of 8470.16 = 228694.32 27 percent of 8470.17 = 228694.59 27 percent of 8470.18 = 228694.86 27 percent of 8470.19 = 228695.13 27 percent of 8470.2 = 228695.4 27 percent of 8470.21 = 228695.67 27 percent of 8470.22 = 228695.94 27 percent of 8470.23 = 228696.21 27 percent of 8470.24 = 228696.48 27 percent of 8470.25 = 228696.75
27 percent of 8470.26 = 228697.02 27 percent of 8470.27 = 228697.29 27 percent of 8470.28 = 228697.56 27 percent of 8470.29 = 228697.83 27 percent of 8470.3 = 228698.1 27 percent of 8470.31 = 228698.37 27 percent of 8470.32 = 228698.64 27 percent of 8470.33 = 228698.91 27 percent of 8470.34 = 228699.18 27 percent of 8470.35 = 228699.45 27 percent of 8470.36 = 228699.72 27 percent of 8470.37 = 228699.99 27 percent of 8470.38 = 228700.26 27 percent of 8470.39 = 228700.53 27 percent of 8470.4 = 228700.8 27 percent of 8470.41 = 228701.07 27 percent of 8470.42 = 228701.34 27 percent of 8470.43 = 228701.61 27 percent of 8470.44 = 228701.88 27 percent of 8470.45 = 228702.15 27 percent of 8470.46 = 228702.42 27 percent of 8470.47 = 228702.69 27 percent of 8470.48 = 228702.96 27 percent of 8470.49 = 228703.23 27 percent of 8470.5 = 228703.5
27 percent of 8470.51 = 228703.77 27 percent of 8470.52 = 228704.04 27 percent of 8470.53 = 228704.31 27 percent of 8470.54 = 228704.58 27 percent of 8470.55 = 228704.85 27 percent of 8470.56 = 228705.12 27 percent of 8470.57 = 228705.39 27 percent of 8470.58 = 228705.66 27 percent of 8470.59 = 228705.93 27 percent of 8470.6 = 228706.2 27 percent of 8470.61 = 228706.47 27 percent of 8470.62 = 228706.74 27 percent of 8470.63 = 228707.01 27 percent of 8470.64 = 228707.28 27 percent of 8470.65 = 228707.55 27 percent of 8470.66 = 228707.82 27 percent of 8470.67 = 228708.09 27 percent of 8470.68 = 228708.36 27 percent of 8470.69 = 228708.63 27 percent of 8470.7 = 228708.9 27 percent of 8470.71 = 228709.17 27 percent of 8470.72 = 228709.44 27 percent of 8470.73 = 228709.71 27 percent of 8470.74 = 228709.98 27 percent of 8470.75 = 228710.25
27 percent of 8470.76 = 228710.52 27 percent of 8470.77 = 228710.79 27 percent of 8470.78 = 228711.06 27 percent of 8470.79 = 228711.33 27 percent of 8470.8 = 228711.6 27 percent of 8470.81 = 228711.87 27 percent of 8470.82 = 228712.14 27 percent of 8470.83 = 228712.41 27 percent of 8470.84 = 228712.68 27 percent of 8470.85 = 228712.95 27 percent of 8470.86 = 228713.22 27 percent of 8470.87 = 228713.49 27 percent of 8470.88 = 228713.76 27 percent of 8470.89 = 228714.03 27 percent of 8470.9 = 228714.3 27 percent of 8470.91 = 228714.57 27 percent of 8470.92 = 228714.84 27 percent of 8470.93 = 228715.11 27 percent of 8470.94 = 228715.38 27 percent of 8470.95 = 228715.65 27 percent of 8470.96 = 228715.92 27 percent of 8470.97 = 228716.19 27 percent of 8470.98 = 228716.46 27 percent of 8470.99 = 228716.73 27 percent of 8471 = 228717 | 1,704 | 4,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-16 | latest | en | 0.840159 |
https://www.askiitians.com/forums/IIT-JEE-Entrance-Exam/a-particle-is-projected-with-velocity-2-gh-so-t_109471.htm | 1,653,544,104,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662601401.72/warc/CC-MAIN-20220526035036-20220526065036-00205.warc.gz | 734,161,327 | 33,776 | # A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show thatthe time of passing between the walls is 2√(h/g)
Yatin Sharma
12 Points
7 years ago
dude, what is the angle between the particle and the horizontal?
the time would only depend on the vertical component, and to compute that we need the angle of projection | 100 | 408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.946838 |
https://myassignmenthelp.com/au/vu/bao6000-accounting-and-financial-management/flexing-the-budget.html | 1,660,416,247,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00624.warc.gz | 393,835,857 | 67,653 | New
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Question 1
The company is well-established and not a start-up, it pays dividends, and the board want a relatively cheap method. Venture capital is not an appropriate method here (why?). This suggests only three methods that were covered in the final textbook chapter / lecture 9, one of which is ‘private placing’ which is not examinable.
Question 2
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Question 4
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### BAO6000 Accounting and Financial Management
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# Many kitchens today are equipped with high-speed electrical gadgets
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Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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24 Dec 2015, 09:43
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Many kitchens today are equipped with high-speed electrical gadgets, such as blenders and food processors, which are able to inflict as serious injuries as those caused by an industrial wood-planing machine.
(A) which are able to inflict as serious injuries as those
(B) which can inflict serious injuries such as those
(C) inflicting injuries as serious as that having been
(D) capable to inflict injuries as serious as that
(E) capable of inflicting injuries as serious as those
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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24 Dec 2015, 09:45
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naumyuk wrote:
Many kitchens today are equipped with high-speed electrical gadgets, such as blenders and food processors, which are able to inflict as serious injuries as those caused by an industrial wood-planing machine.
(A) which are able to inflict as serious injuries as those
(B) which can inflict serious injuries such as those
(C) inflicting injuries as serious as that having been
(D) capable to inflict injuries as serious as that
(E) capable of inflicting injuries as serious as those
Here is official GMAC explanation:
The point of this sentence is the claim that common kitchen appliances can be as dangerous as an industrial wood-planing machine. It makes this point by comparing the injuries (plural) caused by blenders and food processors with those (also plural) caused by the wood-planing machine. An efficient way to make this comparison is to use the idiom capable of, an adjective phrase rather than a relative clause, after blenders and food processors.
##### General Discussion
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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24 Dec 2015, 09:52
naumyuk
HI
You gave the problem at 10.13 PM and also the solution at 10.15 PM. What is the point?? Pl allow some discussion before announcing the result.
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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24 Dec 2015, 16:34
Hi! daagh,
I was confused between 'A' and 'D'. I am unable to see any mistake in 'A' as it says- blenders and food processors are able to inflict as serious injuries as those (injuries) caused by....... (as X as Y)
Could you please help shedding more light on the mistake in this option. Thanks
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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24 Dec 2015, 21:00
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Quote:
I was confused between 'A' and 'D'. I am unable to see any mistake in 'A' as it says- blenders and food processors are able to inflict as serious injuries as those (injuries) caused by....... (as X as Y)
Could you please help shedding more light on the mistake in this option. Thanks
A is ALSO WRONG for the following reason :
A says: Many kitchens today are equipped with high-speed electrical gadgets, such as blenders and food processors, which are able to inflict as serious injuries as those caused by an industrial wood-planing machine.
----->the usage of "WHICH" is WRONG in A; "WHICH" has been used without COMMA in option A .The comma that you are seeing is for the following construction "such as blenders and food processors" AND NOT for "WHICH"
ALSO "ABLE TO" is WRONG in option A ---->it gives a sense that these "electrical gadgets" are doing this INTENTIONALLY !!
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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25 Dec 2015, 12:13
Quote:
I was confused between 'A' and 'D'. I am unable to see any mistake in 'A' as it says- blenders and food processors are able to inflict as serious injuries as those (injuries) caused by....... (as X as Y)
Could you please help shedding more light on the mistake in this option. Thanks
A is ALSO WRONG for the following reason :
A says: Many kitchens today are equipped with high-speed electrical gadgets, such as blenders and food processors, which are able to inflict as serious injuries as those caused by an industrial wood-planing machine.
----->the usage of "WHICH" is WRONG in A; "WHICH" has been used without COMMA in option A .The comma that you are seeing is for the following construction "such as blenders and food processors" AND NOT for "WHICH"
ALSO "ABLE TO" is WRONG in option A ---->it gives a sense that these "electrical gadgets" are doing this INTENTIONALLY !!
Thanks for the explaination aditya8062
Kudos to you!!
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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30 Dec 2015, 05:02
Is the sentence otherwise correct?
"which are able to inflict as serious injuries as those caused by an industrial wood-planing machine."
Isn't it similar to the sentence - A can run as fast as B,
I'm a bit confused with this comparison.
Quote:
I was confused between 'A' and 'D'. I am unable to see any mistake in 'A' as it says- blenders and food processors are able to inflict as serious injuries as those (injuries) caused by....... (as X as Y)
Could you please help shedding more light on the mistake in this option. Thanks
A is ALSO WRONG for the following reason :
A says: Many kitchens today are equipped with high-speed electrical gadgets, such as blenders and food processors, which are able to inflict as serious injuries as those caused by an industrial wood-planing machine.
----->the usage of "WHICH" is WRONG in A; "WHICH" has been used without COMMA in option A .The comma that you are seeing is for the following construction "such as blenders and food processors" AND NOT for "WHICH"
ALSO "ABLE TO" is WRONG in option A ---->it gives a sense that these "electrical gadgets" are doing this INTENTIONALLY !!
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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Updated on: 05 May 2017, 00:29
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This is a comparison between injuries caused by kitchen gadgets and wood planing machines. A says that the gadgets can cause those serious injuries as caused by the planing machines. Whereas E says that, the gadgets can cause many injuries including as the serious ones caused by the machines. The logical follow up is to ask whether the gadgets cannot cause less serious injuries. Yes, they can and, therefore, E wins by practical rationale. Capable of or able to is just a ploy, as I see it.
Of course, The OE throws no more light than announcing the OA, as we often tend to say in our explanations that X is better than Y, rather than say why.
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Originally posted by daagh on 30 Dec 2015, 05:47.
Last edited by daagh on 05 May 2017, 00:29, edited 2 times in total.
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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30 Dec 2015, 06:00
Hi daagh
The meaning is clear, I got it now! Thanks
daagh wrote:
This is a comparison between injuries caused by kitchen gadgets and wood planning machines. A says that the gadgets can cause those serious injuries as caused by the planning machines. Whereas E says that, the gadgets can cause many injuries including as the serious ones caused by the machines. The logical follow up is to ask whether the gadgets cannot cause less serious injuries. Yes, they can and, therefore, E wins by practical rationale. Capable of or able to is just a ploy, as I see it.
Of course, The OE throws no more light than announcing the OA, as we often tend to say in our explanations that X is better than Y, rather than say why.
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30 Dec 2015, 06:05
You mean between A and D?
D has no chance since, it uses the wrong idiom ‘capable to’ and 2, ‘as serious as that’ should be ‘as serious as those’
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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30 Dec 2015, 11:33
daagh wrote:
You mean between A and D?
D has no chance since, it uses the wrong idiom ‘capable to’ and 2, ‘as serious as that’ should be ‘as serious as those’
Thanks for pointing out the mistake. I actually meant between A and E; however with your previous explanation, I am clear with the meaning
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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25 Jan 2016, 23:47
Many Kitchens today are equipped with high-speed electrical gadgets, such as blenders and food processors, which are able to inflict as serious injuries as those caused by an industrial wood planning machine.
A. which are able to inflict as serious injuries as those
B. Which can inflict serious injuries such as those
C. Inflicting injuries as serious as that having been
D. Capable to inflict injuries as serious as that
E. capable of inflicting injuries as serious as those
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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27 Jan 2016, 04:33
friends, why a is wrong?
pls, help
thanks
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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27 Jan 2016, 04:45
a is wrong because, if I do not make a mistake
is correct pattern
we do not have plural noun between as ... as
this point of grammar is too subtle and is purely grammatical and so is not focus of gmat sc. there is one gmatprep question which test this pattern. if this question appear on your test, are are on 750 score already.
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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28 Jan 2016, 09:42
2
thangvietnam wrote:
a is wrong because, if I do not make a mistake
is correct pattern
we do not have plural noun between as ... as
this point of grammar is too subtle and is purely grammatical and so is not focus of gmat sc. there is one gmatprep question which test this pattern. if this question appear on your test, are are on 750 score already.
I am not sure whether the point you mentioned is correct. Take this example:
It is as good a book as any.
They are as good books as any.
The second construction in plural does not seem to be wrong.
However, from concision aspect, option E is better than option A. I shall try to compare with a simpler example:
Option I: using a clause:I love football, which is the national game of Madland.
Option II: using a phrase: I love football, the national game of Madland.
Option II is obviously more concise than option I. This explanation seems to be the point in GMAC answer as mentioned by naumyak. (although the phrase used in the original sentence is an adjective phrase, not a noun phrase as is used here.)
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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Updated on: 25 Mar 2016, 11:44
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Discussing A and E:
One reason why A is not correct because it uses an unidiomatic (in the context) infinitive’ to inflict’ as though the gadgets intentionally cause the injuries. Capable of inflictive, an indicative idiom is better than the one in A.
In addition, one might note a subtle meaning error in A. The idea is to say that the gadgets can inflict injuries as serious as those caused by the woodcutters can. However, A distorts the logical intent by saying that they cause such serious injuries as those caused by the woodcutters. This implies that the gadgets can only cause injuries that are comparable in seriousness with the woodcutters and not less. Now we know the gadgets at worst cause as serious injuries as those by wood cutters and normally they are less serious.
I would say E is good.
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Originally posted by daagh on 28 Jan 2016, 09:47.
Last edited by daagh on 25 Mar 2016, 11:44, edited 1 time in total.
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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25 Mar 2016, 08:25
Why E is better than B?
Is it only because the meaning the Daagh mentioned above?
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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26 Mar 2016, 23:24
In"B" "which" is not required. Also there is a comparison. Hence "as" is correct usage not "such as.".So it is not a case where examples are required. Hence "such as" is not needed in "B."
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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07 Sep 2016, 12:06
For this question, I chose B. I was deciding b/w B and E, but thought that E was lacking a "that are" after the 'processors'. Why was "that are" not required?
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Re: Many kitchens today are equipped with high-speed electrical gadgets [#permalink]
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07 Sep 2016, 13:03
nycgirl212 wrote:
For this question, I chose B. I was deciding b/w B and E, but thought that E was lacking a "that are" after the 'processors'. Why was "that are" not required?
The latter part of the following post explains your query:
many-kitchens-today-are-equipped-212344.html#p1636552
"However, from concision aspect, option E is better than option A (or B). I shall try to compare with a simpler example:
Option I: using a clause:I love football, which is the national game of Madland.
Option II: using a phrase: I love football, the national game of Madland.
"
Re: Many kitchens today are equipped with high-speed electrical gadgets &nbs [#permalink] 07 Sep 2016, 13:03
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 4,605 | 18,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-34 | latest | en | 0.941522 |
https://math.stackexchange.com/questions/728596/fejer-kernel-is-unbounded/728641 | 1,561,480,498,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999853.94/warc/CC-MAIN-20190625152739-20190625174739-00330.warc.gz | 512,264,874 | 35,765 | # Fejer Kernel is Unbounded
Statement: Given the Fejer Kernel $F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2$. Show that $F_n(x)$ is unbounded for $x=0$ as $n\rightarrow \infty$
• why don't you just compute the limit as x -> 0? Is that the right expression for the fejer kernel? wikipedia gives something else... – Tyler Mar 27 '14 at 3:47
• That's not the Fejer kernel. You're missing some squares. – Pedro Tamaroff Mar 27 '14 at 3:54
## 2 Answers
There is a big square missing: $$F_n(x) = \frac{1}{2n\pi}\left( \frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\right)^2\sim_0 \frac{1}{2n\pi}\left( \frac{\frac{nx}{2}}{\frac{x}{2}}\right)^2=\frac n{2\pi}$$
• How did you get those approximation? Also could have I used the fact that $F_n(x) = \frac{1}{n} \frac{1-\cos(nx)}{1-\cos(x)}$ and applied L'Hopital's rule twice to see that the limit at 0 is divergent as $n \rightarrow 0$? – nonameswereavailable Mar 27 '14 at 4:33
• I get them using the fact that $\sin$ has a derivative $=1$ on $x=0$. – mookid Mar 27 '14 at 4:41
Hint
When $a$ goes to $0$, $\sin(a) \simeq a$. Then replacing each sine by its argument, for $x$ going to $0$, you arrive to $$F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2\simeq n$$ If you want to go further, you could use a Taylor expansion built at $x=0$ and obtain $$F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2\simeq n-\frac{1}{12} \left (n-1)n(n+1\right) x^2+O\left(x^3\right)$$ | 574 | 1,511 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-26 | latest | en | 0.778856 |
https://stats.stackexchange.com/questions/572948/what-justifies-the-multiplication-step-in-the-proof-of-the-front-door-adjustment?noredirect=1 | 1,721,055,515,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00635.warc.gz | 485,056,978 | 41,253 | # What justifies the multiplication step in the proof of the front-door adjustment?
$$\newcommand{\doop}{\operatorname{do}}$$ The proofs of the front-door adjustment that I've read take three steps:
1. Show $$P(M|\doop(X))$$ is identifiable
2. Show $$P(Y|\doop(M))$$ is identifiable
3. Multiply the do-free expressions for $$P(M|\doop(X))$$ and $$P(y|\doop(M))$$ to obtain $$P(Y|\doop(X))$$
where $$Y,X,M$$ meet the assumptions for the frontdoor adjustment. A graph meeting these assumptions is:
I'm sure I'm being daft here, but I don't understand what justifies simply multiplying the expressions together to get $$P(Y|\doop(X)).$$
This is like saying:
$$P(Y|\doop(X)) = P(Y|\doop(M)) \cdot P(M|\doop(X))$$
(where perhaps the assumptions for the front-door adjustment are necessary) but I don't recognize this rule in my study of causal inference.
• the law of total probability gets used.
– Ben
Commented Apr 25, 2022 at 14:37
• Thanks for the edit. I'm the worst with proper formatting! Commented Apr 25, 2022 at 15:00
$$\newcommand{\doop}{\operatorname{do}}$$It's a bit more complicated than that. As outlined in Causal Inference in Statistics: A Primer, by Pearl, Glymour, and Jewell, on p. 68, we follow this line of reasoning (variable $$Z$$ in the book changed to your $$M$$ for consistency):
First, we note that the effect of $$X$$ on $$M$$ is identifiable, since there is no backdoor path from $$X$$ to $$M.$$ Thus, we can immediately write $$P(M=m|\doop(X=x))=P(M=m|X=x).\qquad\qquad (3.12)$$ Next we note that the effect of $$M$$ on $$Y$$ is also identifiable, since the backdoor path from $$M$$ to $$Y,$$ namely $$M\leftarrow X\leftarrow U\rightarrow Y,$$ can be blocked by conditioning on $$X.$$ Thus, we can write $$P(Y=y|\doop(M=m))=\sum_x P(Y=y|M=m,X=x)\,P(X=x). \qquad(3.13)$$ Both (3.12) and (3.13) are obtained through the [backdoor, ACK] adjustment formula, the first by conditioning on the null set, and the second by adjusting for $$X.$$
We are now going to chain together the two partial effects to obtain the overall effect of $$X$$ on $$Y.$$ The reasoning goes as follows: If nature chooses to assign $$M$$ the value $$m,$$ then the probability of $$Y$$ would be $$P(Y=y|\doop(M=m)).$$ But the probability that nature would choose to do that, given that we choose to set $$X$$ at $$x,$$ is $$P(M=m|\doop(X=x)).$$ Therefore, summing over all states $$m$$ of $$M,$$ we have $$P(Y=y|\doop(X=x))=\sum_mP(Y=y|\doop(M=m))\,P(M=m|\doop(X=x))\quad (3.14)$$ The terms on the right-hand side of (3.14) were evaluated in (3.12) and (3.13), and we can substitute them to obtain a $$\doop$$-free expression for $$P(Y=y|\doop(X=x)).$$ We also distinguish between the $$x$$ that appears in (3.12) and the one that appears in (3.13), the latter of which is merely an index of summation and might as well be denoted $$x'.$$ The final expression we have is \begin{align*}&P(Y=y|\doop(X=x))=\\&\sum_m\sum_{x'}P(Y=y|M=m,X=x')\,P(X=x')\,P(M=m|X=x)\qquad (3.15)\end{align*} Equation (3.15) is known as the front-door formula.
The authors use the law of total probability in writing (3.13) and again in (3.14).
• My explanation for that would be more intuitive than rigorous: you can think of interventions (do operator) as "wiggling" nodes, and seeing how the disturbance propagates through your network. If you wiggle something upstream, such as $X,$ then it will "wiggle" anything in the direction of arrows downstream (such as $M$ and $Y.$) Does that answer your question, at least a little? Commented Apr 25, 2022 at 22:19 | 1,042 | 3,544 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 42, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.899777 |
https://cses.fi/problemset/task/1083/ | 1,716,691,387,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00239.warc.gz | 153,062,814 | 2,128 | CSES - Missing Number
• Time limit: 1.00 s
• Memory limit: 512 MB
You are given all numbers between 1,2,\ldots,n except one. Your task is to find the missing number.
# Input
The first input line contains an integer n.
The second line contains n-1 numbers. Each number is distinct and between 1 and n (inclusive).
# Output
Print the missing number.
# Constraints
• 2 \le n \le 2 \cdot 10^5
# Example
Input:
5
2 3 1 5
Output:
4 | 131 | 439 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.791019 |
http://www.comissioncontent.com/anaconda-penrith-kkcmd/fe73db-in-parallelogram-lonm%2C-what-is-om | 1,627,170,504,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00504.warc.gz | 56,991,348 | 10,733 | ### in parallelogram lonm, what is om
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A method and an apparatus for providing nanometer precision motion are provided. The supplement of an angle is 12 less than 4 times the complement of the angle. Let TS =x and TQ = 10 - x, since SQ=10. 0M. Calculations include side lengths, corner angles, diagonals, height, perimeter and area of parallelograms. Rochester, N.Y., The Manas press, 1913. In parallelogram LONM, what is OM? ★★★ Correct answer to the question: The trouble t. J gets into, at school was foreshadowed before it happened. Angle TFS is congruent to angle TPQ by alternate interior angles. The circumference of a circle is 36.2 cm. Scribd è il più grande sito di social reading e publishing al mondo. OM and ON OM and ON. Correct answers: 1 question: AReserva nutritiva que se forma en la fecundación de la flor.7 Letras B. Colabora en la polinización y no son insectos tiene 6 letras. Please check your email to find the confirmation of subscription. After how much time is the spring at equilibrium length for the second time? What is the area of RSTU in square inches? proportional to the sines. parallelogram. \qquad m \angle LOM = 4x + 30^\circm∠LOM=4x+30 ∘ m, angle, L, O, M, equals, 4, x, plus, 30, degrees \qquad m \angle MON = 8x + 90^\circm∠MON=8x+90 ∘ m, angle, M, O, N, equals, geometry. There are 3 pairs of congruent angles. If you sketch the parallelogram as described there will be triangles FTS and PTQ within. MC and ON are parallel, we have OCM=p; also OM C = 180 - CMA = 180 - (a + OMC are ~ sin /3) Since the sides of the triangle of the opposite angles, we have. C.Contiene los gametos masculinos de la flor.5 Letras D. Posee un solo organo sexual y tiene 11 letras. What is the value of w? You can now earn points by answering the unanswered questions listed. What is the radius of the circle? 1. Correct answer to the question Can someone me cause this is due today me or im gonna fail. A parallelogram whose angles are all … air is forced into the canal to stop bleeding. 3. The kinematic links (312) may include a high resolution, non-contact encoder to provide position information. BinOGM" -J1. Angle FTS is congruent to angle PTQ by vertical angles. By using this website, you agree to our Cookie Policy. i only need 10 questions answered and you can choose which 10 you do - e-eduanswers.com Construct a parallelogram PQRS in which PQ is 5.2cm PR is 6.8cm QS is 8.2cm - 969013 SOLUTION: In the figure above, RSTU is a parallelogram. Memphis daily appeal. (Points : 5) 18 cm 10.5 cm 19 cm 56 cm 2. A primer of higher space (the fourth dimension) by Claude Bragdon. What is the height? Pete carr is the superintendent of a large county school district with 10 high schools. this year, pete wants to change his greeting. What you need to do is to multiply the length of the base line (i.e. You are allowed to answer only once per question. A is on the top left corner C is on the bottom right corner. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL asked Sep 22, 2018 in Class IX Maths by muskan15 ( -3,443 points) If you did not receive the letter, please check you spam folder. O Scribd é o maior site social de leitura e publicação do mundo. At the end of the first year of operations, yolandi company had $900,000 in sales and accounts receivable of$350,000. Answer: 2 question Alliteration is A)the repetition of themes and ideas. - the answers to estudyassistant.com Ayuda por favor le doy todos mis puntos. I situate the parallelogram as line DC on the bottom and line AB on the top. Parallelogram RSTU is shown below, with = 4 inches, = 7 inches, and = 5 inches. What Is The Area Of A Rhombus Whose Side Is 17 Cm And One Diagonal given isosceles triangle ABC, line AB = BC, and BD is the angle bisector of isosceles triangle ABC prove AB x DC = BC x AD Aren't the two halves congruent? 8.53,ABCD is a parallelogram and E is the mid - point of AD. A. - edu-answer.com Sides R U and S T are horizontal, with S T above, and slightly to the left of side R U. Diagonal S U, which extends from vertex S at the upper left of the parallelogram to vertex U at the lower right of the parallelogram, divides the parallelogram into two triangles, R S U and U S T; and the angle at vertex S into 2 adjacent angles R S U and U S T. be. In Fig. DC=12) by the height (let's say "h" here) of the parallelogram. 16 B. According to the invention, a parallel kinematic micromanipulator (300) is formed using at least three kinematic links (312). A OC. The only parallelogram that would have an area of 24 would have to be a rectangle of dimensions 6x4, and the further you pinch that angle labeled "C" to make is smaller and smaller, you could imagine your height diminishing to almost zero. (Points : 5) 2.5 cm 5 cm 6.5 cm 9 cm 3. Aarushi2043 Aarushi2043 Step-by-step explanation: First elaborate your question and give diagram . E.Unión de gameto masculino y femenino.6 letras. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. Feb 1, 2018 • Comment. the inside of the tooth is carefully cleaned and disinfected. COMPOSITION AND RESOLUTION OF FORCES Let the angle. FS =5 and PQ= 8. D)the repetition of sounds. Provide three pieces of evidence that foreshadowed t. J was likely to get in trouble - … and complete the. The sum of its bases is 36 cm. The perimeter of the pentagon is 40 cm. b.) Which of the following must be true A: the measure of angles R and S are equal B: side ST is parallel to side TU. Bragdon, Claude Fayette, 1866-1946. New questions in Math. Http - Free download as Word Doc (.doc / .docx), PDF File (.pdf), Text File (.txt) or read online for free. 7 cm 17 cm 24 cm 34 cm 2 See answers griffer57 griffer57 Answer: elaborate ur question bro with diagram. xyz’s management has estimated that 1.5% of sales will be uncollectible. C)the repetition of sounds at the ends of words. 17. a. C. o Then and. iinMOC. one of pete's yearly rituals is to call the parents of each high school valedictorian to congratulate them on their child's achievement and to them for investing so much in their child's education and future. An icon used to represent a menu that can be toggled by interacting with this icon. The area of a parallelogram is 342 cm2. c.) the inflamed/infected pulp is removed. Discuss; 238000004519 manufacturing process Methods 0.000 title claims description 97; 239000011347 resins Substances 0.000 claims abstract description 101; 229920005989 resins Po A 10 kg mass is attached to a 490 N/m spring and released from rest when the spring is stretched. ★★★ Correct answer to the question: 1. Calculate certain variables of a parallelogram depending on the inputs provided. A parallelogram is a quadrilateral formed by four points in which no three are collinear having four sides and corresponding four angles. David Recine, Magoosh Tutor. Correct answer to the question: Anyone up to c~hat I’m bored. Given \qquad m \angle LONm∠LONm, angle, L, O, N is a straight angle. 0C. Answer Questions and Earn Points !!! B)the implication of a word meaning. 20 C. 28 D. 35 E. 40 for the end of 2019, after the adjusting entry for bad debts was journalized, what is … A parallelogram is a quadrilateral with opposite sides parallel. OMCN. Each slanted side measures 20 cm. sail document asked in exams [volume] (Memphis, Tenn.) 1847-1886, September 23, 1860, Image 2, brought to you by University of Tennessee, and the National Digital Newspaper Program. OM and ON are the required components. Answer: 1 question Which character from american born chinese experiences a character arc over the course of the novel - the answers to simplyans.com May answer is 25cm. When a person undergoes root canal treatment or another endodontic treatment: [select all that apply] a.) | 2,130 | 7,945 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-31 | latest | en | 0.773707 |
https://anuga.anu.edu.au/changeset/6534/ | 1,713,962,702,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00169.warc.gz | 81,591,624 | 10,816 | # Changeset 6534
Ignore:
Timestamp:
Mar 17, 2009, 5:44:57 PM (15 years ago)
Message:
Added new function is_inside_triangle. This helps the fitting functions.
Location:
anuga_core/source/anuga/utilities
Files:
4 edited
Unmodified
Removed
• ## anuga_core/source/anuga/utilities/numerical_tools.py
r6438 p = num.innerproduct(v1, v2) c = num.innerproduct(v1, normal_vector(v2)) # Projection onto normal # (negative cross product) # (negative cross product) theta = safe_acos(p)
• ## anuga_core/source/anuga/utilities/polygon.py
r6223 return status, value def is_inside_triangle(point, triangle, closed=True, rtol=1.0e-6, atol=1.0e-8, check_inputs=True, verbose=False): """Determine if one point is inside a triangle This uses the barycentric method: Triangle is A, B, C Point P can then be written as P = A + alpha * (C-A) + beta * (B-A) or if we let v=P-A, v0=C-A, v1=B-A v = alpha*v0 + beta*v1 Dot this equation by v0 and v1 to get two: dot(v0, v) = alpha*dot(v0, v0) + beta*dot(v0, v1) dot(v1, v) = alpha*dot(v1, v0) + beta*dot(v1, v1) or if a_ij = dot(v_i, v_j) and b_i = dot(v_i, v) the matrix equation: a_00 a_01 alpha b_0 = a_10 a_11 beta b_1 Solving for alpha and beta yields: alpha = (b_0*a_11 - b_1*a_01)/denom beta = (b_1*a_00 - b_0*a_10)/denom with denom = a_11*a_00 - a_10*a_01 The point is in the triangle whenever alpha and beta and their sums are in the unit interval. rtol and atol will determine how close the point has to be to the edge before it is deemed on the edge. """ if check_inputs is True: triangle = ensure_numeric(triangle) point = ensure_numeric(point, num.Float) # Convert point to array of points msg = 'is_inside_triangle must be invoked with one point only' assert num.allclose(point.shape, [2]), msg # Quickly reject points that are clearly outside if point[0] < min(triangle[:,0]): return False if point[0] > max(triangle[:,0]): return False if point[1] < min(triangle[:,1]): return False if point[1] > max(triangle[:,1]): return False # Start search A = triangle[0, :] B = triangle[1, :] C = triangle[2, :] # Now check if point lies wholly inside triangle v0 = C-A v1 = B-A a00 = num.innerproduct(v0, v0) a10 = a01 = num.innerproduct(v0, v1) a11 = num.innerproduct(v1, v1) denom = a11*a00 - a01*a10 if abs(denom) > 0.0: v = point-A b0 = num.innerproduct(v0, v) b1 = num.innerproduct(v1, v) alpha = (b0*a11 - b1*a01)/denom beta = (b1*a00 - b0*a10)/denom if (alpha > 0.0) and (beta > 0.0) and (alpha+beta < 1.0): return True if closed is True: # Check if point lies on one of the edges for X, Y in [[A,B], [B,C], [C,A]]: res = _point_on_line(point[0], point[1], X[0], X[1], Y[0], Y[1], rtol, atol) if res: return True def is_inside_polygon_quick(point, polygon, closed=True, verbose=False): """Determine if one point is inside a polygon Both point and polygon are assumed to be numeric arrays or lists and no georeferencing etc or other checks will take place. As such it is faster than is_inside_polygon """ # FIXME(Ole): This function isn't being used polygon = ensure_numeric(polygon, num.Float) points = ensure_numeric(point, num.Float) # Convert point to array of points points = points[num.NewAxis, :] msg = 'is_inside_polygon must be invoked with one point only' msg += '\nI got %s and converted to %s' % (str(point), str(points.shape)) assert points.shape[0] == 1 and points.shape[1] == 2, msg indices = num.zeros(1, num.Int) count = _separate_points_by_polygon(points, polygon, indices, int(closed), int(verbose)) #indices, count = separate_points_by_polygon(points, polygon, # closed=closed, # input_checks=False, # verbose=verbose) if count > 0: return True def is_inside_polygon(point, polygon, closed=True, verbose=False): """Determine if one point is inside a polygon raise msg polygon = ensure_absolute(polygon) try: polygon = ensure_absolute(polygon) def in_and_outside_polygon(points, polygon, closed = True, verbose = False): def in_and_outside_polygon(points, polygon, closed=True, verbose=False): """Determine points inside and outside a polygon def separate_points_by_polygon(points, polygon, closed = True, verbose = False): closed=True, check_input=True, verbose=False): """Determine whether points are inside or outside a polygon regarded as belonging to the polygon (closed = True) or not (closed = False) check_input: Allows faster execution if set to False Outputs: """ #if verbose: print 'Checking input to separate_points_by_polygon' #Input checks assert isinstance(closed, bool), 'Keyword argument "closed" must be boolean' assert isinstance(verbose, bool), 'Keyword argument "verbose" must be boolean' try: points = ensure_numeric(points, num.Float) except NameError, e: raise NameError, e except: msg = 'Points could not be converted to Numeric array' raise msg #if verbose: print 'Checking input to separate_points_by_polygon 2' try: polygon = ensure_numeric(polygon, num.Float) except NameError, e: raise NameError, e except: msg = 'Polygon could not be converted to Numeric array' raise msg msg = 'Polygon array must be a 2d array of vertices' assert len(polygon.shape) == 2, msg msg = 'Polygon array must have two columns' assert polygon.shape[1] == 2, msg msg = 'Points array must be 1 or 2 dimensional.' msg += ' I got %d dimensions' %len(points.shape) assert 0 < len(points.shape) < 3, msg if len(points.shape) == 1: # Only one point was passed in. # Convert to array of points points = num.reshape(points, (1,2)) msg = 'Point array must have two columns (x,y), ' msg += 'I got points.shape[1] == %d' %points.shape[0] assert points.shape[1] == 2, msg if check_input: #Input checks assert isinstance(closed, bool), 'Keyword argument "closed" must be boolean' assert isinstance(verbose, bool), 'Keyword argument "verbose" must be boolean' try: points = ensure_numeric(points, num.Float) except NameError, e: raise NameError, e except: msg = 'Points could not be converted to Numeric array' raise msg try: polygon = ensure_numeric(polygon, num.Float) except NameError, e: raise NameError, e except: msg = 'Polygon could not be converted to Numeric array' raise msg msg = 'Polygon array must be a 2d array of vertices' assert len(polygon.shape) == 2, msg msg = 'Polygon array must have two columns' assert polygon.shape[1] == 2, msg msg = 'Points array must be 1 or 2 dimensional.' msg += ' I got %d dimensions' %len(points.shape) assert 0 < len(points.shape) < 3, msg if len(points.shape) == 1: # Only one point was passed in. # Convert to array of points points = num.reshape(points, (1,2)) msg = 'Point array must have two columns (x,y), ' msg += 'I got points.shape[1] == %d' %points.shape[0] assert points.shape[1] == 2, msg msg = 'Points array must be a 2d array. I got %s' %str(points[:30]) assert len(points.shape) == 2, msg msg = 'Points array must have two columns' assert points.shape[1] == 2, msg N = polygon.shape[0] #Number of vertices in polygon M = points.shape[0] #Number of points indices = num.zeros( M, num.Int ) msg = 'Points array must be a 2d array. I got %s' %str(points[:30]) assert len(points.shape) == 2, msg msg = 'Points array must have two columns' assert points.shape[1] == 2, msg N = polygon.shape[0] # Number of vertices in polygon M = points.shape[0] # Number of points indices = num.zeros(M, num.Int) count = _separate_points_by_polygon(points, polygon, indices, class Found(exceptions.Exception): pass polygon = ensure_numeric(polygon) point_in = False while not point_in: point = [x_delta, y_delta] #print "point",point if is_inside_polygon(point, polygon, closed=False): raise Found
• ## anuga_core/source/anuga/utilities/polygon_ext.c
r6189 int __separate_points_by_polygon(int M, // Number of points int N, // Number of polygon vertices double* points, double* polygon, long* indices, // M-Array for storage indices int closed, int verbose) { int N, // Number of polygon vertices double* points, double* polygon, long* indices, // M-Array for storage indices int closed, int verbose) { double minpx, maxpx, minpy, maxpy, x, y, px_i, py_i, px_j, py_j, rtol=0.0, atol=0.0;
• ## anuga_core/source/anuga/utilities/test_polygon.py
r6215 def test_is_inside_polygon_quick(self): """test_is_inside_polygon_quick Test fast version of of is_inside_polygon """ # Simplest case: Polygon is the unit square polygon = [[0,0], [1,0], [1,1], [0,1]] assert is_inside_polygon_quick( (0.5, 0.5), polygon ) assert not is_inside_polygon_quick( (0.5, 1.5), polygon ) assert not is_inside_polygon_quick( (0.5, -0.5), polygon ) assert not is_inside_polygon_quick( (-0.5, 0.5), polygon ) assert not is_inside_polygon_quick( (1.5, 0.5), polygon ) # Try point on borders assert is_inside_polygon_quick( (1., 0.5), polygon, closed=True) assert is_inside_polygon_quick( (0.5, 1), polygon, closed=True) assert is_inside_polygon_quick( (0., 0.5), polygon, closed=True) assert is_inside_polygon_quick( (0.5, 0.), polygon, closed=True) assert not is_inside_polygon_quick( (0.5, 1), polygon, closed=False) assert not is_inside_polygon_quick( (0., 0.5), polygon, closed=False) assert not is_inside_polygon_quick( (0.5, 0.), polygon, closed=False) assert not is_inside_polygon_quick( (1., 0.5), polygon, closed=False) def test_inside_polygon_main(self): assert not is_inside_polygon( (0.5, -0.5), polygon ) assert is_inside_polygon_quick( (0.5, 0.5), polygon ) assert is_inside_polygon_quick( (1, -0.5), polygon ) assert is_inside_polygon_quick( (1.5, 0), polygon ) assert not is_inside_polygon_quick( (0.5, 1.5), polygon ) assert not is_inside_polygon_quick( (0.5, -0.5), polygon ) # Very convoluted polygon assert num.allclose( res, [1,2,3,5,4,0] ) assert count == 3 def test_is_inside_triangle(self): # Simplest case: triangle = [[0, 0], [1, 0], [0.5, 1]] assert is_inside_triangle((0.5, 0.5), triangle) assert is_inside_triangle((0.9, 0.1), triangle) assert not is_inside_triangle((0.5, 1.5), triangle) assert not is_inside_triangle((0.5, -0.5), triangle) assert not is_inside_triangle((-0.5, 0.5), triangle) assert not is_inside_triangle((1.5, 0.5), triangle) # Try point on borders assert is_inside_triangle((0.5, 0), triangle, closed=True) assert is_inside_triangle((1, 0), triangle, closed=True) assert not is_inside_triangle((0.5, 0), triangle, closed=False) assert not is_inside_triangle((1, 0), triangle, closed=False) # Try vertices for P in triangle: assert is_inside_triangle(P, triangle, closed=True) assert not is_inside_triangle(P, triangle, closed=False) # Slightly different triangle = [[0, 0.1], [1, -0.2], [0.5, 1]] assert is_inside_triangle((0.5, 0.5), triangle) assert is_inside_triangle((0.4, 0.1), triangle) assert not is_inside_triangle((1, 1), triangle) # Try vertices for P in triangle: assert is_inside_triangle(P, triangle, closed=True) assert not is_inside_triangle(P, triangle, closed=False) def test_populate_polygon(self): for point in points: assert is_inside_polygon(point, polygon) #Very convoluted polygon assert is_inside_polygon_quick(point, polygon) # Very convoluted polygon polygon = [[0,0], [10,10], [15,5], [20, 10], [25,0], [30,10], [40,-10]] for point in points: assert is_inside_polygon(point, polygon) assert is_inside_polygon_quick(point, polygon) #------------------------------------------------------------- if __name__ == "__main__": suite = unittest.makeSuite(Test_Polygon,'test') suite = unittest.makeSuite(Test_Polygon, 'test') runner = unittest.TextTestRunner() runner.run(suite)
Note: See TracChangeset for help on using the changeset viewer. | 3,244 | 11,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-18 | latest | en | 0.59522 |
https://joaquinbarroso.com/2012/07/27/polarizability-and-hyperpolarizability-in-gaussian/?shared=email&msg=fail | 1,627,579,390,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00362.warc.gz | 336,144,187 | 51,825 | # Polarizability and Hyperpolarizability in Gaussian
Calculating both Polarizability and the Hyperpolarizability in Gaussian is actually very easy and straightforward. However, interpreting the results requires a deeper understanding of the underlying physics of such phenomena. Herein I will try to describe the most common procedures for calculating both quantities in Gaussian09 and the way to interpret the results; if possible I will also try to address some of the most usual problems associated with their calculation.
Introduction
The dipole moment of a molecule changes when is placed under a static electric field, and this change can be calculated as
pe = pe,0 + α:E + (1/2) β:EE + … (1)
where pe,0 is the dipole moment in the absence of an electric field; α is a second rank tensor called the polarizability tensor and β is the first in an infinite series of dipole hiperpolarizabilities. The molecular potential energy changes as well with the influence of an external field in the following way
U = U0 – pe.E – (1/2) α:EE – (1/6) β:EEE – … (2)
.
Route Section Keyword: Polar
This keyword requests calculation of the polarizability and, if available, hyperpolarizability for the molecule under study. This keyword is both available for DFT and HF methods. Hyperpolarizabilities are NOT available for methods that lack analytic derivatives, for example CCSD(T), QCISD, MP4 and other post Hartree-Fock methods.
Frequency dependent polarizabilities may be calculated by including CPHF=RdFreq in the route section and then specifying the frequency (expressed in Hartrees!!!) to which the calculation should be performed, after the molecule specification preceded by a blank line. Example:
```#HF/6-31G(d) Polar CPHF=RdFreq
Title Section
Charge Multiplicity
Molecular coordinates
==blank line==
0.15```
In this example 0.15 is the frequency in Hartrees to which the calculation is to be performed. By default the output file will also include the static calculation, that is, ω = 0.0. Below you can find an example of the output when the CPHF=RdFreq is employed (taken from Gaussian’s website) Notice that the second section is performed at ω = 0.1 Ha
``` SCF Polarizability for W= 0.000000:
1 2 3
1 0.482729D+01
2 0.000000D+00 0.112001D+02
3 0.000000D+00 0.000000D+00 0.165696D+02
Isotropic polarizability for W= 0.000000 10.87 Bohr**3.
SCF Polarizability for W= 0.100000:
1 2 3
1 0.491893D+01
2 0.000000D+00 0.115663D+02
3 0.000000D+00 0.000000D+00 0.171826D+02
Isotropic polarizability for W= 0.100000 11.22 Bohr**3.```
You may have noticed now that the polarizabilities are expressed in volume units (Bohr^3) and the reason is the following:
Consider the simplest case of an atom with nuclear charge Q, radius r, and subjected to an electric field, E, which creates a force QE, and displaces the nucleus by a distance d. According to Gauss’ law this latter force is given by:
(dQ^2)/(4πεr^3) = QE (Hey! WordPress! I could really use an equation editor in here!)
if the polarizability is defined by Qd/E then we can rearrange the previous equation and yield
α = 4πεr^3 which in atomic units yields volume units, r^3, since 4πε = 1. This is why polarizabilities are usually referred to as ‘polarizability volumes’.
****THIS POST IS STILL IN PROGRESS. WILL COMPLETE IT IN SHORT. SORRY FOR ANY INCONVENIENCE****
The use of double zeta quality basis sets is paramount but it also makes these calculations more time consuming. Polarization functions on the basis set functions are a requirement for good results.
As usual, please rate/comment/share this post if you found it useful or if you think someone else might find it useful. Thanks for reading!
Theoretical chemist in his early forties, in love with life and deeply in love with his woman and children. I love science, baseball, literature, movies (perhaps even in that order). I'm passionate about food and lately wines have become a major hobby. In a nutshell I'm filled with regrets but also with hope, and that is called "living".
Posted on July 27, 2012, in Computational Chemistry, Gaussian, Models, Quantum Mechanics, Science, Theoretical Chemistry, White papers and tagged , , , , , , , , . Bookmark the permalink. 32 Comments.
1. Dr. Partha Sarathi Sengupta
higly useful. Thanks
PSS
• Thank you Dr. Partha, if you have any suggestions to improve it I’ll be more than happy to read them
2. anonymous
Hello,
could you explain the calculation of dynamic polarizabilities using 6-10 imaginary frequencies to be able to calculate the casimir-polder integral and to obtain a dispersion coefficient?
• Hi,
Could you be more specific please? I don’t think you can use Gaussian to do that, not even to obtain some of the parameters but it does seem interesting. If you have more information about what you are actually trying to accomplish, please share it.
Have a nice day!
• Abd
COuld you pleas , explain which value from the output file , form polarizability we take there are several values , like 10,87 , 11,22
if i want to compute the polarizability for a molecule how i can get it from output file
thnaks
sincerely
Abd
3. Miguel
Dear Dr. Barroso,
I tried the POLAR keyword in my system’s calculation, in order to to better understand the content of this post. In the output I got the following:
Dipole=-0.2081417,-2.3928717,-0.7711764\P
olar=591.7432869,-8.6548213,411.9016695,13.0911242,7.1150627,479.34551
56\HyperPolar=254.9554434,-31.4068445,-101.616473,-71.4503737,179.6829
ole=16.7928712,-22.3113556,5.5184844,-7.7060051,-45.6514855,13.0795056
\
According to the post, I guess those values for HyperPolar are the ones that you describe as “(…) an infinite series of dipole hyperpolarizabilities”. Would you mind to explain me the meaning of those values? Are they the components of a tensor or the beta, gamma, delta, etc terms of the expresion for the change in the dipole moment?
I guess that for “dipole” those three values correspond to x,y & z values, but I am a bit confused with “Polar” and “HyperPolar”.
Thanks a lot.
Saludos cordiales
4. Melissa Lucero
Nice post. 1) Is there a way to visualize the anisotropy and the higher order polarizabilities since they are volumes? 2) What are the basis set effects/requirements for a good polarizability calculation using the PBC code, or is that not possible for periodic systems? Thank you!
• Jaafar
Dear Sir,I hope to get a detail calculation for the final total value for each of hyperpolarizability and polarizibility of ,for example, H2O molecule
Thank you very much.
J.H.Ali
• Jaafar
Dear Sir,I hope to get a detail calculation for the final total value for each of hyperpolarizability and polarizibility of ,for example, H2O molecule
Thank you very much.
J.H.Ali
5. Ronal Luis
Hello muddy joaquin teacher wants to know that command line can determine the linear and nonlinear optical properties in the gaussian 09. and where in the output file show me this information thanks .. Doctor
6. Varun
Hello, Dr. Joaquin, How can I calculate the imaginary part of Second Hyperpolarizibility using G09? Thanks in Advance.
7. kasa
how to calculate the second hyperpolarizability using G09? how do I use keyword? I dont know which value I have to take from the output to get the second hyperpolarizability? Any suggestions will be helpful
• Dear Kasa
If you are using G09 D.01 you have to use polar=gamma. The output will put a sign pointing to these gamma values and I think it says also “second hyperpolarizabilities”, they are printed along and perpendicular to the dipole moment with their corresponding beta components.
I hope this helps
• Hi
But it did not work keywords and syntax error encountered
———————————-
#cam-b3lyp/6-31g* polar=gamma test
———————————-
QPErr — A syntax error was detected in the input line.
-b3lyp/6-31g* polar=gamma test
Last state=”Pol1″
TCursr=57736 LCursr= 24
Error termination via Lnk1e in d:\g09\l1.exe at Fri Oct 24 16:46:52 2014.
Job cpu time: 0 days 0 hours 0 minutes 0.0 seconds.
File lengths (MBytes): RWF= 1 Int= 0 D2E= 0 Chk= 1
• The problem is your basis set. You cant use a * with 6-31g . Change it to 6-31g(d) and it should eork now unless you have atoms bigger than Kr for which this basis set isnt defined.
I hope this helps
• Dear Casa. The problem is in your bases set. Check it and re-define it.
8. Dear Joaquin
how to conversion Debye-Ang**3 to au?
thanks alot
9. Sudipta
Dear Joaquin,
Currently I am using gaussian 09 to compute the polarizability of a water dimer system. The input keywords those I used for the calculation are following below
%mem=2GB
%nprocshared=4
# polar pop=chelpg mp2/aug-cc-pvdz density=current
Water dimer 1
0 1
O -0.185814 -1.174947 0.766260
H -0.128551 -0.898436 1.680861
H -0.058278 -0.370255 0.263828
O 0.174705 1.105000 -0.724443
H -0.565084 1.313496 -1.294946
H 0.928218 1.065299 -1.313403
The input says that it is single point energy calculation at MP2 level, calculates atomic charge using CHELPG scheme and it also calculates the dipole moment, polarizability and higher order moment using the MP2 level electronic density.
I got the outputs of those numbers from this calculation. The important section of this output is following below
Sum of APT charges= -0.63408
Electronic spatial extent (au): = 9203.7415
Charge= 0.0000 electrons
Dipole moment (field-independent basis, Debye):
X= 2.3376 Y= 0.0000 Z= 0.1359 Tot= 2.3415
XX= -8.6863 YY= -12.4505 ZZ= -13.0426
XY= 0.0002 XZ= -34.5104 YZ= 0.0000
Traceless Quadrupole moment (field-independent basis, Debye-Ang):
XX= 2.7068 YY= -1.0574 ZZ= -1.6494
XY= 0.0002 XZ= -34.5104 YZ= 0.0000
Octapole moment (field-independent basis, Debye-Ang**2):
XXX= 831.5434 YYY= 0.0000 ZZZ= 0.9460 XYY= 39.2300
XXY= -0.0003 XXZ= 23.6280 XZZ= -14.7249 YZZ= 0.0000
YYZ= -0.7517 XYZ= 0.0004
XXXX= -8717.5986 YYYY= -14.6198 ZZZZ= -15.4568 XXXY= 0.0345
XXXZ= -4866.7421 YYYX= 0.0000 YYYZ= 0.0000 ZZZX= -27.0710
ZZZY= 0.0000 XXYY= -1586.7247 XXZZ= -1665.2841 YYZZ= -5.2021
XXYZ= 0.0004 YYXZ= -13.6156 ZZXY= -0.0001
N-N= 2.072555395512D+01 E-N=-4.020567051105D+02 KE= 1.520892110237D+02
Exact polarizability: 18.410 0.000 18.514 -0.524 0.000 18.189
Approx polarizability: 13.876 0.000 13.654 -0.898 0.000 13.732
Generate Potential Derived Charges using the Breneman model, NDens= 1.
Grid spacing= 0.300 Box extension= 2.800
NStep X,Y,Z= 98 25 24 Total possible points= 58800
Number of Points to Fit= 8009
Now, my question is what are Exact and Approx polarizabilities? Are these three numbers correspond polarizabilities along three principal directions. How are those numbers calculated. What are units of Exact and Approx polarizabilities. Please reply me.
Sincerely
10. Mohankumar V
Dear Sir,
How calculate second order hyper polarizabilities of molecule from gaussian09 output.
i was used this key word b3lyp/6-311++g(d,p) Polar=(DCSHG,Cubic) for calculation.
Yours faithfully,
V.Mohankumar
11. kaushik hatua
Using Raman keyword we could get vibrational contribution of polarizability. However is there any reference of this calculation. I think g09 use double harmonic approximation to compute this. Anharmonic terms are not included. Any reference would make it clear
12. kaushik hatua
How could I obtained polarizability and hyperpolarizability as numerical derivative of energy or dipole moment wrt applied electric field.HF DFT are implemented for analytical calculation.
13. How calculate quadrupole moment in Gaussian 09?
14. Ennaceur
Is it possible to calculate second hyperpolarizability values from the crystallographic data (cif file) ? Otherwise, it is possible to show me the method of gait of calculated from software (Gaussian 03W) that I have the chemical formula and the crystallographic data.
N.Ennaceur
ENS-Cachan-France
15. Dear Dr Barroso,
Thanks for the info. What is the best type of first order hyperpolarizability to report? I’ve seen many papers talk about average or magnitude/total, but none specifically differentiates between them.
Also, am I correct with the following descriptions of the beta ‘syntax’:
Beta(0;0,0) – static first order hyperpolarizability
Beta(-w;w,0) – frequency dependent electro-optic Pockels hyperpolarizability
Beta(-2w;w,w) – frequency dependent single harmonic generation (SHG)
I have pasted an excerpt of the input route line and from the output for clarity (dipole direction) below, sorry for the long message. I understand it is common to use the zzz, zxx and zyy values below due to them having the greatest contributions.
#n B3LYP/6-311++G(d,p) POLAR=DCSHG int=UltraFine
(Geom spec)
0.0428 (Ha)
First dipole hyperpolarizability, Beta (dipole orientation).
||, _|_ parallel and perpendicular components, (z) with respect to z axis,
vector components x,y,z. Values do not include the 1/n! factor of 1/2.
(esu units = statvolt**-1 cm**4 , SI units = C**3 m**3 J**-2)
Beta(0;0,0):
(au) (10**-30 esu) (10**-50 SI)
|| (z) -0.686327D+04 -0.592933D+02 -0.220061D+02
_|_(z) -0.228776D+04 -0.197644D+02 -0.733537D+01
x 0.000000D+00 0.000000D+00 0.000000D+00
y 0.000000D+00 0.000000D+00 0.000000D+00
z -0.343163D+05 -0.296466D+03 -0.110031D+03
|| 0.686327D+04 0.592933D+02 0.220061D+02
xxx 0.000000D+00 0.000000D+00 0.000000D+00
xxy 0.000000D+00 0.000000D+00 0.000000D+00
yxy 0.000000D+00 0.000000D+00 0.000000D+00
yyy 0.000000D+00 0.000000D+00 0.000000D+00
xxz 0.224652D+01 0.194082D-01 0.720317D-02
yxz 0.000000D+00 0.000000D+00 0.000000D+00
yyz -0.102060D+05 -0.881722D+02 -0.327243D+02
zxz 0.000000D+00 0.000000D+00 0.000000D+00
zyz 0.000000D+00 0.000000D+00 0.000000D+00
zzz -0.123498D+04 -0.106693D+02 -0.395979D+01
Beta(-w;w,0) w= 1064.6nm:
(au) (10**-30 esu) (10**-50 SI)
|| (z) -0.141076D+07 -0.121879D+05 -0.452342D+04
_|_(z) -0.259084D+07 -0.223828D+05 -0.830716D+04
x 0.000000D+00 0.000000D+00 0.000000D+00
y 0.000000D+00 0.000000D+00 0.000000D+00
z -0.705382D+07 -0.609396D+05 -0.226171D+05
|| 0.141076D+07 0.121879D+05 0.452342D+04
xxx 0.000000D+00 0.000000D+00 0.000000D+00
yxx 0.000000D+00 0.000000D+00 0.000000D+00
yyx 0.000000D+00 0.000000D+00 0.000000D+00
zxx 0.246027D+02 0.212548D+00 0.788852D-01
zyx 0.000000D+00 0.000000D+00 0.000000D+00
zzx 0.000000D+00 0.000000D+00 0.000000D+00
xxy 0.000000D+00 0.000000D+00 0.000000D+00
yxy 0.000000D+00 0.000000D+00 0.000000D+00
yyy 0.000000D+00 0.000000D+00 0.000000D+00
zxy 0.000000D+00 0.000000D+00 0.000000D+00
zyy -0.228705D+06 -0.197583D+04 -0.733311D+03
zzy 0.000000D+00 0.000000D+00 0.000000D+00
xxz 0.188076D+02 0.162483D+00 0.603040D-01
yxz 0.000000D+00 0.000000D+00 0.000000D+00
yyz -0.659044D+07 -0.569363D+05 -0.211313D+05
zxz 0.000000D+00 0.000000D+00 0.000000D+00
zyz 0.000000D+00 0.000000D+00 0.000000D+00
zzz -0.201391D+04 -0.173986D+02 -0.645733D+01
Beta(-2w;w,w) w= 1064.6nm:
(au) (10**-30 esu) (10**-50 SI)
|| (z) -0.818286D+06 -0.706935D+04 -0.262372D+04
_|_(z) -0.176355D+07 -0.152357D+05 -0.565457D+04
x 0.000000D+00 0.000000D+00 0.000000D+00
y 0.000000D+00 0.000000D+00 0.000000D+00
z -0.409143D+07 -0.353468D+05 -0.131186D+05
|| 0.818286D+06 0.706935D+04 0.262372D+04
xxx 0.000000D+00 0.000000D+00 0.000000D+00
yxx 0.000000D+00 0.000000D+00 0.000000D+00
zxx 0.138515D+02 0.119667D+00 0.444130D-01
xyx 0.000000D+00 0.000000D+00 0.000000D+00
yyx 0.000000D+00 0.000000D+00 0.000000D+00
zyx 0.000000D+00 0.000000D+00 0.000000D+00
Best Regards,
Leighton
16. Saisudhakar
Hi,
We can not calculate hyperpolarizabilities using Post-HF methods. Is it the same with DFT also? I was trying to calculate hyperpolarizabilities with B3LYP using Gaussian 09. But I could not get the values for hyperpolarizabilities.
• Edgardo
Hi, yes you can !
You have to use a line like this :
#P B3LYP 6-31G* polar=enonly density=current
The #P keyword to print the Beta tensor 10 values (assuming Kleinman symmetry for the static Betas) at the end of the job output file on the archive session.
Polar calculations in Gaussian seem to have good linear parallel performance increase up to 3 processors so you won’t gain much by using 4 real cores on Xeon or i7 processors.
Hope this helps
Edgardo
17. Saisudhakar
Hi,
How to calculate the polarizability & hyperpolarizability in presence of electric field ?
18. Saisudhakar
What is the default applied electric field in Gaussian 09 when we use the “POLAR” keyword.
19. Edgardo
Hi,
The POLAR calculations are done applying finite fields along Cartesian vectors with default electric fields : w=0 and w=0.1 atomic units
= 0.514220652×10ˆ11 volts / meter (kilogram . meters / ampere . second^3)
= 5 volts / Angstrom.
Not sure how changing this value will impact in the static results. I once tried with w=0.2 au and didn’t observe any meaningful differences for a set of molecules the size of Disperse Red 1
20. Shivaraj Maidur
Is it possible to get second hyperpolarizability (gamma) using the keyword polar=(dcshg, cubic) in the Gaussian program? If so please tell me where can I find the results in the output file. If not, please tell me how to calculate second hyperpolarizability (gamma) using Gaussian. And also please suggest me how to read the output file for these results. Any help would be grateful.
21. Yogesh E
Hi,
To evaluate vibrational contribution to NLO susceptibilities, I have done the computation for molecule using Guassian-09 software, using the following keyword
“#p B3LYP/6-311++G(d,p) freq=raman scrf=(solvent=chloroform)”.
Plz suggest me how to extract the polarizability and hyperpolarizability values from the corresponding output file.
My mail id: erandeyogesh@gmail.com
22. Anup Thomas
Good morning Dear Sir
Sir would you please explain me the finite field static beta and gamma calculations in gaussian. I also would like to calculate nlo properties as in the following paper. https://pubs.acs.org/doi/10.1021/jp046322x | 5,850 | 17,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-31 | latest | en | 0.892616 |
https://askfilo.com/physics-question-answers/in-the-given-circuit-shown-in-figure-it-is-observed-that-the-current-i-is | 1,723,761,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00351.warc.gz | 81,429,389 | 41,207 | Question
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Solving time: 3 mins
# In the given circuit shown in figure, it is observed that the current is independent of the value of resistance . Then, the resistance values must satisfy
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## Text solutionVerified
Since current is independent of the value of , it is clear that the circuit is of a balanced Wheatstone bridge. As per condition of balance, we have
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Question Text In the given circuit shown in figure, it is observed that the current is independent of the value of resistance . Then, the resistance values must satisfy Updated On Oct 17, 2023 Topic Current Electricity Subject Physics Class Class 12 Answer Type Text solution:1 Video solution: 4 Upvotes 491 Avg. Video Duration 14 min | 344 | 1,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.840713 |
http://mathhelpforum.com/calculus/23172-continuous-problem.html | 1,498,271,149,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320209.66/warc/CC-MAIN-20170624013626-20170624033626-00486.warc.gz | 261,082,705 | 10,653 | 1. ## continuous problem
could anyone help me with this problem please?
Prove that the definition of continuous function f: R to R is equivalent to the property that for any open set A in R, the inverse-image of A is open.
So I know the definition of a continuous function is that if for every x sub zero in R,and for every epsilon greater than 0, there exists delta such that the absolute value of f(x)-f(x_0) < epsilon if the absolute value of x-x_0 < delta.
I don't know what kind of open set would have the inverse image is also open.
I don't know how to approach this problem
2. Originally Posted by namelessguy
could anyone help me with this problem please?
