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https://mail.haskell.org/pipermail/haskell-cafe/2006-September/018339.html
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# [Haskell-cafe] Polymorphism/monomorphism (and the restriction)
Aaron McDaid haskell-cafe at aaronmcdaid.com
Thu Sep 21 18:27:38 EDT 2006
```Hi,
I think the following might help a little in understanding the
monomorphic restriction (which I don't fully understand myself). I'm a
bit of a newbie so apologies in advance if I've made a mistake or if my
description isn't as useful to others as it seems to me. I've been
following a thread on haskell at haskell.org and I think the below might
help. I used GHCi, version 6.2.2 (it fails in hugs but that seems to be
because of hugs non-compliance with the standard in this case).
First off, I'm guessing that I'm getting Haskell98 behaviour here and
not some GHCi extension. Please tell me if this is not the case.
Run the code listing at the bottom of this email and you should get the
output which I've also listed below.
This code experiments with "Int", "Float" and "(Num a) => a", and I
tried to print x*2 and x/2 for each. (4::Int)/2 isn't allowed because /
isn't defined for Ints.
You can see that
kN :: (Num a) => a
took two different types depending on what method ( / or * ) was applied
to it.
kN / 2 = 2.0
kN * 2 = 8
kN/2 is a Float (it can't use Int as / isn't defined for Int, so it uses
Float, for which / is defined).
kN*2 is an Int.
The above outputs demonstrates polymorphism, doesn't it? i.e. Not only
has the compiler got a variety of types to choose from, but a variety of
types can be used at runtime?
The output for kI and kF is obvious.
The interesting thing is that k behaves as a Float in both cases. This
is monomorphism isn't it? i.e. the compiler may have a variety of types
to choose from, but it picks one and sticks to it for every usage. In
summary, k didn't give the same outputs as kN.
And the monomorphism restriction is a rule which means that sometimes
things are forced to a monomorphic type (like k as Float here) when it
could have given it a polymorphic type like "kN :: (Num a) => a"
I'm fairly new to these lists, so apologies if I'm covering old ground
again. My first aim is to understand exactly what polymorphism and
monomorphism is and demonstrate corresponding results, before thinking
Thanks,
Aaron
-- The code
kI :: Int
kI = 4
kF :: Float
kF = 4
kN :: (Num a) => a
kN = 4
k = 4
main = do
p "kI * 2" \$ kI * 2
p "kF / 2" \$ kF / 2
p "kF * 2" \$ kF * 2
p "kN / 2" \$ kN / 2
p "kN * 2" \$ kN * 2
p "k / 2" \$ k / 2
p "k * 2" \$ k * 2
p :: (Show a) => String -> a -> IO ()
p s = putStrLn.(s++).(" = "++).show
-- the output - remember kI / 2 is not possible.
kI * 2 = 8
kF / 2 = 2.0
kF * 2 = 8.0
kN / 2 = 2.0
kN * 2 = 8
k / 2 = 2.0
k * 2 = 8.0
-- PS: If you delete the k / 2 line from the program, then k * 2 becomes
simply 8 (not 8.0). It uses Int if possible, and Float if that's not
available.
```
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## reversible vs irreversible expansions
$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$
Elizabeth Bowen 1J
Posts: 53
Joined: Wed Nov 14, 2018 12:20 am
### reversible vs irreversible expansions
If we need to calculate the work of expansion of a system, and we're not told whether it's a reversible or irreversible expansion, how do we know which equation should be used?
thanks
AArmellini_1I
Posts: 107
Joined: Fri Aug 09, 2019 12:15 am
### Re: reversible vs irreversible expansions
Then it's based on your conditions regarding the system. Is there constant pressure? How about constant temperature? etc.
Elizabeth Bowen 1J
Posts: 53
Joined: Wed Nov 14, 2018 12:20 am
### Re: reversible vs irreversible expansions
ok, thanks that makes sense; if we're told that there's not constant pressure, but constant temperature, that means it's reversible, correct?
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https://kalkicode.com/decimal-to-roman-numeral-converter-in-js
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# Decimal to roman numeral converter in node js
Js program for Decimal to roman numeral converter. Here more solutions.
``````// Node Js program for
// Conversion from Decimal to roman number
class Perform
{
// Display roman value of n
show(n)
{
switch (n) {
// Test Cases
case 1:
this.ans += "I";
break;
case 4:
this.ans += "IV";
break;
case 5:
this.ans += "V";
break;
case 9:
this.ans += "IX";
break;
case 10:
this.ans += "X";
break;
case 40:
this.ans += "XL";
break;
case 50:
this.ans += "L";
break;
case 90:
this.ans += "XC";
break;
case 100:
this.ans += "C";
break;
case 400:
this.ans += "CD";
break;
case 500:
this.ans += "D";
break;
case 900:
this.ans += "DM";
break;
case 1000:
this.ans += "M";
break;
}
}
select(number, collection, size)
{
var n = 1;
var i = 0;
for (
i = 0; i < size; ++i)
{
if (number >= collection[i])
{
n = collection[i];
}
else
{
break;
}
}
this.show(n);
// Reduce the value of number
return number - n;
}
roman(number)
{
if (number <= 0)
{
// When is not a natural number
return;
}
this.ans = "";
var num = number;
// Base case collection
var collection = [1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000];
// Get the size of collection
var size = collection.length;
while (number > 0)
{
number = this.select(number, collection, size);
}
console.log(" " + num + " : " +this.ans);
}
}
function main()
{
// Test Case
}
// Start program execution
main();``````
Output
`````` 10 : X
18 : XVIII
189 : CLXXXIX
604 : DCIV
982 : DMLXXXII
3000 : MMM``````
## Comment
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
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Mathbox for Stefan O'Rear < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > dford3lem1 Structured version Visualization version GIF version
Theorem dford3lem1 38119
Description: Lemma for dford3 38121. (Contributed by Stefan O'Rear, 28-Oct-2014.)
Assertion
Ref Expression
dford3lem1 ((Tr 𝑁 ∧ ∀𝑦𝑁 Tr 𝑦) → ∀𝑏𝑁 (Tr 𝑏 ∧ ∀𝑦𝑏 Tr 𝑦))
Distinct variable group: 𝑦,𝑏,𝑁
Proof of Theorem dford3lem1
StepHypRef Expression
1 treq 4892 . . . . 5 (𝑦 = 𝑏 → (Tr 𝑦 ↔ Tr 𝑏))
21cbvralv 3320 . . . 4 (∀𝑦𝑁 Tr 𝑦 ↔ ∀𝑏𝑁 Tr 𝑏)
32biimpi 206 . . 3 (∀𝑦𝑁 Tr 𝑦 → ∀𝑏𝑁 Tr 𝑏)
43adantl 467 . 2 ((Tr 𝑁 ∧ ∀𝑦𝑁 Tr 𝑦) → ∀𝑏𝑁 Tr 𝑏)
5 trss 4895 . . . . . 6 (Tr 𝑁 → (𝑏𝑁𝑏𝑁))
6 ssralv 3815 . . . . . 6 (𝑏𝑁 → (∀𝑦𝑁 Tr 𝑦 → ∀𝑦𝑏 Tr 𝑦))
75, 6syl6 35 . . . . 5 (Tr 𝑁 → (𝑏𝑁 → (∀𝑦𝑁 Tr 𝑦 → ∀𝑦𝑏 Tr 𝑦)))
87com23 86 . . . 4 (Tr 𝑁 → (∀𝑦𝑁 Tr 𝑦 → (𝑏𝑁 → ∀𝑦𝑏 Tr 𝑦)))
98imp 393 . . 3 ((Tr 𝑁 ∧ ∀𝑦𝑁 Tr 𝑦) → (𝑏𝑁 → ∀𝑦𝑏 Tr 𝑦))
109ralrimiv 3114 . 2 ((Tr 𝑁 ∧ ∀𝑦𝑁 Tr 𝑦) → ∀𝑏𝑁𝑦𝑏 Tr 𝑦)
11 r19.26 3212 . 2 (∀𝑏𝑁 (Tr 𝑏 ∧ ∀𝑦𝑏 Tr 𝑦) ↔ (∀𝑏𝑁 Tr 𝑏 ∧ ∀𝑏𝑁𝑦𝑏 Tr 𝑦))
124, 10, 11sylanbrc 572 1 ((Tr 𝑁 ∧ ∀𝑦𝑁 Tr 𝑦) → ∀𝑏𝑁 (Tr 𝑏 ∧ ∀𝑦𝑏 Tr 𝑦))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 382 ∈ wcel 2145 ∀wral 3061 ⊆ wss 3723 Tr wtr 4886 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 This theorem depends on definitions: df-bi 197 df-an 383 df-or 837 df-tru 1634 df-ex 1853 df-nf 1858 df-sb 2050 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-ral 3066 df-rex 3067 df-v 3353 df-in 3730 df-ss 3737 df-uni 4575 df-tr 4887 This theorem is referenced by: dford3lem2 38120 dford3 38121
Copyright terms: Public domain W3C validator
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http://therapeutic-journeys.com/who-invented-hindu-arabic-numerals.html
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| 169,013,604
| 5,198
|
# Who invented hindu arabic numerals. Al 2019-01-06
Who invented hindu arabic numerals Rating: 7,2/10 680 reviews
## The History of the Hindu
Greek numerals The had two important systems of numerals, besides the primitive plan of repeating single strokes, as in for six, and one of these was again a simple grouping system. Many regions still use their own traditional symbols, however the La … tin glyphs are internationally recognised as the standard form of numeration today. This delegation consisted of an astronomer called Kanaka, who carried with him a small library including a book titled Surya Siddhanta and works of Aryabhata and Brahamgupta. In fact, the earliest numerals of which there is a definite record were simple straight marks for the small numbers with some special form for 10. When the Arab traders came to India, they dropped the old Semitic numbering system and adopted the system used in India. Most historians agree that numbers were first invented prehistory in India.
Next
## When were Arabic numerals invented?
The Hindu-Arabic system employs positional notation and the symbol 0 is used as a place holder. By the close of the 8th century, however, some astronomical tables of India are said to have been translated into Arabic at , and in any case the numeral became known to Arabian scholars about this time. Some scholars hold that they appeared even earlier. The Hebrew system is shown in the figure. This is all contained in a work. When we look back at the history of numbers, we can see the societies all had some ability to perform mathematics.
Next
## Who invented the digits 0
In particular, Al-Khwarizmi developed a formula for systematically solving quadratic equations equations involving unknown numbers to the power of 2, or x 2 by using the methods of completion and balancing to reduce any equation to one of six standard forms, which were then solvable. It is capable of expressing very large numbers by means of scientific notation or exponents with the least amount of numbers. Al-Khwarizmi wrote a famous book on the Hindu-Arabic Number System, and that is why he is the most known man regarding this number system, because the book was on Algebra, the course that teaches the Hindu-Arabic Number System. This number system was invented by the Indians who followed the Hindu faith and hence they are sometimes also referred as Hindu … number system. The simplest number system where even number is represented by marks is known as the unary numeral system. There are an infinite amount of numbers. The Romans did not have clocks as we know them.
Next
## History of the Hindu
It is designed for positional notation in a decimal system. Here the binary positional system has been found to have great advantages over the decimal. But it is not just that. The digits 0, 1, …, 19 are, as in the Babylonian, formed by a simple grouping system, in this case to base 5; the groups were written vertically. Al-Khwarizmi wanted to go from the specific problems considered by the Indians and Chinese to a more general way of analyzing problems, and in doing so he created an abstract mathematical language which is used across the world today. Arabic numbers are written from left to right, which is consistent with English but inconsistent with Arabic. However, it was not used consistently and apparently served to hold only interior places, never final places, so that it was impossible to distinguish between 77 and 7,700, except by the.
Next
## Who invented numbers?
Reducing the symbol set to just 10 numeric symbols provided a concise notation that revolutionised computation and mathematics, particularly for those who were solely familiar with Roman numerals which do not easily lend themselves to computation, due to their intrinsically non-positional nature. The dot continued to be used for at least 100 years afterwards, and transmitted to Southeast Asia and Arabia. There is simply fact and we have the ancient inventors of this system to thank for that. Neither you, nor the coeditors you shared it with will be able to recover it again. The Persians took hold of these numerals and even published them in the year 825 in a well-known book. The number 258,458 written in hieroglyphics appears in the figure. The same logic led to the zero, a number with no straight lines.
Next
## How important is the number 0 in Hindu Arabic system
They, amongst other works, contributed to the diffusion of the Indian system of numeration in the and the West. They used the sundial or water clocks or simply looked up at the sky. The complete development of this idea must be attributed to the Hindus, who also were the first to use zero in the modern way. Development of modern numerals and numeral systems The Several different claims, each having a certain amount of justification, have been made with respect to the origin of modern Western numerals, commonly spoken of as but preferably as Hindu-Arabic. This numeral system is most known to be developed by Al-Khwarizmi Persian and Al-Kindi Arab. The language of mathematics that has been created by the Hindu Arabic numerals should be celebrated because it is what can unite us as one people. The cuneiform and the curvilinear numerals occur together in some documents from about 3000 bce.
Next
## Is the 'Arabic' number genuinely an invention of Islam or that of Hindu?
However, see the table for some other modern numeral systems. One of the greatest scientists, Albert Einstein thanked the indians for this. Tallying system have no complex system or place values it was just a line. The Arabs through trading links with India soon took on board the Indian system of counting. The direct influence of for such a long period, the superiority of its numeral system over any other simple one that had been known in before about the 10th century, and the compelling force of tradition explain the strong position that the system maintained for nearly 2,000 years in commerce, in scientific and theological literature, and in. The base 60 still occurs in measurement of time and angles. The principal example of this kind of notation is the , three variants of which are shown in the figure.
Next
## Who Invented Hindu Arabic Numerals
Dis … advantages: Prime numbers can't be worked out in a systematic manner in the same way that number sequences can. This system makes it easier for working in arithmetic rather than other numeral systems. Thereafter Mansoor never broached the subject with him and the Bakhtishu family provided leading physicians and medical teachers for seven generations. Comparison of selected modern systems of numerals Hindu-Arabic 1 2 3 4 5 6 7 8 9 0 Arabic ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩ ٠ Devanagari Hindi १ २ ३ ४ ५ ६ ७ ८ ९ ० Tibetan ༡ ༢ ༣ ༤ ༥ ༦ ༧ ༨ ༩ ༠ Bengali ১ ২ ৩ ৪ ৫ ৬ ৭ ৮ ৯ ০ Thai ๑ ๒ ๓ ๔ ๕ ๖ ๗ ๘ ๙ ๐ The There is one island, however, in which the familiar decimal system is no longer supreme: the electronic. After which, eleven and so forth were represented by compounds eleven being one and ten compounded in a single character. It is interesting to note that while Europe has christened the new system as Arab numerals, the Arabs call them Hindsa Indian numerals giving credit where it is due. Now 3 1 is easy to mistake for thirty-one or even for three thousand and one.
Next
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| 2.828125
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| 0.964357
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https://www.neetprep.com/question/57280-transverse-wave-represented-equation-yysinvtxFor-value--maximum-particle-velocity-equal-two-times-wavevelocity-y-y-y-y/126-Physics--Waves/690-Waves
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| 959,136,067
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# NEET Physics Waves Questions Solved
Watch Physics > Waves Videos
A transverse wave is represented by the equation $y={y}_{0}\mathrm{sin}\frac{2\pi }{\lambda }\left(vt-x\right)$
For what value of λ, the maximum particle velocity equal to two times the wave velocity
(1) $\lambda =2\pi {y}_{0}$
(2) $\lambda =\pi {y}_{0}/3$
(3) $\lambda =\pi {y}_{0}/2$
(4) $\lambda =\pi {y}_{0}$
(4) On comparing the given equation with standard equation $y=a\mathrm{sin}\frac{2\pi }{\lambda }\left(vt-x\right)$. It is clear that wave speed ${\left(v\right)}_{wave}=v$ and maximum particle velocity ${\left({v}_{\mathrm{max}}\right)}_{particle}=a\omega ={y}_{0}×$ co-efficient of t $={y}_{0}×\frac{2\pi v}{\lambda }$
$\because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left({v}_{\mathrm{max}}\right)}_{particle}=2{\left(\omega \right)}_{wave}$$\frac{a×2\pi v}{\lambda }=2v$$\lambda =\pi {y}_{0}$
Difficulty Level:
• 17%
• 23%
• 23%
• 40%
Crack NEET with Online Course - Free Trial (Offer Valid Till September 22, 2019)
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http://gameknot.com/comment.pl?t=2&k=15695
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# chess online
## CHESS PUZZLE #15695
Added by: amr14 Added on: 10-Oct-09 Description: You Are Not Alone Difficulty: online chess puzzle #15695 Attempts: 664 Solved: 59 (8%)
White to move, mate in 3
From Comment
prelate20-Nov-12, 02:57
» Report abuse
Nice puzzle :)
The stalemate of the black pieces due to pins is a pleasant theme - as is the simple transfer of the white Queen from one extreme diagonal to the other. Fun to solve and a good lesson in the strength of pins and zugzwang
mila9220-Nov-12, 03:30
» Report abuse
PUZZLE #15695
There are many solutions
Various solutions don't make a puzzle...
kingdawar20-Nov-12, 05:20
» Report abuse
many solutions?
Unfounded comments don't invalidate a problem
puzzler77
20-Nov-12, 08:35
» Report abuse
1 Qa1
This prevents 2 Qa8 allowing Blacks King to take the pawn and freeing his position somewhat! Mate would still be inevitable, but not in three moves.
20-Nov-12, 11:49
Comment deleted on 20-Nov-12, 11:52.
20-Nov-12, 11:53
» Report abuse
Brilliant
It was easy to see that the queen can give mate on a1 or a8 if the king takes the knight or the pawn respectively, and to understand that any other first move than a queen or king move would spoil that possibility. But it takes some deeper insight to find that a1 and a8 can also be reached by the queen from h8. Whites king just has to move to the right square to open all lines of the mating net, and then blacks zugzwang forces him to play one of the deadly moves in the end
counto20-Nov-12, 12:47
» Report abuse
its a bueatiful puzzle
20-Nov-12, 13:01
» Report abuse
Actually i was wrong, because all other first moves than a king move (and not queen or king), will spoil the possibility of a mate on a1 or a8. So after all, it is somehow very logical that the first move is a king move (and also that it is Kf7)
mila9221-Nov-12, 00:59
» Report abuse
mate in 3 moves starts also with 1.Qa1
1. Qa1 c6
2. Qc3 Kxa3
3. Qxb3#
21-Nov-12, 03:16
Comment deleted on 21-Nov-12, 03:18.
kingdawar21-Nov-12, 03:20
» Report abuse
Third problem in A.C. Palmer, Collection of Chess Problems 1890.
21-Nov-12, 03:20
» Report abuse
Mate in 3 only after 1.Kf7
There is no other mate in 3.
After 1.Qa1 c6 2.Qc3 black just plays Kxa5 (and of course not Kxa3) and white has no mate
Black can also play 1. ... Kxa5 (instead of c6) and then white also has no mate in 3
mila9222-Nov-12, 00:47
» Report abuse
Mate in 3 only after 1.Kf7
You are right yadasampati! And I am wrong....
I forgot Kxa5!
I have so to take back my move.....
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## Horizontal spring, momentum
1. The problem statement, all variables and given/known data
A block of mass m is attached to the end of a horizontally mounted spring as shown.
The spring has a spring constant k and obeys Hooke’s law. The block is given an initial
displacement xo, after which it oscillates back and forth without frictional effects.
(a) Write an expression for the magnitude of the momentum p of the block as a function of its displacement x
from its equilibrium position, given m, k, and xo.
(b) What are the minimum and maximum values of the magnitude of the momentum, and where do they occur in the motion of the block?
2. Relevant equations
I know that the force of the spring on the block is F= -kx, and that the spring has stored the elastic PE= 1/2 kx^2, but i am not sure how to relate this to momentum. I do know that the rate of change (with respect to time) of the momentum is equal to the force, so is it equal to -kx?
any help would be greatly appreciated.
as for part b, im not sure how to attain this answer.
Magnitude is directly related to velocity. If a mass is oscillating back and forth, where is velocity the greatest? Well, starting at the right hand +x side, $$V_{1} = 0$$. When it reaches its maximum on the -x side, $$V_{2} = 0$$ as well. So, this means that velocity will be at a maximum when the mass passes through the center point. If you think about this, it makes sense because as the mass is accelerating towards the center, its velocity in acceleration are in the same direction. After it passes the midpoint, the spring is exerting a force OPPOSITE the motion and thus an OPPOSITE accleration so the mass slows down. Hope that helps. Ill check back later when Im not at work.
Ah, and I forgot, when you displace the spring distance x, and you have PE of 1/2kx², this will be your KE @ 0 which will be 1/2mv² so you can relate x to v which you can relate to p
## Horizontal spring, momentum
ah i think i see what you are saying, but when you said "Magnitude is directly related to velocity." you meant momentum is directly related correct? Also, i do understand your second post, the relation between the PE and KE, but im still confused as to how i would express the momentum of the block as a function of its displacement from its equilibrium position. I know that the momentum is equal to velocity X mass. So do i just substitute p for mv in the KE equation?
But i do see, that when the spring is stretched momentum is at its minimum, 0, because velocity is 0, as well as at the maximum compression of the spring, on the -x side, where velocity is again 0, making the momentum 0. So when the block is at its equilibrium velocity is greatest, and therefore momentum is greatest. But as i express this magnitude, is it just expressed as mv^2?
When the mass is all the way right ( or left ) you have the relationship that PE+KE=0. So, 1/2kx²+1/2mv²=0. Now you can solve for v in terms of x and thusly write p in terms of x also.
ohh ok i understand very well now. Thanks for all your help!!
Quote by dm66048 ohh ok i understand very well now. Thanks for all your help!!
And dont worry about the negative -k that you come up with. Since values for k are always negative, a -k in an equation will actually be positive and youll get a positive energy value.
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0
# help me with this please
Robert works at a car dealership. Each month, he receives a base salary of \$1,617.00, plus a commission of \$284.00 for each vehicle he sells. Which of the following equations could be used to determine Robert's total income each month?
(Let x represent the number of cars sold by Robert and y represent his total monthly income.)
### 2 Answers by Expert Tutors
Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
4.9 4.9 (333 lesson ratings) (333)
1
The total commission is just the product of the commission per car sold and the number of cars he sold.
This would be 284x. It should make sense to you if you look at the units:
284 (dollars / car) * x cars = 284x dollars, since the cars cancel leaving just dollars.
Finally, there is a base salary of \$1617 added to his pay, resulting in 284x+1617.
Thus you get y = 284x + 1617.
Joni S. | UCLA & Pepperdine Alumni - Math & Basketball TutorUCLA & Pepperdine Alumni - Math & Basket...
0
Total monthly income (y) = base salary (1617) + total commissions (based on number of cars sold, which is 284*x) Therefore, y = 1617 + 284x
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# The use and summary of numpy
Keywords: Python
The use of numpy and the summary of common functions
# NumPy
## ndarray understands multidimensional arrays
```import numpy as np
# Generate random multidimensional data of specified dimension
#Mathematical modeling should not be used
data = np.random.rand(2, 3)
print (data)
print (type(data))
#Type is the display data type; shape display dimension; number of ndim dimensions; other functions of type
```
```[[0.46686682 0.68844304 0.76663872]
[0.70747721 0.47887587 0.25943412]]
<class 'numpy.ndarray'>
```
##### ndim, shape and dtype properties
```print ('Dimension number', data.ndim)
print ('Dimensions: ', data.shape)
print ('data type: ', data.dtype)
```
```Number of dimensions 2
Dimensions: (2, 3)
Data type: float64
```
## Create ndarray
```'''1. array Establish'''
# list to ndarray
l = range(10)
data = np.array(l)
print (data)
print (data.shape)
print (data.ndim)
```
```[0 1 2 3 4 5 6 7 8 9]
(10,)
1
```
```# Nested sequence to ndarray
l2 = [range(10), range(10)]
#So it forms an array
data = np.array(l2)
print (data)
print (data.shape)
```
```[[0 1 2 3 4 5 6 7 8 9]
[0 1 2 3 4 5 6 7 8 9]]
(2, 10)
```
```'''2. zeros;ones;empty Establish'''
# np.zeros
zeros_arr = np.zeros((3, 4))
#Pay attention to tuples. Errors are often reported here
# np.ones
ones_arr = np.ones((2, 3))
# np.empty [not all zero, and some random numbers]
empty_arr = np.empty((3, 3))
# np.empty specifies the data type
empty_int_arr = np.empty((3, 3), int)
print (zeros_arr)
print ('-------------')
print (ones_arr)
print ('-------------')
print (empty_arr)
print ('-------------')
print (empty_int_arr)
```
```[[0. 0. 0. 0.]
[0. 0. 0. 0.]
[0. 0. 0. 0.]]
-------------
[[1. 1. 1.]
[1. 1. 1.]]
-------------
[[0.000e+000 0.000e+000 0.000e+000]
[0.000e+000 0.000e+000 2.174e-321]
[0.000e+000 0.000e+000 0.000e+000]]
-------------
[[0 0 0]
[0 0 0]
[0 0 0]]
```
```# np.arange()
#Creating a series of consecutive numbers is a function similar to range in python in numpy
print (np.arange(10))
```
```[0 1 2 3 4 5 6 7 8 9]
```
## Operation nddarray
##### Vectorization
```# Vector and vector operation
arr = np.array([[1, 2, 3],
[4, 5, 6]])
print ("Multiply between elements:")
#Pay attention to the operation between discernible matrices. The vector acid here is the same as the broadcast operation
print (arr * arr)
print (arr + arr)
```
```Multiply between elements:
[[ 1 4 9]
[16 25 36]]
[[ 2 4 6]
[ 8 10 12]]
```
```# Vector and scalar operations
print (1. / arr)
print (2. * arr)
```
```[[1. 0.5 0.33333333]
[0.25 0.2 0.16666667]]
[[ 2. 4. 6.]
[ 8. 10. 12.]]
```
##### Index and slice
```# One-dimensional array
arr1 = np.arange(10)
print (arr1)
print (arr1[2:5])
```
```[0 1 2 3 4 5 6 7 8 9]
[2 3 4]
```
```# Multidimensional array
arr2 = np.arange(12).reshape(3,4)
#To learn how to define multidimensional arrays, range is to form 12 random numbers, and then reshape is to form dimensions
#For example, 3.4.5 = length 4 width 5 height 3
#And points are application functions
print (arr2)
```
```[[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]]
```
```print (arr2[1])
print (arr2[0:2, 2:])
print (arr2[:, 1:3])
```
```[4 5 6 7]
[[2 3]
[6 7]]
[[ 1 2]
[ 5 6]
[ 9 10]]
```
```# Conditional index
# Find out the data after 2015 in data_arr
data_arr = np.random.rand(3,3)
print (data_arr)
year_arr = np.array([[2000, 2001, 2000],
[2005, 2002, 2009],
[2001, 2003, 2010]])
is_year_after_2005 = year_arr >= 2005
#: it will be expanded to an array of the same type
print (is_year_after_2005, is_year_after_2005.dtype)
filtered_arr = data_arr[is_year_after_2005]
filtered_arr = data_arr[year_arr >= 2005]
print (filtered_arr)
#Some statements in the middle can be deleted
#The final result is a one-dimensional array, which is very useful for data filtering
```
```[[0.61482194 0.0249229 0.28525661]
[0.05121173 0.37672803 0.86259463]
[0.22648329 0.4581513 0.18620441]]
[[False False False]
[ True False True]
[False False True]] bool
[0.05121173 0.86259463 0.18620441]
```
```# Multiple conditions &|
filtered_arr = data_arr[(year_arr <= 2005) & (year_arr % 2 == 0)]
print (filtered_arr)
```
```[0.61482194 0.28525661 0.37672803]
```
##### Transpose = = = transpose
```arr = np.random.rand(2,3)
print (arr)
print (arr.transpose())
```
```[[0.01538974 0.47573964 0.90684253]
[0.93683601 0.64306611 0.63846634]]
[[0.01538974 0.93683601]
[0.47573964 0.64306611]
[0.90684253 0.63846634]]
```
```#Conversion of high dimensional array (conversion dimension will be used in the image)
#Don't understand here!!!
arr3d = np.random.rand(2,3,4)
print (arr3d)
print ('----------------------')
print (arr3d.transpose((1,0,2))) # Transposes and definitions of multidimensional arrays do not
```
```[[[0.18074837 0.64652003 0.80527972 0.67800268]
[0.95766577 0.2498768 0.00304503 0.7058178 ]
[0.12523549 0.18796252 0.72463798 0.15352211]]
[[0.38808013 0.31075033 0.53082474 0.32254431]
[0.6861262 0.02999367 0.70980993 0.09099878]
[0.14987301 0.78237398 0.90159408 0.82897071]]]
----------------------
[[[0.18074837 0.64652003 0.80527972 0.67800268]
[0.38808013 0.31075033 0.53082474 0.32254431]]
[[0.95766577 0.2498768 0.00304503 0.7058178 ]
[0.6861262 0.02999367 0.70980993 0.09099878]]
[[0.12523549 0.18796252 0.72463798 0.15352211]
[0.14987301 0.78237398 0.90159408 0.82897071]]]
```
##### ndarray data type conversion = = = astype
```zeros_float_arr = np.zeros((3, 4), dtype=np.float64)
print (zeros_float_arr)
print (zeros_float_arr.dtype)
# astype conversion data type
zeros_int_arr = zeros_float_arr.astype(np.int32)
print (zeros_int_arr)
print (zeros_int_arr.dtype)
```
```[[0. 0. 0. 0.]
[0. 0. 0. 0.]
[0. 0. 0. 0.]]
float64
[[0 0 0 0]
[0 0 0 0]
[0 0 0 0]]
int32
```
```# loadtxt
filename = './presidential_polls.csv'
data_array = np.loadtxt(filename, # file name
delimiter=',', # Specify the element separator inside
dtype=str, # Specify data type
usecols=(0,2,3)) # Specifies the column index number to read
print (data_array, data_array.shape)
```
```[['cycle' 'type' 'matchup']
['2016' '"polls-plus"' '"Clinton vs. Trump vs. Johnson"']
['2016' '"polls-plus"' '"Clinton vs. Trump vs. Johnson"']
...
['2016' '"polls-only"' '"Clinton vs. Trump vs. Johnson"']
['2016' '"polls-only"' '"Clinton vs. Trump vs. Johnson"']
['2016' '"polls-only"' '"Clinton vs. Trump vs. Johnson"']] (10237, 3)
```
```# Loadtext, explicitly specifying the type of data in each column
filename = './presidential_polls.csv'
data_array = np.loadtxt(filename, # file name
delimiter=',', # Separator
skiprows=1,
dtype={'names':('cycle', 'type', 'matchup'),
'formats':('i4', 'S15', 'S50')}, # data type
usecols=(0,2,3)) # Specifies the column index number to read
print (data_array, data_array.shape) # The result of reading is a one-dimensional array, each element is a tuple
```
```[(2016, b'"polls-plus"', b'"Clinton vs. Trump vs. Johnson"')
(2016, b'"polls-plus"', b'"Clinton vs. Trump vs. Johnson"')
(2016, b'"polls-plus"', b'"Clinton vs. Trump vs. Johnson"') ...
(2016, b'"polls-only"', b'"Clinton vs. Trump vs. Johnson"')
(2016, b'"polls-only"', b'"Clinton vs. Trump vs. Johnson"')
(2016, b'"polls-only"', b'"Clinton vs. Trump vs. Johnson"')] (10236,)
```
# Common functions of np
## transpose
```import numpy as np
```
```arr = np.random.rand(2,3)
print (arr)
print (arr.transpose())
```
```[[0.78485041 0.88817969 0.34809014]
[0.32744286 0.97539301 0.94401872]]
[[0.78485041 0.32744286]
[0.88817969 0.97539301]
[0.34809014 0.94401872]]
```
```#Conversion of high dimensional array (conversion dimension will be used in the image)
#Don't understand here!!!
arr3d = np.random.rand(2,3,4)
print (arr3d)
print ('----------------------')
print (arr3d.transpose((1,0,2))) # Transposes and definitions of multidimensional arrays do not
```
```[[[0.28492549 0.60197236 0.45582367 0.21992479]
[0.1747163 0.69201365 0.85460359 0.65311699]
[0.62189644 0.25217555 0.16347156 0.29831219]]
[[0.42826733 0.81396165 0.187138 0.560564 ]
[0.10162186 0.66419751 0.03261665 0.06969256]
[0.55461652 0.55020586 0.50693591 0.31741807]]]
----------------------
[[[0.28492549 0.60197236 0.45582367 0.21992479]
[0.42826733 0.81396165 0.187138 0.560564 ]]
[[0.1747163 0.69201365 0.85460359 0.65311699]
[0.10162186 0.66419751 0.03261665 0.06969256]]
[[0.62189644 0.25217555 0.16347156 0.29831219]
[0.55461652 0.55020586 0.50693591 0.31741807]]]
```
## ceil and floor and rint and isnan
```arr = np.random.randn(2,3)
print (arr)
print (np.ceil(arr))
#Up nearest integer
print (np.floor(arr))
#Down nearest integer
print (np.rint(arr))
#Rounding
print (np.isnan(arr))
#Determine whether the element is NaN
#There are other functions on the notes
```
```[[ 0.262106 -1.33680008 -1.08562543]
[ 0.3990978 0.1410074 0.64278274]]
[[ 1. -1. -1.]
[ 1. 1. 1.]]
[[ 0. -2. -2.]
[ 0. 0. 0.]]
[[ 0. -1. -1.]
[ 0. 0. 1.]]
[[False False False]
[False False False]]
```
## where
```arr = np.random.randn(3,4)
print (arr)
np.where(arr > 0, 1, -1)
#(condition, output satisfied, output not satisfied)
```
```[[ 2.04688394 0.48063737 1.20876913 -0.93412937]
[-0.43427472 -1.47755481 0.36882256 -0.08943138]
[-0.2847686 0.96915893 0.32641235 0.28346922]]
array([[ 1, 1, 1, -1],
[-1, -1, 1, -1],
[-1, 1, 1, 1]])
```
## sum
```arr = np.arange(10).reshape(5,2)
print (arr)
print (np.sum(arr))
print (np.sum(arr, axis=0))
print (np.sum(arr, axis=1))
```
```[[0 1]
[2 3]
[4 5]
[6 7]
[8 9]]
45
[20 25]
[ 1 5 9 13 17]
```
## all and any
```import numpy as np
arr = np.random.randn(2,3)
print (arr)
print (np.any(arr > 0))
#One is right
print (np.all(arr > 0))
#All right
'''
·The purpose is to judge whether or not a group of data===It's kind of Boolean
·This can also be applied to pandas Medium DataFrame in
'''
```
```[[-1.020184 -0.48466272 -0.8496271 ]
[ 0.88815825 -0.81911857 0.64570539]]
True
False
'\n·The purpose is to judge whether or not a group of data===It's kind of Boolean\n·This can also be applied to pandas Medium DataFrame in\n'
```
## unique
```arr = np.array([[1, 2, 1], [2, 3, 4]])
print (arr)
print (np.unique(arr))
```
```[[1 2 1]
[2 3 4]]
[1 2 3 4]
```
Published 15 original articles, won praise 1, visited 39
Posted by V_dirt_God on Fri, 13 Mar 2020 09:23:37 -0700
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# Is it possible to reject noise traveling on the outside of a coaxial transmission line?
Say I have some ordinary coax between a receiver and an antenna. That coax will have three currents in it:
1. the desired signal
2. an exactly equal an opposite current on the inside of the shield (really, also the desired signal)
3. noise on the outside of the shield
Now if this were a balanced transmission line (not coax), I'd connect the pair of conductors to a differential amplifier, which would reject common-mode voltage. I'd be sure the impedance of each side is equal, so that common-mode currents create only common-mode voltages, so my common-mode voltage-rejecting differential amplifier effectively also rejects common-mode currents.
But this is just ordinary coax, with only one center conductor. The impedances of the shield and the center conductor are not equal. Although the signal is trapped inside the coax by the shield, can I maintain this separation of currents when the coax enters my receiver circuitry? In other words, how do I provide a reference for my receiving circuitry that isn't affected by the noise currents (3)? Or, is that not possible?
Note that I'm not asking about alternatives to coax, or other types of coax with multiple shields, etc. I'm also not very concerned about non-ideal noises introduced into the coax by an imperfect shield, etc. A practical example of the situation of concern would be that I have an antenna connected to a receiver by some coax, and I want to receive the signal from the antenna, but not the signal from the coax shield (which can also make a pretty good monopole antenna).
simulate this circuit – Schematic created using CircuitLab
• This sounds like a trick question but because I'm not all that keen on coax (twisted pair rules etc.), I'd say your input amplifier has to reference itself to the screen AND that screen also needs to be grounded or your receiver amp is wobbling up and down with screen noise. The wobbling isn't a problem if you use a transformer coupler or a diff input but i'd not be wanting to use a single ended "ground" referenced input. May 31, 2013 at 20:53
• @Andyaka sure...the question is grounded how? Depending on where, how, and how many ways things are grounded, there could be a lot of noise in the ground, or not. May 31, 2013 at 20:54
• there has never been an option for me - it's a tx coupler or diff amp because I'va fallen foul of this before May 31, 2013 at 21:05
• All you can do in this situation is to use highfrequency chokes en.wikipedia.org/wiki/Choke_(electronics) on a both ends of cable and in places where it was bended. That will reduce parasitic currents. Jul 1, 2013 at 16:43
In some applications where signal purity is critical double shielded coax (or even triple) is used. The inner shield carries the same signal as the center conductor. This makes the capacitance dramatically less and the outer shield is grounded. Essentially this provides a differential to single end signal at the receiver with high common mode noise rejection. The additional shield(s) also helps reduce radiated noise dramatically.
In a single shield system the noise on the shield is suppressed by EMI filters. Sometimes this is simply ferrite beads in series with the grounds or common mode chokes. It depends on the frequency of interest and the type of noise what the best solution is. Remember you only have to spend money and time worrying about filtering out frequencies that could hurt your system.
Here's some good illustrations from murata. And a discussion from stormcable about shielded coax noise sources/types as well as different coax shielding solutions.
EDIT: I have some time to clarify how a multi-shielded coax system works. First off I have to stress that you need to understand your EMI and how your design is sensitive to it. Often this can only be done by testing the real design as coupling paths and component performances are impossible to model completely. So in the process of finding solutions I'm providing you with a broad answer to a broad question.
The center signal benefits from some common-mode and non-common mode noise filtering due to the multiple outer shields. Anyone who has worked with coax knows they are not perfect shields and always leak. The multi-shield solution offers a good balance between both common-mode and non-common-mode EMI rejection (provided they are terminated properly for the application). Adding the differential reception provides more common-mode filtering at a loss of a little non-common mode rejection which Andy Aka asks about.
So how does combining a noisier version of the signal with a cleaner version help? This would be a case of non-common mode noise. In a multi-shielded system the non-common mode noise is much less due to the extra shield. So the noise Andy is curious about less of an issue. However if your system is hyper sensitive to this non-common mode interferer, then using the differential signal will make things worse. It would be best in this case to use the non-differential signal referenced to a filtered version of the outer ground signal, and just putting the inner shielded signal to a terminated load which closely matches the center conductor's impedance load. This is assuming your design would not benefit more by the additional common-mode noise rejection.
The added noise reduction by using the differential signal I'm referring to in the comments is common-mode noise rejection. The center conductor and the inner shield can act as a balanced line. The lines have similar impedance to ground (ideally they would be the same but that's hard to do in a coax system), so the interfering fields or currents induce the same voltage in both wires. Since the receiver responds only to the difference between the wires, it is not influenced by the induced noise voltage.
EMI is a complex subject and the internet has a lot of noisy opinions. For more details on noise and their effects and filtering them both the links I provided are excellent resources based on real EMI trouble shooting.
EDIT #2 (Here is a more specific answer after the chat discussion with Phil): In this analog low power application Phil indicates he has a 50Mhz ADC sampling 7 MHz to 30 MHz with a dynamic range of -55dBm to -110dBm with an unspecified low pass filter preceding it. When he runs an FFT he sees noise sources coming from a direction that is in the null spot of his antenna. The assumption is this must be picked up from the coax, however they could also be from other sources internal to the design or external including the antenna itself because they will receive signals even in null spots. Thus at this point his concern is strictly in-band noise sources. He needs to find the source of these methodically:
1. Replace the antenna with shielded 50 ohm load. Note the spurious levels.
2. Unplug cable put shielded 50 ohm load on ADC. Note the spurious levels.
3. Put the cable back on with 50 ohm load at the antenna site. Add ferrite at the RX end that have Material 31 characteristics for this frequency band. Keep adding (sometimes 5 or 6 can be needed) until you see the levels get close to what you've measured in #2.
4. Connect the antenna. Note the increase in levels, this is what your receiver filters (digital in this case) will have to reject.
Be careful of your dynamic range. If a single signal is higher than -55dBm it could generate what looks like spurious noise at other frequencies mixed by the AGC amplifiers when you are trying to amplify a smaller signal.
If #2 reveals unacceptable high noise then this noise source need to be isolated. It could be power supply, an internal noise source to the PCB, or being picked up inside the room. Shielding, soft ferrite sheets and ferrite beads might be a solution here depending on the source.
If #3 doesn't show improvement, try changing the position of the ferrites along the cable.
Ferrite beads can also be designed into the PCB to separate the grounds on the coax and PCB at the frequency of interest. This will cause a slight loss of power due to a reflection in the pass band, however the noise reduction will more than compensate for the power loss. The muratta link provided above has a lot of discussion on the use of PCB ferrites for noise suppression.
Sometimes as a quick experiment I insert a specially made coax barrel that breaks the ground connection in the shield. This is just 2 female coax connectors with the center pin soldered together. You'll get a power loss and some leakage, but it should quickly tell you if the shield path is a problem or not.
A note about measuring in this band. There are a lot of transient noise sources that come and go. To keep from pulling your hair out while testing use a MAX HOLD function for your FFT. Run this FFT max hold for 20-30 seconds, noting where the transients occur and how long you need to run the max hold to make sure you're seeing everything. Try to run the tests as fast as you can back to back so that the noise source doesn't have time to get turned off and confuse your results. Remember these transients are going to be changing in time, frequency, power, so monitor them closely to understand their source.
FFTS are limited in the resolution based on the input bandwidth and sample rate. Two different spurs that are close together and from different sources can look like one signal. Sometimes multiple transients on the same frequency can be hard to isolate -- you might have an internal noise at 8Mhz at -55dBm and a radiated transient spread over the top at -60dBm. You might eliminate the radiated source with a ferrite and wonder why there is still an 8Mhz noise there and think the ferrite didn't work. It's tricky time consuming business.
One further note about this setup using an FFT. Since there is only one physical low pass filter in place you can not use the FFT to zoom in on a 10Mhz spur at -90dBm while you have other stronger spurs/signals at say at 23Mhz. You will probably violate the dynamic range of the ADC and generate a false spurious noise. Spectrum analyzers have a variety of switched filters to prevent this from happening so that what you see on the screen is the dynamic range of the measurement.
• If inner shield carries same signal as centre conductor, how does a differential signal get produced that doesn't rely on noisy outer screen? Jun 1, 2013 at 9:23
• There are multiple types of noise. The differential method is good for rejecting common-mode noise. The second outer shield provides good isolation from radiated EMI noise. The combination is quite effective in most cases. For single shielded coax the shield is usually decoupled using some type of EMI filter outside the frequencies of interest (see murata link above). Jun 2, 2013 at 18:23
• That hasn't answered what I'd asked Jun 2, 2013 at 18:52
• There are different ways of filtering noise depending on what your circuit is most sensitive to. It is impossible to give you one single answer that fits every design. We need to know your design and/or EMI issues to provide specific answers. If you read the links you could probably be able to answer your own question as it pertains to your design. In general the shield is decoupled in some fashion from the ground rather than directly attached to the PCB ground. Thus you'll see schematics label "RF GND" vs "GND"...etc. Jun 3, 2013 at 4:03
• Also you're missing I4, which is the noise on the center conductor. There are components of I4 and I3 which are in phase/frequency (common-mode) and those that are not (non-common mode). How you filter them depends on what harms your design. Jun 3, 2013 at 4:11
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Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: During simulations, what to do if the regression-based model does not converge?
From "Tiago V. Pereira" To "Tiago V. Pereira" Subject Re: st: During simulations, what to do if the regression-based model does not converge? Date Wed, 20 Jun 2012 12:12:40 -0300 (BRT)
```Thank you very much, Brendan! It worked like a charm!
All the best,
Tiago
--------------
On Tue, Jun 19 2012, Tiago V. Pereira wrote:
> The most intuitive algorithm would be
>
> if [mvmeta b V, reml takes too much time to run] {
>
> break
>
> and continue as
>
> mvmeta b V, mm
> }
Try:
. mvmeta b V, reml iterate(100)
. if e(converged)==0 {
. di in red "NO CONVERGENCE, fitting method of moments model"
. mvmeta b V, mm
. }
This assumes mvmeta takes the standard iterate() option and returns
e(converged), which is the normal Stata practice.
Brendan
--
Brendan Halpin, Department of Sociology, University of Limerick,
Ireland
Tel: w +353-61-213147 f +353-61-202569 h +353-61-338562; Room F1-009 x
3147
------------
Dear statalisters,
I am running some simulations and need to run -mvmeta-. In some cases,
-mvmeta- fails to converge, and the simulation stucks at a certain point
and doesn't finish until I manually cancel it.
Example (data given below):
[simulate data]
mvmeta b V, reml
Note: using method reml
Note: using variables b1 b2
Note: 22 observations on 2 variables
initial: log likelihood = -6.0965061
rescale: log likelihood = -6.0965061
rescale eq: log likelihood = 1.5713054
Iteration 0: log likelihood = 1.5713054
Iteration 1: log likelihood = 1.5735159 (not concave)
Iteration 2: log likelihood = 1.5735286 (not concave)
Iteration 3: log likelihood = 1.5735293 (not concave)
Iteration 4: log likelihood = 1.5735293 (not concave)
Iteration 5: log likelihood = 1.5735293 (not concave)
Iteration 6: log likelihood = 1.5735293 (not concave)
Iteration 7: log likelihood = 1.5735293 (not concave)
.
.
.
.
[neverending]
. mvmeta b V, mm
I get the method-of-moments approach, which would be the best estimate if
the reml option doest not work.
The most intuitive algorithm would be
if [mvmeta b V, reml takes too much time to run] {
break
and continue as
mvmeta b V, mm
}
I have failed to find a solution to that Can you tell me if it is possible?
All the best,
Tiago
*/ ------------ example data ---------------------------------
input b1 b2 V11 V22 V12
.233144 .1990245 .13194766 .00934892 .00169556
-.28375212 .19690602 .21868578 .01441359 .00277669
.00327333 .01059937 .22549555 .01792554 .00327333
.50068392 .1364699 .16447968 .01013253 .00197968
.17374169 .24701389 .24077677 .02068396 .00466566
.50303291 .06834434 .20411383 .01143459 .00219075
-.45014787 .26331932 .14863599 .01185804 .00217135
.65685769 .33265554 .31337754 .01162247 .00226642
1.1248355 .06663273 .44921284 .02497867 .0047684
1.2422897 .2806992 .32947018 .01348576 .00254711
-.48948849 .24811046 .16104793 .01191982 .0022244
.260668 .3046892 .20409032 .01089658 .00207011
.31184814 .11819502 .19728779 .01629585 .00284334
-.34414868 .31684138 .28063149 .01408479 .00285371
-.01790121 .48026602 .21454526 .01923754 .00343415
.11029605 .06801034 .1932147 .01195094 .0023056
.04299536 .2128944 .28981034 .02116782 .00409606
-.30519299 .31617451 .24653605 .02153186 .00367891
.31049675 .10128629 .19748667 .01613425 .00304222
-.22549042 .15365738 .12865038 .00878746 .00166625
-.10485559 .15266178 .13572105 .00932605 .00179248
.62737876 .18297067 .1572774 .00949862 .00172184
end
*/ ------------ example data ---------------------------------
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
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Hello Guest it is September 22, 2020, 04:58:27 AM
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#### oldiron
• 14
« on: November 20, 2013, 07:20:48 PM »
I'm converting my Supermax knee mill from its failing Anilam control to Mach3. I'm using an Intel Atom processor, ethernet SmoothStepper, Cncdrive DG2S16035 drives and AMT10X modular encoders. The encoders have user selectable resolution and will be mounted on the servo motors. I think my math works out like this assuming the encoder set to 1000 and 200"/min rapids.
5 turns per inch lead screw x 1.5 motor to leadscrew ratio = 8 motor turns per inch x 200 inches per min = 1,500 motor turns per min x 4 pulses for quadrature output = 6,000 count per min
6,000 count per min x 1,000 encoder count = 6,000,000 pulses per min = 100,000 pulses per sec or 100 Hz.
First, is my math straight? Second, is 1000 count where I want to be or should I set the encoders to something less? 500, 512, 800 are also available settings.
#### Hood
• 25,846
• Carnoustie, Scotland
##### Re: Encoder count help please
« Reply #1 on: November 21, 2013, 03:39:48 AM »
I am not really clear what you are wanting but our numbers are similar Here is how I would calculate.
1000 count encoders will be 4000 as far as Mach is concerned so with a 1.5:1 reduction that would mean 6000 pulses per rev, you need 5 revs per inch so 30,000 steps per unit.
If your velocity is 200IPM then the frequency required would be (200 x 30,000)/60=100,000Hz or 100KHz.
The SmoothStepper will have no issues with that as it is capable of 40 times more, ie 4MHz. Your CNC4PC breakout and possibly the drives will be the limiting factor so you will have to see what their max frequency is, if however they are not capable of 100KHz then I would be surprised.
The higher the frequency you can have the better your machine should be.
Hood
#### oldiron
• 14
##### Re: Encoder count help please
« Reply #2 on: November 21, 2013, 07:37:09 AM »
Thanks Hood. I wasn't sure what the practical speed limits are. I take it the higher count the better up to the speed limit of the slowest component? I don't have the paper work in front of me but I believe the encoders will go over 2000 counts per turn.
#### Hood
• 25,846
• Carnoustie, Scotland
##### Re: Encoder count help please
« Reply #3 on: November 21, 2013, 09:05:18 AM »
Yes, if the hardware can handle it a higher count will give you better resolution. Also your servo drives will likely be able to respond quicker to position feedback and thus keep things tighter. Obviously that will depend on the update rate of the drives though.
Hood
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# Converting Moles to Atoms, ect. (HS)
• Apr 14th 2009, 01:20 PM
TNB
Converting Moles to Atoms, ect. (HS)
Stoiciometry Qs:
Ho do you
convert a certain mole of a molecule into atoms?
convert molecules to moles?
moles to molecules?
masses to liquid amounts in liters at stp?
convert atoms to moles?
masses to moles?
moles to liquid measures?
A bit clueless, I am.
If you want me to post these separately, OK.
Moliemoliemoliemoliemoliemolie....(Wondering)
• Apr 14th 2009, 03:20 PM
franckherve1
Stoiciometry Qs: Ho do you convert a certain mole of a molecule into atoms? convert
1mol=6.02*10^23 atoms
• Apr 14th 2009, 03:22 PM
franckherve1
masses to liquid amounts in liters at stp?
i mol=22.4liters
and you can't go from mass to liters
unless it's an element and you use the periodic table!
• Apr 14th 2009, 08:17 PM
rangr
No. of Molecules = No. of Moles * Avagadro Number
No. of Atoms = No. of atoms in a molecule * No. of Molecules ( from previous step)
Volume = Mass/ Density
Mass / Molecular weight = No. of Moles (Be sure to use the same units.)
• Apr 15th 2009, 05:43 AM
TNB
Waah, just failed the quiz today with a 60. I'm going to need a LOT more help.
• Apr 15th 2009, 03:30 PM
rangr
Hmmm... It would be easier if you could post the questions you got wrong or other specific questions.
Don't worry about the quiz, you will get better grades if you work on this.
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https://www.hellovaia.com/textbooks/math/essential-calculus-early-transcendentals-2nd/vector-calculus/q11e-a-find-a-function-f-such-that-bff-nabla-f-and-b-use-par/
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Suggested languages for you:
Americas
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Q11E
Expert-verified
Found in: Page 780
### Essential Calculus: Early Transcendentals
Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280
# (a) Find a function $$f$$ such that $${\bf{F}} - \nabla f$$ and (b) use part (a) to evaluate $$\int_C {\bf{F}} \cdot d{\bf{r}}$$ along the given curve$$C$$.$${\bf{F}}(x,y) - x{y^2}{\bf{i}} + {x^2}y{\bf{j}}$$.
(a)
The potential function $$f$$ of vector field $${\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}$$ is$$f(x,y) = \frac{1}{3}{x^3}{y^3}$$
(b)
The value of $$\int_C V f - dr$$ along the curve $$C$$ is $$- 9$$.
See the step by step solution
## Step 1: Given information
The given equation is $${\bf{F}}(x,y) - \left( {y{e^x} + \sin y} \right){\bf{i}} + \left( {{e^x} + x\cos y} \right){\bf{j}}$$
## Step 2: Definition of conservative vector field
Consider vector function $${\bf{r}}(t),a \le t \le b$$ with a smooth curve$$C$$. Consider $$f$$ is a differentiable function two or three variables of gradient function $$\nabla f$$ and is continuous on curve$$C$$. Then,
$$\int_c \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))$$
## Step 3: find the function f of conservative vector field
(a)
Vector field is $${\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}$$ -
Consider$$\nabla f = {f_s}(x,y){\bf{i}} + {f_y}(x,y){\bf{j}}$$.
Write the relation between the potential function $$f$$ and vector field $${\bf{F}}$$.
$$\nabla f = {\bf{F}}$$
Substitute $${f_x}(x,y){\rm{i}} + {f_y}(x,y){\bf{j}}$$ for $$\nabla {f_{{\rm{, }}}}$$,
$${\bf{F}} = {f_x}(x,y){\bf{i}} + {f_y}(x,y){\bf{j}}$$
Compare the equation $${\bf{F}} = {f_x}(x,y){\bf{i}} + {f_y}(x,y){\bf{j}}$$ with $${\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}$$.
\begin{aligned}{l}{f_x}(x,y) = {x^2}{y^3}(1)\\{f_y}(x,y)& = {x^3}{y^2}(2)\end{aligned}
Integrate equation (1) with respect to$$x$$.
\begin{align} & f(x,y)=\int{\left( {{x}^{3}}{{y}^{3}} \right)}dx\\ & ={{y}^{3}}\int{{{x}^{2}}}dx\\ & ={{y}^{3}}\left( \frac{{{s}^{3}}}{3} \right)+g(y)\left\{ \because \int{{{t}^{2}}}dt=\frac{{{t}^{3}}}{3} \right\}\\ & f(x,y)=\frac{1}{3}{{x}^{3}}{{y}^{3}}+g(y)(3)\end{align}
Apply partial differentiation with respect to y on both sides of equation (3).
\begin{align} & {{f}_{y}}(x,y)=\frac{\partial }{\partial y}\left( \frac{1}{3}{{x}^{3}}{{y}^{3}}+g(y) \right) \\ & =\frac{\partial }{\partial y}\left( \frac{1}{3}{{x}^{3}}{{y}^{3}} \right)+\frac{\partial }{dy}(g(y)) \\ & =\frac{1}{3}{{x}^{3}}\frac{\partial }{\partial y}\left( {{y}^{3}} \right)+{{g}^{\prime }}(y) \\ & =\frac{1}{3}{{x}^{3}}\left( 3{{y}^{2}} \right)+{{g}^{\prime }}(y)\left\{ \because \frac{\partial }{id}{{t}^{n}}=n{{l}^{n-1}} \right\} \\ & {{f}_{y}}(x,y)={{x}^{3}}{{y}^{2}}+{{g}^{\prime }}(y)(4) \end{align}
Compare the equations (2) and (4).
$${g^\prime }(y) = 0$$
Apply integration on both sides of equation.
\begin{align} & \int{{{g}^{\prime }}}(y)dy=\int{0}dy \\ & g(y)=K\left\{ \because \int{0}dt=K \right\} \end{align}
Here,
$$K$$ is constant.
Substitute $$K$$ for $$g(y)$$ in equation (3),
$$f(x,y) = \frac{1}{3}{x^3}{y^3} + K(5)$$
Consider the value of $$K$$ as 0 .
Substitute 0 for $$K$$ in equation (5),
$$\begin{array}{c}f(x,y) = \frac{1}{3}{x^3}{y^3} + 0\\ = \frac{1}{3}{x^3}{y^3}\end{array}$$
Thus, the potential function $$f$$ of vector field $${\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}$$ is $$f(x,y) = \frac{1}{3}{x^3}{y^3}$$.
## Step 4: Evaluate the given function along curve c
Consider vector function $${\bf{r}}(t),a \le t \le b$$ with a smooth curve$$C$$. Consider $$f$$ is a differentiable function two or three variables of gradient function $$\nabla f$$ and is continuous on curve$$C$$. Then,
$$\int_c \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))(8)$$
Write the expression of curve.
$${\bf{r}}(t) = \left\langle {{t^3} - 2t,{t^3} + 2t} \right\rangle (7)$$
Substitute 0 for $$t$$ in equation$$(7)$$,
\begin{aligned}r(0) & = \left\langle {{0^3} - 2(0),{0^3} + 2(0)} \right\rangle \\ & = \langle 0,0\rangle \end{aligned}
Hence, the initial point of curve $$(a)$$ is $$(0,0)$$.
Substitute 1 for $$t$$ in equation$$(7)$$,
\begin{aligned}{\bf{r}}(1) & = \left\langle {{1^3} - 2(1),{1^3} + 2(1)} \right\rangle \\ & = \langle 1 - 2,1 + 2\rangle \\ & = \langle - 1,3\rangle \end{aligned}
Hence, the terminal point of curve $$(b)$$ is$$( - 1,3)$$.
Find the common value $$\int_C \nabla f \cdot dr$$ by using equation$$(6)$$.
\begin{aligned} & \int_{C}{\nabla }f\cdot dr=f(-1,3)-f(0,0) \\ & =\left[ \frac{1}{3}{{(-1)}^{3}}{{(3)}^{3}} \right]-\left[ \frac{1}{3}{{(0)}^{3}}{{(0)}^{3}} \right]\ \ \ \left\{ \because f(x,y)=\frac{1}{3}{{x}^{3}}{{y}^{3}} \right\} \\ & =-9-0 \\ & =-9 \end{aligned}
Thus, the value of $$\int_C V f - dr$$ along the curve $$C$$ is $$- 9$$.
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# Algebra
posted by Kim Johnson
An artifact was found and tested for its carbon-14 content. If 74% of the orginal carbon-14 was still present, what is its probable age (to the nearest 100 years)? (Carbon-14 has a half life of 5730 years).
1. Steve
(1/2)^(t/5730) = .74
t = 5730 * log(.74)/log(0.5)
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## We Promise to Make your Math Frustrations Go Away!
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
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Equ. #6:
Equ. #7:
Equ. #8:
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Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
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Author Message
glolcom
Registered: 05.01.2005
From: U.S.
Posted: Thursday 02nd of Aug 07:47 Hi gals and guys I’m really stuck on multiply radicals calculator and would sure desperately need guidance to get me started with rational expressions, multiplying fractions and rational expressions. My math assignment is due soon. I have even thought of hiring a tutor, but they are dear . So any guidance would be greatly treasured.
nxu
Registered: 25.10.2006
From: Siberia, Russian Federation
Posted: Friday 03rd of Aug 15:12 I’m know little in multiply radicals calculator. But , it’s quite complicated to explain it. I may help you answer it but since the solution is complex, I doubt you will really understand the whole process of solving it so it’s recommended that you really have to ask someone to explain it to you in person to make the explaining clearer. Good thing is that there’s this software that can help you with your problems. It’s called Algebrator and it’s an amazing piece of program because it does not only show the answer but it also shows the process of solving it. How cool is that?
DoniilT
Registered: 27.08.2002
From:
Posted: Sunday 05th of Aug 07:49 Hello, I am a mathematics professor . I use this program whenever I get stuck at any problem. Algebrator is undoubtedly a very handy software.
Svizes
Registered: 10.03.2003
From: Slovenia
Posted: Tuesday 07th of Aug 07:42 Algebrator is the program that I have used through several algebra classes - Basic Math, College Algebra and Algebra 2. It is a truly a great piece of math software. I remember of going through difficulties with monomials, midpoint of a line and long division. I would simply type in a problem from the workbook , click on Solve – and step by step solution to my math homework. I highly recommend the program.
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# Solution: ‘The Bulldogs That Bulldogs Fight’
To minimize brain strain when thinking recursively, start simply, look for a pattern and let the pattern do the work.
## Introduction
Our April Insights puzzle explored the magical concept of recursion, a self-referencing process that can create unending complexity from simple beginnings. Recursion, which commonly appears in mathematics, computer science and linguistics, seems especially head-spinning and brain-straining when identical elements are involved, as in our first problem.
## Puzzle 1: The Fighting Bulldogs
One type of recursive sentence uses a grammatical structure called “center-embedding.” An example is the sentence, “The dog the man the maid married owned died.” You can make this somewhat easier to understand by inserting the linking words: “The dog that the man whom the maid married owned died.” Most people can understand who did what here, but it’s already getting tough. Now consider the Yale football cheer “Bulldogs bulldogs bulldogs fight fight fight.” The Rutgers University cognitive philosopher Jerry Fodor once pointed out that this is a grammatically correct triple center-embedded sentence. Your challenge is to try to understand how this cheer works as a real sentence. To make it more specific, imagine that the first set of bulldogs is red, the second brown and the third white. Try to answer the following questions:
1. Whom do the red bulldogs fight?
2. What color bulldogs do the brown bulldogs fight?
3. Which bulldogs fight the brown bulldogs?
4. What color bulldogs do the white bulldogs fight?
This puzzle can make your brain feel like the bruised and battered bulldogs so wonderfully rendered in Dan Page’s illustration. The best way to deal with recursion while minimizing brain strain is as follows:
1. Start as simply as possible.
2. Build on the recursion one element at a time, looking for a pattern.
3. Once you find the pattern, let the pattern do the work.
Let’s apply these techniques to the fighting bulldogs. The simplest possible sentence you can start with by removing the center-embedding is: Bulldogs fight.
Who do they fight? This is not specified — fight is an intransitive verb here, a verb without an object. So the meaning conveyed here is that these bulldogs fight generally or among themselves.
Following our color code, these were the red bulldogs, so the red bulldogs fight generally.
Now let’s add the center-embedding of the second set of bulldogs: Bulldogs [bulldogs fight] fight. The simplest sentence, adding in the colors, becomes: The (red) bulldogs [that the (brown) bulldogs fight] fight.
The verb fight is transitive here and does have an object. The brown bulldogs fight the red bulldogs specifically. The red bulldogs, as far as we know, don’t change their character. They continue to fight generally.
Okay, let’s add the third set: Bulldogs [bulldogs {bulldogs fight} fight] fight. With the colors, we have this sentence: The (red) bulldogs [that the (brown) bulldogs {that the (white) bulldogs fight} fight] fight.
The verb fight is again transitive here, and its object is the brown bulldogs. The white bulldogs fight the brown bulldogs specifically. Again, the description of the first two sets of bulldogs is not changed in any way.
The answers to the questions are:
1. The red bulldogs fight generally.
2. The brown bulldogs fight the red bulldogs.
3. The white bulldogs fight the brown bulldogs.
4. The white bulldogs fight the brown bulldogs.
(You may object that since the brown bulldogs fight the red bulldogs, the red bulldogs must fight them too. But this inference is not necessarily true: The red bulldogs might just keep fighting generally or among themselves, without paying any attention to the brown bulldogs, for all we know.)
As you can see, we arrive at this pattern: Group one fights generally, group two fights group one, group three fights group two, and so on. Thus, we can construct a recursive center-embedded sentence consisting of any number of bulldog groups and very easily determine who fights whom. A hypothetical group four would fight group three, group five would fight group four, and so on. Finding the pattern enables you to correctly answer the questions by offloading the work to the pattern or algorithm, without having to tax your brain’s short-term memory.
This is similar to how mathematicians deal with higher dimensions. No one can visualize the fourth dimension and beyond in the way we can visualize the first three. Not even Einstein could visualize higher dimensions. You have to find patterns or algorithms that generalize to the higher dimensions. Then you can accurately answer questions about them.
Several readers got the right answers, including boymeetswool, plouf, Nqabutho and Danielle Schaper. I enjoyed boymeetswool’s diagram and explanation of the sentence tree — it is perfectly correct. Danielle Schaper also produced some very nice diagrams based on the idea that subsets of bulldogs of each color are being referred to. This is one way to interpret the sentence, but it involves an additional assumption that is not specified explicitly. The most parsimonious approach is to assume that all the bulldogs of a given color do the same thing.
## Puzzle 2: The Lying Legislators
Our second problem is briefly summarized below. For the full description, see the original puzzle column.
On a faraway planet with perfectly logical beings, there was a legislative assembly consisting of 100 members who met every day. Some among them were pathological liars, or “pathos.” A patho was perceived as having a long nose by everybody else, including other pathos. However, a patho was completely unaware that he or she had a long nose and couldn’t hear about it from anyone because of a taboo that prevented others from pointing out or saying anything about anybody else’s long nose. Any patho who logically inferred that he or she was a patho was compelled to resign before the end of the business day. One day an alien leader not bound by the taboo addressed the assembly and said, “I perceive that at least one of you has a long nose.” Many days later, about half of the legislators resigned en masse.
Question 1 (parallel universe situation): What could have caused this strange pattern of resignations?
Let us apply the recursion-solving rules stated above and start small. Suppose that there was only one person in the assembly who had a long nose. Then the alien leader’s statement would allow that one person to infer that he or she was the one with the long nose and cause him or her to resign. Let’s say this happens on day one (more on this choice below).
If there were two people with long noses (let’s say a male, A, and a female, B), each of them would see one person with a long nose. Legislator A infers that B would see nobody with a long nose if he, A, doesn’t have one, or she would see one person if he, A, does. In the first case, B would be compelled to resign on day one. But that wouldn’t happen because she does see A’s long nose, and by an identical reasoning process, she waits to see if he resigns on day one, which would mean that she, B, did not have a long nose. Thus, neither A nor B resign on day one. On day two, both A and B realize that they have long noses and resign.
If there were three people with long noses, each would see two pathos, and they would have to wait two days to see if both resign on day two, which would mean they themselves were not pathos. When that doesn’t happen, then they would all be forced to resign on day three.
The pattern is already becoming clear. If n people in the assembly are pathos, then:
• The mass resignations will happen on day n.
• Everyone’s resignation day is one more than the number of pathos they see. Pathos see n-1 pathos, so their resignation day is n, while all nonpathos see n pathos, so their resignation day (if others don’t resign first) is n+1.
All of the pathos are hoping that the mass resignations happen on day n–1, but when that doesn’t happen, they realize they are pathos themselves and resign next day. Nonpathos see n people with long noses, and they are hoping that the resignations happen on day n, which indeed happens, confirming that they are not pathos.
So if about half of the legislators in the parallel universe resigned, then about half of them must have been pathos to begin with.
We know that the pathos’ days as legislators are numbered, but do we number them starting from zero or one? This point came up in the long discussion between plouf and Sunil Nandella.
We decided above that we should call the day the lone patho resigns “day one.” If we are thinking in terms of days after the alien leader’s visit, then the alien leader must have addressed the assembly after the day’s work was done (say, at an after-dinner speech), so a lone patho would have to resign the next day. On the other hand, if the alien leader addressed the assembly in a morning gathering, then a lone patho would be compelled to resign the same day, which would be day zero. This would change the number of days by 1. This is the basis of the very common off-by-one-error (OBOE or OB1) that can cause bugs in computer software. Whether you want to use the index 1 or 0 for the first element of an array of numbers or first step of a recursion is an arbitrary choice that has to be made initially and applied consistently. Some computer languages force you to use one or the other (usually 0), whereas others allow you to choose, using a declaration such as “Option Base 0” or “Option Base 1”). The first option is commonly used by software engineers; the second is the one that all of us use naturally when counting, and that’s what we used here.
The above parallel universe scenario can take place only if all the legislators meet all the others every day and no one is ever absent. The Quanta universe scenario (questions 2 through 6) considers what happens in a messier situation where people may be absent temporarily (lost at sea and later found), absent indefinitely (slip into a coma) or where a nonpatho can change into a patho. Specifically, the puzzle stated that:
a.) On the 35th day, a nonpatho changes into a patho irreversibly.
b.) On the 43rd day, three legislators, including one patho, were lost at sea. On the 46th evening, one of the pathos was found and rejoined the assembly the next day.
c.) On the 45th day, a legislator who was a patho slipped into a coma.
d.) On the 49th day, there was a mass resignation of a large number of legislators.
Before we answer the remaining questions, let us consider how the absence of a patho from the assembly (as happens in items b and c) should be handled: Should the remaining pathos take the absence into account and resign one day earlier, or should they stick to the original plan? Let’s examine a simple situation.
Assume that there are three pathos in the group. After day one, one of them goes missing, so that as above, only a male A and female B remain. Should they resign on day two now, because there are only two of them actually present (modified plan), or should they follow the original plan and resign on day three?
On day two, legislator A will reason as follows: “Either I have a long nose, or I do not. If I do not, then the only long-nosed person B would have seen is C, and she can assume that the alien leader was referring to C as the one with the long nose. So there is nothing to compel her to resign today. If I do have a long nose, then B will reason in the same way about me that I am reasoning about her. So I cannot be certain that I don’t have a long nose, and neither can she.” Hence neither A nor B can be fully certain that they are pathos, and so neither can logically resign on day two. The modified plan does not work.
On the other hand, everything works if they adhere to the original plan and wait until day three, imagining that C is watching everything from some virtual place and would have followed the plan too if he were present (even if he is actually dead, for all they know). All three pathos had originally seen two people with long noses, and nobody resigned on day two. So they must resign on day three.
What about the addition of a newly created patho, as happens in item a above? In this case, from the point of view of the person who has become a patho, nothing has changed because she never knew whether she was a patho anyway: She still sees the original number of n pathos. The pathos now also see n pathos, and the nonpathos see n+1. This calls for everyone except the affected person to modify their resignation plan to one day later than it originally was. The pathos’ resignation day is now n+1, and the nonpathos’ is n+2. You can see that this situation is indistinguishable from having n+1 pathos from the beginning, provided that all the legislators except the changed one know about the change, which is true in this instance.
So the principle for absences is: Stick to the original plan, and the resignation day remains the same. For additions such as the universally observable change (except to the affected person) of a nonpatho into a patho, add 1 to the resignation day.
With this in mind, let’s answer the remaining questions:
2. Given all the events described, how many pathos were in the assembly originally and how many resigned on the 49th day?
There were 48 pathos originally, and 48 resigned on the 49th day (with the base day understood as day one). The 48 pathos would have all resigned on day 48, but the addition of the new patho in plain sight to everyone except the affected legislator forced them to change the resignation day to day 49. The comatose person was the missing 49th patho. (If the base day was understood as day zero, add 1 to all the answers.)
3. Was the originally truthful legislator who became a patho on the 35th day among those who resigned?
Yes. He had always seen 48 other pathos, so his resignation day was always 49.
4. The legislator who had slipped into a coma recovered and returned to the assembly on the 50th day. He was briefed about everything that had happened. Did he resign?
Yes. When he returned, he asked for the names of the people who had resigned. He saw the name of the legislator who had changed to a patho on the 35th day. He went down the list, desperately scouring it for a name that he knew to have been that of a nonpatho as of the 44th day. He didn’t find any. He resigned.
5. What would have happened if the legislator who had been lost at sea had been found a day later?
Nothing. All the resignations would still have happened on the 49th day. Absences do not affect the set plan.
6. Why was the Truon leader’s visit necessary for the mass resignations to occur? After all, all of the legislators always knew that at least one of them had a long nose.
This question holds the key to this puzzle. This is what is known as a “common knowledge” problem. Common knowledge, in logic, is defined as the knowledge of some truth, p, in a group, such that everybody in the group knows p, they all know they know p, they all know that they all know that they know p, and so on ad infinitum. It is this common, infinitely recursive, web of knowledge that the alien leader’s words impart to the legislators. This web of recursive knowledge remains intact in the parallel universe because the legislators are present every day and are able to carry the recursive reasoning forward one step at a time. Thus, all the pathos are able to reach the same conclusion on the same day. When absences and additions are present, the original plan may need to be modified, and this can be done logically but only among the original legislators who were present at the declaration and therefore knew that everyone knew what they knew. The alien leader’s declaration also gives a common base to count the days starting on the day the declaration was made and the web of common knowledge was constructed. As the days pass, this web is recursively unraveled from the outside in, until resignation day arrives and everyone can logically infer their true nature.
The Quanta prize for this puzzle goes to plouf, who answered both puzzles correctly. Plouf’s answers for the second puzzle counted the base day as zero, so they are 1 off from the ones given here, but they are correct for the assumed base day.
I hope you enjoyed this brain twister. If you need some time to untwist, you have a few weeks before we return next month with more Insights.
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Sunday, 28 July 2013
Home Schooling Early Years ICT, part 4: Creating a spreadsheet.
Teaching how computers are good at calculations. Very impressed with how quickly DD2 absorbed the learning objectives. #blog.mindrocketnow.com #EarlyYearsICT
The second piece of software that I decided to introduce to DD2 was the spreadsheet. Like last time, I taught its use in context of a project, showing how it is a tool to entering, calculating and presenting data.
Objectives:
· Understand how spreadsheets can be used to record, manipulate and present data
· Create a spreadsheet, learn saving strategies, use one type of data presentation
1. Review treasure hunt from last time, to make sure instructions are clear and has spaces to record times for each picture. Print off two copies.
2. Challenge Mummy and DD1 to complete the treasure hunt, timing how long it takes to find each picture.
3. Use google docs to create a spreadsheet recording the time taken for each person to find each clue.
4. Calculate the total time for each person to complete the treasure hunt.
5. Plot a scatter graph of times.
7. Save, print out and present findings to Mummy.
I’d forgotten how bad Google Drive was at printing documents. DD2 had to be patient whilst I tried to figure a workaround. Thankfully, DD2 is a beautifully behaved student, and didn’t mind being patient. Still, it was a waste of both of our time to struggle with something so basic.
The treasure hunt was a roaring success. It had all the ingredients to be engaging to DD2: it was a physical activity outside; the weather was sunny and warm; we were doing this activity as a family; and DD2 was in charge. She gave us all our instructions, and our ready-set-go! By doing the hunt, I realised that our instructions weren’t precise enough – lesson learnt is to be as prescriptive as possible, and try to walk through the activity in advance.
We were exhausted when we got back home, so the rest of the lesson took a well-earned pause for lunch. After the break, entering data into the spreadsheet was the first ICT learning point. Remembering last lesson, I decided to create the structure of the spreadsheet in advance to prevent DD2 fatiguing due to typing. I’m very glad I did, because I did sense her concentration waning towards the end of the list.
The second learning point was how to communicate with the software, in other words, learning the syntax of referencing cells, and entering a formula. DD2 picked that up quite quickly, and was able to enter simple sum and average formulae. She also leant by trial and error, what happens if the syntax is wrong – it just doesn’t work, and you have to figure out where you made the mistake, i.e. debug.
Because she was doing so well, I tried showing her data entry tricks like how to auto-fill cells by dragging, and how to visually check that data ranges were correct. Even though she understood what to do, I found that her fine motor control made the mouse control difficult, until I zoomed in to 150%.
The last learning point was on how to present the data. DD2 has learnt about tally charts at school, so I thought I’d show her bar graphs. She was able to create a simple bar chart, and draw conclusions from the data – that DD1 was better at the treasure hunt than Mummy.
I found that DD2 liked being “tidy”, ensuring that after each step, the highlighted cell was always A1. It’s interesting that she wants to be tidy when learning, but is completely untidy when playing. I was very impressed, once again, with her work and her concentration level. I was much less impressed with Google Drive – it’s just not fit-for-purpose for a novice, which is such a shame because it really ought to be.
More in this series: part 3.
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1. Select the trinomial that is not a perfect square trinomial. A. x2 + 9x + 3 B. 9x2 – 6x + 1 C. 25x2 + 40x + 16 D. x2 + 8x + 16 Hint 2. Factor the binomial c2 – 625. A. (c + 25)(c – 25) B. (c – 25)2 C. prime D. (c – 125)(c + 5) Hint 3. Which of these represents factoring differences of squares? A. a2 + 2ab + b2 = (a + b)(a + b) B. a2 – b2 = (a – b)(a + b) C. a2 – 2ab + b2 = (a – b)(a – b) D. a2 + b2 = (a – b)(a + b) Hint 4. Factor the trinomial 4x2 + 12x + 9. A. (2x + 3)(3x + 2) B. (2x – 3) 2 C. (2x + 3)(2x – 3) D. (2x + 3) 2 Hint 5. The area of a square is x2 – 16x + 64 square units. Find the perimeter. A. (x – 8) units B. (4x – 32) units C. (x + 8) units D. (4x + 32) units Hint
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# Review Games Max And Joel Question Preview (ID: 2018)
### Ch 4,5,6,7. Answers? Play games to get them!
25xy-=150
a) 3
b) 7
c) 6
d) 4
What is \$13.17 divided by 7?
a) \$1.00
b) \$1.02
c) \$1.45
d) \$2.00
What is 246 dived by 6?
a) 43
b) 42
c) 40
d) 41
What is 482 dived by 4?
a) 138 r1
b) 120 r2
c) 123 r3
d) 120 r4
How many degrees are in straight angle?
a) 180
b) 185
c) 190
d) 170
What is a triangle that has three equal sides?
a) isosceles
b) obyuse
c) scalene
d) equilateral
What is the improper fraction of 8 and one eighth?
a) sixty-two eighths
b) sixty-five eighths
c) sixty-six eighths
d) sixty-three eighths
What is the mixed number of forty-eight ninths?
a) five and seven ninths
b) five and eight ninths
c) five and four ninths
d) five and three ninths
22xn=66
a) 5
b) 3
c) 4
d) 7
What is 642 divided by 6
a) 105
b) 109
c) 107
d) 106
Play Games with the Questions above at ReviewGameZone.com
To play games using the questions from above, visit ReviewGameZone.com and enter game ID number: 2018 in the upper right hand corner or click here.