Prove that the definition of continuous function f: R to R is equivalent to the property that for any open set A in R, the inverse-image of A is open.
So I know the definition of a continuous function is that if for every x sub zero in R,and for every epsilon greater than 0, there exists delta such that the absolute value of f(x)-f(x_0) < epsilon if the absolute value of x-x_0 < delta.
I don't know what kind of open set would have the inverse image is also open.
I don't know how to approach this problem
Say $f$ is a continous function on $\mathbb{R}$. Let $S$ be an open set we want to show $f^{-1}(S)$ is also open. Note, if $S=\{ \}$ then the proof is complete, so it is safe to say $S$ is non-empty. Now $f^{-1}(S)$ is open if and only if every point is an interior point, so we will show every point is an interior point. Let $x_0\in f^{-1}(S)$ then $f(x_0) \in S$ and so there exists a $\epsilon > 0$ so that all $y$ satisfing $|y-f(x_0)|<\epsilon$ lie in $S$. But continuity there is a $\delta >0$ so that $|x-x_0|< \delta \implies |f(x)-f(x_0)|< \epsilon$. That means by chosing $\delta$ the set of points satisfing $|x-x_0|<\delta$ all lie in $f^{-1}(S)$. Q.E.D.
3. Now say that $f$ is a function show that the inverse-image of every open set is open. We want to show it is continous. Let $x_0 \in \mathbb{R}$. For $\epsilon > 0$ consider the set $|y - f(x_0)|<\epsilon$ (meaning all $y$ that satisfy this inequality). This set, say $S$, is clearly open. So its inverse image $f^{-1}(S)$ is too open. Now $x_0 \in f^{-1}(S)$ trivially. And so by hypothesis there is a $\delta$ such that $|x-x_0|<\delta \in f^{-1}(S)$ (meaning all $x$ that satisfy this inequality). Thus, $|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon$. Thus $f$ is continous at any point $x_0$. Q.E.D. | 727 | 2,455 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 32, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-26 | longest | en | 0.912488 |
https://riddles360.com/riddle/jumping-frogs | 1,701,671,137,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100525.55/warc/CC-MAIN-20231204052342-20231204082342-00781.warc.gz | 569,571,201 | 7,049 | # Jumping Frogs
Twenty frogs are sitting on a log floating on the surface of a river. Two of them decide to jump off into the water.
How many frogs are there on the log at this moment?
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Artificial Intelligence has crushed all human records in the puzzle game “2048,” achieving a high score of 839,732 and beating the game in only 973 moves without using any undo. | 846 | 3,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-50 | longest | en | 0.981731 |
https://netlib.org/lapack/explore-html/d6/da5/dget34_8f_source.html | 1,670,527,370,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00767.warc.gz | 435,907,925 | 10,573 | LAPACK 3.11.0 LAPACK: Linear Algebra PACKage
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dget34.f
Go to the documentation of this file.
1*> \brief \b DGET34
2*
3* =========== DOCUMENTATION ===========
4*
5* Online html documentation available at
6* http://www.netlib.org/lapack/explore-html/
7*
8* Definition:
9* ===========
10*
11* SUBROUTINE DGET34( RMAX, LMAX, NINFO, KNT )
12*
13* .. Scalar Arguments ..
14* INTEGER KNT, LMAX
15* DOUBLE PRECISION RMAX
16* ..
17* .. Array Arguments ..
18* INTEGER NINFO( 2 )
19* ..
20*
21*
22*> \par Purpose:
23* =============
24*>
25*> \verbatim
26*>
27*> DGET34 tests DLAEXC, a routine for swapping adjacent blocks (either
28*> 1 by 1 or 2 by 2) on the diagonal of a matrix in real Schur form.
29*> Thus, DLAEXC computes an orthogonal matrix Q such that
30*>
31*> Q' * [ A B ] * Q = [ C1 B1 ]
32*> [ 0 C ] [ 0 A1 ]
33*>
34*> where C1 is similar to C and A1 is similar to A. Both A and C are
35*> assumed to be in standard form (equal diagonal entries and
36*> offdiagonal with differing signs) and A1 and C1 are returned with the
37*> same properties.
38*>
39*> The test code verifies these last assertions, as well as that
40*> the residual in the above equation is small.
41*> \endverbatim
42*
43* Arguments:
44* ==========
45*
46*> \param[out] RMAX
47*> \verbatim
48*> RMAX is DOUBLE PRECISION
49*> Value of the largest test ratio.
50*> \endverbatim
51*>
52*> \param[out] LMAX
53*> \verbatim
54*> LMAX is INTEGER
55*> Example number where largest test ratio achieved.
56*> \endverbatim
57*>
58*> \param[out] NINFO
59*> \verbatim
60*> NINFO is INTEGER array, dimension (2)
61*> NINFO(J) is the number of examples where INFO=J occurred.
62*> \endverbatim
63*>
64*> \param[out] KNT
65*> \verbatim
66*> KNT is INTEGER
67*> Total number of examples tested.
68*> \endverbatim
69*
70* Authors:
71* ========
72*
73*> \author Univ. of Tennessee
74*> \author Univ. of California Berkeley
75*> \author Univ. of Colorado Denver
76*> \author NAG Ltd.
77*
78*> \ingroup double_eig
79*
80* =====================================================================
81 SUBROUTINE dget34( RMAX, LMAX, NINFO, KNT )
82*
83* -- LAPACK test routine --
84* -- LAPACK is a software package provided by Univ. of Tennessee, --
85* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
86*
87* .. Scalar Arguments ..
88 INTEGER KNT, LMAX
89 DOUBLE PRECISION RMAX
90* ..
91* .. Array Arguments ..
92 INTEGER NINFO( 2 )
93* ..
94*
95* =====================================================================
96*
97* .. Parameters ..
98 DOUBLE PRECISION ZERO, HALF, ONE
99 parameter( zero = 0.0d0, half = 0.5d0, one = 1.0d0 )
100 DOUBLE PRECISION TWO, THREE
101 parameter( two = 2.0d0, three = 3.0d0 )
102 INTEGER LWORK
103 parameter( lwork = 32 )
104* ..
105* .. Local Scalars ..
106 INTEGER I, IA, IA11, IA12, IA21, IA22, IAM, IB, IC,
107 \$ IC11, IC12, IC21, IC22, ICM, INFO, J
108 DOUBLE PRECISION BIGNUM, EPS, RES, SMLNUM, TNRM
109* ..
110* .. Local Arrays ..
111 DOUBLE PRECISION Q( 4, 4 ), RESULT( 2 ), T( 4, 4 ), T1( 4, 4 ),
112 \$ VAL( 9 ), VM( 2 ), WORK( LWORK )
113* ..
114* .. External Functions ..
115 DOUBLE PRECISION DLAMCH
116 EXTERNAL dlamch
117* ..
118* .. External Subroutines ..
119 EXTERNAL dcopy, dhst01, dlabad, dlaexc
120* ..
121* .. Intrinsic Functions ..
122 INTRINSIC abs, dble, max, sign, sqrt
123* ..
124* .. Executable Statements ..
125*
126* Get machine parameters
127*
128 eps = dlamch( 'P' )
129 smlnum = dlamch( 'S' ) / eps
130 bignum = one / smlnum
131 CALL dlabad( smlnum, bignum )
132*
133* Set up test case parameters
134*
135 val( 1 ) = zero
136 val( 2 ) = sqrt( smlnum )
137 val( 3 ) = one
138 val( 4 ) = two
139 val( 5 ) = sqrt( bignum )
140 val( 6 ) = -sqrt( smlnum )
141 val( 7 ) = -one
142 val( 8 ) = -two
143 val( 9 ) = -sqrt( bignum )
144 vm( 1 ) = one
145 vm( 2 ) = one + two*eps
146 CALL dcopy( 16, val( 4 ), 0, t( 1, 1 ), 1 )
147*
148 ninfo( 1 ) = 0
149 ninfo( 2 ) = 0
150 knt = 0
151 lmax = 0
152 rmax = zero
153*
154* Begin test loop
155*
156 DO 40 ia = 1, 9
157 DO 30 iam = 1, 2
158 DO 20 ib = 1, 9
159 DO 10 ic = 1, 9
160 t( 1, 1 ) = val( ia )*vm( iam )
161 t( 2, 2 ) = val( ic )
162 t( 1, 2 ) = val( ib )
163 t( 2, 1 ) = zero
164 tnrm = max( abs( t( 1, 1 ) ), abs( t( 2, 2 ) ),
165 \$ abs( t( 1, 2 ) ) )
166 CALL dcopy( 16, t, 1, t1, 1 )
167 CALL dcopy( 16, val( 1 ), 0, q, 1 )
168 CALL dcopy( 4, val( 3 ), 0, q, 5 )
169 CALL dlaexc( .true., 2, t, 4, q, 4, 1, 1, 1, work,
170 \$ info )
171 IF( info.NE.0 )
172 \$ ninfo( info ) = ninfo( info ) + 1
173 CALL dhst01( 2, 1, 2, t1, 4, t, 4, q, 4, work, lwork,
174 \$ result )
175 res = result( 1 ) + result( 2 )
176 IF( info.NE.0 )
177 \$ res = res + one / eps
178 IF( t( 1, 1 ).NE.t1( 2, 2 ) )
179 \$ res = res + one / eps
180 IF( t( 2, 2 ).NE.t1( 1, 1 ) )
181 \$ res = res + one / eps
182 IF( t( 2, 1 ).NE.zero )
183 \$ res = res + one / eps
184 knt = knt + 1
185 IF( res.GT.rmax ) THEN
186 lmax = knt
187 rmax = res
188 END IF
189 10 CONTINUE
190 20 CONTINUE
191 30 CONTINUE
192 40 CONTINUE
193*
194 DO 110 ia = 1, 5
195 DO 100 iam = 1, 2
196 DO 90 ib = 1, 5
197 DO 80 ic11 = 1, 5
198 DO 70 ic12 = 2, 5
199 DO 60 ic21 = 2, 4
200 DO 50 ic22 = -1, 1, 2
201 t( 1, 1 ) = val( ia )*vm( iam )
202 t( 1, 2 ) = val( ib )
203 t( 1, 3 ) = -two*val( ib )
204 t( 2, 1 ) = zero
205 t( 2, 2 ) = val( ic11 )
206 t( 2, 3 ) = val( ic12 )
207 t( 3, 1 ) = zero
208 t( 3, 2 ) = -val( ic21 )
209 t( 3, 3 ) = val( ic11 )*dble( ic22 )
210 tnrm = max( abs( t( 1, 1 ) ),
211 \$ abs( t( 1, 2 ) ), abs( t( 1, 3 ) ),
212 \$ abs( t( 2, 2 ) ), abs( t( 2, 3 ) ),
213 \$ abs( t( 3, 2 ) ), abs( t( 3, 3 ) ) )
214 CALL dcopy( 16, t, 1, t1, 1 )
215 CALL dcopy( 16, val( 1 ), 0, q, 1 )
216 CALL dcopy( 4, val( 3 ), 0, q, 5 )
217 CALL dlaexc( .true., 3, t, 4, q, 4, 1, 1, 2,
218 \$ work, info )
219 IF( info.NE.0 )
220 \$ ninfo( info ) = ninfo( info ) + 1
221 CALL dhst01( 3, 1, 3, t1, 4, t, 4, q, 4,
222 \$ work, lwork, result )
223 res = result( 1 ) + result( 2 )
224 IF( info.EQ.0 ) THEN
225 IF( t1( 1, 1 ).NE.t( 3, 3 ) )
226 \$ res = res + one / eps
227 IF( t( 3, 1 ).NE.zero )
228 \$ res = res + one / eps
229 IF( t( 3, 2 ).NE.zero )
230 \$ res = res + one / eps
231 IF( t( 2, 1 ).NE.0 .AND.
232 \$ ( t( 1, 1 ).NE.t( 2,
233 \$ 2 ) .OR. sign( one, t( 1,
234 \$ 2 ) ).EQ.sign( one, t( 2, 1 ) ) ) )
235 \$ res = res + one / eps
236 END IF
237 knt = knt + 1
238 IF( res.GT.rmax ) THEN
239 lmax = knt
240 rmax = res
241 END IF
242 50 CONTINUE
243 60 CONTINUE
244 70 CONTINUE
245 80 CONTINUE
246 90 CONTINUE
247 100 CONTINUE
248 110 CONTINUE
249*
250 DO 180 ia11 = 1, 5
251 DO 170 ia12 = 2, 5
252 DO 160 ia21 = 2, 4
253 DO 150 ia22 = -1, 1, 2
254 DO 140 icm = 1, 2
255 DO 130 ib = 1, 5
256 DO 120 ic = 1, 5
257 t( 1, 1 ) = val( ia11 )
258 t( 1, 2 ) = val( ia12 )
259 t( 1, 3 ) = -two*val( ib )
260 t( 2, 1 ) = -val( ia21 )
261 t( 2, 2 ) = val( ia11 )*dble( ia22 )
262 t( 2, 3 ) = val( ib )
263 t( 3, 1 ) = zero
264 t( 3, 2 ) = zero
265 t( 3, 3 ) = val( ic )*vm( icm )
266 tnrm = max( abs( t( 1, 1 ) ),
267 \$ abs( t( 1, 2 ) ), abs( t( 1, 3 ) ),
268 \$ abs( t( 2, 2 ) ), abs( t( 2, 3 ) ),
269 \$ abs( t( 3, 2 ) ), abs( t( 3, 3 ) ) )
270 CALL dcopy( 16, t, 1, t1, 1 )
271 CALL dcopy( 16, val( 1 ), 0, q, 1 )
272 CALL dcopy( 4, val( 3 ), 0, q, 5 )
273 CALL dlaexc( .true., 3, t, 4, q, 4, 1, 2, 1,
274 \$ work, info )
275 IF( info.NE.0 )
276 \$ ninfo( info ) = ninfo( info ) + 1
277 CALL dhst01( 3, 1, 3, t1, 4, t, 4, q, 4,
278 \$ work, lwork, result )
279 res = result( 1 ) + result( 2 )
280 IF( info.EQ.0 ) THEN
281 IF( t1( 3, 3 ).NE.t( 1, 1 ) )
282 \$ res = res + one / eps
283 IF( t( 2, 1 ).NE.zero )
284 \$ res = res + one / eps
285 IF( t( 3, 1 ).NE.zero )
286 \$ res = res + one / eps
287 IF( t( 3, 2 ).NE.0 .AND.
288 \$ ( t( 2, 2 ).NE.t( 3,
289 \$ 3 ) .OR. sign( one, t( 2,
290 \$ 3 ) ).EQ.sign( one, t( 3, 2 ) ) ) )
291 \$ res = res + one / eps
292 END IF
293 knt = knt + 1
294 IF( res.GT.rmax ) THEN
295 lmax = knt
296 rmax = res
297 END IF
298 120 CONTINUE
299 130 CONTINUE
300 140 CONTINUE
301 150 CONTINUE
302 160 CONTINUE
303 170 CONTINUE
304 180 CONTINUE
305*
306 DO 300 ia11 = 1, 5
307 DO 290 ia12 = 2, 5
308 DO 280 ia21 = 2, 4
309 DO 270 ia22 = -1, 1, 2
310 DO 260 ib = 1, 5
311 DO 250 ic11 = 3, 4
312 DO 240 ic12 = 3, 4
313 DO 230 ic21 = 3, 4
314 DO 220 ic22 = -1, 1, 2
315 DO 210 icm = 5, 7
316 iam = 1
317 t( 1, 1 ) = val( ia11 )*vm( iam )
318 t( 1, 2 ) = val( ia12 )*vm( iam )
319 t( 1, 3 ) = -two*val( ib )
320 t( 1, 4 ) = half*val( ib )
321 t( 2, 1 ) = -t( 1, 2 )*val( ia21 )
322 t( 2, 2 ) = val( ia11 )*
323 \$ dble( ia22 )*vm( iam )
324 t( 2, 3 ) = val( ib )
325 t( 2, 4 ) = three*val( ib )
326 t( 3, 1 ) = zero
327 t( 3, 2 ) = zero
328 t( 3, 3 ) = val( ic11 )*
329 \$ abs( val( icm ) )
330 t( 3, 4 ) = val( ic12 )*
331 \$ abs( val( icm ) )
332 t( 4, 1 ) = zero
333 t( 4, 2 ) = zero
334 t( 4, 3 ) = -t( 3, 4 )*val( ic21 )*
335 \$ abs( val( icm ) )
336 t( 4, 4 ) = val( ic11 )*
337 \$ dble( ic22 )*
338 \$ abs( val( icm ) )
339 tnrm = zero
340 DO 200 i = 1, 4
341 DO 190 j = 1, 4
342 tnrm = max( tnrm,
343 \$ abs( t( i, j ) ) )
344 190 CONTINUE
345 200 CONTINUE
346 CALL dcopy( 16, t, 1, t1, 1 )
347 CALL dcopy( 16, val( 1 ), 0, q, 1 )
348 CALL dcopy( 4, val( 3 ), 0, q, 5 )
349 CALL dlaexc( .true., 4, t, 4, q, 4,
350 \$ 1, 2, 2, work, info )
351 IF( info.NE.0 )
352 \$ ninfo( info ) = ninfo( info ) + 1
353 CALL dhst01( 4, 1, 4, t1, 4, t, 4,
354 \$ q, 4, work, lwork,
355 \$ result )
356 res = result( 1 ) + result( 2 )
357 IF( info.EQ.0 ) THEN
358 IF( t( 3, 1 ).NE.zero )
359 \$ res = res + one / eps
360 IF( t( 4, 1 ).NE.zero )
361 \$ res = res + one / eps
362 IF( t( 3, 2 ).NE.zero )
363 \$ res = res + one / eps
364 IF( t( 4, 2 ).NE.zero )
365 \$ res = res + one / eps
366 IF( t( 2, 1 ).NE.0 .AND.
367 \$ ( t( 1, 1 ).NE.t( 2,
368 \$ 2 ) .OR. sign( one, t( 1,
369 \$ 2 ) ).EQ.sign( one, t( 2,
370 \$ 1 ) ) ) )res = res +
371 \$ one / eps
372 IF( t( 4, 3 ).NE.0 .AND.
373 \$ ( t( 3, 3 ).NE.t( 4,
374 \$ 4 ) .OR. sign( one, t( 3,
375 \$ 4 ) ).EQ.sign( one, t( 4,
376 \$ 3 ) ) ) )res = res +
377 \$ one / eps
378 END IF
379 knt = knt + 1
380 IF( res.GT.rmax ) THEN
381 lmax = knt
382 rmax = res
383 END IF
384 210 CONTINUE
385 220 CONTINUE
386 230 CONTINUE
387 240 CONTINUE
388 250 CONTINUE
389 260 CONTINUE
390 270 CONTINUE
391 280 CONTINUE
392 290 CONTINUE
393 300 CONTINUE
394*
395 RETURN
396*
397* End of DGET34
398*
399 END | 4,445 | 10,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-49 | latest | en | 0.581805 |
https://community.qlik.com/t5/QlikView-Layout-Visualizations/Can-you-do-a-different-aggregate-calculation-in-the-total-of-a/td-p/501732 | 1,548,293,490,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584431529.98/warc/CC-MAIN-20190123234228-20190124020228-00544.warc.gz | 471,075,495 | 45,683 | # QlikView Layout & Visualizations
Discussion Board for collaboration on QlikView Layout & Visualizations.
Not applicable
## Can you do a different aggregate calculation in the total of a pivot table .
Can you do a different aggregate calculation in the total of a pivot table .
I have a pivot table that has a calculation as the measure
I would like to have the total as an avg of the rows
5 Replies
Valued Contributor
## Re: Can you do a different aggregate calculation in the total of a pivot table .
=if( Dimensionality() = 1
, sum(Qty)
, avg(Qty)
)
## Re: Can you do a different aggregate calculation in the total of a pivot table .
You can use the dimensionality() function. Something like if(dimensionality()=0, avg(aggr(sum(Sales),Product)),sum(Sales)) in a pivot table that has Product as dimension and Sales as a measure for the calculations.
talk is cheap, supply exceeds demand
Valued Contributor
## Re: Can you do a different aggregate calculation in the total of a pivot table .
Hi Oliver,
Since the pivot table doesn't have the option to change the Total mode of the expressions in the properties dialog like the straight table, you will have to do that in your expressions. If it's just a simple average of the calculations you can achieve that using the AGGR function (advanced aggregation). For example:
avg(aggr(sum(Value),Dimension))
Cesar
Not applicable
## Re: Can you do a different aggregate calculation in the total of a pivot table .
my calc is a little different
=if (RegMonth =ActivityPeriod30,200,sum(Rev))
the resulting sum will always be the else condition .
Should the if condition be in set analysis . How would the calculation be written?
Not applicable
## Re: Can you do a different aggregate calculation in the total of a pivot table .
select COUNT (*) ,MONTH(c.created_at)
from tbStageF_content (nolock) c
where
creator_source=100
and year(c.created_at) =2013
and MONTH(c.created_at) >= 1
group by
MONTH(c.created_at)
order by 2 | 464 | 2,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-04 | latest | en | 0.818918 |
https://strategiccfo.com/return-investment-roi/ | 1,611,237,408,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524858.74/warc/CC-MAIN-20210121132407-20210121162407-00320.warc.gz | 575,034,249 | 16,653 | # Return on Investment (ROI) Definition
Return on investment (ROI) is the ratio of profit made in a financial year as a percentage of an investment. In other words, ROI reveals the overall benefit (return) of an investment using the gain or loss from the investment along with the cost of the investment.
## Return on Investment (ROI) Explanation
Return on investment is a useful and simple measure of how effective a company generates profits from an investment. Many firms use ROI as a convenient tool to compare the benefit of an investment with the cost of the investment. For example, if a company effectively utilizes an investment and produces gains, ROI will both be high. Whereas if a company ineffectively utilizes an investment and produces losses, ROI will be low. For investors, choosing a company with a good return on investment is important because a high ROI means that the firm is successful at using the investment to generate high returns. Investors will typically avoid an investment with a negative ROI, or if there are other investment opportunities with a positive ROI. Return on investment models are used often because the ROI ratio and inputs can be modified to fit different companies and financial situations.
Similar formulas to calculate profitability include return on equity, return on assets, and return on capital.
## How to Find Return on Investment (ROI)
The return on investment ratio calculates the percentage return (profitability) on an investment. Check out the following ROI formula:
Simple Return on Investment Ratio = (Earnings from Investment – Cost of Investment) ÷ Cost of Investment
One issue with the simple return on investment formula is that it is often used for short-term investments, so it does not account for the time value of money. Thus, it is less accurate for calculating ROI for long-term investments over one year. To measure the long term return on investment for future years, use the discounted ROI formula.
Discounted Return on Investment Ratio = Net present value of benefits ÷ Total present value of costs
= (PV Earnings from Investment – PV Cost of Investment ) ÷ PV Cost of Investment
## Return on Investment Example
For example, this year, ABC company has produced earnings of \$50,000 from an investment. The cost of the investment was \$30,000.
Simple Return on Investment Ratio = (\$50,000 – \$30,000) ÷ \$30,000 = 67%
Based on the result, we assume that ABC company has an annual percentage return on investment of 67%. The benefit (gain) was \$50,000 and the investment cost was \$30,000. | 521 | 2,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-04 | latest | en | 0.922073 |
http://www.autoclaveboiler.com/boiler-heating-surface-25squre-meter/ | 1,607,044,414,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00216.warc.gz | 96,667,213 | 9,485 | Quick inquiry:
I need the quotation of , the fuel is (not for autoclave) , the pressure is , the capacity is , use for . My Name is , my Email is , my phone number is , Please Send the detail information for me!
Keyword: Industrial Autoclave & Boiler Sale , boiler heating surface 25squre meter
# boiler heating surface 25squre meter
• PER SQUARE FOOT (METER) OF HEATING SURFACE . Firetube Boilers Watertube Boilers Boiler Heating Surface Hand fired 5 (24) 6
• ## Boilers – EM & EA
The amount of heating surface of a boiler is expressed in square meters. The larger the heating surface a boiler has, 25 tons/hr and pressures of 17.5
• Boiler has 25 kW of rated power and it's made to heat up 250 square meters of the heating surface. Peletka 25 This boiler is one of Central heating boilers
• 358 Square Feet of Heating Surface Model CB Ohio Special Boiler Mounting Piers Elapsed time meter.
• Determining Boiler Capacity from Heating Surface from heating surfaces on a FT boilers but w/WT boilers steam per square ft of heating surface for
• To calculate the boiler heating surface required; Heating surface area of a boiler is usually expressed in square feet Heating surface calculations.
• ## Exhibit 2 – Texas Department of Licensing and
Exhibit 2 TABLE 1 MINIMUM POUNDS OF STEAM PER HOUR PER SQUARE FOOT (METER) OF HEATING SURFACE Firetube Boilers Watertube Boilers Boiler Heating Surface Hand fired 5 (24)
• Model CB 15-100 HP Boilers Sound Level Meter Five Square Feet of Heating Surface per Boiler hp:
• How many square feet of heating surface You need 100W per square meter Answer There is a lot more to it than square footage. To find out the heat
• = 0.836 sq. meter: 1 square foot =10 ft² of boiler heating surface = 34.5 lb per hour evaporation 3.45 lb. evaporation from & at 212°F) per sq ft heating
• Boiler Regulations. regulation "boiler capacity" means the heat receiving surface of a boiler as specified The heat receiving surface area in square
• ## How many square feet of heating surface is equal
How many square feet of heating surface is equal to 1 Marla = 30.25 sq yards = 25.2928 sq m = 272.25 sq feet A meter is a unit of length. A square foot is a
• of boiler heating surface = 3,348 Btu (ie, 3.45 lb. evaporation from & at 212°F) per sq ft heating surface/hour \$800 per approximately 100 square feet.
• 1003A 861 25.72 94 3.65 3.0 Heating Boilers (Type H) 85% Efficiency Heat Transfer Surface Area, Heat Exchanger Volume and Heat Exchanger Water Content
• BOILER Performance & Design Heating Surface – Fireside: 250.0 Square Feet 274.0 Square Feet ( 25.45 sq. meters) ASME Construction*: Section 1, Power
• BTU/hour square foot Conversion Factors – Heat Flux Density kilocalorie/hour square meter: 1 = 0.003154590778 : kilowatt/square meter: 1 = 0.0003154590778 :
• ## Application for the Registration of Design I.
Fees for Boilers Heating Surface (square meters).. Fee Less than or equal to 23 \$. 245.00. \$. 12.25. \$. 257.25 | 772 | 2,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-50 | longest | en | 0.849425 |
https://physics.stackexchange.com/questions/768511/symmetry-in-the-method-of-images-for-a-charge-and-a-conducting-sphere | 1,716,150,562,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00120.warc.gz | 399,653,831 | 38,524 | # Symmetry in the method of images for a charge and a conducting sphere
This is a doubt on the method of images from J.D Jackson's Classical Electrodynamics first edition, Chapter 2. While trying to find the potential of the configuration of a conducting sphere and charge, he says that by symmetry, the image charge $$q'$$ must lie on the ray joining the center of the sphere and charge $$q$$, where $$q$$ is the original charge and $$q'$$ the image charge. Mathematically, one can show it by using Apollonius' circles, but what is the symmetry that the author is talking about, which ensures that the charges and centre of sphere must be collinear?
## 1 Answer
The set up is symmetric above and below the axis containing the sphere centre and charge, so the solution must also be symmetric about this axis. If we want the simplest solution of one image charge, it must lie on this axis.
More precisely, you could say there is azimuthal symmetry present (the solution should look the same if I rotate the system about the special axis).
• Also, mirror symmetry. Jun 16, 2023 at 12:10
• So can we have solutions where the charge doesn't lie on the axis? Also, we are emphasizing azimuthal symmetry here because the thing we are looking for(the potential) is a scalar quantity right? Jun 16, 2023 at 13:40 | 300 | 1,309 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-22 | latest | en | 0.934524 |
https://www.geeksforgeeks.org/program-cube-sum-first-n-natural-numbers/amp/?ref=rp | 1,607,169,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747774.97/warc/CC-MAIN-20201205104937-20201205134937-00017.warc.gz | 677,115,864 | 21,826 | # Program for cube sum of first n natural numbers
Print the sum of series 13 + 23 + 33 + 43 + …….+ n3 till n-th term.
Examples :
```Input : n = 5
Output : 225
13 + 23 + 33 + 43 + 53 = 225
Input : n = 7
Output : 784
13 + 23 + 33 + 43 + 53 +
63 + 73 = 784
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
A simple solution is to one by one add terms.
`// Simple C++ program to find sum of series ` `// with cubes of first n natural numbers ` `#include ` `using` `namespace` `std; ` ` ` `/* Returns the sum of series */` `int` `sumOfSeries(``int` `n) ` `{ ` ` ``int` `sum = 0; ` ` ``for` `(``int` `x = 1; x <= n; x++) ` ` ``sum += x * x * x; ` ` ``return` `sum; ` `} ` ` ` `// Driver Function ` `int` `main() ` `{ ` ` ``int` `n = 5; ` ` ``cout << sumOfSeries(n); ` ` ``return` `0; ` `} `
`// Simple Java program to find sum of series ` `// with cubes of first n natural numbers ` ` ` `import` `java.util.*; ` `import` `java.lang.*; ` `class` `GFG { ` ` ` ` ``/* Returns the sum of series */` ` ``public` `static` `int` `sumOfSeries(``int` `n) ` ` ``{ ` ` ``int` `sum = ``0``; ` ` ``for` `(``int` `x = ``1``; x <= n; x++) ` ` ``sum += x * x * x; ` ` ``return` `sum; ` ` ``} ` ` ` ` ``// Driver Function ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `n = ``5``; ` ` ``System.out.println(sumOfSeries(n)); ` ` ``} ` `} ` ` ` `// Code Contributed by Mohit Gupta_OMG <(0_o)> `
`# Simple Python program to find sum of series ` `# with cubes of first n natural numbers ` ` ` `# Returns the sum of series ` `def` `sumOfSeries(n): ` ` ``sum` `=` `0` ` ``for` `i ``in` `range``(``1``, n ``+` `1``): ` ` ``sum` `+``=` `i ``*` `i``*``i ` ` ` ` ``return` `sum` ` ` ` ` `# Driver Function ` `n ``=` `5` `print``(sumOfSeries(n)) ` ` ` `# Code Contributed by Mohit Gupta_OMG <(0_o)> `
`// Simple C# program to find sum of series ` `// with cubes of first n natural numbers ` `using` `System; ` ` ` `class` `GFG { ` ` ``/* Returns the sum of series */` ` ``static` `int` `sumOfSeries(``int` `n) ` ` ``{ ` ` ``int` `sum = 0; ` ` ``for` `(``int` `x = 1; x <= n; x++) ` ` ``sum += x * x * x; ` ` ``return` `sum; ` ` ``} ` ` ` ` ``// Driver Function ` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `n = 5; ` ` ``Console.Write(sumOfSeries(n)); ` ` ``} ` `} ` `// This code is contributed by ` `// Smitha Dinesh Semwal `
` `
Output :
```225
```
Time Complexity : O(n)
An efficient solution is to use direct mathematical formula which is (n ( n + 1 ) / 2) ^ 2
```For n = 5 sum by formula is
(5*(5 + 1 ) / 2)) ^ 2
= (5*6/2) ^ 2
= (15) ^ 2
= 225
For n = 7, sum by formula is
(7*(7 + 1 ) / 2)) ^ 2
= (7*8/2) ^ 2
= (28) ^ 2
= 784
```
`// A formula based C++ program to find sum ` `// of series with cubes of first n natural ` `// numbers ` `#include ` `using` `namespace` `std; ` ` ` `int` `sumOfSeries(``int` `n) ` `{ ` ` ``int` `x = (n * (n + 1) / 2); ` ` ``return` `x * x; ` `} ` ` ` `// Driver Function ` `int` `main() ` `{ ` ` ``int` `n = 5; ` ` ``cout << sumOfSeries(n); ` ` ``return` `0; ` `} `
`// A formula based Java program to find sum ` `// of series with cubes of first n natural ` `// numbers ` ` ` `import` `java.util.*; ` `import` `java.lang.*; ` `class` `GFG { ` ` ``/* Returns the sum of series */` ` ``public` `static` `int` `sumOfSeries(``int` `n) ` ` ``{ ` ` ``int` `x = (n * (n + ``1``) / ``2``); ` ` ` ` ``return` `x * x; ` ` ``} ` ` ` ` ``// Driver Function ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `n = ``5``; ` ` ``System.out.println(sumOfSeries(n)); ` ` ``} ` `} ` ` ` `// Code Contributed by Mohit Gupta_OMG <(0_o)> `
`# A formula based Python program to find sum ` `# of series with cubes of first n natural ` `# numbers ` ` ` `# Returns the sum of series ` `def` `sumOfSeries(n): ` ` ``x ``=` `(n ``*` `(n ``+` `1``) ``/` `2``) ` ` ``return` `(``int``)(x ``*` `x) ` ` ` ` ` ` ` `# Driver Function ` `n ``=` `5` `print``(sumOfSeries(n)) ` ` ` `# Code Contributed by Mohit Gupta_OMG <(0_o)> `
`// A formula based C# program to ` `// find sum of series with cubes ` `// of first n natural numbers ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ``// Returns the sum of series ` ` ``public` `static` `int` `sumOfSeries(``int` `n) ` ` ``{ ` ` ``int` `x = (n * (n + 1) / 2); ` ` ` ` ``return` `x * x; ` ` ``} ` ` ` ` ``// Driver Function ` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `n = 5; ` ` ` ` ``Console.Write(sumOfSeries(n)); ` ` ``} ` `} ` ` ` `// Code Contributed by nitin mittal. `
` `
Output:
```225
```
Time Complexity : O(1)
How does this formula work?
We can prove the formula using mathematical induction. We can easily see that the formula holds true for n = 1 and n = 2. Let this be true for n = k-1.
```Let the formula be true for n = k-1.
Sum of first (k-1) natural numbers =
[((k - 1) * k)/2]2
Sum of first k natural numbers =
= Sum of (k-1) numbers + k3
= [((k - 1) * k)/2]2 + k3
= [k2(k2 - 2k + 1) + 4k3]/4
= [k4 + 2k3 + k2]/4
= k2(k2 + 2k + 1)/4
= [k*(k+1)/2]2```
The above program causes overflow, even if result is not beyond integer limit. Like previous post, we can avoid overflow upto some extent by doing division first.
`// Efficient CPP program to find sum of cubes ` `// of first n natural numbers that avoids ` `// overflow if result is going to be withing ` `// limits. ` `#include ` `using` `namespace` `std; ` ` ` `// Returns sum of first n natural ` `// numbers ` `int` `sumOfSeries(``int` `n) ` `{ ` ` ``int` `x; ` ` ``if` `(n % 2 == 0) ` ` ``x = (n / 2) * (n + 1); ` ` ``else` ` ``x = ((n + 1) / 2) * n; ` ` ``return` `x * x; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `n = 5; ` ` ``cout << sumOfSeries(n); ` ` ``return` `0; ` `} `
`// Efficient Java program to find sum of cubes ` `// of first n natural numbers that avoids ` `// overflow if result is going to be withing ` `// limits. ` `import` `java.util.*; ` `import` `java.lang.*; ` `class` `GFG { ` ` ``/* Returns the sum of series */` ` ``public` `static` `int` `sumOfSeries(``int` `n) ` ` ``{ ` ` ``int` `x; ` ` ``if` `(n % ``2` `== ``0``) ` ` ``x = (n / ``2``) * (n + ``1``); ` ` ``else` ` ``x = ((n + ``1``) / ``2``) * n; ` ` ``return` `x * x; ` ` ``} ` ` ` ` ``// Driver Function ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `n = ``5``; ` ` ``System.out.println(sumOfSeries(n)); ` ` ``} ` `} ` `// Code Contributed by Mohit Gupta_OMG <(0_o)> `
`# Efficient Python program to find sum of cubes ` `# of first n natural numbers that avoids ` `# overflow if result is going to be withing ` `# limits. ` ` ` `# Returns the sum of series ` `def` `sumOfSeries(n): ` ` ``x ``=` `0` ` ``if` `n ``%` `2` `=``=` `0` `: ` ` ``x ``=` `(n ``/` `2``) ``*` `(n ``+` `1``) ` ` ``else``: ` ` ``x ``=` `((n ``+` `1``) ``/` `2``) ``*` `n ` ` ` ` ``return` `(``int``)(x ``*` `x) ` ` ` ` ` `# Driver Function ` `n ``=` `5` `print``(sumOfSeries(n)) ` ` ` `# Code Contributed by Mohit Gupta_OMG <(0_o)> `
`// Efficient C# program to find sum of ` `// cubes of first n natural numbers ` `// that avoids overflow if result is ` `// going to be withing limits. ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ``/* Returns the sum of series */` ` ``public` `static` `int` `sumOfSeries(``int` `n) ` ` ``{ ` ` ``int` `x; ` ` ``if` `(n % 2 == 0) ` ` ``x = (n / 2) * (n + 1); ` ` ``else` ` ``x = ((n + 1) / 2) * n; ` ` ``return` `x * x; ` ` ``} ` ` ` ` ``// Driver code ` ` ``static` `public` `void` `Main () ` ` ``{ ` ` ``int` `n = 5; ` ` ``Console.WriteLine(sumOfSeries(n)); ` ` ``} ` `} ` ` ` `// This code is contributed by Ajit. `
` `
Output:
```225
```
This article is contributed by R_Raj. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. | 3,279 | 8,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-50 | latest | en | 0.476761 |
http://reference.wolfram.com/mathematica/ref/StreamPoints.html | 1,394,197,243,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999642519/warc/CC-MAIN-20140305060722-00035-ip-10-183-142-35.ec2.internal.warc.gz | 148,917,172 | 7,919 | BUILT-IN MATHEMATICA SYMBOL
# StreamPoints
StreamPoints
is an option to StreamPlot, ListStreamPlot, and related functions that determines how many streamlines to draw.
## DetailsDetails
• The following settings can be given for StreamPoints:
• None draw no streamlines n draw at most n streamlines Automatic automatically chosen streamlines Coarse automatically chosen coarse set of streamlines Fine automatically chosen fine set of streamlines {p1,p2,...} draw streamlines through the points , , ... {{p1,g1},...} draw streamlines through using graphics directive {spec,d} specify a minimum distance between streamlines {spec,{dstart,dend}} specify minimum distances at start and end {spec,dspec,len} specify a maximum length for any streamline
• When explicit initial points are given in the form , StreamPlot and related functions effectively follow the flow of the vector field forward and backward from each point.
• In the form , the minimum distance d between streamlines is given in ordinary coordinates, and applies in all directions.
• In the form , len gives the total length of the streamlines to generate forward and backward from each explicit initial point specified. The length len is measured in ordinary coordinates.
• The parameters d and len can be specified in the form Scaled[s], representing a fraction of the diagonal of the whole plot.
## ExamplesExamplesopen allclose all
### Basic Examples (2)Basic Examples (2)
Use different numbers of streamlines:
Out[1]=
Specify streamlines that go through a set of seed points:
Out[1]=
Out[2]= | 334 | 1,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2014-10 | longest | en | 0.707812 |
http://www.java-gaming.org/index.php?topic=11240.msg89106 | 1,429,962,564,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246648338.72/warc/CC-MAIN-20150417045728-00092-ip-10-235-10-82.ec2.internal.warc.gz | 573,224,086 | 19,870 | Java-Gaming.org Hi !
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sphere - line intersection (Read 2913 times) 0 Members and 1 Guest are viewing this topic.
g666
Junior Devvie
« Posted 2005-10-24 20:45:08 »
Why doesnt this work it doesnt detect intersections when the line goes above or below the cirlce
/*
*returns true if the specified line segment intersects the specified circle
*/
public final boolean lineIntersectsCircle(
float x, float y, float x1, float y1, //line segment from (x, y) to (x1, y1)
float a, float b, float r){ //circle at (a, b) with radius r
float m = (y-y1)/(x-x1);
//find the part to be sqrted
float sqrtedPart = r*r - ((m*(x-x1)+y1)-b)*((m*(x-x1)+y1)-b);
//check for negativeness
if(sqrtedPart < 0) return false;
//work out the square root part
float sqrtPart = (float) Math.sqrt(sqrtedPart);
float intersectX1 = sqrtPart + a;
float intersectX2 = (-1*sqrtPart) + a;
//sort the x values of the line segment so x < x1
if(x > x1){
float temp = x;
x = x1;
x1 = temp;
}
//check intersection is within line segment bound
return (intersectX1 >= x && intersectX1 < x1)
|| (intersectX2 >= x && intersectX2 < x1);
}
desperately seeking sanity
JAW
Senior Devvie
Medals: 2
« Reply #1 - Posted 2005-10-25 10:23:42 »
I just dont know the maths you need without looking it up somewhere.
But I know that java.awt.geom has Line2d and Ellipse2d which support intersects() methods to check for
Intersection.
So it would make you
Line2D.Double line = new Line2D.Double(x,y,x1,y1);
Ellipse2D.Double ellipse = new Ellipse2D.Double(a-r,b-r,2*r,2*r);
return line.intersects(ellipse.getBounds());
Ellipse is created with defining a surrounding rectangle.
You could check the SourceCode of Ellipse and Line to see if you can optimize the speed of this, but in my experience, it wont be any faster than the code java classes have.
If performed very often, a mathematical soultion might be better.
There should be dozens of collision detection articles in the web, which tell you the maths or algorithms.
-JAW
Anon666
Junior Devvie
aka Abuse/AbU5e/TehJumpingJawa
« Reply #2 - Posted 2005-10-25 12:19:04 »
What you want is :-
1 2 3 4 `if(java.awt.geom.Line2D.ptSegDist(line.x1, line.y1, line.x2, line.y2, circle.x, circle.y) < circle.radius){ //collision has occured}`
Ofcourse, if you need the point of intersection the java2d geom classes are of absolutely no use what-so-ever.
Somebody should post a RFE asking for the functionality of the geom class to be expanded, to include *alot* more functionality, as they realy are barebones atm.
Orangy Tang
JGO Kernel
Medals: 57
Projects: 11
« Reply #3 - Posted 2005-10-25 13:53:35 »
Ofcourse, if you need the point of intersection the java2d geom classes are of absolutely no use what-so-ever.
Somebody should post a RFE asking for the functionality of the geom class to be expanded, to include *alot* more functionality, as they realy are barebones atm.
Thats because they're Java2d classes - unless you're drawing with them they are next to useless (they're also rather slow for the primitive intersection testing that they do have).
What you really want is a proper, seperate set of raw geometry classes. Unfortunatly there doesn't appear to be any existing libraries to do this.
[ TriangularPixels.com - Play Growth Spurt, Rescue Squad and Snowman Village ] [ Rebirth - game resource library ]
g666
Junior Devvie
« Reply #4 - Posted 2005-10-25 20:08:05 »
I would actually like to know *how* to do this rather than just calling some (slow - but thats not the point here) methods. I worked it out by solving the equations of a line and a circle simultaneously. I almost works but when one of the line ends goes above or below the cirlce it doesnt detect the intersection for some reason.
ty
desperately seeking sanity
Anon666
Junior Devvie
aka Abuse/AbU5e/TehJumpingJawa
« Reply #5 - Posted 2005-10-25 22:21:20 »
hmm, I have some code lying around at work that will do what you want - I will have a look for it tomorrow.
Incidentally you could just look at the source code for the method I highlighted above.
Junior Devvie
Java games rock!
« Reply #6 - Posted 2005-10-30 03:46:10 »
Use a parametric representation of the line, i.e.
1 2 `x = x0 + a ty = y0 + b t`
Then you can determine the parameter such that the distance to the center of the line is the smallest possible (this can be done by constructing the perpendicular line protruding from the center of the circle and intersecting our line at a right angle, or it could be done by differentiating the expression for the distance between point and line. The line and circle intersect if and only if the determined value of the parameter t lies within the bounds as specified by the terminating points of the line segment. Both things should just result in a certain formula.
You could, for example, parametrize your line like this:
1 `r = r0 + t (r1 - r0), 0 < t < 1`
where r0 and r1 are the terminating points of the line.
Junior Devvie
Java games rock!
« Reply #7 - Posted 2005-10-30 03:55:11 »
Okay some quick math gives the desired parameter value
1 2 3 4 ` -x0 dx + a dx - dy y0 + dy b T := ---------------------------- 2 2 dx + dy`
Where dx and dy are the differences in x and y, respectively, of the terminating points of the line, (a,b) is the center of the circle, and x0, y0 is one terminating point of the line. I hope I haven't made any typos while doing this. The stylish formula was made with Maple.
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https://nbviewer.jupyter.org/github/CamDavidsonPilon/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers/blob/master/Chapter4_TheGreatestTheoremNeverTold/Ch4_LawOfLargeNumbers_TFP.ipynb | 1,618,412,347,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00138.warc.gz | 511,949,393 | 693,817 | # Probabilistic Programming and Bayesian Methods for Hackers Chapter 4¶
Original content (this Jupyter notebook) created by Cam Davidson-Pilon (@Cmrn_DP)
Ported to Tensorflow Probability by Matthew McAteer (@MatthewMcAteer0), with help from the TFP team at Google ([email protected]).
Welcome to Bayesian Methods for Hackers. The full Github repository is available at github/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers. The other chapters can be found on the project's homepage. We hope you enjoy the book, and we encourage any contributions!
• Dependencies & Prerequisites
• The greatest theorem never told
• The Law of Large Numbers
• Intuition
• How do we compute $Var(Z)$ though?
• Expected values and probabilities
• What does this all have to do with Bayesian statistics?
• The Disorder of Small Numbers
• Example: Aggregated geographic data
• Example: Kaggle's U.S. Census Return Rate Challenge
• Example: How to order Reddit submissions
• Setting up the Praw Reddit API
• Register your Application on Reddit
• Reddit API Setup
• Sorting!
• But this is too slow for real-time!
• Extension to Starred rating systems
• Example: Counting Github stars
• Conclusion
• Appendix
• Exercises
• Kicker Careers Ranked by Make Percentage
• Average Household Income by Programming Language
• References
### Dependencies & Prerequisites¶
Tensorflow Probability is part of the colab default runtime, so you don't need to install Tensorflow or Tensorflow Probability if you're running this in the colab.
If you're running this notebook in Jupyter on your own machine (and you have already installed Tensorflow), you can use the following
• For the most recent nightly installation: pip3 install -q tfp-nightly
• For the most recent stable TFP release: pip3 install -q --upgrade tensorflow-probability
• For the most recent stable GPU-connected version of TFP: pip3 install -q --upgrade tensorflow-probability-gpu
• For the most recent nightly GPU-connected version of TFP: pip3 install -q tfp-nightly-gpu
Again, if you are running this in a Colab, Tensorflow and TFP are already installed
In [1]:
#@title Imports and Global Variables { display-mode: "form" }
"""
The book uses a custom matplotlibrc file, which provides the unique styles for
matplotlib plots. If executing this book, and you wish to use the book's
styling, provided are two options:
1. Overwrite your own matplotlibrc file with the rc-file provided in the
book's styles/ dir. See http://matplotlib.org/users/customizing.html
2. Also in the styles is bmh_matplotlibrc.json file. This can be used to
update the styles in only this notebook. Try running the following code:
import json
matplotlib.rcParams.update(s)
"""
!pip3 install -q praw
!pip3 install -q wget
from __future__ import absolute_import, division, print_function
#@markdown This sets the warning status (default is ignore, since this notebook runs correctly)
warning_status = "ignore" #@param ["ignore", "always", "module", "once", "default", "error"]
import warnings
warnings.filterwarnings(warning_status)
with warnings.catch_warnings():
warnings.filterwarnings(warning_status, category=DeprecationWarning)
warnings.filterwarnings(warning_status, category=UserWarning)
import numpy as np
import os
#@markdown This sets the styles of the plotting (default is styled like plots from [FiveThirtyeight.com](https://fivethirtyeight.com/))
matplotlib_style = 'fivethirtyeight' #@param ['fivethirtyeight', 'bmh', 'ggplot', 'seaborn', 'default', 'Solarize_Light2', 'classic', 'dark_background', 'seaborn-colorblind', 'seaborn-notebook']
import matplotlib.pyplot as plt; plt.style.use(matplotlib_style)
import matplotlib.axes as axes;
from matplotlib.patches import Ellipse
from mpl_toolkits.mplot3d import Axes3D
%matplotlib inline
import seaborn as sns; sns.set_context('notebook')
from IPython.core.pylabtools import figsize
#@markdown This sets the resolution of the plot outputs (retina is the highest resolution)
notebook_screen_res = 'retina' #@param ['retina', 'png', 'jpeg', 'svg', 'pdf']
%config InlineBackend.figure_format = notebook_screen_res
import tensorflow as tf
tfe = tf.contrib.eager
# Eager Execution
#@markdown Check the box below if you want to use [Eager Execution](https://www.tensorflow.org/guide/eager)
#@markdown Eager execution provides An intuitive interface, Easier debugging, and a control flow comparable to Numpy. You can read more about it on the [Google AI Blog](https://ai.googleblog.com/2017/10/eager-execution-imperative-define-by.html)
use_tf_eager = False #@param {type:"boolean"}
# Use try/except so we can easily re-execute the whole notebook.
if use_tf_eager:
try:
tf.enable_eager_execution()
except:
pass
import tensorflow_probability as tfp
tfd = tfp.distributions
tfb = tfp.bijectors
def evaluate(tensors):
"""Evaluates Tensor or EagerTensor to Numpy ndarrays.
Args:
tensors: Object of Tensor or EagerTensors; can be list, tuple,
namedtuple or combinations thereof.
Returns:
ndarrays: Object with same structure as tensors except with Tensor or
EagerTensors replaced by Numpy ndarrays.
"""
if tf.executing_eagerly():
return tf.contrib.framework.nest.pack_sequence_as(
tensors,
[t.numpy() if tf.contrib.framework.is_tensor(t) else t
for t in tf.contrib.framework.nest.flatten(tensors)])
return sess.run(tensors)
class _TFColor(object):
"""Enum of colors used in TF docs."""
red = '#F15854'
blue = '#5DA5DA'
orange = '#FAA43A'
green = '#60BD68'
pink = '#F17CB0'
brown = '#B2912F'
purple = '#B276B2'
yellow = '#DECF3F'
gray = '#4D4D4D'
def __getitem__(self, i):
return [
self.red,
self.orange,
self.green,
self.blue,
self.pink,
self.brown,
self.purple,
self.yellow,
self.gray,
][i % 9]
TFColor = _TFColor()
def session_options(enable_gpu_ram_resizing=True, enable_xla=True):
"""
Allowing the notebook to make use of GPUs if they're available.
XLA (Accelerated Linear Algebra) is a domain-specific compiler for linear
algebra that optimizes TensorFlow computations.
"""
config = tf.ConfigProto()
config.log_device_placement = True
if enable_gpu_ram_resizing:
# allow_growth=True makes it possible to connect multiple colabs to your
# GPU. Otherwise the colab malloc's all GPU ram.
config.gpu_options.allow_growth = True
if enable_xla:
# Enable on XLA. https://www.tensorflow.org/performance/xla/.
config.graph_options.optimizer_options.global_jit_level = (
tf.OptimizerOptions.ON_1)
return config
def reset_sess(config=None):
"""
Convenience function to create the TF graph & session or reset them.
"""
if config is None:
config = session_options()
global sess
tf.reset_default_graph()
try:
sess.close()
except:
pass
sess = tf.InteractiveSession(config=config)
reset_sess()
|████████████████████████████████| 133kB 5.1MB/s
|████████████████████████████████| 204kB 48.0MB/s
Building wheel for wget (setup.py) ... done
WARNING:tensorflow:
The TensorFlow contrib module will not be included in TensorFlow 2.0.
* https://github.com/tensorflow/community/blob/master/rfcs/20180907-contrib-sunset.md
* https://github.com/tensorflow/io (for I/O related ops)
If you depend on functionality not listed there, please file an issue.
## The greatest theorem never told¶
This chapter focuses on an idea that is always bouncing around our minds, but is rarely made explicit outside books devoted to statistics. In fact, we've been using this simple idea in every example thus far.
### The Law of Large Numbers¶
Let $Z_i$ be $N$ independent samples from some probability distribution. According to the Law of Large numbers, so long as the expected value $E[Z]$ is finite, the following holds,
$$\frac{1}{N} \sum_{i=1}^N Z_i \rightarrow E[ Z ], \;\;\; N \rightarrow \infty.$$
In words:
The average of a sequence of random variables from the same distribution converges to the expected value of that distribution.
This may seem like a boring result, but it will be the most useful tool you use.
### Intuition¶
If the above Law is somewhat surprising, it can be made more clear by examining a simple example.
Consider a random variable $Z$ that can take only two values, $c_1$ and $c_2$. Suppose we have a large number of samples of $Z$, denoting a specific sample $Z_i$. The Law says that we can approximate the expected value of $Z$ by averaging over all samples. Consider the average:
$$\frac{1}{N} \sum_{i=1}^N \;Z_i$$
By construction, $Z_i$ can only take on $c_1$ or $c_2$, hence we can partition the sum over these two values: \begin{align} \frac{1}{N} \sum_{i=1}^N \;Z_i & =\frac{1}{N} \big( \sum_{ Z_i = c_1}c_1 + \sum_{Z_i=c_2}c_2 \big) \\ & = c_1 \sum_{ Z_i = c_1}\frac{1}{N} + c_2 \sum_{ Z_i = c_2}\frac{1}{N} \\ & = c_1 \times \text{ (approximate frequency of c_1) } \\ & \;\;\;\;\;\;\;\;\; + c_2 \times \text{ (approximate frequency of c_2) } \\ & \approx c_1 \times P(Z = c_1) + c_2 \times P(Z = c_2 ) \\ & = E[Z] \end{align}
Equality holds in the limit, but we can get closer and closer by using more and more samples in the average. This Law holds for almost any distribution, minus some important cases we will encounter later.
### Example¶
Below is a diagram of the Law of Large numbers in action for three different sequences of Poisson random variables.
We sample sample_size = 100000 Poisson random variables with parameter $\lambda = 4.5$. (Recall the expected value of a Poisson random variable is equal to its parameter.) We calculate the average for the first $n$ samples, for $n=1$ to sample_size.
In [2]:
sample_size_ = 100000
expected_value_ = lambda_val_ = 4.5
N_samples = tf.range(start=1,
limit=sample_size_,
delta=100)
plt.figure(figsize(12.5, 4))
for k in range(3):
samples = tfd.Poisson(rate=lambda_val_).sample(sample_shape=sample_size_)
[ samples_, N_samples_ ] = evaluate([ samples, N_samples ])
partial_average_ = [ samples_[:i].mean() for i in N_samples_ ]
plt.plot( N_samples_, partial_average_, lw=1.5,label="average of $n$ samples; seq. %d"%k)
plt.plot( N_samples_, expected_value_ * np.ones_like( partial_average_),
ls = "--", label = "true expected value", c = "k" )
plt.ylim( 4.35, 4.65)
plt.title( "Convergence of the average of \n random variables to its \
expected value" )
plt.ylabel( "average of $n$ samples" )
plt.xlabel( "# of samples, $n$")
plt.legend();
Looking at the above plot, it is clear that when the sample size is small, there is greater variation in the average (compare how jagged and jumpy the average is initially, then smooths out). All three paths approach the value 4.5, but just flirt with it as $N$ gets large. Mathematicians and statistician have another name for flirting: convergence.
Another very relevant question we can ask is how quickly am I converging to the expected value? Let's plot something new. For a specific $N$, let's do the above trials thousands of times and compute how far away we are from the true expected value, on average. But wait — compute on average? This is simply the law of large numbers again! For example, we are interested in, for a specific $N$, the quantity:
$$D(N) = \sqrt{ \;E\left[\;\; \left( \frac{1}{N}\sum_{i=1}^NZ_i - 4.5 \;\right)^2 \;\;\right] \;\;}$$
The above formulae is interpretable as a distance away from the true value (on average), for some $N$. (We take the square root so the dimensions of the above quantity and our random variables are the same). As the above is an expected value, it can be approximated using the law of large numbers: instead of averaging $Z_i$, we calculate the following multiple times and average them:
$$Y_k = \left( \;\frac{1}{N}\sum_{i=1}^NZ_i - 4.5 \; \right)^2$$
By computing the above many, $N_y$, times (remember, it is random), and averaging them:
$$\frac{1}{N_Y} \sum_{k=1}^{N_Y} Y_k \rightarrow E[ Y_k ] = E\;\left[\;\; \left( \frac{1}{N}\sum_{i=1}^NZ_i - 4.5 \;\right)^2 \right]$$
Finally, taking the square root:
$$\sqrt{\frac{1}{N_Y} \sum_{k=1}^{N_Y} Y_k} \approx D(N)$$
In [3]:
N_Y = tf.constant(250) # use this many to approximate D(N)
N_array = tf.range(1000., 50000., 2500) # use this many samples in the approx. to the variance.
D_N_results = tf.zeros(tf.shape(N_array)[0])
lambda_val = tf.constant(4.5)
expected_value = tf.constant(4.5) #for X ~ Poi(lambda) , E[ X ] = lambda
[
N_Y_,
N_array_,
D_N_results_,
expected_value_,
lambda_val_,
] = evaluate([
N_Y,
N_array,
D_N_results,
expected_value,
lambda_val,
])
def D_N(n):
"""
This function approx. D_n, the average variance of using n samples.
"""
Z = tfd.Poisson(rate=lambda_val_).sample(sample_shape=(int(n), int(N_Y_)))
average_Z = tf.reduce_mean(Z, axis=0)
average_Z_ = evaluate(average_Z)
return np.sqrt(((average_Z_ - expected_value_)**2).mean())
for i,n in enumerate(N_array_):
D_N_results_[i] = D_N(n)
plt.figure(figsize(12.5, 3))
plt.xlabel( "$N$" )
plt.ylabel( "expected squared-distance \nfrom true value" )
plt.plot(N_array_, D_N_results_, lw = 3,
label="expected distance between\n\
expected value and \naverage of $N$ random variables.")
plt.plot( N_array_, np.sqrt(expected_value_)/np.sqrt(N_array_), lw = 2, ls = "--",
label = r"$\frac{\sqrt{\lambda}}{\sqrt{N}}$" )
plt.legend()
plt.title( "How 'fast' is the sample average converging? " );
As expected, the expected distance between our sample average and the actual expected value shrinks as $N$ grows large. But also notice that the rate of convergence decreases, that is, we need only 10 000 additional samples to move from 0.020 to 0.015, a difference of 0.005, but 20 000 more samples to again decrease from 0.015 to 0.010, again only a 0.005 decrease.
It turns out we can measure this rate of convergence. Above I have plotted a second line, the function $\sqrt{\lambda}/\sqrt{N}$. This was not chosen arbitrarily. In most cases, given a sequence of random variable distributed like $Z$, the rate of convergence to $E[Z]$ of the Law of Large Numbers is
$$\frac{ \sqrt{ \; Var(Z) \; } }{\sqrt{N} }$$
This is useful to know: for a given large $N$, we know (on average) how far away we are from the estimate. On the other hand, in a Bayesian setting, this can seem like a useless result: Bayesian analysis is OK with uncertainty so what's the statistical point of adding extra precise digits? Though drawing samples can be so computationally cheap that having a larger $N$ is fine too.
### How do we compute $Var(Z)$ though?¶
The variance is simply another expected value that can be approximated! Consider the following, once we have the expected value (by using the Law of Large Numbers to estimate it, denote it $\mu$), we can estimate the variance:
$$\frac{1}{N}\sum_{i=1}^N \;(Z_i - \mu)^2 \rightarrow E[ \;( Z - \mu)^2 \;] = Var( Z )$$
### Expected values and probabilities¶
There is an even less explicit relationship between expected value and estimating probabilities. Define the indicator function
$$\mathbb{1}_A(x) = \begin{cases} 1 & x \in A \\\\ 0 & else \end{cases}$$
Then, by the law of large numbers, if we have many samples $X_i$, we can estimate the probability of an event $A$, denoted $P(A)$, by:
$$\frac{1}{N} \sum_{i=1}^N \mathbb{1}_A(X_i) \rightarrow E[\mathbb{1}_A(X)] = P(A)$$
Again, this is fairly obvious after a moments thought: the indicator function is only 1 if the event occurs, so we are summing only the times the event occurs and dividing by the total number of trials (consider how we usually approximate probabilities using frequencies). For example, suppose we wish to estimate the probability that a $Z \sim Exp(.5)$ is greater than 5, and we have many samples from a $Exp(.5)$ distribution.
$$P( Z > 5 ) = \frac{1}{N}\sum_{i=1}^N \mathbb{1}_{z > 5 }(Z_i)$$
In [4]:
N = 10000
print("Probability Estimate: ", np.shape(np.where(evaluate(tfd.Exponential(rate=0.5).sample(sample_shape=N)) > 5))[1]/N )
Probability Estimate: 0.0823
### What does this all have to do with Bayesian statistics?¶
Point estimates, to be introduced in the next chapter, in Bayesian inference are computed using expected values. In more analytical Bayesian inference, we would have been required to evaluate complicated expected values represented as multi-dimensional integrals. No longer. If we can sample from the posterior distribution directly, we simply need to evaluate averages. Much easier. If accuracy is a priority, plots like the ones above show how fast you are converging. And if further accuracy is desired, just take more samples from the posterior.
When is enough enough? When can you stop drawing samples from the posterior? That is the practitioners decision, and also dependent on the variance of the samples (recall from above a high variance means the average will converge slower).
We also should understand when the Law of Large Numbers fails. As the name implies, and comparing the graphs above for small $N$, the Law is only true for large sample sizes. Without this, the asymptotic result is not reliable. Knowing in what situations the Law fails can give us confidence in how unconfident we should be. The next section deals with this issue.