TEACHERS / EDUCATORS
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C o n t a c t U s
Media and Product Ball Mill Loading Guide (Percentages are based on total volume of cylinder) NOTE: With media load at 50%, voids are created equal to 20% of cylinder volume. These voids are filled when product is loaded into the mill. Mills can be loaded by volume
Ball Mill Loading (dry milling) When charging a ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product. Charging a Dry Mill The general operation of a grinding mill is to have the product impacted between the balls as
Ball Mill Circulating Load Mineral Processing & Metallurgy
Oct 09, 2015 Calculating a grinding circuit’s circulating loads based on Screen Analysis of its slurries. Compared to %Solids or Density based Circulating load equations, a more precise method of determining grinding circuit tonnages uses the screen size distributions of the pulps instead of the dilution ratios. Pulp samples collected around the ball mill or rod mill and hydrocyclones, screen or
Ball Mill Loading (wet milling) When charging a ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product. Charging a Wet Mill The general operation of a grinding mill is to have the product impacted between the balls as
A guide to maximising ball mill circuit classification
circulating load ratio and ball mill circuit performance makes the optimisation objective seem nebulous to both plant metallurgist and designer. Some of this confusion can be eliminated by taking a step back from the overwhelming detail of sharpness of separation curves, circulating load ratios, cyclone feed pressures, vortex and apex sizes etc
High recirculating loads in Ball Mill Circuits Grinding
Why do ball mills have such high circulating loads?The ball mill grinds. The cyclones separate. But the pump neither grinds nor separates. So what is the purpose of having such a big pump to maintain high circulating loads in our grinding circuit?A recent grinding bulletin addresses this question f
Quick and Easy Black Powder Ball Mill — Skylighter, Inc.
Components for Effective, Safe Use of the Ball Mill To load the mill jar optimally, fill it half full of a dense media such as 1/2-inch lead balls. 50-caliber muzzle loading bullets will work, but they are soft and will both wear down over time and leave lead dust in whatever it is you are milling. It's better to antimony-hardened 1/2-inch
Ball Mills Industry Grinder for Mineral Processing
Jan 30, 2018 Ball Mills Capacity】 From 0.2-90 T/H Advantages】Designed for long service life, minimum maintenance, can grind and homogenize mineral ores down to the nano range, large volume of processing capacity Max Feeding size】 <25mm Discharge size】0.075-0.4mm Types】overflow ball mills, grate discharge ball mills Service】 24hrs quotation, custom made parts, processing
Ball Mill Loading (dry milling) When charging a ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product. Charging a Dry Mill The general operation of a grinding mill is to have the product impacted between the balls as
Ball Mill Loading (wet milling) When charging a ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product. Charging a Wet Mill The general operation of a grinding mill is to have the product impacted between the balls as
Advances in Mechanical Engineering 2021, Vol. 13(2) 1–12
However, the ball mill load is difficult to measure directly. Therefore, the accurate recognition of ball mill load has become the focus of researchers. In the grinding process, the vibration signal gener-ated by ball mill contains useful information related to internal load parameters, which can be used to deter-mine mill load parameters
A guide to maximising ball mill circuit classification
circulating load ratio and ball mill circuit performance makes the optimisation objective seem nebulous to both plant metallurgist and designer. Some of this confusion can be eliminated by taking a step back from the overwhelming detail of sharpness of separation curves, circulating load ratios, cyclone feed pressures, vortex and apex sizes etc
Ball Mill Parameter Selection & Calculation Power
1 Calculation of ball mill capacity. The production capacity of the ball mill is determined by the amount of material required to be ground, and it must have a certain margin when designing and selecting. There are many factors affecting the production capacity of the ball mill, in addition to the nature of the material (grain size, hardness, density, temperature and humidity), the degree of
Quick and Easy Black Powder Ball Mill — Skylighter, Inc.
Components for Effective, Safe Use of the Ball Mill To load the mill jar optimally, fill it half full of a dense media such as 1/2-inch lead balls. 50-caliber muzzle loading bullets will work, but they are soft and will both wear down over time and leave lead dust in whatever it is you are milling. It's better to antimony-hardened 1/2-inch
The grinding balls bulk weight in fully unloaded mill
Apr 11, 2017 The formula for calculating the grinding balls mass in ball mill is given below (we will consider the measurements process and calculations in more detail in our next articles). The formula for calculating the grinding balls mass in ball mill. G the grinding balls mass in ball mill, t; j the mill filling degree with grinding balls, %;
What Are the Requirements for Ball Mill Testing, Start and
Load test is carried out after the no-load test is qualified, which is divided into two stages: half-load test and full-load test. Half-load test run: the filling rate of the medium in the ball mill cylinder is about 20-30%, and appropriate materials and water are given for a continuous test run of 8-10 hours.
Vertical Ball Mill can Grind Coarse and Hard Granular
The vertical ball mill has stable performance, supports dry and wet grinding, and has unique performance for coarse and thickened materials. Vertical ball mill can be widely used for dispersion and grinding of solid or powder materials in paint, medicine, building materials, chemical industry and
Ball Mill for Sale Grinding Machine JXSC Mining
Ball mill is the key equipment for grinding materials. those grinding mills are widely used in the mining process, and it has a wide range of usage in grinding mineral or material into fine powder, such as gold, iron，zinc ore, copper, etc.. JXSC Mining produce reliable effective ball mill for long life and minimum maintenance, incorporate many of the qualities which have made us being
How to Make a Ball Mill: 12 Steps (with Pictures) wikiHow
Aug 10, 2020 Ball mills are a special instrument used to break up hard solids into a fine powder. They are similar to rock tumblers in that the instrument is a rotating container filled with heavy balls to grind the substance into powder. Ceramic material, crystalline compounds, and even some metals can be ground up using a ball mill.
Metcom Expertise Metcom Technologies
Quickly and Accurately Determine Your "Circulating Load Ratio" The typical closed ball milling circuit is shown in Figure 1. The circulating load ratio is known to be an extremely important design/operating parameter, as shown in Figure 2. Figure 1. Standard Ball Mill Circuit Arrangement . Figure 2.
Analysis of ball mill grinding operation using mill power
Mar 01, 2014 Though, there are small differences in the mill speed (N: 0.6 versus 0.7) and ball load (J: 0.5 versus 0.4), in our opinion the main factor is the ball size distribution: a single size ball charge versus a ball charge made up of balls of three different sizes. The void space in the ball charge, the relative movement of different layers of balls
Calculation Of A Ball Mill Load Versus Level
Ball Mill Load feed Page 1 of 1. Dec 03, 2010 Ball Mill Load feed. why the load of main motor of mill reduces with chocking of mill can we calculate how much load will go down by how much chocking tell me actaul calculation what is arm of garvity torrque factor for ball mill rm.
Advances in Mechanical Engineering 2021, Vol. 13(2) 1–12
However, the ball mill load is difficult to measure directly. Therefore, the accurate recognition of ball mill load has become the focus of researchers. In the grinding process, the vibration signal gener-ated by ball mill contains useful information related to internal load parameters, which can be used to deter-mine mill load parameters
Ball Mill Parameter Selection & Calculation Power
1 Calculation of ball mill capacity. The production capacity of the ball mill is determined by the amount of material required to be ground, and it must have a certain margin when designing and selecting. There are many factors affecting the production capacity of the ball mill, in addition to the nature of the material (grain size, hardness, density, temperature and humidity), the degree of
A guide to maximising ball mill circuit classification
circulating load ratio and ball mill circuit performance makes the optimisation objective seem nebulous to both plant metallurgist and designer. Some of this confusion can be eliminated by taking a step back from the overwhelming detail of sharpness of separation curves, circulating load ratios, cyclone feed pressures, vortex and apex sizes etc
Calculation Of A Ball Mill Load Versus Level
Ball Mill Load feed Page 1 of 1. Dec 03, 2010 Ball Mill Load feed. why the load of main motor of mill reduces with chocking of mill can we calculate how much load will go down by how much chocking tell me actaul calculation what is arm of garvity torrque factor for ball mill rm.
Optimization of mill performance by using
Optimization of mill performance by using online ball and pulp measurements by B. Clermont* and B. de Haas* Synopsis Ball mills are usually the largest consumers of energy within a mineral concentrator. Comminution is responsible for 50% of the total mineral processing cost. In today’s global markets, expanding mining groups are trying
The grinding balls bulk weight in fully unloaded mill
Apr 11, 2017 The formula for calculating the grinding balls mass in ball mill is given below (we will consider the measurements process and calculations in more detail in our next articles). The formula for calculating the grinding balls mass in ball mill. G the grinding balls mass in ball mill, t; j the mill filling degree with grinding balls, %;
Quick and Easy Black Powder Ball Mill — Skylighter, Inc.
Components for Effective, Safe Use of the Ball Mill To load the mill jar optimally, fill it half full of a dense media such as 1/2-inch lead balls. 50-caliber muzzle loading bullets will work, but they are soft and will both wear down over time and leave lead dust in whatever it is you are milling. It's better to antimony-hardened 1/2-inch
What Are the Requirements for Ball Mill Testing, Start and
Load test is carried out after the no-load test is qualified, which is divided into two stages: half-load test and full-load test. Half-load test run: the filling rate of the medium in the ball mill cylinder is about 20-30%, and appropriate materials and water are given for a continuous test run of 8-10 hours.
Vertical Ball Mill can Grind Coarse and Hard Granular
The vertical ball mill has stable performance, supports dry and wet grinding, and has unique performance for coarse and thickened materials. Vertical ball mill can be widely used for dispersion and grinding of solid or powder materials in paint, medicine, building materials, chemical industry and
How to Ball Mill Chemicals Safely — Skylighter, Inc.
Homemade Double-Barrel Ball Mill Ball milling replaces potentially unsafe hand grinding of chemicals and compositions. The crushing of the material is accomplished by the repeated falling of heavy balls onto it, over and over, inside the mill jar. So, it sounds like I need a ball mill.
Ball Mills an overview ScienceDirect Topics
Oleg D. Neikov, in Handbook of Non-Ferrous Metal Powders, 2009 Tumbling Ball Mills. Tumbling ball mills or ball mills are most widely used in both wet and dry systems, in batch and continuous operations, and on a small and large scale.. Grinding elements in ball mills travel at different velocities. Therefore, collision force, direction and kinetic energy between two or more elements vary
PROCESS DIAGNOSTIC STUDIES FOR CEMENT MILL
4. MILL LOAD CONTROL- LATEST CONCEPT FOR CEMENT MILL OPTIMISATION 4.1 Concept M/s Holderbank Engineering, Canada has developed a control strategy for ball mills which can maintain a mill production near optimum, with little operator intervention.
Ball Mill Load Test Methods glasvereiser.de
How To Test The Ball Mill Under The No Load Test Running . the no-load test run of the ball mill means idling without the installation of steel balls and ore materials. the idle time is generally not less than four hours, under normal circumstances it takes hours to run. the longer the test run time, the better the running-in of the main bearing and gear tooth surface.
Ball End Mills Grainger Industrial Supply
Ball end mills have a rounded nose and create a round-bottomed groove in milling tasks. Also known as ball-nose end mills, they are rotated against a workpiece to make round-bottomed slots and pockets or mill complex shapes without sharp corners. Each end mill has flutes on its cutting head that carry chips away from the workpiece to prevent
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How to get past level 26 of bcubed?
One is 676534. How do you get past level 26 of bcubed? 1st u have to go l,r,r,l,,d,d,u,l,u What level do all the starter Pokemon evolve?
What is the value of 16 cubed times 15 cubed?
16 cubed times 15 cubed equals 13824000. What is the value of 16 cubed? 16 cubed is equal to 4,096. What is the height width length of a box if it is 16 cubed?
What’s the best way to play B Cubed?
You have conquered every level! Slide the cube using the arrows on your keyboard. Your goal is to pass over every single square on your way to the final square. Checkmate! Play the classic game of strategy. You can challenge the computer, a friend, or join a match against another online player.
You are watching: How to beat level 17 on b cubed
How do you move the Golden Cube in B-Cubed?
Swipe to move the golden cube around the board and remove the cubes. Remove all of the cubes on your way to the red one. Slide the cube using the arrows on your keyboard. Your goal is to pass over every single square on your way to the final square.
One is 676534. How do you get past level 26 of bcubed? 1st u have to go l,r,r,l,,d,d,u,l,u What level do all the starter Pokemon evolve?
What does lillipup evolve into at level 16?
Lillipup evolves at level 16 into Herdierlevel 16. What level does lillpup evolve? level 16Lillipup evolves into Herdier at level 16, and then Herdier evolves into Stoutland at level 32.
How do you get to level 29 on B Cubed?
Scroll down to the bottom of the page until you see “Top Games.” Click on “B-Cubed,” which is number one on the list. Click on the “Start” button then work your way through the levels until you reach level 29. Move the yellow cube four spaces up then one to the right so that it lands on the first light blue cube.
How does B Cubed cool math games work?
B cubed cool math games is an online math game aimed at improving the spacial awareness of children. The game requires you to move a yellow cube over a series of white cubes to reach the red end cube. To complete each level, the yellow cube must move across all of the white cubes before reaching the red cube.
How do you get the Red Cube in Minecraft?
The game requires you to move a yellow cube over a series of white cubes to reach the red end cube. To complete each level, the yellow cube must move across all of the white cubes before reaching the red cube. Level 29 also contains light blue cubes that transport the yellow cube to a different area of the grid.
How does the yellow cube get to the Red Cube?
To complete each level, the yellow cube must move across all of the white cubes before reaching the red cube. Level 29 also contains light blue cubes that transport the yellow cube to a different area of the grid. Open up your Internet browser and navigate to the Cool Math-Games website.
What happens if you do not pass all units for BTEC exteded National Diploma?
However, if you have not completed the Extended Diploma (3 A level sized one) you may well have fulfilled the requirements for a Diploma (2 A level sized one), or failing that the 90 credit diploma (1.5 A level sized one). It all depends on exactly which units you have taken and which ones are compulsory for which qualifications.
See more: How Many Miles Is 2000 Ft ) Converter, 2000 Feet To Miles
How to calculate the number of cubes in a row?
Cubes From 0 3 to 6 3 0 cubed 0 3 0 × 0 × 0 0 1 cubed 1 3 1 × 1 × 1 1 2 cubed 2 3 2 × 2 × 2 8 3 cubed 3 3 3 × 3 × 3 27 4 cubed 4 3 4 × 4 × 4 64
How to solve a perfectly cubed equation in math?
We are solving for x. So, our first set of parentheses gives us x = y when we move the y over by adding it to both sides. To solve our second set of parentheses, we need to use what we know about solving quadratic equations. This particular equation requires the use of the quadratic formula to help us.
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# math
posted by .
find what fractional part the first number is of the second
15;30 18;40 12;28 50;60 thank,jessica
• math -
Just reduce each..15/30=1/2 and so on
• math -
Gggggtrsdhj
## Similar Questions
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# Generating Random Numbers According to a Probability Distribution
by Christoph Schiessl on PostgreSQL
When working with statistics, generating pseudo-random samples distributed according to a given probability distribution is a frequent requirement. For me, generating these numbers directly in PostgreSQL is very convenient and improves performance in some cases. In this article, we'll look at some of the most important continuous probability distributions and implement them as SQL functions.
## Uniform distribution
Uniform distributions are easy because we can use PostgreSQL's built-in `RANDOM()` function. `RANDOM()` returns uniformly distributed `DOUBLE PRECISION` numbers in the `0.0 <= x < 1.0` range. Adjusting that range is simply a question of adding and multiplying with appropriate values:
``````-- random double precision value in the range 1.0 <= x < 20.0
SELECT 1 + 19 * RANDOM();
``````
That being said, it is easy to encapsulate that behavior in an SQL function for later reuse:
``````-- generates `count` uniformly distributed random numbers in the range `min <= x < max`
CREATE OR REPLACE FUNCTION random_uniform(
count INTEGER DEFAULT 1,
min DOUBLE PRECISION DEFAULT 0.0,
max DOUBLE PRECISION DEFAULT 1.0
) RETURNS SETOF DOUBLE PRECISION
RETURNS NULL ON NULL INPUT AS \$\$
BEGIN
RETURN QUERY SELECT min + (max - min) * RANDOM()
FROM GENERATE_SERIES(1, count);
END;
\$\$ LANGUAGE plpgsql;
``````
The function's relative frequency-histogram (based on one million values):
## Exponential distribution
`RANDOM()` is the only way of generating random numbers supported by PostgreSQL. Therefore, the exponential distribution is more challenging because we must somehow derive these numbers from uniformly distributed numbers.
Since the cumulative distribution function of the exponential distribution is invertible, we can use a technique known as inverse transform sampling. All of this boils down to a simple mathematical expression that we can encapsulate in another SQL function:
``````-- generates `count` exponentially distributed random numbers with a given expected value
CREATE OR REPLACE FUNCTION random_exp(
count INTEGER DEFAULT 1,
mean DOUBLE PRECISION DEFAULT 1.0
) RETURNS SETOF DOUBLE PRECISION
RETURNS NULL ON NULL INPUT AS \$\$
DECLARE
u DOUBLE PRECISION;
BEGIN
WHILE count > 0 LOOP
u = RANDOM(); -- range: 0.0 <= u < 1.0
IF u != 0.0 THEN
RETURN NEXT -LN(u) * mean;
count = count - 1;
END IF;
END LOOP;
END;
\$\$ LANGUAGE plpgsql;
``````
The function's relative frequency-histogram (based on one million values):
## Normal distribution
The normal distribution is probably the most important of all continuous distributions.
We must again derive them from uniformly distributed numbers to generate normally distributed numbers. Fortunately, mathematicians have discovered many ways of doing exactly that: Central limit theorem, Box-Muller transform, Marsaglia polar method, ...
For now, I've decided to implement the Marsaglia polar method because it's reasonably simple to implement in SQL:
``````-- generates `count` normally distributed random numbers with a given mean and standard deviation
CREATE OR REPLACE FUNCTION random_normal(
count INTEGER DEFAULT 1,
mean DOUBLE PRECISION DEFAULT 0.0,
stddev DOUBLE PRECISION DEFAULT 1.0
) RETURNS SETOF DOUBLE PRECISION
RETURNS NULL ON NULL INPUT AS \$\$
DECLARE
u DOUBLE PRECISION;
v DOUBLE PRECISION;
s DOUBLE PRECISION;
BEGIN
WHILE count > 0 LOOP
u = RANDOM() * 2 - 1; -- range: -1.0 <= u < 1.0
v = RANDOM() * 2 - 1; -- range: -1.0 <= v < 1.0
s = u^2 + v^2;
IF s != 0.0 AND s < 1.0 THEN
s = SQRT(-2 * LN(s) / s);
RETURN NEXT mean + stddev * s * u;
count = count - 1;
IF count > 0 THEN
RETURN NEXT mean + stddev * s * v;
count = count - 1;
END IF;
END IF;
END LOOP;
END;
\$\$ LANGUAGE plpgsql;
``````
The function's relative frequency-histogram (based on one million values) clearly shows that the generated numbers are indeed normally distributed (with an expected value of 25 and a standard deviation of 7):
That's it for today! Thank you for reading, and see you soon!
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# Two vessels of different materials are similar in size in every respect. The same quantity of ice filled in them gets melted in 20 minutes and 40 minutes respectively. The ratio of thermal conductivities of the materials is
Arun
25757 Points
6 years ago
Q = K* A ($\theta$1 - $\theta$2)* t/L
Because A, L and temp. Difference
Hence
At constant pressure
K1/K2 = t1/t2
K1/K2 = 40/20 = 2
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## Do My Alternating Group Lab
As mentioned over, I utilized to compose an easy as well as simple math lab with only Alternating Group Nonetheless, the simpler you make your lab, the much easier it ends up being to obtain stuck at completion of it, then at the beginning. This can be extremely discouraging, and all this can occur to you since you are using Alternating Group and/or Modular Equations incorrectly.
With Modular Formulas, you are currently using the incorrect equation when you get stuck at the beginning, if not, after that you are most likely in a dead end, as well as there is no feasible escape. This will just worsen as the problem ends up being more intricate, however then there is the inquiry of how to proceed with the problem. There is no other way to correctly set about addressing this sort of math problem without being able to right away see what is taking place.
It is clear that Alternating Group and also Modular Formulas are tough to find out, and it does take practice to develop your own feeling of instinct. However when you wish to address a math trouble, you have to make use of a device, and also the devices for finding out are used when you are stuck, and also they are not made use of when you make the wrong step. This is where lab Aid Service can be found in.
As an example, what is wrong with the concern is incorrect concepts, such as obtaining a partial worth when you do not have sufficient functioning components to complete the entire work. There is a very good factor that this was wrong, and also it is a matter of reasoning, not instinct. Reasoning enables you to comply with a detailed procedure that makes sense, as well as when you make a wrong step, you are generally required to either try to move forward as well as deal with the mistake, or try to step as well as do an in reverse action.
Another instance is when the pupil does not understand an action of a process. These are both sensible failures, as well as there is no other way around them. Also when you are stuck in an area that does not permit you to make any kind of relocation, such as a triangular, it is still vital to recognize why you are stuck, to make sure that you can make a much better relocation and go from the action you are stuck at to the next location.
With this in mind, the very best method to solve a stuck scenario is to just take the advance, as opposed to trying to step. The two procedures are various in their technique, however they have some basic similarities. Nevertheless, when they are tried together, you can promptly inform which one is better at solving the problem, and you can additionally tell which one is more effective.
Let's discuss the initial instance, which associates with the Alternating Group mathematics lab. This is not also complicated, so allow's first go over how to start. Take the adhering to procedure of affixing a part to a panel to be made use of as a body. This would certainly require 3 dimensions, as well as would be something you would certainly require to attach as part of the panel.
Currently, you would have an added measurement, however that doesn't mean that you can simply keep that measurement and also go from there. When you made your primary step, you can easily forget about the measurement, and then you would have to go back and also backtrack your actions.
Nevertheless, instead of keeping in mind the additional dimension, you can use what is called a "mental faster way" to assist you remember that extra dimension. As you make your initial step, imagine yourself taking the dimension and attaching it to the component you intend to attach to, and then see just how that makes you feel when you duplicate the process.
Visualisation is an extremely powerful technique, and also is something that you need to not skip over. Visualize what it would seem like to in fact connect the component as well as have the ability to go from there, without the dimension.
Currently, let's check out the second instance. Let's take the exact same procedure as previously, but now the trainee needs to bear in mind that they are mosting likely to return one step. If you tell them that they have to return one step, yet then you eliminate the idea of needing to return one step, after that they won't know exactly how to proceed with the trouble, they will not understand where to try to find that step, and the procedure will be a mess.
Rather, use a mental faster way like the psychological layout to psychologically reveal them that they are going to move back one step. and also put them in a setting where they can progress from there. without needing to think about the missing out on an action.
## Pay Me To Do Your Alternating Group Lab
" Alternating Group - Need Help with a Math lab?" Regrettably, many pupils have had an issue understanding the concepts of straight Alternating Group. Luckily, there is a new format for linear Alternating Group that can be used to educate straight Alternating Group to pupils that fight with this idea. Trainees can make use of the lab Assist Solution to help them find out brand-new strategies in straight Alternating Group without facing a mountain of issues as well as without needing to take an examination on their ideas.
The lab Assist Service was produced in order to assist having a hard time pupils as they relocate from university and also secondary school to the university and also job market. Several pupils are not able to take care of the tension of the knowing process and can have very little success in grasping the concepts of straight Alternating Group.
The lab Aid Solution was established by the Educational Testing Service, who offers a range of various online examinations that pupils can take and practice. The Examination Help Service has aided lots of students boost their ratings and also can help you improve your scores as well. As students relocate from university as well as senior high school to the college and task market, the TTS will certainly aid make your trainees' change easier.
There are a few different manner ins which you can take advantage of the lab Assist Service. The primary way that trainees use the lab Aid Solution is through the Answer Managers, which can assist students find out techniques in straight Alternating Group, which they can use to help them do well in their programs.
There are a number of troubles that students experience when they initially use the lab Help Service. Pupils are commonly overloaded as well as do not comprehend how much time they will need to dedicate to the Solution. The Answer Managers can assist the trainees examine their principle discovering as well as help them to examine every one of the product that they have currently learned in order to be gotten ready for their following course job.
The lab Assist Solution functions similarly that a professor carries out in regards to aiding students realize the ideas of direct Alternating Group. By supplying your pupils with the devices that they require to discover the essential concepts of direct Alternating Group, you can make your trainees a lot more successful throughout their studies. As a matter of fact, the lab Help Solution is so reliable that several students have actually switched over from traditional mathematics course to the lab Aid Service.
The Job Supervisor is made to help pupils manage their research. The Task Manager can be set up to schedule just how much time the trainee has offered to complete their assigned homework. You can likewise establish a custom period, which is a terrific feature for students who have a hectic schedule or an extremely busy high school. This attribute can aid trainees avoid sensation bewildered with mathematics projects.
An additional beneficial attribute of the lab Assist Solution is the Pupil Aide. The Pupil Assistant assists trainees manage their work and also gives them a place to post their homework. The Student Aide is useful for students who do not intend to obtain overwhelmed with answering numerous inquiries.
As students get more comfortable with their tasks, they are encouraged to connect with the Job Supervisor and also the Trainee Aide to get an on the internet support group. The online support system can help pupils preserve their focus as they address their assignments.
Every one of the tasks for the lab Help Service are included in the bundle. Students can login as well as finish their designated job while having the student help readily available behind-the-scenes to help them. The lab Aid Service can be an excellent help for your students as they start to navigate the difficult college admissions and also task searching waters.
Trainees should be prepared to obtain used to their tasks as quickly as possible in order to reach their primary objective of entering the university. They need to strive sufficient to see outcomes that will certainly allow them to walk on at the next degree of their researches. Obtaining utilized to the process of finishing their jobs is extremely important.
Pupils have the ability to discover different methods to help them find out just how to make use of the lab Help Service. Knowing exactly how to use the lab Assist Solution is important to students' success in university and job application.
## Hire Someone To Take My Alternating Group Lab
Alternating Group is utilized in a lot of schools. Some instructors, nonetheless, do not use it really effectively or use it improperly. This can have an adverse effect on the pupil's knowing.
So, when appointing tasks, make use of a great Alternating Group aid service to aid you with each lab. These solutions supply a selection of valuable services, consisting of:
Jobs may need a great deal of reviewing and browsing on the computer system. This is when using a help solution can be a great benefit. It allows you to get even more work done, boost your understanding, as well as avoid a great deal of stress and anxiety.
These sorts of homework services are an amazing way to begin collaborating with the most effective sort of assistance for your requirements. Alternating Group is among one of the most tough based on grasp for students. Working with a service, you can see to it that your needs are satisfied, you are educated properly, and also you understand the material correctly.
There are numerous manner ins which you can educate yourself to function well with the class and succeed. Make use of a correct Alternating Group assistance solution to lead you and also get the job done. Alternating Group is one of the hardest classes to discover but it can be quickly understood with the best help.
Having a research service also assists to enhance the student's grades. It permits you to include extra credit rating in addition to boost your GPA. Getting extra credit report is typically a huge benefit in many universities.
Students that do not take full advantage of their Alternating Group class will certainly wind up continuing of the rest of the course. The good news is that you can do it with a fast as well as easy service. So, if you wish to continue in your course, use a great assistance service. Something to bear in mind is that if you actually wish to enhance your grade level, your course job requires to obtain done. As much as possible, you need to comprehend and work with all your problems. You can do this with an excellent help solution.
One advantage of having a research solution is that you can aid on your own. If you do not feel great in your ability to do so, then an excellent tutor will certainly be able to assist you. They will have the ability to address the troubles you face as well as help you understand them in order to get a better grade.
When you graduate from secondary school and also get in college, you will require to strive in order to stay ahead of the other trainees. That indicates that you will certainly need to strive on your research. Using an Alternating Group service can assist you get it done.
Maintaining your grades up can be tough since you usually require to study a great deal and take a great deal of tests. You do not have time to deal with your grades alone. Having a great tutor can be a fantastic assistance because they can assist you and your homework out.
An aid service can make it easier for you to manage your Alternating Group course. Furthermore, you can discover more concerning on your own and aid you be successful. Locate the most effective tutoring service as well as you will certainly be able to take your research study abilities to the following level.
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cor_diff: Test the difference between two correlations. In bruceR: Broadly Useful Convenient and Efficient R Functions
Description
Test the difference between two correlations.
Usage
`1` ```cor_diff(r1, n1, r2, n2, n = NULL, rcov = NULL) ```
Arguments
`r1, r2` Correlation coefficients (Pearson's r). `n, n1, n2` Sample sizes. `rcov` [optional] Only for nonindependent rs: `r1` is r(X,Y), `r2` is r(X,Z), then, as Y and Z are also correlated, we should also consider `rcov`: r(Y,Z)
Value
Invisibly return the p value.
Examples
```1 2 3 4 5``` ```# two independent rs (X~Y vs. Z~W) cor_diff(r1=0.20, n1=100, r2=0.45, n2=100) # two nonindependent rs (X~Y vs. X~Z, with Y and Z also correlated [rcov]) cor_diff(r1=0.20, r2=0.45, n=100, rcov=0.80) ```
bruceR documentation built on June 22, 2021, 1:06 a.m.
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# Logical Reasoning - Logical Problems - Discussion
Discussion Forum : Logical Problems - Type 1 (Q.No. 4)
Directions to Solve
Each problem consists of three statements. Based on the first two statements, the third statement may be true, false, or uncertain.
4.
Mara runs faster than Gail.
Lily runs faster than Mara.
Gail runs faster than Lily.
If the first two statements are true, the third statement is
true
false
uncertain
Explanation:
We know from the first two statements that Lily runs fastest. Therefore, the third statement must be false.
Discussion:
10 comments Page 1 of 1.
Sakshi said: 8 years ago
M>G
L>M
G>L
Which means, L>M>G.
(1)
L>M>G>
Ken said: 10 years ago
I agree with answer is B. Together with the reason.
Naresh said: 9 years ago
Explain the logic for this question? I would be easily understand by others.
Chandrima said: 8 years ago
Could anyone please tell me the logic behind this?
Rahul said: 8 years ago
Yes, its B.
Akash nevase said: 7 years ago
Hers, L is grater than M.
Daniel herman said: 7 years ago
How to solve the problem? Not getting the answer. Please explain me.
Basheer said: 7 years ago
Mara his running speed is (30 k/hr ) faster than the speed of Gail are 25 k/hr.
Now second statement Lily running faster 35k/hr than mara speed.
So Lily running fastest and the third statement false (Gail run faster than Lily).
Narender Kashyap said: 1 year ago
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# Thread: Multivariate Normality
1. ## Multivariate Normality
I have a data set I am going to use SEM on. Each of the variables seems to have acceptable skewness and kurtosis (using cutoffs of 3 and 10 on skewness index and kurtosis index respectively).
However, using Mardia's test for multivariate normality I am getting a Mardia's skewness of 15.675, with a test statistic of 1080, p=0. Mardia's kurtosis of 210, with test statistic of 7.77, p<.0001. N=410.
I know that the p-values are sensitive to sample size, so maybe p-value isn't the best to use (although these p-values seem very small) - is there a cutoff for the kurtosis and skewness coefficients themselves that I should go by?
Given that we are attempting to publish this SEM, how should I handle this. Should I try to transform some of my variables? Use an estimation method other than maximum likelihood? Or just ignore the apparent non-normality? Most of the SEM papers I've seen don't really address normality - but that doesn't mean its the right thing to do, I suppose...
2. ## Re: Multivariate Normality
Hi xralphyx,
Certainly, hypothesis testing can be influenced by sample size, though your p values seem to be considerably low. You can either try other multivariate normality test or try by checking the assumption graphically. I would recommend the latter since this way you can also get some idea about what the problem may be. In order to detect multivariate normality with a graph, plot the squared Mahalanobis Distance against the quantiles of a Chi-square distribution with the degrees of freedom equal to the number of variables. There's some info about that graph in this link. With that, you can understand more deeply the distribution of your data.
Now, you are incredibly right when you say that multivariate normality it is a important assumption in SEM and should be addressed. When there are violations, there are other models available, like PLS models, which don't require multivariate normality, as far as I recall. Also, there are corrections that can be applied to the estimators in order to improve the results. And of course, you can transform your data (yet that may affect interpretations). The way to go will depend on what you are measuring and the type of distribution you have.
Hope this helps!
3. ## Re: Multivariate Normality
Hi,
It's incredibly important to understand what your data look like (checking for skew and kurtosis, both uni- and multivariate), but at the end of the day I think it's easiest to just use test statistics that account for non-normality. A lot of programs will provide these just by clicking the appropriate options.
I do SEM in EQS and often use Robust ML (Maximum likelihood; see Satorra-Bentler, 1990) with N~400-500 and Mardia Coeff's anywhere from 3 to 200. I'm a grad student at UCLA and Peter Bentler himself (a faculty member at UCLA) has looked over manuscripts of mine in which I report these stats/approach and it gets his approval. So, for whatever that's worth. For a more formal source to cite, Ullman and Bentler (2013) report data analyzed (albeit as a descriptive example) with ML Robust when Mardia's = 238.65. Citations below.
You could play around with transformations, but as the other post mentions, they may not fix the problem anyways. And in terms of publications, unfortunately I think there's still some stigma around transformations (as though you've done something sneaky to make your data look better, which is completely unfounded). So, again, I think it's easiest just to mention that there was considerable variance among each of your variables but the data were not multivariate normally distributed, so you used robust maximum likelihood estimation (cite Satorra-Bentler, 1990)
Satorra, A., & Bentler, P. M. (1990). Model conditions for asymptotic robustness in the analysis of linear relations. Computational Statistics & Data Analysis, 10, 235-249.
Ullman, J. B., & Bentler, P. M. (2013). Structural equation modeling. In J. A. Schinka & W. F. Velicer (Eds.), Handbook of Psychology, Vol 2: Research Methods in psychology (pp. 661-690). Hoboken NJ: Wiley.
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# An artist wants to make alabaster pyramids using a block of alabaster with a volume of 576 cubic inches. She plans to make each pyramid with a square base area of 3 square inches and a height of 4 inches. At most, how many pyramids can the artist make from the block of alabaster?
2
by jazz4423
2015-02-20T13:28:28-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Pyramid
volume=base area times height times 1/3
or
length times widht times height times 1/3
base area=3
3 times height iems 1/3
height=4
3 times 3 times 1/3=4 times 1/3 times 3=4 times 1
each pyramide=4 in volume
so
x pyramids =alabaster volume
divide by pyramids
x=alabaster/pyrimds
alabaster=576
pyramids=4
divide
576/4=144
the artis can make 144 pyramids
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# Homework tips for statistics subject
When a child is having difficulty in any mathematics subject area, it is important to discover ways to help. Probability, Statistics and Calculus are areas in which even high achievers may find themselves struggling.
Classroom instruction is great for many, but for others students, a 45-minute class and surrounded by the distractions of their classmates, is simply not the ideal way to learn. If students pay at their desks they could not be able to attend to a portion of the lesson for many reasons. And since math works like it is a chain with one concept leading to the next one, a “missed link” causes subsequent content to cease to make sense.
Probability & Statistics
Many students are struggling when it comes to Probability and Statistics because of no cause or fault of their own. For all that, Probability and Statistics has its own unique terminology and rules. However, the stress experienced by Probability and Statistics students can result in a loss self-confidence.
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Calculus
Need assistance with your calculus? You are not alone! Calculus is among the most difficult subjects for many students even students who are typically high achievers. But here’s some good news when a student is acquainted with the particular “language” of calculus, it will become one of the most simple, and even most enjoyable, things for them to do!
Calculus is based on a special type of logic. Once it’s established inside students’ brains it is simple to apply to any type of problem. The problem is that some students learn best at different speeds, and in different contexts, than other.
Geometry
A lot of students struggle with geometry. Geometric shapes can’t be translated well on a blackboard. Students require a more visual imagination to understand concepts such as: lines and points; angles and parallel lines; analytic geometry that includes the coordinate plane midpoints; distance; slope and equations of lines formal proofs and deduction; polygons and triangles; an introduction to triangles; angles of triangles; right triangles pythagorean theorem; congruent triangles; similar triangular; quadrilaterals: trapezoids, parallelograms as well as rectangles, rhombuses and squares; perimeters and areascircles which are lines that connect with circles: radius, diameter, chord and tangent. Secant is the equation of a circle angles related to a circle as well as the circumference and area! Nonetheless geometry is an important part of a well-rounded education.
The tutoring programs and tutorials offered by tutors can aid students in understanding difficult math concepts and perform better in their math class. Locate a tutor in math that will provide a complete tutorial program and cement challenging math concepts with repetition. Hundreds of interactive exercises are provided by tutors to make sure that students master all the formulas and concepts covered in the review. Tutors can also assist with recommending and providing useful math worksheets. Such workbooks provide additional practice and exercises and students may eventually learn to tackle problems on their own.
The experience and the sense of accomplishment that a tutoring program will bring to the student will bring great benefits for many years to come. Not only will it assist students to achieve a top grade at school in Probability and Statistics or Calculus, but it will provide a solid base for success in the future and help with college entrance exams.
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# Show Details On-demand in Excel [Tutorial + Training Program]
Posted on April 7th, 2011 in Charts and Graphs , products , VBA Macros - 48 comments
Yesterday, we have seen a beautiful example of how showing details (like distribution) on-demand can increase the effectiveness of your reports. Today, we will learn how to do the same in Excel.
Before jumping in to the tutorial,
In this post, I have explained one technique of using charts + VBA to dynamically show details for a selected item. There are 4 other ways to do the same – viz. using cell comments, pivot charts, group / un-group feature and hyperlinks. I have made a 45 minute video training explaining all the 5 techniques in detail. Plus there an Excel workbook with all the techniques demoed. You can get both of these for \$17.
Click here to get the video training – Showing on-demand details in Excel
### How does the on-demand details chart work – demo:
This is a replica of yesterday’s chart from Amazon. When you click on any cell inside the Items + Rating table, the corresponding items review break-up is shown in the chart aside.
## Creating this chart in Excel – Step-by-step Instruction
So you are ready to learn how to do this chart? Great, grab a cup of coffee or tea and get started.
### 1. Understanding the data
This is how I have setup the source data for the chart. It has 3 columns – Item name, Reviewer ID and Rating. Each item has several ratings from several different reviewers. And our goal is to summarize all these ratings.
All this data is in the range Table1. We will use structured references [what are they?] in formulas to keep them readable.
### 2. Setting up the Item & Rating Table
The first step is to show a table with all the products we sell and their corresponding average rating. We will then add the circle indicators at the end to visually show the rating.
Calculating the averages using AVERAGEIF() formula:
The formula is quite simple. Assuming the product names are in C5:C13,
We just write =AVERAGEIF(Table1[Item],C5,Table1[Rating]) for first product’s average. Fill the rest by dragging the formula down.
Displaying Circles:
There are no star symbols in the default fonts. But we have circles – a full circle, an empty circle and a donut to indicate half-circle. These symbols are available in Wingdings 2 font. We will use an incell chart to display the circles. Assuming the rating is 2.83, we need to print 2 full circles, one donut and 2 empty circles. [related: inserting symbols in to Excel workbooks]
The formula is quite simple. Since the ratings are in D5:D13, the formula becomes,
=REPT(fullCircleSymbol,INT(D5)) & REPT(donutSymbol,(INT(D5)<>D5)+0) & REPT(emptyCircleSymbol,INT(5-D5))
Naming this grid
Now that we are done with the rating grid, let us name it – rngReviews.
### 3. Finding out which cell is selected
Now comes the macro part.
Before jumping in to the code, take a sip of that coffee. It is getting cold.
When a user selects any cell inside rngReviews, we need to findout which product it is so that we can load corresponding details.
The macro logic is quite straight forward.
1. On Worksheet_SelectionChange, check if the ActiveCell overlaps with rngReviews
2. If so,
1. findout the relative row number of ActiveCell with respect to topmost row in rngReviews (ie the position of selected cell inside rngReviews)
2. Put this value in to a cell on worksheet – say E28
The macro code can be found in the downloaded workbook. Here is an image of macro code.
### 4. Using the macro output to drive…,
We need to use the value E28 to do 2 things.
1. Highlight the corresponding row in the rngReviews using conditional formatting.
2. Findout the corresponding product using INDEX formula.
I am leaving both of these to your imagination.
### 5. Calculating Product – Rating Breakup
In order to show details for the product, we must calculate the corresponding breakup of ratings (ie how many 1 star, 2 star … 5 star reviews the product got).
I am leaving the formulas for this to your imagination. But when you are done, make sure your output looks like this:
(hint: use COUNTIFS formula).
### 6. Create a Chart to show Rating Break-up
This is the last one before we put everything together. Just follow below 5 steps.
1. Select the 3 columns – Rating type, number of reviews, total reviews and create a bar chart (not stacked bar chart). In my workbook, this data is in the range C29:E34 in the sheet “Rating Summary”.
2. Reverse the order of categories as Excel shows them upside down. For this select the vertical axis and hit CTRL+1 (or go to axis options from right click menu). Here check the “Show categories in reverse order” option. Also remove the chart legend.
3. Set both series of the chart such that they completely overlap each other [image]. Adjust the gap width to 50%. Also, adjust the order of the series from Chart’s source data options [image].
4. Remove grid lines, axis line and horizontal axis. Format the chart colors to your pink and translucent green (really!).
5. Re-size the chart, add title, add labels, remove border. You need to use dynamic titles.
### 7. Put everything together
Now is the time to put everything together and test. Move the chart close to the rating table. Test it by clicking on any value inside table.
You can also do some colorful formatting if you prefer.
Finish the coffee and show-off the chart to a colleague or boss. Bask in glory.
Note: You must enable macros to use the file.
Note2: If the file does not open on double-click, just open Excel (2007 or above) and drag the file inside to Excel.
Learn this + 4 other techniques using Video Training,
In this post, I have explained one technique of using charts + VBA to dynamically show details for a selected item. There are 4 other ways to do the same – viz. using cell comments, pivot charts, group / un-group feature and hyperlinks. I have made a 45 minute video training explaining all the 5 techniques in detail. Plus there an Excel workbook with all the techniques demoed. You can get both of these for \$17.
Click here to get the video training – Showing on-demand details in Excel
## How do you like this chart?
Ever since I learned this technique from a good friend, I have been using it in dashboards & complex models to make them more user friendly.
What about you? Did you like this technique? Where are you planning to use it? Please share your views & ideas using comments.
## More Resources to One-up your Chart Awesomeness
Want more, here is more:
Give more details by showing average and distribution [Charting Tips] Auditing Spreadsheets? – Disable Direct Editing Mode to save time [quick tip]
### 48 Responses to “Show Details On-demand in Excel [Tutorial + Training Program]”
1. Arnab says:
Hi Chandoo,
At the beginning of the week,you promised,about a dashboard,celebrating team India's victory at the WC.where is it ?
2. Javier says:
Un fantástico tutorial. Muchas gracias
3. dan l says:
I really like to use click for details in my projects.
The more and more 'complicated' workbooks I do, the more I really think that form controls kind of suck. They have their place, but they really do sort of break up the flow of the workbook. Click to explore feels very 'webish', so it's a good solution.
4. John Kyle says:
This is really cool, Chandoo. I can see how this would be a very helpful part of a dynamic Excel dashboard. Thanks for showing us, once again, how we can have very powerful Excel reporting without too much difficulty.
5. Dave says:
I don't know how to open an excel 07 Zip file. After I unzip there are a number of folders and sub folders, many with xml extensions. Where is the xlsm or xlsx fiel?
6. becha says:
chandoo....U! R! a God!
cool knowledge indeed
7. matt says:
Hello,
How do you move the distribution totals outside of the grap area? Under Lable Position when I choose to move data points "Outside_End" it only move's it outside the distribution bar, not the total. Is there another option?
Thanks
8. Fred says:
Thanks for the great lesson. other than the macro part, which I still have some issues with, how do you get the distribution count on the graph to display outside the ligher color bars?
I tried to format the lable to show outside the "lighter" color bar. However, I can only make them to show up outside the "darker" color bar and so the distribution numbers are not aligned. How do you make them all aligned to the outside of the lighter color bar?
9. Jayant Joshi says:
i am learning lot ... from you thanks a ton
10. ikkeman says:
The use of a half circle whenever the rating is not an integer is I think unnecessary. as it is, 2.01 and 2.99 will receive the same number of Circles. Using the mod formula to test for a specific fraction allows you to use different sized or filled circles to indicate relative score below integer level.
REPT(fullCircleSymbol,INT(D5)) & REPT(donutSymbol,mod(D5,1)<0.5)+0) & REPT(emptyCircleSymbol,5-int(D5)-(mod(D5,1)<0.5))
11. Ulrik says:
Thanks for a good post. It is a very nice integration of table and barchart!
It made me think ..... suppose, for the sake of saving dashboard real estate, you wanted to be able to toggle the visibility of the barcharts. You could then do the following, attaching the macro 'ToggleGroupVisibility' to 'Group 3':
Sub UpdateAfterAction()
Dim topRow As Integer
Sheet1.Shapes("Group 3").Visible = True
'Group 3 being the group with the barcharts
topRow = Range("rngReviews").Cells(1, 1).Row
[valSelItem] = ActiveCell.Row() - topRow + 1
End Sub
Sub ToggleGroupVisibilty()
Sheet1.Shapes("Group 3").Visible = Not Sheet1.Shapes("Group 3").Visible
End Sub
12. Chandoo says:
@All.. Thank you so much for your comments. I am happy you liked this.
@Arnab: I got unusually busy this week. May be next week I will put a dashboard. I have all the data I need for this + made some progress. But I need some more time.
@Dave: Just drag and drop the file in to Excel 2007.
@Matt & Fred: Very simple. First I set the labels for outside series. Then I manually assigned the labels to the inside values using formula bar. If you would like to automate this, you can use Rob Bevey's Chart Labeler - http://www.appspro.com/Utilities/ChartLabeler.htm
@Ikkeman: I agree.
@Ulrik: Very good idea 🙂
13. Cameron says:
@Ikkeman & Chandoo:
A slightly different approach on the "star ranking":
I have arranged my symbols from most filled to least filled in A1:A5
And the formula I've used:
=REPT(\$A\$1,INT(I1)) &
IF(INT(I1)=I1,"",OFFSET(\$A\$1,MAX(1,(CEILING(I1,1)-I1)*6),0))
& REPT(\$A\$5,5-CEILING(I1,1))
I imagine a different formula could be used to generate a better spread between the symbols. Currently the breaks are on .33/.34 .50/.51 .66/.67
14. Cameron says:
@Myself: Duh!
=REPT(\$A\$1,INT(O25))
&IF(INT(O25)=O25,"",OFFSET(\$A\$1,MIN(4,MAX(1,(CEILING(O25,1)-O25)*6)),0))
&REPT(\$A\$5,5-CEILING(O25,1))
The MIN is very important!
15. Cameron says:
...And now I'm amusing myself by making filled-circle Sine waves march across my spreadsheet.
I thought this blog was about PRODUCTIVITY?
16. Hui... says:
@cameron
Look how productive you'll be next time the boss asks for a chart with filled circle sin waves...!
17. ahamed rifkan says:
Hi Chandoo
I would give five stars for this lesson 🙂 and its one of great lesson I learned from you. Thank for bringing up this kind of great excels. I would say it is a attempt to think out of the box .. ...!
18. Istiyak says:
https://cid-c6f0db2b5c8a20d3.skydrive.live.com/redir.aspx?resid=C6F0DB2B5C8A20D3!104&authkey=GMtDjGMi93Q%24
This convert numeric figures to words.
19. Istiyak says:
Sorry For Inconvenience
20. Cameron says:
@Istiyak, it appears that the file or folder would have to be shared in order to work, and it's based off of Windows Live ID/registered email. Probably not going to work here since you would have to share the folder with all the emails of the people who wanted to see it, and the emails remain anonymous on this website.
Perhaps mediafire or some other file hosting option?
21. Hui... says:
@Cameron
Download the file for best results as it doesn't work in Firefox
22. Cameron says:
/This folder might not be shared with you/
-It appears that you don’t have permission to access My Documents. You might be trying to access it with a different account or might need to ask the folder’s owner for permission to access it.
Istiyak, I'd love to see this in action!
Is this in VBA or Excel Formulas?
23. Istiyak says:
@cameron and hui -thanks for ur attension. I will upload it any where else and share it with u. It is just with the help of formula. Give ur email i will mail it to u. Hey hui whats ur veiw about that one?
24. Hui... says:
@Istayak
Well done, its well put together and a good piece of work
Great for One Of's
But as a series of worksheet formulas, it lacks flexibility.
I think it is better handled as a User Defined Function as per the Microsoft link I previously sent you. Using a UDF allows it to be used anywhere and multiple times in a workbook.
25. Cameron says:
@istiyak
vertexvortex
(at)
gmail
26. Istiyak says:
@ Hui : Thanks for ur motivated Comment
@Cameron : Hi i had sent it to ur email id. Plz confirm once received.....
Regards
Istiyak
27. Hui... says:
For anyone trying to use Istiyak's links above and having problems:
2. Paste the link from Post 18 all the way to the %24
28. Dave says:
Chandoo, I'm still a bit confused, what exactly is it that I drag and drop to excel.
I am unzipping and getting quite a few files, mostly with xml extensions. I drag anyone of these to an open workbook and get the schemas but no working file.
if I drag the zip file to excel workbook, I get asci type of data in the workbook.
Thanks and sorry for being so thick on this one.
29. Hui... says:
@Dave
Try renaming the file to *.xlsx
Where the * is the filename
Now try and open in Excel
30. Rahul says:
Hi Chandoo,
Thanks again for the awesome tip. Have a question. When i overlap the two charts, both of them becomes of the colour and i get single colour bars. Anyone on this ???
31. Anushri says:
Hi Chandoo,
Thanks, but I have excel 2003 where some formula do not work. I faced similar issues in "Dynamic Dashboard" file share by you earlier.Is there any alternate for averageif and sumifs in excel 2003..will be a big help for me. I'm a beginner and do not know much about this.
Thanks
32. Hui... says:
@Anushri
AverageIfs and Sumifs can be replicated by using Sumproduct.
.
Sumifs(Range1, Criteria1, Range2, Criteria2)=Sumproduct((Range1=Criteria1)*(Range2=Criteria2))
.
Averageifs(Range1, Range1, Criteria1, Range2, Criteria2)=Sumproduct((Range1=Criteria1)*(Range2=Criteria2),SumRang1)/Sumproduct(1*(Range2=Criteria2))
33. Anushri says:
Thanks Hui!
I tried following formula in the Dynamic Dashboard CH1; All Products & YTD:
=IF(\$C\$21=\$L\$22,SUMPRODUCT((Data!\$C\$2:\$C\$141='CH1'!B24)*Data!\$G\$2:\$G\$141),IF(\$C\$21=\$M\$22,SUMPRODUCT((Data!\$C\$2:\$C\$141='CH1'!B24)*Data!\$H\$2:\$H\$141),IF(\$C\$21=\$N\$22,SUMPRODUCT((Data!\$C\$2:\$C\$141='CH1'!B24)*Data!\$I\$2:\$I\$141),SUMPRODUCT((Data!\$C\$2:\$C\$141='CH1'!B24)*Data!\$J\$2:\$J\$141))))
I pasted KPI parameter names in cells I22 to O22.
It worked fine but its cumbersome specially if I have 10 KPI parameters (here it was only 4)
Please suggest if I could have used any shortcut in Excel 2003 for this
Tx
Anushri
34. Anushri says:
Guess formula did not appear in previous post. Here is what I used
=IF(\$C\$21=\$L\$22,SUMPRODUCT((Data!\$C\$2:\$C\$141='CH1'!B24)*Data!\$G\$2:\$G\$141),IF(\$C\$21=\$M\$22,SUMPRODUCT((Data!\$C\$2:\$C\$141='CH1'!B24)*Data!\$H\$2:\$H\$141),IF(\$C\$21=\$N\$22,SUMPRODUCT((Data!\$C\$2:\$C\$141='CH1'!B24)*Data!\$I\$2:\$I\$141),SUMPRODUCT((Data!\$C\$2:\$C\$141='CH1'!B24)*Data!\$J\$2:\$J\$141))))
Please suggest if there can be a shorter way
35. Hui... says:
@Anushri
Yep, that will get messy with 10 KPI's, except that you can't nest that deep with Excel 2003 anyway.
Is this the only formula like this or is it repeated hundreds of times ?
Can you do all the Sumproducts in named ranges ?
Can you upgrade to 2010 ?
36. Hui... says:
@Anushri
You could try simplifying your formula also to:
`=SUMPRODUCT((Data!\$C\$2:\$C\$141=’CH1?!B24)*OFFSET(Data! \$G\$2:\$G\$141,,IF(\$C\$21=\$L\$22,0,IF(\$C\$21= \$M\$22,1,IF(\$C\$21=\$N\$22,2,3)))))
`You might want to check the logic just in case
37. [...] Using form controls♥ Display on-demand details in excel charts♥ Panel [...]
38. [...] Display on demand details in your excel charts [...]
39. Damian says:
Hola chandoo le escribo desde Argentina. Realmente su pagina es excelente, muchas gracias por su aoporte a la comunidad.
Saludos y muchas gracias!
Damian.
40. Jav says:
Hi Chandoo,
Have a question, I wanted to put 2 charts in same sheet, but it doesn't work.
Something have to do with this "Private Sub Worksheet_SelectionChange(ByVal Target As Range)". can u help?
41. Ahsan says:
Thanks Chandoo for Excel Tricks
42. Shaz says:
Excellent tutorial Chandoo!
This is a silly question, but when adding the data labels I am having difficulty with alignment at the end of the bar. If the data labels are for the distribution series, the "outside end" does not display outside of the "total" series. If they data labels are manually adjusted, they come out of alignment when the selection changes. So I'm wondering how you were able to get the data labels for the distribution series to display in the outside end of the Total series. Probably something real simple that I am overlooking 🙁
43. [...] Adding interactivity using click-able cells [...]
44. Anil says:
Thanks for such wonderful details on Excel!
45. swati says:
HI Chandoo,
I need your help in a coding. Below is the code:
Sub Macro1()
Sheets("P+ Initiative Charter").Select
Range("C3").Cells = "REFM"
Sheets("P+ Initiative Charter").Select
Range("C4").Cells = "RNEA"
Range("C5").Select
Sheets("Data Sheet").Select
Range("B13:E28").Select
Selection.Copy
Sheets("P+ Initiative Charter").Select
Range("B9").Select
Selection.PasteSpecial Paste:=xlPasteValues, Operation:=xlNone, SkipBlanks _
:=False, Transpose:=False
Range("B9").Select
Sheets("P+ Initiative Charter").Select
Range("C3").Cells = "REFM"
Sheets("P+ Initiative Charter").Select
Range("C4").Cells = "RASO"
Range("C5").Select
Sheets("Data Sheet").Select
Range("B2:E12").Select
Selection.Copy
Sheets("P+ Initiative Charter").Select
Range("B9").Select
Selection.PasteSpecial Paste:=xlPasteValues, Operation:=xlNone, SkipBlanks _
:=False, Transpose:=False
Range("B9").Select
End Sub
Above code not switching to another comparison of REFM RASO, it always went back to the first one and showing the same result . In the above code i have two criterias to compare:
1) REFM in cell C3 and RASO in cell C4
2) REFM in cell C3 and RASO in cell C5
So i need data with respect to the above two criterias but every time when i run the above code it always show only first one and one more thing i have more than these two criterias. Totally i have 9 critrias to check.
Please help me on this so that cell comarison goes in a correct manner. and displays appropriate results.
Thanks
46. slick back says:
Great site you have here but I was wanting to know if you knew of any user discussion forums that cover the same topics discussed here?
I'd really like to be a part of community where I can get opinions from other knowledgeable individuals that share the same interest.
If you have any recommendations, please let me know.
Kudos!
Give more details by showing average and distribution [Charting Tips] Auditing Spreadsheets? – Disable Direct Editing Mode to save time [quick tip]
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# Linear set of equations with complex numbers
How can I solve a linear set of equations with complex numbers? I haven't solved a set of equation with complex number before, so I'd like to know if there are particular rules to follow..
Thanks for any help!
-
Could you add a little more detail to your question? What exactly are you looking for? – Sam DeHority Jan 29 '13 at 15:56
You can add, subtract, multiply and divide complex numbers, and these operations satisfy the same basic rules (commutativity, associativity, distributivity, etc.) so the algorithms you learned for real numbers should work just as well here. – Brett Frankel Jan 29 '13 at 15:56
@BrettFrankel nothing else? thanks! – sunrise Jan 29 '13 at 16:14
@DoctorBatmanGod I haven't solved a set of equation with complex number before, so I'd like to know if there are particular rules to follow.. thank you! – sunrise Jan 29 '13 at 16:40
@sunrise: mention this in your question (just append the content of your comment to DoctorBatmanGod to your question). – robjohn Jan 29 '13 at 16:49
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# GMAT Test M06 Master Thread - All discussions
Author Message
GMAT Club team member
Joined: 16 Mar 2009
Posts: 115
Kudos [?]: 996 [1], given: 19
Location: Bologna, Italy
### Show Tags
15 May 2009, 02:28
1
KUDOS
Kudos [?]: 996 [1], given: 19
Current Student
Status: Up again.
Joined: 31 Oct 2010
Posts: 528
Kudos [?]: 528 [0], given: 75
Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42
### Show Tags
19 Feb 2011, 15:33
I found this test very tough. Got 18 wrong
_________________
My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html
Kudos [?]: 528 [0], given: 75
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Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 190
Kudos [?]: 15 [0], given: 12
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### Show Tags
30 May 2011, 01:11
yeah this was tough, i got 12 incorrect, q 7,8,9 where unthinkable
_________________
Cheers !!
Quant 47-Striving for 50
Verbal 34-Striving for 40
Kudos [?]: 15 [0], given: 12
Manager
Joined: 23 Jan 2011
Posts: 128
Kudos [?]: 118 [0], given: 13
### Show Tags
01 Jul 2011, 04:05
I got 14 wrong..
One question- how do I start to comment on a question in m06 that has not been discussed so far?
Kudos [?]: 118 [0], given: 13
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1964
Kudos [?]: 2050 [0], given: 376
### Show Tags
01 Jul 2011, 05:19
Chetangupta wrote:
I got 14 wrong..
One question- how do I start to comment on a question in m06 that has not been discussed so far?
Please do a search using few words from the question. It is highly likely that the question has already been discussed. Just go to the discussion and reply.
If not discussed already, go ahead and create a new thread in gmat-club-tests-59/ Please make sure you put the subject of the topic in following format:
m06q32
m07q01
m22q37
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Kudos [?]: 2050 [0], given: 376
Re: GMAT Test M06 Master Thread - All discussions [#permalink] 01 Jul 2011, 05:19
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# GMAT Test M06 Master Thread - All discussions
Moderator: Bunuel
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®.
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### Boundary and Eigenvalue Problems (Summer Semester 2018)
• Lecturer:
• Classes: Lecture (0157500), Problem class (0157600)
• Weekly hours: 4+2
This Lecture provides an introduction to boundary value problems for second order elliptic boundary problems arising in Mathematical Physics. The students should have the basics in analysis (Calculus, Lebesgue integration, ODE theory) as well as functional analysis. Moreover, some interest in partial differential equations is desirable.
Seminar Announcement
Seminar Announcement with details
Schedule
Lecture: Tuesday 8:00-9:30 SR 3.69
Friday 8:00-9:30 SR 2.67
Problem class: Friday 14:00-15:30 SR 3.69
Lecturers
Lecturer PD Dr. Rainer Mandel
Office hours: by appointment
Room -1.019 Kollegiengebäude Mathematik (20.30)
Email: rainer.mandel@kit.edu
Problem classes M.Sc. Peter Rupp
Office hours: Wednesday, 14.00-15:30
Room 3.026 Kollegiengebäude Mathematik (20.30)
Email: peter.rupp@kit.edu
The whole course deals with the linear theory of 2nd order elliptic PDEs in the framework of Sobolev spaces.
1. Introduction (examples and motivation for eigenvalue and boundary value problems)
2. Boundary value problems for ODEs
3. Boundary value problems for 2nd order elliptic PDEs
4. Eigenvalue problems for 2nd order elliptic PDEs
The 3rd part basically treats the following topics:
• Weak derivatives and weak formulation of boundary value problems
• Sobolev spaces: Poincaré's inequality, Sobolev's imbedding theorem, Extension Theorem, Trace Theorem
• Solvability of elliptic boundary value problems via Fredholm operator theory
• Qualitative properties of solutions: regularity (=smoothness) and positivity
The 4th part deals with:
• Spectral theory for compact selfadjoint operators in Hilbert spaces
• Existence of an orthonormal basis of eigenfunctions
# References
• L. C. Evans: Partial Differential Equations
• D. Gilbarg, N. Trudinger: Elliptic Partial Differential Equations of Second Order
• R. A. Adams, J. F. Fournier: Sobolev Spaces
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# Number 2021110203
### Properties of number 2021110203
Cross Sum:
Factorization:
3 * 7 * 7 * 113 * 281 * 433
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 16 (Hexadecimal):
7877b1bb
Base 32:
1s7fcdr
sin(2021110203)
0.84123675551399
cos(2021110203)
-0.54066692258016
tan(2021110203)
-1.5559242120814
ln(2021110203)
21.426912802827
lg(2021110203)
9.30558999449
sqrt(2021110203)
44956.759258203
Square(2021110203)
### Number Look Up
Look Up
2021110203 which is pronounced (two billion twenty-one million one hundred ten thousand two hundred three) is a very special figure. The cross sum of 2021110203 is 12. If you factorisate the number 2021110203 you will get these result 3 * 7 * 7 * 113 * 281 * 433. The figure 2021110203 has 48 divisors ( 1, 3, 7, 21, 49, 113, 147, 281, 339, 433, 791, 843, 1299, 1967, 2373, 3031, 5537, 5901, 9093, 13769, 16611, 21217, 31753, 41307, 48929, 63651, 95259, 121673, 146787, 222271, 342503, 365019, 666813, 851711, 1027509, 1555897, 2397521, 2555133, 4667691, 5961977, 7192563, 13749049, 17885931, 41247147, 96243343, 288730029, 673703401, 2021110203 ) whith a sum of 3181108896. The figure 2021110203 is not a prime number. The figure 2021110203 is not a fibonacci number. The number 2021110203 is not a Bell Number. The number 2021110203 is not a Catalan Number. The convertion of 2021110203 to base 2 (Binary) is 1111000011101111011000110111011. The convertion of 2021110203 to base 3 (Ternary) is 12012212002000222110. The convertion of 2021110203 to base 4 (Quaternary) is 1320131323012323. The convertion of 2021110203 to base 5 (Quintal) is 13114401011303. The convertion of 2021110203 to base 8 (Octal) is 17035730673. The convertion of 2021110203 to base 16 (Hexadecimal) is 7877b1bb. The convertion of 2021110203 to base 32 is 1s7fcdr. The sine of the number 2021110203 is 0.84123675551399. The cosine of the number 2021110203 is -0.54066692258016. The tangent of the figure 2021110203 is -1.5559242120814. The root of 2021110203 is 44956.759258203.
If you square 2021110203 you will get the following result 4084886452670701209. The natural logarithm of 2021110203 is 21.426912802827 and the decimal logarithm is 9.30558999449. that 2021110203 is very impressive figure!
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# The mean life of a sample of 60 bulbs was 650
Question:
The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.
Solution:
Given the mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours
Now we have to find the overall standard deviation
As per given criteria, in first set of samples,
Number of sample bulbs, n1=60
Standard deviation, s1=8hrs
Mean life, $\bar{x}_{1}=650$
And in second set of samples,
Number of sample bulbs, $\mathrm{n}_{2}=80$
Standard deviation, $s_{2}=7 \mathrm{hr}$ s
Mean life, $\bar{x}_{2}=660$
We know the standard deviation for combined two series is
S. D $(\sigma)=\sqrt{\frac{n_{1} s_{1}^{2}+n_{2} s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$
Substituting the corresponding values, we get
S. $D(\sigma)=\sqrt{\frac{(60)(8)^{2}+(80)(7)^{2}}{60+80}+\frac{(60 \times 80)(650-660)^{2}}{(60+80)^{2}}}$
$\mathrm{S} . \mathrm{D}(\sigma)=\sqrt{\frac{(60) 64+(80) 49}{140}+\frac{(4800)(10)^{2}}{(140)^{2}}}$
S. D $(\sigma)=\sqrt{\frac{(60) 64+(80) 49}{140}+\frac{(4800) 100}{19600}}$
On simplifying we get
S. D $(\sigma)=\sqrt{\frac{(6) 64+(8) 49}{14}+\frac{(4800) 1}{196}}$
S. D $(\sigma)=\sqrt{\frac{384+392}{14}+\frac{4800}{196}}$
S. D $(\sigma)=\sqrt{\frac{388}{7}+\frac{1200}{49}}$
S. D $(\sigma)=\sqrt{\frac{388 \times 7+1200}{49}}$
S. D $(\sigma)=\sqrt{\frac{2716+1200}{49}}$
S. D $(\sigma)=\sqrt{\frac{3916}{49}}$
Or, σ=8.9
Hence the standard deviation of the set obtained by combining the given two sets is 8.9
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Search a number
36357532564 = 2261208971329
BaseRepresentation
bin100001110111000100…
…111110101110010100
310110211211010000112011
4201313010332232110
51043430012020224
624411420034004
72424654633133
oct416704765624
9113754100464
1036357532564
1114467974381
127068092904
1335755537a2
141a8c92591a
15e2bd37394
hex87713eb94
36357532564 has 24 divisors (see below), whose sum is σ = 64700589800. Its totient is φ = 17871943680.
The previous prime is 36357532507. The next prime is 36357532619. The reversal of 36357532564 is 46523575363.
It can be written as a sum of positive squares in 4 ways, for example, as 21548164 + 36335984400 = 4642^2 + 190620^2 .
It is a congruent number.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 474052 + ... + 545380.
Almost surely, 236357532564 is an apocalyptic number.
It is an amenable number.
36357532564 is a deficient number, since it is larger than the sum of its proper divisors (28343057236).
36357532564 is a wasteful number, since it uses less digits than its factorization.
36357532564 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 73483 (or 73481 counting only the distinct ones).
The product of its digits is 6804000, while the sum is 49.
The spelling of 36357532564 in words is "thirty-six billion, three hundred fifty-seven million, five hundred thirty-two thousand, five hundred sixty-four".
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# Math Limerick
Question: Why is this a mathematical limerick?
( (12 + 144 + 20 + 3 Sqrt[4]) / 7 ) + 5*11 = 92 + 0 .
A dozen, a gross, and a score,
plus three times the square root of four, divided by seven, plus five times eleven, is nine squared and not a bit more.
---Jon Saxton (math textbook author)
Presentation Suggestions:
Challenge students to invent their own math limerick!
The Math Behind the Fact:
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Resources:
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# English to Kannada Meaning of conjugate - ಕಾಂಜುಗೇಟ್
Conjugate :
ಕಾಂಜುಗೇಟ್
ಕಾಂಜುಗೇಟ್
ಕಾಂಜುಗೇಟ್, ಲಿಂಕ್, ತುಣುಕು ಒಟ್ಟಾಗಿ, ಲಗತ್ತಿಸಬಹುದು, ಬಾರು, ತಗುಲಿಕೊಂಡಿರು, ಸೂಚಿಸುತ್ತವೆ, ಅವನತಿ
ಕಾಂಜುಗೇಟ್, ಸಾಂದರ್ಭಿಕ, ಬೇರ್ಪಡಿಸಲಾಗದ, ಸಂಬಂಧಿತ, ಸಂಪರ್ಕ, ಪಕ್ಕದ, ಸಮೀಪದ, ಪ್ರತಿಸ್ಪರ್ಧಿ
ಸಂಯುಕ್ತವಾಗಿದೆಸಂಯೋಜಿತಸಂಯುಕ್ತಗಳನ್ನುಸಂಯುಕ್ತಗಳ
Definitions of conjugate in English
Noun(1) a mixture of two partially miscible liquids A and B produces two conjugate solutions: one of A in B and another of B in A
Verb(1) unite chemically so that the product is easily broken down into the original compounds(2) add inflections showing person, number, gender, tense, aspect, etc.(3) undergo conjugation
Adjective(1) joined together especially in a pair or pairs(2) (of a pinnate leaflet(3) formed by the union of two compounds(4) of an organic compound; containing two or more double bonds each separated from the other by a single bond
Examples of conjugate in English
(1) And don't worry, even French four students occasionally forget how to conjugate verbs.(2) The quartic in y must factor into two quadratics with real coefficients, since any complex roots must occur in conjugate pairs.(3) Geometry optimization was performed by the use of steepest descent and conjugate gradient algorithms.(4) Oh yeah, I'm trying to learn how to conjugate verbs in Japanese now.(5) The principles governing the seismic behavior of structures are the conjugate laws of equilibrium and compatibility, and force-displacement relationships of structural components.(6) The most common buffers are mixtures of weak acids and their conjugate bases.(7) The energy of the final ÔÇÿannealedÔÇÖ structure was then minimized using the conjugate gradient algorithm.(8) In other words, the term pK a is that pH at which an equivalent distribution of acid and conjugate base (or base and conjugate acid) exists in solution.(9) Of course, one can change the pH of the buffer by selecting other concentrations of acid and conjugate base, but the range of pH values over which a given buffer functions most effectively are close to the pK a of the acid.(10) The protein concentration of the protein conjugate and the degree of labeling were calculated from the following equations according to the instructions of the manufacturer.(11) One secondary terminal is connected directly to the spark plug of the parent cylinder while the other is connected to the second spark plug of the conjugate cylinder.(12) Compounds containing this group are enols, and their conjugate bases - the C = COH anion - are enolates.(13) The second tube was stained with the secondary conjugate alone and served as a measure of background fluorescence.(14) Its chromophore structure is also very simple: the conjugate base of p-hydroxythiocinnamate.(15) The conjugate phrase, ÔÇÿoperates to a significant extent for the benefitÔÇÖ, directs attention to certain features of the Trust.(16) We administered a solution containing bioactive rhodamine nerve growth factor conjugate via pressure injection and monitored the dispersion in the striatal region of the coronal slices.
Related Phrases of conjugate
(1) conjugate base ::
ಕಾಂಜುಗೇಟ್ ಬೇಸ್
(2) complex conjugate ::
ಸಂಕೀರ್ಣ ಕಾಂಜುಗೇಟ್
(3) conjugate acid ::
ಕಾಂಜುಗೇಟ್ ಆಮ್ಲ
Synonyms
Adjective
1. conjugated ::
ಸಂಯೋಜಿತ
Different Forms
conjugate, conjugated, conjugates, conjugating
English to Kannada Dictionary: conjugate
Meaning and definitions of conjugate, translation in Kannada language for conjugate with similar and opposite words. Also find spoken pronunciation of conjugate in Kannada and in English language.
Tags for the entry 'conjugate'
What conjugate means in Kannada, conjugate meaning in Kannada, conjugate definition, examples and pronunciation of conjugate in Kannada language.
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Order Of Operations Worksheets With Answers. Order of Operations With Hints Worksheets. Parentheses Exponents Multiplication Division Addition Subtraction 1. Knowing how to solve order of operations problems is a critical prerequisite for algebra and using PEMDAS as your mental cue to get the order of operations. 2 x 18 7 2.
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You can also find order of operations worksheets with negative numbers and order of operations worksheets with comparisons at these other worksheet pages. These Order of Operations Worksheets are a great resource for children in Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade and 5th Grade. Add subtract multiply divide 6 terms 62. 15 x 18 12. If you are looking for order of operations worksheets that test knowledge of the PEMDAS rules these math worksheets are a good start. Many many more variations to choose from all free and ready to print math worksheet PDFs.
### Our Order of Operations Worksheets are free to download easy to use and very flexible.
PEMDAS is a mnemonic for Parentheses Exponents Multiplication Division Addition and Subtraction. Parentheses Exponents Multiplication Division Addition Subtraction 1. The printable order of operations worksheets below will get you mastering your PEMDAS skills in no time. 2-Step Order of Operations Worksheet 1. Print Worksheet 1 of 6 with answers on the second page of the PDF. Sample Grade 4 Order of Operations Worksheet.
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Order of Operations Without Exponents 1. Our Order of Operations Worksheets are free to download easy to use and very flexible. Order Of Operations Worksheet 4th Grade. Knowing how to solve order of operations problems is a critical prerequisite for algebra and using PEMDAS as your mental cue to get the order of operations. Order of Operations Without Exponents 2.
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PEMDAS is a mnemonic for Parentheses Exponents Multiplication Division Addition and Subtraction. Free Math Help Grade 11. Add subtract multiply 6 terms 321 x 6- 24-4 x 2. All of our worksheets come with answer keys and are available as an individual order of operations worksheet PDF for easy downloading and printing. Order of Operations With Exponents Worksheet1.
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Add subtract multiply 6 terms 321 x 6- 24-4 x 2. You can also find order of operations worksheets with negative numbers and order of operations worksheets with comparisons at these other worksheet pages. Order of Operations Worksheets. 15 x 18 12. 2 of PDF Worksheet 1.
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2-Step Order of Operations with Whole Numbers 3-Step Order of Operations with Whole Numbers 4-Step Order of Operations with Whole Numbers 5-Step Order of Operations with Whole Numbers 6-Step Order of Operations with Whole. Order of Operations Pemdas Worksheet 8. Order of operations help with parenthesis. 10 9 x 24. Order of Operations Without Exponents 2.
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02092019 The Corbettmaths Practice Questions on the Order of Operations. Print Worksheet 1 of 6 with answers on the second page of the PDF. The Order Operations With from Order Of Operations Worksheet Answers source. Add subtract multiply 6 terms 321 x 6- 24-4 x 2. No need to create an account.
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Order of Operations Worksheets. Order of operations-Missing Operation Worksheet1. Science Tech Math Science Math Social Sciences Computer Science Animals. The Order Operations With from Order Of Operations Worksheet Answers source. You can also find order of operations worksheets with negative numbers and order of operations worksheets with comparisons at these other worksheet pages.
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Add subtract multiply 6 terms 321 x 6- 24-4 x 2. 2 - 3 x 3 6 x 2. Order of Operations Pemdas Worksheet 7. Order of Operations Without Exponents 1. Add subtract multiply divide.
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03092019 Order of Operations worksheets. No need to create an account. Order of Operations Pemdas Worksheet 5. This is the order in which operations are applied to solve more complex math problems that have multiple terms and multiple operations. 3 9 3.
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Worksheet 1 of 6 Answers on Pg. No need to create an account. 10 9 x 24. 15 x 18 12. Add subtract multiply divide 6 terms 62.
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Order of Operations Without Exponents 1. They also feature Prodigy Epics your students will love. Order of operations-Missing Operation Worksheet1. 16 x 7 x 15 11 17 7. Order of operations help with parenthesis.
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GEMS is a foolproof order of operations strategy where G stands for Groupings. Order Of Operations Worksheet 4th Grade. If you are looking for order of operations worksheets that test knowledge of the PEMDAS rules these math worksheets are a good start. These Order of Operations Worksheets are a great resource for children in Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade and 5th Grade. Order of operations-Missing Operation Worksheet1.
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Order of Operations With Hints Worksheets. Many many more variations to choose from all free and ready to print math worksheet PDFs. 8 x 4 9 9 18 4. All of our worksheets come with answer keys and are available as an individual order of operations worksheet PDF for easy downloading and printing. Add subtract multiply 6 terms 321 x 6- 24-4 x 2.
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The printable order of operations worksheets below will get you mastering your PEMDAS skills in no time. Order of Operations With Exponents Worksheet1. All of our worksheets come with answer keys and are available as an individual order of operations worksheet PDF for easy downloading and printing. Evaluating Numerical Expressions with Exponents. Using order of operations activity worksheets is the best way to consolidate knowledge on the topicOnce youve explained the method to the class set some questions from the worksheet for pupils to attemptYou can either go through a few together or set the questions as independent workOnce everyones completed the order of operations activity worksheets youll be able to go through answers.
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Order of Operations With Exponents Worksheet 2. Order of Operations Pemdas Worksheet 7. 2-Step Order of Operations Worksheet 1. 2 of PDF Worksheet 1. 16 x 7 x 15 11 17 7.
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Order of Operations Worksheets. Order of Operations Without Exponents 2. 2 with 8 2 11. Order of Operations With Exponents Worksheet 2. Order of operations worksheets pdf.
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8 x 4 9 9 18 4. GEMS is a foolproof order of operations strategy where G stands for Groupings. All of our worksheets come with answer keys and are available as an individual order of operations worksheet PDF for easy downloading and printing. Add subtract multiply 6 terms 321 x 6- 24-4 x 2. Free Math Games For Kids.
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Order of Operations Pemdas Worksheet 5. Add subtract multiply divide. Order of operations-Missing Operation Worksheet2. Sample Grade 4 Order of Operations Worksheet. Evaluating Numerical Expressions with Exponents.
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This is the order in which operations are applied to solve more complex math problems that have multiple terms and multiple operations. Order of Operations With Exponents Worksheet1. Knowing how to solve order of operations problems is a critical prerequisite for algebra and using PEMDAS as your mental cue to get the order of operations. They also feature Prodigy Epics your students will love. Order of Operations Pemdas Worksheet 7.
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The limits of friction
March 29, 2018, University of Konstanz
Friction is created when two surfaces slide one on top of the other. Since this consumes additional energy, this so-called sliding friction is considered an irksome yet inevitable aspect of dynamic processes. However, to set a stationary object in motion, its static friction must be overcome first. In collaboration with their Italian colleagues, researchers from the University of Konstanz have demonstrated how to entirely suppress static friction between two surfaces. This means that even a minuscule force suffices to set objects in motion. Especially in micromechanical parts, where only small forces are at play, a vanishing static friction can lead to hugely improved levels of efficiency. These findings have been published in the current edition of the online journal Physical Review X (PRX).