### The Disorder of Small Numbers¶
The Law of Large Numbers is only valid as $N$ gets infinitely large: never truly attainable. While the law is a powerful tool, it is foolhardy to apply it liberally. Our next example illustrates this.
### Example: Aggregated geographic data¶
Often data comes in aggregated form. For instance, data may be grouped by state, county, or city level. Of course, the population numbers vary per geographic area. If the data is an average of some characteristic of each the geographic areas, we must be conscious of the Law of Large Numbers and how it can fail for areas with small populations.
We will observe this on a toy dataset. Suppose there are five thousand counties in our dataset. Furthermore, population number in each state are uniformly distributed between 100 and 1500. The way the population numbers are generated is irrelevant to the discussion, so we do not justify this. We are interested in measuring the average height of individuals per county. Unbeknownst to us, height does not vary across county, and each individual, regardless of the county he or she is currently living in, has the same distribution of what their height may be:
$$\text{height} \sim \text{Normal}(\text{mu}=150, \text{sd}=15 )$$
We aggregate the individuals at the county level, so we only have data for the average in the county. What might our dataset look like?
In [5]:
plt.figure(figsize(12.5, 4))
std_height = 15.
mean_height = 150.
n_counties = 5000
smallest_population = 100
largest_population = 1500
pop_generator = np.random.randint
norm = np.random.normal
population_ = pop_generator(smallest_population, largest_population, n_counties)
# Our strategy to vectorize this problem will be to end-to-end concatenate the
# number of draws we need. Then we'll loop over the pieces.
d = tfp.distributions.Normal(loc=mean_height, scale= 1. / std_height)
x = d.sample(np.sum(population_))
average_across_county_array = []
seen = 0
for p in population_:
average_across_county_array.append(tf.reduce_mean(x[seen:seen+p]))
seen += p
average_across_county =tf.stack(average_across_county_array)
# locate the counties with the apparently most extreme average heights.
[
average_across_county_,
i_min,
i_max
] = evaluate([
average_across_county,
tf.argmin(average_across_county),
tf.argmax(average_across_county)
])
#plot population size vs. recorded average
plt.scatter( population_, average_across_county_, alpha = 0.5, c=TFColor[6])
plt.scatter( [ population_[i_min], population_[i_max] ],
[average_across_county_[i_min], average_across_county_[i_max] ],
s = 60, marker = "o", facecolors = "none",
edgecolors = TFColor[0], linewidths = 1.5,
label="extreme heights")
plt.xlim( smallest_population, largest_population )
plt.title( "Average height vs. County Population")
plt.xlabel("County Population")
plt.ylabel("Average height in county")
plt.plot( [smallest_population, largest_population], [mean_height, mean_height], color = "k", label = "true expected \
height", ls="--" )
plt.legend(scatterpoints = 1);
What do we observe? Without accounting for population sizes we run the risk of making an enormous inference error: if we ignored population size, we would say that the county with the shortest and tallest individuals have been correctly circled. But this inference is wrong for the following reason. These two counties do not necessarily have the most extreme heights. The error results from the calculated average of smaller populations not being a good reflection of the true expected value of the population (which in truth should be $\mu =150$). The sample size/population size/$N$, whatever you wish to call it, is simply too small to invoke the Law of Large Numbers effectively.
We provide more damning evidence against this inference. Recall the population numbers were uniformly distributed over 100 to 1500. Our intuition should tell us that the counties with the most extreme population heights should also be uniformly spread over 100 to 1500, and certainly independent of the county's population. Not so. Below are the population sizes of the counties with the most extreme heights.
In [6]:
print("Population sizes of 10 'shortest' counties: ")
print(population_[ np.argsort( average_across_county_ )[:10] ], '\n')
print("Population sizes of 10 'tallest' counties: ")
print(population_[ np.argsort( -average_across_county_ )[:10] ])
Population sizes of 10 'shortest' counties:
[175 134 156 430 219 106 116 221 103 175]
Population sizes of 10 'tallest' counties:
[106 144 138 162 113 188 160 160 112 115]
Not at all uniform over 100 to 1500. This is an absolute failure of the Law of Large Numbers.
### Example: Kaggle's U.S. Census Return Rate Challenge¶
Below is data from the 2010 US census, which partitions populations beyond counties to the level of block groups (which are aggregates of city blocks or equivalents). The dataset is from a Kaggle machine learning competition some colleagues and I participated in. The objective was to predict the census letter mail-back rate of a group block, measured between 0 and 100, using census variables (median income, number of females in the block-group, number of trailer parks, average number of children etc.). Below we plot the census mail-back rate versus block group population:
In [7]:
reset_sess()
import wget
url = 'https://raw.githubusercontent.com/CamDavidsonPilon/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers/master/Chapter4_TheGreatestTheoremNeverTold/data/census_data.csv'
filename
Out[7]:
'census_data.csv'
In [8]:
plt.figure(figsize(12.5, 6.5))
delimiter= ",")
plt.scatter( data_[:,1], data_[:,0], alpha = 0.5, c=TFColor[6])
plt.title("Census mail-back rate vs Population")
plt.ylabel("Mail-back rate")
plt.xlabel("population of block-group")
plt.xlim(-100, 15e3 )
plt.ylim( -5, 105)
i_min = tf.argmin( data_[:,0] )
i_max = tf.argmax( data_[:,0] )
[ i_min_, i_max_ ] = evaluate([ i_min, i_max ])
plt.scatter( [ data_[i_min_,1], data_[i_max_, 1] ],
[ data_[i_min_,0], data_[i_max_,0] ],
s = 60, marker = "o", facecolors = "none",
edgecolors = TFColor[0], linewidths = 1.5,
label="most extreme points")
plt.legend(scatterpoints = 1);
The above is a classic phenomenon in statistics. I say classic referring to the "shape" of the scatter plot above. It follows a classic triangular form, that tightens as we increase the sample size (as the Law of Large Numbers becomes more exact).
I am perhaps overstressing the point and maybe I should have titled the book "You don't have big data problems!", but here again is an example of the trouble with small datasets, not big ones. Simply, small datasets cannot be processed using the Law of Large Numbers. Compare with applying the Law without hassle to big datasets (ex. big data). I mentioned earlier that paradoxically big data prediction problems are solved by relatively simple algorithms. The paradox is partially resolved by understanding that the Law of Large Numbers creates solutions that are stable, i.e. adding or subtracting a few data points will not affect the solution much. On the other hand, adding or removing data points to a small dataset can create very different results.
For further reading on the hidden dangers of the Law of Large Numbers, I would highly recommend the excellent manuscript The Most Dangerous Equation.
### Example: How to order Reddit submissions¶
You may have disagreed with the original statement that the Law of Large numbers is known to everyone, but only implicitly in our subconscious decision making. Consider ratings on online products: how often do you trust an average 5-star rating if there is only 1 reviewer? 2 reviewers? 3 reviewers? We implicitly understand that with such few reviewers that the average rating is not a good reflection of the true value of the product.
This has created flaws in how we sort items, and more generally, how we compare items. Many people have realized that sorting online search results by their rating, whether the objects be books, videos, or online comments, return poor results. Often the seemingly top videos or comments have perfect ratings only from a few enthusiastic fans, and truly more quality videos or comments are hidden in later pages with falsely-substandard ratings of around 4.8. How can we correct this?
Consider the popular site Reddit (I purposefully did not link to the website as you would never come back). The site hosts links to stories or images, called submissions, for people to comment on. Redditors can vote up or down on each submission (called upvotes and downvotes). Reddit, by default, will sort submissions to a given subreddit by Hot, that is, the submissions that have the most upvotes recently.
How would you determine which submissions are the best? There are a number of ways to achieve this:
1. Popularity: A submission is considered good if it has many upvotes. A problem with this model is that a submission with hundreds of upvotes, but thousands of downvotes. While being very popular, the submission is likely more controversial than best.
2. Difference: Using the difference of upvotes and downvotes. This solves the above problem, but fails when we consider the temporal nature of submission. Depending on when a submission is posted, the website may be experiencing high or low traffic. The difference method will bias the Top submissions to be the those made during high traffic periods, which have accumulated more upvotes than submissions that were not so graced, but are not necessarily the best.
3. Time adjusted: Consider using Difference divided by the age of the submission. This creates a rate, something like difference per second, or per minute. An immediate counter-example is, if we use per second, a 1 second old submission with 1 upvote would be better than a 100 second old submission with 99 upvotes. One can avoid this by only considering at least t second old submission. But what is a good t value? Does this mean no submission younger than t is good? We end up comparing unstable quantities with stable quantities (young vs. old submissions).
4. Ratio: Rank submissions by the ratio of upvotes to total number of votes (upvotes plus downvotes). This solves the temporal issue, such that new submissions who score well can be considered Top just as likely as older submissions, provided they have many upvotes to total votes. The problem here is that a submission with a single upvote (ratio = 1.0) will beat a submission with 999 upvotes and 1 downvote (ratio = 0.999), but clearly the latter submission is more likely to be better.
I used the phrase more likely for good reason. It is possible that the former submission, with a single upvote, is in fact a better submission than the later with 999 upvotes. The hesitation to agree with this is because we have not seen the other 999 potential votes the former submission might get. Perhaps it will achieve an additional 999 upvotes and 0 downvotes and be considered better than the latter, though not likely.
What we really want is an estimate of the true upvote ratio. Note that the true upvote ratio is not the same as the observed upvote ratio: the true upvote ratio is hidden, and we only observe upvotes vs. downvotes (one can think of the true upvote ratio as "what is the underlying probability someone gives this submission a upvote, versus a downvote"). So the 999 upvote/1 downvote submission probably has a true upvote ratio close to 1, which we can assert with confidence thanks to the Law of Large Numbers, but on the other hand we are much less certain about the true upvote ratio of the submission with only a single upvote. Sounds like a Bayesian problem to me.
One way to determine a prior on the upvote ratio is to look at the historical distribution of upvote ratios. This can be accomplished by scraping Reddit's submissions and determining a distribution. There are a few problems with this technique though:
1. Skewed data: The vast majority of submissions have very few votes, hence there will be many submissions with ratios near the extremes (see the "triangular plot" in the above Kaggle dataset), effectively skewing our distribution to the extremes. One could try to only use submissions with votes greater than some threshold. Again, problems are encountered. There is a tradeoff between number of submissions available to use and a higher threshold with associated ratio precision.
2. Biased data: Reddit is composed of different subpages, called subreddits. Two examples are r/aww, which posts pics of cute animals, and r/politics. It is very likely that the user behaviour towards submissions of these two subreddits are very different: visitors are likely friendly and affectionate in the former, and would therefore upvote submissions more, compared to the latter, where submissions are likely to be controversial and disagreed upon. Therefore not all submissions are the same.
In light of these, I think it is better to use a Uniform prior.
With our prior in place, we can find the posterior of the true upvote ratio. The Python script below will scrape the best posts from the showerthoughts community on Reddit. This is a text-only community so the title of each post is the post.
#### Setting up the Praw Reddit API¶
Use of the praw package for retrieving data from Reddit does require some private information on your Reddit account. As such, we are not releasing the secret keys and reddit account passwords that we originally used for the code cell below. Fortunately, we've provided detailed information on how to set up the next code cell with your custom information.
#### Register your Application on Reddit¶
2. Click the down arrow to the right of your name, then click the Preferences button.
1. Click the app tab.
1. Click the create another app button at the bottom left of your screen.
1. Hit the create app button once you have populated all fields. You should now have a script which resembles the following:
NOTE: Certain components of the reddit = praw.Reddit("BasyesianMethodsForHackers") code have been intentionally omitted. This is because praw requires a user ID for accessing Reddit. the praw function follows the following format:
reddit = praw.Reddit(client_id='PERSONAL_USE_SCRIPT_14_CHARS', \
client_secret='SECRET_KEY_27_CHARS ', \
user_agent='YOUR_APP_NAME', \
For help with creating a Reddit instance, visit https://praw.readthedocs.io/en/latest/code_overview/reddit_instance.html.
For help on configuring PRAW, visit https://praw.readthedocs.io/en/latest/getting_started/configuration.html.
In [9]:
#@title Reddit API setup
import sys
import numpy as np
from IPython.core.display import Image
import praw
reset_sess()
enter_client_id = 'ZhGqHeR1zTM9fg' #@param {type:"string"}
enter_client_secret = 'keZdvIa1Ge257NKEm3v-eGEdv8M' #@param {type:"string"}
enter_user_agent = "bayesian_app" #@param {type:"string"}
subreddit_name = "showerthoughts" #@param ["showerthoughts", "todayilearned", "worldnews", "science", "lifeprotips", "nottheonion"] {allow-input: true}
reddit = praw.Reddit(client_id=enter_client_id,
client_secret=enter_client_secret,
user_agent=enter_user_agent,
subreddit = reddit.subreddit(subreddit_name)
# go by timespan - 'hour', 'day', 'week', 'month', 'year', 'all'
# might need to go longer than an hour to get entries...
timespan = 'day' #@param ['hour', 'day', 'week', 'month', 'year', 'all']
top_submissions = subreddit.top(timespan)
#adding a number to the inside of int() call will get the ith top post.
ith_top_post = 2 #@param {type:"number"}
n_sub = int(ith_top_post)
i = 0
while i < n_sub:
top_submission = next(top_submissions)
i += 1
top_post = top_submission.title
contents = []
for sub in top_submissions:
try:
ratio = sub.upvote_ratio
ups = int(round((ratio*sub.score)/(2*ratio - 1))
if ratio != 0.5 else round(sub.score/2))
contents.append(sub.title)
except Exception as e:
continue
print("Post contents: \n")
print(top_post)
Post contents:
If giraffes were stealthy meat-eating predators with a taste for human flesh, they would stalk apartment complexes at night and wait for people to walk by their open windows.
Above is the top post as well as some other sample posts:
In [10]:
"""
contents: an array of the text from the last 100 top submissions to a subreddit
"""
submissions = tfd.Uniform(low=float(0.), high=float(n_submissions_)).sample(sample_shape=(4))
submissions_ = evaluate(tf.cast(submissions,tf.int32))
print("Some Submissions (out of %d total) \n-----------"%n_submissions_)
for i in submissions_:
print('"' + contents[i] + '"')
Some Submissions (out of 98 total)
-----------
"You are free to commit crime but you just have to face the consequences."
"You can burn yourself with water."
"Telling someone that she was made for you goes from romantic to selfish the more you think about it."
"Just need to flip the Flat Earth over so we can chill on the cool side for awhile"
For a given true upvote ratio $p$ and $N$ votes, the number of upvotes will look like a Binomial random variable with parameters $p$ and $N$. (This is because of the equivalence between upvote ratio and probability of upvoting versus downvoting, out of $N$ possible votes/trials). We create a function that performs Bayesian inference on $p$, for a particular submission's upvote/downvote pair.
In [0]:
def joint_log_prob(upvotes, N, test_upvote_ratio):
"""
Args:
test_upvote_ratio: hypothesized value for true value of upvote ratio
Returns:
Joint log probability optimization function to compute true upvote ratio.
"""
tfd = tfp.distributions
# use a uniform prior
rv_upvote_ratio = tfd.Uniform(name="upvote_ratio", low=0., high=1.)
rv_observations = tfd.Binomial(name="obs",
total_count=float(N),
probs=test_upvote_ratio)
return (
rv_upvote_ratio.log_prob(test_upvote_ratio)
)
in some cases we might want to run someting like an HMC for multiple, or a variable number, of inputs. Loops are common examples of this. Here we define our function for setting up an HMC that can take in different numbers of upvotes and/or downvotes.
In [0]:
def posterior_upvote_ratio(upvotes, downvotes):
reset_sess()
burnin = 5000
# Initialize the step_size. (It will be automatically adapted.)
with tf.variable_scope(tf.get_variable_scope(), reuse=tf.AUTO_REUSE):
step_size = tf.get_variable(
name='step_size',
initializer=tf.constant(0.5, dtype=tf.float32),
trainable=False,
use_resource=True
)
# Set the chain's start state.
initial_chain_state = [
0.5 * tf.ones([], dtype=tf.float32, name="init_upvote_ratio")
]
# Since HMC operates over unconstrained space, we need to transform the
# samples so they live in real-space.
unconstraining_bijectors = [
tfp.bijectors.Sigmoid()
]
# Define a closure over our joint_log_prob.
unnormalized_posterior_log_prob = lambda *args: joint_log_prob(upvotes, N, *args)
hmc=tfp.mcmc.TransformedTransitionKernel(
inner_kernel=tfp.mcmc.HamiltonianMonteCarlo(
target_log_prob_fn=unnormalized_posterior_log_prob,
num_leapfrog_steps=2,
step_size=step_size,
bijector=unconstraining_bijectors)
)
# Sample from the chain.
[
posterior_upvote_ratio
], kernel_results = tfp.mcmc.sample_chain(
num_results=20000,
num_burnin_steps=burnin,
current_state=initial_chain_state,
kernel=hmc)
# Initialize any created variables.
init_g = tf.global_variables_initializer()
init_l = tf.local_variables_initializer()
evaluate(init_g)
evaluate(init_l)
return evaluate([
posterior_upvote_ratio,
kernel_results,
])
In [13]:
plt.figure(figsize(11., 8))
posteriors = []
colours = ["#5DA5DA", "#F15854", "#B276B2", "#60BD68", "#F17CB0"]
for i in range(len(submissions_)):
j = submissions_[i]
plt.hist( posteriors[i], bins = 10, normed = True, alpha = .9,
histtype="step",color = colours[i], lw = 3,
plt.hist( posteriors[i], bins = 10, normed = True, alpha = .2,
histtype="stepfilled",color = colours[i], lw = 3, )
plt.legend(loc="upper left")
plt.xlim( 0, 1)
plt.title("Posterior distributions of upvote ratios on different submissions");
WARNING:tensorflow:From /usr/local/lib/python3.6/dist-packages/tensorflow_probability/python/distributions/uniform.py:182: add_dispatch_support.<locals>.wrapper (from tensorflow.python.ops.array_ops) is deprecated and will be removed in a future version.
Instructions for updating:
Use tf.where in 2.0, which has the same broadcast rule as np.where
Some distributions are very tight, others have very long tails (relatively speaking), expressing our uncertainty with what the true upvote ratio might be.
#### Sorting!¶
We have been ignoring the goal of this exercise: how do we sort the submissions from best to worst? Of course, we cannot sort distributions, we must sort scalar numbers. There are many ways to distill a distribution down to a scalar: expressing the distribution through its expected value, or mean, is one way. Choosing the mean is a bad choice though. This is because the mean does not take into account the uncertainty of distributions.
I suggest using the 95% least plausible value, defined as the value such that there is only a 5% chance the true parameter is lower (think of the lower bound on the 95% credible region). Below are the posterior distributions with the 95% least-plausible value plotted:
In [14]:
N = posteriors[0].shape[0]
lower_limits = []
for i in range(len(submissions_)):
j = submissions_[i]
plt.hist( posteriors[i], bins = 20, normed = True, alpha = .9,
histtype="step",color = colours[i], lw = 3,
plt.hist( posteriors[i], bins = 20, normed = True, alpha = .2,
histtype="stepfilled",color = colours[i], lw = 3, )
v = np.sort( posteriors[i] )[ int(0.05*N) ]
plt.vlines( v, 0, 30 , color = colours[i], linestyles = "--", linewidths=3 )
lower_limits.append(v)
plt.legend(loc="upper left")
plt.legend(loc="upper left")
plt.title("Posterior distributions of upvote ratios on different submissions");
order = np.argsort( -np.array( lower_limits ) )
print(order, lower_limits)
[2 1 3 0] [0.7836006, 0.9008505, 0.9514801, 0.8460546]
The best submissions, according to our procedure, are the submissions that are most-likely to score a high percentage of upvotes. Visually those are the submissions with the 95% least plausible value close to 1.
Why is sorting based on this quantity a good idea? By ordering by the 95% least plausible value, we are being the most conservative with what we think is best. When using the lower-bound of the 95% credible interval, we believe with high certainty that the 'true upvote ratio' is at the very least equal to this value (or greater), thereby ensuring that the best submissions are still on top. Under this ordering, we impose the following very natural properties:
1. given two submissions with the same observed upvote ratio, we will assign the submission with more votes as better (since we are more confident it has a higher ratio).
2. given two submissions with the same number of votes, we still assign the submission with more upvotes as better.
#### But this is too slow for real-time!¶
I agree, computing the posterior of every submission takes a long time, and by the time you have computed it, likely the data has changed. I delay the mathematics to the appendix, but I suggest using the following formula to compute the lower bound very fast.
$$\frac{a}{a + b} - 1.65\sqrt{ \frac{ab}{ (a+b)^2(a + b +1 ) } }$$
where \begin{align} & a = 1 + u \\ & b = 1 + d \\ \end{align} $u$ is the number of upvotes, and $d$ is the number of downvotes. The formula is a shortcut in Bayesian inference, which will be further explained in Chapter 6 when we discuss priors in more detail.
In [15]:
def intervals(u, d):
std_err = 1.65 * tf.sqrt((a * b) / ((a + b) ** 2 * (a + b + 1.)))
return (mu, std_err)
print("Approximate lower bounds:")
lb = posterior_mean - std_err
print(lb)
print("\n")
print("Top 40 Sorted according to approximate lower bounds:")
print("\n")
[ order ] = evaluate([tf.nn.top_k(lb, k=lb.shape[0], sorted=True)])
ordered_contents = []
for i, N in enumerate(order.values[:40]):
ordered_contents.append( contents[i] )
print("-------------")
Approximate lower bounds:
[0.99269927 0.99397486 0.9954533 0.99409133 0.99016213 0.9921388
0.9932303 0.9950698 0.9924482 0.9905308 0.99219704 0.9883038
0.98586273 0.98920876 0.98701847 0.9825861 0.9784089 0.98187244
0.9896374 0.98088 0.9845036 0.9757879 0.98391163 0.97651654
0.9760173 0.96944773 0.97180194 0.98139995 0.984723 0.98333776
0.97501504 0.98007494 0.9776715 0.9744575 0.97023726 0.9856403
0.97080845 0.97544485 0.9791857 0.961427 0.9745684 0.9577916
0.9674608 0.97471726 0.9576553 0.96101296 0.9536104 0.9696114
0.9635106 0.9534545 0.95889676 0.9733387 0.9485553 0.95074564
0.9599505 0.9509203 0.9458036 0.95018905 0.9485553 0.9536757
0.94073695 0.95903665 0.95053595 0.9481962 0.96678776 0.9522802
0.96910393 0.95128834 0.94423944 0.9413886 0.9619237 0.9421574
0.94563764 0.9444311 0.9643911 0.94299465 0.9511917 0.9458798
0.95053595 0.9304495 0.93613225 0.93222207 0.94465697 0.93661344
0.9331166 0.94067496 0.9303111 0.9245648 0.9280292 0.938496
0.92397803 0.93731874 0.9207582 0.9331846 0.9306861 0.9498649
0.9209626 0.9306861 ]
Top 40 Sorted according to approximate lower bounds:
4750 648 Dragons would probably fear us as we can create water in our mouth
-------------
3983 254 Having a car the same age as you is lame. Having a car the same age as your parents is awesome.
-------------
3747 116 All pebbles seem small and insignificant until you get one get inside your shoes
-------------
2898 121 If you listen to slightly older music, you are out of the loop. But if you listen to really old music, you are listening to the classics.
-------------
-------------
2007 106 At some point in your life, you've probably made eye contact with a murderer, without even knowing it.
-------------
2208 92 The reason ghosts always make lights flash is because they are amazed by modern technology so just keep pressing the switch like a child.
-------------
2199 45 Most kids today will never have to face the awkwardness of calling a girl at home and having her dad answer the phone.
-------------
1777 74 If we could view life in the third person, we would probably spend a lot less time on our phones and watching TV because we would actually see ourselves mindlessly watching and/or scrolling.
-------------
893 28 The gasses found in Jupiter's atmosphere (ammonia, methane, etc) can cause brain damage, so those who go to Jupiter do, indeed, get more stupider.
-------------
914 19 Humans are probably the most consistently noisy animal to most other animals.
-------------
748 31 You'll never be in the same place in the universe ever again
-------------
727 46 If public libraries didn’t already exist, they’d be thought of now as the most impractical and unrealistic idea ever.
-------------
697 22 We tend to think that the knowledge we have today will be preserved forever. But actually the way we store information today is way less secure than ancient stone carvings.
-------------
618 26 A minor disadvantage to being older is that you have to scroll down further to select your birth year.
-------------
612 53 You can burn yourself with water.
-------------
544 74 We just walk around pretending it's not weird that one of our hands is better at stuff than the other.
-------------
513 39 "Pull Up" still reads as pull up backwards.
-------------
510 10 If you’re a short person, people are always looking down on you at a flattering selfie angle.
-------------
354 19 "Space" button on keyboard takes a lot of space for just a " ".
-------------
335 10 Telling someone that she was made for you goes from romantic to selfish the more you think about it.
-------------
321 28 A male mermaid would probably be a merbutler not a merman
-------------
322 10 If hydras were real, we’d probably farm them by hacking off their heads for an infinite food source
-------------
306 23 "Nice guys finish last" takes on a whole new meaning in a bedroom setting.
-------------
387 43 “You a real one” and “You are alone” is the same sentence, just with differently placed spaces
-------------
330 63 With self-driving Cars on the horizon, we’ll see some people suddenly die while in the car for any particular reason, and just arrive at their destination dead.
-------------
261 26 For adults, Halloween is a candy tax day in which children come around like irs agents to collect what you owe.
-------------
251 8 The dinosaur game is the only browser game where when your internet is better it gets laggier.
-------------
254 5 Empathy is a great emotion to have, but it can make you feel horrible
-------------
232 5 Some of the most beautiful singing voices will never be heard because the singers are too self conscious.
-------------
209 11 Songs that you like immediately usually don't last long, but songs that take a while to like usually last a long time.
-------------
207 6 We are all side characters in other people’s lives.
-------------
217 9 The reason a ghost will try to kill you is because it needs a friend.
-------------
204 11 We'll soon see gamers passing away of old age before being able to play the next installment of their favorite game franchise.
-------------
211 18 Once the internet disappears and all the books printed on acid-based paper the past 100 years decay, nothing will be left of our civilization except the highway system, skyscrapers, and billions of discarded water bottles.
-------------
193 2 Every house has an odor and it is kind of like the house’s personality.
-------------
162 9 People who are sensitive to smells could just say that they’re scentsitive
-------------
154 5 Every time you recall a memory you spend a portion of your current life reliving the life you already lived.
-------------
151 3 The worse you are the easier it is to improve.
-------------
162 20 Just need to flip the Flat Earth over so we can chill on the cool side for awhile
-------------
We can view the ordering visually by plotting the posterior mean and bounds, and sorting by the lower bound. In the plot below, notice that the left error-bar is sorted (as we suggested this is the best way to determine an ordering), so the means, indicated by dots, do not follow any strong pattern.
In [16]:
r_order = order.indices[::-1][-40:]
ratio_range_ = evaluate(tf.range( len(r_order)-1,-1,-1 ))
r_order_vals = order.values[::-1][-40:]
plt.errorbar( r_order_vals,
np.arange( len(r_order) ),
xerr=std_err[r_order], capsize=0, fmt="o",
color = TFColor[0])
plt.xlim( 0.3, 1)
plt.yticks( ratio_range_ , map( lambda x: x[:30].replace("\n",""), ordered_contents) );
` | 11,975 | 47,584 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-17 | latest | en | 0.820324 |
https://curriculum.illustrativemathematics.org/MS/students/1/8/16/index.html | 1,723,513,959,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00512.warc.gz | 144,580,159 | 27,632 | # Lesson 16
Box Plots
Let's explore how box plots can help us summarize distributions.
### 16.1: Notice and Wonder: Puppy Weights
Here are the birth weights, in ounces, of all the puppies born at a kennel in the past month.
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• 18
• 18
• 18
• 18
• 18
• 19
• 20
What do you notice and wonder about the distribution of the puppy weights?
### 16.2: Human Box Plot
Your teacher will give you the data on the lengths of names of students in your class. Write the five-number summary by finding the data set's minimum, Q1, Q2, Q3, and the maximum.
Twenty people participated in a study about blinking. The number of times each person blinked while watching a video for one minute was recorded. The data values are shown here, in order from smallest to largest.
• 3
• 6
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• 11
• 11
• 13
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• 14
• 14
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• 20
• 20
• 20
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• 32
• 36
• 51
1. Use the grid and axis to make a dot plot of this data set.
2. Find the median (Q2) and mark its location on the dot plot.
3. Find the first quartile (Q1) and the third quartile (Q3). Mark their locations on the dot plot.
4. What are the minimum and maximum values?
1. A box plot can be used to represent the five-number summary graphically. Let’s draw a box plot for the number-of-blinks data. On the grid, above the dot plot:
1. Draw a box that extends from the first quartile (Q1) to the third quartile (Q3). Label the quartiles.
2. At the median (Q2), draw a vertical line from the top of the box to the bottom of the box. Label the median.
3. From the left side of the box (Q1), draw a horizontal line (a whisker) that extends to the minimum of the data set. On the right side of the box (Q3), draw a similar line that extends to the maximum of the data set.
2. You have now created a box plot to represent the number of blinks data. What fraction of the data values are represented by each of these elements of the box plot?
1. The left whisker
2. The box
3. The right whisker
Suppose there were some errors in the data set: the smallest value should have been 6 instead of 3, and the largest value should have been 41 instead of 51. Determine if any part of the five-number summary would change. If you think so, describe how it would change. If not, explain how you know.
### Summary
A box plot represents the five-number summary of a data set.
It shows the first quartile (Q1) and the third quartile (Q3) as the left and right sides of a rectangle or a box. The median (Q2) is shown as a vertical segment inside the box. On the left side, a horizontal line segment—a “whisker”—extends from Q1 to the minimum value. On the right, a whisker extends from Q3 to the maximum value.
The rectangle in the middle represents the middle half of the data. Its width is the IQR. The whiskers represent the bottom quarter and top quarter of the data set.
Earlier we saw dot plots representing the weights of pugs and beagles. The box plots for these data sets are shown above the corresponding dot plots.
We can tell from the box plots that, in general, the pugs in the group are lighter than the beagles: the median weight of pugs is 7 kilograms and the median weight of beagles is 10 kilograms. Because the two box plots are on the same scale and the rectangles have similar widths, we can also tell that the IQRs for the two breeds are very similar. This suggests that the variability in the beagle weights is very similar to the variability in the pug weights.
### Glossary Entries
• box plot
A box plot is a way to represent data on a number line. The data is divided into four sections. The sides of the box represent the first and third quartiles. A line inside the box represents the median. Lines outside the box connect to the minimum and maximum values.
For example, this box plot shows a data set with a minimum of 2 and a maximum of 15. The median is 6, the first quartile is 5, and the third quartile is 10.
• interquartile range (IQR)
The interquartile range is one way to measure how spread out a data set is. We sometimes call this the IQR. To find the interquartile range we subtract the first quartile from the third quartile.
For example, the IQR of this data set is 20 because $$50-30=20$$.
22 29 30 31 32 43 44 45 50 50 59 Q1 Q2 Q3
• median
The median is one way to measure the center of a data set. It is the middle number when the data set is listed in order.
For the data set 7, 9, 12, 13, 14, the median is 12.
For the data set 3, 5, 6, 8, 11, 12, there are two numbers in the middle. The median is the average of these two numbers. $$6+8=14$$ and $$14 \div 2 = 7$$.
• quartile
Quartiles are the numbers that divide a data set into four sections that each have the same number of values.
For example, in this data set the first quartile is 30. The second quartile is the same thing as the median, which is 43. The third quartile is 50.
22 29 30 31 32 43 44 45 50 50 59 Q1 Q2 Q3
• range
The range is the distance between the smallest and largest values in a data set. For example, for the data set 3, 5, 6, 8, 11, 12, the range is 9, because $$12-3=9$$. | 1,488 | 5,154 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.7801 |
http://8foxes.blogspot.com/2010/11/fox-315.html | 1,600,646,874,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00189.warc.gz | 2,627,426 | 22,340 | ## Wednesday, November 17, 2010
### Fox 315
Polar Fox: I have a problem with this one!
Red Fox: What's wrong with it?
Polar Fox: Nothing moves continuously?
Red Fox: Come again?!
Polar Fox: The universe is discrete! THERE IS NO CONTINUUM.
Red Fox: Then how do the things move, flow, or slide?
Polar Fox: Nothing moves! Absolutely nothing moves!!
Red Fox: But the time ticks away, no? Clock arms DO advance.
Polar Fox: No, they don't! But they die in one instance and resurrect in the next one. In between no measurable time passes. Matter oscillates between existence and non-existence continuously.
Red Fox: Continuously? Isn't that ironic?
Polar Fox: Between any two existence, there is nothing but emptiness.
Red Fox: I am having the feeling that your intelligence fall into non-existing state just now.
Polar Fox: Hold on. I think I am about to jump back into the reality. Wait a sec.
1. So we have,
x + y + z = 360 and sinx*siny*sinz
-> f(x,y) = sin(x)*sin(y)*sin(360-x-y)
-> f(x,y) = sin(x)*sin(y)*-sin(x+y)
Now to find the critical points of f(x,y), we just need to find the partial derivative with respect to x and y and solve for 0.
f_x(x,y) =
sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)
solve for 0.
sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)=0
-> tan(x) = -tan(x+y)
Since the function is symmetric, we should get the same partial derivative for y.
-> tan(y) = -tan(y+x)
-> tan(x)=tan(y)
-> x = y or they are opposites. However, if they are opposites, the original function just becomes 0. Thus, x = y.
Now substitute in x for y in the original equation and find its critical points.
Eventually, you will get sin(3x)=0
x = 120 degrees
y = 120 degrees
z = 120 degrees
2. Flawless execution! It's nice to solve this analytically.
Thank you six.
3. ah ah! may I respectfully object?
the 3 clock arms move "continuously" but not "independently". Unless otherwise stated the clock is not broken, so the 3 angles should also correspond to a correct time of day, which isn't the case for the optimum solution.
from quick informal search, the biggest figure I could get is at 06:09:50.002858 (+/-1 microsecond) where angles are 121.983, 118.983 and 119.034. This is 99,879% from optimum. But this may not be the "absolute" max.
Maybe somone could come up with an analytic solution for this max?
bleaug
4. http://www.wolframalpha.com/input/?i=max%28sinx-siny-sin%282pi-x-y%29%29
5. Bleaug,
Are you telling us that you, too, are believing in the nothing-moves theory as stated by Polar Fox. If you are assuming that Second-hand advances 6 degrees every second DISCRETELY (as most clocks do), then you may be right. Otherwise, I can't see why 120-120-120 perfect split is not possible.
Let's try this way. Suppose the clock is at 00:00:00. About 21 minutes later (say around 00:21:00) alpha is very close to 120...
Wait a second.
Wait a second.
I am thinking.
Wooow..
I think you are right Bleaug! There is a chance that when the Second-hand reaches 120 degrees (beta=120), then alpha might have already advanced beyond 120 degrees. Actually there is a chance that 120-120-120 may never be possible!
Woow, are we saying that 3 clock arms (discrete or continuous) can NEVER be equally separated???
6. my informal search above was based on wrong calculations. However the conclusion still holds.
If t is the time of day expressed in seconds, the maximum we look for must be an extremum of the function:
f(t)=sin(11.pi.t./21600).sin(708.pi.t/21600).sin(719.pi.t/21600)
some properties:
- this function has a period of 43200 = 12h which makes sense
- if T is an extremum, 43200-T is also an extremum
- extremums can be split in two sets: those which approach a 120-120-120 angle separation (max) and those which approach a 60-60-240 angle separation (min)
Based on this, the maximum is achieved for the two following timestamps:
a) 2:54:34.56169
b) 9:05:25.43831 = 12:00 - 2:54:34.56169
modulo 12:00, i.e. AM/PM of course.
This max is 99,999139% from theoretical max = 3.sqrt(3)/8
bleaug
7. @8foxes
bleaug
PS: you can call me "Polar Fox" ;-)
8. Yes, Bleaug is absolutely correct. Thank you for pointing that out.
If we have
11xmod12 (minute hand)
719xmod12 (second hand)
This should give us the distances from the hour hand, where the hour hand is just xmod12. So if we want a clock with perfect 60 degree angles between each clock, then these set of equations should be satisfied.
11x = 4mod12
719x = 8mod12
x can't be an irrational number, for then the product would be irrational and not 4mod12 or 8mod12. However, we also see that since 11 and 719 are prime, the only rational numbers that can work for this expression are n/11 and t/719, respectively. However, 719 and 11 are coprime, therefore x must be an integer.
i.e. Denominator must be 719*11, but say, divide 719 by 719*11, and you will see that x must be a multiple of 11 in order for the product to be an integer, vice-versa for 11 divided by 719*11.
Since x must be an integer, there are no solutions to this system of equations.
The general solution for the first is 8 + 12n, and the general solution for the second is
4 + 12k.
9. Ah, and I forgot to check the case where the second hand is 4 units away from the hour hand. But the results should be the same.
10. So here we are:
Bleaug and his alter-ego Polar Fox, rightfully objected the claim and proved their case. But the amount of work in this effort should not be wasted. Here's what I am suggesting:
1. update the options for this fox (so that the truth shall be spoken all over the land)
2. construct a new fox for which six's good analytic solution remains valid, and
3. construct a new fox to coin the claim that "Clock arms can never be equally-separated".
These should be completed within this week.
Any objections? Please ring the bell, otherwise, thank you for the great work!
11. At 3h, 37.21m, 58.18s
The 3 arms of the clock will form the Benz mark - separated by 120 degrees.
■ The hour hand： 3600/33 = 109.09...degrees
■ The minute hand: 2520/11 = 229.09...degrees
■ The second time: 3840/11 = 349.09...degrees
See the 120-degree separation.
12. Do not use all of these Private Money Lender here.They are located in Nigeria, Ghana Turkey, France and Israel.My name is Mrs.Ramirez Cecilia, I am from Philippines. Have you been looking for a loan?Do you need an urgent personal or business loan?contact Fast Legitimate Loan Approval he help me with a loan of \$78.000 some days ago after been scammed of \$19,000 from a woman claiming to be a loan lender from Nigeria but i thank God today that i got my loan worth \$78.000.Feel free to contact the company for a genuine financial call/whats-App Contact Number +918929509036 Email:(fastloanoffer34@gmail.com) | 1,856 | 6,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-40 | latest | en | 0.910101 |
http://math.stackexchange.com/questions/237456/lebesgue-integral-on-measurable-succesion | 1,469,808,162,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831769.86/warc/CC-MAIN-20160723071031-00174-ip-10-185-27-174.ec2.internal.warc.gz | 166,728,315 | 18,101 | # Lebesgue integral on measurable succesion
Let $(\Omega,\Sigma,\mu)$ a measure space and $f_n:\Omega \rightarrow [-\infty;+\infty]$ measurable functions. Supposing $g\leq f_n, \qquad \forall n \in N$, a measurable function such that his negative part $g^-$ in integrable.
Show that: $\int_\Omega \lim \inf_{n \rightarrow \infty} f_n d\mu \leq \lim \inf_{n \rightarrow \infty} \int_\Omega f_n d\mu$ and that the integral in the formula bove have sense.
This is an exercise I'm trying to solve, I started working on the demonstration that the integral have sense and then I'll try to prove the formula.
I know it's similar to the Fatou Lemma and his reverse but $f_n$ is not nonnegative and g is not dominating so I'm trying to work in that direction.
So first of all I tried to prove the existence of the limit on the right. Let $g=g^+ - g^-,\qquad g^+ - g^-\leq f_n$ so $g^-\geq g^+-f_n$ since $g^-$ is integrable I can assume that $g^+ - f_n$ is measurable too.Now can I tell that also $f_n$ is integrable as a consequence of it?
For the second integral $f_n$ in measurable,the liminf of the sucesion is measurable too but I don't know how to get to the result.
-
We have $g\geqslant -g^-$ so $f_n\geqslant -g^-$ and $f_n+g^-\geqslant 0$. We apply Fatou's lemma to the sequence $\{f_n+g^-\}$.
How do you get that $f_n$ are integrable from what you wrote? – Davide Giraudo Nov 15 '12 at 21:23
I'm sorry,I have a problem with the connection, I wanted to write: If I apply the Fatou Lemma to ${f_n+g^-}$ and to $g^-$ I got the final result as you told but I'm wondering if the things I wrote above are right to justify that the integral have sense or if I'm missing something. If I use the fact that $f_N\geq -g^-$ I get that $g^-\geq -f_n$ so since $\int g^-$ is finished also $\int (-f_n)$ is finished and $\int f_n$ too.Is it right?What about the liminf? – Laura Nov 15 '12 at 21:26
It depends: on $\Bbb R_+$, $-1\leqslant e^{—x}$ but the map in the LHS is not integrable. – Davide Giraudo Nov 15 '12 at 21:28
I meant that the fact that $g^-\geqslant -f_n$ doesn't imply that $f_n$ integrable. If we had had $g^-\geqslant -f_n\geqslant 0$, it would be ok. The integral of $f_n+g^-$ is possibly infinite, but makes sense in $\overline{\Bbb R}$ (for example when $g(x)=-e^{-x^2}$ and $f_n=n$). "Make sense here is that not both $f_n^+$ and $f_n^-$ have infinite integral. – Davide Giraudo Nov 15 '12 at 21:41 | 781 | 2,413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-30 | latest | en | 0.90798 |
https://www.geeksforgeeks.org/binary-search-algorithms-the-c-standard-template-library-stl/?ref=lbp | 1,620,796,529,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00597.warc.gz | 823,086,930 | 19,392 | Related Articles
Binary Search in C++ Standard Template Library (STL)
• Difficulty Level : Basic
• Last Updated : 17 Feb, 2021
Binary search is a widely used searching algorithm that requires the array to be sorted before search is applied. The main idea behind this algorithm is to keep dividing the array in half (divide and conquer) until the element is found, or all the elements are exhausted.
It works by comparing the middle item of the array with our target, if it matches, it returns true otherwise if the middle term is greater than the target, the search is performed in the left sub-array.
If the middle term is less than the target, the search is performed in the right sub-array.
The prototype for binary search is :
```binary_search(startaddress,
Parameters :
element of the array.
location of the last element of the array.
valuetofind: the target value which we have
to search for.
Returns :
true if an element equal to valuetofind is found, else false.```
## CPP
`// CPP program to implement``// Binary Search in``// Standard Template Library (STL)``#include ``#include `` ` `using` `namespace` `std;`` ` `void` `show(``int` `a[], ``int` `arraysize)``{`` ``for` `(``int` `i = 0; i < arraysize; ++i)`` ``cout << a[i] << ``","``;``}`` ` `int` `main()``{`` ``int` `a[] = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };`` ``int` `asize = ``sizeof``(a) / ``sizeof``(a[0]);`` ``cout << ``"\nThe array is : \n"``;`` ``show(a, asize);`` ` ` ``cout << ``"\n\nLet's say we want to search for "``;`` ``cout << ``"\n2 in the array So, we first sort the array"``;`` ``sort(a, a + asize);`` ``cout << ``"\n\nThe array after sorting is : \n"``;`` ``show(a, asize);`` ``cout << ``"\n\nNow, we do the binary search"``;`` ``if` `(binary_search(a, a + 10, 2))`` ``cout << ``"\nElement found in the array"``;`` ``else`` ``cout << ``"\nElement not found in the array"``;`` ` ` ``cout << ``"\n\nNow, say we want to search for 10"``;`` ``if` `(binary_search(a, a + 10, 10))`` ``cout << ``"\nElement found in the array"``;`` ``else`` ``cout << ``"\nElement not found in the array"``;`` ` ` ``return` `0;``}`
Output
```The array is :
1,5,8,9,6,7,3,4,2,0,
Let's say we want to search for
2 in the array So, we first sort the array
The array after sorting is :
0,1,2,3,4,5,6,7,8,9,
Now, we do the binary search
Element found in the array
Now, say we want to search for 10 | 795 | 2,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | latest | en | 0.695287 |
http://www.solutioninn.com/suppose-you-write-a-1year-cashornothing-put-with-a-strike | 1,506,321,068,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690340.48/warc/CC-MAIN-20170925055211-20170925075211-00264.warc.gz | 574,561,689 | 7,330 | # Question: Suppose you write a 1 year cash or nothing put with a strike
Suppose you write a 1-year cash-or-nothing put with a strike of \$50 and a 1-year cash-or-nothing call with a strike of \$215, both on stock A.
a. What is the 1-year 99% VaR for each option separately?
b. What is the 1-year 99% VaR for the two written options together?
c. What is the 1-year 99% tail VaR for each option separately and the two together?
Sales0
Views127 | 126 | 444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-39 | latest | en | 0.954547 |
http://www.jiskha.com/display.cgi?id=1321490819 | 1,462,438,189,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860126377.4/warc/CC-MAIN-20160428161526-00220-ip-10-239-7-51.ec2.internal.warc.gz | 603,174,736 | 3,573 | Thursday
May 5, 2016
# Homework Help: algebra
Posted by lulu on Wednesday, November 16, 2011 at 7:46pm.
For the sequence an=6+5*(n-1),
its first term is__??__ ;
its second term is__??__ ;
its third term is__??__ ;
its fourth term is__??__ ;
its fifth term is__??_ ;
its common difference d= __??__ . | 100 | 302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-18 | longest | en | 0.950897 |
https://catcostaclujul.ro/isometric-paper-printable-pdf.html | 1,670,445,858,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711218.21/warc/CC-MAIN-20221207185519-20221207215519-00573.warc.gz | 184,510,882 | 15,684 | Isometric Paper Printable Pdf
Isometric Paper Printable Pdf. Dots graph paper pdf generator check out our many other free graph/grid paper styles. (no spam, ever!) subscribe (free!) this paper is available free for download.
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Once it's finished downloading, unpack the archived pdf and open it in the pdf viewer program or application of your choice. This type of graph paper has 1/10 inch squares, which makes it. Free printable 1/10 inch graph paper with grid lines in portrait orientation.
Free Isometric Paper Dots PDF 1 Page(s) Isometric paper
Available in.pdf format, just download it, open it in a program that can display pdf files, and print. Isometric paper, math charts, grids, graph paper. Isometric graph paper or 3d graph paper is a triangular graph paper which uses a series of three guidelines forming a 60° grid of small triangles. So, this was the basic introduction about the grid papers which you have received in our article.
Source: www.pinterest.de
10 squares per inch graph paper pdf printable has 10 squares per inch. Once it's finished downloading, unpack the archived pdf and open it in the pdf viewer program or application of your choice. Following a bumpy launch week that saw frequent server trouble and bloated player queues, blizzard has announced that over 25 million overwatch 2 players have logged on in its first 10 days.sinc Hex # letter 8.5 x 11 8.5 x 11 a4 11 x 17 a3. 4th grade math lesson on factor trees. I'm going to guess you're really into pens. As we know that graphs are made up of dots and grid is a part of a graph then similarly it is also linked up with the dot paper too. Get 50 of our best printable papers in one convenient download for \$19 For example, this worksheet contains such problems as 2 x 9, 2 x 2, and 2 x 3. Grid paper is nothing more than one of the types of dot paper too.
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It creates a 3d effect and can be used for isometric illustrations or designs. 10 squares per inch graph paper pdf printable has 10 squares per inch. Subscribe to the free printable newsletter. I'm going to guess you're really into pens. When students become active doers of mathematics, the greatest gains of their mathematical thinking can be realized. It is perfect for graphing and drawing small diagrams. These are all of our printable paper templates available for a4 paper. If you do not have a.pdf document reader, we recommend the free. Once it's finished downloading, unpack the archived pdf and open it in the pdf viewer program or application of your choice. How to print this handwriting paper template.
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It can be used for a variety of purposes, from sketching and drawing to math and science. When students become active doers of mathematics, the greatest gains of their mathematical thinking can be realized. It creates a 3d effect and can be used for isometric illustrations or designs. Hex # letter 8.5 x 11 8.5 x 11 a4 11 x 17 a3. Dots graph paper pdf generator check out our many other free graph/grid paper styles. Isometric graph paper can be defined as the triangular graph paper and is made using the 60° grid in the form of small triangles. Grid paper is nothing more than one of the types of dot paper too. Isometric paper, math charts, grids, graph paper. Once it's finished downloading, unpack the archived pdf and open it in the pdf viewer program or application of your choice. Graph paper, coordinate paper, grid paper, or squared paper is writing paper that is printed with fine lines making up a regular grid.
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Each style is available in.pdf documents that are formatted to print on standard sheets of 8 ½ x 11 paper. It creates a 3d effect and can be. (no spam, ever!) subscribe (free!) this paper is available free for download. How to print this handwriting paper template. It can be used for a variety of purposes, from sketching and drawing to math and science. Sometric graph paper is just the other form of the graph paper, which is used for some specific purposes. Following a bumpy launch week that saw frequent server trouble and bloated player queues, blizzard has announced that over 25 million overwatch 2 players have logged on in its first 10 days.sinc Isometric graph paper or 3d graph paper is a triangular graph paper which uses a series of three guidelines forming a 60° grid of small triangles. When students become active doers of mathematics, the greatest gains of their mathematical thinking can be realized. 3d projections use the primary qualities of an object's basic shape to create a map of.
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(no spam, ever!) subscribe (free!) this paper is available free for download. These projections rely on visual perspective and aspect analysis to project a complex object for viewing capability on a simpler plane. Following a bumpy launch week that saw frequent server trouble and bloated player queues, blizzard has announced that over 25 million overwatch 2 players have logged on in its first 10 days.sinc I'm going to guess you're really into pens. How to print this handwriting paper template. Download free grid paper in pdf. For example, this worksheet contains such problems as 2 x 9, 2 x 2, and 2 x 3. This printable is the first in this series that uses the same factor—in this case, the number 2—in each problem. Once it's finished downloading, unpack the archived pdf and open it in the pdf viewer program or application of your choice. Well, in the above section we have already.
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These projections rely on visual perspective and aspect analysis to project a complex object for viewing capability on a simpler plane. 1/10 inch graph paper pdf. Hex # letter 8.5 x 11 8.5 x 11 a4 11 x 17 a3. When students become active doers of mathematics, the greatest gains of their mathematical thinking can be realized. Sometric graph paper is just the other form of the graph paper, which is used for some specific purposes. Grid paper is nothing more than one of the types of dot paper too. It can be used for a variety of purposes, from sketching and drawing to math and science. Available in.pdf format, just download it, open it in a program that can display pdf files, and print. After you click the download button, save the archived pdf file on your pc. 3d projections use the primary qualities of an object's basic shape to create a map of.
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So, this was the basic introduction about the grid papers which you have received in our article. These are all of our printable paper templates available for a4 paper. Once it's finished downloading, unpack the archived pdf and open it in the pdf viewer program or application of your choice. The triangles are arranged in groups of six to make hexagons. Dots graph paper pdf generator check out our many other free graph/grid paper styles. It is perfect for graphing and drawing small diagrams. For example, this worksheet contains such problems as 2 x 9, 2 x 2, and 2 x 3. From there on, you can probably find the print option listed under the file menu. As we know that graphs are made up of dots and grid is a part of a graph then similarly it is also linked up with the dot paper too. 10 squares per inch graph paper pdf printable has 10 squares per inch.
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Well, in the above section we have already. 1/10 inch graph paper pdf. So, this was the basic introduction about the grid papers which you have received in our article. These are all of our printable paper templates available for a4 paper. Isometric graph paper or 3d graph paper is a triangular graph paper which uses a series of three guidelines forming a 60° grid of small triangles. (no spam, ever!) subscribe (free!) this paper is available free for download. It creates a 3d effect and can be used for isometric illustrations or designs. It can be used for a variety of purposes, from sketching and drawing to math and science. Different types of graph papers are shared in the a4 size, you can choose to download any graph paper of your choice from here and start using it. When students become active doers of mathematics, the greatest gains of their mathematical thinking can be realized. | 1,849 | 8,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.936298 |
http://math.stackexchange.com/questions/135621/how-does-the-standard-representation-restrict-to-the-cyclic-group-generated-by/135638 | 1,469,329,944,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00201-ip-10-185-27-174.ec2.internal.warc.gz | 155,973,171 | 19,582 | # How does the standard representation restrict to the cyclic group generated by (1234).
So we have the group $S_4$ which has the standard representation. We also have the subgroup generated by permutation (1234). This is isomorphic to $C_4$ which has four irreducible representation. How does Res$^{S_4}_{C_4}(V)$ decompose into a direct sum of the four irreducibles of $C_4$. I know that for the trivial (and alt.) representation of $S_4$, it restricts to the trivial (and alt.) rep'n of $C_4$, but I can't seem to figure out how to do this for the other three irreducibles of $S_4$, and am hoping if you teach me how to do it for the standard I can do the rest.
I feel like the problem is that the standard rep'n is confusing in that I cannot even tell if there is a subspace of $V$ that is fixed when I only act on it with the elements of $C_4$ i.e. those generated by $(1234)$.
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Have you written down the matrix corresponding to $(1,2,3,4)$ under the standard representation with respect to some basis? – Alex B. Apr 23 '12 at 4:12
I just did, I got {{0,0,-1},{1,0,-1},{0,1,-1}} in Mathematica notation. – Steven-Owen Apr 23 '12 at 4:28
which has eigenvalues $-1,-i,$ and $i$. So does the standard representation decompose into one of each irreducible of $C_4$ except the trivial? – Steven-Owen Apr 23 '12 at 4:30
$\def\triv{\boldsymbol{1}}$ One way of getting the answer is to just write down the matrix of $(1,2,3,4)$ under the standard representation with respect to your favourite basis and compute eigenvalues, as you have done. Here is another way, using some more machinery, but with almost no calculation:
The standard representation can be written as $\rho=\mathrm{Ind}_{S_4/S_3}\triv_{S_3} - \triv_{S_4}$, i.e. as the 4-dimensional permutation representation $\mathbb{C}[S_4/S_3]$, but without the trivial summand. So, using Mackey decomposition and the fact that $S_4 = S_3C_4$, $S_3\cap C_4 = \{1\}$, we get $$\mathrm{Res}_{S_4/C_4}\rho = \mathrm{Res}_{S_4/C_4}(\mathrm{Ind}_{S_4/S_3}\triv_{S_3} - \triv_{S_4}) = \mathrm{Ind}_{C_4/1}\triv - \triv_{C_4},$$ i.e. the sum of all non-trivial representations of $C_4$.
Similar tricks work for all the other representations of $S_4$, since it just so happens, that any irreducible character of $S_n$, $n\in \mathbb{N}$, is a linear combination of permutation characters.
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$\newcommand{\Res}{{\text{Res}}} \newcommand{\Ind}{{\text{Ind}}} \newcommand{\ds}{{\displaystyle}} \newcommand{\inv}{{^{-1}}}$By Frobenius reciprocity, if $W$ and $U$ are irreducible representations of $H < G$, respectively, thenthe number of times $U$ appears in $\Ind W$ is the same as the number of times $W$ appears in $\Res U$. Let $W$ be the one dimensional representation (thus obviously irreducible) of the subgroup $H$ of $G$ generated by $(1234)$ in which $(1234)\cdot w=iw$. If we want to decompose $\Ind W$ into irreducible representations of $G=S_4$ i.e how many times an irreducible representation $U$ appears in $\Ind W$, we just need to find how many times $W$ appears in $\Res U$. The trivial and alternating representations restrict to trivial representation of $C_4$ and the alternating representation of $C_4$. We know that the induced representation is $\bigoplus_{\sigma \in S_4/C_4} \sigma W$, and since $W$ is one dimensional, the induced representation has dimension $[S_4:C_4]$ which is 6. Let us begin by writing the matrix corresponding to $(1234)$ in the standard representation of $S_4$. We get this by knowing that the standard representation is just the regular representation "minus" the trivial representation. (1234) in the regular representation is just the $$\begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix}$$ And we can write this in the basis $\begin{pmatrix} 1\\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1\\ -1 \\ 0 \\0 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 1 \\ -1 \end{pmatrix}$ and we would get $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1\\ \end{pmatrix}$$ And thus we have that the element (1234) acts in the standard representation by the matrix $$\begin{pmatrix} 0 & 0 & -1 \\ 1 & 0 & -1 \\ 0 & 1 & -1\\ \end{pmatrix}$$ which has eigenvalues $i, -i,$ and $-1$, so that the standard representation restricted to the group generated by (1234) decomposes into one of each of the irreducibles of the $C_4$ representations except for the trivial. We would find the same to be true for the standard representation tensor the alternating representation. Now since $\Res V$ and $\Res V'$ each contain a copy of $W$, we know that $\Ind W$ contains a copy of $V$ and $V'$, and since $\dim(V)=\dim(V')=3$ we need not even check the other irreducible representation of $S_4$, as $\Ind W = V \oplus V'$.
We do precisely the same for $(123)$. Let $W$ be the one dimensional representation (thus obviously irreducible) of the subgroup $H$ of $G$ generated by $(123)$ in which $(123)\cdot w=e^{2 \pi i/3} w$. We know the dimension of $\Ind W$ will be $[S_4:C_3]= 8$. In this case, We have that both the trivial and alternating representation restrict to the trivial of $C_3$.
Let us begin by writing the matrix corresponding to $(123)$ in the standard representation of $S_4$. We get this by knowing that the standard representation is just the regular representation "minus" the trivial representation. (1234) in the regular representation is just the $$\begin{pmatrix} 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
And we can write this in the basis $\begin{pmatrix} 1\\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1\\ -1 \\ 0 \\0 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 1 \\ -1 \end{pmatrix}$ and we would get $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 1\\ \end{pmatrix}$$
And thus we have that the element (1234) acts in the standard representation by the matrix $$\begin{pmatrix} 0 & -1 & 1 \\ 1 & -1 & 1 \\ 0 & 0 & 1\\ \end{pmatrix}$$ which has eigenvalues $1, e^{2 \pi i/3}, e^{4 \pi i/3}$, so that the standard representation restricted to the group generated by (123) decomposes into each of the irreducibles of the $C_4$ representations We would find the same to be true for the standard representation tensor the alternating representation. Now since $\Res V$ and $\Res V'$ each contain a copy of $W$, we know that $\Ind W$ contains a copy of $V$ and $V'$, and since $\dim(V)=\dim(V')=3$ we need not even check the other irreducible representation of $S_4$, as $\Ind W = V \oplus V' \oplus Q$ just by dimensionality where $Q$ is the representation of the quotient group, it is labeled as "Another $W$" in the text by Fulton and Harris.
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A minor terminology nitpick: the regular representation of a group $G$ is $\mathbb{C}[G]$, which is $|G|$-dimensional, i.e. 24-dimensional in the case of $S_4$. That's not what you meant. See also my answer. The beginning of the second paragraph is what you meant to write. – Alex B. Apr 23 '12 at 5:21
Good catch; I still haven't mastered the terminology as of yet. – Steven-Owen Apr 23 '12 at 5:23 | 2,238 | 7,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2016-30 | latest | en | 0.876527 |
https://www.weegy.com/?ConversationId=E53DD69A | 1,624,036,361,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00501.warc.gz | 966,809,851 | 10,728 | Multiply these binomials: (18x + 20)(–12x – 6)=
(18x + 20)(–12x – 6) = -216x^2 - 240x - 108x - 120 = -216x^2 - 348x - 120
s
Question
Updated 7/23/2014 3:50:01 PM
Edited by andrewpallarca [7/23/2014 3:46:50 PM]
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(18x + 20)(–12x – 6)
= -216x^2 - 240x - 108x - 120
= -216x^2 - 348x - 120
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,449 | 3,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-25 | latest | en | 0.745583 |
https://elsagames.org/are-there-one-player-card-games/ | 1,709,155,771,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00041.warc.gz | 233,627,986 | 15,286 | # Are There One Player Card Games?
By Ralph Torres
If you love playing card games but can’t always find someone to play with, you might be wondering if there are any one player card games out there. The good news is that yes, there are plenty of great one player card games to choose from. In this article, we’ll introduce you to some of the best ones and explain how to play them.
## 1. Solitaire
Solitaire is perhaps the most well-known one player card game. It’s been around for over 200 years and has countless variations. The goal of Solitaire is to move all the cards from the tableau (the main playing area) to the foundation (usually four piles in the top right corner) in ascending order, starting with Ace and ending with King.
There are many different ways to play Solitaire, but most versions involve dealing cards from a shuffled deck into a tableau layout and then moving cards around according to specific rules.
### 2. Freecell
Freecell is another popular one player card game that’s similar to Solitaire but with a few key differences. In Freecell, all 52 cards are dealt face up at the beginning of the game, so you can see everything from the start.
The goal of Freecell is to move all the cards into four foundation piles in ascending order by suit. You can also move cards between eight columns on the tableau, as long as they’re in descending order and alternating colors.
### 3. Pyramid
Pyramid is a fun and challenging one player card game that requires some strategy and planning ahead. The goal of Pyramid is to remove all the cards from the pyramid-shaped tableau by pairing them up with other cards that add up to 13.
To play Pyramid, you start by dealing out a pyramid shape made up of 28 cards. You can then match cards that add up to 13 and remove them from the pyramid. You can also draw cards from a stock pile and use them to make pairs.
## 4. Spider
Spider is a one player card game that’s similar to Solitaire but with some added complexity. In Spider, you start with ten tableau piles of cards, with the first four piles having six cards each and the remaining six piles having five cards each.
The goal of Spider is to build eight foundation piles in descending order by suit, starting with King and ending with Ace. You can move cards between tableau piles as long as they’re in descending order and alternating colors.
### 5. Clock Patience
Clock Patience is a one player card game that’s played using a deck of 52 cards and a clock face layout. The goal of Clock Patience is to move all the cards from the deck onto the clock face in numerical order.
To play Clock Patience, you place the Ace of Spades at the top of the clock face, followed by two through ten in clockwise order around the edge of the clock face. You then deal out the rest of the deck into four rows around the clock face, making sure each row has 13 cards.