To move a block of wood across a table one needs to pull it. When Leonardo da Vinci examined this deceptively simple relationship systematically more than 500 years ago, he discovered the basic laws of sliding friction. Since sliding friction usually generates heat, one must pull constantly on the chunk of wood to make up for frictional losses. However, to generate motion in the first place, it is not sliding friction but static friction that must be overcome. Static friction is typically larger than sliding friction and a result of the atomic structure of the contact surfaces locking into place. The surfaces can only free themselves and move against each other once the applied force has reached adequate levels.
Working with physicists from the Universities of Milan and Trieste, a University of Konstanz working group led by Professor Clemens Bechinger was able to conduct experiments and numerical simulations confirming a prediction made by the physicist Serge Aubry in the 1980s: He postulated that, if the lattice spacing between particles in one surface were to differ slightly from the lattice spacing in the other, friction between the two surfaces should disappear entirely. This is even expected to apply if the two surfaces are pressed together. In practical terms, this would mean that a randomly small force would suffice to move a chunk of wood weighing tons across a surface.
This effect can be observed particularly well in ideal contacts, where both surfaces are perfectly flat against each other. It is these kinds of surfaces that Clemens Bechinger and his team were able to create in a model system: Using laser beams and glass spheres in the mircometre range, so-called colloids, they were able to create a two-dimensional model of two surfaces rubbing against each other. Since the electrically charged spheres repel each other, they position themselves in a periodically ordered flat layer. This colloid monolayer forms one of the two surfaces. The researchers created the second beneath the layer of colloids by using three . As a result of their superposition a light crystal forms, which is a sort of optical egg crate with recesses and ridges. "In comparison to real surfaces, these optical surfaces have the added advantage of being entirely transparent, which means that we can directly observe the processes at work between them using a microscope," says Thorsten Brazda, the doctoral researcher who conducted the experiments in Bechinger's group for his doctoral thesis.
While Aubry restricted his prediction to one-dimensional contacts at zero-point temperatures, the research collaboration was able to prove that extended, two-dimensional contacts at room temperature can also be put in motion without static friction. "We were able to turn Aubry's artificial one-dimensional set-up into a realistic situation and demonstrate that his idea remains valid in two-dimensional systems and at finite temperatures," Clemens Bechinger says.
Observing particle movements directly also allowed the researchers to understand the disappearance of static friction between the colloid monolayer and the light crystal: It turns out that the colloidal monolayer twists ever so slightly in relation to the optical grid. That way, the particles do not lock on to the recesses of the substrate, which they would not find it easy to escape from. Instead, some of them position themselves around the ridges. If external force is applied, these particles do not have to escape from the recesses but are free to move straight away as soon as a minimum amount of is exerted. Static friction disappears.
These results, which are in excellent agreement with the numerical simulations performed by the Italian team, show that static friction cannot only be suppressed, but also generated as desired if the contact pressure between the two surfaces is increased. This is important insofar as static friction – in contrast to sliding – is often a desired phenomenon. It enables us to grasp objects safely and ensures that wheels have sufficient grip. This way of varying creates new opportunities for moving objects easily across surfaces and to lock them into place safely. This would be of immense advantage in micro- and nanomechanical gearboxes or couplings, since, here, typically only very small forces are at play.
More information: T. Brazda et al. Experimental Observation of the Aubry Transition in Two-Dimensional Colloidal Monolayers, Physical Review X (2018). DOI: 10.1103/PhysRevX.8.011050
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# Thread: Any other solution to this inequality ?
1. ## Any other solution to this inequality ?
Hi everyone !
I found that inequality that i proved , but i wanted another solution .
Show that for all a,b and c positive real numbers , the following inequality holds:
$\sum_{cyc}\frac{a}{(b+c)^2}\geq \frac{9}{4(a+b+c)}$
And please , don't tell me it is pre-calculus !
Anybody with another solution ?
(if you need mine , say it )
Moderator edit: Moved from Pre-algebra and Algebra. Closed until OP reads and follows the subforum rules for posting here.
2. Maybe it's a matter of notation in different regions, bet personally I am not exactly sure of summation
$\displaystyle $\sum\limits_{cyc} {}$$
3. Originally Posted by Pranas
Maybe it's a matter of notation in different regions, bet personally I am not exactly sure of summation
$\displaystyle $\sum\limits_{cyc} {}$$
cyc=cyclic
4. Can i give a hint ?
5. I still don't understand what $\displaystyle\sum_{cyc}\frac{a}{(b+c)^2}$ is. Is it $\displaystyle\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\ frac{c}{(a+b)^2}$?
6. Yes it is !
7. Well , you must use both Cauchy-Schwartz and Nesbitt inequalities .
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https://www.shaalaa.com/question-bank-solutions/particle-nature-light-photon-monochromatic-light-wavelength-6328-nm-produced-helium-neon-laser-power-emitted-942-mw_11574
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# Solution - Monochromatic Light of Wavelength 632.8 Nm is Produced by A Helium-neon Laser. the Power Emitted is 9.42 Mw. - Particle Nature of Light: the Photon
ConceptParticle Nature of Light: the Photon
#### Question
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
#### Solution
Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 × 10−9 m
Power emitted by the laser, P = 9.42 mW = 9.42 × 10−3 W
Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
Mass of a hydrogen atom, m = 1.66 × 10−27 kg
(a)The energy of each photon is given as:
E = (hc)/lambda
= (6.626 xx 10^(-34) xx 3xx 10^8)/632.8 xx 10^(-9) = 3.141 xx 10^(-19) J
The momentum of each photon is given as:
P = h/lambda
= (6.626 xx 10^(-34))/632.8 = 1.047 xx 10^(-27) kg ms^(-1)
(b)Number of photons arriving per second, at a target irradiated by the beam = n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence, the equation for power can be written as:
P = nE
:. n = P/E
= (9.42 xx 10^(-3))/3.141 xx 10^(-19) ~~ 3 xx 10^16 "photon/s"
(c)Momentum of the hydrogen atom is the same as the momentum of the photon,
p = 1.047 xx 10^(-27) kg ms^(-)
Momentum is given as:
p = mv
Where,
v = Speed of the hydrogen atom
:. v = p/m
= (1.047 xx 10^(-27))/(1.66 xx 10^(-27)) = 0.621 "m/s"
Is there an error in this question or solution?
#### APPEARS IN
NCERT Physics Textbook for Class 12 Part 2
Chapter 11: Dual Nature of Radiation and Matter
Q: 4 | Page no. 407
#### Reference Material
Solution for question: Monochromatic Light of Wavelength 632.8 Nm is Produced by A Helium-neon Laser. the Power Emitted is 9.42 Mw. concept: Particle Nature of Light: the Photon. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)
S
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# Problem of the Month: Once Upon a Time
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2 clock. Their task involves counting up and understanding how a clock measures time. At Level B, the students are asked to convert their age from years into seasons, months, and weeks. Students are also asked to determine what day number the current date is in the year. In Level C, students are given a problem that requires an understanding of divisibility and may be determined by using knowledge of relatively prime factors. In Level D, the student is presented with a problem that involves three different sized alarm clocks that ring at varied intervals. The task is to determine if or when the three clocks chime simultaneously. Some students may use an understanding of modular division to solve the problem. In Level E, students are asked to determine the times in a day that the clock hands form an angle of 48 degrees. Mathematical Concepts This POM involves cyclical patterns and measurement. The most common cyclical pattern is time measurement. Time measurement and circles are intertwined. The fact that circles have 360 degrees has its roots in the original calendars that determined a year was 360 days. Cyclical relationships are often represented by circles and their component parts, such as angles and sectors. Cyclical relationships are common throughout mathematics with links between number, geometry, algebra, trigonometry, and measurement.
3 Problem of the Month Level A: When it is four o clock, how many minutes must pass before the big hand (minute hand) gets to where the little hand (hour hand) was at four o clock? How did you figure it out? When it is six-thirty, how many minutes must pass before the big hand (minute hand) gets to where the little hand (hour hand) was at six-thirty? Explain the way you figured it out. Problem of the Month Page 1 This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 Unported License (
5 Level C: I met a man who said, If you can guess my age, I will pay you one dollar for each year that I have lived. I will also give you two hints. If you take my age and divide it by any odd number greater than 1 and less than 9, you will get a remainder of 1. But if you take my age and divide it by any even number greater than 1 and less than 9, you will not get a remainder of 1. How much money could you earn? Explain your solution and how you know it is the only correct answer. Problem of the Month Page 3 This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 Unported License (
6 Level D: An eccentric clockmaker built three different clocks. The first clock was a five-minute clock designed with an alarm set to sound each time the hand reached the number 2. The second clock was a six-minute clock designed to sound each time the hand reached the number 3. The third clock was a seven-minute clock designed to sound each time the hand reached the number 4. The clockmaker started the clocks simultaneously one day, and each clock began to sound at its appropriate time. Was there a time when all three clocks sounded their alarms together? If so, tell when it occurred and explain why. If not, explain why not Problem of the Month Page 4 This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 Unported License (
7 Level E: The minute hand and the hour hand on a clock form a 48 angle. What time is it? At what other times during the day do the hands on the clock form a 48 angle? How many times in a day (24 hour period) do the hands form a 48 angle? Explain your reasoning. Problem of the Month Page 5 This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 Unported License (
8 Problem of the Month Primary Version Level A Materials: A large demonstration clock, a small clock per group Discussion on the rug: Teacher holds up the clock set at 4:00. What time does this clock say? Teacher solicits answers from students. Which is the minute hand? Which is the hour hand? Teacher solicits answers from students; teacher may refer to hands as big and small. How do the hands move around the clock? Students demonstrate. Which hand moves faster? In small groups: Each group has an individual clock. The teacher states the following: Set your clock to 4 o clock. Which direction do the hands move? Which is the minute hand (big hand)? How many minutes must pass before the minute hand gets to where the hour hand (little hand) is now? Draw a picture or use words to explain how you know. Set your clock to 6:30. Which direction do the hands move? Which is the minute hand (big hand)? How many minutes must pass before the minute hand gets to where the hour hand (little hand) is now? Draw a picture or use words to explain how you know. At the end of the investigation have students either discuss or dictate a response to these summary questions. Problem of the Month Page 6 This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 Unported License (
11 Problem of the Month Task Description Level C This task challenges a student to determine a man s age and to explain the solution and how it is the only correct solution. Two hints are given. A student will use basic operations and understand remainders in a real life context. A student will use an understanding of divisibility and the knowledge of relatively prime factors to work on this task. Common Core State Standards Math Content Standards Operations and Algebraic Thinking Gain familiarity with factors and multiples. 4.OA.4 Find all factor pairs for a whole number in the range Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range is a multiple of a given one digit number. Determine whether a given whole number in the range is prime or composite. Analyze patterns and relationships. 5.OA.3 Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane. For example, given the rule Add 3 and the starting number 0, and given the rule Add 6 and the starting number 0, generate terms in the resulting sequences, and observe that the terms in one sequence are twice the corresponding terms in the other sequence. Explain informally why this is so. The Number System Compute fluently with multi digit numbers and find common factors and multiples. 6.NS.2 Fluently divide multi digit numbers using the standard algorithm. 6.NS.4 Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Common Core State Standards Math Standards of Mathematical Practice MP.7 Look for and make use of structure. Mathematically proficient students look closely to discern a pattern or structure. Young students, for example, might notice that three and seven more is the same amount as seven and three more, or they may sort a collection of shapes according to how many sides the shapes have. Later, students will see 7 x 8 equals the well remembered 7 x x 3, in preparation for learning about the distributive property. In the expression x 2 + 9x + 14, older students can see the 14 as 2 x 7 and the 9 as They recognize the significance of an existing line in a geometric figure and can use the strategy of drawing an auxiliary line for solving problems. They also can step back for an overview and shift perspective. They can see complicated things, such as some algebraic expressions, as single objects or as being composed of several objects. For example, they can see 5 3(x y) 2 as 5 minus a positive number times a square and use that to realize that its value cannot be more than 5 for any real numbers x and y. MP.8 Look for and express regularity in repeated reasoning. Mathematically proficient students notice if calculations are repeated, and look both for general methods and for shortcuts. Upper elementary students might notice when dividing 25 by 11 that they are repeating the same calculations over and over again, and conclude they have a repeating decimal. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1, 2) with slope 3, middle school students might abstract the equation (y 2)/(x 1) = 3. Noticing the regularity in the way terms cancel when expanding (x 1)(x+1),(x 1)(x2+x+1),and(x 1)(x3 +x2+x+1)might lead them to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of their intermediate results.
12 Problem of the Month: Task Description Level D This task challenges a student to determine when the alarms on three different modular clocks that ring at varied intervals will all sound at the same time. A student will need a deep understanding of place value and quantities for working on this task. Some students may use an understanding of modular division to solve the problem. Common Core State Standards Math Content Standards Operations and Algebraic Thinking Gain familiarity with factors and multiples. 4.OA.4 Find all factor pairs for a whole number in the range Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range is a multiple of a given one digit number. Determine whether a given whole number in the range is prime or composite. Analyze patterns and relationships. 5.OA.3 Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane. For example, given the rule Add 3 and the starting number 0, and given the rule Add 6 and the starting number 0, generate terms in the resulting sequences, and observe that the terms in one sequence are twice the corresponding terms in the other sequence. Explain informally why this is so. The Number System Compute fluently with multi digit numbers and find common factors and multiples. 6.NS.2 Fluently divide multi digit numbers using the standard algorithm. 6.NS.4 Find the greatest common factor of two whole numbers less than or equal to 12. Expressions and Equations Solve real life and mathematical problems using numerical and algebraic expressions and equations. 7.EE.4 Use variables to represent quantities in a real world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. Analyze and solve linear equations and pairs of simultaneous linear equations. 8.EE.7 Solve linear equations in one variable. Common Core State Standards Math Standards of Mathematical Practice MP.7 Look for and make use of structure. Mathematically proficient students look closely to discern a pattern or structure. Young students, for example, might notice that three and seven more is the same amount as seven and three more, or they may sort a collection of shapes according to how many sides the shapes have. Later, students will see 7 x 8 equals the wellremembered 7 x x 3, in preparation for learning about the distributive property. In the expression x 2 + 9x + 14, older students can see the 14 as 2 x 7 and the 9 as They recognize the significance of an existing line in a geometric figure and can use the strategy of drawing an auxiliary line for solving problems. They also can step back for an overview and shift perspective. They can see complicated things, such as some algebraic expressions, as single objects or as being composed of several objects. For example, they can see 5 3(x y) 2 as 5 minus a positive number times a square and use that to realize that its value cannot be more than 5 for any real numbers x and y. MP.8 Look for and express regularity in repeated reasoning. Mathematically proficient students notice if calculations are repeated, and look both for general methods and for shortcuts. Upper elementary students might notice when dividing 25 by 11 that they are repeating the same calculations over and over again, and conclude they have a repeating decimal. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1, 2) with slope 3, middle school students might abstract the equation (y 2)/(x 1) = 3. Noticing the regularity in the way terms cancel when expanding (x 1)(x+1),(x 1)(x2+x+1),and(x 1)(x3 +x2+x+1)might lead them to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of their intermediate results.
13 Problem of the Month Task Description Level E This task challenges a student to determine the times in a day when the hands of an analog clock will form a 48 degree angle. A student will use their understanding of angle and angular measures to work on this task. Common Core State Standards Math Content Standards Geometry Draw and identify lines and angles, 4.G.1 Draw points, lines, line segments, rays, angles (right, acute, obtuse), Ratios and Proportional Relationships 6.RP Understand ratio concepts and use ratio reasoning to solve problems. 6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak. For every vote candidate A received, candidate C received nearly three votes. 6.RP.3 Use ratio and rate reasoning to solve real world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations. 6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities. The Number System Compute fluently with multi digit numbers and find common factors and multiples. 6.NS.2 Fluently divide multi digit numbers using the standard algorithm. 6.NS.4 Find the greatest common factor of two whole numbers less than or equal to 12. Expressions and Equations Solve real life and mathematical problems using numerical and algebraic expressions and equations. 7.EE.4 Use variables to represent quantities in a real world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. Analyze and solve linear equations and pairs of simultaneous linear equations. 8.EE.7 Solve linear equations in one variable. High School Algebra Creating Equations Create equations that describe numbers or relationships. A CED.1 Create equations and inequalities in one variable and use them to solve problems. High School Geometry Congruence Experiment with transformations in the plane G CO.1 Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc. G CO.4 Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. Make geometric constructions G CO.12 Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Common Core State Standards Math Standards of Mathematical Practice MP.7 Look for and make use of structure. Mathematically proficient students look closely to discern a pattern or structure. Young students, for example, might notice that three and seven more is the same amount as seven and three more, or they may sort a collection of shapes according to how many sides the shapes have. Later, students will see 7 x 8 equals the well remembered 7 x x 3, in preparation for learning about the distributive property. In the expression x 2 + 9x + 14, older students can see the 14 as 2 x 7 and the 9
14 as They recognize the significance of an existing line in a geometric figure and can use the strategy of drawing an auxiliary line for solving problems. They also can step back for an overview and shift perspective. They can see complicated things, such as some algebraic expressions, as single objects or as being composed of several objects. For example, they can see 5 3(x y) 2 as 5 minus a positive number times a square and use that to realize that its value cannot be more than 5 for any real numbers x and y. MP.8 Look for and express regularity in repeated reasoning. Mathematically proficient students notice if calculations are repeated, and look both for general methods and for shortcuts. Upper elementary students might notice when dividing 25 by 11 that they are repeating the same calculations over and over again, and conclude they have a repeating decimal. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1, 2) with slope 3, middle school students might abstract the equation (y 2)/(x 1) = 3. Noticing the regularity in the way terms cancel when expanding (x 1)(x+1),(x 1)(x2+x+1),and(x 1)(x3 +x2+x+1)might lead them to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of their intermediate results.
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# awaterpon
Senior Members
72
## Posts posted by awaterpon
1. ### massive human body can stand on its weak feet toes
I'm glad this can be tested that easily.
weight of 60 kg is 60*9.8 or 588 Newton which is a huge number.
Try to pull toes with your hands and compare it with lifting a 60 kg mass you will see how these muscle are weak and won't bear even far less that mass
2. ### massive human body can stand on its weak feet toes
46 minutes ago, Ghideon said:
If forces are not balanced then the body is accelerating down through the earth. Does that match your observations?
The force which represent weight is one direction downwards opposed with small force upwards.Theses forces are not equal .The body will be at rest and no acceleration since it actually opposed by earth surface . in the case for instance the object accelerating into water , this is applied according to the calculation of what I suggested. Taking in consideration acceleration as well as water resistance
3. ### massive human body can stand on its weak feet toes
Hello ,
This appears to violate physics but it is only a body phenomenon :
A human can stand on toes tips, toes , or a compilation of toes and feet balls .It is possible for a human to stand on toes. But if a person put load equivalent to his body mass on toes " and without wearing shoes " the load will break joints.
My hypothesis is:
Human body mass can be determined by comparing the body with another mass in scale" nothing new here ".The gravity of the human body is its mass times acceleration “nothing change here "
When a person stands on surface. The gravity force of body on a surface is far greater than the normal force by the surface on the body upwards.
The gravity force of body a” weight” appears in a spring of scale by compressing it which shows body mass times acceleration" nothing new". But the normal force upwards is less than that value. That why the pressure on feet toes and feet soles appears very small. So always the force of surface upwards is far smaller than force downwards " weight" .
The pressure on toes or soles is as a result of two equals forces , one is the force of ground upwards “ normal force “and the other is force downwards far less than weight.
The values: weight and normal force upwards related to each other mathematically .The massive the body is the force downwards “weight “and the force upwards are big
The explanation of the phenomenon of a person lifting ones’ massive body with weak foot and calf’s muscles when trying to pick a fruit on a tree vixra is :
The small normal force I mentioned of earth surface upwards on the body is equal to the force lifting the body, so the force lifting a body is small, far less than force downwards “weight”
4. ### A mass can be be lifted with force less than its weight
13 minutes ago, Bufofrog said:
I find it quite believable and quite unremarkable.
Thanks you.
5. ### A mass can be be lifted with force less than its weight
45 minutes ago, Strange said:
Also, the total force on the ground will be exactly equal to the weight. And the force of the ground on the person will also be equal to that. So Newton is redeemed once again.
Thanks
It's unbelievable that I carry up my 60 kg body with only my feet when trying to pick a fruit on a tree.
6. ### A mass can be be lifted with force less than its weight
46 minutes ago, pzkpfw said:
For one thing, a body is made of parts.
If you lift a 60 kg weight with your arm, the arm has to cope will all 60 kg.
A 60 kg person lifting themselves, does it in parts. Their neck lifts their head. Their core lifts their torso plus neck plus head. Their legs lift all the above.
It's not an apples to apples comparison. A 60 kg person doing a one-handed pull up or press up, would be a fairer comparison.
Thanks.
7. ### A mass can be be lifted with force less than its weight
According to classical mechanics for a force to lift a mass it should be slightly greater than its weight .
My hypothesis is that a human body can lift itself by a force far less than its weight .
It is obvious phenomenon that when lifting an object of 60 kg up , it would be extremely hard than lifting one's body " 60 kg" .while standing.
This applied to many phenomenon .A body will seem to have inertia far less than its actual mass inertia , moving and walking effortlessly , standing effortlessly , lifting one's body parts easily.
In this special case the Newtonian equations doesn't apply , however we could measure the ratio between the force lifting a body and the force lifting an object both body and the object have the same mass.
8. ### Is the gravitational equation valid?
F=GmM/r²
This equation is not valid and wrong.
What is wrong with this equation?
Mathematics equations are a away to measure quantities related to other quantities .
Equations works for finite values, I measure r = 4 meters and find out F to be x Newtons
Equations won't work for infinite values , that is equations themselves works finitely but not infinitely .
Let say F decreases without bound while r increases without bound, then the equation itself as a functional tool will approach infinity and the equation limit as r approaches ∞ is ∞, the equation is undefined and wrong.
If the equation definition is the tool that finds values of F for each value of r, then I can't find all values of F for all values of r , then the equation won't work for all values of F and r and it is useless and invalid.That means the equation itself approaches ∞ in measuring the quantities and undefined
I can say at F=0 , r doesn't exist and equals ∞, so the equation won't work and undefined in such case and invalid, but some scientists might say F won't reach 0 ever , so I presented the explanation above.
9. ### A question about gravity
If space-time is infinite ,how gravity extends to infinity? we know infinity is unreachable because it continues forever and no-one reach a finite point.How gravity extends to infinite distances while infinity is unreachable? how gravity bends and curve space-time everywhere while space time end is unreachable? for gravity to bend space-time everywhere it should reach its end , how gravity bends space-time end while this end is unreachable?
10. ### reducing torque with high ratios
Two set of gear box one with 1:14 and the other of 1:56 the total is 784
Torque is first reduced by 1:14 then stored in spring then the spring release its energy to the gearbox 1:56.
11. ### reducing torque with high ratios
Hello,
I was able to build a machine to reduce torque with high ratios up to 1:700 gear box.
What applications in mechanical engineering I could use my new invention in?
Thanks,
12. ### Gravity is limited to a range extendable with the speed of light c
8 hours ago, swansont said:
So you're claiming that Newton's first law is wrong. That an object in uniform motion will not remain in uniform motion in the absence of an external force. It will instead come to rest as it displaces space, which exerts a force on it.
Do you have any evidence that this is the case?
An object at uniform motion is different from accelerated object my idea applies to accelerated objects gravitational waves also appears for accelerated objects
12 hours ago, Ghideon said:
Which theory is that? Movement of mass and curvature makes your description sound like detection of gravity waves, currently part of established science.
New range caused by a rotating mass is different from gravitational waves this is a prediction for my theory and now it could be tested
13. ### Gravity is limited to a range extendable with the speed of light c
In my thread here and according to this quote"
"When an object moves , it tries to displace space in front of it " making non existence in space " space will resist its motion " inertia of an object" logically space can't be displaced without exerting the same force against the mass .Exerting force continuously will result in displacing space continuously and making non-existence in space continuously .The two forces " force exerted on an object and resistance force by space will make energy creation "as energy and mass are interchangeable then the process actually creates new mass amount"
I can predict and test my theory as follows:
instead of mass popping out from nowhere we have energy or mass/energy coming from original energy transferred by force , when an object moves its mass increases causing new curvature in space-time doubling the curvature already there .I can predict that new range of gravity will start from time t equals zero and extend this also could be tested
14. ### gravity force : A new view
When an object moves , it tries to displace space in front of it " making non existence in space " space will resist its motion " inertia of an object" logically space can't be displaced without exerting the same force against the mass .Exerting force continuously will result in displacing space continuously and making non-existence in space continuously .The two forces " force exerted on an object and resistance force by space will make energy creation "as energy and mass are interchangeable then the process actually creates new mass amount I can conclude to this:
If space exerts force on mass "inertia" then non-space will also exert the same force on space.
if non-space exerts force on space between two masses and this space exerts force on the other mass then attraction or gravity will appear according to my view.
All these entities " space, non-space , mass " exert forces on each other mutually.
1 hour ago, Ghideon said:
What is your exact definition of the "space" that the mass "fills"? What is "mass" in your model?
Space is everything that is not anything mass is what has weight.
15. ### gravity force : A new view
3 hours ago, swansont said:
What predictions can you make, based on your idea? How can it be tested?
3 hours ago, swansont said:
What does this mean?
I made a mistake mistake:
Mass m is attracted to non-existence of space inside mass M and mass M is attracted to non-existence of space in mass m.
3 hours ago, swansont said:
That’s not what space is in mainstream physics.
I made another mistake mass and non-existence of space are opposites if there is mass then something is missing by the existence of mass " this thing is space itself" causing non-existence of space
16. ### gravity force : A new view
5 hours ago, Ghideon said:
Ok, so the idea is about particle physics. How do you define a particle, radius and space in your idea? Is your definition compatible with quantum physics or are you suggesting something different? Radius, volume, density are not the most usable concepts at the small scales of subatomic particles, how does your model look? Is there a distinction between elementary particles such as electrons or quarks, which have no known internal structure*, versus composite particles such as protons, which do have internal structure?
Whether there is internal structure or point particle in general we have mass that fills a place of space and we have space that fills the place where there is not mass.
17. ### gravity force : A new view
11 minutes ago, Ghideon said:
A quick comment; a lot more details and math is required since the idea seems to say that Newton and GR is incorrect. Example: As far as I know a star that have run out of fuel may collapse. If for instance the result is a neutron star we have about the same mass in less volume, ok? What is your definition of "space" in this context? Does the star loose a lot of gravity when it turns* into a neutron star, according to your model? How does your idea match observations? There are already mainstream models for this.
Space doesn't mean measurements from mass edges .There is still gaps between particles. Gravity is particle to particle attraction the collection is mass m.
18. ### gravity force : A new view
Well , I'm not against GR this is a new view added to GR view of space time curvature of positive and negative attraction concept.
Does space exist inside mass m? No since mass m occupied the place of space.
Does mass m exist in place of the space I mentioned? No space is nothingness there is not mass occupied in it
So mass m occupy space , this space is equivalent to mass m.
If mass m occupy space " there is not space in the place of the mass" then mass m causes Non-existence of space in that place
If we have to masses M and m they will attract each other This is due to mass m is attracted to non-existence of mass M and mass M is attracted to non-existence of mass m.
Mass tends to fulfill the non-existence of space caused by other masses .The two " space and matter" are opposite mass means no space , space means no mass so they have such interaction of fulfillment .
The fulfillment tendency causes potential between masses and force we have such concept in electricity where positive charges are attracted to negative charges.
Non-existence of space could be like a negative charge and mass could be like a positive charge.
Notice the medium here for the force is space itself so I'm not against GR .
I mentioned the space which has been occupied is equivalent to the mass m so in the equation of Newton gravitation mass m and M can represent both masses and equivalent space
Fulfillment tendency happens for sure. If mass interacts with space and curve it according to GR then mass"positive " will also interact with non-space "negative" causes attraction between two masses
19. ### Infinite gravity
7 minutes ago, Strange said:
You don't seem to understand how science works. Here is a summary:
1. Develop a mathematical model to describe what we observe
2. Make quantifiable and testable predictions based on that model
3. Test those predictions by means of observation/experiment
The infinite reach of gravity is not testable.
Neither of these is applied to the fact that gravity is infinite then no one can scientifically say gravity is infinite.
20. ### Infinite gravity
1 hour ago, swansont said:
All tests of gravity are consistent with the GR model, so we have a lot of confidence that the model is correct.
Not gravity in general but the fact that it should be infinite, how physics determines it should be infinite ?
21. ### Infinite gravity
I post a thread here in speculation proposing that gravity has a limited range I would like to ask some questions about the existing fact.
My questions are :
What does science think of infinite gravity " gravity being everywhere" ?
How it is mathematically described "giving the idea that Newtonian gravitational law doesn't work for infinity distance" ?
Is the idea of infinite gravity part of GR ? how it is described according to GR?
Is there a proof for gravity being infinite in range ?
Thanks,
22. ### Gravity is limited to a range extendable with the speed of light c
On 4/23/2019 at 11:23 AM, Ghideon said:
start by opening a new thread with some relevant question in the cosmology section!
I have started one .
Thanks.
24. ### Gravity is limited to a range extendable with the speed of light c
2 minutes ago, Ghideon said:
Detection of gravity waves, with finite speed c, and hence reaching finite distance in finite time, got a Nobel Prize: https://www.nobelprize.org/prizes/physics/2017/press-release/ So it is considered a well established part of science.
I mean gravity is not infinitely throughout space. It is limited to a range extendable with speed of light c.
What if there is not new math, is that possible ?and it's only an addition to GR and just the equation I presented which shows limited time instead of infinite distance ?
25. ### Gravity is limited to a range extendable with the speed of light c
14 minutes ago, Ghideon said:
Again, nothing new.
Is there any point of continuing this discussion?
Just one last thing:
What if there is not new math, is that possible ?and it's only an addition to GR and just the equation I presented which shows limited time instead of infinite distance ?
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How much power is produced if a voltage of 6 V is applied to a circuit with a resistance of 54 Omega?
Mar 25, 2016
$P = \frac{2}{3}$ watts or $0.667$ watts
Explanation:
Here:
$V = 6$
$R = 54$
P=?
We know,
$P = {V}^{2} / R$
$P = {6}^{2} / 54$
$P = \frac{36}{54}$
$P = \frac{2}{3}$ watts or $0.667$ watts
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# Which one of the following statements CANNOT be true?
ankita96 on August 15, 2020
Hi, Can I get an explanation for this answer. I am unable to eliminate B. Thanks in advance!
Replies
shunhe on August 17, 2020
Hi @ankita96,
Thanks for the question! So remember, we know that S is a labrador, exactly one labrador wins a ribbon. We also know that S places ahead of Q and T, and since S finishes ahead of two dogs and only two dogs don’t get ribbons, S has to get a ribbon. So S is a labrador who has to get a ribbon. And exactly one female dog gets a ribbon, so S is the female labrador that gets a ribbon. U is a labrador, and since only one labrador gets a ribbon, U doesn’t get a ribbon. So now U, Q, and T are behind S, and P and R are ahead of S. Well, that means that S is in third place, and P and R are in first and second place, though we don’t know which is which. But that means first and second are both greyhounds. So there can’t be a labrador in second place. And that’s how we can eliminate (B).
Hope this helps! Feel free to ask any other questions that you might have.
ankita96 on August 24, 2020
Yes that makes sense, thanks!!
akinyiwilliams on December 24, 2020
So there is no scenario with Q and T in, where either P or R is out. Both P and T have to in the 4 with ribbons in any scenario, right?
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Question Video: Finding the Equation of a Line with Given 𝑥𝑦-Intercepts and Calculating the Area of a Triangle | Nagwa Question Video: Finding the Equation of a Line with Given 𝑥𝑦-Intercepts and Calculating the Area of a Triangle | Nagwa
# Question Video: Finding the Equation of a Line with Given 𝑥𝑦-Intercepts and Calculating the Area of a Triangle Mathematics • Third Year of Preparatory School
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Find the equation of the line with 𝑥-intercept 3 and 𝑦-intercept 7, and calculate the area of the triangle on this line and the two coordinate axes.
04:17
### Video Transcript
Find the equation of the line with 𝑥-intercept three and 𝑦-intercept seven and calculate the area of the triangle on this line and the two coordinate axes.
So this question has two parts. We’re first asked to find the equation of a line and then we’re asked to calculate the area of this triangle. I think a diagram would be helpful here in order to visualize the situation. So we have a pair of coordinate axes. We’re told that this line has 𝑥-intercept three, which means it cuts the 𝑥-axis at three. We’re also told the line has 𝑦-intercept seven, so it cuts the 𝑦-axis at seven. By connecting these two points, I have the line that I’m looking to find the equation of and I can see the triangle that I’m asked to find the area of. It’s this triangle here.
So let’s start with the first part of this question, which asks to find the equation of this line. I’m going to do this using the slope-intercept form, 𝑦 equals 𝑚𝑥 plus 𝑐. I can work out one of these two values straight away. Remember 𝑐 represents the 𝑦-intercept of the line. And I’m told in the question that this is equal to seven. So the equation of the line is 𝑦 equals 𝑚𝑥 plus seven. I now need to work out the slope of this line. And in order to do so, I need the coordinates of two points on the line. Well, I can use the coordinates of these points, the 𝑥-intercept and the 𝑦-intercept.
The slope of the line remember is calculated as the change in 𝑦 divided by the change in 𝑥. So looking at my diagram and using these two points, I’m gonna find the change in 𝑦 first of all. I can see that as I move from left to right across the diagram, the 𝑦-coordinate changes from seven to zero, which is a change of negative seven. It’s really important that you consider this change in 𝑦 as negative seven, not seven. The line is sloping downwards from left to right, and therefore it has a negative gradient. Now, let’s look at the change in 𝑥. I can see that as I move from left to right across this diagram, the 𝑥-coordinate changes from zero to three, which gives me a change in 𝑥 of positive three.
Now, I can substitute the change in 𝑦 and the change in 𝑥 into my calculation for the slope of this line. And we have that the slope of the line is equal to negative seven over three. Finally, in order to complete the first part of the question and find the equation of the line, I need to substitute this value for 𝑚 into the equation. I have then that the equation of this line is 𝑦 equals negative seven over three 𝑥 plus seven. Now, sometimes you may be asked to give your answer in a slightly different format, for example, a format that doesn’t involve fractions. So you’d need to multiply the equation there by three, but as it hasn’t been specified here I’m going to leave my answer as it is now. So that’s the first part of the question completed.
The second part asked me to calculate the area of the triangle formed by this line and the two coordinate axes. Now from the diagram, we can see that this is a right-angled triangle because the 𝑥- and 𝑦-axes meet at a right angle. To find the area of a right-angled triangle, we need to multiply the base by the perpendicular height and then divide by two. So looking at the diagram, I can see that the base of this triangle is that measurement of three units. The height of the triangle is seven units. Now, we refer to this as negative seven when we’re calculating the slope of the line because the direction was important. For when we’re just looking at the length of that line in order to calculate an area, we’ll take its positive value of seven. So our calculation for the area is three multiplied by seven divided by two. And this gives us an answer of 10.5 square units for the area of this triangle.
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Broad Topics > 2D Geometry, Shape and Space > Pythagoras' theorem
### Some(?) of the Parts
##### Stage: 4 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Xtra
##### Stage: 4 and 5 Challenge Level:
Find the sides of an equilateral triangle ABC where a trapezium BCPQ is drawn with BP=CQ=2 , PQ=1 and AP+AQ=sqrt7 . Note: there are 2 possible interpretations.
### Circle Packing
##### Stage: 4 Challenge Level:
Equal circles can be arranged so that each circle touches four or six others. What percentage of the plane is covered by circles in each packing pattern? ...
### At a Glance
##### Stage: 4 Challenge Level:
The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?
### Pythagoras
##### Stage: 2 and 3
Pythagoras of Samos was a Greek philosopher who lived from about 580 BC to about 500 BC. Find out about the important developments he made in mathematics, astronomy, and the theory of music.
### Are You Kidding
##### Stage: 4 Challenge Level:
If the altitude of an isosceles triangle is 8 units and the perimeter of the triangle is 32 units.... What is the area of the triangle?
### Napkin
##### Stage: 4 Challenge Level:
A napkin is folded so that a corner coincides with the midpoint of an opposite edge . Investigate the three triangles formed .
### Isosceles
##### Stage: 3 Challenge Level:
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
### Liethagoras' Theorem
##### Stage: 2 and 3
Liethagoras, Pythagoras' cousin (!), was jealous of Pythagoras and came up with his own theorem. Read this article to find out why other mathematicians laughed at him.
### Pareq Calc
##### Stage: 4 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Floored
##### Stage: 3 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Medallions
##### Stage: 4 Challenge Level:
Three circular medallions fit in a rectangular box. Can you find the radius of the largest one?
##### Stage: 4 Challenge Level:
The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### Squaring the Circle and Circling the Square
##### Stage: 4 Challenge Level:
If you continue the pattern, can you predict what each of the following areas will be? Try to explain your prediction.
### Two Circles
##### Stage: 4 Challenge Level:
Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap?
##### Stage: 4 Challenge Level:
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Star Gazing
##### Stage: 4 Challenge Level:
Find the ratio of the outer shaded area to the inner area for a six pointed star and an eight pointed star.
### Hex
##### Stage: 3 Challenge Level:
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
### Fitting In
##### Stage: 4 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
### Get Cross
##### Stage: 4 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
### Circle Scaling
##### Stage: 4 Challenge Level:
You are given a circle with centre O. Describe how to construct with a straight edge and a pair of compasses, two other circles centre O so that the three circles have areas in the ratio 1:2:3.
### The Dangerous Ratio
##### Stage: 3
This article for pupils and teachers looks at a number that even the great mathematician, Pythagoras, found terrifying.
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### The Fire-fighter's Car Keys
##### Stage: 4 Challenge Level:
A fire-fighter needs to fill a bucket of water from the river and take it to a fire. What is the best point on the river bank for the fire-fighter to fill the bucket ?.
### Six Discs
##### Stage: 4 Challenge Level:
Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?
### All Tied Up
##### Stage: 4 Challenge Level:
A ribbon runs around a box so that it makes a complete loop with two parallel pieces of ribbon on the top. How long will the ribbon be?
### The Spider and the Fly
##### Stage: 4 Challenge Level:
A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?
### Holly
##### Stage: 4 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
### Compare Areas
##### Stage: 4 Challenge Level:
Which has the greatest area, a circle or a square inscribed in an isosceles, right angle triangle?
### Circle Box
##### Stage: 4 Challenge Level:
It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?
### Inscribed in a Circle
##### Stage: 4 Challenge Level:
The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?
### Grid Lockout
##### Stage: 4 Challenge Level:
What remainders do you get when square numbers are divided by 4?
### Squ-areas
##### Stage: 4 Challenge Level:
Three squares are drawn on the sides of a triangle ABC. Their areas are respectively 18 000, 20 000 and 26 000 square centimetres. If the outer vertices of the squares are joined, three more. . . .
### Rectangular Pyramids
##### Stage: 4 and 5 Challenge Level:
Is the sum of the squares of two opposite sloping edges of a rectangular based pyramid equal to the sum of the squares of the other two sloping edges?
### Ball Packing
##### Stage: 4 Challenge Level:
If a ball is rolled into the corner of a room how far is its centre from the corner?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### Under the Ribbon
##### Stage: 4 Challenge Level:
A ribbon is nailed down with a small amount of slack. What is the largest cube that can pass under the ribbon ?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Tennis
##### Stage: 3 Challenge Level:
A tennis ball is served from directly above the baseline (assume the ball travels in a straight line). What is the minimum height that the ball can be hit at to ensure it lands in the service area?
### Equilateral Areas
##### Stage: 4 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
### Semi-detached
##### Stage: 4 Challenge Level:
A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Take a Square
##### Stage: 4 Challenge Level:
Cut off three right angled isosceles triangles to produce a pentagon. With two lines, cut the pentagon into three parts which can be rearranged into another square.
### Corridors
##### Stage: 4 Challenge Level:
A 10x10x10 cube is made from 27 2x2 cubes with corridors between them. Find the shortest route from one corner to the opposite corner.
### Pythagorean Triples
##### Stage: 3 Challenge Level:
How many right-angled triangles are there with sides that are all integers less than 100 units?
### Three Four Five
##### Stage: 4 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
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# A MEMOIR ON MATHEMATICAL AND STATISTICAL CHARACTERISTICS OF NONLINEAR REGRESSION MODELS
## Material Information
Title:
A MEMOIR ON MATHEMATICAL AND STATISTICAL CHARACTERISTICS OF NONLINEAR REGRESSION MODELS
Physical Description:
Memoir
Creator:
Overman, Allen
Publication Date:
## Subjects
Subjects / Keywords:
models
regression
elements of probability
## Notes
Abstract:
This memoir is focused on mathematical and statistical characteristics of nonlinear regression models, and includes a discussion on elements of probability. Particular models are chosen to illustrate various aspects of the procedures. A simple exponential model with two parameters is chosen as the first example. The model is rearranged to a linear form by performing logarithms on the response variable Y. This is referred to as the ‘linearized’ form of the model. Linear regression is then performed on ln Y vs. X (the control variable) to obtain estimates for the exponential parameter b along with the linear correlation coefficient r. Since the correlation coefficient is a measure of system response to the input variable X and reflects scatter in the response data, a decision is then made as to whether the linearized model is adequate or whether nonlinear regression is then needed. The ‘least squares criterion’ is used to determine ‘goodness of fit’ of the model to the data. Second order Newton-Raphson procedure is then selected to minimize the error sum of squares of deviations E between measured and estimated values of the response variable and to obtain optimum estimates of model parameters. In addition standard errors of parameter estimates are calculated using the Hessian matrix for the 2nd derivatives of E with respect to the parameters. This requires the inverse of the Hessian matrix along with the variance of the estimate, from which the standard errors of parameter estimates is obtained. The nonlinear correlation coefficient R is also used as a measure of goodness of fit of the model to the data. Contours of equal probability are then estimated for various levels of uncertainty using the Fisher F statistic. The memoir includes extensive discussion of elements of probability using the binomial expansion first for the natural numbers and then for n equal to fractions and negative values. For the natural numbers the expansion leads to finite series, whereas for fractions and negative values it leads to infinite series. All of this was established by Isaac Newton before he invented the calculus and for which he was appointed the second Lucasian professor of mathematics at Cambridge University, and led to his first memoir On the Analysis of Infinite Series. Coupling between discrete and continuous distributions are illustrated using the simple pegboard for linear, triangular, and rectangular configurations. This approach provides a logical foundation for the continuous Gaussian distribution of mathematical statistics. The procedure is further applied to role of dice for a single, two, three, and four dice. The pegboard is judged to be a simpler procedure to grasp and use in practice.
Acquisition:
Collected for University of Florida's Institutional Repository by the UFIR Self-Submittal tool. Submitted by Allen Overman.
Publication Status:
Unpublished
## Record Information
Source Institution:
University of Florida Institutional Repository
Holding Location:
University of Florida
Rights Management:
System ID:
IR00003500:00001
## Material Information
Title:
A MEMOIR ON MATHEMATICAL AND STATISTICAL CHARACTERISTICS OF NONLINEAR REGRESSION MODELS
Physical Description:
Memoir
Creator:
Overman, Allen
Publication Date:
## Subjects
Subjects / Keywords:
models
regression
elements of probability
## Notes
Abstract:
This memoir is focused on mathematical and statistical characteristics of nonlinear regression models, and includes a discussion on elements of probability. Particular models are chosen to illustrate various aspects of the procedures. A simple exponential model with two parameters is chosen as the first example. The model is rearranged to a linear form by performing logarithms on the response variable Y. This is referred to as the ‘linearized’ form of the model. Linear regression is then performed on ln Y vs. X (the control variable) to obtain estimates for the exponential parameter b along with the linear correlation coefficient r. Since the correlation coefficient is a measure of system response to the input variable X and reflects scatter in the response data, a decision is then made as to whether the linearized model is adequate or whether nonlinear regression is then needed. The ‘least squares criterion’ is used to determine ‘goodness of fit’ of the model to the data. Second order Newton-Raphson procedure is then selected to minimize the error sum of squares of deviations E between measured and estimated values of the response variable and to obtain optimum estimates of model parameters. In addition standard errors of parameter estimates are calculated using the Hessian matrix for the 2nd derivatives of E with respect to the parameters. This requires the inverse of the Hessian matrix along with the variance of the estimate, from which the standard errors of parameter estimates is obtained. The nonlinear correlation coefficient R is also used as a measure of goodness of fit of the model to the data. Contours of equal probability are then estimated for various levels of uncertainty using the Fisher F statistic. The memoir includes extensive discussion of elements of probability using the binomial expansion first for the natural numbers and then for n equal to fractions and negative values. For the natural numbers the expansion leads to finite series, whereas for fractions and negative values it leads to infinite series. All of this was established by Isaac Newton before he invented the calculus and for which he was appointed the second Lucasian professor of mathematics at Cambridge University, and led to his first memoir On the Analysis of Infinite Series. Coupling between discrete and continuous distributions are illustrated using the simple pegboard for linear, triangular, and rectangular configurations. This approach provides a logical foundation for the continuous Gaussian distribution of mathematical statistics. The procedure is further applied to role of dice for a single, two, three, and four dice. The pegboard is judged to be a simpler procedure to grasp and use in practice.
Acquisition:
Collected for University of Florida's Institutional Repository by the UFIR Self-Submittal tool. Submitted by Allen Overman.
Publication Status:
Unpublished
## Record Information
Source Institution:
University of Florida Institutional Repository
Holding Location:
University of Florida
Rights Management:
System ID:
IR00003500:00001
Full Text
PAGE 1
A MEMOIR ON MATHEMATICAL AND STATISTICAL CHARACTERISTICS OF NONLINEAR REGRESSION MODELS Allen R. Overman Agricultural and Biological Engineering University of Florida Copyright 201 3 Allen R. Overman
PAGE 2
Statistics and Nonlinear Regression Allen R. Overman i ABSTRACT This memoir is focused on mathematical and statistical characteristics of nonlinear regression models, and includes a discussion on elements of probability. Particular models are chosen to illustrate various aspects of the procedures. A simple exponential model with two parameters is chosen as the first example. The model is rearranged to a linear form by performing logarithms on the response variable Y This is referred to as the linearized form of the model. Linear regression is then performed on ln Y vs. X ( the control variable ) to obtain estimates for the exponential parameter b along with the linear correlation coefficient r Since the correlation coefficient is a measure of system response to the input variable X and reflects scatter in the response data, a decision is then made as to whether the linearized model is adequate or whether nonlinear regression is then needed. The least squares criterion is used to determine goodness of fit of the model to the data. Second order NewtonRaphson procedure is then s elected to minimize the error sum of squares of deviations E between measured and estimated values of the response variable and to obtain optimum estimates of model parameters. In addition standard errors of parameter estimates are calculated using the Hes sian matrix for the 2nd derivatives of E with respect to the parameters. This requires the inverse of the Hessian matrix along with the variance of the estimate, from which the standard errors of parameter estimates is obtained. The nonlinear correlation c oefficient R is also used as a measure of goodness of fit of the model to the data. Contours of equal probability are then estimated for various levels of uncertainty using the Fisher F statistic. The memoir includes extensive discussion of elements of pr obability using the binomial expansion first for the natural numbers ,3,2,1n and then for n equal to fractions and negative values. For the natural numbers the expansion leads to finite series, whereas for fractions and negative values it l eads to infinite series. All of this was established by Isaac Newton before he invented the calculus and for which he was appointed the second Lucasian professor of mathematics at Cambridge University and led to his first memoir On the Analysis of Infinite Series Coupling between discrete and continuous distributions are illustrated using the simple pegboard for linear, triangular, and rectangular configurations. This approach provides a logical foundation for the continuous Gaussian distribution of mathe matical statistics. The procedure is further applied to role of dice for a single, two, three, and four dice. The pegboard is judged to be a simpler procedure to grasp and use in practice. Keywords : Models, regression, elements of probability. Acknowled gement: The author expresses appreciation to Amy G. Buhler, Associate University Librarian, University of Florida, for assistance with preparation of this memoir as part of the UF digital library.
PAGE 3
Statistics and Nonlinear Regression Allen R. Overman 1 Table of Contents Introduction Mathematical Char acteristics Nonlinear Model Linearized Form of the Model Least Squares Criterion Newton Raphson Procedure for Nonlinear Regression Standard Errors of the Estimates Equal Probability Contours of Parameters Error Sum of Squares Near the Optimum Maxim um Likelihood and Least Squares Analysis Statistical Analysis of the Model Linearized Form of the Model Nonlinear Regression Standard Errors of the Estimates Equal Probability Contours Dependence of E on b Near Minimum E Summary References Tables 1. Dependence of a response variable ( Y ) to a control variable ( X ). 2. Newton Raphson iterations for nonlinear regression of the exponential model. 3. Newton Raphson iterations of the exponential model for initial b = 0.5000. 4. Combinations of A and b to satisfy equal probability equation near minimum E 5. Combinations of A and b to satisfy equal probability equation for 75% probability. 6. Combinations of A and b to satisfy equal probability equation for 95% probability. 7. Combi nations of A and b to satisfy equal probability equation for 99% probability. 8. Correlation of E with b using a parabolic model. Figures 1. Dependence of response variable ( Y ) on control variable ( X ). Data from Table 1. Curve drawn from Eq. ( 35 ). 2. Dependence of ln Y on X Data from Table 1. Line drawn from Eq. (33). 3. Scatter plot for estimated response variable ( Y ) vs. measured response variable ( Y ). Line represents the 45% diagonal. 4. Equal probability contours between parameters A and b. Contours drawn from Table 5 (75%), Table 6 (95%), and Table 7 (99%) probability levels, respectively. Optimum and standard error values of 075.0016.5A and 0129.05161.0b are also shown. 5. Dependence of erro r sum of squares ( E ) on exponential parameter ( b) for linear parameter ( A = 5.016). Parabola drawn from Eq. (67).
PAGE 4
Statistics and Nonlinear Regression Allen R. Overman 2 Introduction Scientific analysis generally involves two essential components: (1) a set of data (measurements or observation s ) and (2) a co nceptual model. The process of drawing inference about the system involves uncertainty in both the data and in the model. In the case of an algebraic mathematical model, regression analysis is used to evaluate the parameters in the model. In regression ana lysis it is common to minimize the sum of squares of deviations between measured and estimated values of the response variable as the criterion of goodness of fit of the model to the data. If all the parameters in the model occur in linear form (such as linear, quadratic, cubic, etc.) then the procedure is called linear regression If one or more of the parameters in the model occur in nonlinear form (such as exponential), then the procedure is called nonlinear regression. Linear regression is the simple r of the two since it involves linear algebra, whereas nonlinear regression involves an iterative procedure to estimate the parameters. Both methods are illustrated in this memoir. A variety of statistical measures are used to describe the quality of a mo del with a particular set of data. The first step is optimization of the model to obtain best estimates of the parameters. The next step is to calculate standard errors of the parameter estimates to determine uncertainty in the parameters. Relative error o f an estimate is then calculated as standard error divided by the estimate. A scatter plot of estimated vs. measured response variable is often included to illustrate scatter of values and any evidence of bias in the model. It is also possible to draw cont ours of equal probability (uncertainty) between two parameters by use of Fishers F statistic. Many of these points have been addressed in a previous publication (Overman et al., 1990) describing cr op response to applied nitrogen with a logistic model. In this document a simple exponential model with one linear and one nonlinear parameter is applied to a set of data. Mathematical and statistical characteristics are discussed in detail to illustrate the various steps involved. Mathematical Characteristics Nonlinear Model Consider the nonlinear regression model bXAYexp (1) where X is the control variable Y is the response variable A is the linear model parameter, and b is the exponential model parameter. For this discussion we conside r X to be positive ( 0X ) and Y to be positive ( 0 Y ). It follows that parameter A must be positive as well. Parameter b can be positive or negative. For positive b it turns out that 0 / dX dY A Y whereas fo r negative b we have 0/,dXdYAY Equation (1) is considered nonlinear in the regression sense because of the exponential parameter b. Linearized Form of the Model Now Eq. (1) can be converted to a linear form by performing logarithms on Y
PAGE 5
Statistics and Nonlinear Regression Allen R. Overman 3 bX a bX A Y Z ln ln (2) where a = ln A Parameters a and b can then be estimated by linear regression of Z vs. X In Eq. (2) ln represents the natural logarithm. In some cases these estimates of parameters may be deemed sufficient for the purpose at hand In other cases a more rigorous procedure may be desired such as nonlinear regression. Least Squares Criterion For regression analysis we begin with a criterion for goodness of fit of the model to the data. Define the error sum of squares of devi ations ( E ) between data and model by n i i iY Y E1 2 (3) where n is the number of observations, Yi is the observed value, and iY is the estimated value from the model. The goal is to choose parameters A and b to minimize E T his is call ed the least squares criterion for goodness of fit of the model to the data. For the exponential model this takes the form niiibXAYE12exp (4) For regression purposes think of E as a function of A and b, say E = E ( A, b ). At the minimum value of E it can be shown from calculus that 0dbbEdAAEdE (5) To minimize E w.r.t. (with respect to) A and b requires that bEAE0 (6) simultaneously. This is called the necessary condition for a minimum. To insure a mini mum, the sufficient condition from calculus is 0 and 02 2 2 2 b E A E (7)
PAGE 6
Statistics and Nonlinear Regression Allen R. Overman 4 The partial derivatives can be obtained from Eq. (4) and are given by the equations bX A bX Y A E 2 exp exp 2 (8) bXAE2exp222 (9) bX X A bX XY A b E b A E 2 exp 2 exp 22 2 (10) bX X A bX XY A b E 2 exp exp 2 (11) bXXAbXYXAbE2exp2exp22222 (12) The subscripts have been omitted for convenience and the cross derivative s have been included for later use in the analysis. The derivative in Eq. (8) can be set to zero, which leads to bXbXYA2expexp (13) Equation (13) gives the optimum estimate of linear parameter A for an assumed value of b. Setting Eq. (11) to zero leads to an implicit equation in parameter b. An iterative procedure is needed to find b which will cause Eq. (11) to vanish. The second order Newton Raphson procedure is chosen for this purpose (Adby and Dempster, 1974) Newton Raphson Procedure for Nonlinear Regression An initial estimate of parameter b is chosen in the neighborhood of minimum E Since we can treat E as a continuous function of b, the derivative at a new value, say bb can be related to the derivative at b by Taylor series expansion 3 4 4 2 3 3 2 2) ( 3 1 ) ( 2 1 b b E b b E b b E b E b Eb b b b b (14) It is implicitly assumed that the series represented by Eq. (14) con verges to a finite value. The strategy is to set this new derivative to zero and to truncate t he series with the linear term in b which leads to b bb E b E b2 2/ / (15)
PAGE 7
Statistics and Nonlinear Regression Allen R. Overman 5 A new estimate of parameter b is obtained from b b b (16) New estimates are then obtained for A bE/ 2 2/ b E b and b The procedure is repeated until the criterion is met bb (17) where is typically chosen as 103 to 105. The final values obtained by this procedure are chosen as optimum b b A A and minimum E E assuming that the procedure converges It i s necessary to choose the initial value of b near the optimum value to insure convergence of the procedure. Convergence requires that the second derivative in Eq. (15) be positive Standard Errors of the Estimates The next step is to calculate the stan dard errors of the estimates of the parameters. This procedure requires calculation of the Hessian matrix [ H ] of the second order derivatives given by AbbbbAAbAAbEAbEbAEAEHHHHH,222222 (18) where the derivatives are evaluated at ( A b ). Since th e cross derivatives are equal, it follows that the Hessian matrix is symmetric The inverse of the Hessian matrix yields the elements 11111bbbAAbAAHHHHH (19) where the inverse Hessian is also symmetric. The variance of the estimate 2 X YS is defined by pnYYSiiXY22 (20) where p is the number of parameters in the model. The standard errors of the estimates are then given by 2 / 1 1 2 AA X YH S A (21)
PAGE 8
Statistics and Nonlinear Regression Allen R. Overman 6 2/112bbXYHSb (22) The covariance of the estimate is give n by 12,COVAbXYHSbA (23) Standard errors of the estimates provide a measure of uncertainty in the parameter estimates for a given model and a particular set of data. Equal Probability Contours of Parameters The next mathematical characteristi c which we explore is equal probability contours of A vs. b around the optimum for a chosen level of uncertainty. Note that minimum error is calculated from n i i iX b A Y b A E1 2exp (24) Now the error at some level of probability, q, is related to E by (Draper and Smith, 1981, p. 472) q p n p F p n p b A E b A E , 1 , (25) w here p is the number of parameters in the model, q is the probability level, and F is taken from tables for Fishers analysis of variance ( F statistic) The goal is to obtain combinations of parameters A and b which satisfy bAEbXAYniii,exp12 = constant (26) This leads to a plot of A vs. b which satisfies Eq. (26), and leads to an equal probability contour. Error Sum of Squares Near the Optimum The final charact eristic which we explore is to examine E vs. b at fixed value of A A For the case of a linear model this result follows a parabola. Does this relationship hold for the nonlinear exponential model? If so, then we should obtain the parabola 2bbE (27) where , are estimated from values of E vs. b near the optimum (minimum)
PAGE 9
Statistics and Nonlinear Regression Allen R. Overman 7 Maximum Likelihood and Least Squares Analysis This section focuses on the connection between the maximum likelihood method of Fisher and that of the least squares criterion (Frieden, 1983, chapter 14 ). The challenge is to calibrate a mathematical model to relate the dependent variable ( y ) to the independent variable ( x ). Assume that the error in the measurements of yi follow s a Gaussian probability density function 2 22 ) ( exp 2 1 ) ( i iy y p (28) with mean of and variance of 2 If we further assume that the error in the predicted values ( iy ) from the model also follow this same error law with the same mean and variance, then the probability density funct ion for the error between measured and estimated y is given by 222)(exp21),(iiiiyyyyp (29) For n observations we can assume the joint probability ( p ) given by the product of individual terms n i i i n i iy y p y y p1 2 2 12 ) ( exp 2 1 ) ( (30) This is referred to as the maximum likelihood principle when the parameters of the model have been chosen to maximize the function given by Eq. (30). Such a choice will also maximize the logar ithm of p n i i i n i i i ny y n y y p1 2 2 1 2 2 2 1 2 ln 2 1 2 1 ln ln (31) In order for ln p to be a maximum, it follows that the error ( E ) defined by n i i iy y E1 2 (32) must be a minimum. Equation (32) therefore defines the least squares error between measured and predicted va lues of y based on the assumptions stated.
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Statistics and Nonlinear Regression Allen R. Overman 8 Statistical Analysis of the Model Linearized Form of the Model In this section the procedure is applied to the particular set of data listed in Table 1. The first step is to plot the data to see the trend a nd scatter (see Figure 1). T he decrease in Y with increase in X appears t o follow an exponential pattern with negative b. T he next step is to plot ln Y vs. X to test this hypothesis (see Figure 2). Since Figure 2 appears to follow a straight line, linear r egression of ln Y vs. X leads to the regression equation X bX a Y Z 5524 0 670 1 ln r = 0.9945 ( 33) with a correlation coefficient of r = 0.9945. This leads to the prediction equation XY5524.0exp31.5 ( 34) It should be noted that Eq. ( 34) does not minimize E for Eq. (1), but instead minimizes the error sum of squares for Z For some purposes Eq. ( 34) may be deemed adequate for analysis. A more rigorous procedure follows nonlinear regression. The value of parameter b = 0.5524 is then use d as a first estimate in the iteration procedure. Nonlinear Regression We now outline the nonlinear regression procedure in detail, as given in Table 2. 5524 0 b 1141.53508.2/0858.122exp/expbXbXYA 4083.35643.11411.53737.81411.522expexp2/bXXAbXXYAbE 7920 115 7548 2 1411 5 2 0640 17 1411 5 2 2 exp 2 exp 2 /2 2 2 2 bX X A bX Y X A b E 0294.07920.115/4083.3///22bEbEb 5230 0 0294 0 5524 0 b b b 5230.0b 0409.54476.2/3381.12A 7008 0 7382 1 0409 5 8316 8 0409 5 2 / b E 5555 160 1844 3 0409 5 2 1792 16 0409 5 2 /2 2 b E 0044 0 5555 160 / 7008 0 b 5186 0 0044 0 5230 0 b
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Statistics and Nonlinear Regression Allen R. Overman 9 5186.0b 0251.54631.2/3773.12A 2619.07666.10251.59034.80251.52/bE 4710 164 2559 3 0251 5 2 3575 16 0251 5 2 /2 2 b E 0016.04710.164/2619.0b 5170.00016.05186.0b 5170 0 b 0194.54687.2/3915.12A 1017 0 7770 1 0194 5 9296 8 0194 5 2 / b E 9168 165 2823 3 0194 5 2 4228 16 0194 5 2 /2 2 b E 0006.09168.165/1017.0b 5164 0 0006 0 5170 0 b 5164.0b 0173 5 4708 2 / 3968 12 A 0380.07810.10173.59396.80173.52/bE 4853 166 2925 3 0173 5 2 4478 16 0173 5 2 /2 2 b E 0002 0 4853 166 / 0380 0 b 5162 0 0002 0 5164 0 b 5162.0b 0165 5 4715 2 / 3984 12 A 0200 0 7822 1 0165 5 9424 8 0165 5 2 / b E 6255 166 2954 3 0165 5 2 4550 16 0165 5 2 /2 2 b E 00012 0 6255 166 / 0200 0 b 5161.00001.05162.0b 5161 0 b 0161.547194.2/39956.12A 00645.078305.10161.594460.80161.52/bE 7581.16629766.30161.5246050.160161.52/22bE 000039 0 7581 166 / 00645 0 b 5161.00000.05161.0b 410 000076 0 5161 0 000039 0 b b
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Statistics and Nonlinear Regression Allen R. Overman 10 The iterations are terminated at this point and lead to the estimation equatio n X Y 5161 0 exp 016 5 ( 35) Note that the procedure converged to the final values in seven steps. The regression curve i n Figure 1 is drawn from Eq. ( 35). A scatter plot of Y vs. Y is shown in Figure 3. The question now arises as to convergence if the first estimate is greater than the true value, say b = 0.5000. The steps are outlined below for this case and are summarized in Table 3. 5000 0 b 9566.45311.2/5457.12A 4253.08937.19566.43435.99566.42/bE 9699.1815804.39566.421369.179566.42/22bE 0023 0 9699 181 / 4253 0 b 5023.00023.05000.0b 5023 0 b 9653.45224.2/5244.12A 4465.18733.19653.41757.99653.4/bE 5627 179 5381 3 9653 4 2 0537 17 9653 4 2 /2 2 b E 0081 0 5627 179 / 4465 1 b 5104.00081.05023.0b 5104.0b 9951 4 4927 2 / 4513 12 A 5872.08215.19951.40398.99951.4/bE 0235 172 3950 3 9951 4 2 6975 16 9951 4 2 /2 2 b E 0034 0 0235 172 / 5872 0 b 5138 0 0034 0 5104 0 b 5138 0 b 0077.54802.2/4201.12A 2348 0 7985 1 0077 5 9829 8 00771 5 / b E 8986 168 3369 3 0077 5 2 5565 16 0077 5 2 /2 2 b E 0014 0 8986 168 / 2348 0 b 5152.00014.05138.0b
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Statistics and Nonlinear Regression Allen R. Overman 11 5152.0b 0130.54751.2/4077.12A 0908 0 7891 1 0130 5 9597 8 0130 5 / b E 6221 167 3131 3 0130 5 2 4984 16 0130 5 2 /2 2 b E 0005.06221.167/0908.0b 5157.00005.05152.0b 5157 0 b 0147.54734.2/4033.12A 0376.07858.10147.59515.80147.5/bE 1526 167 3047 3 0147 5 2 4779 16 0147 5 2 /2 2 b E 0002.01526.167/0376.0b 5159 0 0002 0 5157 0 b 5159.0b 0154.54726.2/4012.12A 0118.07843.10154.59478.80154.5/bE 9252 166 3008 3 0154 5 2 4684 16 0154 5 2 /2 2 b E 00007.09252.166/0118.0b 5160 0 00007 0 5159 0 b 5160.0b 0157.54724.2/4007.12A 00744 0 7839 1 0157 5 9468 8 0157 5 / b E 8813 166 2999 3 0157 5 2 4659 16 0157 5 2 /2 2 b E 000045.08813.166/00744.0b 5161.0000045.05160.0b 410 000087 0 5160 0 000045 0 b b The proc edure again converges to Eq. ( 35 ).
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Statistics and Nonlinear Regression Allen R. Overman 12 Another measure of quality of fit of the model to the data is given by the nonlinear correlation coefficient defined by (Cornell and Berger, 1987) 99674.0396364.23152098.01(12/12/122YYYYRiii ( 36) which shows excellent agreement between model and data The final derivatives are given by 000012.047194.20161.539956.1222expexp2bXAbXYAE ( 37) 0 94388 4 47194 2 2 2 exp 22 2 bX A E ( 38) 0 8866 17 78305 1 0161 5 2 9446 8 2 2 exp 2 exp 22 2 bX X A bX XY A b E b A E ( 39) 0 00645 0 78305 1 0161 5 94460 8 0161 5 2 2 exp exp 2 bX X A bX XY A b E ( 40) 07581.16629766.30161.5246050.160161.522exp2exp22222bXXAbXYXAbE ( 41) Note that the first derivatives are approximately zero and the second derivatives are positive as required. Standard Errors of the Estimates The second derivatives allow calculation of the Hessian matrix 7581 166 8866 17 8866 17 94388 42 2 2 2 2 2b E A b E b A E A E H H H H Hbb bA Ab AA ( 42)
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Statistics and Nonlinear Regression Allen R. Overman 13 It follows that the inverse Hessian matrix becomes 0097995332.040354540022.040354540022.073305402955.011111bbbAAbAAHHHHH ( 43) which is symmetric as required. The variance of the estimate is calculated from 016900.0211152098.022pnYYSiiXY ( 44) It follows that the standard errors of the estimates and the covariance become 0747.073305402955.0016900.02/12/112AAXYHSA ( 45) 0129 0 0097995332 0 016900 02 / 1 2 / 1 1 2 bb X YH S b ( 46) 000599.040354540022.0016900.0,COV12AbXYHSbA ( 47) Under ideal circumstances the covariance would be zero to signify that the model parameters were uncorr el ated Final estimates o f parameters are 075.0016.5A ( 48) 0129 0 5161 0 b ( 49) with relative errors of %49.10149.0016.50747.0AA ( 50) %50.20250.05161.00129.0bb ( 51) which are relatively small as desired. A check of the Hessian inverse shows that 100159999999808.000000000002.02060000000000.099999999999.00097995332.040354540022.040354540022.073305402955.07581.1668866.178866.1794388.41HH within roundoff as required. We also note that the determinant of the Hessian
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Statistics and Nonlinear Regression Allen R. Overman 14 0 501575868 504 930459560 319 432035428 824 7581 166 8866 17 8866 17 94388 4 is positive definite as required for convergence. Equal Probability C ontours In this section we examine combinations of parameters A and b which lead to equal values of the error sum of squares E The optimized model is described by XXbAY5161.0exp016.5exp ( 52) which leads to the minimum error sum of squares of niiiiiiXYYYE1111221521.05161.0exp016.5 ( 53) Values of E are calculated for 5161.0,075.0016.5bA and for 0129.05161.0,016.5bA Results for these values are listed in Table 4. Other combinations of A and b which lead to the same E are also given. These results provide a contour of equal probabilities which pass through the standard errors for A and b. From the table we note that for A = 5.016 we obtain ( b, E ) = ( 0.5032, 0.1666) and ( b, E ) = ( 0.5290, 0.1652). This analysis is now extended to various levels of probability. 75 % probability contour It can be shown that the error at some level of probability, say 75%, can be calculated from 1112exp2069.062.1211211521.0%75,,1,iiibXAYpnpFpnpEbAE ( 54) where the value of F is obtained from statistical tables as F (2,9,75%) = 1.62. Combinations of A and b which satisfy Eq. ( 54) are listed in Table 5. A graph of the pr obability contour is shown in Figure 4, from which estimates are made of ( b, E ) = ( 0.491, 0.2069) and ( b, E ) = ( 0.543, 0.2069).
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Statistics and Nonlinear Regression Allen R. Overman 15 95 % probability contour It can be shown that the error at the level of pro bability of 95% can be calculated from 11 1 2exp 2961 0 26 4 2 11 2 1 1521 0 % 95 , 1 ,i i ibX A Y p n p F p n p E b A E ( 55) where the value of F is obtained from statistical tables as F (2,9,95%) = 4.26. Combinations of A and b which satisfy Eq ( 55) are listed in Table 6 and shown in Figure 4. Estimates are made of ( b, E ) = ( 0.477, 0.2961) and ( b, E ) = ( 0.560, 0.2961). 99 % probability contour It can be shown that the error at the level of pro bability of 99% can be calculated from 11 1 2exp 4232 0 02 8 2 11 2 1 1521 0 % 99 , 1 ,i i ibX A Y p n p F p n p E b A E ( 56) where the value of F is obtained from statistical tables as F (2,9,99%) = 8.02. Combinati ons of A and b which satisfy Eq. ( 56) are listed in Table 7 and shown in Figure 4. Estimates are made of ( b, E ) = ( 0.466, 0.4232 ) and ( b, E ) = ( 0.577, 0.4232). Dependence of E on b Near M inimum E A summary of E vs. b for 016.5AA whi ch satisfy various probability levels is given Table 8. The question is whether or not this relationship follows a parabola given by 2bbE ( 57) To optimize the parabolic model requires that 2 4 3 2 3 2 2Eb Eb E b b b b b b b b n ( 58) where again n is the number of observations used in the analysis. Calculations are carried out for different numbers of points listed in Table 8.
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Statistics and Nonlinear Regression Allen R. Overman 16 n = 9 4 6312315359 0 21038451 1 3363 2 9 6648422154 0 8 2660996762 1 42264121 2 8 2660996762 1 42264121 2 6581 4 42264121 2 6581 4 9 ( 59) which leads to the regression equation 25133 75 3473 78 4827 20 b b E ( 60) The minimum of the parabola occurs at 1608.0,5188.005133.7523473.78EbbbE ( 61) which is inconsistent with 1521 0 5161 0 E b Correlation between E and E is given by E E 9166 0 0216 0 r = 0.99837 ( 62) n = 7 44000958010.077084361.04902.125084421553.089753885152.087626821.189753885152.087626821.16191.387626821.16191.37 ( 63) whi ch leads to the regression equation 25410 83 5400 86 5631 22 b b E ( 64) The minimum of the parabola occurs at 1514 0 5179 0 0 5410 83 2 5400 86 E b b b E ( 65) which is inconsistent with 1521 0 5161 0 E b Correlation between E and E is given by E E 9958 0 00088 0 r = 0.99778 ( 66) While this is better than using all nine values, it is still off a bit.
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Statistics and Nonlinear Regression Allen R. Overman 17 n = 5 4 2398675041 0 46378791 0 8980 0 8 3583277494 0 8 6912411822 0 33513921 1 8 6912411822 0 33513921 1 5821 2 33513921 1 5821 2 5 ( 67) which leads to the regression equation 27507.804693.837221.21bbE ( 68) The minimum of the parabola occurs at 1522.0,5168.007507.8024693.83EbbbE ( 69) which is near 1521.0,5161.0Eb Correlation between E and E is given by EE99722.000046.0 r = 0.99905 ( 70) n = 3 4 1289835871 0 24985331 0 4842 0 1 2132717681 0 8 4127674042 0 79920921 0 8 4127674042 0 79920921 0 5481 1 79920921 5481 1 3 ( 71) which leads to the regression equation 25409 82 2091 85 1430 22 b b E ( 72) The minimum of the parabola occurs at 1520.0,5162.005409.8222091.85EbbbE ( 73) which virtually agrees with 1521 0 5161 0 E b Correlation between E and E is given by E E 00000 1 00010 0 r = 1 ( 74) Values of E vs. b are shown in Figure 5, where th e parabola is drawn from Eq. ( 72). As b is changed away from 5161.0b the values deviate from the parabola. Note for b = 0.5524 that E is within the parabolic envelope. For b = 0.5000 note that E is also within the parabolic
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Statistics and Nonlinear Regression Allen R. Overman 18 envelope. In both cases the procedure converges toward the minimum 1521 0 E at 5161.0b Note that it is important to carry a large number of digits to avoid roundoff errors in the m atrix computational procedure. Summary This memoir has focused on the mathematical and statistical characteristics of a nonlinear regression model. The model assumed an exponential relationship between the control variable ( X ) and the response variable ( Y ) as described by Eq. (1). Analysis was performed for a given set of data (Table 1). The first step was to linearize the model to the form of Eq. (2). A plot of the data supported this step as shown in Figure 2. Linear regression of Eq. (2) led to a firs t estimate of the parameters A and b. This estimate of b was then used to perform nonlinear regression of the model on the data to optimize the values of A and b in order to minimize the error sum of squares ( E ) between measured and estimated values of Y It was shown that the Newton Raphson procedure converged rapidly to the minimum E Standard errors of the parameters were then estimated which showed low relative errors in the parameters. A further measure of uncertainty in the parameters was illustra ted by the equal probability contours for A vs. b for various levels of probability (see Figure 4). The cross in Figure 4 represents the most probable values of parameters A and b for the exponential model and for this particular set of data, i.e. the valu es which minimize the error sum of squares between measured and predicted values of response variable y Vertical bars represent the standard error in parameter b around optimum value b while horizontal bars represent the standard er ror in parameter A around optimum value A These are equivalent to the standard deviation around the mean of a set of measurements which follow a Gaussian distribution. The contours in Figure 4 represent combinations of parameters A a nd b which produce various levels of uncertainty Following Fishers maximum likelihood method, these represents combinations of equal probability. It was further shown that E vs. b at optimum A in the neighborhood of minimum E followed parabolic depend ence. At this point it seems appropriate to call attention to several general points about data analysis and mathematical models. R. A. Fisher called attention to two elements of uncertainty in this process in his classic article of 1922 (see Bennet, 1971). Uncertainty in data led to his analysis of variance (ANOVA) procedure, while uncertainty in a model led to a subject called Fisher Information (see Frieden, 1998). In her biography of her father Joan Fisher Box noted that the passion of Fi shers life was the subject of inference (see Fisher Box, 1978, p. 447) i.e. drawing inference about a system from analysis of a specific set of data. Scientific research can be divided into two approaches: bottom/up and top/down. In the bottom/up approach data (me asurements or observations) are examined in order to identify a unifying theory or model (from specific to general), which is commonly referred to as a process of induction. In the top/down approach a general principle is postulated and the consequences of these are developed (from general to specific), which is commonly referred to as the process of deduction. Most research appears to have followed the bottom/up approach. The top/down approach was championed by Einstein and by Paul Dirac (Farmelo, 2009, p. 2, 94, 382). Following a series of lectures by Murray Gell Mann it appears that Dirac gained increased respect for the bottom/up approach, which Gell Mann had followed. The work described in this
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Statistics and Nonlinear Regression Allen R. Overman 19 memoir has followed the bottom/up approach. Of course we cant be certain that the simple exponential model is the very best model possible for the given data. There is always a level of uncertainty. Science progresses by assuming a theory or model and then checking the consequences of the theory through measureme nts. A final point has to do with pursuit of knowledge and understanding of how nature really works. I will call this the battle between subjective and objective criteria for judging the values of ideas in science. According to James Glanz (see Chang, 2000, p. 354) the theore tical physicist Steven Weinberg has battled with thinkers and philosophers of science over this issue. Today, one of his major battles is with postmodernist thinkers and philosophers of science who maintain that scientific theories reflect not objective reality but social negotiations among scientists. In its rawest form, this philosophy would say that the theories of the most persuasive or political ly powerful scientists become accepted fact. I was trained on the belief in objective criteria, and I still hold to this view. Otherwise, it becomes a battle for power and control of ideas in science based on personalities Unfortunately I have observed an increasing trend to cite experts as the source of truth in the ev aluation of sc ientific ideas. Some e ditors and reviewers seem to find this an attractive alternative in the peer review process. References Adby, P.R. and M.A.H. Dempster. 1974. 1974. Introduction to Optimization Methods. John Wiley & Sons. New York, NY. Bennett, J.H 1971. Collected Papers of R.A. Fisher. Vol 1 ( 19121924). University of Adelaide. Chang, L. 2000. Scientists at Work: Profiles of Todays Groundbreaking Scientists from Science Times. McGraw Hill. New York, NY. Cornell, J.A. and R.D. Berger. 1987. Fac tors that influence the value of the coefficient of determination in simple linear and nonlinear regression models. Phytopathology 77:6370. Draper, N.R. and H. Smith. 1981. Applied Regression Analysis John Wiley & Sons. New York, NY. Farmelo, G. 2009. Th e Strangest Man: The Hidden Life of Paul Dirac, Mystic of the Atom. Basic Books. New York, NY. Fisher Box, J. 1978. R.A. Fisher: The Life of a Scientist John Wiley & Sons. New York, NY. Frieden, B.R. 1983. Probability, Statistical Optics, and Data Testing. Springer Verlag. New York, NY. Frieden, B.R. 1998. Physics from Fisher Information: A Unification. Cambridge University Press. New York, NY. Overman, A.R., F.G. Martin, and S.R. Wilkinson. 1990. A logistic equation for yield response of forage grass to n itrogen. Commun. Soil Science and Plant Analysis 21:595609.