## Conclusion
In conclusion, there are plenty of great one player card games out there for anyone who loves playing cards by themselves. Whether you prefer Solitaire or something more challenging like Spider or Pyramid, there’s sure to be a one player card game that suits your taste. So next time you’re looking for some solo entertainment, why not give one of these games a try | 704 | 3,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.958285 |
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| accepted | 1,658 | 4,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-05 | latest | en | 0.752927 |
http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Rate_Laws/Reactions/Elementary_Reaction | 1,371,629,513,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708144156/warc/CC-MAIN-20130516124224-00068-ip-10-60-113-184.ec2.internal.warc.gz | 48,354,808 | 12,536 | More
# Elementary Reactions
An elementary reaction is a single step reaction having a single transition state and no intermediates
### Introduction
Elementary reactions add up to complex reactions. In other words, non-elementary reactions can be explained in multiple elementary reaction steps. A set of elementary reactions comprises a reaction mechanism, which predicts the elementary steps involved in a complex reaction. Below are two reaction coordinates of two reactions. One describes an elementary reaction while the other describes a non-elementary reaction.
Elementary Reaction Complex Reaction $\text{Reactants} \rightarrow \text{Products}$ This is a sample reaction coordinate of an elementary reaction. Note that there is one transition state and no intermediates. Again, elementary steps cannot be explained in simpler reactions. $\text{Reactants} \rightarrow \text{Intermediates} \rightarrow \text{Products}$ This is a sample reaction coordinate of a complex reaction. Note that it involves an intermediate and multiple transition states. Again, a complex reaction can be explained in terms of elementary reactions.
### Types of Elementary Reactions
The molecularity of a reaction refers to the number of reactant particles involved in the reaction. As there can only be discrete number of particles, the molecularity must be an integer. Molecularity can be described as either unimolecular, bimolecular, or termolecular. There are no known elementary reactions involving four or more molecules.1The following table summarizes the three known types of elementary reactions:
Molecularity Elementary Step Rate Law Example
Unimolecular $$A \rightarrow Products$$ $$rate = k[A]$$ $$N_2O_{4(g)} \rightarrow 2NO_{2(g)}$$
Bimolecular $$A + A \rightarrow Products$$ $$rate = k[A]^2$$ $$2NOCl \rightarrow 2NO_{(g)} + CO_{2(g)}$$
$$A + B \rightarrow Products$$ $$rate = k[A][B]$$ $$CO_{(g)} + NO_{3(g)} \rightarrow NO_{2(g)} + CO_{2(g)}$$
Termolecular $$A + A +A \rightarrow Products$$ $$rate = k[A]^3$$
$$A + A + B \rightarrow Products$$ $$rate = k[A]^2[B]$$ $$2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}$$ 2
$$A + B + C \rightarrow Products$$ $$rate = k[A][B][C]$$ $$H + O_{2(g)} + M \rightarrow HO_{2(g)} + M$$ 3
#### Unimolecular Reaction
unimolecular reaction occurs when a molecule rearranges itself to produce one or more products. An example of this is radioactive decay, in which particles are emitted from an atom. Other examples include "cis-trans isomerization, thermal decomposition, ring opening, and racemization."1 The rate at which a substance decomposes is dependent on its concentration. Unimolecular reactions are often first-order reactions as explained by Frederick Alexander Lindemann in 1922.3 To read further about Lindemann and why unimolecular reactions are not second-order, read here or here.
#### Bimolecular Reaction
bimolecular reaction involves the collision of two particles. Bimolecular reactions are common in organic chemical reactions such as nucleophilic subsitution. The rate of reaction depends on the product of both species involved, which makes bimolecular reactions second-order reactions.
#### Termolecular Reaction
termolecular reaction requires the collision of three particles at the same place and time. This type of reaction is very uncommon as all three reactants must simultaneously collide with each other, with sufficient energy and correct orientation, in order to produce a reaction. There are three ways termolecular reactions can react, and all are third order.
### Practice Problems
1. How are non-elementary steps and elementary steps related?
2. Choose the correct statements.
a. An elementary step has 0 intermediates.
b. An elementary step has 1 intermediate.
c. An elementary step has 2 intermediates.
d. An elementary step has 0 transition states.
e. An elementary step has 1 transition state.
f. An elementary step has 2 transition states.
3. Which of the following is a termolecular reaction?
a. $$A + 2B + C \rightarrow D$$
b. $$A + B + B \rightarrow C$$
c. all of the above
d. $$2A + 2B + 2C \rightarrow 2D$$
e. b and d
f. none of the above
4. Which rate law corresponds to a bimolecular reaction?
a. $$rate = k[A][A]^2$$
b. $$rate = k[A][B]$$
c. all of the above
d. $$rate = k[A]^2$$
e. b and d
f. none of the above
5. Give an example of a reaction with a molecularity of 1/2.
6. True or False: Given species A and B inside a container, instruments detect that three (3) collisions occured before product was formed. That is, we know a reaction occured after detecting three collisions in a box. We can conclude that the reaction is a termolecular reaction (as the reaction could have been produced from A+A+B or A+B+B).
#### Solutions
1. Non-elementary steps, or complex reactions, are sets of elementary reactions. That is to say, the addition of elementary steps produces complex, non-elementary reactions.
2. The correct statements are "a" and "e". By definition of elementary reactions they have 0 intermediates because they cannot be broken down. Again by definition of an elementary reaction, a single-step reaction will have 1 transition state. There is no reaction with 0 transition states. Having 2 transition states implies having 1 intermediate, making the reaction non-elementary.
"a" is not a termolecular reaction as it involves A + B + B + C, or 4 molecules
"b" is a termolecular reaction as it involves 3 particles: A + B + B
"c" is incorrect because "a" is incorrect
"d" is a termolecular reaction as the coefficients cancel out to give the reaction: $$A + B + C \rightarrow D$$, which involves 3 particles (A + B + C)
"e" is the correct answer as "b" and "d" are correct
"a" is incorrect because the rate law describes a third-order reaction, which is true for termolecular reactions
"b" is a possible rate law for the bimolecular reaction: \(A + B \rightarrow Products") }}
"c" is incorrect because "a" is incorrect
"d" is a possible rate law for the bimolecular reaction: \(A + A \rightarrow Products") }}
"e" is the correct answer as "b" and "d" are correct
5. Impossible. The molecularity of a reaction MUST be an integer as there cannot be "a half of a particle" producing a reaction.
6. False, we cannot conclude anything. While a termolecular reaction requires the collision of three particles, the reverse logic is not necessarily true. That is, having three collisions is not sufficient for a termolecular reaction.
1. For example, particles A + A + B collide with each other at the same place and time. However, particle B was in the wrong orientation, so no reaction occurred. Instead, the two A particles were in the correct orientation and produced a reaction, which is a bimolecular reaction.
2. Consider a second example: two collisions between particles A + A + B occured, but there was not enough energy to produce a reaction. Instead, a third collision between A and B had the sufficient energy and correct orientation to produce a reaction. Such a reaction is, again, only bimolecular.
3. A last example: particle A collides twice with a wall, and then once with B to produce a reaction. Such a reaction involving three collisions at different places and different time is only a bimolecular reaction.
### References
1. Chang, Raymond. "Chemical Kinetics." Physical Chemistry for the Biosciences. Sansalito, CA: University Science, 2005. 325-328. Print.
2. Olbregts, J. (1985), Termolecular reaction of nitrogen monoxide and oxygen: A still unsolved problem. International Journal of Chemical Kinetics, 17: 835–848. doi: 10.1002/kin.550170805.
3. Kerr, James Alistair. "Notes on the Tables." CRC Handbook of bimolecular and termolelucar gas reactions . Boca Raton, Florida: CRC Press, 1987. 2. Print.
4. Baer, Tomas, and William L. Hase. "Introduction." Unimolecular reaction dynamics: theory and experiments. New York: Oxford University Press, 1996. 4-5. Print.
### Contributors
• Tho Nguyen (UCD), Minh Ngo (UCD)
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# Lect20_L6_L7_handout - lecture 20 Hypothesis Testing II...
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lecture 20, Hypothesis Testing II z Tests about a Population Mean: σ Known Testing a “Greater Than” Alternative Critical Value Rule The p -value Approach Testing a “Less Than” Alternative Critical Value Rule The p -value Approach Summary of Testing a One-Sided Alternative lecture 20, Hypothesis Testing II Outline 1 z Tests about a Population Mean: σ Known 2 Testing a “Greater Than” Alternative Critical Value Rule The p -value Approach 3 Testing a “Less Than” Alternative Critical Value Rule The p -value Approach 4 Summary of Testing a One-Sided Alternative
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lecture 20, Hypothesis Testing II z Tests about a Population Mean: σ Known Testing a “Greater Than” Alternative Critical Value Rule The p -value Approach Testing a “Less Than” Alternative Critical Value Rule The p -value Approach Summary of Testing a One-Sided Alternative lecture 20, Hypothesis Testing II z Tests about a Population Mean: σ Known z Tests about a Population Mean: σ Known Test hypotheses about a population mean using the normal distribution: Called z tests Require that the population standard deviation σ is known In most real-world situations, σ is not known — But often is estimated from s of a single sample — When σ is unknown, test hypotheses about a population mean using the t distribution —to be talked about later Here, assume that we know σ Two approaches: (i) Critical Value Rule (sometimes also called Rejection Point Rule) (ii) p -value
lecture 20, Hypothesis Testing II z Tests about a Population Mean: σ Known Testing a “Greater Than” Alternative Critical Value Rule The p -value Approach Testing a “Less Than” Alternative Critical Value Rule The p -value Approach Summary of Testing a One-Sided Alternative lecture 20, Hypothesis Testing II Testing a “Greater Than” Alternative Critical Value Rule Steps to Apply the Critical Value Rule 1 State the null and alternative hypotheses 2 Specify the significance level α 3 Select a test statistic 4 Determine the critical value rule for deciding whether or not to reject H 0 . Use the specified α to find the critical value 5 Collect the sample data and calculate the value of the test statistic 6 Decide whether to reject H 0 by using the test statistic and the critical value rule 7 Interpret the statistical results in real-world terms and assess their practical importance
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lecture 20, Hypothesis Testing II z Tests about a Population Mean: σ Known Testing a “Greater Than” Alternative Critical Value Rule The p -value Approach Testing a “Less Than” Alternative Critical Value Rule The p -value Approach Summary of Testing a One-Sided Alternative lecture 20, Hypothesis Testing II Testing a “Greater Than” Alternative Critical Value Rule Testing a “Greater Than” Alternative: Trash Bag Case Tests show the trash bag has a mean breaking strength μ close to but not exceeding 50 lbs The new bag’s mean breaking strength is not known and is in question, but it is hoped it is stronger than the current one Assume that σ = 1 . 65
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Ask a homework question - tutors are online | 805 | 3,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-13 | latest | en | 0.764012 |
https://momentumchevrolet.com/auto-parts/why-motors-are-connected-in-star-delta.html | 1,643,048,007,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00330.warc.gz | 452,534,340 | 19,085 | # Why motors are connected in star Delta?
Contents
During starting the motor windings are connected in star configuration and this reduces the voltage across each winding 3. … They are used in an attempt to reduce the start current applied to the motor during start as a means of reducing the disturbances and interference on the electrical supply.
## Are motors connected in star or delta?
Comparison between Star and Delta Connections
STAR Connection (Y) DELTA Connection (Δ)
The speeds of Star connected motors are slow as they receive 1/√3 voltage. The speeds of Delta connected motors are high because each phase gets the total of line voltage.
## Why do motor run in Delta not in Star?
The coil of the delta contactor is burnt out or the contactor is physically stuck. The motor is not connected correctly, it is possible to connect the motor in such a way that it will only run in the star connection. If the system worked before and no connections were changed then obviously his can’t be the problem.
## Can we run Delta motor in Star?
The answer to your question is no. One end of each star winding is connected internally to the neutral point. Since the neutral connection is internal, it cannot be opened to make a delta.
## What is Y connection?
wye (Y) connection
A method of connecting the ends of the windings of a poly-phase transformer; each of the three windings are joined at a common point; the other ends of the windings provide the line-to-line voltages. Compare with delta connection.
## Is current higher in Star or Delta?
In star connection, the line current is equal to the phase current, whereas in delta connection the line current is equal to root three times of the phase current. … The amount of insulation required in star connection is low and in delta connection high insulation level is required.
## Can I run a motor in Star?
So if it’s a motor with 230V windings (and doesn’t need any special starting – perhaps because it drives a very light mechanical load) then running in star would be fine.
## Why 3 phase motor is star delta connection?
Star/Delta starters are probably the most common reduced voltage starters. They are used in an attempt to reduce the start current applied to the motor during start as a means of reducing the disturbances and interference on the electrical supply.
## What happens if motor is connected in Delta?
If you have 415V, you connect the motor in Star. If you have 240V and you connect it in Delta, the effective violtage across the windings will become 240/1.732 or only 138V. That means the motor output torque will be 33% of normal. So unless your load is drastically reduced, your motor will stall.
## Which motors are connected in star?
Star connected motors have a central connected point, called a short circuit point or star point and each winding receives phase voltages (230volts), star connected motor only run at one third of the motor rated torque and power, whereas delta connected motors have no connected point and each winding receives line …
IMPORTANT: How much does it cost to bump out a dent in a car?
## Does Star or Delta have more torque?
The motor will speed up according to the vfd ramp up time and setpoint frequency. Delta can also provide higher torque as opposed to star.
## What is 3phase Delta?
The Delta configuration has the three phases connected like a triangle. Delta systems have four wires total: three hot wires and one ground wire. Wye systems utilize a star configuration, with all three hot wires connected at a single neutral point.
## What is Delta Delta Connection?
A delta connection is a connection used in a three-phase electrical system in which three elements in series form a triangle, the supply being input and output at the three junctions.
## What is wye and delta connection?
Delta connection is used for shorter distances, whereas wye connection is used for power transmission networks for longer distances. Delta was primarily used at small industrial facilities that had a relatively large (240 VAC) motor load but only a small need for convenience outlets and lighting. | 847 | 4,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-05 | latest | en | 0.947091 |
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• ## Related Books
0 answersPost by julius mogyorossy on December 16, 2013x=+-2i, so what that is really saying is, x=2, after you square it, make it negative, is that it?
### Roots and Zeros
• A polynomial of degree n has n roots.
• If the coefficients of the polynomial are real, then complex roots occur as conjugate pairs.
• Learn and understand Descartes’ rule of signs for the roots of a polynomial.
• Use synthetic substitution to find the roots of a polynomial of degree 3 or more.
### Roots and Zeros
Solve x3 − 5x2 + 6x = 0
• Find the GCF
• GCF = x
• x(x2 − 5x + 6) = 0
• Find two numbers such that when multiplied = 6; and when added = − 5
• Those two numbers are − 3 and − 2
• x(x − 3)(x − 2) = 0
• Using Zero Product Property we have
• x = 0, x − 3 = 0, and x − 2 = 0
• Solve
x = 0, x = 3, and x = 2
Solve x3 − 3x2 − 4x = 0
• Find the GCF
• GCF = x
• x(x2 − 3x − 4) = 0
• Find two numbers such that when multiplied = − 4; and when added = − 3
• Those two numbers are − 4 and 1
• x(x − 4)(x + 1) = 0
• Using Zero Product Property we have
• x = 0, x − 4 = 0, and x + 1 = 0
• Solve
x = 0, x = 4, and x = − 1
Solve 2x3 − 2x2 − 40x = 0
• Find the GCF
• GCF = 2x
• 2x(x2 − x − 20) = 0
• Find two numbers such that when multiplied = − 20; and when added = − 1
• Those two numbers are 4 and − 5
• 2x(x + 4)(x − 5) = 0
• Using Zero Product Property we have
• 2x = 0, x + 4 = 0, and x − 5 = 0
• Solve
x = 0, x = − 4, and x = 5
Solve x4 − 17x2 + 16 = 0
• Recall that factoring a 4th power polynomial is the same as factoring a 2nd degree polynomial.
• Find two numbers m and n such that when multiplied, m*n = 16 and when added m + n = − 17
• Those numbers are − 16 and − 1
• (x2 − 16)(x2 − 1) = 0
• Notice that nowe have two difference of squares.
• (x − 4)(x + 4)(x − 1)(x + 1) = 0
• Solve using Zero Product Property
• x − 4 = 0, x + 4 = 0, x − 1 = 0, and x + 1 = 0
• Solve
x = 4, x = − 4, x = 1, and x = − 1
Solve x4 − 17x2 + 16 = 0
• Recall that factoring a 4th power polynomial is the same as factoring a 2nd degree polynomial.
• Find two numbers m and n such that when multiplied, m*n = 16 and when added m + n = − 17
• Those numbers are − 16 and − 1
• (x2 − 16)(x2 − 1) = 0
• Notice that nowe have two difference of squares.
• (x − 4)(x + 4)(x − 1)(x + 1) = 0
• Solve using Zero Product Property
• x − 4 = 0, x + 4 = 0, x − 1 = 0, and x + 1 = 0
• Solve
x = 4, x = − 4, x = 1, and x = − 1
Solve x4 − 3x2 − 54 = 0
• Recall that factoring a 4th power polynomial is the same as factoring a 2nd degree polynomial.
• Find two numbers m and n such that when multiplied, m*n = − 54 and when added m + n = − 3
• Those numbers are 6 and − 9
• (x2 + 6)(x2 − 9) = 0
• Notice that we have one difference of squares.
• (x2 + 6)(x − 3)(x + 3) = 0
• Solve using Zero Product Property
• x2 + 6 = 0, x − 3 = 0, and x + 3 = 0
• Solve
x = ±i√6 , x = 3, x = − 3
Determine the possible combinations of possible real roots, negative real roots and complex roots.
3x5 − 15x4 − 29x3 + 145x2 + 40x − 200 = 0
• Type of Root: Total Roots
How to Find It: You will find the total number of roots by the Degree of the Polynomial
• Type of Root: Total Positive Real Roots
How to Find It: Find the number of sign changes or is less than this by an even number. If number of sign changes is 5, then there could be, 5, 3, 1 positive real roots.
• Type of Root: Total Negative Real Roots
How to Find It: Compute f(-x). Count the number of sign changes, or is less than this by an even number.
• Type of root: Complex Roots
How to Find It: ComplexRoots = Total Roots - Positve Real Roots - Negative Real Roots
• What is the total possible number of roots?
• Total Possible roots is 5 because Degree is 5.
• How Many Positive Real Root? How many Sign Changes are there?
• According to Descarte's Change of Signs Rule there could be 3 Positive Real Roots, or less than an even number
• Positive Real Roots = 3 or 1
• How many Negative Real Roots are there? Compute f( − x) then count the number of sign changes
• f( − x) = 3( − x)5 − 15( − x)4 − 29( − x)3 + 145( − x)2 + 40( − x) − 200
• f( − x) = − 3(x)5 − 15x4 + 29x3 + 145x2 − 40x − 200
• There couuld be 2 Negative Real Roots because there were only 2 sign changes, or less than that by an even number.
• Negative Real Roots, 2 or 0
• Complex Roots can be found by adding the different combinations of + Real and - Real Roots as follows
These are all the possible combinations of positive real, negative real and complex roots:
+Real -Real Complex Total 3 2 0 5 3 0 2 5 1 2 2 5 1 0 4 5
Determine the possible combinations of possible real roots, negative real roots and complex roots.
2x5 + 10x4 − 3x3 − 15x2 − 35x − 175 = 0
• Type of Root: Total Roots
How to Find It: You will find the total number of roots by the Degree of the Polynomial
• Type of Root: Total Positive Real Roots
How to Find It: Find the number of sign changes or is less than this by an even number. If number of sign changes is 5, then there could be, 5, 3, 1 positive real roots.
• Type of Root: Total Negative Real Roots
How to Find It: Compute f(-x). Count the number of sign changes, or is less than this by an even number.
• Type of root: Complex Roots
How to Find It: ComplexRoots = Total Roots - Positve Real Roots - Negative Real Roots
• What is the total possible number of roots?
• Total Possible roots is 5 because Degree is 5.
• How Many Positive Real Root? How many Sign Changes are there?
• According to Descarte's Change of Signs Rule there could be 1 Positive Real Roots, or less than an even number
• Positive Real Roots = 1
• How many Negative Real Roots are there? Compute f( − x) then count the number of sign changes
• f( − x) = 2( − x)5 + 10( − x)4 − 3( − x)3 − 15( − x)2 − 35( − x) − 175
• f( − x) = − 2(x)5 + 10(x)4 + 3(x)3 − 15(x)2 + 35(x) − 175
• There could be 4 Negative Real Roots because there were only 4 sign changes, or less than that by an even number.
• Negative Real Roots, 4, 2 or 0
• Complex Roots can be found by adding the different combinations of + Real and - Real Roots as follows
These are all the possible combinations of positive real, negative real and complex roots:
+Real -Real Complex Total 1 4 0 5 1 2 2 5 1 0 4 5
Determine the possible combinations of possible real roots, negative real roots and complex roots.
10x5 − 4x4 + 5x3 − 2x2 − 50x + 20 = 0
• Type of Root: Total Roots
How to Find It: You will find the total number of roots by the Degree of the Polynomial
• Type of Root: Total Positive Real Roots
How to Find It: Find the number of sign changes or is less than this by an even number. If number of sign changes is 5, then there could be, 5, 3, 1 positive real roots.
• Type of Root: Total Negative Real Roots
How to Find It: Compute f(-x). Count the number of sign changes, or is less than this by an even number.
• Type of root: Complex Roots
How to Find It: ComplexRoots = Total Roots - Positve Real Roots - Negative Real Roots
• What is the total possible number of roots?
• Total Possible roots is 5 because Degree is 5.
• How Many Positive Real Root? How many Sign Changes are there?
• + 10x5 − 4x4 + 5x3 − 2x2 − 50x + 20 = 0
• According to Descarte's Change of Signs Rule there could be 4 Positive Real Roots, or less than an even number
• Positive Real Roots = 4 or 2 or 0
• How many Negative Real Roots are there? Compute f( − x) then count the number of sign changes
• f( − x) = 10( − x)5 − 4( − x)4 + 5( − x)3 − 2( − x)2 − 50( − x) + 20
• f( − x) = − 10(x)5 − 4(x)4 − 5(x)3 − 2(x)2 + 50(x) + 20
• There could be 1 Negative Real Roots because there were only 1 sign changes, or less than that by an even number.
• Negative Real Roots: 1
• Complex Roots can be found by adding the different combinations of + Real and − Real Roots as follows
These are all the possible combinations of positive real, negative real and complex roots:
+Real -Real Complex Total 4 1 0 5 2 1 2 5 0 1 4 5
Determine the possible combinations of possible real roots, negative real roots and complex roots.
64x6 − 1 = 0
• Type of Root: Total Roots
How to Find It: You will find the total number of roots by the Degree of the Polynomial
• Type of Root: Total Positive Real Roots
How to Find It: Find the number of sign changes or is less than this by an even number. If number of sign changes is 5, then there could be, 5, 3, 1 positive real roots.
• Type of Root: Total Negative Real Roots
How to Find It: Compute f(-x). Count the number of sign changes, or is less than this by an even number.
• Type of root: Complex Roots
How to Find It: ComplexRoots = Total Roots - Positve Real Roots - Negative Real Roots
• What is the total possible number of roots?
• Total Possible roots is 6 because Degree is 6.
• How Many Positive Real Root? How many Sign Changes are there?
• + 64x6 − 1 = 0
• According to Descarte's Change of Signs Rule there could be 1 Positive Real Roots, or less than an even number
• Positive Real Roots = 1
• How many Negative Real Roots are there? Compute f( − x) then count the number of sign changes.
• f( − x) = 64( − x)6 − 1
• f( − x) = 64x6 − 1
• There could be 1 Negative Real Roots because there were only 1 sign changes, or less than that by an even number.
• Negative Real Roots: 1
• Complex Roots can be found by adding the different combinations of + Real and − Real Roots as follows
These are all the possible combinations of positive real, negative real and complex roots:
+Real -Real Complex Total 1 1 4 6
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Roots and Zeros
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Number of Roots 0:08
• Not Nature of Roots
• Example: Real and Complex Roots
• Descartes' Rule of Signs 2:05
• Positive Real Roots
• Example: Positve
• Negative Real Roots
• Example: Negative
• Finding the Roots 9:59
• Example: Combination of Real and Complex
• Conjugate Roots 13:18
• Example: Conjugate Roots
• Example 1: Solve Polynomial 16:03
• Example 2: Solve Polynomial 18:36
• Example 3: Possible Combinations 23:13
• Example 4: Possible Combinations 27:11
### Transcription: Roots and Zeros
Welcome to Educator.com.0000
We are going to continue our lesson on polynomials by discussing roots and zeroes.0002
The number of roots can be determined by looking at the degree of the polynomial equation,0008
because a polynomial equation of degree n has n roots.0014
This tells you only the number of roots; it doesn't tell you about the nature of the roots; the roots may be real or complex.0018
For example, if I am given a polynomial such as 5x6 + 2x3 - x2 + 9x - 2 = 0,0025
a polynomial equation, I see that the degree equals 6; this tells me that there are 6 roots for this equation.0038
Now, they can be real; and for example, these roots could be something like 1, -2...those are rational numbers.0053
They actually could be irrational numbers, like the square root of 3.0065
You also can have complex numbers as roots.0069
Recall that complex numbers have two parts--something like 2 + 3i; and the first part here is real, and the second part is imaginary.0073
And these are in the form a + bi.0086
We are going to talk, in a few minutes, about how these complex roots occur as conjugate pairs.0091
For example, 2 + 3i and 2 - 3i would be a conjugate pair--the same values here, but opposite signs; or 4 + i and 4 - i.0097
So, I know that here I have 6 roots; but I have no idea if they are real, rational, irrational, or complex, or what combination of those.0111
However, I can determine at least some of that through using Descartes' Rule of Signs.0119
And there are two sections to this rule: the first section helps you determine the possible numbers of positive real roots.0125
Then, we will talk about determining the possible number of negative real roots.0136
So, first we are just looking at the positive real roots: if we let p(x) be a polynomial with real coefficients (no imaginary coefficients--0141
just real coefficients), arranging in descending powers, recall that descending powers would be something like0154
f(x) = x6 + 5x5 + 7x4 - x3 + 6x2 - 2x + 4.0164
This is a degree 6 polynomial; and I have 6, 5, 4, and on down.0180
If you are trying to work with Descartes' Rule of Signs, the first thing to do is check the polynomial.0187
And if it is not arranged in descending powers, you need to put it that way.0192
OK, so I have this arranged in descending powers; the number of positive real roots0196
is the number of changes in sign of the coefficients, or is less than this by an even number.0202
Let's look at what that means: I want to look for the number of sign changes of the coefficient.0209
So, here I have...this coefficient is a 1, and that is positive; 5 is positive; 7 is positive; -1--so there is a sign change.0217
This is positive; this is negative; so that is one sign change.0230
OK, now here, I am going from -1 to positive 6; again, I have a sign change.0237
Here, I am going from a positive to a negative--another sign change; negative to positive--another sign change.0245
That gives me 1, 2, 3, 4--4 sign changes means there are 4 positive real roots, or less than this by an even number.0254
So, less than 4 by an even number would mean 4 - 2 (would give me 2), or another even number, 4: 4 - 4 would give me 0.0275
There are 4, 2, or 0 positive real roots.0287
So again, take the polynomial; arrange it in descending powers; and then look at the number of sign changes.0295
I have 1, 2, 3, 4 sign changes; that tells me that I will have, at most, four positive real roots.0302
However, I may have less than this by an even number (4 - 2: 2; 4 - 4; 0).0312
So, I have three possibilities: I may have 4 positive real roots, 2 positive real roots, or 0 real roots.0319
And I have, since the degree equals 6, a total of 6 roots.0327
I have a total of 6; of these, 4, 2, or 0 may be positive real roots.0338
OK, the second part of this is looking at the number of negative real roots.0343
The number of negative real roots is the number of changes in sign of the coefficients of the terms p(-x), or is less than this by an even number.0350
Let's continue on with the example that we just looked at, where we were given0361
f(x) = x6 + 5x5 + 7x4 - x3 + 6x2 - 2x + 4.0366
Now, I found there were four sign changes, which means 4, 2, or 0 positive real roots.0383
For the negative real roots, I have to look at f(-x).0388
So, I need to change these x's to -x and be very careful with the signs; OK.0394
So, this is going to give me a coefficient here of -1.0416
But if I take a negative to an even power, it is going to become positive.0421
-1 to the sixth power is going to become positive, so this is going to give me f(-x); here it is just going to be x6.0425
Here, I have -1 times x; you can look at it that way--it is -1 times x5.0435
Well, if I take a -1 to an odd power, it is going to remain negative; so, -x5 times 5 is going to give me -5x5.0443
Here, I have a negative coefficient to an even power; it is going to become positive, so this is really 7x4.0457
Here, I have a negative coefficient to an odd power, so it will remain negative; so this is -x3,0467
but it is times a negative: so -x3 times -1 becomes + x3.0475
OK, I have a -x; this actually should be outside...(-x)2 is going to give me -x times -x; that is going to give me a positive.0485
So, this is going to be plus 6x2.0501
Here, I have -x times -2; that is +2x; and my constant remains positive.0507
OK, number of sign changes: I am going to look for the changes in sign.0516
This is positive out here; so a positive to a negative--that is 1; a negative to a positive--that is 2.0523
This stays positive, positive, positive, positive; the number of sign changes equals 2.0533
OK, so the number of negative real roots is the number of changes in sign of the coefficients of the term p(-x), or less than this by an even number.0539
Therefore, I am going to have two negative real roots, or less than this by an even number.0552
Well, 2 minus 2 is 0; I can't go any lower than that for the number of roots; so there are 2 or 0 negative real roots.0559
Again, this power is 6; I have 6 total roots; I have 2 or 0 negative real roots; and last slide, we talked about having 4, 2, or 0 positive real roots.0575
So, we covered the real roots; there also may be complex roots, so let's talk now about the total roots and the different combinations that you could have.0589
We can use Descartes' Rule of Signs to determine the possible combinations of the real and complex roots.0600
So, in that example above, f(x) = x6 + 5x5 + 7x4 - x3 + 6x2 - 2x + 4,0607
the total roots (this is real and complex) equal 6.0624
Positive real roots: using Descartes' Rule of Signs, I found that I could have 4, 2, or 0.0632
Negative real roots: I found f(-x) and looked for the sign changes, and found two of those.0645
So, the number of negative real roots could be 2 or 0.0650
Now, I can figure out the combinations: I know that they need to total 6, and I have0661
my positive real roots' possibilities and my negative real roots; and the last part of this is complex roots.0669
So, I have positive real roots, the number of negative, and the number of complex; and these need to total 6; the total must equal 6.0681
So, I said that, for positive real roots, I could have 4; then, I may have 2 negative real roots.0693
4 and 2 is 6, so that leaves me with no complex.0700
I could have 4 positive real roots; I could have 0 negative real roots; 4 and 0 is 4; to total 6, I will have to have 2 complex roots.0703
OK, another possibility: I have 2 positive real roots and 2 negative real roots: 2 and 2 is 4; 2 more complex roots will give me 6.0717
Or I could have 2 positive real roots and 0 negative real roots.0731
2 and 0 is 2; to total 6, I will have 4 complex roots.0737
OK, finally, I may have 0 positive real roots and 2 negative real roots.0744
0 and 2 is 2, so I have to have 4 complex roots.0753
Then, I may have 0 positive real roots and 0 negative real roots, and that gives me 0 and 0.0760
So, to total 6, I would have to have 6 right here.0774
And since 3 times 2 is 6, I expect 6 combinations, and that is what I have here.0780
So, Descartes' Rule of Signs, and knowing that the degree is 6 (so I have 6 total roots)0785
allows me to figure out the possible combinations of the roots of a polynomial.0790
Conjugate roots: we talked a bit about complex conjugates and the fact that there are complex roots.0798
But getting into a bit more detail: if the polynomial p(x) has real coefficients...0806
the coefficients are not imaginary, which is what we will be working with--real coefficients--for now...0812
the complex roots of p(x) occur as complex conjugates.0819
And you saw this earlier on, when we worked with quadratic equations and quadratic functions.0823
Thinking about something like this: x2 + 4 = 0, if I wanted to find the solutions for this, I could say, "OK, x2 = -4."0830
Now, using the square root property, I take the square root of both sides; and this gives me x = ±√-4.0851
Recall that the square root of -1 equals i; then I can rewrite this as x = ±√-1, times √4,0861
or x = ±i√4, or x equals ±2i.0875
And we can look at this, also, in a different way: we can say x = 0 + 2i, or x = 0 - 2i.0886
And you can then see that this is a pair of complex conjugates.0896
And it is because of this, where we are finding the square root, that we end up with plus or minus some number.0902
And therefore, these complex roots of polynomials occur as complex conjugates.0909
Another example of complex conjugates would be, say, 8 + 5i and 8 - 5i--the same values, but switch the sign--or 2 + 7i and 2 - 7i.0918
And this explains why, when we talked about the number of negative real roots and positive real roots,0935
we said it was that number (say, 4), or less than that by 2.0943
And the reason it goes down by pairs is because the complex conjugates occur in pairs.0946
So, they could take up two of the spots for the roots; and so they are going to decrease the number of the real roots by 2 or multiples of 2.0951
Here, we are asked to solve this polynomial equation, x3 - x2 - 6x = 0.0965
I am going to solve this by factoring; and the first step is to factor out the greatest common factor.0975
And here, I see I have a common factor of x; there is an x in each of these.0980
So, factor that out; this is going to give me x times x2, minus x, minus 6, equals 0.0984
Don't forget to bring this x along as you factor this, because this is going to give us one of the solutions.0995
I have x2 - x - 6 that I need to factor; this is in the form (x + a constant) (x - a constant), because the sign is negative.1002
Factors of 6: 1 and 6, 2 and 3; and I need these to add up to -1, and their signs are going to be opposite.1015
These (2 and 3) are close together, so if I make the larger one negative and the smaller one positive, I am going to get a -1.1026
Therefore, 3 is negative, and 2 is positive.1037
Now, according to the zero product property, if any of these terms (x, x + 2, or x - 3) is 0, then this product will be 0.1042
And it will equal the right side of the equation.1053
Therefore, x equals 0 (and make sure you don't leave this one out, or you will be missing one of the solutions), x + 2 = 0, or x - 3 = 0.1056
So here, I don't have to do anything further with this.1068
I just have that one of the solutions here is x = 0.1070
Here, I have to subtract 2 from both sides to get x = -2; and here, I need to add 3 to both sides.1075
So, solutions: x = 0, x = -2, and x = 3; these are all solutions to this equation.1083
And I solved this by factoring out the greatest common factor, x, factoring this trinomial,1098
and then using the zero product property to solve for x in each of these expressions' terms.1107
Here, I am asked to solve x4 - 256 = 0.1119
This is actually the difference of two squares; this is in the form a2 - b2.1126
So, it is going to factor out to (a + b) (a - b).1133
And if you look at it and think about it this way, x2, squared, is x4.1137
And if you look at this one, 256, and take the square root of that, it is actually 16.1143
Therefore, this is telling me that a equals x2, and b equals 16.1155
So, I can factor it as follows: a + b (that is x2 + 16), times a - b (or x2 - 16).1161
OK, and looking at what I have, I can't do anything else with x2 + 16.1177
But I recognized again, here: I have the difference of two squares; this time, a equals x, and b equals 4.1182
So, it is going to factor out to (x + 4) (x - 4); and these are set equal to 0.1189
Now, I use the zero product property, which is going to tell me that I could have x2 + 16 = 0, x + 4 = 0, or x - 4 = 0.1200
And let's work with these simpler ones first.1214
Simply subtract 4 from both sides to give me x = -4.1217
Add 4 to both sides: x = 4; I have two of my solutions.1224
Now, looking over here, it is a little bit more complex: subtract 16 from both sides--that gives me x2 = -16.1229
Now, I am going to take the square root of both sides, and you can see that this ends up being ±√-16.1239
And this is a negative number; and since I know that √-1 equals i, I can rewrite this as x = ±i√16.1247
Well, the square root of 16 is 4, so this gives me a complex conjugate pair, plus or minus 4i.1264
So, I have four solutions: x = 4i, x = -4i, x = 4, and x = -4.1276
And let's just think about Descartes' Rule of Signs and show that it predicted the possibilities for the type of roots that I could get.1293
Since I have x4 - 256, if I look at the number of sign changes for this, this is positive to negative (one sign change).1303
This tells me that I am going to have one positive real root, or less than that by an even number;1320
but I can't go there, because then I would be going into negative numbers,1327
and I can't say there are -1 real roots; that wouldn't make sense.1330
So, it is just one positive real root.1333
Now, looking at f(-x): this gives me -x4 - 256.1340
Well, this -1, when you take it to the fourth power, is just going to become positive; so this gives me this.1350
And again, I have one sign change; so this tells me that I have one negative real root.1359
Since the degree here is 4, I have 4 total roots.1370
So, this is going to leave me with one positive real, one negative real, and two complex roots,1376
which is exactly what I see: a positive real, a negative real, and the set of complex conjugates.1387
Determine the possible combination of positive real roots, negative real roots, and complex roots.1395
We will use Descartes' Rule of Signs to determine this, and we will start out by thinking about the total.1402
Since the degree is 3, there are 3 total roots.1410
Now, using the rule of signs, I am going to look for f(x) and the sign changes: this is 2x3 - 3x2 + 4x - 5.1421
Number of sign changes: well, this is positive, and this is negative--that is one.1432
I am going from negative here to positive here; that is two; from positive to negative--that is 3.1440
So, the number of sign changes equals 3.1448
Therefore, the number of positive real roots is 3, or less than this by an even number (3 - 2 is 1).1450
You can't go any lower than that; if I subtracted by 2 again, I would get a negative number.1465
So, I have either 3 or 1 positive real roots.1470
All right, now let's look at the negative scenario for f(-x) to figure out the negative real roots.1477
2 times -x cubed, minus 3 times -x squared, plus 4 times -x, minus 5:1484
OK, this is going to give me f(-x) =...this is going to remain negative, so this is going to give me -2x3.1500
A negative and a negative (squared) is going to give me a positive, so this will become x2.1512
This is negative here, though, so it is -3x2.1519
4 times -x is -4x, minus 5.1524
OK, the number of sign changes: none, none, none: the signs are all negative--the coefficients of f(-x).1529
So, the number of sign changes equals 0; so there are 0 negative real roots.1542
OK, so let's figure out what we have going on here.1551
We have positive real roots, negative real roots, and complex.1554
And remember: these need to total 3.1567
Positive real: I could have 3; negative real: 0; I need them to total 3--this already does, so the complex is going to be 0, since this totals 3.1571
OK, another possibility is that I have one positive real root, 0 negative real roots.1586
1 and 0 is 1; I need it to total 3; so there must be 2 complex real roots.1594
So, the possible combination of positive real roots, negative real roots, and complex roots is 3, 0, 0, or 1, 0, 2.1601
And I know I have three total roots, because the degree is 3.1612
I use Descartes' Rule of Signs to tell me that I had either 3 or 1 positive real roots; I have 0 negative real roots.1615
And then, figuring out what is lacking to get my total of 3, I could fill it in with complex roots.1623
OK, determine the possible combinations of positive real roots, negative real roots, and complex roots.1632
Degree equals 4, so I have four total roots.1639
f(x) = -3x4 - 4x3 + 2x2 + 6x + 7.1649
So, positive real roots: let's look for sign changes between these coefficients.1659
A negative to a negative; a negative to a positive (that is 1); a positive to a positive--no sign change; a positive to a positive again.1667
So, the positive real roots equals 1; and I can't go by less than that,1678
because if I subtracted 2, I would go into negative numbers.1685
So, the number of positive real roots is 1.1688
Now, looking at negative real roots: I am going to take f(-x) to get1691
-3 times -x to the fourth, minus 4 times -x to the third, times 2 times -x squared, plus 6 times -x, plus 7.1700
OK, this gives me f(-x) =...well, -1 to the fourth power--this will become positive; so I have -3x41718
Here, -x cubed...this is going to remain negative as a coefficient...times -4; that is going to become positive, so this is going to give me + 4x3.1733
-x squared is going to give me positive x2 times 2; so that is + 2x2; 6 minus -x is -6x; plus 7.1748
OK, negative real roots is going to be determined by the sign changes of the coefficients of f(-x), according to Descartes' Rule of Signs.1760
A negative to a positive; that is one sign change; that stays positive; a positive to a negative: 2; a negative to a positive: 3.1772
So, here I have 3 or less than that by an even number: 3 - 2 is 1.1784
You can't go any less than that.1796
I have a total of 4 roots; I have one positive real root, and either 3 or 1 negative real roots.1799
So, let's look at the possibilities: positive real roots, negative real roots, and complex.1810
Positive real: 1; then I could have a negative real root totaling 3: 1 + 3 is 4; I only have 4; so complex must be 0.1825
Or I could have 1 positive real root and 1 negative real root; and I am going to total those to get 2.1838
But I should have 4; so in this case, there would be two complex roots--a pair of complex conjugates.1847
OK, so I determined there were four total roots, because the degree here is 4.1855
Using Descartes' Rule of Signs, the positive case, f(x), I found that there is one positive real root.1861
f(-x) gave me three sign changes, so there are either three or one negative real roots.1869
And then, to get a total of four, I had zero complex roots in this case, and two in this case.1875
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# How many bulbs can be connected in parallel
I need to know how many 15W bulbs can be connected in parallel If I use an extension cord in the 5A socket? If I use the 15 A socket then how many bulbs can I connect in parallel?
Assuming the voltage to be 220 V,
The current drawn by a single bulb is 15/220=0.068 A
So the no. of bulbs which can be connected in parallel = 5/.068=73 (avoiding the fractional part)
If you use 15A, then 220 bulbs can be connected. (follow the same procedure)
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https://tesckt.com/2sc5200-transistor/ | 1,656,504,736,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00474.warc.gz | 619,671,598 | 20,299 | ## 2SC5200 transistor
The 2sc5200 is an NPN BJT transistor used for power amplifier applications, this transistor had more variety of features at its working. We explain every electrical characteristic of the transistor from the 2sc5200 datasheet.
Then also we explain about circuit application aspects of the 2sc5200 transistor that include equivalent transistors and different circuits using 2sc5200.
## 2sc5200 pinout
Pin01 – Base
Pin02 – Collector
Pin03 – Emitter
## TO-264 package
For this 2sc5200 transistor, to-264 is the package used in it. It is a package made with plastic and epoxy material.
This package had different applications at IGBT, MOSFET and integrated circuit, specialty of this packages is that it can easily attach to the heat sinks.
## 2SC5200 datasheet
In this section, we include information’s about electrical characteristics shown by the 2sc5200 transistor, so we can use these values to make circuits, and also we can get a proper idea about the 2sc5220 transistor.
## 2sc5200 voltage
The collector-base voltage is 230v and Collector – emitter voltage is 230v, both these values will indicate the high voltage of the 2sc5200 transistor, then also the emitter-base voltage is the triggering voltage it is only 5v.
• Collector – Base Voltage (VCB) = 230V
• Collector-Emitter voltage (VCE) = 230V
• Emitter – Base voltage (VEB) = 5V
## 2sc5200 transistor ampere
The collector current is 15A, it is the current value that indicates the maximum current capacity of the transistor, then the Base current is 1.5A, they are the maximum base current of the transistor.
The DC current gain is mainly utilized at the amplifier application, so the gain values are in two collector current conditions.
• Collector current (IC) = 15A
• Base current (IB)= 1.5A
• DC current gain = VCE = 5V, IC = 1A
= 55 to 160 HFe
= VCE = 5V, IC = 7A
= 35 to 60 HFe
The power dissipation of the 2sc5200 transistor is 150w, and the maximum temperature capacity of the transistor is around -55 to 150˚C.
The frequency response or transition frequency of the transistor device is 30MHz, and we know the importance of collector output capacitance in a transistor, here at 2sc5200 the value is 200pf.
Collector power dissipation (PC) = 150w
Junction temperature = 150˚C
Storage temperature range = -55 to 150˚C
Transition frequency (FT) = 30MHz
Collector output capacitance = 200pf
## 2sc5200 equivalent
The transistor equivalent for 2sc5200 is 2sc3320, 2sc5242, 2sd1313 and FJL4315
## 2sc5200 replacement
The perfect replacement for the 2sc5200 transistor is TTC5200, MJL3281A, MJl13002A, and KTC5242.
## 2sc5200 complementary
The PNP complementary for 2sc5200 is 2sa1943
## 2sc5200 SMD version
The sc5200 is a power transistor, so they don’t have an SMD version transistor.
## 2sc5200 amplifier
The main amplifier circuit using a 2sc5200 transistor is with its complementary PNP transistor that is 2sa1943.
## 2sc5200 amplifier circuit
The main components at this circuit are the 2sc5200 and 2sa1943 complementary transistor combination, and a BC547 transistor is used at the amplifier circuit.
For this amplifier circuit, we can set single and dual power supply, the circuit in the figure works with a dual power supply.
The BC547 transistor is used to set a fixed voltage for the push-pull transistor arrangement, then the audio signal reaches the 2sc5200 and 2sa1943. Both the transistor’s combined operation makes a high power output for the amplifier circuit.
## 2sc5200 transistor uses
• High power audio circuits
• AF circuits
• Audio frequency amplifiers
• Low slew rate device
• Push-pull configuration circuits
• High current circuits
• Used at medium power circuits
• High fidelity audio circuits | 976 | 3,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.873707 |
https://math.stackexchange.com/questions/2929877/mean-value-property-on-ball | 1,631,819,626,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053717.37/warc/CC-MAIN-20210916174455-20210916204455-00603.warc.gz | 448,157,748 | 37,494 | # Mean Value Property on Ball
$$\textbf{Problem}$$ Let $$\Omega$$ be open and connected in $$\mathbb{R}^n$$ for $$n\geq 2$$, and suppose that $$u \in C^2(\Omega)$$. Prove that the following statements are all equivalent. \begin{align*} (\textrm{i}) \quad u \textrm{ is harmonic in } \Omega\\ (\textrm{ii}) \quad \textrm{If } \overline{B_r(x)} \subset \Omega, \textrm{ then}\\ &u(x)=\frac{1}{\textrm{Vol} \partial B_r(x)}\int _{\partial B_r(x)} u(y)dS(y).\\ (\textrm{iii}) \quad \textrm{If } \overline{B_r(x)} \subset \Omega, \textrm{ then}\\ &u(x)=\frac{1}{\textrm{vol}B_r(x)}\int_{B_r(x)} u(y) dy. \end{align*}
I proved (i) and (ii) are equivalent. I want to know how to prove (i) and (iii) are equivalent.
I knew that this problem was already uproaded. However, I don't know the Poisson's Integral fromula on $$\mathbb{R}^n$$....
Any help is appreciated..
Let's see that $$1)$$ implies $$3)$$: \begin{align}\frac{1}{\textrm{vol}B_r(x)}\int_{B_r(x)} u(y) dy&=\frac{1}{\textrm{vol}B_r(x)} \int_0^r\int_{\partial B_s(x)}u(y)dS(y)ds\\ &=\frac{1}{\textrm{vol}B_r(x)} \int_0^r\textrm{vol}\partial B_s(x)u(x)ds\\ &=u(x)\frac{1}{\textrm{vol}B_r(x)}\textrm{vol}B_r(x)=u(x) .\end{align} The second equality is given by the already proved equivalence between $$1)$$ and $$2)$$.
Let's see that $$3)$$ implies $$1)$$:
Define $$\phi(r)=\frac{1}{\textrm{vol}\partial B_r(x)}\int_{\partial B_r(x)}u(y)dS(y).$$ You can check (and probably you already did proving the first equivalence) that $$\phi'(r)=\frac{r}{n\textrm{vol}B_r(x)}\int_{B_r(x)}\Delta u(y)dy.$$ Assume that $$\Delta u$$ is not identically $$0$$, we can find a point $$x$$ and an open ball $$B_\rho(x)$$ such that $$\Delta u(y)>0$$ in $$B_\rho(x)$$ (you can use the same argument if $$\Delta u(y)<0$$). This implies that $$\phi'(s)>0$$ for $$0, thus $$\phi$$ is strictly increasing in this interval. Finally, you can conclude \begin{align}\frac{1}{\textrm{vol}B_\rho(x)}\int_{B_\rho(x)} u(y) dy&=\frac{1}{\textrm{vol}B_\rho(x)} \int_0^\rho\int_{\partial B_s(x)}u(y)dS(y)ds\\ &=\frac{1}{\textrm{vol}B_\rho(x)} \int_0^\rho\textrm{vol}\partial B_s(x)\phi(s)ds\\ &>\phi(0)=u(x),\end{align} which is a contradiction.
• Nice answer, is it possible to prove separately that $(ii) \implies (iii)$? Aug 2 '20 at 5:12 | 868 | 2,262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-39 | latest | en | 0.712106 |
https://capitalsportsbet.com/ZuluCode5/betting-odds-explained-each-way.html | 1,569,116,118,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00049.warc.gz | 401,707,582 | 4,263 | The last format we want to look at is fractional odds. Personally, we aren’t a huge fan of fractional odds because they’re the most challenging to work with. The formula is almost the same as with decimal odds, but it gives your profit instead of total money returned. It also requires you to solve a fraction, which may be a nightmare for a lot of people. Regardless, we are going to walk you through how to do it with the same bet we’ve been working with.
It is a very good idea to shop around to find the best line when you are betting on sports. Books may offer slightly different lines, and you might be able to gain a point or half a point in your favor on certain markets. Consider an NBA game between the Heat and the Lakers – one firm has the Heat 9.5 point favorites and another have the Heat as 8.5 point favorites. You back the Heat with the second firm and they win by exactly nine points. Here you’ve gone from a loss to a win simply by shopping around, and in the long run this will make a big difference to your bottom line.
The most important takeaway is the actual pointspread, which is seven points in this example. The plus sign is always in front of the spread for the underdog and the minus sign is used to signify the favorite. Next to the pointspread in this example is (-110). This number reflects the actual commission (or juice) that the sportsbook is charging to book this bet. If you wager \$100 on New England as the favorite and the Patriots go on to win my more than seven points, you would win \$100. If they won by fewer than seven points or lost the game outright, you would owe this betting outlet \$110. If New England wins by exactly seven points, the bet is considered to be a PUSH and no money exchanges hands. You only pay the 10 percent commission on losing bets.
```Earlier, we explained how the implied probability of -240 is 70.59% and how the implied probability of +210 is 32.36%. Notice these two probabilities total 102.95%. The extra 2.95% is the bookmaker's advantage. It's called vig, and it's basically a commission that they charge customers for placing wagers. By removing the vig, you can see what the fair odds on the game would be.
```
The first part of each line tells you who you are betting on. The first line is a bet on Mike Tyson and the second line is a bet on Old Man River. Next, you’ll see a plus or a minus sign. The plus sign signifies the fighter who is an underdog, and the minus sign signifies the fighter who is the favorite. As you can see, Mike Tyson is the favorite and Old Man River is the underdog.
### As you might guess, you’re mainly going to see American odds in American sportsbooks. The other two formats are much more prominent in Europe and Asia. If you’re betting online, most sportsbooks give you the ability to change all of the odds on the site into the format that you prefer. No format is different regarding payouts; it’s just a different way of presenting the information.
The last format we want to look at is fractional odds. Personally, we aren’t a huge fan of fractional odds because they’re the most challenging to work with. The formula is almost the same as with decimal odds, but it gives your profit instead of total money returned. It also requires you to solve a fraction, which may be a nightmare for a lot of people. Regardless, we are going to walk you through how to do it with the same bet we’ve been working with.
First, however, a word of caution: Sports betting can be a fun and profitable venture. However, like most good things in life there are pitfalls to be aware of. You should be able to enjoy many positive experiences as long as you bet in moderation and under control. We know you have heard this before but it definitely bears repeating: don’t bet money you can’t afford to lose, either emotionally or financially. If you or someone you know shows signs of compulsive gambling, one place to find help is Gamblers Anonymous.
Here in this point spread example for the NFL, the Falcons are playing the Panthers. Atlanta has been set as a three-point favorite on the betting line. That means that for Atlanta to cover the spread that has been set, they will need to win by at least four points. And for Carolina to cover the point spread, they can do so with a loss by two points or less, or obviously a win straight up. If the Falcons win by exactly three points, the bet would result in a push with no payouts.
# Now that we’ve covered a lot of the basics concerning moneyline bets, let’s talk about the fun stuff – how much you’re going to make on your next correct moneyline bet. Remember, most online sportsbooks will automatically calculate the amount you are going to make on a moneyline bet before you even make the bet. You’re able to put in the amount you want to bet, and they will tell you immediately how much you would win from a correct pick.
Bets on “Winner of Point”, “Scorer of Goal" and similar offers refer to the participant winning the listed occurrence. For the settlement of these offers, no reference to events happening prior to the listed occurrence will be taken into consideration. Should the listed event not be won within the stipulated time frame, all bets will be declared void, unless otherwise stated.
```This is a very common occurrence in sports betting and sportsbooks have the full right to shift the spread or odds for any given match prior to it starting. Many factors can influence a change of the spread such as injuries, the number of bets coming in for either team or the weather, to name a few. Depending on the timing of placing the bet, the bettor can also have an advantage or a disadvantage depending on which way the spread has shifted.
``` | 1,231 | 5,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-39 | latest | en | 0.960241 |
https://darkessays.com/2022/05/30/according-to-the-central-limit-theorem-if-a-sample-of-size-64-is/ | 1,701,460,075,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100304.52/warc/CC-MAIN-20231201183432-20231201213432-00648.warc.gz | 224,459,253 | 23,275 | # According to the central limit theorem, if a sample of size 64 is
Question 1
According to the central limit theorem, if a sample of size 64 is drawn from a population with a mean of 56, the mean of all sample means would equal _________ .
7.00
56.00
64.00
0.875
128.00
Question 2
According to the central limit theorem, if a sample of size 100 is drawn from a population with a standard deviation of 80, the standard deviation of sample means would equal __________ .
0.80
8
80
800
0.080
Question 3
According to the central limit theorem, for samples of size 64 drawn from a population with μ = 800 and σ = 56, the standard deviation of the sampling distribution of sample means would equal __________ .
7
8
100
800
80
Question 4
Suppose a population has a mean of 90 and a standard deviation of 28. If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean of more than 95 is _________ .
0.1056
0.3944
0.4286
0.8944
1.0000
Question 5
Suppose a population has a mean of 90 and a standard deviation of 28. If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean between 85 and 95 is _________ .
0.1056
0.3944
0.7888
0.2112
0.5000
Question 6
Suppose a population has a mean of 90 and a standard deviation of 28. If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean between 80 and 100 is _________ .
0.9876
0.0124
0.4938
0.0062
1.0000
Question 7
Suppose a population has a mean of 400 and a standard deviation of 24. If a random sample of size 144 is drawn from the population, the probability of drawing a sample with a mean of more than 404.5 is __________ .
0.0139
0.4861
0.4878
0.0122
0.5000
Question 8
Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers. Accordingly, he plans a sample of waiting times for walk-in customers. If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert’s sample of 64 will have a mean less than 14 minutes is _________ .
0.4772
0.0228
0.9772
0.9544
1.0000
Question 9
Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers. Accordingly, he plans a sample of waiting times for walk-in customers. If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert’s sample of 64 will have a mean between 13.5 and 16.5 minutes is __________ .
0.9974
0.4987
0.9772
0.4772
0.5000
Question 10
Suppose 40% of the population possess a given characteristic. If a random sample of size 300 is drawn from the population, then the probability that 44% or fewer of the samples possess the characteristic is __________ .
0.0793
0.4207
0.9207
0.9900
1.0000
Question 11
Suppose 30% of a population possess a given characteristic. If a random sample of size 1200 is drawn from the population, then the probability that less than 348 possess that characteristic is __________ .
0.2236
0.2764
0.2900
0.7764
0.3336
Question 12
If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89?
0.1064
0.2700
0.3936
0.6064
0.9000
Question 13
Suppose 40% of all college students have a computer at home and a sample of 100 is taken. What is the probability that more than 50 of those in the sample have a computer at home?
0.4793
0.9793
0.0207
0.5207
0.6754
Question 14
Pinky Bauer, Chief Financial Officer of Harrison Haulers, Inc., suspects irregularities in the payroll system. If 10% of the 5,000 payroll vouchers issued since January 1, 2000, have irregularities, the probability that Pinky’s random sample of 200 vouchers will have a sample proportion of between .06 and .14 is ___________ .
0.4706
0.9706
0.0588
0.9412
0.8765
Question 15
Catherine Chao, Director of Marketing Research, needs a sample of Kansas City households to participate in the testing of a new toothpaste package. If 40% of the households in Kansas City prefer the new package, the probability that Catherine’s random sample of 300 households will have a sample proportion between 0.35 and 0.45 is __________ .
0.9232
0.0768
0.4616
0.0384
0.8976
Question 16
Kristen Ashford purchased the subscribers list for Wind Surfing magazine. She plans to survey a sample of the subscribers before using the list in her mail order business. She randomly selects the fourth name as a starting point and then selects every 50th subsequent name (54, 104, 154, etc.). Her sample is a ___________ .
simple random sample
stratified sample
systematic sample
convenience sample
cluster sample
Question 17
Pinky Bauer, Chief Financial Officer of Harrison Haulers, Inc., suspects irregularities in the payroll system. She knows that 2,500 payroll vouchers have been issued since January 1, 2000, and her staff doesn’t have time to inspect each voucher. So, she randomly selects 53 as a starting point and orders her staff to inspect the 53rd voucher and each voucher at an increment of 100 (53, 153, 253, etc.). Her sample is a ____________ .
stratified sample
simple random sample
convenience sample
cluster sample
systematic sample
Question 18
Financial analyst Larry Potts needs a sample of 100 securities listed on the New York Stock Exchange. In the current issue of the Wall Street Journal, 2,531 securities are listed in the “New York Exchange Composite Transactions,” an alphabetical listing of all securities traded on the previous business day. Larry uses a table of random numbers to select 100 numbers between 1 and 2,531. His sample is a ____________ .
quota sample
simple random sample
systematic sample
stratified sample
cluster sample
Question 19
On Saturdays, cars arrive at David Zebda’s Scrub and Shine Car Wash at the rate of 80 cars per hour during the ten-hour shift. David wants a sample of 40 Saturday customers to answer the long version of his quality service questionnaire. He instructs the Saturday crew to select the first 40 customers. His sample is a __________ .
convenience sample
simple random sample
systematic sample
stratified sample
cluster sample
Question 20
A carload of palletized aluminum castings has arrived at Mansfield Motor Manufacturers. The car contains 1,000 pallets of 100 castings each. Mario Munoz, manager of Quality Assurance, directs the receiving crew to deliver the 127th and 869th pallets to his crew for 100% inspection. Mario randomly selected 127 and 869 from a table of random numbers. Mario’s sample of 200 castings is a ____________ .
simple random sample
systematic sample
stratified sample
cluster sample
convenience sample
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 2,102 | 8,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-50 | latest | en | 0.905582 |
https://mathoverflow.net/questions/221447/how-to-realize-the-descent-data-of-qcoh-as-a-pseudo-limit-in-cat | 1,571,002,311,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986647517.11/warc/CC-MAIN-20191013195541-20191013222541-00301.warc.gz | 676,703,044 | 25,628 | # How to realize the descent data of Qcoh as a (pseudo)-limit in Cat?
It is well-know that $Qcoh$ is a fibered category on $Sch$. In more details let $\mathcal{C}$ be the category $(Sch/S)$ of schemes over a fixed base scheme S. For each scheme $U$ we define $Qcoh(U)$ to be the category of quasi-coherent sheaves on $U$. Given a morphism $f : U \to V$, we have a functor $f^* : Qcoh(V) \to Qcoh(U)$. However we don't have $(gf)^*=f^*g^*$ on the nose, but there is a canonical natural equivalence between them. See Vistoli's notes Section 3.1 and 3.2.1.
Now let $X$ be a scheme and $U_i$ be an open cover of $X$. We have the following cosimplicial diagram of categories $$\prod Qcoh(U_i)\rightrightarrows\prod Qcoh(U_i\times_X U_j)\text{triple arrows}\prod Qcoh(U_i\times_X U_j\times_X U_k)\ldots$$ Keep in mind that this diagram is only a pseudo diagram in $Cat$, the $2$-category of categories, i.e. the cosimplicial identities only hold up to canonical natural equivalence.
It is also well-known that the descent data of quasi-coherent sheaves is given by a collection $(\xi_i,\phi_{ij})$, where $\xi_i$ is a quasi-coherent sheaf on each $U_i$ and $\phi_{ij}$ is an isomorphism $pr_2^*\xi_j\to pr_1^*\xi_i$ in $Qcoh(U_{i}\times_X U_j)$ which satisfies the cocycle condition $$pr_{13}^*\phi_{ik}=pr_{12}^*\phi_{ij}\circ pr_{23}^*\phi_{jk}.$$ on $U_i\times_X U_j\times_X U_k$. See Vistoli's notes Section 4.1.2.
The above descent data is given "by hand". On the other hand, I've heard that descent data is a kind of (homotopy) limit. Nevertheless Vistoli's notes doesn't consider homotopy limit.
$\textbf{My question}$ is: is there any reference which studies the (pseudo) homotopy limit (I'm not sure whether we should call it 2-categorical limit) of the above cosimplicial pseudo-diagram and show that it coincides with the descent data given in the literature?
• If you truncate the cosimplicial diagram above degree 2 and unfold the definition of pseudocone you will see that it is more or less the same as the definition of descent data. So there's not really much to prove at all. If that seems obscure, you may want to first think about defining the set of matching families of elements of a presheaf as a limit of a cosimplicial diagram truncated above degree 1. – Zhen Lin Oct 21 '15 at 7:34 | 668 | 2,307 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-43 | latest | en | 0.811959 |
https://www.coursehero.com/tutors-problems/Accounting/468000-Harris-Company-manufactures-and-sells-a-single-product-A-partically/ | 1,540,122,797,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514005.65/warc/CC-MAIN-20181021115035-20181021140535-00542.warc.gz | 897,164,122 | 21,108 | View the step-by-step solution to:
# Harris Company manufactures and sells a single product.
Harris Company manufactures and sells a single product. A partically completed schedule of the company's total and per unit cost over the relevant range of 30,000 to 50,000 per units produced and sold are:
United produced and Sold: 30,000 ; 40,000; 50,000
Total costs:
Variable costs: @ 30,000 = \$180,000
Fixed costs: @ 30,000 = 300,000
Total costs: = \$480,000
Cost per unit:
Variable costs: @ 30,000 = \$6
Fixed costs: @ 30,000 = \$10
Total per units costs: @ 30,00 = \$16
What are total costs for 40,000 and 50,000?
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Browse Documents | 256 | 999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-43 | longest | en | 0.900892 |
https://au.mathworks.com/matlabcentral/cody/problems?action=index&controller=problems&sort=&term=tag%3A%22vectors%22+difficulty_rating_bin%3Amedium | 1,670,408,700,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00631.warc.gz | 145,696,019 | 21,049 | Problems
1 – 36 of 36
Problem Title Likes Solvers Difficulty
Problem 44398. ベクトルの値が増加しているかを調べよう
Created by: mizuki
2
253
Problem 381. Angle between two vectors
Created by: Aurelien Queffurust
2
574
Problem 341. count to vector
Created by: Chris
0
85
Created by: Bert
0
20
Problem 1737. The sum of individual numbers...
Created by: Chris E.
3
75
Problem 163. Love triangles
Created by: the cyclist
4
554
Problem 763. Find the elements of a matrix according to a defined property.
Created by: Julio
Tags vectors, matrix
3
81
Problem 44439. Remove the air bubbles from a vector
Created by: Shaul Salomon
Tags vectors, 22105
4
182
Problem 762. Matrix which contains the values of an other matrix A at the given locations.
Created by: Julio
Tags vectors, matrix
1
221
Problem 42612. Convert ColorSpec string to RGB triplet
Created by: Matthew Eicholtz
1
40
Created by: Tim
Tags vectors
1
56
Problem 42755. Angle bisectors
Created by: Daniel Pereira
2
23
Problem 2199. Pairwise Euclidean Distance
Created by: Marios
1
43
Problem 2244. Compute hamming distance between two binary vectors represented using lists of 1-byte numbers
Created by: Ryszard Maciol
1
39
Problem 44443. Matrix to vector transformation
Created by: jean claude
1
34
Problem 44929. Plot Line Specifications
Created by: Pooja Lalan
Tags vectors, plot
2
88
Problem 867. Replicate elements in vectors
Created by: Jean-Marie Sainthillier
11
1088
Problem 44500. Find the starting index of a consecutive condition
Created by: Shaul Salomon
2
48
Problem 45440. Create a vector of the first n odd numbers (★)
Created by: Anirban Pal
2
126
Problem 45444. Create a vector with n repeated values of a number x (★★)
Created by: Anirban Pal
Tags vectors
0
105
Problem 45857. Doubling elements in a vector (★★)
Created by: Anirban Pal
Tags vectors
0
85
Problem 6. Select every other element of a vector
Created by: Cody Team
161
27132
Problem 838. Check if number exists in vector
Created by: Nichlas
45
8236
Problem 42613. Convert RGB triplet to ColorSpec string
Created by: Matthew Eicholtz
0
31
Problem 576. Return elements unique to either input
Created by: James
6
722
Problem 45441. Create a vector of n alternating ones and zeros (★★)
Created by: Anirban Pal
Tags vectors
0
101
Problem 45397. Assess the scatter of wind turbines in a field
Created by: Karl Ezra Pilario
2
48
Problem 44948. Calculate a Damped Sinusoid
Created by: Pooja Lalan
45
6553
Problem 10. Determine whether a vector is monotonically increasing
Created by: Cody Team
116
16427
Problem 44483. Separate even from odd numbers in a vector - without loops
Created by: Shaul Salomon
1
383
Problem 44416. Sum of adjacent elements in a vector
Created by: Shaul Salomon
Tags vectors, 22105
1
560
Problem 16. Return the largest number that is adjacent to a zero
Created by: Cody Team
Tags vectors
25
4605
Problem 44490. Vector pop
Created by: Shaul Salomon
Tags vectors, 22105
2
379
Problem 44484. Separate even from odd numbers in a vector - with a loop
Created by: Shaul Salomon
1
443
Problem 29. Nearest Numbers
Created by: Cody Team
39
3894
Problem 38. Return a list sorted by number of occurrences
Created by: Cody Team
13
2022
1 – 36 of 36 | 937 | 3,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-49 | latest | en | 0.746725 |
https://www.compadre.org/Physlets/electromagnetism/illustration26_1.cfm | 1,631,792,040,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053493.41/warc/CC-MAIN-20210916094919-20210916124919-00716.warc.gz | 757,004,122 | 4,977 | ## Illustration 26.1: Microscopic View of a Capacitor
Please wait for the animation to completely load.
The animation shows a parallel-plate capacitor (at the top) connected to a battery (at the bottom). This Illustration shows you what happens when the battery is connected and the blue electrons are separated from the positively charged atoms. This animation only shows what happens for ten charge pairs. Restart.
As the electrons pile up on the left plate, what is the direction of the electric field between the plates? The electric field always points from positive charges and toward negative charges; therefore, the electric field points to the left. Charges move until the electric potential between the two plates matches the electric potential of the battery.
Now change the configuration to add a thin dielectric between the capacitor. What happens to the atoms on the dielectric material between the plates (represented by the circles)? Due to the electric field created by the charges on the capacitor plates, the charges in the dielectric are polarized. Since positive and negative charges experience forces in opposite directions in the same electric field, the electrons move to the right and the positively charged atoms move to the left. Note that these charges are not free to completely separate and move like the charges in the plates and wires. They only polarize. Because these charges are still bound together, we call them bound charges.
What is the direction of the electric field due to the separation of charge (the bound charge) in the dielectric? It is in the opposite direction from the initial electric field. Thus, the overall electric field between the plates (for the same number of charges) is smaller. Since the potential across the plates matches the potential of the battery, is this battery bigger or smaller than the first one? If the charge on the capacitor is the same in both animations, and the capacitance is increased with the inclusion of the dielectric (C = k ε0A/d), then ΔV = Q/C shows us that the electric potential difference is reduced. In addition, since the electric field is reduced, the electric potential is similarly reduced (for a constant electric field ΔV = -Ed).
If the battery was the same in the two animations (ΔV would be the same between the plates), the set of plates with a dielectric between them would be able to hold more charge. Using ΔV = -Ed, the electric field between the plates would have to be the same in the two animations. Because the bound charge of the dielectric reduces the electric field between the plates, there would have to be more charge on the plates with the dielectric present than without. This explains why the capacitance of the capacitor is greater with a dielectric.
Illustration authored by Anne J. Cox and Mario Belloni.
Script authored by Morten Brydensholt.
Applet authored by Vojko Valencic.
Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College. | 608 | 3,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-39 | latest | en | 0.930642 |
https://brainly.in/question/307434 | 1,485,294,055,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285289.45/warc/CC-MAIN-20170116095125-00107-ip-10-171-10-70.ec2.internal.warc.gz | 795,791,889 | 10,049 | # Answer these questions- 1. 7-8/(-2)+3*(-4) 2.9-[7-24/(8+6*2-16)]
2
by gbalaji
2016-03-20T15:20:37+05:30
Using BODMAS Theorem,we can solve these:
1) 7-[8/-2]+3x(-4)
=7+4-12
=-1
2) 9-{7-24/(8+6x2-16)}
=9-[7-24/(20-16)]
=9-[7-6]
=9-1
=8
2016-03-20T15:32:37+05:30
Hi there,
1.7+4+-12=-1
2.9-97-6)=9-1=8
i hope this helps u... | 191 | 325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-04 | latest | en | 0.542018 |
http://mathoverflow.net/questions/108011?sort=newest | 1,369,028,082,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698354227/warc/CC-MAIN-20130516095914-00052-ip-10-60-113-184.ec2.internal.warc.gz | 166,150,332 | 11,462 | ## Countably many random vectors and related problems.
Say I have a countably infinite number of iid random vectors $X_i:i\in\mathbb{N}$, each uniformly distributed on $[0,1]^k$ with say, $k=2$.
I need to evaluate stuff like:
$E_{X_0,X_1,\ldots}[\int_{[0,1]^k}\inf_{i\in\mathbb{N}}\|X_i-y\|^2dy]$,
which I "believe" is equal to $0$ (this is indeed a random vector quantizer). Regarding my belief, I can at least show that
$E_{X_0,X_1,\ldots,X_n}[\int_{[0,1]^k}\min_{0 \leq i \leq n}\|X_i-y\|^2dy] \rightarrow 0$ as $n\rightarrow\infty$.
Remark: Obviously writing the first expectation above does not make sense unless one defines how to do it; I do not know how to do it.
Is there any reference that might be relevant and covers similar problems?
-
The first expectation makes perfect sense; the underlying probability space is the product of countably many copies of $[0,1]^k$. Unless I'm overlooking something all you need here is the dominated convergence theorem. – Mark Meckes Sep 25 at 0:13
1) Yes, you can define properly the first expectation, see e.g. http://planetmath.org/encyclopedia/TotallyFiniteMeasure.html
2) Then you have with your notations $$\mathbb E_{X_i : i\in\mathbb N}\int_{[0,1]^k}\inf_{i\in\mathbb N}\|X_i-y\|^2dy\leq \mathbb E_{X_i : i\in\mathbb N}\int_{[0,1]^k}\min_{1\leq i \leq N}\|X_i-y\|^2dy = \mathbb E_{X_1,\ldots,X_N}\int_{[0,1]^k}\min_{1\leq i \leq N}\|X_i-y\|^2dy$$ and use what you know.
Anyway, questions like this should be first asked on math stack exchange, they are not "research level questions".
-
Wait - why so complicated ? Don't you have that the sequence $(X_n)$ is a.s. everywhere dense in the cube ? Then the inf in the expectation you want to compute is always equal to $0$ ... | 556 | 1,741 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2013-20 | latest | en | 0.827836 |
http://stackoverflow.com/questions/14249153/how-to-search-an-area-of-the-same-values-in-a-matrix/14249211 | 1,441,015,135,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065910.17/warc/CC-MAIN-20150827025425-00041-ip-10-171-96-226.ec2.internal.warc.gz | 228,637,103 | 20,300 | # How to search an area of the same values in a matrix?
I have a matrix of integers like this:
``````1 1 1 0 3 3 3 0 2 2
1 1 1 0 3 3 3 0 2 2
0 0 0 0 3 3 3 0 0 0
2 2 0 0 3 3 3 0 4 4
0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 4 4 4 0
``````
where I have different areas of the same value that are separated by "0"s and I need to count how many areas are in the matrix. My algorithm is based on the "0"s and every time a "0" is found a new area is about to come so I increment the counter. The problem is that I search row by row and enter the same area more than once. I only need to count the rectangular areas.
-
Showing one example would make your question more understandable. – User 104 Jan 10 '13 at 0:58
How are the matrices stored? – Ben Jan 10 '13 at 0:59
Are the areas guaranteed to be rectangles? – GManNickG Jan 10 '13 at 0:59
the answer for this matrix is 7 – Buradi Jan 10 '13 at 1:00
@GManNickG yes they are – Buradi Jan 10 '13 at 1:02
One simple, fast algorithm is to iterate over all entries and set each region to zeroes when it is encountered. This takes O(N*M) runtime (each entry visited at most twice) and O(1) additional memory.
To set a region to zeroes, just note the column where it begins, then iterate to the rightmost column. Then iterate over the rows below from the left to right columns, setting each entry to zero.
Working code:
``````int count_regions( int *arr, int rows, int cols ) {
int region_count = 0;
for ( int first_index = 0; first_index != rows * cols; ++ first_index ) {
if ( arr[ first_index ] == 0 ) continue;
++ region_count;
int first_row = first_index / cols, first_col = first_index % cols;
int last_col;
for ( last_col = first_col;
last_col != cols && arr[ first_row * cols + last_col ] != 0;
++ last_col ) ;
for ( int last_row = first_row;
last_row != rows && arr[ last_row * cols + first_col ] != 0;
++ last_row ) {
for ( int col = first_col; col != last_col; ++ col ) {
arr[ last_row * cols + col ] = 0;
}
}
}
return region_count;
}
``````
-
This will work for rectangular entries. The general version I posted works for entries of any shape. – Mel Nicholson Jan 10 '13 at 1:11
@MelNicholson Yep, I made that assumption. Come to think of it, the bitmap fill is easier to compute if you have an original, unmodified copy sitting around… – Potatoswatter Jan 10 '13 at 1:14
could you give me a pseudocode as an example, it would be easier for me to understand – Buradi Jan 10 '13 at 1:16
@Buradi even better… – Potatoswatter Jan 10 '13 at 1:36
You modified the input array... add a copy – Mel Nicholson Jan 10 '13 at 1:57
The painting technique works well. The idea is that you drop a paint bomb that completely covers an area. Pseudocode:
``````for (each space) {
if (space has been painted or is a border) continue;
else {
numAreas++;
drop paint bomb
}
}
``````
And a paint bomb works like this:
``````paint space;
if (space has been painted or space is a border) continue;
else {
paint space;
drop another bomb;
}
}
``````
-
in the case i am not allowed to modify the original matrix, is it more convinient to make a copy of the matrix and apply this method or to create another boolean matrix of the same size? – Buradi Jan 10 '13 at 1:09
Creating a boolean matrix of the same size would work. Copying the matrix doesn't help. – Mel Nicholson Jan 10 '13 at 1:10
@Buradi It would be easiest to make your own copy. Converting to Boolean or changing the format entirely could improve efficiency, but would certainly be more work. – Potatoswatter Jan 10 '13 at 1:11
@Potatoswatter the boolean matrix starts as false (unpainted) and each is marked as true when painted. That avoids the need for specially marking the starting column or making assumptions about the size or shape of the region. – Mel Nicholson Jan 10 '13 at 1:15
One way is to use the Flood Fill algorithm to detect contiguous areas, like this:
Assuming that the initial matrix has only positive numbers,
• Prepare a `counter`, set it to `-1`
• For each point in the matrix:
• If the cell is zero or negative, skip it; otherwise
• Flood fill starting at the cell with the negative value of the `counter`
• Decrement the counter after the flood fill is over
When you reach the end of your matrix, `counter` will have the negative number equal to the number of contiguous numbered areas minus one, i.e. the result that you need is `-(counter+1)`.
-
I always liked the Union/Find algorithm for isolating groups. But that's cos I had an old implementation kicking around that I had written years ago. It's not a big deal to implement, but it might be more appropriate to use Flood Fill.
- | 1,298 | 4,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2015-35 | latest | en | 0.82502 |
https://www.simscale.com/docs/content/validation/3DPunch/3DPunch.html?_ga=2.10717108.1649374746.1493829183-1857285549.1493829183/ | 1,550,946,605,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249508792.98/warc/CC-MAIN-20190223162938-20190223184938-00451.warc.gz | 955,146,591 | 6,170 | # 3D Punch (Rounded Edges) (NAFEMS Contact Benchmark 2)¶
## Overview¶
The aim of this test case is to validate the following functions:
• penalty contact
• augmented lagrange contact
• friction
The simulation results of SimScale were compared to the results presented in [NAFEMS_R94]. The mesh used in (A) was taken from [SSNA122] and extruded in 3D hexahedral elements. The mesh in (B) was created with the tetrahedralization-with-refinements tool on the SimScale platform.
Import validation project into workspace
## Geometry¶
Geometry of the sphere
Due to the symmetry of the problem only a quarter of the model is used for the simulation. The punch (DGHJKL) is on top of the foundation (ABCDEF).
Punch diameter 100 mm Punch height 100 mm Foundation diameter 200 mm Foundation height 200 mm Fillet radius at the edge of the punch contact 10 mm
## Analysis type and Domain¶
Tool Type : Code_Aster
Analysis Type : Dynamic
Mesh and Element types :
Case Mesh type Number of nodes Element type
(A) linear hexahedral 7803 3D isoparametric
(B) linear tetrahedral 19353 3D isoparametric
Mesh used for the SimScale case (A)
Mesh used for the SimScale case (B)
## Simulation Setup¶
Important
All displacement and load components are referred to the coordinate system in the figure of the geometry section.
Material:
• Punch: isotropic: E = 210 GPa, $$\nu$$ = 0.3
• Foundation: isotropic: E = 70 GPa, $$\nu$$ = 0.3
Constraints:
• Face ABC fixed
• Face ACDF and DHJL zero x-displacement
• Face ABDE and DGJK zero y-displacement
• Uniform pressure $$p$$ = 10⁸ Pa on face JKL
Contact I Contact II Contact III
Solution method Augmented Lagrange Penalty Penalty
Contact smoothing on on on
Coefficient of contact 100 10¹⁴ 10¹⁴
Coefficient of friction $$\mu$$ 0 0 0.1
## Results¶
Comparison of the displacements and the normal pressure of the nodes on edge DE. The values of the reference [MARC] in all figures were extracted from [NAFEMS_R94] with WebPlotDigitizer.
Comparison of the axial displacement on edge DE
Comparison of the radial displacement on edge DE
Normal pressure on edge DE
## References¶
[MARC] The results calculated with MSC.MARC as stated in [NAFEMS_R94]
[NAFEMS_R94]_ 2006 “NAFEMS Advanced Finite Element Contact Benchmarks”, Benchmark 2, NAFEMS Publication Ref: R0094 | 607 | 2,308 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-09 | latest | en | 0.774247 |
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# If 65 percent of x is 195, what is 75 percent of x?
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If 65 percent of x is 195, what is 75 percent of x? [#permalink]
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24 Dec 2018, 23:06
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If 65 percent of x is 195, what is 75 percent of x?
A. 215
B. 225
C. 235
D. 250
E. 260
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Re: If 65 percent of x is 195, what is 75 percent of x? [#permalink]
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24 Dec 2018, 23:18
1
Given : 65 percent of x is 195 ==> 0. 65 x = 195 ==> x=300
Question : what is 75 percent of x?
0.75 x = 0.75 x 300 = 225
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Re: If 65 percent of x is 195, what is 75 percent of x? [#permalink]
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25 Dec 2018, 07:44
Bunuel wrote:
If 65 percent of x is 195, what is 75 percent of x?
A. 215
B. 225
C. 235
D. 250
E. 260
65/100 * x = 195
x= 300
300*75/100 = 225 IMO B
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Re: If 65 percent of x is 195, what is 75 percent of x? [#permalink] 25 Dec 2018, 07:44
Display posts from previous: Sort by | 644 | 1,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-26 | latest | en | 0.846784 |
http://feet.car4car.top/grid-method-multiplication-worksheet/ | 1,600,633,203,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198652.6/warc/CC-MAIN-20200920192131-20200920222131-00064.warc.gz | 44,470,531 | 31,009 | # Grid Method Multiplication Worksheet
In Free Printable Worksheets168 views
4.16 / 5 ( 141votes )
Top Suggestions Grid Method Multiplication Worksheet :
Grid Method Multiplication Worksheet When setting out your multiplication as an area model draw a grid and partition each number base 10 or counters to help you complete the worksheet using the area models provided or draw Let s take a look at an example of solving a multiplication problem in your a 2 digit number by a 1 digit number using a written method with this interactive guide you only need to play The most common method uses a look up table lut also known as a resistance expression of variables that involves only the operations of addition subtraction multiplication and non negative.
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Grid Method Multiplication Worksheet
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Top | 1,853 | 9,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-40 | latest | en | 0.788869 |
https://scidesign.github.io/designbook/fractional-factorial-designs.html | 1,606,605,386,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00345.warc.gz | 453,363,303 | 15,248 | 12 Fractional factorial designs
A $$2^k$$ full factorial requires $$2^k$$ runs. Full factorials are seldom used in practice for large k (k>=7). For economic reasons fractional factorial designs, which consist of a fraction of full factorial designs are used. There are criteria to choose “optimal” fractions.
12.1 Example - Effect of five factors on six properties of film in eight runs
The following example is taken from Box, Hunter, and Hunter (2005).
Five factors were studied in 8 runs. The factors were:
• Catalyst concentration (A)
• Amounts of three emulsifiers (C, D, E)
Polymer solutions were prepared and spread as a film on a microscope slide. Six different responses were recorded.
run A B C D E y1 y2 y3 y4 y5 y6
1 -1 -1 -1 1 -1 no no yes no slightly yes
2 1 -1 -1 1 1 no yes yes yes slightly yes
3 -1 1 -1 -1 1 no no no yes no no
4 1 1 -1 -1 -1 no yes no no no no
5 -1 -1 1 -1 1 yes no no yes no slightly
6 1 -1 1 -1 -1 yes yes no no no no
7 -1 1 1 1 -1 yes no yes no slightly yes
8 1 1 1 1 1 yes yes yes yes slightly yes
• The eight run design was constructed beginning with a standard table of signs for a $$2^3$$ design in the factors A, B, C.
• The column of signs associated with the BC interaction was used to accommodate factor D, the ABC interaction column was used for factor E.
• A full factorial for the five factors A, B, C, D, E would have needed $$2^5 = 32$$ runs.
• Only 1/4 were run. This design is called a quarter fraction of the full $$2^5$$ or a $$2^{5-2}$$ design (a two to the five minus two design). In general a $$2^{k-p}$$ design is a $$\frac{1}{2^p}$$ fraction of a $$2^k$$ design using $$2^{k-p}$$ runs.
12.2 Effect Aliasing and Design Resolution
A chemist in an industrial development lab was trying to formulate a household liquid product using a new process. The liquid had good properties but was unstable. The chemist wanted to synthesize the product in hope of hitting conditions that would give stability, but without success. The chemist identified four important influences: A (acid concentration), B (catalyst concentration), C (temperature), D (monomer concentration). His 8 run fractional factorial design is shown below.
test A B C D y
1 -1 -1 -1 -1 20
2 1 -1 -1 1 14
3 -1 1 -1 1 17
4 1 1 -1 -1 10
5 -1 -1 1 1 19
6 1 -1 1 -1 13
7 -1 1 1 -1 14
8 1 1 1 1 10
The signs of the ABC interaction is used to accommodate factor D. The tests were run in random order. He wanted to achieve a stability value of at least 25.
The factorial effects and Normal, half-Normal, and Lenth plots are below.
library(FrF2)
fact.prod <- lm(y~A*B*C*D,data = tab0602)
fact.prod1 <- aov(y~A*B*C*D,data = tab0602)
round(2*fact.prod$coefficients,2) (Intercept) A B C D A:B 29.25 -5.75 -3.75 -1.25 0.75 0.25 A:C B:C A:D B:D C:D A:B:C 0.75 -0.25 NA NA NA NA A:B:D A:C:D B:C:D A:B:C:D NA NA NA NA DanielPlot(fact.prod,half = F) DanielPlot(fact.prod,half = T) LenthPlot(fact.prod1) alpha PSE ME SME 0.050000 1.125000 4.234638 10.134346 Even though the stability never reached the desired level of 25, two important factors, A and B, were identified. This Normal and half-Normal plots indicate the importance of these factors, although factor B is not significant according to the Lenth plot. What information could have been obtained if a full $$2^5$$ design had been used? Factors Number of effects Main 5 2-factor 10 3-factor 10 4-factor 5 5-factor 1 There are 31 degrees of freedom in a 32 run design. But, are 16 used for estimating three factor interactions or higher. Is it practical to commit half the degrees of freedom to estimate such effects? According to effect hierarchy principle three-factor and higher not usually important. Thus, using full factorial wasteful. It’s more economical to use a fraction of full factorial design that allows lower order effects to be estimated. Consider a design that studies five factors in 16 run. A half fraction of a $$2^5$$ or $$2^{5-1}$$. Run B C D E Q 1 -1 1 1 -1 -1 2 1 1 1 1 -1 3 -1 -1 1 1 -1 4 1 -1 1 -1 -1 5 -1 1 -1 1 -1 6 1 1 -1 -1 -1 7 -1 -1 -1 -1 -1 8 1 -1 -1 1 -1 9 -1 1 1 -1 1 10 1 1 1 1 1 11 -1 -1 1 1 1 12 1 -1 1 -1 1 13 -1 1 -1 1 1 14 1 1 -1 -1 1 15 -1 -1 -1 -1 1 16 1 -1 -1 1 1 The factor E is assigned to the column BCD. But, the column for E is used to estimate the main effect of E and also for BCD. So, this design cannot distinguish between E and BCD. The main factor E is said to be aliased with the BCD interaction. This aliasing relation is denoted by $E = BCD \thinspace or \thinspace I = BCDE,$ where $$I$$ denotes the column of all +’s. This uses same mathematical definition as the confounding of a block effect with a factorial effect. Aliasing of the effects is a price one must pay for choosing a smaller design. The $$2^{5-1}$$ design has only 15 degrees of freedom for estimating factorial effects, it cannot estimate all 31 factorial effects among the factors B, C, D, E, Q. The equation $$I = BCDE$$ is called the defining relation of the $$2^{5-1}$$ design. The design is said to have resolution IV because the defining relation consists of the “word” BCDE, which has “length” 4. Multiplying both sides of $$I = BCDE$$ by column B $B = B \times I = B \times BCDE = CDE$, the relation $$B = CDE$$ is obtained. B is aliased with the CDE interaction. Following the same method all 15 aliasing relations can be obtained. $B = CDE, C = BDE, D = BCE, E = BCD, \\ BC = DE, BD = CE, BE = CD, \\ Q = BCDEQ, BQ = CDEQ, CQ = BDEQ, DQ = BCEQ, \\ EQ = BCDQ, BCQ = DEQ, BDQ = CEQ, BEQ = CDQ$ Each of the four main effects $$B, C, D, E$$ is respectively aliased with $$CDE, BDE, BCE,BCD$$. Therefore, the main effects of $$B,C,D,E$$ are estimable only if the aforementioned three-factor interactions are negligible. The other factorial effects have analogous aliasing properties. 12.3 Example - Leaf Spring Experiment The following example is from Wu and Hamada (2009). An experiment to improve a heat treatment process on truck leaf springs. The heat treatment that forms the camber in leaf springs consists of heating in a high temperature furnace, processing by forming a machine , and quenching in an oil bath. The free height of an unloaded spring has a target value around 8in. The goal of the experiment is to make the variation about the target as small as possible. Five factors were studied in this $$2^{5-1}$$ design. Factor Level B. Temperature 1840 (-), 1880 (+) C. Heating time 23 (-), 25 (+) D. Transfer time 10 (-), 12 (+) E. Hold down time 2 (-), 3 (+) Q. Quench oil temperature 130-150 (-), 150-170 (+) B C D E Q y -1 1 1 -1 -1 7.7900 1 1 1 1 -1 8.0700 -1 -1 1 1 -1 7.5200 1 -1 1 -1 -1 7.6333 -1 1 -1 1 -1 7.9400 1 1 -1 -1 -1 7.9467 -1 -1 -1 -1 -1 7.5400 1 -1 -1 1 -1 7.6867 -1 1 1 -1 1 7.2900 1 1 1 1 1 7.7333 -1 -1 1 1 1 7.5200 1 -1 1 -1 1 7.6467 -1 1 -1 1 1 7.4000 1 1 -1 -1 1 7.6233 -1 -1 -1 -1 1 7.2033 1 -1 -1 1 1 7.6333 The factorial effects are estimated as before. library(FrF2) fact.leaf <- lm(y~B*C*D*E*Q,data = leafspring) fact.leaf2 <- aov(y~B*C*D*E*Q,data = leafspring) round(2*fact.leaf$coefficients,2)
(Intercept) B C D E Q
15.27 0.22 0.18 0.03 0.10 -0.26
B:C B:D C:D B:E C:E D:E
0.02 0.02 -0.04 NA NA NA
B:Q C:Q D:Q E:Q B:C:D B:C:E
0.08 -0.17 0.05 0.03 NA NA
B:D:E C:D:E B:C:Q B:D:Q C:D:Q B:E:Q
NA NA 0.01 -0.04 -0.05 NA
C:E:Q D:E:Q B:C:D:E B:C:D:Q B:C:E:Q B:D:E:Q
NA NA NA NA NA NA
C:D:E:Q B:C:D:E:Q
NA NA
Notice that the factorial effects are missing for effects that are aliased. The Normal, half-Normal, and Lenth plots are below.
DanielPlot(fact.leaf,half = F)
DanielPlot(fact.leaf,half = T)
LenthPlot(fact.leaf2,cex.fac = 0.5)
alpha PSE ME SME
0.0500000 0.0606000 0.1557773 0.3162503
Consider a factorial design to study the effects of the amounts of three factors on the taste of chocolate chip cookies.
Factor Amount
Butter 10g (-1), 15g (+1)
Sugar 1/2 cup (-1), 3/4 cup (+1)
Baking powder 1/2 teaspoon (-), 1 teaspoon (+)
Taste will be measured on a scale of 1 (poor) to 10 (excellent). A full factorial will require $$2^3 = 8$$ runs.
The 8 runs of the full factorial design tell the experimenter how to set the levels of the different ingredients (factors).
Run butter sugar powder
1 -1 -1 -1
2 1 -1 -1
3 -1 1 -1
4 1 1 -1
5 -1 -1 1
6 1 -1 1
7 -1 1 1
8 1 1 1
In the first run the experimenter will bake chocolate chip cookies with 10g butter, 1/2 cup sugar, and 1/2 teaspoon of baking powder; the second run will use 15g butter, 1/2 cup sugar, and 1/2 teaspoon of baking powder; etc.
But, the experimenter decides to also study baking time on taste, but can’t afford to do more than 8 runs since each run requires a different batch of cookie dough. He wants to test if a baking time of 12 minutes versus 16 minutes has an impact on taste
So, he decides to use the three factor interaction between butter, sugar, and powder to assign if baking time will be 12 minutes (-1) or 16 minutes (+1) in each of the 8 runs.
Run butter sugar powder baking time
1 -1 -1 -1 -1
2 1 -1 -1 1
3 -1 1 -1 1
4 1 1 -1 -1
5 -1 -1 1 1
6 1 -1 1 -1
7 -1 1 1 -1
8 1 1 1 1
In this factorial design with four factors in 8 runs the experimenter will bake the cookies with 10g butter, 1/2 cup sugar, 1/2 teaspoon of baking powder, and baking time 12 minutes in the first run; in the second run use 15g butter, 1/2 cup sugar, and 1/2 teaspoon of baking powder, and 16 minutes baking time; etc.
Let $$A$$=butter, $$B$$=sugar, $$C$$=powder, $$D$$=baking time. But, $D = ABC$ since he used the three factor interaction to assign baking times to each run. In other words $$D$$ is aliased with $$ABC$$.
This type of design is called a $$2^{4-1}$$ fractional factorial design. Instead of using a full factorial or $$2^4 = 16$$ runs to study 4 factors we are using $$\frac{1}{2}2^4 = 8$$ runs.
The aliasing relation $D = ABC \Rightarrow I = ABCD,$ where $$I$$ is the column of $$+1$$s.
The aliasing relation also means that other factors in the design have aliases. We can find these aliases by multiplying $$I = ABCD$$ by all the possible main and interaction effects.
$A = BCD, B = ACD, C = ABD, D = ABC, AB = CD, AC = BD, BC = AD$
All the main effects are aliased with three factor interactions, and two factor interactions with two factor interactions.
Suppose that the experimental runs were conducted and the following results were obtained.
Run butter sugar powder baking time taste (y)
1 -1 -1 -1 -1 9
2 1 -1 -1 1 4
3 -1 1 -1 1 7
4 1 1 -1 -1 1
5 -1 -1 1 1 2
6 1 -1 1 -1 5
7 -1 1 1 -1 3
8 1 1 1 1 10
The factorial effect of baking time ($$D$$) is really the effect of $$D+ABC$$. In other words the effects of $$D$$ and $$ABC$$ are confounded. They cannot be separately estimated which is why $$ABC$$ is called an alias of $$D$$.
This means that effect of $$D$$
$\frac{1}{4}\left(-y_1+y_2+y_3-y_4+y_5-y_6-y_7+y_8\right)=-2.5$
is really the effect of $$D+ABC$$ or baking time plus the interaction of butter, sugar, and baking powder. In order for this to be equal to the effect of baking time ($$D$$) we must assume the three-way interaction ($$ABC$$) is small enough to be ignored (i.e., the factorial estimate of $$ABC$$ is close to 0).
If the experimenter were to use a full factorial then he would require $$2^4 = 16$$ different batches of cookies. In a full $$2^4$$ design he would be estimating 4 main effects, 6 two-way interactions, 4 three-way interactions, and 1 four-way interaction. If we assume that we can ignore three-factor and higher order interactions then a 16 run design is being used to estimate then a 16 run design is being used to estimate 10 effects. Fractional factorials use these redundancies by arranging that lower order effects are confounded with higher order interactions that are assumed negligible.
12.5 Questions
1. Consider the table of contrasts from a $$2^{k-p}$$ design. The four factors investigated are A,B,C,D.
## C B A D
## 1 -1 -1 -1 -1
## 2 1 -1 -1 1
## 3 -1 1 -1 1
## 4 1 1 -1 -1
## 5 -1 -1 1 1
## 6 1 -1 1 -1
## 7 -1 1 1 -1
## 8 1 1 1 1
1. What are the values of $$k$$ and $$p$$?
2. What is the defining relation? What is the resolution of this design?
3. What are the aliasing relations?
1. There are 4 factors so $$k=4$$ in 8 runs so, $$p=1$$.
2. By inspection of the table we can see that $$D=ABC$$. The defining relation is $$I=ABCD$$. The defining relation has 4 letters so the resolution is IV.
3. The aliasing relations are: $$D=ABC, A=BCD,B=ACD,C=ABD, AB=CD,AC=BD, AD=BC$$. | 4,173 | 12,814 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-50 | latest | en | 0.943211 |
https://www.physicsforums.com/threads/why-div-grad-1-r-4pd-r-vec.919278/ | 1,511,384,878,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806660.82/warc/CC-MAIN-20171122194844-20171122214844-00670.warc.gz | 861,385,622 | 15,954 | # I Why div [grad(1/r)]=-4πδ(r vec)
Tags:
1. Jul 3, 2017
### cristianbahena
I don't understand why grad(1/r)= (-1/r^2)ê_r. I know its derivate, but in cero the function is not definite and you can't find its derivative in cero.
And, why div [grad(1/r)]=-4πδ(r vec) ?
2. Jul 3, 2017
### Math_QED
This is simply applying definitions. Also,
$\frac{d|x|}{dx} = \frac{|x|}{x} = \begin{cases} 1 \quad \mathrm{if} \quad x >0\\-1 \quad \mathrm{if} \quad x <0 \end{cases}$
3. Jul 3, 2017
### hilbert2
You're probably supposed to show that
$\lim_{\epsilon \rightarrow 0+}\int_{\mathbb{R}^3}\nabla \cdot \left(\nabla \left(\frac{1}{|\mathbf{r}|+\epsilon}\right)\right) f(\mathbf{r})dV \propto f(\mathbf{0})$
for a relatively arbitrary function $f( \mathbf{r})$. Note that $\mathbf{r}$ is a position vector, $|\mathbf{r}|$ its norm and $dV$ a volume element. | 324 | 859 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-47 | longest | en | 0.651502 |
https://hoopercharles.wordpress.com/category/sql/page/4/ | 1,527,350,066,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867559.54/warc/CC-MAIN-20180526151207-20180526171207-00475.warc.gz | 567,436,082 | 40,633 | ## Matching the Expected Output – Analytic RANK, ROW_NUMBER, DENSE_RANK, or Something Different?
16 11 2011
November 16, 2011
I noticed an interesting request on the comp.databases.oracle.misc usenet group, and I thought that I would make an attempt at solving the request. The original poster (OP) stated that he had a table with data similar to the following:
```TX ID DEPT LOCATION LOAD
1 99 A NY 12
2 99 A LA 10
3 99 B LA 05
4 77 B LA 15
5 77 C NY 12
6 77 D LA 11 ```
He would like to obtain the following output:
```TX ID DEPT DEPT_RANK LOCATION LOC_RANK LOAD
1 99 A 1 NY 2 12
2 99 A 1 LA 1 10
3 99 B 2 LA 1 05
4 77 B 1 LA 1 15
5 77 C 2 NY 2 12
6 77 D 3 LA 1 11 ```
The rankings are to be determined as follows:
DEPT_RANK for ID 99 is 1 for A because sum(LOAD) = 22 is the max
LOC_RANK for ID 99 is 1 for LA because sum(LOAD) = 15 is the max
At first glance, that request seems to be reasonably easy to accomplish. Let’s start by creating a table with the sample data (ideally, the OP should have provided the DDL and DML to create and populate this table):
```DROP TABLE T1 PURGE;
CREATE TABLE T1 (
TX NUMBER,
ID NUMBER,
DEPT VARCHAR2(1),
LOCATION VARCHAR2(2),
INSERT INTO T1 VALUES (1,99,'A','NY',12);
INSERT INTO T1 VALUES (2,99,'A','LA',10);
INSERT INTO T1 VALUES (3,99,'B','LA',05);
INSERT INTO T1 VALUES (4,77,'B','LA',15);
INSERT INTO T1 VALUES (5,77,'C','NY',12);
INSERT INTO T1 VALUES (6,77,'D','LA',11); ```
The first step, if we were to think about creating the solution in logical steps, is to find a way to calculate the SUM values that were mentioned by the OP. So, as a starting point, we might try this:
```SELECT
TX,
ID,
DEPT,
LOCATION,
FROM
T1
ORDER BY
TX;
--- --- - ----------- -- ----------------- -----
1 99 A 22 NY 12 12
2 99 A 22 LA 15 10
3 99 B 5 LA 15 5
4 77 B 15 LA 26 15
5 77 C 12 NY 12 12
6 77 D 11 LA 26 11 ```
If I am understanding the OP’s request correctly, the above is a good starting point (even though the alias for the first analytic function could have been better selected).
We are then able to take the above SQL statement and push it into an inline view to hopefully produce the output that is expected by the OP (note that the PARTITION clause differs for the LOC_RANK column from what is specified in the inline view for the function that is used to create that column):
```SELECT
TX,
ID,
DEPT,
RANK() OVER (PARTITION BY ID,DEPT ORDER BY SUM_LOAD_ID DESC) DEPT_RANK,
LOCATION,
RANK() OVER (PARTITION BY ID,DEPT ORDER BY SUM_LOAD_LOCATION DESC) LOC_RANK,
FROM
(SELECT
TX,
ID,
DEPT,
LOCATION,
FROM
T1)
ORDER BY
TX;
TX ID D DEPT_RANK LO LOC_RANK LOAD
--- --- - ---------- -- ---------- -----
1 99 A 1 NY 2 12
2 99 A 1 LA 1 10
3 99 B 1 LA 1 5
4 77 B 1 LA 1 15
5 77 C 1 NY 1 12
6 77 D 1 LA 1 11 ```
Let’s compare the above output with what the OP requested:
```TX ID DEPT DEPT_RANK LOCATION LOC_RANK LOAD
1 99 A 1 NY 2 12
2 99 A 1 LA 1 10
3 99 B 2 LA 1 05
4 77 B 1 LA 1 15
5 77 C 2 NY 2 12
6 77 D 3 LA 1 11```
Almost for the LOC_RANK column (not even close for the DEPT_RANK column), but not quite right. The problem is that when attempting to calculate the RANK columns in the above output, we should only PARTITION on the ID column, not the ID column and some other column, as was the case when we used the SUM analytic function.
Let’s fix the PARTITION clause and try again:
```SELECT
TX,
ID,
DEPT,
RANK() OVER (PARTITION BY ID ORDER BY SUM_LOAD_ID DESC) DEPT_RANK,
LOCATION,
RANK() OVER (PARTITION BY ID ORDER BY SUM_LOAD_LOCATION DESC) LOC_RANK,
FROM
(SELECT
TX,
ID,
DEPT,
LOCATION,
FROM
T1)
ORDER BY
TX;
TX ID D DEPT_RANK LO LOC_RANK LOAD
--- --- - ---------- -- ---------- -----
1 99 A 1 NY 3 12
2 99 A 1 LA 1 10
3 99 B 3 LA 1 5
4 77 B 1 LA 1 15
5 77 C 2 NY 3 12
6 77 D 3 LA 1 11 ```
Let’s compare the above output with what the OP requested:
```TX ID DEPT DEPT_RANK LOCATION LOC_RANK LOAD
1 99 A 1 NY 2 12
2 99 A 1 LA 1 10
3 99 B 2 LA 1 05
4 77 B 1 LA 1 15
5 77 C 2 NY 2 12
6 77 D 3 LA 1 11```
Notice in the above that when two rows have the same SUM_LOAD_ value, the displayed rank is correct, but that repeated rank value then causes a rank value to be skipped (compare the DEPT_RANK column value on row 3, the LOC_RANK column value on row 1, and the LOC_RANK column value on row 5).
Now what? The ROW_NUMBER function could be used to produce sequential rank numbers without gaps, for example:
```SELECT
TX,
ID,
DEPT,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY SUM_LOAD_ID DESC) DEPT_RANK,
LOCATION,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY SUM_LOAD_LOCATION DESC) LOC_RANK,
FROM
(SELECT
TX,
ID,
DEPT,
LOCATION,
FROM
T1)
ORDER BY
TX;
TX ID D DEPT_RANK LO LOC_RANK LOAD
--- --- - ---------- -- ---------- -----
1 99 A 1 NY 3 12
2 99 A 2 LA 2 10
3 99 B 3 LA 1 5
4 77 B 1 LA 1 15
5 77 C 2 NY 3 12
6 77 D 3 LA 2 11 ```
The above output, as mentioned, does not match the output requested by the OP, since the OP’s requested output specifes that equal values for different rows should show the same rank value.
One more try using the DENSE_RANK analytic function:
```SELECT
TX,
ID,
DEPT,
DENSE_RANK() OVER (PARTITION BY ID ORDER BY SUM_LOAD_ID DESC) DEPT_RANK,
LOCATION,
DENSE_RANK() OVER (PARTITION BY ID ORDER BY SUM_LOAD_LOCATION DESC) LOC_RANK,
FROM
(SELECT
TX,
ID,
DEPT,
LOCATION,
FROM
T1)
ORDER BY
TX;
TX ID D DEPT_RANK LO LOC_RANK LOAD
--- --- - ---------- -- ---------- -----
1 99 A 1 NY 2 12
2 99 A 1 LA 1 10
3 99 B 2 LA 1 5
4 77 B 1 LA 1 15
5 77 C 2 NY 2 12
6 77 D 3 LA 1 11 ```
Let’s compare the above output with what the OP requested:
```TX ID DEPT DEPT_RANK LOCATION LOC_RANK LOAD
1 99 A 1 NY 2 12
2 99 A 1 LA 1 10
3 99 B 2 LA 1 05
4 77 B 1 LA 1 15
5 77 C 2 NY 2 12
6 77 D 3 LA 1 11```
As best as I am able to determine, the above SQL statement will satisfy the OP’s request.
—–
Part 2 of the Challenge
If the OP has the following SQL statement:
```SELECT
ORG_UNIT_CODE,
RANK () OVER (PARTITION BY LOAD_YEAR, CLASSIF_CODE ORDER BY SUM (FTE_DAYS) DESC) ORG_RANK,
CLASSIF_CODE,
RANK () OVER (PARTITION BY LOAD_YEAR, ORG_UNIT_CODE ORDER BY SUM (FTE_DAYS) DESC) CLASSIF_RANK,
SUM (FTE_DAYS) FTE
FROM
GROUP BY
ORG_UNIT_CODE,
CLASSIF_CODE; ```
And the above SQL statement produces the following output:
```YEAR ORG_UNIT_CODE ORG_RANK CLASSIF_CODE CLASSIF_RANK FTE
2010 A46 1 HEW3 1 59
2010 A42 2 HEW3 1 13
2010 A42 1 HEW4 1 13
2010 A46 2 HEW4 2 12```
And the OP wants the output to look like this:
```YEAR ORG_UNIT_CODE ORG_RANK CLASSIF_CODE CLASSIF_RANK FTE
2010 A46 1 HEW3 1 59
2010 A42 2 HEW3 1 13
2010 A42 2 HEW4 2 13
2010 A46 1 HEW4 2 12 ```
Write the DDL and DML statements to create the source table and populate it with the non-aggregated original data, and then produce the output requested by the OP (without looking at the updated usenet thread).
## Simple SQL – Finding the Next Operation
3 11 2011
November 3, 2011
An interesting request came in from an ERP mailing list – how would you write a SQL statement that indicates the next operation in a manufacturing process. Sounds like an easy requirement. Let’s take a look at a graphical view of one example (the graphical view is created using a program that I wrote probably 7 years ago):
The cyan colored rectangles in the above picture are the various operations (found in the OPERATION table) in the manufacturing process. The light green colored rectangles are the material requirements (found in the REQUIREMENT table) that are used by the operation that is immediately above the sequence of light green rectangles. The white rectangles are the sub-assembly headers for the manufacturing process. The white rectangle at the far left is considered the main sub-assembly 0 (WORKORDER_SUB_ID = ‘0’), the middle white rectangle in this case is sub-assembly 1 (WORKORDER_SUB_ID = ‘1’), and the right-most white rectangle in this case is sub-assembly 200 (WORKORDER_SUB_ID = ‘200’). All sub-assemblies except the main sub-assembly 0 are tied to another “parent” operation that consumes the sub-assembly just as if it were another material requirement; therefore a dummy row is added to the material requirements table (REQUIREMENT) with the SUBORD_WO_SUB_ID column set to the sub-assembly number of the connected operation.
In the above picture, on the main sub-assembly 0, operation 888 (at resource ID INSPECT) is the first operation, and operation 999 (at resource ID SHIP) is the second operation. On sub-assembly 1, operation 10 (at resource ID 92) is the first operation, operation 20 (at resource ID 62) is the second operation, and operation 541 (at resource ID KRW) is the third operation. On sub-assembly 200, operation 10 (at resource ID WSD70-TACK) is the first operation, operation 20 (at resource ID WELD-CBA) is the second operation, operation 40 (at resource ID BLAST) is the third operation, and operation 50 (at resource ID PAINT) is the fourth operation.
Since I am working with Oracle Database, I can construct a SQL statement using the LEAD analytic function to find the next operation number and the next resource ID:
```SELECT
*
FROM
(SELECT
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID,
LEAD(O.SEQUENCE_NO,1) OVER (PARTITION BY O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID
ORDER BY O.SEQUENCE_NO) AS NEXT_SEQUENCE_NO,
LEAD(O.RESOURCE_ID,1) OVER (PARTITION BY O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID
ORDER BY O.SEQUENCE_NO) AS NEXT_RESOURCE_ID
FROM
OPERATION O
WHERE
O.WORKORDER_TYPE='W'
AND O.WORKORDER_BASE_ID='13000'
AND O.WORKORDER_LOT_ID='50'
AND O.WORKORDER_SPLIT_ID='0') O
ORDER BY
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO;
WORKORDER_BASE_ID WOR WOR WOR SEQUENCE_NO RESOURCE_ID NEXT_SEQUENCE_NO NEXT_RESOURCE_I
-------------------- --- --- --- ----------- --------------- ---------------- ---------------
13000 50 0 0 888 INSPECT 999 SHIP
13000 50 0 0 999 SHIP
13000 50 0 1 10 92 20 62
13000 50 0 1 20 62 541 KRW
13000 50 0 1 541 KRW
13000 50 0 200 10 WSD70-TACK FIX 20 WELD-CBA
13000 50 0 200 20 WELD-CBA 40 BLAST
13000 50 0 200 40 BLAST 50 PAINT
13000 50 0 200 50 PAINT```
Looks like a pretty easy solution… unless we recognize that after the last operation on sub-assembly 200 the next operation is really the parent operation of that sub-assembly (operation 10 on sub-assembly 1). Likewise, after the last operation on sub-assembly 1, the next operation is really the parent operation of that sub-assembly (operation 888 on sub-assembly 0). There is no next operation after operation 999 on the main sub-assembly 0. How can we fix this problem with the NULL next operations in the previous output? We just need an outer join to the dummy row in the REQUIREMENT table to pick up the parent’s sub-assembly number, and then join that dummy row back to a second reference of the OPERATION table. The NVL2 and COALESCE functions are used to handle the cases where the original next operation and next resource ID would have been output as NULL values:
```SELECT
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID,
NVL2(O.NEXT_SEQUENCE_NO,O.WORKORDER_SUB_ID,O2.WORKORDER_SUB_ID) NEXT_SUB_ID,
COALESCE(O.NEXT_SEQUENCE_NO,O2.SEQUENCE_NO) NEXT_SEQUENCE_NO,
COALESCE(O.NEXT_RESOURCE_ID,O2.RESOURCE_ID) NEXT_RESOURCE_ID
FROM
(SELECT
O.WORKORDER_TYPE,
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID,
LEAD(O.SEQUENCE_NO,1) OVER (PARTITION BY O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID
ORDER BY O.SEQUENCE_NO) AS NEXT_SEQUENCE_NO,
LEAD(O.RESOURCE_ID,1) OVER (PARTITION BY O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID
ORDER BY O.SEQUENCE_NO) AS NEXT_RESOURCE_ID
FROM
OPERATION O
WHERE
O.WORKORDER_TYPE='W'
AND O.WORKORDER_BASE_ID='13000'
AND O.WORKORDER_LOT_ID='50'
AND O.WORKORDER_SPLIT_ID='0') O,
REQUIREMENT R,
OPERATION O2
WHERE
O.WORKORDER_TYPE=R.WORKORDER_TYPE(+)
AND O.WORKORDER_BASE_ID=R.WORKORDER_BASE_ID(+)
AND O.WORKORDER_LOT_ID=R.WORKORDER_LOT_ID(+)
AND O.WORKORDER_SPLIT_ID=R.WORKORDER_SPLIT_ID(+)
AND O.WORKORDER_SUB_ID=R.SUBORD_WO_SUB_ID(+)
AND R.WORKORDER_TYPE=O2.WORKORDER_TYPE(+)
AND R.WORKORDER_BASE_ID=O2.WORKORDER_BASE_ID(+)
AND R.WORKORDER_LOT_ID=O2.WORKORDER_LOT_ID(+)
AND R.WORKORDER_SPLIT_ID=O2.WORKORDER_SPLIT_ID(+)
AND R.WORKORDER_SUB_ID=O2.WORKORDER_SUB_ID(+)
AND R.OPERATION_SEQ_NO=O2.SEQUENCE_NO(+)
ORDER BY
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO;
WORKORDER_BASE_ID WOR WOR WOR SEQUENCE_NO RESOURCE_ID NEX NEXT_SEQUENCE_NO NEXT_RESOURCE_I
-------------------- --- --- --- ----------- --------------- --- ---------------- ---------------
13000 50 0 0 888 INSPECT 0 999 SHIP
13000 50 0 0 999 SHIP
13000 50 0 1 10 92 1 20 62
13000 50 0 1 20 62 1 541 KRW
13000 50 0 1 541 KRW 0 888 INSPECT
13000 50 0 200 10 WSD70-TACK FIX 200 20 WELD-CBA
13000 50 0 200 20 WELD-CBA 200 40 BLAST
13000 50 0 200 40 BLAST 200 50 PAINT
13000 50 0 200 50 PAINT 1 10 92
```
In case you are wondering, the execution plan for the above SQL statement looks like this:
```Plan hash value: 1263351280
-----------------------------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Starts | E-Rows | A-Rows | A-Time | Buffers | OMem | 1Mem | Used-Mem |
-----------------------------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | | 9 |00:00:00.01 | 43 | | | |
| 1 | SORT ORDER BY | | 1 | 9 | 9 |00:00:00.01 | 43 | 2048 | 2048 | 2048 (0)|
| 2 | NESTED LOOPS OUTER | | 1 | 9 | 9 |00:00:00.01 | 43 | | | |
| 3 | NESTED LOOPS OUTER | | 1 | 9 | 9 |00:00:00.01 | 23 | | | |
| 4 | VIEW | | 1 | 9 | 9 |00:00:00.01 | 9 | | | |
| 5 | WINDOW BUFFER | | 1 | 9 | 9 |00:00:00.01 | 9 | 2048 | 2048 | 2048 (0)|
| 6 | TABLE ACCESS BY INDEX ROWID| OPERATION | 1 | 9 | 9 |00:00:00.01 | 9 | | | |
|* 7 | INDEX RANGE SCAN | SYS_C0021734 | 1 | 9 | 9 |00:00:00.01 | 3 | | | |
| 8 | TABLE ACCESS BY INDEX ROWID | REQUIREMENT | 9 | 1 | 7 |00:00:00.01 | 14 | | | |
|* 9 | INDEX RANGE SCAN | X_REQUIREMENT_5 | 9 | 1 | 7 |00:00:00.01 | 13 | | | |
| 10 | TABLE ACCESS BY INDEX ROWID | OPERATION | 9 | 1 | 7 |00:00:00.01 | 20 | | | |
|* 11 | INDEX UNIQUE SCAN | SYS_C0021734 | 9 | 1 | 7 |00:00:00.01 | 13 | | | |
-----------------------------------------------------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
7 - access("O"."WORKORDER_TYPE"='W' AND "O"."WORKORDER_BASE_ID"='13000' AND "O"."WORKORDER_LOT_ID"='50' AND
"O"."WORKORDER_SPLIT_ID"='0')
9 - access("O"."WORKORDER_TYPE"="R"."WORKORDER_TYPE" AND "O"."WORKORDER_BASE_ID"="R"."WORKORDER_BASE_ID" AND
"O"."WORKORDER_LOT_ID"="R"."WORKORDER_LOT_ID" AND "O"."WORKORDER_SPLIT_ID"="R"."WORKORDER_SPLIT_ID" AND
"O"."WORKORDER_SUB_ID"="R"."SUBORD_WO_SUB_ID")
filter("R"."SUBORD_WO_SUB_ID" IS NOT NULL)
11 - access("R"."WORKORDER_TYPE"="O2"."WORKORDER_TYPE" AND "R"."WORKORDER_BASE_ID"="O2"."WORKORDER_BASE_ID" AND
"R"."WORKORDER_LOT_ID"="O2"."WORKORDER_LOT_ID" AND "R"."WORKORDER_SPLIT_ID"="O2"."WORKORDER_SPLIT_ID" AND
"R"."WORKORDER_SUB_ID"="O2"."WORKORDER_SUB_ID" AND "R"."OPERATION_SEQ_NO"="O2"."SEQUENCE_NO")
Note
-----
- cardinality feedback used for this statement ```
The above takes less that 0.01 seconds to execute, in part because I was careful to preserve the leading columns in the primary key indexes’ columns when building the inline view.
But, what if the OP (as he mentioned) was running with an older version of SQL Server that did not support the LEAD analytic function? Is the OP out of luck (it could be argued that is a trick question)?
Let’s start again, this time we will self-join the OPERATION table rather than using an analytic function to determine the next operation, and just for fun we will use ANSI join syntax:
```SELECT
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
MIN(O2.SEQUENCE_NO) AS NEXT_SEQUENCE_NO
FROM
OPERATION O
LEFT OUTER JOIN
OPERATION O2
ON
O.WORKORDER_TYPE=O2.WORKORDER_TYPE
AND O.WORKORDER_BASE_ID=O2.WORKORDER_BASE_ID
AND O.WORKORDER_LOT_ID=O2.WORKORDER_LOT_ID
AND O.WORKORDER_SPLIT_ID=O2.WORKORDER_SPLIT_ID
AND O.WORKORDER_SUB_ID=O2.WORKORDER_SUB_ID
AND O.SEQUENCE_NO < O2.SEQUENCE_NO
WHERE
O.WORKORDER_TYPE='W'
AND O.WORKORDER_BASE_ID='13000'
AND O.WORKORDER_LOT_ID='50'
AND O.WORKORDER_SPLIT_ID='0'
GROUP BY
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO
ORDER BY
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO;
WORKORDER_BASE_ID WOR WOR WOR SEQUENCE_NO NEXT_SEQUENCE_NO
-------------------- --- --- --- ----------- ----------------
13000 50 0 0 888 999
13000 50 0 0 999
13000 50 0 1 10 20
13000 50 0 1 20 541
13000 50 0 1 541
13000 50 0 200 10 20
13000 50 0 200 20 40
13000 50 0 200 40 50
13000 50 0 200 50 ```
Now, we will slide the above SQL statement into an inline view and use the COALESCE function to return the first non-NULL value in a list of columns, and the Oracle only NVL2 function to return one of two column values depending on whether or not a third column contains a NULL value:
```SELECT
O.WORKORDER_TYPE,
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID,
NVL2(O.NEXT_SEQUENCE_NO,O.WORKORDER_SUB_ID,R.WORKORDER_SUB_ID) AS NEXT_SUB_ID,
COALESCE(O.NEXT_SEQUENCE_NO,R.OPERATION_SEQ_NO) AS NEXT_SEQUENCE_NO
FROM
(SELECT
O.WORKORDER_TYPE,
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID,
MIN(O2.SEQUENCE_NO) AS NEXT_SEQUENCE_NO
FROM
OPERATION O
LEFT OUTER JOIN
OPERATION O2
ON
O.WORKORDER_TYPE=O2.WORKORDER_TYPE
AND O.WORKORDER_BASE_ID=O2.WORKORDER_BASE_ID
AND O.WORKORDER_LOT_ID=O2.WORKORDER_LOT_ID
AND O.WORKORDER_SPLIT_ID=O2.WORKORDER_SPLIT_ID
AND O.WORKORDER_SUB_ID=O2.WORKORDER_SUB_ID
AND O.SEQUENCE_NO < O2.SEQUENCE_NO
WHERE
O.WORKORDER_TYPE='W'
AND O.WORKORDER_BASE_ID='13000'
AND O.WORKORDER_LOT_ID='50'
AND O.WORKORDER_SPLIT_ID='0'
GROUP BY
O.WORKORDER_TYPE,
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID) O
LEFT OUTER JOIN
REQUIREMENT R
ON
O.WORKORDER_TYPE=R.WORKORDER_TYPE
AND O.WORKORDER_BASE_ID=R.WORKORDER_BASE_ID
AND O.WORKORDER_LOT_ID=R.WORKORDER_LOT_ID
AND O.WORKORDER_SPLIT_ID=R.WORKORDER_SPLIT_ID
AND O.WORKORDER_SUB_ID=R.SUBORD_WO_SUB_ID
ORDER BY
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO;
W WORKORDER_BASE_ID WOR WOR WOR SEQUENCE_NO RESOURCE_ID NEX NEXT_SEQUENCE_NO
- -------------------- --- --- --- ----------- --------------- --- ----------------
W 13000 50 0 0 888 INSPECT 0 999
W 13000 50 0 0 999 SHIP
W 13000 50 0 1 10 92 1 20
W 13000 50 0 1 20 62 1 541
W 13000 50 0 1 541 KRW 0 888
W 13000 50 0 200 10 WSD70-TACK FIX 200 20
W 13000 50 0 200 20 WELD-CBA 200 40
W 13000 50 0 200 40 BLAST 200 50
W 13000 50 0 200 50 PAINT 1 10 ```
Almost there – we still need the resource ID of the next operation. Sliding the above into another level of inline view, we end up with the following:
```SELECT
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.RESOURCE_ID,
O.NEXT_SUB_ID,
O.NEXT_SEQUENCE_NO,
O2.RESOURCE_ID NEXT_RESOURCE_ID
FROM
(SELECT
O.WORKORDER_TYPE,
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID,
NVL2(O.NEXT_SEQUENCE_NO,O.WORKORDER_SUB_ID,R.WORKORDER_SUB_ID) AS NEXT_SUB_ID,
COALESCE(O.NEXT_SEQUENCE_NO,R.OPERATION_SEQ_NO) AS NEXT_SEQUENCE_NO
FROM
(SELECT
O.WORKORDER_TYPE,
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID,
MIN(O2.SEQUENCE_NO) AS NEXT_SEQUENCE_NO
FROM
OPERATION O
LEFT OUTER JOIN
OPERATION O2
ON
O.WORKORDER_TYPE=O2.WORKORDER_TYPE
AND O.WORKORDER_BASE_ID=O2.WORKORDER_BASE_ID
AND O.WORKORDER_LOT_ID=O2.WORKORDER_LOT_ID
AND O.WORKORDER_SPLIT_ID=O2.WORKORDER_SPLIT_ID
AND O.WORKORDER_SUB_ID=O2.WORKORDER_SUB_ID
AND O.SEQUENCE_NO < O2.SEQUENCE_NO
WHERE
O.WORKORDER_TYPE='W'
AND O.WORKORDER_BASE_ID='13000'
AND O.WORKORDER_LOT_ID='50'
AND O.WORKORDER_SPLIT_ID='0'
GROUP BY
O.WORKORDER_TYPE,
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO,
O.RESOURCE_ID) O
LEFT OUTER JOIN
REQUIREMENT R
ON
O.WORKORDER_TYPE=R.WORKORDER_TYPE
AND O.WORKORDER_BASE_ID=R.WORKORDER_BASE_ID
AND O.WORKORDER_LOT_ID=R.WORKORDER_LOT_ID
AND O.WORKORDER_SPLIT_ID=R.WORKORDER_SPLIT_ID
AND O.WORKORDER_SUB_ID=R.SUBORD_WO_SUB_ID) O
LEFT OUTER JOIN
OPERATION O2
ON
O.WORKORDER_TYPE=O2.WORKORDER_TYPE
AND O.WORKORDER_BASE_ID=O2.WORKORDER_BASE_ID
AND O.WORKORDER_LOT_ID=O2.WORKORDER_LOT_ID
AND O.WORKORDER_SPLIT_ID=O2.WORKORDER_SPLIT_ID
AND O.NEXT_SUB_ID=O2.WORKORDER_SUB_ID
AND O.NEXT_SEQUENCE_NO=O2.SEQUENCE_NO
ORDER BY
O.WORKORDER_BASE_ID,
O.WORKORDER_LOT_ID,
O.WORKORDER_SPLIT_ID,
O.WORKORDER_SUB_ID,
O.SEQUENCE_NO;
WORKORDER_BASE_ID WOR WOR WOR RESOURCE_ID NEX NEXT_SEQUENCE_NO NEXT_RESOURCE_I
-------------------- --- --- --- --------------- --- ---------------- ---------------
13000 50 0 0 INSPECT 0 999 SHIP
13000 50 0 0 SHIP
13000 50 0 1 92 1 20 62
13000 50 0 1 62 1 541 KRW
13000 50 0 1 KRW 0 888 INSPECT
13000 50 0 200 WSD70-TACK FIX 200 20 WELD-CBA
13000 50 0 200 WELD-CBA 200 40 BLAST
13000 50 0 200 BLAST 200 50 PAINT
13000 50 0 200 PAINT 1 10 92 ```
The above SQL statement has the following execution plan:
```Plan hash value: 3374377097
---------------------------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Starts | E-Rows | A-Rows | A-Time | Buffers | OMem | 1Mem | Used-Mem |
---------------------------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | | 9 |00:00:00.01 | 41 | | | |
| 1 | SORT ORDER BY | | 1 | 9 | 9 |00:00:00.01 | 41 | 2048 | 2048 | 2048 (0)|
| 2 | NESTED LOOPS OUTER | | 1 | 9 | 9 |00:00:00.01 | 41 | | | |
| 3 | VIEW | | 1 | 9 | 9 |00:00:00.01 | 19 | | | |
| 4 | HASH GROUP BY | | 1 | 9 | 9 |00:00:00.01 | 19 | 781K| 781K| 1103K (0)|
|* 5 | HASH JOIN OUTER | | 1 | 9 | 13 |00:00:00.01 | 19 | 851K| 851K| 653K (0)|
|* 6 | HASH JOIN OUTER | | 1 | 9 | 9 |00:00:00.01 | 16 | 927K| 927K| 651K (0)|
| 7 | TABLE ACCESS BY INDEX ROWID| OPERATION | 1 | 9 | 9 |00:00:00.01 | 9 | | | |
|* 8 | INDEX RANGE SCAN | SYS_C0021734 | 1 | 9 | 9 |00:00:00.01 | 3 | | | |
|* 9 | TABLE ACCESS BY INDEX ROWID| REQUIREMENT | 1 | 1 | 2 |00:00:00.01 | 7 | | | |
|* 10 | INDEX RANGE SCAN | SYS_C0021842 | 1 | 1 | 18 |00:00:00.01 | 3 | | | |
|* 11 | INDEX RANGE SCAN | SYS_C0021734 | 1 | 1 | 9 |00:00:00.01 | 3 | | | |
| 12 | TABLE ACCESS BY INDEX ROWID | OPERATION | 9 | 1 | 8 |00:00:00.01 | 22 | | | |
|* 13 | INDEX UNIQUE SCAN | SYS_C0021734 | 9 | 1 | 8 |00:00:00.01 | 14 | | | |
---------------------------------------------------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
5 - access("O"."WORKORDER_SUB_ID"="O2"."WORKORDER_SUB_ID" AND "O"."WORKORDER_SPLIT_ID"="O2"."WORKORDER_SPLIT_ID" AND
"O"."WORKORDER_LOT_ID"="O2"."WORKORDER_LOT_ID" AND "O"."WORKORDER_BASE_ID"="O2"."WORKORDER_BASE_ID" AND
"O"."WORKORDER_TYPE"="O2"."WORKORDER_TYPE")
filter("O"."SEQUENCE_NO"<"O2"."SEQUENCE_NO")
6 - access("O"."WORKORDER_SUB_ID"="R"."SUBORD_WO_SUB_ID" AND "O"."WORKORDER_SPLIT_ID"="R"."WORKORDER_SPLIT_ID" AND
"O"."WORKORDER_LOT_ID"="R"."WORKORDER_LOT_ID" AND "O"."WORKORDER_BASE_ID"="R"."WORKORDER_BASE_ID" AND
"O"."WORKORDER_TYPE"="R"."WORKORDER_TYPE")
8 - access("O"."WORKORDER_TYPE"='W' AND "O"."WORKORDER_BASE_ID"='13000' AND "O"."WORKORDER_LOT_ID"='50' AND
"O"."WORKORDER_SPLIT_ID"='0')
9 - filter("R"."SUBORD_WO_SUB_ID" IS NOT NULL)
10 - access("R"."WORKORDER_TYPE"='W' AND "R"."WORKORDER_BASE_ID"='13000' AND "R"."WORKORDER_LOT_ID"='50' AND
"R"."WORKORDER_SPLIT_ID"='0')
11 - access("O2"."WORKORDER_TYPE"='W' AND "O2"."WORKORDER_BASE_ID"='13000' AND "O2"."WORKORDER_LOT_ID"='50' AND
"O2"."WORKORDER_SPLIT_ID"='0')
13 - access("O"."WORKORDER_TYPE"="O2"."WORKORDER_TYPE" AND "O"."WORKORDER_BASE_ID"="O2"."WORKORDER_BASE_ID" AND
"O"."WORKORDER_LOT_ID"="O2"."WORKORDER_LOT_ID" AND "O"."WORKORDER_SPLIT_ID"="O2"."WORKORDER_SPLIT_ID" AND
"O"."NEXT_SUB_ID"="O2"."WORKORDER_SUB_ID" AND "O"."NEXT_SEQUENCE_NO"="O2"."SEQUENCE_NO")
Note
-----
- cardinality feedback used for this statement ```
So, now that we have two different SQL statements that both solve the problem that was presented by the OP, which version of the SQL statement is the most efficient, and which would you use in production?