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Statistics and Nonlinear Regression Allen R. Overman 20 Table 1. Dependence of a response variable ( Y ) on a control variable ( X ). X Y ln Y 0.0 5.0 1.609 0.5 4.0 1.386 1.0 2.8 1.030 1.5 2.2 0.788 2.0 2.0 0.693 2.5 1.5 0.405 3.0 1.0 0.000 3.5 0.9 0.105 4.0 0.6 0.511 4.5 0.4 0.916 5.0 0.3 1.204 Table 2 Newton Raphson ite rations of the exponential model for initial b = 0.5524. X Y exp ( bX ) Y 0.0 5.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.00000 5.016 0.5 4.0 0.7587 0.7699 0.7716 0.7722 0.7724 0.7725 0.77256 3.875 1.0 2.8 0.5756 0.5927 0.5954 0.5963 0.5967 0.5968 0.59684 2.994 1.5 2.2 0.4367 0.4564 0.4594 0.4605 0.4609 0.4610 0.46110 2.313 2.0 2.0 0.3313 0.3513 0.3544 0.3556 0.3560 0.3561 0.35622 1.787 2.5 1.5 0.2513 0.2705 0.2735 0.2746 0.2750 0.2751 0.27520 1.380 3.0 1.0 0.1907 0.2082 0.2110 0.2120 0.2124 0.2125 0.21261 1.066 3.5 0.9 0.1447 0.1603 0.1628 0.1637 0.1641 0.1642 0.16425 0.824 4.0 0.6 0.1097 0.1234 0.1256 0.1264 0.1267 0.1268 0.12689 0.637 4.5 0.4 0.0833 0.0950 0.0969 0.0976 0.0979 0.0980 0.09803 0.492 5.0 0.3 0.0632 0.0732 0.0748 0.0754 0.0756 0.0757 0.07574 0.380 b 0.5524 0.5230 0.5186 0.5170 0.5164 0.5162 0.5161 0.5161 A 5.1411 5.0409 5.0251 5.0194 5.0173 5.0165 5.0161 5.0161 b E / 3.4083 0.7008 0.2619 0.1017 0.0380 0.0200 0.00645 22/bE 115.7920 160.5555 164.4710 165.9168 166.4853 166 .6255 166.7581 b 0.0294 0.0044 0.0016 0.0006 0.0002 0.0001 0.00004 b 0.5230 0.5186 0.5170 0.5164 0.5162 0.5161 0.51614
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Statistics and Nonlinear Regression Allen R. Overman 21 Table 3 Newton Raphson iterations of the exponential model for initial b = 0.5000. X Y exp ( bX ) 0.0 5.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.5 4.0 0.7788 0.7779 0.7748 0.7734 0.7729 0.7727 0.7726 1.0 2.8 0.6065 0.6051 0.6003 0.5982 0.5974 0.5971 0.5970 1.5 2.2 0.4724 0.4707 0.4651 0.4627 0.4617 0.4614 0.4612 2.0 2.0 0.3679 0.3662 0.3603 0.3579 0.3569 0.3565 0.3564 2.5 1.5 0.2865 0.2849 0.2792 0.2768 0.2758 0.2755 0.2753 3.0 1.0 0.2231 0.2216 0.2163 0.2141 0.2132 0.2129 0.2127 3.5 0.9 0.1738 0.1724 0.1676 0.1656 0.1648 0.1645 0.1644 4.0 0.6 0.1353 0.1341 0.1298 0.1281 0.1274 0.1271 0.1270 4.5 0.4 0.1054 0.1043 0.1006 0.0991 0.0984 0.0982 0.0981 5.0 0.3 0.0821 0.0811 0.0779 0.0766 0.0761 0.0759 0.0758 b 0.5000 0.5023 0.5104 0.5138 0.5152 0.5157 0.5159 A 4.9566 4.9653 4.9951 5.0077 5.0130 5.0147 5.0154 b E / +0.4253 +1.4465 +0.5872 +0.2348 +0.0908 +0.0376 +0.0118 22/bE 181.9699 179.5627 172.0235 168.8986 167.6221 167.1526 166.9252 b 0.0023 0.0081 0.0034 0.0014 0.0005 0.0002 0.00007 b 0.5023 0.5104 0.5138 0.5152 0.5157 0.5159 0.5160 Table 3 ( Continued ) X Y exp ( bX ) Y YY 0.0 5.0 1.000000 1.000000 1.000000 1.000000 1.0000000 5.016 0.016 0.5 4.0 0.772607 0.772589 0.772583 0.772580 0.7725794 3.875 +0.125 1.0 2.8 0.596921 0.596894 0.596884 0.596880 0.5968789 2.994 0.194 1.5 2.2 0.461185 0.461154 0.461143 0.461138 0.4611364 2.313 0.113 2.0 2.0 0.356315 0.356283 0.356271 0.356266 0.3562644 1.787 +0.213 2.5 1.5 0.275291 0.275260 0.275249 0.275244 0.2752426 1.381 +0.119 3.0 1.0 0.212692 0.212663 0.212652 0.212648 0.2126467 1.067 0.067 3.5 0.9 0.164327 0.164301 0.164292 0.164288 0.1642865 0.824 +0.076 4.0 0.6 0.126960 0.126938 0.126929 0.126925 0.1269244 0.637 0.037 4.5 0.4 0.098090 0.098071 0.098063 0.098060 0.0980591 0.492 0.092 5.0 0.3 0.075785 0.075768 0.075762 0.075759 0.0757585 0.380 0.080 b 0.51597 0.516015 0.516032 0.516039 0.516041 0.5160 A 5.01566 5.01582 5.01588 5.01591 5.0159173 5.0160 bE/ +0.007444 +0.002792 +0.001101 +0.000301 +0.000256 22/bE 166.881252 166.840459 166.825228 166.819155 166.817429 b 0.000045 0.000017 0.000007 0.000002 0.0000015 b 0.516015 0.516032 0.516039 0.516041 0.516042
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Statistics and Nonlinear Regression Allen R. Overman 22 Table 4. Combinations of A and b to satisfy equal probability equation near minimum E X Y Y 0.0 5.0 4.941 5.091 5.016 5.016 4.970 5.110 5.060 4.920 0.5 4.0 3.817 3.933 3.900 3.850 3.828 3.936 3.921 3.813 1.0 2.8 2.950 3.039 3.033 2.955 2.949 3.032 3.039 2.954 1.5 2.2 2.278 2.347 2.358 2.269 2.271 2.335 2.355 2.289 2.0 2.0 1.760 1.814 1.834 1.741 1.750 1.799 1.825 1.774 2.5 1.5 1.360 1.401 1.426 1.337 1.348 1.386 1.414 1.375 3.0 1.0 1.051 1.082 1.109 1.026 1.038 1.067 1.096 1.065 3.5 0.9 0.812 0.836 0.862 0.788 0.800 0.822 0.849 0.826 4.0 0.6 0.627 0.646 0.670 0.604 0.616 0.633 0.658 0.640 4.5 0.4 0.484 0.499 0.521 0.464 0.474 0.488 0.510 0.496 5.0 0.3 0.374 0.386 0.405 0.356 0.365 0.376 0.395 0.384 b 0.5161 0.5161 0.5032 0.5290 0.522 0.522 0.510 0.510 A 4.941 5.091 5.016 5.016 4.97 5.11 5.06 4.92 E 0.1660 0.1665 0.1664 0.1657 0.1647 0.1668 0.1653 0.1673 Target is E = 0.1660 Table 4. (Continued). X Y Y 0.0 5.0 5.110 4.930 5.017 5.017 4.970 5.050 5.080 0.5 4.0 3.922 3.833 3.901 3.851 3.871 3.871 3.894 1.0 2.8 3.011 2.981 3.033 2.956 3.014 2.967 2.984 1.5 2.2 2.311 2.318 2.359 2.269 2.348 2.274 2.287 2.0 2.0 1.774 1.802 1.834 1.742 1.828 1.743 1.753 2.5 1.5 1.362 1.401 1.426 1.337 1.424 1.336 1.344 3.0 1.0 1.045 1.090 1.109 1.026 1.109 1.024 1.030 3.5 0.9 0.802 0.847 0.862 0.788 0.864 0.785 0.789 4.0 0.6 0.616 0.659 0.670 0.605 0.673 0.601 0.605 4.5 0.4 0.473 0.512 0.521 0.464 0.524 0.461 0.464 5.0 0.3 0.363 0.398 0.405 0.356 0.408 0.353 0.355 b 0.5290 0.5032 0.5032 0.5290 0.5000 0.5320 0.5320 A 5.11 4.93 5.017 5.017 4.97 5.05 5.08 E 0.1663 0.1650 0.1666 0.1652 0.1662 0.1658 0.1648 Target is E = 0.1660
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Statistics and Nonlinear Regression Allen R. Overman 23 Table 5. Combinations of A and b to satisfy equal probability equation for 75% probability. X Y Y 0.0 5.0 5.160 4.870 5.210 4.960 5.080 4.830 5.140 5.130 0.5 4.0 3.986 3.762 3.987 3.796 3.956 3.762 3.904 3.897 1.0 2.8 3.080 2.907 3.051 2.905 3.081 2.930 2.966 2.960 1.5 2.2 2.379 2.246 2.335 2.223 2.400 2.282 2.253 2.248 2.0 2.0 1.838 1.735 1.787 1.701 1.869 1.777 1.711 1.708 2.5 1.5 1.420 1.340 1.368 1.302 1.455 1.384 1.300 1.297 3.0 1.0 1.097 1.035 1.047 0.996 1.134 1.078 0.987 0.985 3.5 0.9 0.848 0.800 0.801 0.763 0.883 0.839 0.750 0.748 4.0 0.6 0.655 0.618 0.613 0.584 0.688 0.654 0.570 0.568 4.5 0.4 0.506 0.477 0.469 0.447 0.535 0.509 0.433 0.432 5.0 0.3 0.391 0.369 0.359 0.342 0.417 0.396 0.329 0.328 b 0.5161 0.5161 0.535 0.535 0.500 0.500 0.550 0.550 A 5.16 4.87 5.21 4.96 5.08 4.83 5.14 5.13 E 0.2035 0.2052 0.2087 0.2064 0.2044 0.2062 0.2082 0.2080 E = 0.2069 target Table 5. (Continued). X Y Y 0.0 5.0 5.000 4.840 4.920 4.960 4.860 5.050 4.830 4.910 0.5 4.0 3.914 3.788 3.861 3.796 3.813 3.943 3.771 3.855 1.0 2.8 3.063 2.965 3.029 2.905 2.992 3.078 2.944 3.026 1.5 2.2 2.398 2.321 2.377 2.223 2.348 2.403 2.299 2.376 2.0 2.0 1.877 1.817 1.865 1.701 1.842 1.876 1.795 1.865 2.5 1.5 1.469 1.422 1.463 1.302 1.446 1.465 1.401 1.464 3.0 1.0 1.150 1.113 1.148 0.996 1.134 1.144 1.094 1.149 3.5 0.9 0.900 0.871 0.901 0.763 0.890 0.893 0.854 0.902 4.0 0.6 0.704 0.682 0.707 0.584 0.698 0.697 0.667 0.708 4.5 0.4 0.551 0.534 0.555 0.447 0.548 0.544 0.521 0.556 5.0 0.3 0.431 0.418 0.435 0.342 0.430 0.425 0.407 0.437 b 0.490 0.490 0.485 0.485 0.485 0.495 0.495 0.484 A 5.00 4.84 4.92 4.96 4.86 5.05 4.83 4.91 E 0.2051 0.2042 0.2047 0.2064 0.2077 0.2074 0.2052 0.2077 E = 0.2069 target
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Statistics and Nonlinear Regression Allen R. Overman 24 Table 5. (Continued). X Y Y 0.0 5.0 4.840 5.130 4.900 5.190 4.990 5.210 5.040 5.190 0.5 4.0 3.754 3.979 3.769 3.992 3.809 3.977 3.838 3.952 1.0 2.8 2.912 3.087 2.899 3.070 2.908 3.036 2.922 3.009 1.5 2.2 2.259 2.394 2.229 2.361 2.220 2.318 2.225 2.292 2.0 2.0 1.752 1.857 1.715 1.816 1.695 1.769 1.695 1.745 2.5 1.5 1.359 1.441 1.319 1.397 1.294 1.351 1.290 1.329 3.0 1.0 1.054 1.118 1.014 1.074 0.988 1.031 0.983 1.012 3.5 0.9 0.818 0.867 0.780 0.826 0.754 0.787 0.748 0.770 4.0 0.6 0.634 0.672 0.600 0.636 0.575 0.601 0.570 0.587 4.5 0.4 0.492 0.522 0.462 0.489 0.439 0.459 0.434 0.447 5.0 0.3 0.382 0.405 0.355 0.376 0.335 0.350 0.330 0.340 b 0.508 0.508 0.525 0.525 0.540 0.540 0.545 0.545 A 4.84 5.13 4.90 5.19 4.99 5.21 5.04 5.19 E 0.2095 0.2074 0.2095 0.2054 0.2089 0.2095 0.2068 0.2058 E = 0.2069 target Table 5. (Continued). X Y Y 0.0 5.0 4.930 5.200 5.080 5.170 5.016 5.016 5.016 0.5 4.0 3.782 3.989 3.862 3.931 3.924 3.823 3.922 1.0 2.8 2.902 3.061 2.937 2.989 3.070 2.914 3.067 1.5 2.2 2.226 2.348 2.233 2.272 2.402 2.221 2.398 2.0 2.0 1.708 1.802 1.698 1.728 1.879 1.693 1.875 2.5 1.5 1.310 1.382 1.291 1.314 1.470 1.291 1.466 3.0 1.0 1.005 1.060 0.981 0.999 1.150 0.984 1.146 3.5 0.9 0.771 0.814 0.746 0.759 0.900 0.750 0.896 4.0 0.6 0.592 0.624 0.567 0.577 0.704 0.572 0.701 4.5 0.4 0.454 0.479 0.431 0.439 0.551 0.436 0.548 5.0 0.3 0.348 0.367 0.328 0.334 0.431 0.332 0.429 b 0.530 0.530 0.548 0.548 0.491 0.543 0.492 A 4.93 5.20 5.08 5.17 5.016 5.016 5.016 E 0.2068 0.2056 0.2071 0.2062 0.2086 0.2220 0.2037 E = 0.2069 target
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Statistics and Nonlinear Regression Allen R. Overman 25 Table 6. Combinations of A and b to satisfy equal probability equation for 95% probability. X Y Y 0.0 5.0 5.260 4.780 5.310 4.860 5.170 4.730 5.320 4.980 0.5 4.0 4.064 3.693 4.064 3.719 4.026 3.684 4.031 3.773 1.0 2.8 3.139 2.853 3.110 2.846 3.136 2.869 3.054 2.859 1.5 2.2 2.425 2.204 2.380 2.178 2.442 2.234 2.314 2.166 2.0 2.0 1.874 1.703 1.821 1.667 1.902 1.740 1.753 1.641 2.5 1.5 1.448 1.315 1.394 1.276 1.481 1.355 1.328 1.243 3.0 1.0 1.118 1.016 1.067 0.976 1.154 1.055 1.006 0.942 3.5 0.9 0.864 0.785 0.816 0.747 0.898 0.822 0.763 0.714 4.0 0.6 0.667 0.607 0.625 0.572 0.700 0.640 0.578 0.541 4.5 0.4 0.516 0.469 0.478 0.438 0.545 0.499 0.438 0.410 5.0 0.3 0.398 0.362 0.366 0.335 0.424 0.388 0.332 0.310 b 0.5161 0.5161 0.535 0.535 0.500 0.500 0.555 0.555 A 5.26 4.78 5.31 4.86 5.17 4.73 5.31 4.98 E 0.2986 0.2900 0.2946 0.2897 0.2949 0.2956 0.2932 0.2931 E = 0.2961 target Table 6. (Continued). X Y Y 0.0 5.0 5.120 4.710 5.050 4.710 4.900 5.070 5.290 5.320 4.910 0.5 4.0 4.007 3.687 3.972 3.705 3.874 3.822 3.988 4.051 3.739 1.0 2.8 3.137 2.885 3.125 2.914 3.062 2.882 3.007 3.085 2.847 1.5 2.2 2.455 2.258 2.458 2.293 2.421 2.172 2.267 2.349 2.168 2.0 2.0 1.922 1.768 1.934 1.803 1.914 1.638 1.709 1.789 1.651 2.5 1.5 1.504 1.384 1.521 1.419 1.513 1.235 1.288 1.362 1.257 3.0 1.0 1.177 1.083 1.196 1.116 1.196 0.931 0.971 1.037 0.957 3.5 0.9 0.921 0.848 0.941 0.878 0.946 0.702 0.732 0.790 0.729 4.0 0.6 0.721 0.663 0.740 0.691 0.748 0.529 0.552 0.601 0.555 4.5 0.4 0.564 0.519 0.582 0.543 0.591 0.399 0.416 0.458 0.423 5.0 0.3 0.442 0.406 0.458 0.427 0.467 0.301 0.314 0.349 0.322 b 0.490 0.490 0.480 0.480 0.470 0.565 0.565 0.545 0.545 A 5.12 4.71 5.05 4.71 4.90 5.07 5.29 5.32 4.91 E 0.2926 0.2989 0.2981 0.2987 0.2921 0.2944 0.2930 0.2912 0.2944 E = 0.2961 target
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Statistics and Nonlinear Regression Allen R. Overman 26 Table 6. (Continued). X Y Y 0.0 5.0 4.720 5.000 4.740 5.140 5.250 4.860 4.770 5.200 5.200 0.5 4.0 3.722 3.943 3.747 3.865 3.948 3.846 3.775 3.907 3.908 1.0 2.8 2.935 3.109 2.963 2.907 2.969 3.044 2.987 2.935 2.936 1.5 2.2 2.315 2.452 2.342 2.186 2.233 2.409 2.364 2.205 2.207 2.0 2.0 1.825 1.934 1.852 1.644 1.679 1.906 1.871 1.656 1.658 2.5 1.5 1.440 1.525 1.464 1.236 1.263 1.508 1.480 1.244 1.246 3.0 1.0 1.135 1.203 1.157 0.930 0.950 1.194 1.172 0.935 0.936 3.5 0.9 0.895 0.948 0.915 0.699 0.714 0.945 0.927 0.702 0.704 4.0 0.6 0.706 0.748 0.723 0.526 0.537 0.748 0.734 0.528 0.529 4.5 0.4 0.557 0.590 0.572 0.395 0.404 0.592 0.581 0.396 0.397 5.0 0.3 0.439 0.465 0.452 0.297 0.304 0.468 0.459 0.298 0.299 b 0.475 0.475 0.470 0.570 0.570 0.468 0.468 0.572 0.5715 A 4.72 5.00 4.74 5.14 5.25 4.86 4.77 5.20 5.20 E 0.2948 0.2960 0.2942 0.2967 0.2952 0.2821 0.2887 0.2994 0.2960 E = 0.2961 target Table 6. (Continued). X Y Y 0.0 5.0 5.190 5.180 5.170 4.800 4.810 4.910 5.320 4.810 5.290 0.5 4.0 3.909 3.901 3.894 3.799 3.806 3.739 4.051 3.700 4.069 1.0 2.8 2.944 2.938 2.933 3.006 3.012 2.847 3.085 2.845 3.129 1.5 2.2 2.217 2.213 2.209 2.379 2.384 2.168 2.349 2.188 2.407 2.0 2.0 1.670 1.667 1.663 1.883 1.886 1.651 1.789 1.683 1.851 2.5 1.5 1.258 1.255 1.253 1.490 1.493 1.257 1.362 1.295 1.424 3.0 1.0 0.947 0.945 0.944 1.179 1.181 0.957 1.037 0.996 1.095 3.5 0.9 0.713 0.712 0.711 0.933 0.935 0.729 0.790 0.766 0.842 4.0 0.6 0.537 0.536 0.535 0.738 0.740 0.555 0.601 0.589 0.648 4.5 0.4 0.405 0.404 0.403 0.584 0.585 0.423 0.458 0.453 0.498 5.0 0.3 0.305 0.304 0.304 0.462 0.463 0.322 0.349 0.348 0.383 b 0.567 0.567 0.567 0.468 0.468 0.545 0.545 0.525 0.525 A 5.19 5.18 5.17 4.80 4.81 4.91 5.32 4.81 5.29 E 0.2813 0.2748 0.2756 0.2809 0.2800 0.2944 0.2912 0.2940 0.2991 E = 0.2961 target
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Statistics and Nonlinear Regression Allen R. Overman 27 Table 6. (Continued). X Y Y 0.0 5.0 4.820 4.810 4.810 4.810 4.820 4.800 4.790 4.750 5.220 0.5 4.0 3.815 3.807 3.807 3.808 3.816 3.800 3.793 3.685 4.049 1.0 2.8 3.020 3.014 3.013 3.015 3.022 3.009 3.003 2.858 3.141 1.5 2.2 2.390 2.386 2.385 2.387 2.392 2.382 2.377 2.217 2.436 2.0 2.0 1.892 1.888 1.888 1.890 1.894 1.886 1.882 1.720 1.890 2.5 1.5 1.498 1.495 1.494 1.497 1.500 1.493 1.490 1.334 1.466 3.0 1.0 1.186 1.183 1.182 1.185 1.187 1.182 1.180 1.035 1.137 3.5 0.9 0.938 0.937 0.936 0.938 0.940 0.936 0.934 0.803 0.882 4.0 0.6 0.743 0.741 0.741 0.743 0.744 0.741 0.740 0.623 0.684 4.5 0.4 0.588 0.587 0.586 0.588 0.589 0.587 0.586 0.483 0.531 5.0 0.3 0.465 0.464 0.464 0.466 0.467 0.465 0.464 0.375 0.412 b 0.4675 0.4675 0.4677 0.4670 0.4670 0.4670 0.4670 0.508 0.508 A 4.82 4.81 4.81 4.81 4.82 4.80 4.79 4.75 5.22 E 0.2818 0.2829 0.2813 0.2853 0.2846 0.2864 0.2882 0.2932 0.2918 E = 0.2961 target Table 7. Combinations of A and b to satisfy equal probability equation for 99% probability. X Y Y 0.0 5.0 5.350 4.680 5.280 4.640 5.160 4.600 5.420 4.780 0.5 4.0 4.133 3.616 4.112 3.614 4.059 3.618 4.138 3.649 1.0 2.8 3.193 2.793 3.202 2.814 3.193 2.846 3.158 2.786 1.5 2.2 2.467 2.158 2.494 2.192 2.512 2.239 2.411 2.126 2.0 2.0 1.906 1.667 1.942 1.707 1.976 1.761 1.841 1.623 2.5 1.5 1.472 1.288 1.513 1.329 1.554 1.385 1.406 1.239 3.0 1.0 1.137 0.995 1.178 1.035 1.223 1.090 1.073 0.946 3.5 0.9 0.879 0.769 0.806 0.763 0.962 0.857 0.819 0.722 4.0 0.6 0.679 0.594 0.715 0.628 0.756 0.674 0.625 0.551 4.5 0.4 0.524 0.459 0.557 0.489 0.595 0.530 0.477 0.421 5.0 0.3 0.405 0.354 0.433 0.381 0.468 0.417 0.364 0.321 b 0.5161 0.5161 0.500 0.500 0.480 0.480 0.540 0.540 A 5.35 4.68 5.28 4.64 5.16 4.60 5.42 4.78 E 0.4274 0.4311 0.4301 0.4193 0.4285 0.4259 0.4248 0.4254 E = 0.4232 target
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Statistics and Nonlinear Regression Allen R. Overman 28 Table 7. (C ontinued). X Y Y 0.0 5.0 5.440 4.890 4.960 5.440 5.040 5.420 5.360 5.160 5.220 0.5 4.0 4.111 3.696 3.730 4.091 3.771 4.056 3.991 3.842 4.086 1.0 2.8 3.107 2.793 2.805 3.076 2.822 3.035 2.971 2.860 3.198 1.5 2.2 2.349 2.111 2.109 2.314 2.112 2.271 2.212 2.130 2.503 2.0 2.0 1.775 1.596 1.586 1.740 1.580 1.699 1.647 1.586 1.959 2.5 1.5 1.341 1.206 1.193 1.308 1.182 1.271 1.226 1.180 1.533 3.0 1.0 1.014 0.911 0.897 0.984 0.885 0.951 0.913 0.879 1.200 3.5 0.9 0.766 0.689 0.675 0.740 0.662 0.712 0.680 0.654 0.939 4.0 0.6 0.579 0.521 0.507 0.556 0.495 0.533 0.506 0.487 0.735 4.5 0.4 0.438 0.393 0.382 0.418 0.371 0.399 0.377 0.363 0.576 5.0 0.3 0.331 0.297 0.287 0.315 0.277 0.298 0.281 0.270 0.450 b 0.560 0.560 0.570 0.570 0.580 0.580 0.590 0.590 0.490 A 5.44 4.89 4.96 5.44 5.04 5.42 5.36 5.16 5.22 E 0.4193 0.4209 0.4188 0.4239 0.4221 0.4251 0.4241 0.4231 0.4220 E = 0.4232 target Table 7. (C ontinued). X Y Y 0.0 5.0 5.280 4.610 4.990 4.800 4.600 5.080 4.620 4.630 4.930 0.5 4.0 3.925 3.663 3.965 3.837 3.637 4.016 3.616 3.696 3.927 1.0 2.8 2.918 2.910 3.150 3.067 2.875 3.175 2.830 2.944 3.128 1.5 2.2 2.169 2.312 2.503 2.451 2.273 2.510 2.215 2.345 2.491 2.0 2.0 1.613 1.837 1.989 1.959 1.797 1.984 1.734 1.868 1.984 2.5 1.5 1.199 1.460 1.580 1.566 1.421 1.569 1.357 1.488 1.581 3.0 1.0 0.891 1.160 1.255 1.252 1.123 1.240 1.062 1.185 1.259 3.5 0.9 0.663 0.921 0.997 1.001 0.888 0.980 0.831 0.944 1.003 4.0 0.6 0.493 0.732 0.792 0.800 0.702 0.775 0.651 0.752 0.799 4.5 0.4 0.366 0.582 0.630 0.639 0.555 0.613 0.509 0.599 0.636 5.0 0.3 0.272 0.462 0.500 0.511 0.439 0.484 0.399 0.477 0.507 b 0.593 0.460 0.460 0.448 0.470 0.470 0.490 0.455 0.455 A 5.28 4.61 4.99 4.80 4.60 5.08 4.62 4.63 4.93 E 0.4207 0.4213 0.4264 0.4222 0.4192 0.4222 0.4171 0.4188 0.4252 E = 0.4232 target
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Statistics and Nonlinear Regression Allen R. Overman 29 Table 7. (C ontinued). X Y Y 0.0 5.0 4.730 5.390 4.830 5.440 5.430 4.660 5.310 4.720 5.380 0.5 4.0 3.629 4.135 3.669 4.132 4.124 3.615 4.119 3.630 4.138 1.0 2.8 2.784 3.173 2.787 3.139 3.133 2.804 3.195 2.792 3.183 1.5 2.2 2.136 2.434 2.117 2.384 2.380 2.175 2.478 2.148 2.448 2.0 2.0 1.639 1.867 1.608 1.811 1.807 1.687 1.922 1.652 1.883 2.5 1.5 1.257 1.433 1.221 1.375 1.373 1.309 1.491 1.270 1.448 3.0 1.0 0.965 1.099 0.928 1.045 1.043 1.015 1.157 0.977 1.114 3.5 0.9 0.740 0.843 0.705 0.794 0.792 0.787 0.897 0.751 0.857 4.0 0.6 0.568 0.647 0.535 0.603 0.602 0.611 0.696 0.578 0.659 4.5 0.4 0.436 0.496 0.407 0.458 0.457 0.474 0.540 0.445 0.507 5.0 0.3 0.334 0.381 0.309 0.348 0.347 0.368 0.419 0.342 0.390 b 0.530 0.530 0.550 0.550 0.550 0.508 0.508 0.525 0.525 A 4.73 5.39 4.83 5.44 5.43 4.66 5.31 4.72 5.38 E 0.4346 0.4174 0.4246 0.4301 0.4159 0.4221 0.4174 0.4191 0.4259 E = 0.4232 target Table 7. (C ontinued). X Y Y 0.0 5.0 4.700 4.680 4.670 4.660 4.860 4.720 4.700 4.750 0.5 4.0 3.753 3.737 3.729 3.721 3.881 3.773 3.757 3.799 1.0 2.8 2.997 2.984 2.978 2.971 3.099 3.016 3.003 3.038 1.5 2.2 2.393 2.383 2.378 2.373 2.474 2.410 2.400 2.429 2.0 2.0 1.911 1.903 1.899 1.895 1.976 1.927 1.919 1.943 2.5 1.5 1.526 1.519 1.516 1.513 1.578 1.540 1.534 1.554 3.0 1.0 1.218 1.213 1.211 1.208 1.260 1.231 1.226 1.243 3.5 0.9 0.973 0.969 0.967 0.965 1.006 0.984 0.980 0.994 4.0 0.6 0.777 0.774 0.772 0.770 0.803 0.786 0.783 0.795 4.5 0.4 0.620 0.618 0.616 0.615 0.641 0.629 0.626 0.635 5.0 0.3 0.495 0.493 0.492 0.491 0.512 0.502 0.500 0.508 b 0.450 0.450 0.450 0.450 0.450 0.448 0.448 0.447 A 4.70 4.68 4.67 4.66 4.86 4.72 4.70 4.75 E 0.4063 0.4139 0.4183 0.4229 0.4280 0.4159 0.4200 0.4226 E = 0.4232 target
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Statistics and Nonlinear Regression Allen R. Overman 30 Table 8. Correlation of E with b using a parabolic model with 016 5 A b E E E E E 0.577 0.4232 0.4169 0.560 0.2961 0.2892 0.2992 0.543 0.2069 0.2051 0.2039 0.2075 0.529 0.1657 0.1689 0.1616 0.1642 0.1656 0.5161 0.1521 0.1612 0.1517 0.1523 0.1520 0.503 0.1664 0.1796 0.1701 0.1677 0.1663 0.491 0.2069 0.2190 0.2121 0.2061 0.477 0.2961 0.2925 0.2915 0.462 0.4232 0.4041
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Statistics and Nonlinear Regression Allen R. Overman 31 List of Figures 1. Dependence of response variable ( Y ) on control variable ( X ). Data from Table 1. Curve drawn fro m Eq. ( 35). 2. Dependence of ln Y on X Data from Table 1. Line drawn from Eq. ( 33). 3. Scatter plot for estimated response variable ( Y ) vs. measured response variable ( Y ). Line represents the 45% diagonal. 4. Equal probability contours between parameters A and b. Contours drawn from Table 5 (75%), Table 6 (95 %) and Table 7 (99%) probability levels, respectively. Optimum and standard error values of 075 0 016 5 A and 0129.05161.0b are also shown. 5. Dependence of error sum of squares ( E ) on exponential parameter ( b) for linear parameter A = 5.016. Parabola drawn from Eq. ( 72).
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Statistics and Nonlinear Regression Allen R. Overman 32 ABE 6933 Special Topics Mathematical and Statistical Characteristics of Nonlinear Regression Models A. R. Overman I. Elements of Probability and Calculus A. Arithmetic the process of c ounting B. Natural numbers positive integers ( ,2,1,0 ) C. Rational numbers ratio of two integers ( ,3/2,3/1,2/1,,1/2,1/1 ) D. Irrational numbers (such as e, 2 etc.) E. Complex nu mbers z = x + i y with i = 1 F. Binomial theorem and Pascals triangle (a + b)0 = 1 (a + b)1 = a1 + b1 (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 (a + b)7 = a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab 6 + b7 (a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3 + 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8 (a + b)9 = a9 + 9a8b + 36a7b2 + 84a6b3 + 126a5b4 + 126a4b5 + 84a3b6 + 36a2b7 + 9ab8 + b9 Note symmetry in the distribution of coefficients for each expansion.
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Statistics and Nonlinear Regression Allen R. Overman 33 Pascals triangle for binomial coefficients 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 Note the pattern in the coefficients, including symmetry. G. Frequency distributions 1. Discrete distribution Consider the problem of a peg board. This is a two state system a cell (hole) is either filled or empty. Eac h cell holds one and only one object (peg), which can be viewed as a type of exclusion principle Define n as the total number of cells and x as the number of filled cells (pegs). Cells (holes) are indistinguishable (all alike), as are the objects (pegs). Order of filling the cells is irrelevant. Note that a peg board can be linear, triangular, rectangular (Eigen and Winkler, 1993, p. 40; Polster, 2004, p. 33), or even 3dimensional. The number of distinguishable combinations which are possible for each x, xnc, can be calculated from (Ruhla, 1992, p. 18; Watkins, 2000, p. 22) )!(!!,xnxnxnc (1)
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Statistics and Nonlinear Regression Allen R. Overman 34 where n n 3 2 1 and is called n factorial. Note that n can assume positive integers ( 3 2 1 n ) and x can also assume positive integers ( n x , 2 1 0 ). For small values of n it is easy to estimate c by intuition, but for larger n calculations of c are best performed on a pocket calculator or computer with the algorithm for computations (Eq. (1)) built in. The total number of combinations C for the system is defined as the sum of c values for all values of x and can be calculated from C = 2n. The frequency distribution of c values is then calculated from f = c/C. Cumulative frequency is calculated from the cumulative sum fF (2) so that F is normalized 10F It should be noted that F forms a discrete set of numbers for a particular case. 2. Continuous distribution The next step is to compare the discret e distribution to a continuous Gaussian distribution where x is considered a continuous variable and the cumulative distribution is described by 2 erf 1 2 1 x F (3) where and are the mean and spread of the distribution. The e rror function is defined by 202)exp(22erfxduux (4) where )exp(2u represents the Gaussian distribution (bell shaped curve). Values of the erf can be obtained from mathematical tables (cf. Abramowitz and Stegun, 1965, chp. 7). Some pr operties of the error function should be noted: erf (0) = 0, erf ( ) = 1, erf (x ) = erf (+ x ), erf ( ) = 1 Equation (3) can be rearranged to the linear form xFZ21212erf1 (5) where erf 1 is the inverse error function. For example, 00.1)8427.0(erf1 Linear regression of Z vs. x leads to values of the parameters and With these parameters now known the frequency distribution for f vs. x can be calculated for the cont inuous distribution from
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Statistics and Nonlinear Regression Allen R. Overman 35 22exp21xf (6) The procedure can now be applied to a linear peg board, triangular peg board, and square peg board. It can even be applied to a 3dimensional system. This analysis falls within a branch of mathematic s known as group theory Values of the error function can be calculated from the series approximation (Abramowitz and Stegun, 1965, p. 299) 4 4 3 2] 078108 0 000972 0 230389 0 278393 0 1 [ 1 1 erf x x x x x (7) for 8 1 0 x For the case where erf x is given, the inverse erf 1 an d therefore x can be obtained on a scientific calculator or computer using the solver routine Note that for the case F < 0.5 and 2F 1 < 0 (negative) the procedure is to change the value from to +, solve for the inverse by Eq. (7) and change the sign from + x to x Equation (7) does not work directly for x because the power series in Eq. (7) is not symmetric. H. Symmetry and conservation principle In all of the discrete and continuous Gaussian distributions we note symmetry in the distributions around a mean point. A mathematical consequence of this property is that something is conserved (remains constant) in the system. Note that the number of filled cells is defined by x Since this is a two state (binary) system (cells are either empty of filled), it follows that the number of unfilled cells is n x The total capacity of the system is the sum of filled and unfilled cells so that total capacity is = x + n x = n While this is obvious for our case, it illustrates the connection between symmetry and conservation. This property turns out to be very important in the various models of physics (including mechanics, electromagnetism, relativity, and quantum mechanics). It also shows up in chemistry and biology.