## Simple SQL with and without Inline Views 2
26 10 2011
October 26, 2011
In the previous article of this series we examined a couple of moderately simple data retrieval requirements, and used inline views and/or analytic functions to solve those data retrieval requirements. I recently saw another opportunity to help a person on the ERP mailing list with a SQL statement. The original poster (OP) provided a screen capture that contained information similar to the following:
```CUST_ORDER_ID PART_ID TYPE CLASS WAREHOUSE_ID LOCATION_ID QTY_ON_HAND TRANSACTI
--------------- --------------- ---- ----- ------------ ------------ ----------- ---------
ZQHZG 2J-2908 O I FIB AB7A10 30 21-JUN-10
ZQHZZ 2J-2909 O I FIB AB7A10 1 21-JUN-10
ZIHKR 2J0836 O I HIBB A77Y3 1092 30-NOV-07
ZQKZH 2J0836 O I HIBB A77Y3 1092 10-JUN-08
ZQIHZ 2J0836 O I HIBB A77Y3 1092 23-NOV-09
ZKOHI 2J4120 O I PIPABTONA AB7A10 2 16-NOV-95
ZKOHQ 2J4129 O I PIPABTONA AB7A10 2 16-NOV-95
ZKGQG 2J9785 O I HIBB AB7A10 2 25-SEP-95
ZKGQG 2J9785 O I HIBB AB7A10 2 25-SEP-95
ZKGQG 2J9785 O I HIBB AB7A10 2 26-OCT-95
ZKGQG 2J9785 O I HIBB AB7A10 2 26-OCT-95
ZKGQG 2J9785 O I HIBB AB7A10 2 12-DEC-95
ZKGQG 2J9785 O I HIBB AB7A10 2 12-DEC-95
ZKGQG 2J9785 O I HIBB AB7A10 2 27-FEB-96
ZKGQG 2J9785 O I HIBB AB7A10 2 04-APR-96
ZKGQG 2J9785 O I HIBB AB7A10 2 04-APR-96
ZKGQG 2J9785 O I HIBB AB7A10 2 31-MAY-96
ZKGQG 2J9785 O I HIBB AB7A10 2 31-MAY-96
ZKGQG 2J9785 O I HIBB AB7A10 2 06-SEP-96
ZKGQG 2J9785 O I HIBB AB7A10 2 30-SEP-96
ZKGQG 2J9785 O I HIBB AB7A10 2 30-SEP-96
ZKGQG 2J9785 O I HIBB AB7A10 2 30-SEP-96
ZKGQG 2J9785 O I HIBB AB7A10 2 12-JAN-98
ZKGQG 2J9785 O I HIBB AB7A10 2 12-FEB-98
ZKGQG 2J9785 O I HIBB AB7A10 2 04-MAR-98
ZKGQG 2J9785 O I HIBB AB7A10 2 05-MAY-98
ZKGQG 2J9785 O I HIBB AB7A10 2 05-MAY-98
ZKGQG 2J9785 O I HIBB AB7A10 2 18-AUG-98
ZKGQG 2J9785 O I HIBB AB7A10 2 18-AUG-98
ZKGQG 2J9785 O I HIBB AB7A10 2 29-SEP-98
ZKGQG 2J9785 O I HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 O I HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 O I HIBB AB7A10 2 02-DEC-98
ZIRZQ 3J8547 O I HIBB A77Y2 252 18-OCT-07
ZZUQG 4G0007 O I HIBB AB7A10 2 08-NOV-99
ZZUQG 4G0007 O I HIBB AB7A10 2 30-NOV-99
...```
Initially, the OP requested that the most recent TRANSACTION_DATE (the last column in the above output) for the PART_ID value to be shown on each row of the output, rather than the TRANSACTION_DATE that is directly associated with a specific row. However, there are two problems:
• The OP did not mention what tables were used as the data source for the included screen capture.
• The OP did not mention whether a recent release of Oracle Database or some other brand of database was in use.
For times like these, it helps a lot to be familiar with the data model used in the database. The INVENTORY_TRANS table in this particular ERP system has columns that match several of the columns that were included in the OP’s screen capture: CUST_ORDER_ID, PART_ID, TYPE, CLASS, WAREHOUSE_ID, LOCATION_ID, and TRANSACTION_DATE. The question remains, where did the QTY_ON_HAND column come from – that column is not in the INVENTORY_TRANS table. We could find out, at least on Oracle Database, with a SQL statement similar to the following:
```SELECT
DTC.TABLE_NAME,
DT.TABLE_NAME
FROM
DBA_TAB_COLUMNS DTC,
DBA_TABLES DT
WHERE
DTC.COLUMN_NAME='QTY_ON_HAND'
AND DTC.OWNER='SCHEMA_OWNER_HERE'
AND DTC.OWNER=DT.OWNER(+)
AND DTC.TABLE_NAME=DT.TABLE_NAME(+)
ORDER BY
DTC.TABLE_NAME;
TABLE_NAME TABLE_NAME
------------------------------ ----------
CR_EC_LINE_PART
CR_OM_CO_LINE_PART
CR_PART_PO_LINE
CR_WO_PART
PART PART
PART_USER_DEFINED_VIEW ```
Note that in the above, I had to join the DBA_TAB_COLUMNS dictionary view with the DBA_TABLES dictionary view (I could have used DBA_OBJECTS instead) to determine that 5 of the 6 rows returned from the DBA_TAB_COLUMNS view are in fact not tables, but views. Thus, the QTY_ON_HAND column in the OP’s screen capture must have originated from the PART table (unless, of course, the column in the screen capture is actually an aliased calculation).
Now that the data sources are known for the query, we can take a best guess approach at reproducing the output that the OP provided in the screen capture:
```SELECT
IT.CUST_ORDER_ID,
IT.PART_ID,
TYPE,
CLASS,
IT.WAREHOUSE_ID,
IT.LOCATION_ID,
P.QTY_ON_HAND,
IT.TRANSACTION_DATE
FROM
INVENTORY_TRANS IT,
PART P
WHERE
IT.CUST_ORDER_ID IS NOT NULL
AND IT.TYPE='O'
AND IT.CLASS='I'
AND IT.PART_ID=P.ID
AND P.QTY_ON_HAND>0
ORDER BY
IT.PART_ID,
IT.TRANSACTION_DATE; ```
Next, there are a couple of ways to have the most recent transaction date appear on each row of the output (partially dependent on the database version). First, let’s build a query that returns the most recent transaction date where the TYPE is O and the CLASS is I and the CUST_ORDER_ID column does not contain a NULL:
```SELECT
PART_ID,
MAX(TRANSACTION_DATE) AS LAST_TRANSACTION_DATE
FROM
INVENTORY_TRANS
WHERE
TYPE='O'
AND CLASS='I'
AND CUST_ORDER_ID IS NOT NULL
GROUP BY
PART_ID
ORDER BY
PART_ID;
PART_ID LAST_TRAN
--------------- ---------
0000009 29-DEC-00
005035008005 29-MAY-98
005035008006 29-MAY-98
00576649-0080 20-OCT-11
00576649-0088 08-JUN-11
00576649-8005 14-JAN-11
007580398 17-JAN-03
0099229940 01-NOV-95
0099229990 01-NOV-95
0108556 28-APR-00
01550867KM 15-NOV-05
01552316KM 02-OCT-00
01552346KM 30-APR-03
01552369KM 30-APR-01
01558943M 07-JAN-10
01561230KM 30-OCT-01
01563001M 08-MAR-10
01563882M 01-APR-10
01566790KM 08-SEP-08
01569945M 01-APR-10
01583508TO 11-FEB-05
01780151TO 19-JUL-07
...```
Next, we will just plug the above inline view into the FROM clause of our original SQL statement and add an additional predicate into the WHERE clause to tell the query optimizer how to associate the inline view with the other tables in the SQL statement:
```SELECT
IT.CUST_ORDER_ID,
IT.PART_ID,
IT.QTY,
IT.WAREHOUSE_ID,
IT.LOCATION_ID,
P.QTY_ON_HAND,
MIT.LAST_TRANSACTION_DATE
FROM
INVENTORY_TRANS IT,
PART P,
(SELECT
PART_ID,
MAX(TRANSACTION_DATE) AS LAST_TRANSACTION_DATE
FROM
INVENTORY_TRANS
WHERE
TYPE='O'
AND CLASS='I'
AND CUST_ORDER_ID IS NOT NULL
GROUP BY
PART_ID) MIT
WHERE
IT.CUST_ORDER_ID IS NOT NULL
AND IT.TYPE='O'
AND IT.CLASS='I'
AND IT.PART_ID=P.ID
AND P.QTY_ON_HAND>0
AND IT.PART_ID=MIT.PART_ID
ORDER BY
IT.PART_ID,
IT.CUST_ORDER_ID,
IT.TRANSACTION_DATE;
CUST_ORDER_ID PART_ID QTY WAREHOUSE_ID LOCATION_ID QTY_ON_HAND LAST_TRAN
--------------- ------------ ---------- ------------ ------------ ----------- ---------
ZQHZG 2J-2908 1 FIB AB7A10 30 21-JUN-10
ZQHZZ 2J-2909 1 FIB AB7A10 1 21-JUN-10
ZIHKR 2J0836 48 HIBB A77Y3 1092 23-NOV-09
ZQIHZ 2J0836 72 HIBB A77Y3 1092 23-NOV-09
ZQKZH 2J0836 111 HIBB A77Y3 1092 23-NOV-09
ZKOHI 2J4120 2 PIPABTONA AB7A10 2 16-NOV-95
ZKOHQ 2J4129 2 PIPABTONA AB7A10 2 16-NOV-95
ZKGQG 2J9785 9 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 1 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 7 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 2 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 6 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 4 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 10 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 8 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 4 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 10 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 4 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 12 HIBB AB7A10 2 02-DEC-98
...```
Since I am using Oracle Database 11.2.0.2, I can take advantage of analytic functions to further simplify the SQL statement:
```SELECT
IT.CUST_ORDER_ID,
IT.PART_ID,
IT.QTY,
IT.WAREHOUSE_ID,
IT.LOCATION_ID,
P.QTY_ON_HAND,
MAX(IT.TRANSACTION_DATE) OVER (PARTITION BY IT.PART_ID) AS LAST_TRANSACTION_DATE
FROM
INVENTORY_TRANS IT,
PART P
WHERE
IT.CUST_ORDER_ID IS NOT NULL
AND IT.TYPE='O'
AND IT.CLASS='I'
AND IT.PART_ID=P.ID
AND P.QTY_ON_HAND>0
ORDER BY
IT.PART_ID,
IT.CUST_ORDER_ID,
IT.TRANSACTION_DATE;
CUST_ORDER_ID PART_ID QTY WAREHOUSE_ID LOCATION_ID QTY_ON_HAND LAST_TRAN
--------------- ------------ ---------- ------------ ------------ ----------- ---------
ZQHZG 2J-2908 1 FIB AB7A10 30 21-JUN-10
ZQHZZ 2J-2909 1 FIB AB7A10 1 21-JUN-10
ZIHKR 2J0836 48 HIBB A77Y3 1092 23-NOV-09
ZQKZH 2J0836 111 HIBB A77Y3 1092 23-NOV-09
ZQIHZ 2J0836 72 HIBB A77Y3 1092 23-NOV-09
ZKOHI 2J4120 2 PIPABTONA AB7A10 2 16-NOV-95
ZKOHQ 2J4129 2 PIPABTONA AB7A10 2 16-NOV-95
ZKGQG 2J9785 9 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 1 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 7 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 2 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 6 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 4 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 10 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 8 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 4 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 10 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 4 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 12 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 2 HIBB AB7A10 2 02-DEC-98
ZKGQG 2J9785 11 HIBB AB7A10 2 02-DEC-98
...```
—–
Let’s assume that the OP’s intended result changes – the OP just wants to know which PART_IDs have not had any transactions in the last 89 day. So, the OP is interested in those PART_IDs with a last transaction that was either 90 or more days ago, or there has never been a transaction for the PART_ID. Oh, scrap… start over.
First, we will identify the list of PART_IDs that have had transactions in the last 89 days (CLASS values of ‘A’ are adjustments, not true inventory transactions, so rows with those CLASS values will be excluded). Once we have that list, we can simply subtract the list from the entire list of PART_IDs:
```SELECT DISTINCT
PART_ID
FROM
INVENTORY_TRANS
WHERE
TRANSACTION_DATE > TRUNC(SYSDATE-90)
AND PART_ID IS NOT NULL
AND CLASS<>'A';
PART_ID
---------
2J-2908
9F4810
867-8649
90772357L
90772341L
90772297L
8475047
8145432
90772326L
9340730
90772357
90772347
782-9484
...```
The OP is interested in seeing the actual warehouse location containing the PART_IDs, so we need to use the PART_LOCATION table, not the PART table as we did before. We then need to create a left outer join between the above list and the PART_LOCATION table – essentially, we will find cases where there is a row in the PART_LOCATION table without a corresponding part ID in the above list. ANSI syntax will be used here for the outer join, because it is still unknown whether the OP is running an Oracle Database backend or a SQL Server backend:
```SELECT
PL.PART_ID,
PL.WAREHOUSE_ID,
PL.LOCATION_ID,
IT.PART_ID,
PL.QTY
FROM
PART_LOCATION PL
LEFT OUTER JOIN
(SELECT DISTINCT
PART_ID
FROM
INVENTORY_TRANS
WHERE
TRANSACTION_DATE > TRUNC(SYSDATE-90)
AND PART_ID IS NOT NULL
AND CLASS<>'A') IT
ON
PL.PART_ID=IT.PART_ID
AND IT.PART_ID IS NULL
WHERE
PL.QTY<>0
ORDER BY
PL.PART_ID,
PL.WAREHOUSE_ID,
PL.LOCATION_ID;
PART_ID WAREHOUSE_ID LOCATION_ID PART_ID QTY
---------------------- ------------ ------------ ------------ ----------
#2 HIBB AB7A10 4
#46954 HIBB TOOL CRAB 15
#4HWG LHRK GROUND WIRK HIBB TOOL CRAB 250
#4THHN HIBB AB7A10 2460
#5 HIBB AB7A10 3
#6 HIBB TOOL CRAB 4
#649 HIBB TOOL CRAB 2
#655 HIBB TOOL CRAB 4
#709 HIBB TOOL CRAB 2
#889 HIBB TOOL CRAB 1
#89 KJTKNSION HIBB TOOL CRAB 3
#90 HIBB AB7A10 4
#92 KJTKNSION HIBB TOOL CRAB 2
*G-WM370H-MT HIBB AB7A10 2
.235-H522 BURN AB7A10 .5
.235RD-9K0045 BURN AB7A10 .7
.260RDJ.963-HSTMH57 BURN AB7A10 2.5625
.35-H596-845 PIPABTONA AB7A10 4.039
...```
In the above, notice that there is a seeminly odd ON clause: “PL.PART_ID=IT.PART_ID AND IT.PART_ID IS NULL” – that simply means that when resolving the left outer join, a value must be found on the left side, but a corresponding value should not be found on the right side of the join. This is the “minus” syntax that I selected to use here (the SQL MINUS syntax could not be used because the WAREHOUSE_ID, LOCATION_ID, and QTY column values returned by the query of the INVENTORY_TRANS table would have to exactly match the values found in the PART_LOCATION table… and that is more difficult to accomplish).
If I was certain that the OP was running with an Oracle Database backend, I would have suggested the following syntax, rather than the ANSI syntax as shown above:
```SELECT
PL.PART_ID,
PL.WAREHOUSE_ID,
PL.LOCATION_ID,
IT.PART_ID,
PL.QTY
FROM
PART_LOCATION PL,
(SELECT DISTINCT
PART_ID
FROM
INVENTORY_TRANS
WHERE
TRANSACTION_DATE > TRUNC(SYSDATE-90)
AND PART_ID IS NOT NULL
AND CLASS<>'A') IT
WHERE
PL.PART_ID=IT.PART_ID(+)
AND IT.PART_ID IS NULL
AND PL.QTY<>0
ORDER BY
PL.PART_ID,
PL.WAREHOUSE_ID,
PL.LOCATION_ID; ```
Why the NULL values in the fourth column (the second PART_ID column) of the output? That NULL simply means that the query is working exactly as intended. That fourth column would be showing only those PART_IDs that have had transactions in the last 89 days (and NULL if no transactions in the last 89 days) if the restriction “AND IT.PART_ID IS NULL” was NOT included in the ON clause – but such rows should not be included in the output if the output is to be used as intended.
## Simple SQL with and without Inline Views
21 10 2011
October 21, 2011
Sometimes it is interesting to take a step back from some of the more difficult to construct SQL statements, and just put together something simple. The following request recently arrived through an ERP mailing list:
“I want to find the last ship date for all the open part numbers in Customer Order Entry. The last ship date is not necessarily from the Open order in Customer Order Entry…”
The original poster (OP) provided more information, and even provided a description of what was already tried as a solution. It is not exceptionally clear what the OP wanted, but because I have several years of experience working with this particular ERP system, I had a feeling that the OP might be headed in the wrong direction. This ERP system uses tables that are well normalized, which would make it a bit more difficult to locate the most recent ship date for a part number that is recorded with one customer order’s tables’ rows when another customer order’s tables’ rows are reported. Unless, of course, we remember that there is also a transaction table that records every inventory movement of part numbers through the system – we just need to know what those rows in the transaction table look like. In this case, the rows have the following characteristics:
• The CUST_ORDER_ID column value must not be NULL
• The TYPE column must have a value of ‘O’
• The CLASS column must have a value of ‘I’
With that knowledge, we could then find the most recent date that any part number shipped with a SQL statement similar to the following:
```SELECT
PART_ID,
MAX(TRANSACTION_DATE) AS LAST_TRANSACTION_DATE
FROM
INVENTORY_TRANS
WHERE
CUST_ORDER_ID IS NOT NULL
AND TYPE='O'
AND CLASS='I'
GROUP BY
PART_ID;```
In the above, the AS keyword is optional on an Oracle Database platform, but it is not optional on some other database platforms.
With an Oracle Database backend, inline views may be used to permit the above SQL statement to be used to retrieve additional information for a parent SQL statement. If we wrap the above SQL statement in ( ) and place it in the FROM clause of the SQL statement, Oracle will treat the data returned by the above SQL statement quite similar to how it would handle a regular table (the Oracle query optimizer might decide to rewrite the combined SQL statement into an equivalent form that no longer contains an inline view). There is always a chance that a part number that is on order may have never shipped to a customer, so we will need to outer join (in this case using a left outer join) to the newly created inline view:
```SELECT
COL.CUST_ORDER_ID,
COL.PART_ID,
LS.LAST_TRANSACTION_DATE
FROM
CUSTOMER_ORDER CO,
CUST_ORDER_LINE COL,
(SELECT
PART_ID,
MAX(TRANSACTION_DATE) AS LAST_TRANSACTION_DATE
FROM
INVENTORY_TRANS
WHERE
CUST_ORDER_ID IS NOT NULL
AND TYPE='O'
AND CLASS='I'
GROUP BY
PART_ID) LS
WHERE
CO.ID=COL.CUST_ORDER_ID
AND CO.STATUS IN ('U','F','R')
AND COL.LINE_STATUS='A'
AND COL.ORDER_QTY>COL.TOTAL_SHIPPED_QTY
AND COL.PART_ID=LS.PART_ID(+);```
In the above, I joined the CUSTOMER_ORDER and CUST_ORDER_LINE tables to the inline view that I aliased as LS. A third table, CUST_LINE_DEL, that optionally contains the delivery schedule for some of the rows in the CUST_ORDER_LINE table, could have also been outer joined to the CUST_ORDER_LINE table.
Running SQL Server, or just feeling ANSI? If so, the above may be rewritten as follows:
```SELECT
COL.CUST_ORDER_ID,
COL.PART_ID,
LS.LAST_TRANSACTION_DATE
FROM
CUSTOMER_ORDER CO
JOIN
CUST_ORDER_LINE COL
ON
CO.ID=COL.CUST_ORDER_ID
LEFT OUTER JOIN
(SELECT
PART_ID,
MAX(TRANSACTION_DATE) AS LAST_TRANSACTION_DATE
FROM
INVENTORY_TRANS
WHERE
CUST_ORDER_ID IS NOT NULL
AND TYPE='O'
AND CLASS='I'
GROUP BY
PART_ID) LS
ON
COL.PART_ID=LS.PART_ID
WHERE
CO.STATUS IN ('U','F','R')
AND COL.LINE_STATUS='A'
AND COL.ORDER_QTY>COL.TOTAL_SHIPPED_QTY;```
Unfortunately, the OP is actually using a SQLBase database backend that does not support inline views. I remember the feeling before I discovered that Oracle Database supported inline views… in that case I would do something like the following:
1. Create a statically defined view.
2. Join to that statically defined view just as if the view were a table.
```CREATE VIEW CUST_ORDER_PART_LS AS
SELECT
PART_ID,
MAX(TRANSACTION_DATE) AS LAST_TRANSACTION_DATE
FROM
INVENTORY_TRANS
WHERE
CUST_ORDER_ID IS NOT NULL
AND TYPE='O'
AND CLASS='I'
GROUP BY
PART_ID;
SELECT
COL.CUST_ORDER_ID,
COL.PART_ID,
LS.LAST_TRANSACTION_DATE
FROM
CUSTOMER_ORDER CO,
CUST_ORDER_LINE COL,
CUST_ORDER_PART_LS LS
WHERE
CO.ID=COL.CUST_ORDER_ID
AND CO.STATUS IN ('U','F','R')
AND COL.LINE_STATUS='A'
AND COL.ORDER_QTY>COL.TOTAL_SHIPPED_QTY
AND COL.PART_ID=LS.PART_ID(+);```
I guess that it is a good exercise once in a while to practice simple SQL.
A second example from the ERP mailing list showed a SQL Server solution for a particular problem. The particular problem that the SQL Server solution set out to solve is as follows:
“Someone posed an interesting question to me – How can you do a mass insert of Operations? Well most of us know how to manage that without too much trouble but this one came with a couple of caveats – The Master’s first operation (lowest numbered) has to have a particular ResourceID and then a different but specific ResourceID must be on a subsequent Operation. It is after the second operation where the insert must be placed.
Sounds like fun – So I figured that it could be done in a single SQL statement – well after a 4-way join of the Operation table to itself I got it to work how I expected…”
Interesting, I think that there are times in the past that I have used multiple self-joins to a table in order to solve similar SQL problems. However, there is an easier way using analytic functions. The following SQL statement attempts to indicate: the sequence of the operation within the work order sub ID, the sequence number that follows sequentially, the second sequence number that follows sequentially, and the resource ID of the next operation in sequential order by the operation number:
```SELECT
WORKORDER_BASE_ID,
WORKORDER_LOT_ID,
WORKORDER_SPLIT_ID,
WORKORDER_SUB_ID,
SEQUENCE_NO,
RESOURCE_ID,
ROW_NUMBER() OVER (PARTITION BY WORKORDER_BASE_ID, WORKORDER_LOT_ID, WORKORDER_SPLIT_ID, WORKORDER_SUB_ID
ORDER BY RESOURCE_ID) AS RN,
LEAD(RESOURCE_ID,1) OVER (PARTITION BY WORKORDER_BASE_ID, WORKORDER_LOT_ID, WORKORDER_SPLIT_ID, WORKORDER_SUB_ID
ORDER BY RESOURCE_ID) AS NEXT_RESOURCE_ID,
LEAD(SEQUENCE_NO,1) OVER (PARTITION BY WORKORDER_BASE_ID, WORKORDER_LOT_ID, WORKORDER_SPLIT_ID, WORKORDER_SUB_ID
ORDER BY RESOURCE_ID) AS NEXT_SEQ_NO,
LEAD(SEQUENCE_NO,2) OVER (PARTITION BY WORKORDER_BASE_ID, WORKORDER_LOT_ID, WORKORDER_SPLIT_ID, WORKORDER_SUB_ID
ORDER BY RESOURCE_ID) AS NEXT_NEXT_SEQ_NO
FROM
OPERATION
WHERE
WORKORDER_TYPE='M'; ```
Next, we only care about those rows when our resource ID of interest (69 in my example) is specified in the first operation on the work order sub ID and the second operation on that work order sub ID specifies the other resource ID that is of interest (FW in my example). If we are to avoid primary key violations, we should select a new sequence number that is half way between the next-next operation and the next operation:
```SELECT
WORKORDER_BASE_ID,
WORKORDER_LOT_ID,
WORKORDER_SPLIT_ID,
WORKORDER_SUB_ID,
SEQUENCE_NO,
NEXT_SEQ_NO,
NEXT_NEXT_SEQ_NO,
NEXT_SEQ_NO + ROUND((COALESCE(NEXT_NEXT_SEQ_NO,NEXT_SEQ_NO+10) - NEXT_SEQ_NO)/2) NEW_SEQUENCE_NO,
'NEW RESOURCE' RESOURCE_ID
FROM
(SELECT
WORKORDER_BASE_ID,
WORKORDER_LOT_ID,
WORKORDER_SPLIT_ID,
WORKORDER_SUB_ID,
SEQUENCE_NO,
RESOURCE_ID,
ROW_NUMBER() OVER (PARTITION BY WORKORDER_BASE_ID, WORKORDER_LOT_ID, WORKORDER_SPLIT_ID, WORKORDER_SUB_ID
ORDER BY RESOURCE_ID) AS RN,
LEAD(RESOURCE_ID,1) OVER (PARTITION BY WORKORDER_BASE_ID, WORKORDER_LOT_ID, WORKORDER_SPLIT_ID, WORKORDER_SUB_ID
ORDER BY RESOURCE_ID) AS NEXT_RESOURCE_ID,
LEAD(SEQUENCE_NO,1) OVER (PARTITION BY WORKORDER_BASE_ID, WORKORDER_LOT_ID, WORKORDER_SPLIT_ID, WORKORDER_SUB_ID
ORDER BY RESOURCE_ID) AS NEXT_SEQ_NO,
LEAD(SEQUENCE_NO,2) OVER (PARTITION BY WORKORDER_BASE_ID, WORKORDER_LOT_ID, WORKORDER_SPLIT_ID, WORKORDER_SUB_ID
ORDER BY RESOURCE_ID) AS NEXT_NEXT_SEQ_NO
FROM
OPERATION
WHERE
WORKORDER_TYPE='M')
WHERE
RN=1
AND RESOURCE_ID='69'
AND NEXT_RESOURCE_ID='FW';
WORKORDER_ WOR WOR WOR SEQUENCE_NO NEXT_SEQ_NO NEXT_NEXT_SEQ_NO NEW_SEQUENCE_NO RESOURCE_ID
---------- --- --- --- ----------- ----------- ---------------- --------------- ------------
231610 0 0 0 10 777 888 833 NEW RESOURCE
237680 0 0 0 10 777 888 833 NEW RESOURCE
32018917X 0 0 0 10 777 888 833 NEW RESOURCE
3201B8920 0 0 0 10 777 888 833 NEW RESOURCE
3201C8765 0 0 0 10 777 888 833 NEW RESOURCE
3201G8639 0 0 0 10 777 888 833 NEW RESOURCE
3201G9003 0 0 0 10 777 888 833 NEW RESOURCE
3201J8772 0 0 0 10 777 888 833 NEW RESOURCE
3201J8850 0 0 0 10 777 888 833 NEW RESOURCE
3201K8669 0 0 0 10 777 888 833 NEW RESOURCE
3201M8281 0 0 0 10 777 888 833 NEW RESOURCE
3201M8489 0 0 0 10 777 888 833 NEW RESOURCE
3201N8516 0 0 0 10 777 888 833 NEW RESOURCE
3201N8776 0 0 0 10 777 888 833 NEW RESOURCE
3201Q8545 0 0 0 10 777 888 833 NEW RESOURCE
3201T8964 0 0 0 10 777 888 833 NEW RESOURCE
3201T8964X 0 0 0 10 20 30 25 NEW RESOURCE
3201V8524 0 0 0 10 777 888 833 NEW RESOURCE
3201V8966 0 0 0 10 777 888 833 NEW RESOURCE
3201V8966X 0 0 0 10 20 30 25 NEW RESOURCE
3201W8967 0 0 0 10 777 888 833 NEW RESOURCE
3201W8967X 0 0 0 10 20 30 25 NEW RESOURCE
3201Y8423 0 0 0 10 777 888 833 NEW RESOURCE
3201Z8996 0 0 0 10 777 888 833 NEW RESOURCE
24 rows selected. ```
With the above, we are well on our way to perfoming a mass insert into this table. If I recall correctly, the above must first be inserted into another table (a temp table, preferrably) and then the rows may be inserted into the OPERATION table (the table that is the row source for the analytic functions used in the SQL statement).
## Brain Teaser: Why is this Query Performing a Full Table Scan
14 09 2011
September 14, 2011
While taking a look at the OTN threads this morning, I found an interesting test case, where the OP was asking why a query used a full table scan and not an index range scan. The OP would like for the query to use the index without using a hint in the query. So, why doesn’t the OP’s test case use an index range scan? Just for fun I will state that my first two initial guesses were not quite on target.
A slightly modified table creation script of the setup for the OP’s test case:
```DROP TABLE T1 PURGE;
CREATE TABLE T1 AS
SELECT
*
FROM
DBA_OBJECTS
WHERE
STATUS='VALID';
UPDATE T1 SET STATUS='INVALID' WHERE ROWNUM=1;
COMMIT;
CREATE INDEX IND_T1_STATUS ON T1(STATUS); ```
Let’s take a look at the data distribution in the table:
```SELECT
STATUS,
CNT,
ROUND((RATIO_TO_REPORT(CNT) OVER ())*100,6) PERCENT
FROM
(SELECT
STATUS,
COUNT(*) CNT
FROM
T1
GROUP BY
STATUS);
STATUS CNT PERCENT
------- ---------- ----------
INVALID 1 .001513
VALID 66095 99.998487 ```
99.998% of the table’s rows have a STATUS of VALID with just a single row having a STATUS of invalid.
Now let’s collect the statistics for the table and index and check the execution plan:
```ANALYZE INDEX IND_T1_STATUS COMPUTE STATISTICS;
ANALYZE TABLE T1 COMPUTE STATISTICS;
SET AUTOTRACE TRACEONLY EXPLAIN
SET PAGESIZE 1000
SET LINESIZE 140
SELECT
*
FROM
T1
WHERE
STATUS='INVALID';
Execution Plan
----------------------------------------------------------
Plan hash value: 3617692013
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 33048 | 3227K| 265 (1)| 00:00:04 |
|* 1 | TABLE ACCESS FULL| T1 | 33048 | 3227K| 265 (1)| 00:00:04 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("STATUS"='INVALID') ```
If you were assigned to help the OP achieve his task with the test case, what would you do?
## The New Order Oracle Coding Challenge 4 – Tic Tac Toe
12 08 2011
Tic Tac Toe, the game of X’s and O’s, was an oddly popular game in elementary school. When playing the game you quickly learn a couple of rules:
• Because X always places his mark first (alternating between X and O), there is an unfair advantage for the player placing the X marks.
• The player placing his mark in the center square typically has an advantage in the game.
• Once a player scores three marks in a row horizontally, vertically, or diagonally, there is no point in continuing the game.
Consider the above game matches. In the first two matches both X and O score three in a row, but which player won the match – it depends on the order in which the marks were placed. In the third match, X won with a diagnal three in a row. The final match resulted in a tie (neither player won). You can use a SQL statement similar to the following to output all of the roughly 362,000 combinations (note that not all combinations are unique – rotating the board 90 or 180 degrees to generate a unique combination probably is not fair, and the order in which the marks are placed could matter):
```WITH
N AS
(SELECT
ROWNUM N
FROM
DUAL
CONNECT BY
LEVEL<=9),
C AS
(SELECT
'XOX'||'OXO'||'XOX' C
FROM
DUAL)
SELECT
SUBSTR(C.C,N1.N,1) ||
SUBSTR(C.C,N2.N,1) ||
SUBSTR(C.C,N3.N,1) || CHR(10) ||
SUBSTR(C.C,N4.N,1) ||
SUBSTR(C.C,N5.N,1) ||
SUBSTR(C.C,N6.N,1) || CHR(10) ||
SUBSTR(C.C,N7.N,1) ||
SUBSTR(C.C,N8.N,1) ||
SUBSTR(C.C,N9.N,1) GAME
FROM
N N1,
N N2,
N N3,
N N4,
N N5,
N N6,
N N7,
N N8,
N N9,
C
WHERE
N1.N<>N2.N
AND N1.N<>N3.N
AND N1.N<>N4.N
AND N1.N<>N5.N
AND N1.N<>N6.N
AND N1.N<>N7.N
AND N1.N<>N8.N
AND N1.N<>N9.N
AND N2.N<>N3.N
AND N2.N<>N4.N
AND N2.N<>N5.N
AND N2.N<>N6.N
AND N2.N<>N7.N
AND N2.N<>N8.N
AND N2.N<>N9.N
AND N3.N<>N4.N
AND N3.N<>N5.N
AND N3.N<>N6.N
AND N3.N<>N7.N
AND N3.N<>N8.N
AND N3.N<>N9.N
AND N4.N<>N5.N
AND N4.N<>N6.N
AND N4.N<>N7.N
AND N4.N<>N8.N
AND N4.N<>N9.N
AND N5.N<>N6.N
AND N5.N<>N7.N
AND N5.N<>N8.N
AND N5.N<>N9.N
AND N6.N<>N7.N
AND N6.N<>N8.N
AND N6.N<>N9.N
AND N7.N<>N8.N
AND N7.N<>N9.N
AND N8.N<>N9.N; ```
The output of the above SQL statement should appear similar to the following:
```GAME
----
...
XXX
OOO
OXX
XXX
OOO
OXX
...
XOX
XOO
OXX
OXX
XOO
OXX
XXO
XOO
XOX
XXO
XOO
XOX
XOX
XOO
XOX
...```
The big problem with the above SQL statement is that it is not clear which player won in all cases. Ideally, the N1.N value would be used to output an X mark in the specified position, the N2.N value would be used to output an O mark in a specified position, the N3.N value would be used to output an X in the specified position, etc. until one of the players places three marks in a row. For example, if N1.N is 5, an X would be placed in the center square, if N2.N is 9, an O would be placed in the bottom right square.
Now that we know that the order in which the marks are placed is important, how would be know when a player wins? You could experiment with the following:
Vertical win, with three positions having the same resulting values:
```SELECT
MOD(position - 1, 3) + 1 V
FROM
DUAL; ```
Horizontal win, with three positions having the same resulting values:
```SELECT
TRUNC((position - 1) / 3) + 1 V
FROM
DUAL;```
Diagnal win \ when V=0 in three positions:
```SELECT
(MOD(position - 1, 3) + 1) - (TRUNC((position - 1) / 3) + 1) V
FROM
DUAL;```
Diagnal win / when V=0 (watch out, could end up in a V pattern) or V=2:
```SELECT
ABS((MOD(position - 1, 3) + 1) - (TRUNC((position - 1) / 3) + 1)) V
FROM
DUAL; ```
———–
OK, now that I have explained the game, given you a couple of SQL statements to possibly help you with the solution… on to the challenge. With the help of Oracle Database build a tic tac toe solver that will help a player win at tic tac toe. Provided a current description of the board, for example ‘_X_’||’O_X’||’OX_’, display all board solutions that allows player O (or player X if his turn is next) to win. (Side note: I have not yet completed the solution to this challenge – it might be possible to accomplish this challenge with just a SQL statement.)
## The New Order Oracle Coding Challenge 3 – Mind Boggle
5 08 2011
August 5, 2011 (Modified August 7, 2011)
In part 1 of this series the challenge was to simply reverse the order of digits in the numbers from 1 to 1,000,000 to find that cases where the numbers formed by the reverse ordered digits evenly divided into the original number. In part 2 of this series the challenge required examining all of the numbers between 1000 and 9999, where arranging the digits of the original number into any of 23 valid combinations resulted in a new number that cleanly divided into the original four digit number. There were several different solutions provided to the two challenges, so now it is time to move on to part three of the series.
In part 1 of this blog article series I mentioned playing a game years ago that used letters on the face of dice – the dice were rolled, and then the challenge was to find all words that could be completely spelled using the letters on the top of the dice. I was not very good at the game, so I enlisted the help of a computer. One such dice game is called Boggle, and that game’s name is probably fitting for today’s challenge. Imagine that you played this game and the following letters appeared on the top of the dice:
One of the rules of the game requires that words must be at least 3 letters in length, for example: you say melee eye (I) see elfs file some mail (OK, the word I is too short, but we can have some fun with the words that are found). As you might be able to guess, there are a lot of possible combinations of the 16 letters found on the dice, some of which are valid words. If we just consider the 5 letter, 4 letter, and 3 letter combinations of the dice, there are more than a half million possible combinations (in the following table, multiply the numbers across and add the results for each row) – no wonder I needed the computer’s help with these puzzles.
16 15 14 13 12 = 16! / 11! 16 15 14 13 = 16! / 12! 16 15 14 = 16! / 13! = 571,200
To make the full challenge of finding words a little easier, let’s break the challenge into a couple of parts:
Part 1: Consider the 2 x 2 letter arrangement at the left. With the help of Oracle Database, list all of the three letter combinations of those four letters. There will be 4 * 3 * 2 = 24 possible combinations of the letters.
Part 2: Consider the 4 x 4 letter arrangement at the left. With the help of Oracle Database, list all of the four letter combinations of those 16 letters. There will be 16 * 15 * 14 * 13 = 43,680 possible combinations of the letters.
Part 3: Consider the 4 x 4 letter arrangement above. With the help of Oracle Database, list all of the three, four, five, and six letter combinations of those 16 letters. If you see any seven letter words in the above set of letters, you might as well retrieve those letter combinations also. How many letter combinations do you have in total for part 3?
Part 4: Extra Credit: How many of the letter combinations generated in part 3 above are valid U.S. or U.K. English words? List the words.
Part 5: Extra, Extra Credit: List any words found in the letters at the left that have any connection to Oracle Corporation. Remember that a letter can only be used as many times in a single word as it appears at the left (if you can form a word with three letter A’s that have a connection to Oracle Corp., go for it.).
Added August 7, 2011:
When I put together this challenge I did not think that it was possible to complete Part 4 Extra Credit using just SQL. I was fairly certain that there were some interesting techniques to retrieve HTML content with the help of PL/SQL, but I had not worked out a solution that utilized that technique. As I write this, Radoslav Golian in the comments section appears to have both a PL/SQL and a SQL solution that uses the dictionary.reference.com website to validate the words (only 6 words to avoid a denial of service type attack on the dictionary.reference.com website). One of the approaches that I considered, but did not develop, is something similar to how Radoslav verified the words, but I would use a VBS script to submit the request and check the result as is demonstrated in these two articles: Submit Input to an ASP Web Page and Retrieve the Result using VBS and Use VBS to Search for Oracle Books using Google’s Book Library.
The solution that I put together for Part 4 Extra Credit started with an Excel macro that I posted in another blog article, which was then converted to PL/SQL. I then transformed the PL/SQL for use in this article, and generated a new Excel macro from the PL/SQL code. The Excel macro (along with the calling code looks like this:
```Sub StartBoggle()
Call Boggle("ESOIMEFOALEUSAYE", 6, 3)
End Sub
Sub Boggle(strCharacters As String, intMaxWordLength As Integer, intMinWordLength As Integer)
Dim i As Integer
Dim strCharacter(20) As String
Dim intCharacterIndex(20) As Integer
Dim intCharacters As Integer
Dim intCharactersMax As Integer
Dim intCharactersMin As Integer
Dim intNumberOfSuppliedCharacters As Integer
Dim intAdjustmentPosition As Integer
Dim intFlag As Integer
Dim strOutput As String
Dim strWords(10000) As String
Dim intWordCount As Integer
Dim intFilenum As Integer
intFilenum = FreeFile
Open "C:\Words " & strCharacters & ".txt" For Output As #intFilenum
If intMaxWordLength = 0 Then
intCharactersMax = Len(strCharacters)
Else
If intMaxWordLength <= Len(strCharacters) Then
intCharactersMax = intMaxWordLength
Else
intCharactersMax = Len(strCharacters)
End If
End If
If intMinWordLength = 0 Then
intCharactersMin = 3
Else
If intMaxWordLength < intMinWordLength Then
intCharactersMin = intCharactersMax
Else
intCharactersMin = intMinWordLength
End If
End If
intNumberOfSuppliedCharacters = Len(strCharacters)
For i = 1 To intNumberOfSuppliedCharacters
strCharacter(i) = Mid(strCharacters, i, 1)
Next i
intCharacters = intCharactersMin - 1
intWordCount = 0
Do While intCharacters < intCharactersMax
intCharacters = intCharacters + 1
For i = 1 To intCharacters
intCharacterIndex(i) = i
Next i
Do While intAdjustmentPosition > 0
intFlag = 0
For i = 1 To intAdjustmentPosition - 1
If intCharacterIndex(i) = intCharacterIndex(intAdjustmentPosition) Then
' Found a duplicate index position in the other values to the left
intFlag = 1
Exit For
End If
Next i
If intFlag = 1 Then
' Try the next index position in this element
Else
If intAdjustmentPosition = intCharacters Then
' Output
strOutput = ""
For i = 1 To intCharacters
strOutput = strOutput & strCharacter(intCharacterIndex(i))
Next i
intFlag = 0
For i = intWordCount To 1 Step -1
If strOutput = strWords(i) Then
intFlag = 1
Exit For
End If
Next i
If intFlag = 0 Then
If Application.CheckSpelling(Word:=UCase(strOutput)) <> 0 Then
intWordCount = intWordCount + 1
strWords(intWordCount) = strOutput
Print #intFilenum, strOutput
Debug.Print strOutput
End If
End If
If intCharacterIndex(intAdjustmentPosition) = intNumberOfSuppliedCharacters Then
' No more available values in the last position
If intAdjustmentPosition > 0 Then
End If
Else
End If
Else
' No duplicate so prepare to check the next position
End If
End If
Do While (intAdjustmentPosition > 0) And (intCharacterIndex(intAdjustmentPosition) > intNumberOfSuppliedCharacters)
' Roll back one index position as many times as necessary
If intAdjustmentPosition > 0 Then
End If
Loop ' (intAdjustmentPosition > 0) And
Loop 'intAdjustmentPosition > 0
Loop 'intCharacters < intCharactersMax
Close #intFilenum
End Sub ```
The Excel macro builds letter combinations that are between the minimum and maximum length, and then tests those letter combinations using the built-in dictionary that is in Excel. I had a little bit of difficulty coming up with a way to generate the letter combinations of variable length, so I settled on a custom developed technique – I would simply keep track of the original character positions, manipulate those original character positions, and then output the corresponding characters. The challenge is then how does one verify that the same character position is not used more than once in a single word?
The method that I came up with is as follows, which assumes that we are trying to build four letter words from the supplied 16 letters. We can start with the seed combination 1,2,3,4. The idea is to work from left to right, and then back to the left. Every time to make it to the right, we output a word, when we make it all the way back to the left (just before the number 1 in the above), we are done. The rules are simple:
• Increment the number in a position, and if that number does not appear in a position to the left, move one position to the right.
• When the maximum character number (16 in this example) is exceeded in a position, reset the number to 1, move one position to the left, and increment the value in the new position by 1.
• In the last position the character number should be incremented as many times as necessary to reach the maximum character number – each time a potential new combination will be generated.
But there is a problem with this approach – it does not use Oracle Database!
Let’s go back to the PL/SQL function from which I created the Excel function (I have not worked much with pipelined functions – so there may be one or two errors):
```CREATE OR REPLACE FUNCTION BOGGLE_VAR_LENGTH(strCHARACTERS IN VARCHAR2, intMaxWordLength IN NUMBER, intMinWordLength IN NUMBER) RETURN SYS.AQ\$_MIDARRAY PIPELINED
AS
TYPE NUMBER_ARRAY IS TABLE OF NUMBER INDEX BY PLS_INTEGER;
TYPE CHARACTER_ARRAY IS TABLE OF VARCHAR(1) INDEX BY PLS_INTEGER;
strCharacter CHARACTER_ARRAY;
intCharacterIndex NUMBER_ARRAY;
intCharacters NUMBER;
intCharactersMax NUMBER;
intCharactersMin NUMBER;
intNumberOfSuppliedCharacters NUMBER;
intFlag NUMBER;
intI NUMBER;
strOutput VARCHAR2(100);
BEGIN
IF intMaxWordLength IS NULL THEN
intCharactersMax := LENGTH(strCHARACTERS);
ELSE
IF intMaxWordLength <= LENGTH(strCHARACTERS) THEN
intCharactersMax := intMaxWordLength;
ELSE
intCharactersMax := LENGTH(strCHARACTERS);
END IF;
END IF;
IF intMinWordLength IS NULL THEN
intCharactersMin := 3;
ELSE
IF intMaxWordLength < intMinWordLength THEN
intCharactersMin := intCharactersMax;
ELSE
intCharactersMin := intMinWordLength;
END IF;
END IF;
intNumberOfSuppliedCharacters := LENGTH(strCHARACTERS);
FOR I IN 1.. intNumberOfSuppliedCharacters LOOP
strCharacter(I) := SUBSTR(strCHARACTERS, I, 1);
END LOOP;
intCharacters := intCharactersMin - 1;
WHILE intCharacters < intCharactersMax LOOP
intCharacters := intCharacters + 1;
FOR I IN 1 .. intCharacters LOOP
intCharacterIndex(I) := I;
END LOOP;
WHILE intAdjustmentPosition > 0 LOOP
intFlag := 0;
FOR I IN 1 .. intAdjustmentPosition - 1 LOOP
IF intCharacterIndex(I) = intCharacterIndex(intAdjustmentPosition) Then
-- Found a duplicate index position in the other values to the left
intFlag := 1;
END IF;
END LOOP;
IF intFlag = 1 Then
-- Try the next index position in this element
ELSE
IF intAdjustmentPosition = intCharacters Then
-- Output
strOutput := '';
FOR i IN 1 .. intCharacters LOOP
strOutput := strOutput || strCharacter(intCharacterIndex(i));
END LOOP;
PIPE ROW (strOutput);
IF intCharacterIndex(intAdjustmentPosition) = intNumberOfSuppliedCharacters THEN
-- No more available values in the last position
IF intAdjustmentPosition > 0 THEN
END IF;
ELSE
END IF;
ELSE
-- No duplicate so prepare to check the next position
END IF;
END IF;
WHILE (intAdjustmentPosition > 0) And (intCharacterIndex(intAdjustmentPosition) > intNumberOfSuppliedCharacters) LOOP
-- Roll back one index position as many times as necessary
IF intAdjustmentPosition > 0 THEN
END IF;
END LOOP;
END LOOP;
END LOOP;
END;
/ ```
We are able to call the function from a SQL statement like this:
```SELECT
*
FROM
TABLE(BOGGLE_VAR_LENGTH('ESOIMEFOALEUSAYE', 6, 3)); ```
Remember that there are more than a half million character combinations for just the 3, 4, and 5 letter combinations – the above will as for 6,336,960 letter combinations to be generated. But there is a problem with this approach – it does not verify that the letter combinations are actual words!
For fun, let’s see how many possible combinations will result if we allow 3, 4, 5, 6, 7, and 8 letter combinations:
Len Combinations 8 16 15 14 13 12 11 10 9 518,918,400 = 16! / 8! 7 16 15 14 13 12 11 10 57,657,600 = 16! / 9! 6 16 15 14 13 12 11 5,765,760 = 16! / 10! 5 16 15 14 13 12 524,160 = 16! / 11! 4 16 15 14 13 43,680 = 16! / 12! 3 16 15 14 3,360 = 16! / 13! 582,912,960 582,912,960
That is more than a half billion combinations! Warning, significant database server CPU consumption will result when generating all combinations.
Let’s take a look at the final solution that I created for Part 4 Extra, Extra Credit. The solution is an Excel macro that calls the PL/SQL function through a SQL statement:
```Sub StartBoggleOracle()
Call BoggleOracle("ESOIMEFOALEUSAYE", 8, 3)
End Sub
Sub BoggleOracle(strCharacters As String, intMaxWordLength As Integer, intMinWordLength As Integer)
Dim strSQL As String
Dim strUsername As String
Dim strPassword As String
Dim strDatabase As String
Dim intFilenum As Integer
Dim intCharacters As Integer
Dim intCharactersMax As Integer
Dim intCharactersMin As Integer
Dim strOutput As String
Dim dbDatabase As ADODB.Connection
Dim snpData As ADODB.Recordset
Set dbDatabase = New ADODB.Connection
Set snpData = New ADODB.Recordset
strDatabase = "MyDatabase"
dbDatabase.ConnectionString = "Provider=OraOLEDB.Oracle;Data Source=" & strDatabase & ";User ID=" & strUsername & ";Password=" & strPassword & ";FetchSize=5000;"
dbDatabase.Open
intFilenum = FreeFile
Open "C:\WordsOracle " & strCharacters & ".txt" For Output As #intFilenum
If intMaxWordLength = 0 Then
intCharactersMax = Len(strCharacters)
Else
If intMaxWordLength <= Len(strCharacters) Then
intCharactersMax = intMaxWordLength
Else
intCharactersMax = Len(strCharacters)
End If
End If
If intMinWordLength = 0 Then
intCharactersMin = 3
Else
If intMaxWordLength < intMinWordLength Then
intCharactersMin = intCharactersMax
Else
intCharactersMin = intMinWordLength
End If
End If
strSQL = "SELECT DISTINCT" & vbCrLf
strSQL = strSQL & " *" & vbCrLf
strSQL = strSQL & "FROM" & vbCrLf
strSQL = strSQL & " (SELECT" & vbCrLf
strSQL = strSQL & " *" & vbCrLf
strSQL = strSQL & " FROM" & vbCrLf
strSQL = strSQL & " TABLE(BOGGLE_VAR_LENGTH('" & strCharacters & "', " & Format(intCharactersMax) & ", " & Format(intCharactersMin) & ")))" & vbCrLf
strSQL = strSQL & "ORDER BY" & vbCrLf
strSQL = strSQL & " 1"
snpData.Open strSQL, dbDatabase
If snpData.State = 1 Then
Do While Not snpData.EOF
strOutput = snpData(0)
If Application.CheckSpelling(Word:=UCase(strOutput)) <> 0 Then
Print #intFilenum, strOutput
Debug.Print strOutput
End If
snpData.MoveNext
Loop
snpData.Close
End If
Close #intFilenum
dbDatabase.Close
Set snpData = Nothing
Set dbDatabase = Nothing
End Sub```
The words found appear to depend on the version of Excel – Excel 2010 seems to find more words than Excel 2007.
• The 799 word list from Excel 2007 for word lengths between 3 and 8 characters, including the timing information to show when the SQL statement was submitted, when the first 5,000 combinations were retrieved from the database, and when the Excel spell check finished. Words Oracle_ESOIMEFOALEUSAYE.txt
• The 2,179 word list from Excel 2007 for word lengths between 3 and 8 characters, including the timing information to show when the SQL statement was submitted, when the first 5,000 combinations were retrieved from the database, and when the Excel spell check finished. Words Oracle_OSERIEFAARLNCAYL.txt
Excel found Ellison in the second word list. For Part 5 Extra, Extra Credit, what other words connected to Oracle Corporation were found? | 23,738 | 81,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-22 | latest | en | 0.681837 |
http://www.velocityreviews.com/forums/t964944-re-tail-recursion-to-while-iteration-in-2-easy-steps.html | 1,398,359,009,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00436-ip-10-147-4-33.ec2.internal.warc.gz | 891,637,357 | 8,589 | Velocity Reviews > Re: Tail recursion to while iteration in 2 easy steps
# Re: Tail recursion to while iteration in 2 easy steps
Terry Reedy
Guest
Posts: n/a
10-02-2013
On 10/2/2013 8:31 AM, http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> On Tue, Oct 1, 2013, at 17:30, Terry Reedy wrote:
>> Part of the reason that Python does not do tail call optimization is
>> that turning tail recursion into while iteration is almost trivial, once
>> you know the secret of the two easy steps. Here it is.
>
> That should be a reason it _does_ do it - saying people should rewrite
> their functions with loops means declaring that Python is not really a
> multi-paradigm programming language but rather rejects functional
> programming styles in favor of imperative ones.
It is true that Python does not encourage the particular functional
style that is encouraged by auto optimization of tail recursion. A
different functional style would often use reduce (or fold) instead.
Some other points I left out in a post of medium length yet brief for
the topic.
1. If one starts with body recursion, as is typical, one must consider
commutativity (possibly associativity) of the 'inner' operator in any
conversion.
2. Instead of converting to tail recursion, one might convert to while
iteration directly.
3. One often 'polishes' the while form in a way that cannot be done
automatically.
4. While loops are actually rare in idiomatic Python code. In Python,
for loops are the standard way to linearly process a collection. The
final version I gave for a factorial while loop,
def fact_while(n):
if n < 0 or n != int(n):
raise ValueError('fact input {} is not a count'.format(n))
fac = 1
while n > 1:
fac *= n
n -= 1
return fac
should better be written with a for loop:
def fact_for(n):
if n < 0 or n != int(n):
raise ValueError('fact input {} is not a count'.format(n))
fac = 1:
for i in range(2, n):
fac *= n
When the input to a function is an iterable instead of n, the iterable
should be part of the for loop source expression. For loops are
integrated with Python's iterator protocol in much the same way that
recursion is integrated with list first:rest pattern matching in some
functional languages. It is true that Python encourages the use of for
loops and for clauses in comprehensions (a functional construct).
5. Conversion of apparent recursion to iteration assumes that the
function really is intended to be recursive. This assumption is the
basis for replacing the recursive call with assignment and an implied
internal goto. The programmer can determine that this semantic change is
correct; the compiler should not assume that. (Because of Python's late
name-binding semantics, recursive *intent* is better expressed in Python
with iterative syntax than function call syntax. )
--
Terry Jan Reedy
88888 Dihedral
Guest
Posts: n/a
10-04-2013
On Thursday, October 3, 2013 5:33:27 AM UTC+8, Terry Reedy wrote:
> On 10/2/2013 8:31 AM, (E-Mail Removed) wrote:
>
> > On Tue, Oct 1, 2013, at 17:30, Terry Reedy wrote:
>
> >> Part of the reason that Python does not do tail call optimization is
>
> >> that turning tail recursion into while iteration is almost trivial, once
>
> >> you know the secret of the two easy steps. Here it is.
>
> >
>
> > That should be a reason it _does_ do it - saying people should rewrite
>
> > their functions with loops means declaring that Python is not really a
>
> > multi-paradigm programming language but rather rejects functional
>
> > programming styles in favor of imperative ones.
>
>
>
> It is true that Python does not encourage the particular functional
>
> style that is encouraged by auto optimization of tail recursion. A
>
> different functional style would often use reduce (or fold) instead.
>
>
>
> Some other points I left out in a post of medium length yet brief for
>
> the topic.
>
>
>
> 1. If one starts with body recursion, as is typical, one must consider
>
> commutativity (possibly associativity) of the 'inner' operator in any
>
> conversion.
>
>
>
> 2. Instead of converting to tail recursion, one might convert to while
>
> iteration directly.
>
>
>
> 3. One often 'polishes' the while form in a way that cannot be done
>
> automatically.
>
>
>
> 4. While loops are actually rare in idiomatic Python code. In Python,
>
> for loops are the standard way to linearly process a collection. The
>
> final version I gave for a factorial while loop,
>
>
>
> def fact_while(n):
>
> if n < 0 or n != int(n):
>
> raise ValueError('fact input {} is not a count'.format(n))
>
> fac = 1
>
> while n > 1:
>
> fac *= n
>
> n -= 1
>
> return fac
>
>
>
> should better be written with a for loop:
>
As I pointed out before, an accelerated version without the limit
of the stack depth for computing
facotrials can be obtained
by storing a list of products of primes
first.
Of course integer divisions are
required to transform the to stack
depth problem into the size of the
32-64 bit heap space. | 1,244 | 4,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2014-15 | latest | en | 0.878514 |
http://eulersprint.org/problem/81 | 1,531,840,287,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589752.56/warc/CC-MAIN-20180717144908-20180717164908-00560.warc.gz | 125,937,893 | 3,044 | Problem 81
Path sum: two ways
In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by only moving to the right and down, is indicated in bold red and is equal to 2427.
131 673 234 103 18 201 96 342 965 150 630 803 746 422 111 537 699 497 121 956 805 732 524 37 331
Find the minimal path sum, in matrix.txt (right click and 'Save Link/Target As...'), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by only moving right and down.
These problems are part of Project Euler and are licensed under CC BY-NC-SA 2.0 UK | 179 | 587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | longest | en | 0.834341 |
https://newpathworksheets.com/math/grade-4/number-words-and-place-value/montana-standards | 1,627,420,192,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153491.18/warc/CC-MAIN-20210727202227-20210727232227-00521.warc.gz | 437,324,922 | 8,363 | ## ◂Math Worksheets and Study Guides Fourth Grade. Number Words and Place Value
### The resources above correspond to the standards listed below:
#### Montana Content Standards
MT.CC.4.NBT. Number and Operations in Base Ten
Generalize place value understanding for multi-digit whole numbers.
4.NBT.1. Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. For example, recognize that 700 ÷ 70 = 10 by applying concepts of place value and division.
4.NBT.2. Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. | 174 | 780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-31 | latest | en | 0.818574 |
http://dyerlab.github.io/applied_population_genetics/map-projections.html | 1,585,450,417,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00541.warc.gz | 60,530,278 | 14,022 | # 32 Map Projections
A spatial projection is a mathematical representation of a coordinate space used to identify geospatial objects. Because the earth is both non-flat and non-spheroid, we must use mathematical approaches to describe the shape of the earth in a coordinate space. We do this using an ellipsoid—a simplified model of the shape of the earth. Common ellipsoids include:
• NAD27 (North American Datum of 1927) based upon land surveys
• NAD83 based upon satellite data measuring the distance of the surface of the earth to the center of the plant. This is also internationally known as GRS80 (Geodetic Reference System 1980) internationally.
• WGS84 (World Geodetic System 1984) is a refinement of GRS80 done by the US military that was used in the development of GPS systems (and subsequently for all of us).
Onto this ellipsoid, we must define a set of reference locations (in 3-space) called datum that help describe the precise shape of the surface.
## 32.1 Projections
A projection onto an ellipsoid is a way of converting the spherical coordinates, such as longitude and latitude, into 2-dimensional coordinates we can use. There are three main types of approaches that have been used to develop various projections. (see wikipedia for some example imagery of different projections).
These include:
• Azimuthal: An approach in which each region of the earth is projected onto a plane tangential to the surface, typically at the pole or equator. Cylindrical: This approach projects the surface of the earth onto a cylinder, which is ‘unrolled’ like a large map. This approach ‘stretches’ distances in a east-west fashion, which is why Greenland looks so large…
• Conic: Another ‘unrolling’ approach, though this time instead of a cylinder, it is projected onto a cone.
All projections produce bias in area, distance, or shape (some do so in more than one), so there is no ‘optimal’ projection. To give you an idea of the consequences of these projections, I’ll use the United States map as an example and we can visualize how it is projected onto a 2-dimensional space using different projections.
### 32.1.1 Equatorial Projections
These are projections centered on the Prime Meridian (Longitude=0)
Mercator Projection
library(maps)
map("state",proj="mercator")
MollWeide Projection
map("state",proj="mollweide")
Gilbert Projection
map("state",proj="gilbert")
Cylequalarea Projection
Some projections require additional parameters, this one is based upon equally spaced and straight meridians, equal area, and true centered on a particular Latitude. I used the centroid of the US.
map("state",proj="cylequalarea",par=39.83)
### 32.1.2 Azimuth Projections
These projections are centered on the North Pole with parallels making concentric circles. Meridians are equally spaced radial lines.
Orthographic Projection
map("state",proj="orthographic")
Stereographic Projection
map("state",proj="orthographic")
Perspective Projection
Here the parameter is the distance (in earth radii) the observer is looking.
map("state",proj="perspective", param=8)
Gnomonic Projection
map("state",proj="gnomonic")
### 32.1.3 Polar Conic Projections
Here projections are symmetric around the Prime Meridian with parallel as segments of concentric circles with meridians being equally spaced.
map("state",proj="conic",par=39.83)
map("state",proj="lambert",par=c(30,40))
### 32.1.4 Miscellaneous Projections
Square Projection
map("state",proj="square")
Hexagon Projection
map("state",proj="hex")
Bicentric Projection
map("state",proj="bicentric", par=-98)
Guyou Projection
map("state",proj="guyou")
There are a lot of ways to project a 3-dimensional surface onto a 2-dimensional representation. Be aware of what you are using and how you are using it when plotting materials.
### 32.1.5 Reprojecting Rasters
When working with rasters, we can reproject these onto other projections rather easily. Here is an example from the worldclim elevation tile we used previously (see 31.1).
library(raster)
alt <- raster("./spatial_data/alt.tif")
e <- extent( c(-115,-109,22,30) )
baja_california <- crop( alt, e )
baja_california
## class : RasterLayer
## dimensions : 960, 720, 691200 (nrow, ncol, ncell)
## resolution : 0.008333333, 0.008333333 (x, y)
## extent : -115, -109, 22, 30 (xmin, xmax, ymin, ymax)
## coord. ref. : +proj=longlat +datum=WGS84 +no_defs +ellps=WGS84 +towgs84=0,0,0
## data source : in memory
## names : alt
## values : -202, 2263 (min, max)
We can now project it to another projection, lets say Lambert Conic Conformal.
library(rgdal)
projection(baja_california)
## [1] "+proj=longlat +datum=WGS84 +no_defs +ellps=WGS84 +towgs84=0,0,0"
baja_lcc <- projectRaster( baja_california, crs="+proj=lcc +lat_1=48 +lat_2=33 +lon_0=-100 +ellps=WGS84")
baja_lcc
## class : RasterLayer
## dimensions : 1060, 878, 930680 (nrow, ncol, ncell)
## resolution : 845, 935 (x, y)
## extent : -1610637, -868727, 2790826, 3781926 (xmin, xmax, ymin, ymax)
## coord. ref. : +proj=lcc +lat_1=48 +lat_2=33 +lon_0=-100 +ellps=WGS84
## data source : in memory
## names : alt
## values : -202, 2205.154 (min, max)
These two projections influence the region as shown below.
### 32.1.6 Projecting In GGPlot
As usual, there is probably a way to plot these values in ggplot to make the output just a little bit more awesome. Projections of data in ggplot displays can be manipulated by appending a coord_* object to the plot. Here are two examples using a mercator and azimuth equal area projection of the state maps.
library(ggplot2)
states <- map_data("state")
map <- ggplot( states, aes(x=long,y=lat,group=group))
map <- map + geom_polygon( fill="white",color="black")
map <- map + xlab("Longitude") + ylab("Latitude")
map + coord_map("mercator")
Conversely, we can plot it using the equal area Azimuth projection
map + coord_map("azequalarea")
or fisheye
map + coord_map("fisheye",par=3)
or any other projection available listed in the mapproject() function.
## 32.2 Coordinate Systems
In R, we typically are dealing with a combination of data that we’ve collected and that we’ve attained from some other provider. In most GIS applications, the coordinate systems we encounter are either:
• UTM (Universal Transverse Mercator) measuring the distance from the prime meridian for the x-coordinate and the distance from the equator (often called northing in the northern hemisphere) for the y-coordinate. These distances are in meters and the globe is divided into 60 zones, each of which is 6 degrees in width. Geographic coordinate systems use longitude and latitude. For historical purposes these are unfortunately reported in degrees, minutes, seconds, a temporal abstraction that is both annoying and a waste of time (IMHO).
• Decimal degrees, while less easy to remember, are easier to work with in R.
• State Planar coordinate systems are coordinate systems that each US State has defined for their own purposes. They are based upon some arbitrarily defined points of reference and another pain to use (IMHO). Given the differential in state area, some states are also divided into different zones. Maps you get from municipal agencies may be in this coordinate system. If your study straddles different zones or even state lines, you have some work ahead of you…
It is best to use a system that is designed for your kind of work. Do not, for example, use a state plane system outside of that state as you have bias associated with the distance away from the origin. That said, Longitude/Latitude (decimal degrees) and UTM systems are probably the easiest to work with in R.
In R, we use rgdal to project points. Here I load in the coordinates of the populations in the Arapatus attenuatus data set and make a SpatialPoints object out of it. Setting the proj4string() here does not project the data, I am just specifying that the data are already in the lat/long WGS84 format.
library(sp)
library(gstudio)
data(arapat)
coords <- strata_coordinates( arapat )
pts <- SpatialPoints( coords[,2:3] )
proj4string(pts) <- CRS("+proj=longlat +datum=WGS84 +no_defs +ellps=WGS84 +towgs84=0,0,0")
pts
## class : SpatialPoints
## features : 39
## extent : -114.2935, -109.1263, 23.0757, 29.32541 (xmin, xmax, ymin, ymax)
## coord. ref. : +proj=longlat +datum=WGS84 +no_defs +ellps=WGS84 +towgs84=0,0,0
The CRS() function holds the definition of the projection and interfaces between the PROJ.4 and RGDAL libraries. To project a set of data points into a new coordinate systems, we use spTransform() and pass it the definition of the new system to use.
### 32.2.1 Changing Datum
We will first begin by looking at differences in the actual datum used to record the loation of plots. Here I compare the decimal Longitude/Latitude we’ve used thus far with that from the Universal Transverse Mercator (UTM).
pts.utm <- spTransform(pts, CRS("+proj=utm +zone=12 +datum=WGS84"))
summary( pts.utm )
## Object of class SpatialPoints
## Coordinates:
## min max
## Longitude 180128 686925.2
## Latitude 2552540 3248545.0
## Is projected: TRUE
## proj4string :
## [+proj=utm +zone=12 +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0]
## Number of points: 39
You can see the tranformations in the coordinate system by comparing the plots below. The relative position of each point is the same. | 2,422 | 9,408 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-16 | latest | en | 0.917023 |
https://umc-golitcino.ru/casinonav/what-odds-does-0-pay-on-roulette-1279.php | 1,610,773,054,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703500028.5/warc/CC-MAIN-20210116044418-20210116074418-00267.warc.gz | 597,234,294 | 5,995 | # What odds does 0 pay on roulette
How Much Do You Win on 0 or 00 in Roulette? If you bet on 0 or 00 on an American roulette wheel, the odds against you winning at 37/1, thanks to the addition of the extra number. This means that the expected value of betting \$1 on either 0 or 00 on an American roulette wheel is -\$0.053, which is significantly worse than on a European roulette wheel. Roulette Odds, Probability and Payout Chart for All Bets The payout for this bet is 17:1, and the odds of it winning are 5.41% for European, and 5.26% for American roulette. Street is a three-number bet, where the players bet on a row of numbers – for example, 4, 5, and 6, or 19, 20, and 21. Roulette Odds and Payouts – Beat Roulette This page explains the various roulette odds and bets for both American double zero and European single zero roulette. Below is a quick reference chart: Bet: Payout: European Roulette Odds (Chance of Winning) American Roulette Odds (Chance of Winning) Reds / Blacks (colour) Play on European Single 0 wheels where possible: The house edge is ... What Odds Does 0 Pay On Roulette - topbonusslotcasino.loan
## Roulette Basics | HowStuffWorks
Roulette Payouts and Odds - Online Roulette Odds & Payouts Chart So how do roulette payouts and odds work? ... Can You Win Playing Roulette? .... A straight-up bet requires 1 of 38 numbers (the 1-36 numbers along with 0 and ... European Roulette vs American Roulette | Guide | Betsson Jun 27, 2017 ... However, because the '0' is green, the house's odds of winning are slightly ... Secondly, when playing on a European roulette wheel you can ... Roulette 101 - Basics Of The Game - Online casinos
### Bet, Payout, European Roulette Odds (Chance of Winning), European Roulette House Edge .... Series 0/2/3: A bet on the orange “Series 0/2/3” area is: Streets (3 ...
What difference does this make you may ask, and the answer would be quite a lot...in fact the presence of the double zero makes the odds on each numberThis game is very similar to American Roulette, with the exception that the wheel is configured differently and features a single zero, instead of the... Roulette odds red hitting | Safe gambling free&paid Roulette Odds & Probabilities All the even bets in roulette are: red or black, even or odd, or All of those bets have a chance of winning close to 50%. What's the probability of the results of 5 spins of the roulette wheel being red? On the roulette charts above I have used; ratio odds. Roulette - Wikiwand | Bet odds table Roulette is a casino game named after the French word meaning little wheel. In the game, playersBet odds table. The expected value of a \$1 bet (except for the special case of Top line bets), forHowever, the house also has an edge on inside bets because the pay outs (including the original... How To Play Roulette - Guide To Rules, Odds & Bets + Pictures Roulette rules, bets and payouts explained. Check out our illustrated tutorial and learn how roulette works. Play roulette like a pro. | Bojoko | 715 | 3,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | latest | en | 0.876921 |
https://www.physicsforums.com/threads/simple-hyperbolic-calculus.721741/ | 1,508,636,889,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00514.warc.gz | 945,366,151 | 17,069 | # Simple hyperbolic calculus
1. Nov 9, 2013
### synkk
q: http://gyazo.com/297417b9665206ae8e38cb8b5d930a83
I'm stuck trying to find the value of x when TN is a minimum
here's what I've tried so far:
Let T be the point (a,0) and N be the point (b,0)
line of tangent through P:
$y = sinh(x)(x-a)$
line of normal through P $y = \dfrac{-1}{sinh(x)}(x-b)$
my plan was to rearrange for a and b and find b-a and try differentiate that and set the derivative to be 0 and solve for x:
$a = x - \dfrac{y}{sinh(x)}$
$b = y(sinh(x)) + x$
$b - a = y(sinh(x) + \dfrac{1}{sinh(x)})$
$\dfrac{d(b-a)}{dx} = \dfrac{dy}{dx} (sinh(x) + \dfrac{1}{sinh(x)}) + y(cosh(x) - coth(x)cosech(x)) = sinh^2(x) + 1 + y(cosh(x) - coth(x)cosech(x)) = 0$ but I can't seem to solve that
any ideas on where I went wrong?
2. Nov 9, 2013
### tiny-tim
hi synkk!
erm
y isn't a variable, y is coshx !
3. Nov 9, 2013
### synkk
oh thank you, but is my method correct?
following on from this I get $x = arsinh(\dfrac{1}{\sqrt{2}})$ is this correct?
4. Nov 9, 2013
### tiny-tim
yup!
5. Nov 10, 2013
### FeDeX_LaTeX
You could have saved yourself differentiating $\sinh(x) + \frac{1}{\sinh(x)}$ by instead adding the two terms and noting the hyperbolic trig identity -- otherwise, all correct. | 470 | 1,271 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-43 | longest | en | 0.844193 |
https://it.mathworks.com/matlabcentral/cody/players/6171655-sune/solved | 1,590,622,838,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396163.18/warc/CC-MAIN-20200527204212-20200527234212-00088.warc.gz | 392,487,902 | 21,731 | Cody
# SuNe
Rank
Score
1 – 50 of 71
#### Problem 189. Sum all integers from 1 to 2^n
Created by: Dimitris Kaliakmanis
#### Problem 1087. Magic is simple (for beginners)
Created by: Jean-Marie Sainthillier
#### Problem 568. Number of 1s in a binary string
Created by: Srivardhini
Tags binary, sum
#### Problem 645. Getting the indices from a vector
Created by: Doug Hull
#### Problem 2015. Length of the hypotenuse
Created by: Tanya Morton
#### Problem 641. Make a random, non-repeating vector.
Created by: Doug Hull
#### Problem 1035. Generate a vector like 1,2,2,3,3,3,4,4,4,4
Created by: Binbin Qi
Tags zeros, for, ones
#### Problem 23. Finding Perfect Squares
Created by: Cody Team
#### Problem 25. Remove any row in which a NaN appears
Created by: Cody Team
#### Problem 43215. Convert radians to degrees
Created by: Jamil Kasan
#### Problem 135. Inner product of two vectors
Created by: AMITAVA BISWAS
#### Problem 247. Arrange Vector in descending order
Created by: Vishwanathan Iyer
#### Problem 11. Back and Forth Rows
Created by: Cody Team
Tags matrices
#### Problem 42298. Repeat The Components of Matrix
Created by: Vishal
Tags matrix
#### Problem 2350. What is Sum Of all elements of Matrix
Created by: Vishal
#### Problem 42652. Test Problem; Create a 5x5 array containing all ones
Created by: Scott Robinson
Tags easy, arrays, basics
#### Problem 2568. Squaring Matrix
Created by: Marisa
#### Problem 2814. Remove the Zero
Created by: Juan Moreta
#### Problem 350. Back to basics 7 - Equal NaNs
Created by: Alan Chalker
#### Problem 2402. Area of a Square
Created by: Shravankumar P
#### Problem 77. Clean the List of Names
Created by: Cody Team
Tags strings
#### Problem 2572. Counting down
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#### Problem 33. Create times-tables
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#### Problem 34. Binary numbers
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Tags strings
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#### Problem 105. How to find the position of an element in a vector without using the find function
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Tags indexing, find
1 – 50 of 71 | 1,096 | 3,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-24 | latest | en | 0.706567 |
http://tenminutetutor.com/maths/a-level/pure/coordinate-systems/ellipse-parametric-equation/ | 1,603,592,249,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885126.36/warc/CC-MAIN-20201025012538-20201025042538-00623.warc.gz | 107,056,391 | 3,491 | # Parametric equation of ellipse
Martin McBride, 2020-09-14
Tags ellipse major axis minor axis
Categories coordinate systems pure mathematics
The parametric equation of an ellipse is:
\begin{align} x = a \cos{t}\newline y = b \sin{t} \end{align}
## Understanding the equations
We know that the equations for a point on the unit circle is:
\begin{align} x = \cos{t}\newline y = \sin{t} \end{align}
Multiplying the $x$ formula by $a$ scales the shape in the x direction, so that is the required width (crossing the x axis at $x = a$). In this example, $a > 1$ so the circle is stretched in the x direction:
Multiplying the $y$ formula by $b$ scales the shape in the y direction, so that is the required height (crossing the y axis at $y= b$). In this example, $b < 1$ so the circle is compressed in the y direction:
Copyright (c) Axlesoft Ltd 2020 | 240 | 854 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-45 | latest | en | 0.773455 |
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-10-infinite-sequences-and-series-section-10-3-the-integral-test-exercises-10-3-page-586/9 | 1,576,130,973,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00082.warc.gz | 740,785,764 | 13,858 | ## Thomas' Calculus 13th Edition
Let us consider $f(x)=\dfrac{x^2}{e^{x/3}}$ Here, the function $f(x)$ is positive, continuous for $x \geq 1$ and $f(x)$ is decreasing for $x \gt 7$ Then $\int_7^\infty \dfrac{x^2}{e^{x/3}}dx= \lim\limits_{k \to \infty} \int_7^k \dfrac{x^2}{e^{x/3}}dx$ or, $\lim\limits_{k \to \infty} [\dfrac{-54}{e^{k/3}}+\dfrac{327}{e^{7/3}}]=\dfrac{327}{e^{7/3}}$ Thus, the sequence $\Sigma_{n=1}^\infty \dfrac{n^2}{e^{n/3}}$ is convergent. | 208 | 460 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-51 | latest | en | 0.514854 |
https://discourse.matplotlib.org/t/how-to-visualize-a-b-results-of-x-y-variables/19374 | 1,670,061,100,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00242.warc.gz | 240,425,008 | 6,735 | # How to visualize a,b results of x,y variables
I have a bunch of experimental data points each of which has 2 variables (x,y) and 2 results (a,b). Each pair or x,y values produces a pair of a,b resultant values.
There is a single optimal pair of a,b values and I’d like to figure out a way to illustrate the data to show the relationship between each x,y pair and how close each a,b pair is to the ideal.
I’m thinking about a dual surface/contour plot with 2 different z-axes. Ideally I would center both z-axes at the ideal values. I don’t know if this is possible. Might be kinda messy.
Any other thoughts? I’m sure there must be other examples where this is a problem.
In the x,y plane, could you overlay contours of a with contours of b?
-Sterling
···
On Jul 8, 2015, at 8:19PM, Jonno <jonnojohnson@...287...> wrote:
I have a bunch of experimental data points each of which has 2 variables (x,y) and 2 results (a,b). Each pair or x,y values produces a pair of a,b resultant values.
There is a single optimal pair of a,b values and I'd like to figure out a way to illustrate the data to show the relationship between each x,y pair and how close each a,b pair is to the ideal.
I'm thinking about a dual surface/contour plot with 2 different z-axes. Ideally I would center both z-axes at the ideal values. I don't know if this is possible. Might be kinda messy.
Any other thoughts? I'm sure there must be other examples where this is a problem.
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I was thinking of doing that or having 2 surface plots but I think it would be visually quite confusing.
I was trying to think of an example since I’m sure someone has come up with a nice way to display this kind of data.
Imagine if the data was average temperature (a) and average rainfall (b) for a region in the world (lat/long = x,y). The goal is to display the data such that it’s obvious where the locations are that have closest to the ideal temp/rain combination.
How would you go about that?
···
On Thu, Jul 9, 2015 at 12:28 AM, Sterling Smith <smithsp@…3304…> wrote:
In the x,y plane, could you overlay contours of a with contours of b?
-Sterling
On Jul 8, 2015, at 8:19PM, Jonno <jonnojohnson@…287…> wrote:
I have a bunch of experimental data points each of which has 2 variables (x,y) and 2 results (a,b). Each pair or x,y values produces a pair of a,b resultant values.
There is a single optimal pair of a,b values and I’d like to figure out a way to illustrate the data to show the relationship between each x,y pair and how close each a,b pair is to the ideal.
I’m thinking about a dual surface/contour plot with 2 different z-axes. Ideally I would center both z-axes at the ideal values. I don’t know if this is possible. Might be kinda messy.
Any other thoughts? I’m sure there must be other examples where this is a problem.
GigeNET’s Cloud Solutions provide you with the tools and support that
https://www.gigenetcloud.com/_______________________________________________
Matplotlib-users mailing list
Matplotlib-users@lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/matplotlib-users
It might just have to be 2 separate contour/surface plots side by side, perhaps with a linked cursor between them.
The other thing I considered was combining the a,b data into a single value (combined % deviation from ideal?) but that reduces the data which I’d rather not do if possible.
···
On Thu, Jul 9, 2015 at 9:40 AM, Jonno <jonnojohnson@…287…> wrote:
I was thinking of doing that or having 2 surface plots but I think it would be visually quite confusing.
I was trying to think of an example since I’m sure someone has come up with a nice way to display this kind of data.
Imagine if the data was average temperature (a) and average rainfall (b) for a region in the world (lat/long = x,y). The goal is to display the data such that it’s obvious where the locations are that have closest to the ideal temp/rain combination.
How would you go about that?
On Thu, Jul 9, 2015 at 12:28 AM, Sterling Smith <smithsp@…3304…> wrote:
In the x,y plane, could you overlay contours of a with contours of b?
-Sterling
On Jul 8, 2015, at 8:19PM, Jonno <jonnojohnson@…287…> wrote:
I have a bunch of experimental data points each of which has 2 variables (x,y) and 2 results (a,b). Each pair or x,y values produces a pair of a,b resultant values.
There is a single optimal pair of a,b values and I’d like to figure out a way to illustrate the data to show the relationship between each x,y pair and how close each a,b pair is to the ideal.
I’m thinking about a dual surface/contour plot with 2 different z-axes. Ideally I would center both z-axes at the ideal values. I don’t know if this is possible. Might be kinda messy.
Any other thoughts? I’m sure there must be other examples where this is a problem.
GigeNET’s Cloud Solutions provide you with the tools and support that
https://www.gigenetcloud.com/_______________________________________________
Matplotlib-users mailing list
Matplotlib-users@lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/matplotlib-users
It's not an easy thing to visualize in general. You might want to look at approaches to visualizing complex functions (i.e., functions whose input and output are both complex variables). These essentially map pairs (a, b) to pairs (x, y) as in your situation, and mathematicians have come up with various ways to visualize them. Some are described at https://www.pacifict.com/ComplexFunctions.html and the wikipedia article at https://en.wikipedia.org/wiki/Complex_analysis has some links in the references to web pages for graphing such functions.
If the data are measured at (or can be reasonably reduced to) discrete points (as temp/rainfall are likely to be), another possibility is a scatterplot using, say, the color and size of the markers as indicators of the two variables (e.g., red/blue for hot/cold temp, larger/smaller circles for higher/lower rainfall).
In some cases, like your example with temperature and rainfall, you may instead be able to combine the two output dimensions into a single one that somehow captures the overall "distance" from the ideal point. That is, for a given point, if your goal is to show how close it is to the ideal *combination* of temp and rain, you may not need to display how close it is on each dimension separately, but just how close it is to the ideal overall. Exactly how to compute this would vary based on the data (e.g., standardizing the values and taking the euclidean distance from the ideal).
Your temp/rainfall example caught my eye because a few years ago I did a blog post on a similar topic, considering temperature and humidity (http://iq.brenbarn.net/2011/11/18/good-days-mate/). There I decided to graph just a single variable, namely the number of days on which either temperature *or* humidity is outside a "comfortable" range. Obviously this approach may not make sense for every situation. But what I mean is that, in some cases, you can use domain-specific knowledge about what the dimensions mean to combine them into one dimension that approximates what it is you're trying to illustrate with the graph.
···
On 2015-07-09 07:40, Jonno wrote:
I was thinking of doing that or having 2 surface plots but I think it
would be visually quite confusing.
I was trying to think of an example since I'm sure someone has come up
with a nice way to display this kind of data.
Imagine if the data was average temperature (a) and average rainfall (b)
for a region in the world (lat/long = x,y). The goal is to display the
data such that it's obvious where the locations are that have closest to
the ideal temp/rain combination.
How would you go about that?
--
Brendan Barnwell
"Do not follow where the path may lead. Go, instead, where there is no path, and leave a trail."
--author unknown
Maybe you could plot the ratio? That should give you rainfall per degree Celsius.
···
On Thu, Jul 9, 2015 at 12:28 AM, Sterling Smith <smithsp@…3304…> wrote:
In the x,y plane, could you overlay contours of a with contours of b?
-Sterling
On Jul 8, 2015, at 8:19PM, Jonno <jonnojohnson@…287…> wrote:
I have a bunch of experimental data points each of which has 2 variables (x,y) and 2 results (a,b). Each pair or x,y values produces a pair of a,b resultant values.
There is a single optimal pair of a,b values and I’d like to figure out a way to illustrate the data to show the relationship between each x,y pair and how close each a,b pair is to the ideal.
I’m thinking about a dual surface/contour plot with 2 different z-axes. Ideally I would center both z-axes at the ideal values. I don’t know if this is possible. Might be kinda messy.
Any other thoughts? I’m sure there must be other examples where this is a problem.
GigeNET’s Cloud Solutions provide you with the tools and support that
https://www.gigenetcloud.com/_______________________________________________
Matplotlib-users mailing list
Matplotlib-users@lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Thanks for all the ideas.
···
On Thu, Jul 9, 2015 at 8:09 PM, Joy merwin monteiro <joy.merwin@…287…> wrote:
Maybe you could plot the ratio? That should give you rainfall per degree Celsius.
On 9 Jul 2015 20:11, “Jonno” <jonnojohnson@…985…> wrote:
I was thinking of doing that or having 2 surface plots but I think it would be visually quite confusing.
I was trying to think of an example since I’m sure someone has come up with a nice way to display this kind of data.
Imagine if the data was average temperature (a) and average rainfall (b) for a region in the world (lat/long = x,y). The goal is to display the data such that it’s obvious where the locations are that have closest to the ideal temp/rain combination.
How would you go about that?
GigeNET’s Cloud Solutions provide you with the tools and support that
https://www.gigenetcloud.com/
Matplotlib-users mailing list
Matplotlib-users@lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/matplotlib-users
On Thu, Jul 9, 2015 at 12:28 AM, Sterling Smith <smithsp@…3304…> wrote:
In the x,y plane, could you overlay contours of a with contours of b?
-Sterling
On Jul 8, 2015, at 8:19PM, Jonno <jonnojohnson@…287…> wrote:
I have a bunch of experimental data points each of which has 2 variables (x,y) and 2 results (a,b). Each pair or x,y values produces a pair of a,b resultant values.
There is a single optimal pair of a,b values and I’d like to figure out a way to illustrate the data to show the relationship between each x,y pair and how close each a,b pair is to the ideal.
I’m thinking about a dual surface/contour plot with 2 different z-axes. Ideally I would center both z-axes at the ideal values. I don’t know if this is possible. Might be kinda messy.
Any other thoughts? I’m sure there must be other examples where this is a problem.
GigeNET’s Cloud Solutions provide you with the tools and support that | 2,735 | 11,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-49 | latest | en | 0.891338 |
https://mathematica.stackexchange.com/questions/13717/how-to-efficiently-take-complement-of-two-big-lists | 1,716,519,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00081.warc.gz | 323,783,809 | 40,043 | # How to efficiently take complement of two big lists?
I have a giant list l1 and some smaller lists l21, l21, ..., l2n. The l1 is the superset and has $\tilde{}10^7$ elements. l2? is subset of l1 (if my other code does correct job). All lists only contain numbers (again, if my code does correct job). I need to take complement of l1 and each of the l2?. The number of subsets (i.e. n) is several hundreds. What's the most memory/time efficient way of doing this? I run this on 4GB machine and usually have 1.* GB of free memory when the program reaches that stage. I'd probably save numbers in file and use some command-line tool like grep to do this. But the original lists are all prepared in a Mathematica program. If there is a good solution in Mathematica then i'd like to avoid going out.
• You've tried Complement[] already? Oct 27, 2012 at 2:33
• If your lists of integers are reasonably dense and entries are non-repeating, you can use arbitrary-length integers interpreted as bit vectors as representation of these lists. These integers can be constructed using repetitive BitSet, complement performed using BitAnd[l1,BitNot[l2?]], and values collected through BitGet. This is elegant, but slow. For a faster kludge, see [mathematica.stackexchange.com/a/13708/3056]. Both methods consume roughly one bit (not byte) per integer (present or not in the list) in list range. (I believe Complement should be sufficient though.) Oct 27, 2012 at 13:45
• Small addition: unless you create the lists in special fashion from the start, you don't really get space benefit over Complement in the large scheme of things. If you had billions of integers, optimization could make sense, but otherwise... Oct 27, 2012 at 13:54
This depends.. How many numbers are in the subsets? In which range are you numbers? What do you want to do with each complement? Can you give a small example using RandomInteger to create sample-data? Generally, you could first try to calculate one complement by using something like this
l1 = RandomInteger[{0, 10^6}, {10^7}];
l21 = RandomInteger[{0, 10^6}, {10^5}];
compl = Complement[l1, l21];
and see whether you memory is sufficient.
To see how much memory is used you can tryMemoryInUse[]. ByteCount[expr] is able to find out how much memory is used by a variable (or expression in general). After the above command, I have wasted
MemoryInUse[]/2^20.
(* Out[9]= 101.952 *) | 602 | 2,411 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.920348 |
http://www.enotes.com/homework-help/explain-why-graph-f-x-3x-cos-1-2x-lies-below-x-328059 | 1,477,488,665,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720945.67/warc/CC-MAIN-20161020183840-00470-ip-10-171-6-4.ec2.internal.warc.gz | 422,680,597 | 10,415 | Explain why the graph of f(x)=(3x−π)cos(1/2x) lies below the x-axis for −πI get that for x the interval [pi/3,pi], 3x-pi lies in the interval [0,2pi] so 3x-pi is positive. Also for x in the...
Explain why the graph of f(x)=(3x−π)cos(1/2x) lies below the x-axis for −πI get that for x the interval [pi/3,pi], 3x-pi lies in the interval [0,2pi] so 3x-pi is positive. Also for x in the interval [pi/3,pi] 1/2(x) lies in the interval [pi/6,pi/2] so cos(1/2(x)) is positive. Hence the graph of y= (3x-pi)cos(1/2(x)) is above the x axis on the interval [pi/3,pi]. For x in the interval [-pi,pi/3] 3x-pi lies in the interval [-4pi,0] so 3x-pi is negative. Also for x in the interval [-pi,pi/3] 1/2(x) lies in the interval [-pi/2,pi/6]. . . . Pi/6 is positive not negative! help!
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You need to solve the equation `(3x−pi)cos(x/2) = 0` such that:
`3x - pi = 0 =gt 3x = pi =gt x = pi/3`
`cos (x/2) = 0 `
You need to remember that the values of cosine function rest positive in 1 and 4 quadrants, such that:
`x/2 = arccos 0 =gt x/2 = pi/2 =gt x = pi`
`x/2 = 3pi/2 =gt x = 3pi`
You need to substitute x by a value in interval `[pi;3pi]` moving in positive direction over trigonometric quadrants.
You need to notice that the graph of function intercepts axis at `x=pi/3 in [pi,3pi]`
You need to consider a value `x = pi - pi/3 = 2pi/3` such that:
`3x - pi = 2pi - pi = pi`
`cos 2pi/6 = cos pi/3 = 1/2`
You need to consider a value `x = pi+ pi/3 = 4pi/3` such that:
`4pi-pi = 3pi`
`cos 4pi/6 = cos 2pi/3 = -cos pi/3 = -1/2`
Notice that if `x in [pi,3pi/2]` , the values of function are negative (for `x=4pi/3 =gt f(x) = -3pi/2` ), hence the graph of function goes below x axis (consider `pi~~3.14` ) | 683 | 1,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2016-44 | latest | en | 0.757052 |
https://friesian.com/century.htm | 1,708,530,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00842.warc.gz | 264,496,007 | 29,849 | # The Century and the Millennium;Cardinalists vs. Ordinalists
SCULLY: And besides, 2001 is actually the start of the new millennium.
MULDER: Nobody likes a math geek, Scully.
If you didn't know there was a year zero in astronomy, let me respectfully suggest you're not informed enough to tell others when to begin their centuries.
There are two kinds of people, those who think there are two kinds of people, and those who don't.
I am among those who think there are two kinds of people, and that they are people who think that the new century and millennium began on January 1, 2000, and those who think that they began on January 1, 2001.
A definitive discussion of this issue may be found in an essay of the late, great Stephen Jay Gould, "Dousing Diminutive Dennis's Debate (or DDDD=2000)," which is collected in Dinosaur in a Haystack [Harmony Books, 1995]. Gould, as in his defense of the name Brontosaurus ("Bully for Brontosaurus"), displays towering good sense. (For more on "two kinds of people," see "Psychological Types")
The advocates of "2001" usually say that since a century is 100 years, a millennium is 1000 years, and the calendar began with year 1, therefore all subsequent centuries will begin with a year 1 (e.g. 1901), and all millennia will begin with a year 1 (e.g. 2001). There is no year zero.
Fair enough. However, when calendars were invented that numbered the years, whether the regal years of Egyptian Kings or a continuous count like the Seleucid Era, the number systems used did not contain the number zero. There could be no year zero when there was no zero. Also, years were thought of as ordinals: the first year of a reign was thus year 1. People who still think in these terms I will call "Ordinalists."
The number zero, conceived in India, was introduced into Western mathematics by the mathematician al-Khuwârizmî (c.780-850). The Arabs still call this system "Indian" (Hindî) numbers, while Europeans, etc., call it "Arabic" numerals. The number zero answers the question of cardinal numbers, "How many?" rather than the question of ordinal numbers, "Which one?" Mathematical questions are usually about cardinals rather than ordinals.
When we say that it is the year 1997 of the "Annô Domini" or "Common" Era, does this mean that it is the 1997th year of the Era ("Which one?"), or that 1997 years have elapsed ("How many?") since a Benchmark? Well, it can mean both. If it is the 1997th year of the Era, then the Era began on January 1st, 1 AD. On the other hand, if 1997 years have elapsed since a Benchmark, then the Benchmark was January 1, 0 AD. January 1, 1 AD, would mean that 1 year has elapsed since the Benchmark. That makes the calendar begin with the year 0 AD, not with the year 1 AD. People who think in these terms I will call "Cardinalists" [note].
If the question is about when the calendar really "begins," then of course the truth is that the calendar did not begin either in 1 AD or 0 AD. The Julian Calendar began in 46 BC, the Gregorian Calendar began in 1582 AD, and the "AD" numbering of the years was proposed, although not extensively used until later, by Dionysius Exiguus in the 6th Century AD. Even the calendar of the French Revolution was introduced in October 1793, a year and a month after the beginning of its own year "1" (September 1792), which was identified retrospectively -- and the calendar soon enough ceased to be used (Napoleon abolished it in year 12, or 1804, when he crowned himself Emperor). When a calendar or a reckoning "begins" thus usually says nothing about whether a calendar might be reckoned from a year 1 or a year 0.
The bald statement by "2001" partisans that "there is no year zero" is now, as a matter of usage, simply false. Calculations by astronomers and chronologists conveniently use zeros for years, months, and days. Perusing the Astronomical Almanac for the year 1997 [U.S. Government Printing Office & Her Majesty's Stationery Office, 1996], one finds days like "January 0." This is not surprising. Astronomers and chronologists do their calculations with Arabic numerals, which contain the number zero. This introduces a Cardinalist bias.
Even though traditional usage for years was ordinal, while modern mathematical and scientific use tends to be cardinal, there is one common numbering usage that is cardinal: the numbering of personal age. When someone is born in the United States, they may be starting their 1st year, but they are not already "1 year old." That comes a year later. Age is thus seen as elapsed time, starting from 0. That a different usage is possible should be obvious. Indeed, the traditional Chinese reckoning of age is ordinal, so that "1 year old" means the 1st year of life. This can be very confusing in places where both Chinese and Western reckoning may be used along side each other, as in Hawaii.
The real choice between 2001 and 2000 is an aesthetic preference: 2001 is consistent with tradition and thus the conservative, traditionalist choice. 2000 is the natural result of the introduction of the number zero, which made the mathematical power of science possible, and thus the modernistic, progressive choice. There is no particular reason why one should be seen as really superior to the other, if we are to honor both tradition and innovation in human affairs. It would be appropriate to celebrate both to indicate that we both look ahead (2000) and look back (2001) in our worldview.
On the other hand, the worst thing about the Ordinalists is their customary dogmatism and arrogance: they just know that there was no year zero, which means that people who begin the century with a zero are vulgar, ignorant, and can't add the number 100 (I've had at least one completely indignant, rude, hostile, and blockheaded correspondent display this attitude -- I think he wished he could have me arrested). "Vulgar" is a significant component of their judgment, since their pronouncements are often delivered with a sneer and a lofty, superior air (if not the aforesaid hostility). The cognitive psychologist and linguist Steven Pinker says that some things are "gotcha! material for pedants and know-it-alls (the kind of people who insist that the millennium begins January 1, 2001)" [Words and Rules, Basic Books, 1999, p.54]. Indeed, Ordinalists use the issue to prove how superior they are compared to the gaucherie of everyone else. For this they should, at least about this turn of the century, be savagely ridiculed. They seem to be among the mathematical illiterates who have never heard of actually using the number zero. Or perhaps they are racists who don't want to use some newfangled thing that comes from India or the Arabs. Such charges would at least serve to drive them out of their arrogance, if not silence them completely (since we know that anyone accused of being a racist, who denies it, therefore is a racist).
Sadly, the great libertarian economist Walter Williams, who always displays towering good sense much more than Stephen Jay Gould on economic and political issues, nevertheless has revealed himself to be an Ordinalist (cf. "Feelings are more important than facts," Conservative Chronicle, November 3, 1999, p.31). Although he can cite the U.S. Naval Observatory in his defense, Williams nevertheless betrays his unfamiliarity with the chronological use of zero on a issue unrelated to the century and the millennium: He says that "the new millennium starts at 12:01 a.m. 2001." Saying that the day starts at "12:01" is natural for an Ordinalist, and perhaps also for someone just using a 12 hour clock. Anyone using a 24 hour clock, especially a digital clock, knows, however, that the day starts at 00:00, "zero hundred hours" -- at 00:01 a minute has already passed. This betrays for us the characteristic Ordinalist lack of attention, or familiarity, with the modern use of zero.
Unfortunately for the Ordinalists, everyone celebrates the New Year at the stroke of Midnight, not at 12:01, and, much worse, the meaning of 00:00 hours January 1, 2000 was hightened by anxieties and fears about the damage that could be done by the Y2K computer bug. It was thought that various essential public services could stop because older computers might lock up when their internal clocks showed the year as 00 and this was interpreted as 1900 rather than 2000. A made-for-television disaster movie anticipated riots and anarchy, and doomsday theorists were ready to take to the hills with enough supplies to surivive the End of Civilization. It was especially of concern that the Russians might lose control of their nuclear missles. Few expected the Russians to upgrade their computers in time to be free of any possible Y2K bugs. As it happened, little or nothing went wrong at Midnight of January 1, 2000, either where the day began, in the western Pacific, in Russia, or anywhere else. But there was not going to be anything like same kind of anxiety about 2001.
Traditional English Names of Full Moons, and the "Blue Moon"
### The Century and the Millennium; Cardinalists vs. Ordinalists; Note 1
In 1999 Dr. Dutch posted a generally excellent webpage about the millennium. Unfortunately, it is now disappeared, but I have retained the following critique.
Unfortunately, despite an insightful and helpful discussion, the page is marred at the end with the following diagram, which I reproduce:
```199 BC 99 BC 4 BC 3 BC 2 BC 1 BC | 1 AD 2AD
+---------+----...-----+---------+---------+---------+---------+....
-200 -150 -100 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2
| Year -2 | Year -1 | Year -0 | Year +0 | Year +1 |
|Century-1| Century -0 | Century +0```
Here we see the year 1 BC identified as "Year +0," 2 BC as "Year -0," 3 BC as "Year -1," and 4 BC as "Year -2." However, "+0" and "-0" are not different numbers (adding or substracting 0 does not change any number). What comes before the number 1 in the sequence of integers is 0, and what comes before 0 is -1. Thus, while 1 BC is indeed 0 AD, 2 BC is already -1 AD, 3 BC is -2 AD, and 4 BC is -3 AD.
I have brought this problem to Dr. Dutch's attention, and he insists that this is correct because the "0" refers to the integer portion of the decimal, while, with the numbers listed in the middle of the diagram (showing years and half-years), 2 BC does begin with -1.
Now, such a construction is reasonable enough, with a zero point at the boundary between 1 BC and 2 BC, but it is confusing and does not clearly represent the way years are actually numbered in historical usage, whether Ordinalist or Cardinalist. Or, for that matter, centuries, where the 1st century AD and the first century BC are not helpfully labeled "+0" and "-0" any more than the years.
It did not add to Dr. Dutch's explanation when he wrote to me that this was like "0oN" and "0oS" latitude -- when in fact 0o latitude is the Equator, and there is no difference between North and South there. Again, he was apparently using this construction to mean the integer portion within the first degees of latitude, North and South, despite the confusing way that will appear. A location at 30' (minutes) North latitude will be identified as 30'N, or 0o30'N, or 0.5oN, never as just "0oN." That is no actual, specific location of latitude -- except as equivalent to the Equator, where the "North" part is moot.
It is a bad idea to employ "+0" and "-0" as though they are different numbers, when they are not, especially where readers might get the idea that the year before 0 AD is "-0 AD" and that 4 BC corresponds to -2 AD rather than -3 AD. The way that Dutch uses these numbers, while it may be reasonably motivated and explicable, displays a prima facie numerical fallacy and is far too easily misunderstood or misapplied. When the problem of the century and the millennium is that the Ordinalists don't use zero at all, it doesn't help to introduce a usage where we see more zeroes than there actually are in the sequence of integers.
Come to think of it, I'm going too easy on Dr. Dutch. Since +0 and -0 are not different numbers, it is wrong to use them as though they are, whatever the explanation. And if the explanation for this involves decimals, this is also bad, since calendars use integers and fall under discrete or modular mathematics -- a whole branch of mathematics -- with deals with integers alone. If Dr. Dutch doesn't approve of this, I can just see him trying to lecture Gauss, who substantially developed that form of mathematics and who wrote modular formulae for calendars.
### The Century and the Millennium; Cardinalists vs. Ordinalists; Note 2
Ordinalism has its own permutations. When the Julian Calendar was instituted in 46 BC, with the rule that a day should be added every fourth year, a curious misunderstanding ensued. The priests (pontifices) in charge of the calendar inserted the leap day every three years, not every four years. They did this because they were counting "inclusively," i.e. the fourth year from the previous cycle counted as the first year of the next. This kept up from 46 BC to 9 BC, when the error was realized. The Emperor Augustus then brought things back in order by omitting sixteen leap years until 8 AD [cf. E.J. Bickerman, Chronology of the Ancient World, Cornell U. Press, 1968, 1980, 1982, p.47].
Ordinalists now tend to count ordinals "exclusively," as Augustus began doing in 8 AD, counting the first year of the cycle as the one after the fourth year. This practice, however, in effect leaves the fourth year of the cycle as the zero year of the next, which means that inclusive counting is really more conformable to ordinals, while the exclusive counting of ordinals is commensurable with cardinals.
The inclusive counting of ordinals can also be found in the Bible, where Jesus is said to have risen on the "third day" after his burial [Matthew 16:21, Mark 9:32, etc.], i.e. Easter Sunday. Once I saw an evangelist on television (I think it was Herbert W. Armstrong, 1892-1986) who protested that Jesus could not have been crucified and buried on a Friday and resurrected on a Sunday, because that was only two days later, not three. It is, of course, the third day if you count Friday as the first. Of such stuff are heresies and schisms made.
Inclusive counting is something also seen in music. The "octave" (octava, "eighth") is the next note in the scale of the same kind. Thus, beginning with C, we go through D, E, F, G, A, B, and finally C again. One notices, however, that there are only seven notes. To get eight going from C to C, one must count inclusively. This is the case with other intervals on the scale, the second, the fifth, etc.
Thus, ordinal counting includes varities of practice that may not trouble the cardinalists.
### The Century and the Millennium; Cardinalists vs. Ordinalists; Note 3
How do we calculate what Olympiad it is? If we multiply 75 by 4 (=300) and substract from 776, we get 476, which is not 480 BC. What went wrong? We are confusing cardinals with ordinals. The 1st Olympiad, in 776 BC is not four years (1x4) after the First Olympiad. We can multiply 75 by 4 and subtract it if we use the "0th" Olympiad, which would be 780 BC: 780-(75x4)=480.
How do we determine which year is the first year of the Olympiad? 480 is convenientaly divisible by 4. But dividing by four leaves a remainder of zero, while we might be prefer to have a remainder of 1 for the first year. No problem. All we need to do is use dates in the AD era. 480 BC is -479 AD, as 1 BC is 0 AD. The previous year, -480 AD (481 BC), is now evenly divisible by 4, which means that the first year of the Olympiad is -480+1=-479 AD=480 BC, which is what we are looking for.
The "0th" year of the "0th" Olympiad is -780 AD. We can get the zero year of any Olympiad by adding 4 times the number of the Olympiad to this year. Or we can substract this from the zero year of any Olympiad, divide by 4, and get the number of the Olympiad.
Using years of the AD era means that we can conveniently calculate modern Olympiads. 2008 is evenly divisible by 4 (it is a leap year on the Julian and Gregorian calendars), which means that: 2008-(-780)/4=697. 2008 is the zero year of the 697th Olympiad. The Olympic Games would be in 2009, the first year of the Olympiad. 2011 is the 3rd year of the Olympiad.
Oh, Oh. The Modern (Summer) Olympic Games were in 2008, not 2009. The Modern Olympic Games use the zero year of the Olympiad! Where were the Ordinalists when we needed them? But perhaps the creators of the Modern Olympics simply didn't know that years evenly divisible by 4 in the AD era would also be the "0th" year of an Olympiad.
Since 1994, the Winter Olympics (which have been held since 1924) have been offset two years from the Summer Olympics. They are thus like the Pythian Games, at Delphi, which were held two years after the Olympic Games -- which now puts the Winter Olympics in the 2nd rather than the (Pythian) 3rd year of the Olympiad. The Nemean and Isthmian Games were held every two years and were less prestigious.
Greek historians like Polybius dated events using Olympiads, but the system was never used as a continuous era, like the Seleucid Era, in a Greek civil calendar.
# Traditional English Names of Full Moons,and the "Blue Moon"
While the division of the year is similar to the calendar of the French Revolution, where the year begins with the Autumnal Equinox, the French months were conventionally set to a length of 30 days each.
Full MoonZodiacal
Period
Starting DateLength
Yule, Winter Solstice89d
1. Moon after Yule CapricornDecember 2229d
2. Wolf Moon AquariusJanuary 2030d
13. Blue MoonThe third Full Moon of any Season with four
3. Lenten Moon PiscesFebruary 1930d
First Day of Spring, Vernal Equinox92d
4. Egg Moon
(Paschal Moon)
AriesMarch 2130d
5. Milk Moon TaurusApril 2031d
13. Blue MoonThe third Full Moon of any Season with four
6. Flower Moon GeminiMay 2131d
The Long Day, Summer Solstice94d
7. Hay Moon CancerJune 2132d
8. Grain Moon LeoJuly 2331d
13. Blue MoonThe third Full Moon of any Season with four
9. Fruit Moon VirgoAugust 2331d
Summer's End, Autumnal Equinox90d
10. Harvest Moon LibraSeptember 2330d
11. Hunter's Moon ScorpioOctober 2330d
13. Blue MoonThe third Full Moon of any Season with four
12. Moon before
Yule
SagittariusNovember 2230d
The zodiacal periods do not have conventional lengths but instead mark astronomical periods based on 30 degree increments in the Sun's longitude on the ecliptic (the apparent path of the Sun in the sky). In the table at right, the dates for the beginning of the zodiacal periods, and their lengths, will not be accurate for every year. Each zodiacal period may start a day earlier, as the year is reset by the addition of a leap day and then runs slightly fast for three years. Thus, while March 21 is the traditional date of the Vernal Equinox, the equinox actually occurs on March 20 three years out of four. The periods and seasons are so different in length because, according to Kepler's Second Law, the Earth travels faster the closer it is to the Sun. Between January 2 and 4, the Earth reaches Perihelion, its closest approach to the Sun, and travels the fastest. Winter in the Northern Hemisphere (89 days) is thus shorter than Summer (94 days).
This strict positioning of phases of the Moon in relation to the actual longitude of the Sun is also characteristic of the Chinese Calendar, though it positions New Moons, which mark the beginning of the Chinese year and months, rather than Full Moons. The first New Moon after the Vernal Equinox was also the basis of the Babylonian calendar; but the Jewish calendar, although similarly starting each month with the New Moon, was structured to position the month of Niisân so that its Full Moon would be the first Full Moon after the Vernal Equinox. This set the date of Passover and was inherited by Christianity in the determination of Easter. If the "Egg Moon" is allowed to occur on the Vernal Equinox itself (rather than, for instance, only after), then it is equivalent to the Paschal Moon defined for Easter. Easter calculation, however, always defines the Vernal Equinox as March 21, so there may be some years, when the Equinox is on March 20, that the Paschal Moon differs from an astronomical calculation. Although the "Egg Moon" may be pre-Christian, it does suggest Easter eggs, and the prior Full Moon, the "Lenten Moon," definitely shows Christian influence. A popup window with this table, for reference, can be created with this link.
Every two or three years thirteen Full Moons will occur from one Winter Solstice to another, and two of these will necessarily occur in the same zodiacal period. Wherever four Full Moons occur in the same season, the third is called a "Blue Moon." Seven such moons will occur in a nineteen year period -- the "Metonic" cycle of the Babylonian, Jewish, and Chinese calendars.
The detailed workings of this system had become very obscure knowledge until recently. A "Blue Moon" is now commonly said to be a second Full Moon in a calendar month, though this means that in some years, as in 1999, there are two months, January and March, with Blue Moons, while February contains no Full Moon at all. This leaves only 11 non-Blue Moons in the year to which the 12 standard moon names would need to be assigned. If the moon names are all assigned by calendar month, then presumably the absence of a Full Moon in February would mean that there is no "Wolf Moon" -- though that loss of such an omnious name might be seen as auspicious. Assigning the "Egg Moon" always to April, on the other hand, severs its relationship to the Paschal Moon. In 1999 the Paschal Moon is on March 31 (with Easter on April 4), but the April Full Moon falls on April 30, a month later. In 1999 the Paschal Moon itself would be, by the calendrical month rule, a Blue Moon, and the Lenten Moon would coincide with, indeed, the astronomical Lenten Moon on March 2.
The recent meaning of "Blue Moon" occurred because the old moon names have mostly fallen out of use, and even been forgotten. Although I have found close to accurate definitions of the "Harvest Moon" and "Hunter's Moon" in a 1962 World Book Encyclopedia and in the 1997 Microsoft Encarta Encyclopedia (using almost the same language), and I had long heard that such traditional names existed for all the Full Moons of the year, I never saw a list of all of them until an article on "Blue Moons" in the March 1999 issue of Sky & Telescope magazine ("Once in a Blue Moon," by Philip Hiscock). That article contained an inset by Donald W. Olson and Roger W. Sinnott, "Blue-Moon Mystery Solved?", where they reproduced a page from the 1937 Maine Farmer's Almanac (see at right), that identified the Full Moon occuring on August 21, 1937 as a "Blue Moon" and provided a list of all the moon names. This was mysterious to them because the Full Moon on August 21, 1937 was not a second Full Moon in the month, and the Almanac really didn't explain why this particular Full Moon was a Blue Moon. The list of moon names, however, clearly established their astronomical character, and the only explanation for August 21 is the astronomical one. Both Sky & Telescope pieces reveal, however, how obscure the matter has become and how difficult it has been to find older references. The encyclopedia articles, also, don't seem to get it quite right. The World Book said that the nearest Full Moon to the equinox was the "Harvest Moon," while Encarta said that it was the one right before the equinox. The interpretation of the Almanac, however, requires that it be in the Autumn and after (or perhaps on) "Summer's End," i.e. the equinox.
A follow up Sky & Telescope article, "What's a Blue Moon?" by Donald W. Olson, Richard Tresch Fienberg, and Roger W. Sinnott, in the May 1999 issue, seems to have cleared up the problems. After the questions raised by the original article, an examination of more than a century of copies of the Maine Farmers' Almanac revealed the rule they were using for Blue Moons. Rather than using the actual position of the sun, the mean position of the sun was used, March 21 was always used for the Vernal Equinox, and the principle that the third Full Moon out of four in a season was discerned. Thus, the Lenten Moon would always occur in Lent and the Egg Moon would always be the Paschal Moon. Using the mean longitude of the sun would make the seasons of equal length. The error that the second Full Moon in a calendar month was a Blue Moon was traced to a specific issue of Sky & Telescope magazine in March 1946. What does not seem to have been revealed is the source used by the Maine Farmers' Almanac for its method and information.
Reading the original article, my impression was that the rule for Blue Moons was probably the second Full Moon in a zodiacal period. This was unsatisfactory, since it would mean that the Full Moon in Lent might be a Blue Moon rather than the Lenten Moon. It is therefore satisfying to discover that the actual rule preserves the Lenten Moon for Lent. However, the new article lists me as one of the people who proposed that the Blue Moon was the second Full Moon under a "given astrological sign." I think this is a bit of a misrepresentation, since the longitude of the sun defining the zodiacal period does not necessarily have anything to do with astrology. The use of the mean position of the sun also means that different methods, and different results, could be obtained for assigning the Full Moon names. I have discussed a purely astronomical determination. It will be interesting to see when this will diverge from the results of the mean longitude of the sun.
Full Moons
1937 Farmer's Almanac2014 Farmer's Almanac2014 Alternates
Yule, Winter Solstice
0.Moon after YuleWolf MoonSnow Moon
1.Wolf MoonSnow MoonHunger Moon
2.Lenten MoonWorm MoonCrow/Crust/Sap Moon
First Day of Spring, Vernal Equinox
3.Egg/Paschal MoonPink MoonSprouting
Grass/Fish Moon
4.Milk MoonFlower MoonCorn Planting Moon
5.Flower MoonStrawberry MoonRose Moon
The Long Day, Summer Solstice
6.Hay MoonBuck MoonThunder Moon
7.Grain MoonSturgeon MoonGreen Corn/Red Moon
8.Fruit MoonCorn/Harvest Moon
Summer's End, Autumnal Equinox
9.Harvest MoonHunter's/Harvest MoonBlood Moon
10.Hunter's MoonBeaver MoonFrosty Moon
11.Moon before
Yule
Cold/Long Nights Moon
After a fashion, all hell has broken lose, and the presumably authoritative list that the Almanac gave in 1937 has now been thoroughly revised and complicated. Some of the names of 1937 are not even mentioned (e.g. the Fruit Moon), while most of the others are relegated to mention as alternatives, often at the end of the list, as afterthoughts (the second column in the table gives the names as they occur in the headings of the Almanac treatment). This is improper, if not dishonest, given the 1937 list. And while the older treatment was presented in terms of English tradition, with a key role for the Lenten Moon, the Old Farmer's Almanac has now shifting to prefering a narrative in which the names "date back to Native Americans." Since American Indians did not originally observe Easter, the Lenten Moon obviously would have had no meaning.
There is no mention on the 2014 Farmer's Almanac page about the Sky & Telescope discussion of the rule for Blue Moons. Indeed, the Full Moon names are identified by calendar month. The only reference to the astronomical benchmarks is the assertion that the Harvest Moon can occur in September or October because it is supposed to be the closest Full Moon to the Autumnal Equinox. If such a Moon occurs before the Equinox, of course, it is in the Summer rather than the Autumn, according to the Babylonian system used by Western astronomy. There is no reference cited for this rule, which would confuse the principle for Blue Moons as determined by Sky & Telescope, which names the Moons by the astronomical seasons. Indeed, the aspect of the rule that the Blue Moon is the third moon of the season, preserving the standard name of the third moon as the last moon of the season, only makes sense when we realize that the Lenten Moon should be immediately followed by the Full Moon associated with Easter. Again, this consideration would not arise if our frame of reference is pre-Christian American Indians. Much of the point of the Sky & Telescope discussion thus voided.
As of 2017, Sky & Telescope still explains the rule for Blue Moons as the third Full Moon in a season, something the Old Farmer's Almanac has never done; and the only Blue Moon since 2014, in 2015, was both the second Full Moon in a month (July) and the third in its season. So we have not seen a falsifying test of whatever rules the Almanac is using. Otherwise, I do not see that Sky & Telescope has addressed the question of the actual Moon names. Instead, Sky & Telescope actually links to the webpage of the Almanac about the names, which has not been included in the print editions since 2015. I do not see any reference sources cited, either in print or on line, for the Almanac treatment of the names. Instead, we must wonder about what sources would be referring back to Algonquin [2015 editon, p.275] calendar or astronomical knowledge or practices. We would need historical or at least anthropological sources, since the Algonquin did not have written records. While we might also wonder about the sources for the 1937 names, there is no doubt that we have them at least from 1937, which antedate all these recent discussions and claims.
A regularity we might notice in the revised names is that the 1937 names sometimes occur a month earlier in the 2014 version. This is the case with the Wolf Moon, the Flower Moon, the Harvest Moon, and the Hunter's Moon (as in the Weather Channel name for the October 2013 Full Moon). Of course, the 1937 names often do occur in the previous month, precisely because they are following the equinoxes and solstices rather than the (Gregorian) calendar months. Given the curious treatment of the Harvest Moon, perhaps a general confusion about this has crept into the Almanac account. In fact, I like some of the new names better. The "Moon before Yule" and the "Moon after Yule" tell us nothing about the nature of the season or the weather. "Cold Moon" and "Snow Moon" are much more evocative. However, the whole business is a mess when the whole basis of the treatment shifts from "English ancestors" to "Native Americans" without explanation, and even without coherence (e.g. over the Lenten Moon). In the tables below, the 1937 names are used because that is what I have been using since 1999.
1999
Full MoonZodiacal by Month
2 January 2h 49mMoon after YuleMoon after Yule
31 January 16h 6mWolf MoonBlue Moon
2 March 6h 58mLenten MoonLenten Moon
31 March 22h 49mEgg/Paschal MoonBlue Moon
30 April 14h 55mMilk MoonEgg Moon
30 May 6h 40mFlower MoonMilk Moon
28 June 21h 37mHay MoonFlower Moon
28 July 11h 25mGrain MoonHay Moon
26 August 23h 48mFruit MoonGrain Moon
25 September 10h 51mHarvest MoonFruit Moon
24 October 21h 2mHunter's MoonHarvest Moon
23 November 7h 4mMoon before YuleHunter's Moon
22 December 17h 31mMoon after YuleMoon before Yule
1999 is an unusual year for Full Moons. They are all listed in the table at left, with their exact occurrence in Universal Time (or "Terrestrial [Dynamical] Time," TDT or TT), as given in the The Astronomical Almanac for the Year 1999 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 1999], which is what Greenwich Mean Time is now called. There are 13 Full Moons in the calendar year, but only 12 in the tropical year (solstice to solstice). Thus, there would be a Blue Moon when reckoning by months but not when reckoning by seasons. More unusal is that there is no Full Moon in the calendar month of February, the only month where that is possible, since it is only 28 days long (and the synodic month, from New Moon to New Moon, is 29.5 days). This means that the Full Moon that might have occurred in February, the Wolf Moon, must be in some other month instead, so that, with 13 months in the calendar year, two months will have two Full Moons. Those are January and March, which, with two Full Moons, both have a Blue Moon by the monthly reckoning. This will happen again in 2018.
2000
Full MoonZodiacal by Month
21 January 4h 40mWolf MoonMoon after Yule
19 February 16h 27mBlue MoonWolf Moon
20 March 4h 44mLenten MoonLenten Moon
18 April 17h 41mEgg/Paschal MoonEgg Moon
18 May 7h 34mMilk MoonMilk Moon
16 June 22h 27mFlower MoonFlower Moon
16 July 13h 55mHay MoonHay Moon
15 August 05h 13mGrain MoonGrain Moon
13 September 19h 37mFruit MoonFruit Moon
13 October 8h 53mHarvest MoonHarvest Moon
11 November 21h 15mHunter's MoonHunter's Moon
11 December 9h 3mMoon before YuleMoon before Yule
The Full Moons of 2000 are completely conventional if we reckon by month but are a nice illustration of the principle of the system if we reckon by seasons. There are only twelve Full Moons in the calendar year, but the Moon after Yule has already occurred late in 1999, so by seasons the year begins with the Wolf Moon -- which experienced a spectacular Total Eclipse conveniently visible from North America. The actual Paschal Moon occurs on April 18th (Gregorian Easter is April 23rd), so we end up with four Full Moons during the Winter of 1999-2000. The actual Full Moon during Lent is March 20th. This leaves the odd Full Moon as that of February 19. So it's a Blue Moon. The Blue Moon that actually didn't occur during the 13 Full Moons of the calendar year of 1999, we get at the beginning of 2000.
2001
Full MoonZodiacal by Month
9 January 20h 24mMoon after YuleMoon after Yule
8 February 07h 12mWolf MoonWolf Moon
9 March 17h 23mLenten MoonLenten Moon
8 April 03h 22mEgg/Paschal MoonEgg Moon
7 May 13h 52mMilk MoonMilk Moon
6 June 01h 39mFlower MoonFlower Moon
5 July 15h 4mHay MoonHay Moon
4 August 05h 56mGrain MoonGrain Moon
2 September 21h 43mFruit MoonFruit Moon
2 October 13h 49mHarvest MoonHarvest Moon
1 November 05h 41mHunter's MoonHunter's Moon
30 November 20h 49mMoon before YuleBlue Moon
30 December 10h 40mMoon after YuleMoon before Yule
There are thirteen Full Moons in calendar 2001. None of these is, properly speaking, a Blue Moon. But two of them do occur in one month, November, which then would have a Blue Moon by the monthly reckoning.
2002
Full MoonZodiacal by Month
28 January 22h 50mWolf MoonMoon after Yule
27 February 9h 17mLenten MoonWolf Moon
28 March 18h 25mEgg/Paschal MoonLenten Moon
27 April 03h 00mMilk MoonEgg Moon
26 May 11h 51mFlower MoonMilk Moon
24 June 21h 42mHay MoonFlower Moon
24 July 09h 7mGrain MoonHay Moon
22 August 22h 29mBlue MoonGrain Moon
21 September 13h 59mFruit MoonFruit Moon
21 October 07h 20mHarvest MoonHarvest Moon
20 November 01h 34mHunter's MoonHunter's Moon
19 December 19h 10mMoon before YuleMoon before Yule
There are twelve Full Moons in calendar 2002. The "Moon after Yule," however, already occurred in 2001, leaving only eleven names. One of the Full Moons therefore, that of August 22nd, ends up as a proper Blue Moon -- there are four Full Moons between the Summer Solstice and the Autumnal Equinox. As they should be, the Lenten Moon is in Lent, which begins February 13th, and the Egg Moon is the Paschal Moon, with (Gregorian) Easter falling on March 31st.
This table, like the previous ones, is taken from The Astronomical Almanac, in this case for the year 2002 [published as for 1999 above].
2003
Full MoonZodiacal by Month
18 January 10h 48mMoon after YuleMoon after Yule
16 February 23h 51mWolf MoonWolf Moon
18 March 10h 35mLenten MoonLenten Moon
16 April 19h 36mEgg/Paschal MoonEgg Moon
16 May 03h 36mMilk MoonMilk Moon
14 June 11h 16mFlower MoonFlower Moon
13 July 19h 21mHay MoonHay Moon
12 August 04h 48mGrain MoonGrain Moon
10 September 16h 36mFruit MoonFruit Moon
10 October 07h 27mHarvest MoonHarvest Moon
09 November 01h 13mHunter's MoonHunter's Moon
08 December 20h 37mMoon before YuleMoon before Yule
There are twelve Full Moons in calendar 2003, matching up completely with the reckoning by month. As they should be, the Lenten Moon is in Lent, which begins March 5th, and the Egg Moon is the Paschal Moon, with (Gregorian, Western) Easter falling on April 20th (Julian, Eastern Easter on April 27th).
This table, like the previous ones, is taken from The Astronomical Almanac, in this case for the year 2003 [published as for 1999 above, 2001].
2004
Full MoonZodiacal by Month
7 January 15h 40mMoon after YuleMoon after Yule
6 February 08h 47mWolf MoonWolf Moon
6 March 23h 14mLenten MoonLenten Moon
5 April 11h 03mEgg/Paschal MoonEgg Moon
4 May 20h 33mMilk MoonMilk Moon
3 June 04h 20mFlower MoonFlower Moon
2 July 11h 09mHay MoonHay Moon
31 July 18h 05mGrain MoonBlue Moon
30 August 02h 22mFruit MoonGrain Moon
28 September 13h 09mHarvest MoonFruit Moon
28 October 03h 07mHunter's MoonHarvest Moon
26 November 20h 07mMoon before YuleHunter's Moon
26 December 15h 06mMoon after YuleMoon before Yule
There are thirteen Full Moons in calendar 2004, but no true Blue Moon, since the Full Moon of December occurs after the Winter Solstice and so is the Moon after Yule, adding a thirteenth name for the thirteenth moon. Reckoning by month, there are two Full Moons in July.
Lent begins on February 25th, and the Lenten Moon is on March 6th. Both Gregorian and Julian Easter in 2004 fall on April 11th, after the Egg or Paschal Moon on April 5th.
This table, with universal times for the Full Moons, is taken from The Astronomical Almanac for the year 2004 [Washington, U.S. Government Printing office; London, The Stationery Office, 2002].
2005
Full MoonZodiacal by Month
25 January 10h 32mWolf MoonMoon after Yule
24 February 04h 54mLenten MoonWolf Moon
25 March 20h 58mEgg/Paschal MoonLenten Moon
24 April 10h 06mMilk MoonEgg Moon
23 May 20h 18mFlower MoonMilk Moon
22 June 04h 14mHay MoonFlower Moon
21 July 11h 00mGrain MoonHay Moon
19 August 17h 53mBlue MoonGrain Moon
18 September 02h 01mFruit MoonFruit Moon
17 October 12h 14mHarvest MoonHarvest Moon
16 November 00h 57mHunter's MoonHunter's Moon
15 December 16h 15mMoon before YuleMoon before Yule
There are twelve Full Moons in calendar 2005, but like 2002 the "Moon after Yule" already occurred (in 2004), leaving only eleven names. One of the Full Moons therefore, that of August 19th, ends up as a proper Blue Moon -- there are four Full Moons between the Summer Solstice and the Autumnal Equinox. As they should be, the Lenten Moon is in Lent, which begins February 9th, and the Egg Moon is the Paschal Moon, with (Gregorian) Easter falling on March 27th. While this Gregorian (Western) Easter occurs relatively early this year, Easter by the Julian (Eastern) reckoning occurs very late, on May 1st, five weeks later.
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2005 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2003].
2006
Full MoonZodiacal by Month
14 January 9h 48mMoon after YuleMoon after Yule
13 February 04h 44mWolf MoonWolf Moon
14 March 23h 35mLenten MoonLenten Moon
13 April 16h 40mEgg/Paschal MoonEgg Moon
13 May 6h 51mMilk MoonMilk Moon
11 June 18h 03mFlower MoonFlower Moon
11 July 03h 02mHay MoonHay Moon
9 August 10h 54mGrain MoonGrain Moon
7 September 18h 42mFruit MoonFruit Moon
7 October 03h 13mHarvest MoonHarvest Moon
5 November 12h 58mHunter's MoonHunter's Moon
5 December 00h 25mMoon before YuleMoon before Yule
There are twelve Full Moons in calendar 2006, each matching up to their expected month. As they should be, the Lenten Moon is in Lent, which begins March 1st, and the Egg Moon is the Paschal Moon, with (Gregorian) Easter falling on April 16th.
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2006 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2004].
2007
Full MoonZodiacal by Month
3 January 13h 57mMoon after YuleMoon after Yule
2 February 05h 45mWolf MoonWolf Moon
3 March 23h 17mLenten MoonLenten Moon
2 April 17h 15mEgg/Paschal MoonEgg Moon
2 May 10h 09mMilk MoonMilk Moon
1 June 01h 04mFlower MoonFlower Moon
30 June 13h 49mHay MoonBlue Moon
30 July 00h 48mGrain MoonHay Moon
28 August 10h 35mFruit MoonGrain Moon
26 September 19h 45mHarvest MoonFruit Moon
26 October 04h 52mHunter's MoonHarvest Moon
24 November 14h 30mMoon before YuleHunter's Moon
24 December 01h 16mMoon after YuleMoon before Yule
There are thirteen Full Moons in calendar 2007, but no true Blue Moon, since the Full Moon of December occurs after the Winter Solstice and so is the Moon after Yule, adding a thirteenth name for the thirteenth moon. Reckoning by month, there are two Full Moons in June, giving us a Blue Moon by monthly reckoning.
As they should be, the Lenten Moon is in Lent, which begins February 21st, and the Egg Moon is the Paschal Moon, with (Gregorian) Easter falling on April 8th.
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2007 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2005].
2008
Full MoonZodiacal by Month
22 January 13h 35mWolf MoonMoon after Yule
21 February 03h 30mLenten MoonWolf Moon
21 March 18h 40mEgg/Paschal MoonLenten Moon
20 April 10h 25mMilk MoonEgg Moon
20 May 02h 11mBlue MoonMilk Moon
18 June 17h 30mFlower MoonFlower Moon
18 July 07h 59mHay MoonHay Moon
16 August 21h 16mGrain MoonGrain Moon
15 September 09h 13mFruit MoonFruit Moon
14 October 20h 02mHarvest MoonHarvest Moon
13 November 06h 17mHunter's MoonHunter's Moon
12 December 16h 37mMoon before YuleMoon before Yule
There are twelve Full Moons in calendar 2008, but like 2005 the "Moon after Yule" already occurred (in 2007), leaving only eleven names. One of the Full Moons therefore, that of May 20th, ends up as a proper Blue Moon -- there are four Full Moons between the Vernal Equinoz and the Summer Solstice. As they should be, the Lenten Moon is in Lent, which begins February 6th, and the Egg Moon is the Paschal Moon, with (Gregorian) Easter falling on March 23rd. While this Gregorian (Western) Easter occurs relatively early this year, Easter by the Julian (Eastern) reckoning occurs late, on April 27th, five weeks later.
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2008 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2006].
2009
Full MoonZodiacalby Month
11 January 03h 27mMoon after YuleMoon after Yule
09 February 14h 49mWolf MoonWolf Moon
11 March 02h 38mLenten MoonLenten Moon
09 April 14h 56mEgg/Paschal MoonEgg Moon
09 May 04h 01mMilk MoonMilk Moon
07 June 18h 12mFlower MoonFlower Moon
07 July 09h 21mHay MoonHay Moon
06 August 00h 55mGrain MoonGrain Moon
04 September 16h 03mFruit MoonFruit Moon
04 October 06h 10mHarvest MoonHarvest Moon
02 November 19h 14mHunter's MoonHunter's Moon
02 December 07h 30mMoon before YuleMoon before Yule
31 December 19h 13mMoon after YuleBlue Moon
There are thirteen Full Moons in calendar 2009, but like 2007, there is no true Blue Moon. This is the first time in the years covered that two Full Moons occur in December. So the names of the Moons all match until the last Full Moon.
Despite the good work of Sky & Telescope in clearing up the issues of Blue Moons, I notice in December, 2009, that the Weather Channel is still uncritically touting the Full Moon of 31 December as being a "Blue Moon," according to the principle that it is the second Full Moon in the calendar month. Someone did not do their homework.
The Full Moons on this table, in Universal Time (GMT) are taken from The Astronomical Almanac, in this case for the year 2009 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2007].
2010
Full MoonZodiacalby Month
30 January 06h 18mWolf MoonMoon after Yule
28 February 16h 38mLenten MoonWolf Moon
30 March 02h 25mEgg/Paschal MoonLenten Moon
28 April 12h 18mMilk MoonEgg Moon
27 May 23h 07mFlower MoonMilk Moon
26 June 11h 30mHay MoonFlower Moon
26 July 01h 37mGrain MoonHay Moon
24 August 17h 05mFruit MoonGrain Moon
23 September 09h 17mHarvest MoonFruit Moon
23 October 01h 37mHunter's MoonHarvest Moon
21 November 17h 27mBlue MoonHunter's Moon
21 December 08h 13mMoon before YuleMoon before Yule
There are twelve Full Moons in calendar 2010; but there is a true Blue Moon, since the Moon after Yule occurred in the previous calendar year. Easter falls on April 4th this year by both Julian and Gregorian reckoning. Lent began on February 17th.
The Full Moons on this table, in Universal Time (GMT) are taken from The Astronomical Almanac, in this case for the year 2010 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2008].
2011
Full MoonZodiacal by Month
19 January 21h 21mMoon after YuleMoon after Yule
18 February 08h 36mWolf MoonWolf Moon
19 March 18h 10mLenten MoonLenten Moon
18 April 02h 44mEgg/Paschal MoonEgg Moon
17 May 11h 09mMilk MoonMilk Moon
15 June 20h 14mFlower MoonFlower Moon
15 July 06h 40mHay MoonHay Moon
13 August 18h 57mGrain MoonGrain Moon
12 September 09h 27mFruit MoonFruit Moon
12 October 02h 06mHarvest MoonHarvest Moon
10 November 20h 16mHunter's MoonHunter's Moon
10 December 14h 36mMoon before YuleMoon before Yule
There are twelve Full Moons in calendar 2011, and no true Blue Moon. The months and the moon all match up for the first time since 2006.
The Full Moons on this table, in Universal Time (GMT) are taken from The Astronomical Almanac, in this case for the year 2011 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2010].
2012
Full MoonZodiacalby Month
09 January 07h 30mMoon after YuleMoon after Yule
07 February 21h 54mWolf MoonWolf Moon
08 March 09h 39mLenten MoonLenten Moon
06 April 19h 19mEgg/Paschal MoonEgg Moon
06 May 03h 35mMilk MoonMilk Moon
04 June 11h 12mFlower MoonFlower Moon
03 July 18h 52mHay MoonHay Moon
02 August 03h 27mGrain MoonGrain Moon
31 August 13h 58mFruit MoonBlue Moon
30 September 03h 19mHarvest MoonFruit Moon
29 October 19h 49mHunter's MoonHarvest Moon
28 November 14h 46mMoon before YuleHunter's Moon
28 December 10h 21mMoon after YuleMoon before Yule
There are thirteen Full Moons in calendar 2012, but like 2009, there is no true Blue Moon. There are two proper Moons after Yule. Since two Full Moons do occur in calendar August, the monthly reckoning shows a Blue Moon there.
The Full Moons on this table, in Universal Time (GMT) are taken from The Astronomical Almanac, in this case for the year 2012 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2011].
2013
Full MoonZodiacal by Month
27 January 04h 38mWolf MoonMoon after Yule
25 February 20h 26mLenten MoonWolf Moon
27 March 09h 27mEgg/Paschal MoonLenten Moon
25 April 19h 57mMilk MoonEgg Moon
25 May 04h 25mFlower MoonMilk Moon
23 June 11h 32mHay MoonFlower Moon
22 July 18h 16mGrain MoonHay Moon
21 August 01h 45mBlue MoonGrain Moon
19 September 11h 13mFruit MoonFruit Moon
18 October 23h 38mHarvest MoonHarvest Moon
17 November 15h 16mHunter's MoonHunter's Moon
17 December 09h 28mMoon before YuleMoon before Yule
There are twelve Full Moons in calendar 2013; but there is a true Blue Moon, since the Moon after Yule occurred in the previous calendar year and there are four Full Moons in the Summer. Gregorian Easter falls on March 31st this year. Lent began on February 13th.
The Full Moons on this table, in Universal Time (GMT) are taken from The Astronomical Almanac, in this case for the year 2013 [U.S. Government Printing Office, Washington, and the United Kingdom Hydrographic Office, London, 2012].
Note that the Weather Channel called the October 2013 Full Moon the "Hunter's Moon" rather than the Harvest Moon. This has happened because of shifting ideas about the proper names of the Full Moons, as I have discussed above.
2014
Full MoonZodiacal by MonthAlmanac
16 January 04h 52mMoon after YuleMoon after YuleWolf Moon
14 February 23h 53mWolf MoonWolf MoonSnow Moon
16 March 17h 08mLenten MoonLenten MoonWorm Moon
15 April 07h 42mEgg/Paschal MoonEgg MoonPink Moon
14 May 19h 16mMilk MoonMilk MoonFlower Moon
13 June 04h 11mFlower MoonFlower MoonStrawberry Moon
12 July 11h 25mHay MoonHay MoonBuck Moon
10 August 18h 09mGrain MoonGrain MoonSturgeon Moon
09 September 01h 38mFruit MoonFruit MoonHarvest Moon
8 October 10h 51mHarvest MoonHarvest MoonHunter's Moon
06 November 22h 23mHunter's MoonHunter's MoonBeaver Moon
06 December 12h 27mMoon before YuleMoon before YuleCold Moon
There are twelve Full Moons in calendar 2014 and no Blue Moon by any reckoning.
The basic names that are now in the Farmer's Almanac have been added here for comparison. In these names, the relation of the series to Easter has obviously been purged, apparently on the principle that these are "Native American" names, where of course Christianity was not originally a factor. At the same, time there are no astronomical or scientific records present, let alone common, to all Native Americans. So the Farmer's Almanac names are in part a fantasy exercise, perhaps even part of the current cultural and political attack on Christianity in the United States -- ironically coupled with a complacent or apologetic attitude towards the terrors and horrors of radical Islâm. The Full Moon of September 9th was reported as the Weather Channel as the "Harvest Moon." See discussion above for more details.
The Full Moons on this table, in Universal Time (GMT) are taken from The Astronomical Almanac, in this case for the year 2014 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2013].
2015
Full MoonZodiacal by MonthFarmer's
Almanac
05 January 04h 53mMoon after
Yule
Moon after YuleWolf Moon
03 February 23h 09mWolf MoonWolf MoonSnow Moon
05 March 18h 05mLenten MoonLenten MoonWorm Moon
04 April 12h 06mEgg/Paschal MoonEgg MoonPink Moon
04 May 03h 42mMilk MoonMilk MoonFlower Moon
02 June 16h 19mFlower MoonFlower MoonStrawberry
Moon
02 July 02h 20mHay MoonHay MoonBuck Moon
31 July 10h 43mGrain MoonBlue MoonBlue Moon
29 August 18h 35mFruit MoonGrain MoonSturgeon
Moon
28 September 02h 50mHarvest MoonFruit MoonHarvest Moon
27 October 12h 05mHunter's MoonHarvest MoonHunter's Moon
25 November 22h 44mMoon before YuleHunter's MoonBeaver Moon
25 December 11h 11mMoon after YuleMoon before YuleCold Moon
There are thirteen Full Moons in calendar 2015. However, there is no true Blue Moon. The thirteenth Full Moon occurs after the Winter Solstice and so belongs to the following season. A Blue Moon by monthly reckoning occurs in July.
The Almanac names for 2015 have been supplied directly from the Old Farmer's Almanac [Yankee Publishing Inc., Dublin, New Hampshire, 2014, p.274]. This repeats the information discussed above. The Moon names, with variations, are given in terms of months, with the proviso that "The Harvest Moon is always the full Moon closest to the autumnal equinox. If the Harvest Moon occurs in October, the September full Moon is usually called the Corn Moon" [ibid.]. We also get the account here that the Moon names are due to "Native Americans" of the northeastern United States and that a name applied "to the entire month in which it occurred," which would seem to mean that the Native Americans used the Julian or Gregorian Calendars -- if "the entire month" means the calendar month -- which seems unlikely. And if "month" means the period of the Moon from New Moon to New Moon, that needs to be stated. The specific names in the Almanac, however, are now said to be those "used by the Algonquin tribes from New England to Lake Superior" [p.275], with no source for this information cited, nor any account of how the Algonquin would have reckoned the months. This increases one's curiosity about the rule for the "closest" Moon the the autumnal equinox, since this would bespeak an astronomical and calendrical tradition, and an awareness of the equinoxes and solstices, about which we otherwise hear nothing. The ability of a people like the Algonquin to determine the equinoxes and solstices, let alone be aware of them, is something, given their technology and environment (i.e. forests), is something about which some skepticism may be in order.
Since there are thirteen Full Moons in the year, some provision will need to be made for an intercalation or a Blue Moon. The Old Farmer's Almanac, although noting two Full Moons in July, makes no mention of Blue Moons and does not identify the extra Moon by name. This grave oversight seems characteristic of the carelessness and fantasy of the whole treatment. Meanwhile, a rival 2015 Farmer's Almanac, published by the "Almanac Publishing Company" of Lewiston, Maine [2014 -- this one says it has been published since 1818 -- is it the "Maine Farmer's Almanac"? -- while the Old Farmer's Almanac claims an origin in 1792] repeats the names given by the other Almanac but does identity the second Full Moon of July as a "Blue Moon" [p.107]. Both Almanacs, having constructed their own fantasy system of Moon names, without bothering to consider how "Native Americans" reckoned the months, recognized the equinoxes and solstices, or constructed their own calendars, thus have tossed and ignored the history of these questions from the Sky & Telescope examinations, let alone the history and system evident in the 1937 Maine Farmer's Almanac. That the reckoning of Lent and Easter has been purged from the matter is only part of its carelessness and shamefulness.
The Full Moons on this table, in Universal Time (GMT) are taken from The Astronomical Almanac, in this case for the year 2015 [U.S. Government Printing Office, Washington, and Her Majesty's Stationery Office, London, 2014].
The final column again is from the Old Farmer's Almanac, now for 2016 [Yankee Publishing Inc., Dublin, New Hampshire, 2015, pp.129-151]. This repeats the names, month by month, from 2015. However, this edition of the Almanac has completely dropped the section "The Origin of Full Moon Names" that was featured in the 2015 edition [p.275].
Hopefully, this deletion was out of embarrassment over the attribution of the names to "Native Americans," who, of course, did not use the months of the Gregorian Calendar and did not possess, in any references I have seen, a calendrical system, like the Babylonian or the Chinese, to adjust lunar months to the solar year, or any accurate tradition of observing the Equinoxes and Solstices. "Native Americans" who did have a sophisticated astronomical tradition, namely the Mayans, nevertheless did not use a lunar or a luni-solar calendar and so did not have canonical names for the Full Moons.
What the "Native American" names obviously continue to do is to purge the system of any relation to Lent or Easter. One might see in this an example of the politically correct War on Religion, particularly Christianity, that has been discussed elsewhere in these pages. For the 1937 names "by Month," I have added the alternatives "Cold" and "Snow" Moon for the Moons before and after Yule, which are not very descriptive or evocative.
The final column again is from the Old Farmer's Almanac, now for 2017 [Yankee Publishing Inc., Dublin, New Hampshire, 2016, pp.143-169]. This repeats the names from 2016 but with some changes. We've lost the Hunter's Moon altogether. The Harvest Moon has moved up to October, leaving the previously unseen Corn Moon back in September. This is an application of the rule the Almanac has mentioned that the Harvest Moon should be the closest to the Equinox. As in 2016, we do not see a section "The Origin of Full Moon Names," as was featured in the 2015 edition. The assault, of course, on Christianity continues, with no indication of the Lenten or Paschal Moons, based on the pretext of putative "Native American" calendars.
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2017 [U.S. Government Printing Office, Washington, and now The U.K. Hydrographic Office, Taunton Somerset, 2016].
The final column again is from the Old Farmer's Almanac, now for 2018 [Yankee Publishing Inc., Dublin, New Hampshire, 2017, pp.145-167]. This repeats the names from 2017 but with some changes. We've regained the Hunter's Moon in October; and the Harvest Moon is back in September. As in 2016 and 2017, we do not see a section "The Origin of Full Moon Names," as was featured in the 2015 edition. The assault, of course, on Christianity continues, with no indication of the Lenten or Paschal Moons, based on the pretext of putative "Native American" calendars -- I have put these Almanac names in gray for their doubtless ahistorical nature. What we see here that was not in 2017 is the "Old" Moon and the "Sap" Moon. The "Old" Moon perhaps is their equivalent of the Blue Moon. The "Sap" Moon is for the Egg or Paschal Moon. But I also get the impression here that they are making things up, year by year, with some care not to give hints of Christianity (as even the neutral "Egg" Moon might -- in 2014 the "Sap" Moon was an alternative for the Lenten, not the Egg Moon).
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2018 [U.S. Government Printing Office, Washington, and now The U.K. Hydrographic Office, Taunton Somerset, 2017].
The final column again is from the Old Farmer's Almanac, now for 2019 [Yankee Publishing Inc., Dublin, New Hampshire, 2018, pp.124-147]. As in 2016, 2017, and 2018, we do not see a section "The Origin of Full Moon Names," as was featured in the 2015 edition.
The war on Christianity continues, with no indication of the Lenten or Paschal Moons, based on the pretext of putative "Native American" calendars -- I have put these Almanac names in gray for their suspiciously politically correct nature. What we see here that was not in 2018 is the "Pink" Moon instead of the "Sap" Moon. I get the impression here again that the Almanac staff are making things up, year by year, with some care not to give hints of Christianity (as even the neutral "Egg" Moon might). I am still waiting to see who is the Native American Ptolemy or Gregory XIII who is responsible for the names and calendar calculations for all this.
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2019 [U.S. Government Printing Office, Washington, and now The U.K. Hydrographic Office, Taunton Somerset, 2018].
The final column again is from the Old Farmer's Almanac, now for 2020 [Yankee Publishing Inc., Dublin, New Hampshire, 2019, pp.125-147]. This repeats the names from 2014 but with one change to accommodate the 13 months, i.e. the Harvest and Hunter's Moons are separated. As in recent editions, we do not see a section "The Origin of Full Moon Names," as was featured in the 2015 edition. Thus, there is absolutely no explanation of how the Full Moon names are assigned, or from where they derive. The assault, of course, on Christianity continues, with no indication of the Lenten or Paschal Moons, based on the pretext, as previously claimed, of putative "Native American" calendars. I get the impression here that the Almanac is making things up, year by year, with some care not to give hints of Christianity (as even the neutral "Egg" Moon might).
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2020 [U.S. Government Printing Office, Washington, and now The U.K. Hydrographic Office, Taunton Somerset, 2019]. I used to have, since the 1970's, a standing order, originally for The American Ephemeris and Nautical Almanac, with the United States Superintendant of Documents. A few years ago, however, Office began to neglect to fill my order. They would send it out promptly on request, now by e-mail; but I finally decided just to order on line, if they couldn't manage to remember my standing order.
2021
Full MoonZodiacal by MonthFarmer's
Almanac
28 January 19h 16mWolf MoonMoon after YuleWolf Moon
27 February 08h 17mLenten MoonWolf MoonSnow Moon
28 March 18h 48mEgg/Paschal MoonLenten MoonWorm Moon
27 April 03h 32mMilk MoonEgg MoonPink Moon
26 May 11h 14mFlower MoonMilk MoonFlower Moon
24 June 18h 40mHay MoonFlower MoonStrawberry Moon
24 July 02h 37mGrain MoonHay MoonBuck Moon
22 August 12h 02mBlue MoonGrain MoonSturgeon Moon
20 September 23h 55mFruit MoonFruit MoonHarvest Moon
20 October 14h 57mHarvest MoonHarvest MoonHunter's Moon
19 November 08h 57mHunter's MoonHunter's MoonBeaver Moon
19 December 04h 35mMoon before YuleMoon before YuleCold Moon
There are twelve Full Moons in calendar 2021. There are four Full Moons in Summer, so it contains a true Blue Moon.
The Old Farmer's Almanac for 2021 [Yankee Publishing, Dublin, NH, 2020] follows its rules for the names of the Full Moons. The publishers, however, do not seem to have noticed that they have included four Full Moons for the Summer season of 2021. This would call for a Blue Moon, but they have not provided one. By the simple reckoning of months, 2020 did contain a Blue Moon in October, as 2021 will contain by Zodiacal reckoning a Blue Moon in August. So the Almanac goes through 25 months without a Blue Moon. In fact, I only find them identifying one Blue Moon in all the years since I began paying attention to them in 2014. This reinforces my impression that the Almanac is following no true calendar or astronomy but is simply making this up as they go along.
They are still not giving the time of day to Easter. Gregorian Easter for 2021 is April 4th, following the Paschal or Egg Moon on March 28th. If the Almanac were deliberately intending to be insulting or dismissive about Easter, they could not do better than what they do, which is to call the Paschal Moon the "Worm" Moon. They also catch Passover with that, since the Jewish celebration begins with the Full Moon on March 28th. So, we might say, Christians and Jews are "worms." Perhaps they enjoy laughing about that in Dublin, New Hampshire. This what America, and the "Old Farmer's Almanac," has come to.
The Full Moons on this table, in Universal Time (GMT), are taken from The Astronomical Almanac, in this case for the year 2021 [U.S. Government Printing Office, Washington, and now The U.K. Hydrographic Office, Taunton Somerset, 2020]. In some cases, that is still the previous calendar day in the time zones of the United States.
The final column again is from the Old Farmer's Almanac, now for 2022 [Yankee Publishing Inc., Dublin, New Hampshire, 2021, pp.125-147]. This repeats the names from 2021. As before, we do not see a section "The Origin of Full Moon Names," as was featured in the 2015 edition. The assault, of course, on Christianity continues, with no indication of the Lenten or Paschal Moons, based on the pretext of putative "Native American" calendars. News sources, like the Weather Channel, continue to use these names uncritically.
The Full Moons on this table, in Universal Time (GMT), like the previous ones, are taken from The Astronomical Almanac, in this case for the year 2022 [U.S. Government Printing Office, Washington, and now The U.K. Hydrographic Office, Taunton Somerset, 2021].
The final column again is from the Old Farmer's Almanac, now for 2023 [Yankee Publishing Inc., Dublin, New Hampshire, 2022, pp.125-147]. This repeats the names from 2014 but with one change to accommodate the 13 months, i.e. the sturgeon and Harvest Moons are separated, by a Blue Moon.
As in recent editions, we do not see a section "The Origin of Full Moon Names," as was featured in the 2015 edition. Thus, there is absolutely no explanation of how the Full Moon names are assigned, or from where they derive.
The assault, of course, on Christianity continues, with no indication of the Lenten or Paschal Moons, based on the pretext, as previously claimed, of putative "Native American" calendars. Again, I get the impression here that the Almanac is making things up, year by year, with some care not to give hints of Christianity (as even the neutral "Egg" Moon might). Yet the title page identifies 2023 as "the year of Our Lord," which seems inconsistent with the deletion of Christianity from the Moon names.
The Full Moons on this table, in Universal Time (GMT), are taken from The Nautical Almanac for the Year 2023 [U.S. Government Publishing Office, Washington, and now The U.K. Hydrographic Office, Taunton Somerset, 2022]. Previously, of course, the times were from The Astronomical Almanac. However, The Almanac for the Year 2023 is not yet in print. I inquired of the Government Publishing Office and was informed, the "Astronomical Almanac 2023 should be in stock on February 17, 2023." I hope so. The 2022 Almanac was late last year, but not this late. If anyone wanted Almanac information for January, they are already out of luck.
The final column again is from the Old Farmer's Almanac, now for 2024 [Yankee Publishing Inc., Dublin, New Hampshire, 2023, pp.125-147]. This repeats the names from where I began listing them in 2014.
As in recent editions, we do not see a section "The Origin of Full Moon Names," as was featured in the 2015 edition. Thus, there is absolutely no explanation of how the Full Moon names are assigned, or from where they derive. Since this all seems to have been a fantasy exercise anyway, perhaps on the authority of a "Pretendian" source, that is just as well.
The Almanac's assault on Christianity continues, of course, with no indication of the Lenten or Paschal Moons, based on the pretext, as previously claimed, of putative "Native American" calendars (with no reference cited). Again, I get the impression here that the Almanac is making things up, year by year, with some care not to give hints of Christianity (as even the neutral "Egg" Moon might). Yet the title page identifies 2024 as "the year of Our Lord," which seems inconsistent with the deletion of Christianity from the Moon names.
The Full Moons on this table, in Universal Time (GMT), are taken from The Nautical Almanac for the Year 2024 [U.S. Government Publishing Office, Washington, and now The U.K. Hydrographic Office, Taunton Somerset, 2023]. Previously, of course, the times were from The Astronomical Almanac. However, The Astronomical Almanac for the Year 2024 is not yet in print. The 2023 Almanac was late last year, only becoming available in March. If anyone wanted Almanac information for January or February, they are already out of luck. Amazon.com now lists an "Astronomical Almanac, 2024 Edition," in paperback, for \$440. This is ridiculous. The 2023 edition was not cheap, but it was in hardback and not astronomically expensive.
The Occurrence of the Solar Terms in 1995-2013
# The French Revolutionary Calendar
The French Revolutionary Era, which was reckoned to begin on 22 September 1792 (roughly the Autumnal Equinox), was based on a year of 12 months of 30 days each, with five or six intercalary days at the end of the year. This was identical in form and position to the Coptic version of the ancient Egyptian Calendar, though with the year starting about ten days later. The months were poetically named for the seasons (in the northern hemisphere), which immediately inspired an English parody. I have added some Chinese characters, with Chinese proununciation (below) and
NametranslationParodyRoman
Month
Coptic
Month
1. VendémiairevintageWheezeySeptemberThout
2. Brumairemist SneezyOctoberPaape
3. Frimairefrost FreezyNovemberHator
4. Nivôsesnow SlippyDecemberKiahk
5. Pluviôserain DrippyJanuaryTobe
6. Ventôsewind NippyFebruaryMshir
7. GerminalseedShoweryMarchParmhat
8. FloréalblossomFloweryAprilParmute | 17,720 | 68,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-10 | longest | en | 0.969603 |
https://solvedlib.com/n/7-two-machines-are-used-to-fill-plastic-bottles-with-dishwashing,685474 | 1,695,903,068,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00329.warc.gz | 589,357,832 | 19,854 | # 7_ Two machines are used to fill plastic bottles with dishwashing detergent: The standard deviations of fill volumes are known
###### Question:
7_ Two machines are used to fill plastic bottles with dishwashing detergent: The standard deviations of fill volumes are known to be 0= 0.10 and 0= 0.15 fluid ounces for the two machines, respectively. Two random samples of n = 12 bottles from machine and n= 10 bottles from machine 2 are selected, and the sample mean fill volumes are 30.61 fluid ounces from machine and 30.34 fluid ounces for machine 2_ 4) Test the hypothesis that both machines do not fill to the same mean volume. Use 0=0.05 b) What is the P-value? Interpret this value_ c) Construct a 90% two-sided Cl on the mean difference. Interpret this interval : d) What iS the B-error of the test if the true difference in mean fill volumes is 1.2 fluid ounces?
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##### 5.3.Find3* 3-* lim 230 3* + 3-b) Find1 F 2 cos % + cos 2x lim x-0 22movaviScreen Recorder
5.3. Find 3* 3-* lim 230 3* + 3- b) Find 1 F 2 cos % + cos 2x lim x-0 22 movavi Screen Recorder... | 2,552 | 8,822 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.814739 |
https://prepinsta.com/all-about-c-language/prime-number-between-intervals/ | 1,685,656,211,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648209.30/warc/CC-MAIN-20230601211701-20230602001701-00697.warc.gz | 534,391,393 | 25,227 | # Prime Number between Intervals
## Prime numbers
The numbers which are divisible only by 1 and by the number itself are known to be prime numbers. Prime numbers are always greater than 1. The only even prime number is 2 while the rest of the prime numbers are odd.We will learn how to find prime number between intervals.
## Print prime number between intervals:
Step 1 : Take the values of lower limit and upper limit of the interval as input.
Step 2 : Initialize a variable flag with value 0.
Step 3 : Iterate a for loop from lower limit to the upper limit.
Step 4 : Take the value as num.
Step 5 : Now iterate a for loop from 0 to num/2.
Step 6 : If the num is divisible by loop iterator,then increment flag.
Step 7 : If flag = 0 , then prime
else , not prime
Step 8 : Back to Step 4.
Step 9 : End
## Example 1:
Run
```#include<stdio.h>
int main()
{
int a, b, i, j, flag;
printf("Enter lower bound of the interval: ");
scanf("%d", &a);
printf("\nEnter upper bound of the interval: ");
scanf("%d", &b);
printf("\nPrime numbers between %d and %d are: ", a, b);
for (i = a; i <= b; i++) {
if (i == 1 || i == 0)
continue;
flag = 1;
for (j = 2; j <= i / 2; ++j) {
if (i % j == 0) {
flag = 0;
break;
}
}
if (flag == 1)
printf("%d ", i);
}
return 0;
}
```
## Output:
```Enter lower bound of the interval: 1
Enter upper bound of the interval: 15
Prime numbers between 1 and 10 are: 2 3 5 7 11 13
```
## Example 2:
Run
```#include<stdio.h>
int main() {
int low, high, i, flag;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
printf("Prime numbers between %d and %d are: ", low, high);
while (low < high) {
flag = 0;
if (low <= 1) {
++low;
continue;
}
for (i = 2; i <= low / 2; ++i) {
if (low % i == 0) {
flag = 1;
break;
}
}
if (flag == 0)
printf("%d ", low);
++low;
}
return 0;
}
```
## Output:
```Enter two numbers(intervals): 20
50
Prime numbers between 20 and 50 are: 23 29 31 37 41 43 47
```
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Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others | 704 | 2,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.716146 |
https://www.cazoommaths.com/teaching-resource/operation-grid-blanks-4-x-4/ | 1,725,862,569,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00724.warc.gz | 658,300,235 | 64,896 | # Operation Grid Blanks 4 x 4
Suitable for Year groups: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
## Operation Grid Blanks 4 x 4 resource description
Here is a page of blank 4 x 4 operation grids which can be used for addition, multiplication and other operations for topics such as number, algebra or probability sample space diagrams.
# Operation Grid Blanks
## What is the resource?
This PDF offers blank grids in two sizes – a classic 'hundred square' and a simple 3x3. It targets primary learners practising operations (addition, subtraction, etc.).
## Why are operation grids important?
They turn operations into a visual puzzle, making maths meaningful:
• Number patterns: Kids fill the grid doing 'hops' of +10, -2, etc. Spotting rules emerge.
• Place value: Each row shows tens/ones clearly, boosting number sense.
• Missing numbers: Blank cells act as a fun challenge – what goes here? Builds reasoning.
• Error detection: When a pattern breaks, students have to analyse, and fix mistakes.
## Why is this resource helpful?
The blank format encourages a focus on maths thinking alongside number facts:
• Flexible tool: Teachers fill in starting numbers to set the task difficulty.
• Hands-on: Kids can colour, cross off, and visualise as they fill the grid – more than a worksheet.
• Supports exploration: Open-ended prompts spark discovery like- What if I start here?
• Printable PDF: Offers endless copies for drills, games, and puzzles for individual kids or a group.
These grids help bridge the gap between rote calculation and deep number understanding. It encourages problem-solving through play.
#### Get 20 FREE MATHS WORKSHEETS
Fill out the form below to get 20 FREE maths worksheets! | 398 | 1,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.855653 |
http://mathhelpforum.com/calculus/40695-trigonometric-differentiation.html | 1,513,314,679,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948563629.48/warc/CC-MAIN-20171215040629-20171215060629-00324.warc.gz | 185,392,732 | 10,840 | 1. ## Trigonometric Differentiation
Given that x = 4sin(2y+6), find (dy/dx) in terms of x.
i have managed to get:
(dy/dx) = 1/8cos(2y+6)
where do i go from here?
2. Originally Posted by Stylis10
Given that x = 4sin(2y+6), find (dy/dx) in terms of x.
i have managed to get:
(dy/dx) = 1/8cos(2y+6)
where do i go from here?
$\cos(2y+6) = \pm \sqrt{1 - \sin^2 (2y+6)} = \pm \sqrt{1 - \left(\frac{x}4\right)^2}$
I dont know what to do with the sign of cos though. It can be both plus and minus...
3. Originally Posted by Isomorphism
$\cos(2y+6) = \pm \sqrt{1 - \sin^2 (2y+6)} = \pm \sqrt{1 - \left(\frac{x}4\right)^2}$
I dont know what to do with the sign of cos though. It can be both plus and minus...
ok i under stand the first step of that, but can u please explain to me how you managed to get (x/4)^2 from sin^2(2y+6)
4. ok i understand how you got that, thanxs
5. Hello, Stylis10!
Given: . $x \:= \:4\sin(2y+6)$, .find $\frac{dy}{dx}$ in terms of $x.$
i have managed to get: . $\frac{dy}{dx} \:= \:\frac{1}{8\cos(2y+6)}$ . . . . Good!
where do i go from here?
We are given: . $4\sin(2y+6) \:=\:x\quad\Rightarrow\quad \sin(2y+6) \:=\:\frac{x}{4} \:=\:\frac{opp}{hyp}$
So $(2y+6)$ is an angle in a right triangle with: . $opp = x,\;hyp = 4$
. . From Pythagorus, we have: . $adj = \sqrt{16-x^2}$
Hence: . $\cos(2y+6) \:=\:\frac{\sqrt{16-x^2}}{4}$
Substitute into your equation: . $\frac{dy}{dx} \;=\;\frac{1}{8\!\cdot\!\frac{\sqrt{16-x^2}}{4}}$
Therefore: . $\boxed{\frac{dy}{dx}\;=\;\frac{1}{2\sqrt{16-x^2}}}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For the fun of it, I did it another way.
. . (I've got to get a life!)
We have: . $4\sin(2y+6) \;=\;x \quad\Rightarrow\quad \sin(2y+6) \:=\:\frac{x}{4}$
. . . $2y + 6 \;=\;\sin^{-1}\left(\frac{x}{4}\right) \quad\Rightarrow\quad y \;=\;\frac{1}{2}\sin^{-1}\left(\frac{x}{4}\right) - 3$
Then: . $\frac{dy}{dx} \;\;=\;\;\frac{1}{2}\cdot\frac{\frac{1}{4}}{\sqrt{ 1-\left(\frac{x}{4}\right)^2}} \;\;=\;\; \frac{1}{8}\!\cdot\!\frac{1}{\sqrt{1-\frac{x^2}{16}}}$
. . . . . $\frac{dy}{dx} \;\;=\;\;\frac{1}{8}\!\cdot\!\frac{1}{\sqrt{\frac{ 16-x^2}{16}}} \;\;=\;\;\frac{1}{8}\!\cdot\!\frac{1}{\frac{\sqrt{ 16-x^2}}{4}}$
Therefore: . $\frac{dy}{dx} \;=\;\frac{1}{2\sqrt{16-x^2}}\quad\hdots$ ta-DAA! | 973 | 2,284 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-51 | longest | en | 0.596142 |
http://math.stackexchange.com/questions/266084/whats-the-name-for-a-bijection-where-pairs-of-elements-map-to-each-other | 1,441,440,061,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645396463.95/warc/CC-MAIN-20150827031636-00168-ip-10-171-96-226.ec2.internal.warc.gz | 157,104,775 | 19,538 | # What's the name for a bijection where pairs of elements map to each other?
First, a bit of context. About a quarter of an hour ago I came across one of those "Internet math puzzles" on Facebook that stated:
If 1 = 5, 2 = 10, 3 = 15, and 4 = 20, then 5 = ?
The answer was supposed to be 1, as we had already stated. But of course, assuming the rule is x = 5x, then 5 is equal to both 1 and 25, as well as 125, 625, 3125, etc.
This got me thinking, can we define a relation like the one the question asks for so that 5 isn't mapped to 25, but x is still mapped to 5x in general? What I got was the following:
$x = \left\{ \begin{matrix} 5x & \text{the prime factorization of } x \text{ has an even power of 5} \\ x/5 & \text{the prime factorization of } x \text{ has an odd power of 5} \end{matrix} \right.$
I'm pretty sure this is a bijection. But if it is, I don't know if there's a special name for bijections like this, that map pairs of values to each other.
Basically, what's the name for the type of bijection $f: D \to D$, such that for any $a, b \in D$, if $f(a) = b$, then $f(b) = a$, and $a \neq b$?
-
This comment doesn't address your question directly but you may be intersted in the Garsia-Milne involution principle if you don't already know about it - particularly the discussion in Wilf's "Lectures on integer partitions" – alancalvitti Dec 27 '12 at 21:02
Such a bijection is called an involution. It is a bijection $f$ such that $f(f(x))=x$.
Note that the premise $f(f(x))$ is only true for a bijection - that is, if $f$ is any function such that $f(f(x))=x$ then $f$ is naturally a bijections, since its left- and right-inverse is itself.
I don't think there is a name for an involution that has no fixed points - that is, where $f(x)\neq x$ for all $x$, so that you always get a pair.
You might want to call it a "free involution." In involution can be seen as a group action of $\mathbb Z_2$ on a set $X$. A "free" group action is one with no stabilizers - that is, for each $x\in X$, $\{g:gx=x\}=\{1\}$.
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I had a hunch that it might have been an involution; I looked it up, but it wasn't exactly what I was looking for. It seems like "fixed-point-free involution" is the closest we have so far. – Joe Z. Dec 27 '12 at 19:56
The basic non-fixed involution is in boolean algebra, where we send a set $A$ to it's complement $A^c$. – Thomas Andrews Dec 27 '12 at 20:04
So the "complement" operator is the same as the "conjugate" operator in that case? – Joe Z. Dec 27 '12 at 20:05
Yeah, maybe I misremembered "complement" as "conjugate." Never mind. Edited my comment above to remove that wrongness. :) – Thomas Andrews Dec 27 '12 at 20:08
You might want to call it a "free involution." An involution can be seen as a group action of $\mathbb Z_2$ on a set $X$. A "free" group action is one with no stabilizers - that is, for each $x\in X$, $\{g:gx=x\}=\{1\}$. – Thomas Andrews Dec 27 '12 at 20:13
A map such that $f(f(a))=a$ for every $a$ is sometimes called an involution, and an involution certainly has the property that whenever $f(a)=b$, $f(b)=a$.
I am not requiring $f$ to be a bijection, but it turns out to be one anyway, since it is its own inverse function.
Involutions are well-studied in ring theory, where rings with involutions pop up naturally, especially in operator algebras. One such involution is given by the transposition map on a full ring of square matrices. Another good example is the complex conjugate mapping in the ring of complex numbers.
Update: as for your addition to the question of having no fixed points, I haven't heard a special name used for them. As you can see from the two examples I listed, it's quite common and unalarming for an involution to have fixed points. What is interesting is that the involution is the identity map when restricted to the fixed points.
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By definition, the restriction of any map to fixed points is the identity: $f(x)=x$, so not just involutions, but also idempotents (where the fixed point set is often called normal form) etc. – alancalvitti Dec 27 '12 at 21:00
@alancalvitti Yes, that's a good comment. I did not mean to make it look like this was a special property of involutions. – rschwieb Dec 27 '12 at 21:09 | 1,232 | 4,234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2015-35 | longest | en | 0.950854 |
https://frontendmasters.com/courses/algorithms/maps/ | 1,721,688,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00405.warc.gz | 224,320,212 | 14,694 | The Last Algorithms Course You'll Need
# Maps
## ThePrimeagen
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### Check out a free preview of the full The Last Algorithms Course You'll Need course
The "Maps" Lesson is part of the full, The Last Algorithms Course You'll Need course featured in this preview video. Here's what you'd learn in this lesson:
ThePrimeagen discusses an overview of map terminology, including load factor, key-value, and collision.
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### Transcript from the "Maps" Lesson
[00:00:00]
>> Fantastic, where does this move on, all right? I'm ready, are you ready? All right, so, there's a common saying in interviews, just use a HashMap, right? I think Joma Tech even did it. Again, this the second time I represent him. He always does these shorts, and one of them was HashMap, right?
[00:00:16]
Prepares for an interview for a month, right, when the person asks a question his answer is HashMaps, right? It's just always the answer. So what are maps? It's too easy for us to think of either squirrely braces or new map and just assume we know what they are, right?
[00:00:33]
That's what we do. The reality is, we probably know how to use them, we probably don't know what they actually are. And more so, this is not even a map, right? That's not a map, it's an object that has different properties associated with it. So, just things to think about, everything's a little bit different.
[00:00:51]
So we are gonna go through, and we are actually going to explain what a map actually is. So here's some terms first. There's the load factor. That's the amount of data points versus the amount of storage that we have, yeah? Yeah, so if we had seven items in our map, and we had ten available units of storage, we'd have a load factor of 0.7.
[00:01:13]
A key, a key is the thing that is hashable, right? We use the key to create something in which we can access our value. A value of course is the thing that's associated with the key, and a collision is when two keys map to the same cell. Okay, there we go, we've made some progress.
[00:01:33]
All right, so let's whiteboard all these things. This is where things get kind of fun cuz we're kind of finally done with this whole graph. Everything's a LinkedList, everything's a graph kind of section. We're kind of just into something new. Actually, joke's on you, it's an ArrayList. All right, so again, we're back down to an ArrayList.
[00:01:51]
Somehow these foundational structures that we discussed just keep coming back up. They're inside of a heap. We use a heap inside of Dijkstra's shortest path. We are now on to Maps, guess what, ArrayList, right? It always keeps on being the same thing, always down. So effectively how a Map works is that, we have a key and it needs to map to value.
[00:02:10]
And what the key is mapping to a value has to be, is a consistent hash, consistent hash. All right, so what this means is that when I give case of 0, right, some key, that is uniquely defined, it will always gimme the same answer back out. If it doesn't gimme the same answer back out, that's an inconsistent hash.
[00:02:33]
What is it good for? Absolutely nothing. Okay, that's great song. But, so now what we have to do is, we have to come up with a way to hash. Now, JavaScript is particularly difficult when it comes to creating these structures, because we can take say a string, and walk through every character in the string and create a hash from the string, right?
[00:02:51]
We can have a number and use the number as a way to kind of create a hash for the string. But what we don't have is an object. If you just hand me an object, I can't create some sort of unique identifier for that object in the sense that if you create an object that's like foo:5, right, my goodness, great five.
[00:03:10]
Now let's say foo:3, and we'll call this thing const F, right? And then you create another one, t, and it has the exact same value. You can't tell the difference of that in JavaScript, right, you don't have a way to be able to say, hey, object, what's your unique ID?
[00:03:26]
And in Java you do, you can call like get hash ID, or get some sort of ID to the object in which produces a unique identifier for that specific object. This may have changed in JavaScript. But last time I checked, this is not a thing, someone can prove me wrong.
[00:03:41]
I'm sure there's someone that could probably prove me wrong, but I'm quite positive you just can't uniquely identify objects in this sort of sense. You can only uniquely identify them based on properties and their associated values. So if you have two with the identical, good luck. Which means that our key lookup really breaks down in this case, right?
[00:04:00]
Because you could actually have two separate objects that are meant to be two separate objects, but mapped to the same value. So for our case if we do implement one, we'll simply just use strings or numbers as our entrance into it, which I don't think we're gonna actually implement one.
[00:04:15]
So, let's just talk about how this works. So, we can do something like this, is that we need a hashing function, right, that takes in the key and needs to produce out some sort of unique number, right? And this is the important part, is that we need a number.
[00:04:30]
And the reason why is, let's just say k comes in and it returns out some great number, and it can be a really big number. It can be a really small number if there's not anything specific to it. Then what we can do is, we can take it, and we can modulo it based on say, the length of our data storage, right, so say 10.
[00:04:50]
And what that means is that if we have a 10-item data, okay, say we have this many items, I realized I can't count as fast as I can, right? That means our key, whatever we give it, can uniquely map to one of the bins inside of our array structure.
[00:05:05]
And we can just say, hey, we'll store the key plus the value inside of that structure right there. Then we map another one which say, goes over here, and then we put the key and the value. But every now and then, you're gonna map two very different keys.
[00:05:30]
And since we say only have ten slots, the chance of them colliding is uniformal. If they uniformly go to any number, it would be a chance of one over ten, right? It wouldn't be that great. So if you had a perfectly uniform hash generator, you get a 10% chance of collision.
[00:05:45]
Which means that what happens when we collide here? Well, we need a way to be able to store both. Now when I went to college, I know, I can't believe I'm saying this whippersnappers. What we used to do, at least how I learned it is that you'd actually do something called linear or exponential back off.
[00:06:01]
Meaning we would literally go down to the next slot and put it in there. And we always store the key with the value, and we'll see why on retrieval. Which means that if you did this, you will fill up more and more slots. So as things become less and less efficient, or more and more full, you're gonna need a larger amount of storage with a smaller amount of load factor to prevent collisions, right?
[00:06:25]
Cuz the moment you have five items in here, the chance of you hitting it and then having to linearly go down gets greater, and greater, and greater. So in more modern ways or what I've seen is that, instead you use a list underneath the hood to be able to add in items.
[00:06:41]
So you have an ArrayList, that has an ArrayList, that can add an items and then we just walk the ArrayList. Sounds good. We could technically also use a LinkedList underneath the hood because you do linearly walk it every single time. And so long as the keys sufficiently map to a bunch of them, what we're gonna see is little sub amounts with only just one item in it.
[00:07:01]
Maybe sometimes two, right? Long as you have a good hashing function, you're gonna have very few. So how does retrieval work? Well, we go through our magical hashing function, right? We pass in a key, we go through a hashing function and produces an index, we then modulo that with the size of our storage area, and we now have an item into it.
[00:07:19]
Well, the problem is, is there could be multiple items inside of any one storage unit, correct? So that's why we store the key plus the value. We can use our key and say hey key, are you equal to that key? You're the same key? Well guess what, that's your associated value then, get on out, right?
[00:07:38]
We can actually do that and get it on out, and we can use that specific value. It's a little bit tricky, but then as you think about it becomes really simple. Maps tend to be pretty simple creations, and that's really all you're doing. Deletion, exact same thing, you go to that spot within your ArrayList, you find the associated key with the value, remove it from the entry, and there you go.
[00:08:02]
You now have it removed, of course you have to decrement length. But there is always one problem. So, I believe the ideal load factor is 0.7 at this point. That's kind of what I believe is considered the ideal one long as you do this ArrayLists or this LinkedList style storing.
[00:08:18]
If you do that, once you exceed that load factor, we need a way to be able to redo this, right? So something we can do is that, we can take our current data storage, what it is, and we can iterate through all of the keys that we can find.
[00:08:35]
And then we can rehash them and restore them into a larger storage arena, thus cutting down our load factor, say by half if we double the storage, and then we have a bunch more in which we can store. And this is the effective way in which maps operate on systems.
[00:08:53]
There we go, they're not actually all that complicated, they're actually surprisingly uncomplicated. It's actually surprisingly boring, kind of, it doesn't feel that cool. But now that you know what ArrayLists are and LinkedLists, this is very, very easy. If you didn't know those two terms, this might be a bit harder to kind of understand how this works.
[00:09:09]
So an ArrayList allows you to kind of just simply increase the length. But usually you just use raw arrays underneath the hood, and then double the size when you need to or increase the size by 50%, whatever you think is good. Always a trade off between how much do you want to store versus how efficient you want to be with memory.
[00:09:26]
So it's always a game. That's why often you'll see maps come with some sort of capacity. Because they want you to give them the hint of how big we should be so that way the storage and retrieval is as efficient as possible. Cuz the more amount of collisions, the less o of one it is.
[00:09:42]
And so why is it o of one? Well, we assume that, again, the key has some sort of length that is not infinity, right? You're not storing large text files as the key. Instead, you're storing some small string, right? You could hash a text file, and then boom, you have this key that you use all the time to get that value.
[00:10:00]
That's kind of how the internet works. You do those kind of things, it has some sort of hash associated with it. You don't store gigantic pieces of data, you still have small things that point to other things. And then usually the thing you point to is actually the big thing, so we assume hashing takes over one.
[00:10:14]
I'm sure there's a great proof out there that says, this is why it is. And so therefore, we have this. And we only have to do that plus accessing the data, and then going through a relatively small amount. So long as our collisions don't gather up into one single point, you will always have effectively one item in there, two items in there, also again considered constant.
[00:10:34]
And so that's why it has a constant lookup time. There was a pretty funny bug that somebody found that some high performance piece of code, no one could figure out how to make it fast in this one condition. And someone went through and looked at the buckets of the map.
[00:10:48]
And effectively all the keys somehow were just mapping to this one bucket cuz they had a bad hashing algorithm. And that caused this one thing to just operate at linear time when you assumed it was operating at constant time. And it was just super slow when it should have been super fast.
[00:11:03]
And a simple change improve the speed of the program by an enormous amount of time. And so maps can be quite tricky. In JavaScript land, of course you get zero understanding of what's happening underneath the hood, right? You have no idea you are in an abstract land, and you just hope that it's well implemented.
[00:11:17]
The people who made the V8 are clearly so talented they created a engine that can run a slow language swiftly. So they probably did it right, I assume they did it right. And yeah, that's my argument for why this happens. So yeah, that's really all a map is.
[00:11:31]
I didn't really wanna implement one, I think it's kinda boring. Plus we don't really have the tools we can to really implement it well in JavaScript. So we're just gonna move on.
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http://physics.stackexchange.com/questions/61073/why-the-direction-of-dipole-moment-is-from-negative-charge-to-positive-charge | 1,462,506,552,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861718132.40/warc/CC-MAIN-20160428164158-00016-ip-10-239-7-51.ec2.internal.warc.gz | 214,013,592 | 20,094 | # Why the direction of dipole moment is from negative charge to positive charge?
An electric dipole moment is defined as $p = q\times 2d$. How to understand it physically? Why the direction of the electric dipole moment is from negative charge to positive charge?
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Important: there shouldn't be a factor of 2 in that formula. Two point charges +q and -q separated by a distance d will have a dipole moment p = qd. – David H Apr 14 '13 at 15:28
There are two separate issues here. (1) Why does it make sense to consider a dipole moment as a vector? (2) Given that it's a vector, why does it make sense to say that it points in this particular direction, rather than the opposite direction.
1. Intuitively, it makes sense to define a dipole as a vector because when we put it in a field, it aligns itself with the field like a little arrow. Fundamentally, we treat things as vectors when they transform as vectors. We have monopoles, dipoles, quadrupoles, ... Monopoles (electric charges) don't change under rotation, so they're scalars. Dipoles reverse themselves under 180 degree rotation, so they're vectors. Quadrupoles reverse themselves under 90 degree rotation, so they're tensors.
2. This is purely a matter of convention. According to the usual convention, the potential energy of an electric dipole is $-\mathbf{p}\cdot \mathbf{E}$. Historically, whoever first defined the dipole moment could have defined it with the opposite sign. Then the energy would have been $+\mathbf{p}\cdot \mathbf{E}$. The sign would also have been reversed in every other equation, e.g., $\boldsymbol{\tau}=\mathbf{p}\times \mathbf{E}$ would have become $\boldsymbol{\tau}=\mathbf{E}\times \mathbf{p}$.
There are many, many arbitrary choices of sign like this in physics. If Ben Franklin had made the opposite choice for the sign of the charge of cat fur rubbed on glass (or whatever it was he used as a standard), then we'd say today that electrons had positive charge. The direction of the magnetic field is also arbitrary and could have been defined as pointing the opposite way (in which case some of the signs in Maxwell's equations would flip).
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I don't agree with Ben Crowell. I think that the reason that the moment is directed from negative to positive is because of the definition of moment:
$$\mathbf{p} = \sum\nolimits q*\mathbf{d}$$ q: charge, d:distance from the origin of coordinates to the carge
If you have a negative and positive charge, this relation gives you the direction from negative to positive.
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There is no particular way of describing the direction of dipole rather it is hypothetic.But one should know the fact that if lenth is taken as 2L,means it has a specific origion(No matter for ease it is taken).Now as possible,draw a graph of dipole and its axial line co-insided with the X-axis.Therefore we will find the -ve charge takes the position on left of the origion shows charge is also negative.Now leave all the quadrants except when all trig.functions are +ve..i.e Quad.1st....Draw a point on axis on this quadrant,now predict the direction,we say it is from centre 'O' towards that point.i.e we are supposing its direction from origion to positive....i.e from left to right if direction is obtained from the pont left of origion to the point under consideration,we have to follow same rules.Same is in dipole direction i.e from -ve to +ve......Other all stories are hypothetical...
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Aasif, excuse me, not being offensive, you have a bunch of typos and grammar error. like "origin." – Idear Mar 26 '14 at 2:56
The dipole moment is in the direction, in whch unit test charge moves, wn put on axial line outside the dipole.
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Please correct the form of you answer, especially after the second comma. – Valter Moretti Jan 8 '14 at 15:51
While finding direction of torque , right hand screw rule is used in cross product.Obviously it has been assumed that to satisfy that rule, we have to assume direction of dipole moment from negative to positive. $\text{Torque} = p \times E$
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There is no need of a big theory to explain the direction of a dipole moment. It's just because the charge of an electron is negative (by convention).
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Electron has less mass and therefore it start accelerating first therefore it is considered that dipole moment from negative to positive when placed in electric field
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The direction is wrt the applied field....dipole opposes the field hence p has direction from negative to positive...
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## protected by Kyle KanosOct 11 '15 at 17:32
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