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Statistics and Nonlinear Regression Allen R. Overman 36 Frequency distributions for a linear peg board Case 1 x 2 In this case n = 1 x 2 = 2 and x = 0, 1, 2. Table A1 Frequency distribution for the linear peg board with a 1 x 2 array. x c f F Z f c 0.0000 0 1 0.2500 0.2167 0.867 0.2500 0.4767 1 2 0.5000 0.5379 2.152 0.7500 +0.4767 2 1 0.2500 0.2167 0.867 1.0000 C 4 = 22 Note symmetry in the frequency distribution. x x F Z 9534 0 9534 0 2 1 2 1 2 erf 1 r = 1 000.1,0489.12 220489.100.1exp5379.02exp21 xxf f f 2848 1 1045 0 r = 1 fc4
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Statistics and Nonlinear Regression Allen R. Overman 37 Case 1 x 3 In this case n = 1 x 3 = 3 and x = 0, 1, 2, 3. Table A2 Frequency distribution for the linear peg board with a 1 x 3 array. x c f F Z f c 0.0000 0 1 0.1250 0.1037 0.83 0.1250 0.8142 1 3 0.3750 0.3888 3.11 0.5000 0.0000 2 3 0.3750 0.3888 3.11 0.8750 +0.8142 3 1 0.1250 0.1037 0.83 1.0000 C 8 = 23 Note symmetry in the frequency distribution. x x F Z 8142 0 2213 1 2 1 2 1 2 erf 1 r = 1.0000 500.1,2282.12 222282.150.1exp4594.02exp21xxf ff1404.103885.0 r = 1.0000 fc8
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Statistics and Nonlinear Regression Allen R. Overman 38 Case 1 x 4 In this case n = 1 x 4 = 4 and x = 0, 1, 2, 3, 4. Table A3 Frequency distribution for the linear peg board with a 1 x 4 array. x c f F Z f c 0.0000 0 1 0.0625 0.0512 0.82 0.0625 1.0842 1 4 0.2500 0.2419 3.87 0.3125 0.3452 2 6 0.3750 0.4060 6.50 0.6875 +0.3452 3 4 0.2500 0.2419 3.87 0.9375 +1.0842 4 1 0.0625 0.0512 0.82 1.0000 C 16 = 24 Note symmetry in the frequency distribution. x x F Z 7196 0 4391 1 2 1 2 1 2 erf 1 r = 0. 999909 000.2,3897.12 223897.100.2exp4060.02exp21xxf f f 1052 1 0226 0 r = 0.99710 f c 16
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Statistics and Nonlinear Regression Allen R. Overman 39 Case 1 x 5 In this case n = 1 x 5 = 5 and x = 0, 1, 2, 3, 4, 5. Table A4 Frequency distribution for the linear peg board wit h a 1 x 5 array. x c f F Z f c 0.00000 0 1 0.03125 0.02590 0.83 0.03125 1.3148 1 5 0.15625 0.14143 4.53 0.18750 0.6277 2 10 0.31250 0.33050 10.58 0.50000 0.0000 3 10 0.31250 0.33050 10.58 0.81250 +0.6277 4 5 0.15625 0.14143 4.53 0.96875 +1.3148 5 1 0.03125 0.02590 0.83 1.00000 C 32 = 25 Note symmetry in the frequency distribution. x x F Z 65146 0 6286 1 2 1 2 1 2 erf 1 r = 0.99983 500.2,5350.12 225350.150.2exp3675.02exp21xxf f f 0882 1 01543 0 r = 0.99723 fc32
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Statistics and Nonlinear Regression Allen R. Overman 40 Case 1 x 6 In this case n = 1 x 6 = 6 and x = 0, 1, 2, 3, 4, 5, 6. Table A5 Frequency distribution for the linear peg board with a 1 x 6 array x c f F Z f c 0.000000 0 1 0.015625 0.013247 0.85 0.015625 1.5213 1 6 0.093750 0.080181 5.13 0.109375 0.8703 2 15 0.234375 0.236181 15.11 0.343750 0.2841 3 20 0.312500 0.338560 21.67 0.656250 +0.2841 4 15 0.234375 0.236181 15.11 0.890625 +0.8703 5 6 0.093750 0.080181 5.13 0.984375 +1.5213 6 1 0.015625 0.013247 0.85 1.000000 C 64 = 26 Note symmetry in the frequency distribution. x x F Z 60009 0 8003 1 2 1 2 1 2 erf 1 r = 0.99975 000 3 66643 1 2 2266643.100.3exp33856.02exp21xxf f f 0809 1 01187 0 r = 0.99730 f c 64
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Statistics and Nonlinear Regression Allen R. Overman 41 Case 1 x 7 In this case n = 1 x 7 = 7 and x = 0, 1, 2, 3, 4, 5, 6, 7. Table A6 Frequency distribution for the linear peg board with a 1 x 7 array. x c f F Z f 0.0000000 0 1 0.0078125 0.008580 0.0078125 1.7123 1 7 0.0546875 0.049286 0.0625000 1.0842 2 21 0.1640625 0.158080 0.2265625 0.5305 3 35 0.2734375 0.283108 0.5000000 0.0000 4 35 0.2734375 0.283108 0.7734375 +0.5305 5 21 0.1640625 0.158080 0.9375000 +1.0842 6 7 0.0546875 0.049286 0.9921875 +1.7123 7 1 0.0078125 0.008580 1.0000000 C 128 = 27 Note symmetry in the frequency distribution. x x F Z 53978 0 88923 1 2 1 2 1 2 erf 1 r = 0.999963 500 3 8526 1 2 2 28526 1 50 3 exp 3045 0 2 exp 2 1 x x f f f 0689 1 011875 0 r = 0.999198
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Statistics and Nonlinear Regression Allen R. Overman 42 Case 1 x 8 In this case n = 1 x 8 = 8 and x = 0, 1, 2, 3, 4, 5, 6, 7, 8. Table A7 Frequency distribution for the linear peg board with a 1 x 8 array. x c f F Z f 0.000000 0 1 0.003906 0.004626 0.5 0.003906 1.8934 0.012167 1 8 0.031250 0.028129 1.5 0.035156 1.2777 0.057167 2 28 0.109375 0.102126 2.5 0.144531 0.7504 0.160373 3 56 0.218750 0.221375 3.5 0.363281 0.2470 0.268612 4 70 0.273438 0.286500 4.5 0.636719 +0.2470 0.268612 5 56 0.218750 0.221375 5.5 0.855469 +0.7504 0.160373 6 28 0.109375 0.102126 6.5 0.964844 +1.2777 0.057167 7 8 0.031250 0.028129 7.5 0.996094 +1.8934 0.012167 8 1 0.003906 0.004626 1.000000 C 256 = 28 Note symmetry in the frequency distribution. xxFZ5078.00312.221212erf1 r = 0.999946 0000.4,9692.12 229692.100.4exp2865.02exp21xxf f f 0589 1 008700 0 r = 0.999084
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Statistics and Nonlinear Regression Allen R. Overman 43 Frequency distributions of a triangular peg board Case 1: 1 Cell ( n = 1) For this case n = 1 and x = 0 or 1. The distribution is given in Table A8. Table A8 Frequency distribution for a triangular peg board with 1 cell. x c f F 0.000 0 1 0.500 0.500 1 1 0.500 1.000 C 2 = 21 Case 2: 3 Cells ( n = 3) For this case n = 3 and x = 0, 1, 2, or 3. The pegboard is shown in the diagram. The corresponding distribution is given in Table A9. Table A9 Frequency distribution for a triangular peg board with 3 cell s x c f F Z f 0.0000 0 1 0.1250 0.1034 0.5 0.1250 0.8142 0.2367 1 3 0.3750 0.3892 1.5 0.5000 0.0000 0.4594 2 3 0.3750 0.3892 2.5 0.8750 +0.8142 0.2367 3 1 0.1250 0.1034 1.0000 C 8 = 23 xxFZ8142.02213.1212)12(erf1 r = 1.0000 500 1 2282 1 2 222282.1500.1exp4594.02exp21xxf f f 1432 1 03950 0 r = 1.000000
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Statistics and Nonlinear Regression Allen R. Overman 44 This system is an example of group theory in mathematics, which links principles of symmetry and conservation. Note symmetry of c around the mean value of x ( 50.1 ). Conservation comes from Filled cells + Unfilled cells = x + ( n x ) = n = total capacity of the system (number of cells) Case 3: 6 Cells ( n = 6) For this case n = 6 and x = 0, 1, 2, 3, 4, 5, 6. The distribution is given in Table A10. Table A10 Frequency distri bution for a triangular peg board with 6 cell s x c f F Z f 0.000000 0 1 0.015625 0.013246 0.015625 1.5213 1 6 0.093750 0.080177 0.109375 0.8703 2 15 0.234375 0.236178 0.343750 0.2841 3 20 0.312500 0.338560 0.656250 +0.2841 4 15 0.234375 0.236178 0.890625 +0.8703 5 6 0.093750 0.080177 0.984375 +1.5213 6 1 0.015625 0.013246 1.000000 C 64 = 26 xxFZ6001.08003.1212)12(erf1 r = 0.999748 000 3 6664 1 2 226664.100.3exp33856.02exp21xxf f f 0809 1 01187 0 r = 0.99730
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Statistics and Nonlinear Regression Allen R. Overman 45 Case 4: 10 Cells ( n = 10) For this case n = 10 and x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The distribution is given in Table A11. Table A11 Frequency distribution for a triangula r peg board with 10 cell s x c f F Z f 0.0000000 0 1 0.0009766 0.001342 0.0009766 1 10 0.0097656 0.008921 0.0107422 1.6260 2 45 0.0439453 0.038930 0.0546875 1.1311 3 120 0.1171875 0.111513 0.1718750 0.6700 4 210 0.2050781 0.209679 0.3769531 0.2215 5 252 0.2460938 0.258800 0.6230469 +0.2215 6 210 0.2050781 0.209679 0.8281250 +0.6700 7 120 0.1171875 0.111513 0.9453125 +1.1311 8 45 0.0439453 0.038930 0.9892578 +1.6260 9 10 0.0097656 0.008921 0.9990234 10 1 0.0009766 0.001342 1.0000000 C 1024 = 210 xxFZ4588.02939.2212)12(erf1 r = 0.99988 000.5,1797.22 221797.2000.5exp2588.02exp21xxf f f 0383 1 00352 0 r = 0.99894
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Statistics and Nonlinear Regression Allen R. Overman 46 Case 5: 15 Cells ( n = 15) For this case n = 15 and x = 15 , 3 2 1 0 The distribution is given in Table A12. Table A12 Frequency distribution for a triangular peg board with 15 cell s x c f F Z f 0.0000000 0 1 0.0000305 0.000096 0.0000305 1 15 0.0004578 0.000652 0.0004883 2 105 0.0032043 0.003355 0.0036926 3 455 0.0138855 0.013138 0.0175781 1.4875 4 1365 0.0416565 0.039158 0.0592346 1.1032 5 3003 0.0916443 0.088829 0.1508789 0.7309 6 5005 0.1527405 0.153358 0.3036194 0.3630 7 6435 0.1963806 0.201504 0.5000000 0.0000 8 6435 0.1963806 0.201504 0.6963806 +0.3630 9 5005 0.1527405 0.153358 0.8491211 +0.7309 10 3003 0.0916443 0.088829 0.9407654 +1.1032 11 1365 0.0416565 0.039158 0.9824219 +1.4875 12 455 0.0138855 0.013138 0.9963074 13 105 0.0032043 0.003355 0.9995117 14 15 0.0004578 0.000652 0.9999695 15 1 0.0000305 0.000096 1.0000000 C 32768 = 215 xxFZ3695.07711.2212)12(erf1 r = 0.999972 500.7,7065.22
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Statistics and Nonlinear Regression Allen R. Overman 47 2 27065 2 500 7 exp 2085 0 2 exp 2 1 x x f ff02423.1002046.0 r = 0.999637
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Statistics and Nonlinear Regression Allen R. Overman 48 Case 6: 21 Ce lls ( n = 21) For this case n = 21 and x = 21 , 3 2 1 0 The distribution is given in Table A13. Table A13 Frequency distribution for a triangular peg board with 21 cell s x c f F Z f 0.000000 0 1 0.000000 0.000004 0.000000 1 21 0.000010 0.000028 0.000010 2 210 0.000100 0.000159 0.000110 3 1,330 0.000634 0.000749 0.000744 4 5,985 0.002854 0.002913 0.003598 5 20,349 0.009703 0.009333 0.013301 1.5668 6 54,264 0.025875 0.024629 0.039176 1.2429 7 116,280 0.055447 0.053525 0.094623 0.9287 8 203,490 0.097032 0.095807 0.191655 0.6168 9 293,930 0.140157 0.141240 0.331812 0.3071 10 352,716 0.168188 0.171489 0.500000 0.0000 11 352,716 0.168188 0.171489 0.668188 +0.3071 12 293,930 0.140157 0.141240 0.808345 +0.6168 13 203,490 0.097032 0.095807 0.905377 +0.9287 14 116,280 0.055447 0.053525 0.960824 +1.2429 15 54,264 0.025875 0.024629 0.986699 +1.5668 16 20,349 0.009703 0.009333 0.996402 17 5,985 0.002854 0.002913 0.999256 18 1,330 0.000634 0.000749 0.999890
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Statistics and Nonlinear Regression Allen R. Overman 49 19 210 0.000100 0.000159 0.999990 20 21 0.000010 0.000028 1.000000 21 1 0.000000 0.000004 1.000000 C 2,097,152 = 221 x x F Z 3115 0 2707 3 2 1 2 ) 1 2 ( erf 1 r = 0.999982 500 10 2103 3 2 222103.3500.10exp1757.02exp21xxf ff0183.100135.0 r = 0.99980
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Statistics and Nonlinear Regression Allen R. Overman 50 Case 7: 28 Cells ( n = 28) For this case n = 28 and x = 28 , 3 2 1 0 The distribution is given in Table A14. Table A14 Frequency distribution for a triangular peg board with 28 cell s x c f F Z f 0.0000000 0 1 0.0000000 0.000000 0.0000000 1 28 0.0000001 0.000001 0.0000001 2 378 0.0000014 0.000004 0.0000015 3 3,276 0.0000122 0.000022 0.0000137 4 20,475 0.0000763 0.000104 0.0000900 5 98,280 0.0003661 0.000414 0.0004561 6 376,740 0.0014035 0.001431 0.0018596 7 1,184,040 0.0044109 0.004273 0.0062705 1.7706 8 3,108,105 0.0115786 0.011029 0.0178491 1.4830 9 6,906,900 0.0257302 0.024603 0.0435793 1.2079 10 13,123,110 0.0488874 0.047433 0.0924667 0.9378 11 21,474,180 0.0799975 0.079034 0.1724642 0.6684 12 30,421,755 0.1133299 0.113814 0.2857941 0.3996 13 37,442,160 0.1394829 0.141652 0.4252770 0.1335 14 40,116,600 0.1494460 0.152370 0.5747230 +0.1335 15 37,442,160 0.1394829 0.141652 0.7142059 +0.3996 16 30,421,755 0.1133299 0.113814 0.8275358 +0.6684 17 21,474,180 0.0799975 0.079034 0.9075333 +0.9378 18 13,123,110 0.0488874 0.047433 0.9564207 +1.2079
PAGE 53
Statistics and Nonlinear Regression Allen R. Overman 51 19 6,906,900 0.0257302 0.024603 0.9821509 +1.4830 20 3,108,105 0.0115786 0.011029 0.9937295 +1.7706 21 1,184,040 0.0044109 0.004273 0.9981404 22 376,740 0.0014035 0.001431 0.9995439 23 98,280 0.0003661 0.000414 0.9999100 24 20,475 0.0000763 0.000104 0.9999863 25 3,276 0.0000122 0.000022 0.9999985 26 378 0.0000014 0.000004 0.9999999 27 28 0.0000001 0.000001 1.0000000 28 1 0.0000000 0.000000 1.0000000 C 268,435,456 = 228 xxFZ27007.07810.3212)12(erf1 r = 0.999974 000.14,70275.32 2270275.300.14exp15237.02exp21xxf f f 01674 1 000995 0 r = 0.99985
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Statistics and Nonlinear Regression Allen R. Overman 52 Frequency distributions for a square peg board. Case 1: 2 x 2 For this case n = 2 x 2 = 4 and x assumes values of 0, 1, 2, 3, and 4. Corresponding values of x n c are calculated from Eq. (1). Results are given in Table A15 Table A15 Frequency dist ribution for the square peg board with a 2 x 2 array. x c f F Z f 0.0000 0 1 0.0625 0.05117 0.0625 1.0842 1 4 0.2500 0.24191 0.3125 0.3452 2 6 0.3750 0.40600 0.6875 +0.3452 3 4 0.2500 0.24191 0.9375 +1.0842 4 1 0.0625 0.05117 1.0000 C 16 = 24 Note the symmetry in the discrete frequency distribution. x x F Z 7196 0 4391 1 2 1 2 1 2 erf 1 r = 0.999909 000.2,3897.12 223897.100.2exp4060.02exp21xxf f f 1053 1 02262 0 r = 0.9971
PAGE 55
Statistics and Nonlinear Regression Allen R. Overman 53 Case 2: 3 x 3 For this case n = 3 x 3 = 9 and x assumes values of 0, 1, 2, 3, 9, Corresponding values of xnc, are calculated from Eq. (1). Results are given in Table A16. Table A16 Frequency distribution for the square peg board with a 3 x 3 array. x c f F Z f 0.000000 0 1 0.001953 0.002497 0.001953 1 9 0.017578 0.015920 0.019531 1.4568 2 36 0.070312 0.063869 0.089843 0.9491 3 84 0.164063 0.161257 0.253906 0.4680 4 126 0.246094 0.256230 0.500000 0.0000 5 126 0.246094 0.256230 0.746094 +0.4680 6 84 0.164063 0.161257 0.910157 +0.9491 7 36 0.070312 0.063869 0.980469 +1.4568 8 9 0.017578 0.015920 0.998047 9 1 0.001953 0.002497 1.000000 Total 512 = 29 Again note the symmetry in the discrete frequency distribution. xxFZ4812.01653.221212erf1 r = 0.999919 500.4,0782.22 2 20782 2 50 4 exp 2715 0 2 exp 2 1 x x f ff0369.100373.0 r = 0.99895
PAGE 56
Statistics and Nonlinear Regression Allen R. Overman 54 Case 3: 4 x 4 For this case n = 4 x 4 = 16 and x assumes values of 0, 1, 2, 3, 16, Corresponding values of xnc, are calculated f rom Eq. (1). Results are given in Table A17. Table A17 Frequency distribution for the square peg board with a 4 x 4 array. x c f F Z f 0.0000000 0 1 0.0000153 0.000044 0.0000153 1 16 0.0002441 0.000319 0.0002594 2 120 0.0018311 0.001966 0.0020905 3 560 0.0085449 0.008102 0.0106354 1.6287 4 1820 0.0277710 0.025807 0.0384064 1.2493 5 4368 0.0666504 0.063543 0.1050568 0.8868 6 8008 0.1221924 0.120946 0.2272492 0.5288 7 11440 0.1745605 0.177953 0.4018097 0.1759 8 12870 0.1963806 0.202400 0.5981903 +0.1759 9 11440 0.1745605 0.177953 0.7727508 +0.5288 10 8008 0.1221924 0.120946 0.8949432 +0.8868 11 4368 0.0666504 0.063543 0.9615936 +1.2493 12 1820 0.0277710 0.025807 0.9893646 +1.6287 13 560 0.0085449 0.008102 0.9979095 14 120 0.0018311 0.001966 0.9997406 15 16 0.0002441 0.000369 0.9999847 16 1 0.0000153 0.000053 1.0000000 C 65536 = 216 Note the symmetry in the frequency distribution.
PAGE 57
Statistics and Nonlinear Regression Allen R. Overman 55 x x F Z 3588 0 8703 2 2 1 2 1 2 erf 1 r = 0.999958 000.8,7872.22 2 27872 2 00 8 exp 2024 0 2 exp 2 1 x x f ff0248.100194.0 r = 0.99962 Several characteristics should be noted from these calculations. First, note the symmetry in the distributions in the tables. Second, note that the continuous Gaussian function approximates the discrete distributions rather well. Third, as the number o f values increases, agreement between the discrete and continuous distributions improves.
PAGE 58
Statistics and Nonlinear Regression Allen R. Overman 56 Case 3: 5 x 5 For this case n = 5 x 5 = 25 and x assumes values of 0, 1, 2, 3, 25 Corresponding values of xnc, are calculated from Eq. (1). Results are given in Table A18. Table A18 Frequency distribution for the square peg board with a 5 x 5 array. x c f F Z f 0.00000000 0 1 0.00000003 0.000000 0.00000003 1 25 0.00000075 0.000003 0.00000078 2 300 0.00000894 0.000020 0.00000972 3 2,300 0.00006855 0.000101 0.00007827 4 12,650 0.00037700 0.000439 0.00045527 5 53,130 0.00158340 0.001624 0.00203867 6 177,100 0.00527799 0.005101 0.00731666 1.7297 7 480,700 0.01432598 0.013604 0.02164264 1.4266 8 1,081,575 0.03223345 0.030810 0.05387609 1.1362 9 2,042,975 0.06088540 0.059254 0.11476149 0.8503 10 3,268,760 0.09741664 0.096770 0.21217813 0.5650 11 4,457,400 0.13284087 0.134200 0.34501900 0.2816 12 5,200,300 0.15498102 0.158037 0.50000002 0.0000 13 5,200,300 0.15498087 0.158037 0.65498089 +0.2816 14 4,457,400 0.13284087 0.134200 0.78782176 +0.5650 15 3,268,760 0.09741664 0.096770 0.88523840 +0.8503 16 2,042,975 0.06088540 0.059254 0.94612380 +1.1362 17 1,081,575 0.03223345 0.030810 0.97835725 + 1.4266 18 480,700 0.01432598 0.013604
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Statistics and Nonlinear Regression Allen R. Overman 57 0.99268323 +1.7297 19 177,100 0.00527799 0.005101 0.99796122 20 53,130 0.00158340 0.001624 0.99954462 21 12,650 0.00037700 0.000439 0.99992162 22 2,300 0.00006855 0.000101 0.99999017 23 300 0.00000894 0.000020 0.99999911 24 25 0.00000075 0.000003 0.99999986 25 1 0.00000003 0.000000 0.99999989 C 33,554,432 = 225 Note the symmetry in the frequency distribution. x x F Z 2859 0 5740 3 2 1 2 1 2 erf 1 r = 0.999974 50 12 4975 3 2 224975.350.12exp1613.02exp21xxf ff017789.1001129.0 r = 0.999827
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Statistics and Nonlinear Regression Allen R. Overman 58 Case 4: 6 x 6 For this case n = 6 x 6 = 36 and x assumes values of 0, 1, 2, 3, 36 Corresponding values of xnc, are calculated from Eq. (1). Results are given in Table A19. Table A19 Frequency distribution for the square peg board with a 6 x 6 array. x c f F Z f 0.000000 0 1 0.000000 1 36 0.00000000 0.000000 0.00000000 2 630 0.00000001 0.000000 0.00000001 3 7,140 0.00000010 0.000000 0.00000010 4 58,905 0.00000086 0.000002 0.00000096 5 376,992 0.00000549 0.000010 0.00000645 6 1,947,792 0.00002834 0.000039 0.00003479 7 8,347,680 0.00012147 0.000144 0.00015626 8 30,260,340 0.00044035 0.000472 0.00059661 9 94,143,280 0.00136997 0.001380 0.00196658 10 254,186,856 0.00369891 0.003606 0.00566549 1.7970 11 600,805,296 0.00874287 0.008415 0.01440836 1.5442 12 1,251,677,700 0.01821431 0.017541 0.03262267 1.3013 13 2,310,789,600 0.03362641 0.032657 0.06624908 1.0633 14 3,796,297,200 0.05524340 0.054303 0.12149248 0.8263 15 5,567,902,560 0.08102365 0.080647 0.20251613 0.5890 16 7,307,872,110 0.10634354 0.106974 0.30885967 0.3524 17 8,597,496,600 0.12511004 0.126733 0.43396971 0.1179 18 9,075,135,300 0.13206060 0.134100
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Statistics and Nonlinear Regression Allen R. Overman 59 0.56603031 +0.1179 19 8,597,496,600 0.12511004 0.126733 0.69114035 +0.3524 20 7,307,872,110 0.10634354 0.106974 0.79748389 +0.5890 21 5,567,902,560 0.08102365 0.080647 0.87850754 +0.8263 22 3,796,297,200 0.05524340 0.054303 0.93375094 +1.0633 23 2,310,789,600 0.03362641 0.032657 0.96737735 +1.3013 24 1,251,677,700 0.01821431 0.017541 0.98559166 +1.5442 25 600,805,296 0.00874287 0.008415 0.99433453 +1.7970 26 254,186,856 0.00369891 0.003606 0.99803344 27 94,143,280 0.00136997 0.001380 0.99940341 28 30,260,340 0.00044035 0.000472 0.99984376 29 8,347,680 0.00012147 0.000144 0.99996523 30 1,947,792 0.00002834 0.000039 0.99999357 31 376,992 0.00000549 0.000010 0.99999906 32 58,905 0.00000086 0.000002 0.99999992 33 7,140 0.00000010 0.000000 1.00000002 34 630 0.00000001 1.00000003 35 36 0.00000000 1.00000003 36 1 0.00000000 1.00000003 C 68,719,476,736 = 236 Note the symmetry in the frequency distribution.
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Statistics and Nonlinear Regression Allen R. Overman 60 x x F Z 2377 0 2786 4 2 1 2 1 2 erf 1 r = 0.999981 00.18,2070.42 222070.400.18exp1341.02exp21xxf ff01356.1000723.0 r = 0.999909
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Statistics and Nonlinear Regression Allen R. Overman 61 Case 4: 7 x 7 For this case n = 7 x 7 = 49 and x assumes values of 0, 1, 2, 3, 49 Corr esponding values of xnc, are calculated from Eq. (1). Results are given in Table A20. Table A20 Frequency distribution for the square peg board with a 7 x 7 array. x c f F Z f 0.000000 0 1 0.000000 0.000000 1 49 0.000000 0.000000 0.000000 2 1,176 0.000000 0.000000 0.000000 3 18,424 0.000000 0.000000 0.000000 4 211,876 0.000000 0.000002 0.000000 5 1,906,884 0.000000 0.000000 0.000000 6 13,983,816 0.000000 0.000000 0.000000 7 85,900,584 0.000000 0.000000 0.000000 8 450,978,066 0.000001 0.000002 0.000001 9 2,054,455,634 0.000004 0.000006 0.000005 10 8,217,822,536 0.000015 0.000020 0.000020 11 29,135,916,264 0.000052 0.000064 0.000072 12 92,263,734,836 0.000164 0.000185 0.000236 13 262,596,783,764 0.000466 0.000497 0.000702 14 675,248,872,536 0.001199 0.001228 0.001901 15 1,575,580,703,000 0.002799 0.002795 0.004700 16 3,348,108,993,000 0.005947 0.005859 0.010647 1.6285 17 6,499,270,398,000 0.011545 0.011315 0.022192 1.4192 18 11,554,258,486,000 0.020524 0.020124
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Statistics and Nonlinear Regression Allen R. Overman 62 0.042716 1.2145 19 18,851,684,898,000 0.033487 0.032967 0.076203 1.0119 20 28,277,527,346,000 0.050231 0.049740 0.126434 0.8093 21 39,049,918,716,000 0.069367 0.069120 0.195801 0.6061 22 49,699,896,548,000 0.088285 0.088467 0.284086 0.4032 23 58,343,356,817,000 0.103639 0.104288 0.387725 0.2016 24 63,205,303,219,000 0.112275 0.113230 0.500000 0.0000 25 63,205,303,219,000 0.112275 0.113230 0.612275 +0.2016 26 58,343,356,817,000 0.103639 0.104288 0.715914 +0.4032 27 49,699,896,548,000 0.088285 0.088467 0.804199 +0.6061 28 39,049,918,716,000 0.069367 0.069120 0.873566 +0 .8093 29 28,277,527,346,000 0.050231 0.049740 0.923797 +1.0119 30 18,851,684,898,000 0.033487 0.032967 0.957284 +1.2145 31 11,554,258,486,000 0.020524 0.020124 0.977808 +1.4192 32 6,499,270,398,000 0.011545 0.011315 0.989353 +1.6285 33 3,348,108,993,000 0.005947 0.005859 0.995300 34 1,575,580,703,000 0.002799 0.002795 0.998099 35 675,248,872,536 0.001199 0.001228 0.999298 36 262,596,783,764 0.000466 0.000497 0.999764 37 92,263,734,836 0.000164 0.000185 0.999928 38 29,135,916,264 0.000052 0.000064 0.999980 39 8,217,822,536 0.000015 0.000020 0.999995 40 2,054,455,634 0.000004 0.000006 0.999999 41 450,978,066 0.000001 0.000002
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Statistics and Nonlinear Regression Allen R. Overman 63 1.000000 42 85,900,584 0.000000 0.000000 43 13,983,816 44 1,906,884 45 211,876 46 18,424 47 1,176 48 49 49 1 C 562,949,953,421,000 = 249 Note the symmetry in the frequency distribution. x x F Z 20281 0 9687 4 2 1 2 1 2 erf 1 r = 0.9999962 50.24,9308.42 2 29308 4 50 24 exp 1144 0 2 exp 2 1 x x f f f 00727 1 000345 0 r = 0.999965
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Statistics and Nonlinear Regression Allen R. Overman 64 Case 5: 8 x 8 For this case n = 8 x 8 = 64 and x assumes values of 0, 1, 2, 3, 64 Corresponding values of xnc, are calculated from Eq. (1). Results are given in Table A2 1. Table A21 Frequency distribution for the square peg board with a 8 x 8 array. x c f F Z f 0.000000 0 1 0.000000 0.000000 1 64 0.000000 0.000000 0.000000 2 3 4 7 8 0.00000000 1018 9 0.00000003 1018 10 0.00000015 1018 11 0.00000074 1018 0.0000000 0.0000000 12 0.00000328 1018 0.0000002 0.000000 0.0000002 13 0.00001314 1018 0.0000007 0.000001 0.0000009 14 0.00004786 1018 0.0000026 0.000004 0.0000035 15 0.00015952 1018 0.0000086 0.000011 0.0000121 16 0.00048853 1018 0.0000265 0.000032 0.0000386 17 0.00137937 1018 0.0000748 0.000084 0.0001134 18 0.00360169 1018 0.0001952 0.000210 0.0003086 19 0.00871988 1018 0.0004727 0.000491 0.0007813
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Statistics and Nonlinear Regression Allen R. Overman 65 20 0.01961973 1018 0.0010636 0.001079 0.0018449 21 0.04110800 1018 0.0022285 0.002224 0.0040734 22 0.08034745 1018 0.0043556 0.004306 0.0084390 1.6919 23 0.14672143 1018 0.0079538 0.007828 0.0163828 1.5077 24 0.25064910 1018 0.0135877 0.013363 0.0299705 1.3279 25 0.40103857 1018 0.0217403 0.021421 0.0517108 1.1504 26 0.60155785 1018 0.0326105 0.032244 0.0843213 0.9736 27 0.84663698 1018 0.0458963 0.045576 0.1302176 0.7965 28 1.11877029 1018 0.0606487 0.060493 0.1908663 0.6189 29 1.38881829 1018 0.0752880 0.075395 0.2661543 0.4412 30 1.62028801 1018 0.0878360 0.088238 0.3539903 0.2645 31 1.77709008 1018 0.0963362 0.096971 0.4503265 0.0887 32 1.83262414 1018 0.0993468 0.100070 0.5496733 +0.0887 33 1.77709008 1018 0.0963362 0.096971 0.6460095 +0.2645 34 1.62028801 1018 0.0878360 0.088238 0.7338455 +0.4412 35 1.38881829 1018 0.0752880 0.075395 0.8091335 +0.6189 36 1.11877029 1018 0.0606487 0.060493 0.8697822 +0.7965 37 0.84663698 1018 0.0458963 0.045576 0.9156785 +0.9736 38 0.60155785 1018 0.0326105 0.032244 0.9482890 +1.1504 39 0.40103857 1018 0.0217403 0.021421 0.9700293 +1.3279 40 0.25064910 1018 0.0135877 0.013363 0.9836170 +1.5077 41 0.14672143 1018 0.0079538 0.007828 0.9915708 +1.6919 42 0.08034745 1018 0.0043556 0.004306 0.9959264
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Statistics and Nonlinear Regression Allen R. Overman 66 43 0.04110800 1018 0.0022285 0.002224 0.9981549 44 0.01961973 1018 0.0010636 0.001079 0.9992185 45 0.00871988 1018 0.0004727 0.000491 0.9996912 46 0.00360169 1018 0.0001952 0.000210 0.9998864 47 0.00137937 1018 0.0000748 0.000084 0.9999612 48 0.00048853 1018 0.0000265 0.000032 0.9999877 49 0.00015952 1018 0.0000086 0.000011 0.9999963 50 0.00004786 1018 0.0000026 0.000004 0.9999989 51 0.00001314 1018 0.0000007 0.000001 0.9999996 52 0.00000328 1018 0.0000002 0.000000 0.9999998 53 0.00000074 1018 0.0000000 0.9999998 54 0.00000015 1018 55 0.00000003 1018 56 0.00000000 1018 57 58 59 60 61 62 63 64 64 1 C 18.4467441 1018 = 264
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Statistics and Nonlinear Regression Allen R. Overman 67 Note the symmetry in the frequency distribution. x x F Z 17737 0 67576 5 2 1 2 1 2 erf 1 r = 0.9999963 000.32,6380.52 226380.500.32exp10007.02exp21xxf f f 00622 1 000252 0 r = 0.999978
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Statistics and Nonlinear Regression Allen R. Overman 68 Case 6: 9 x 9 For this case n = 9 x 9 = 81 and x assumes values of 0, 1, 2, 3, 81 Corresponding values of xnc, are calculated from Eq. (1). Results are given in Table A22. Table A22 Frequency distribution for the square p eg board with a 9 x 9 array. x c f F Z f 0.000000 0 1 0.000000 0.000000 1 81 0.000000 0.000000 0.000000 2 3 4 5 6 14 0.00000002 1023 15 0.00000008 1023 16 0.00000034 1023 0.0000000 17 0.00000128 1023 0.0000000 0.0000000 18 0.00000457 1023 0.0000002 0.000000 0.0000002 19 0.00001514 1023 0.0000006 0.000001 0.0000008 20 0.00004694 1023 0.0000019 0.000003 0.0000027 21 0.00013636 1023 0.0000056 0.000007 0.0000083 22 0.00037190 1023 0.0000154 0.000018 0.0000237 23 0.00095400 1023 0.0000395 0.000044 0.0000632 24 0.00230549 1023 0.0000953 0.000103
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Statistics and Nonlinear Regression Allen R. Overman 69 0.0001585 25 0.00525652 1023 0.0002174 0.000228 0.0003759 26 0.01132173 1023 0.0004683 0.000480 0.0008442 27 0.02306279 1023 0.0009539 0.000961 0.0017981 28 0.04447825 1023 0.0018396 0.001834 0.0036377 29 0.08128783 1023 0.0033620 0.003329 0.0069997 1.7415 30 0.14089890 1023 0.0058274 0.005750 0.0128271 1.5769 31 0.23180142 1023 0.0095871 0.009449 0.0224142 1.4162 32 0.36218972 1023 0.0149798 0.014777 0.0373940 1.2579 33 0.53779686 1023 0.0222428 0.021988 0.0596368 1.1008 34 0.75924263 1023 0.0314015 0.031133 0.0910383 0.9439 35 1.01955439 1023 0.0421678 0.041944 0.1332061 0.7866 36 1.30276395 1023 0.0538811 0.053771 0.1870872 0.6288 37 1.58444264 1023 0.0655310 0.065592 0.2526182 0.4709 38 1.83461779 1023 0.0758780 0.076134 0.3284962 0.3136 39 2.02278372 1023 0.0836604 0.084087 0.4121566 0.1571 40 2.12392290 1023 0.0878434 0.088370 0.5000000 0.0000 41 2.12392290 1023 0.0878434 0.088370 0.5878434 +0.1571 42 2.02278372 1023 0.0836604 0.084087 0.6715038 +0.3136 43 1.83461779 1023 0.0758780 0.076134 0.7473818 +0.4709 44 1.58444264 1023 0.0655310 0.065592 0.8129128 +0.6288 45 1.30276395 1023 0.0538811 0.053771 0.8667939 +0.7866 46 1.01955439 1023 0.0421678 0.041944 0.9089617 +0.9439 47 0.75924263 1023 0.0314015 0.031133
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Statistics and Nonlinear Regression Allen R. Overman 70 0.9403632 +1.1008 48 0.53779686 1023 0.0222428 0.021988 0.9626060 +1.2579 49 0.36218972 1023 0.0149798 0.014777 0.9775858 +1.4162 50 0.23180142 1023 0.0095871 0.009449 0.9871729 +1.5769 51 0.14089890 1023 0.0058274 0.005750 0.9930003 +1.7415 52 0.08128783 1023 0.0033620 0.003329 0.9963623 53 0.04447825 1023 0.0018396 0.001834 0.9982019 54 0.02306279 1023 0.0009539 0.000961 0.9991558 55 0.01132173 1023 0.0004683 0.000480 0.9996241 56 0.00525652 1023 0.0002174 0.000228 0.9998415 57 0.00230549 1023 0.0000953 0.000103 0.9999368 58 0.00095400 1023 0.0000395 0.000044 0.9999763 59 0.00037190 1023 0.0000154 0.000018 0.9999917 60 0.00013636 1023 0.0000056 0.000007 0.9999973 61 0.00004694 1023 0.0000019 0.000003 0.9999992 62 0.00001514 1023 0.0000006 0.000001 0.9999998 63 0.00000457 1023 0.0000002 0.000000 1.0000000 64 0.00000128 1023 0.0000000 0.000000 1.0000000 65 0.00000034 1023 0.0000000 1.0000000 66 0.00000008 1023 0.000000 1.0000000 67 0.00000002 1023 0.000000 1.0000000 68 0.00000000 1023 0.000000 1.0000000 69 0.00000000 1023 0.000000 1.0000000 C 24.1785164 1023 = 281
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Statistics and Nonlinear Regression Allen R. Overman 71 Note the symmetry in the frequency distribution. x x F Z 157606 0 38305 6 2 1 2 1 2 erf 1 r = 0.9999962 5000 40 34493 6 2 2234493.650.40exp08892.02exp21xxf f f 00595 1 000232 0 r = 0.999985
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Statistics and Nonlinear Regression Allen R. Overman 72 Case 7: 10 x 10 For this case n = 10 x 10 = 100 and x assumes values of 0, 1, 2, 3, 100 Corresponding values of xnc, are calculated from Eq. (1). Results are given in Table A23. Table A23 Frequency distribution for the square peg board with a 10 x 10 array. x c f F Z f 0.000000 0 1 0.000000 0.000000 1 100 0.000000 0.000000 0.000000 2 3 4 5 6 20 0.00000001 1029 21 0.00000002 1029 22 0.00000007 1029 23 0.00000025 1029 0.0000000 0.0000000 24 0.00000080 1029 0.0000001 0.0000001 25 0.00000243 1029 0.0000002 0.000000 0.0000003 26 0.00000700 1029 0.0000006 0.000001 0.0000009 27 0.00001917 1029 0.0000015 0.000002 0.0000024 28 0.00004999 1029 0.0000039 0.000005 0.0000063 29 0.00012411 1029 0.0000098 0.000011 0.0000161 30 0.00029372 1029 0.0000232 0.000026
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Statistics and Nonlinear Regression Allen R. Overman 73 0.0000393 31 0.00066325 1029 0.0000523 0.000056 0.0000916 32 0.00143012 1029 0.0001128 0.000118 0.0002044 33 0.00294692 1029 0.0002325 0.000239 0.0004369 34 0.00580717 1029 0.0004581 0.000465 0.0008950 35 0.01095067 1029 0.0008639 0.000867 0.0017589 36 0.01977205 1029 0.0015597 0.001553 0.0033186 37 0.03420030 1029 0.0026979 0.002673 0.0060165 1.7815 38 0.05670049 1029 0.0044729 0.004420 0.0104894 1.6325 39 0.09013924 1029 0.0071107 0.007020 0.0176001 1.4871 40 0.13746234 1029 0.0108439 0.010708 0.0284440 1.3440 41 0.20116440 1029 0.0158691 0.015692 0.0443131 1.2024 42 0.28258809 1029 0.0222923 0.022088 0.0666054 1.0613 43 0.38116533 1029 0.0300686 0.029866 0.0966740 0.9202 44 0.49378236 1029 0.0389526 0.038790 0.1356266 0.7787 45 0.61448471 1029 0.0484743 0.048394 0.1841009 0.6367 46 0.73470998 1029 0.0579584 0.057996 0.2420593 0.4946 47 0.84413487 1029 0.0665905 0.066763 0.3086498 0.3529 48 0.93206559 1029 0.0735270 0.073826 0.3821768 0.2119 49 0.98913083 1029 0.0780287 0.078417 0.4602055 0.0711 50 1.00891345 1029 0.0795892 0.080010 0.5397947 +0.0711 51 0.98913083 1029 0.0780287 0.078417 0.6178234 +0.2119 52 0.93206559 1029 0.0735270 0.073826 0.6913504 +0.3529 53 0.84413487 1029 0.0665905 0.066763
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Statistics and Nonlinear Regression Allen R. Overman 74 0.7579409 +0.4946 54 0.73470998 1029 0.0579584 0.057996 0.8158993 +0.6367 55 0.61448471 1029 0.0484743 0.048394 0.8643736 +0.7787 56 0.49378236 1029 0.0389526 0.038790 0.9033262 +0.9202 57 0.38116533 1029 0.0300686 0.029866 0.9333948 +1.0613 58 0.28258809 1029 0.0222923 0.022088 0.9556871 +1.2024 59 0.20116440 1029 0.0158691 0.015692 0.9715562 +1.3440 60 0.13746234 1029 0.0108439 0.010708 0.9824001 +1.4871 61 0.09013924 1029 0.0071107 0.007020 0.9895108 +1.6325 62 0.05670049 1029 0.0044729 0.004420 0.9939837 +1.7815 63 0.03420030 1029 0.0026979 0.002673 0.9966816 64 0.01977205 1029 0.0015597 0.001553 0.9982413 65 0.01095067 1029 0.0008639 0.000867 0.9991052 66 0.00580717 1029 0.0004581 0.000465 0.9995633 67 0.00294692 1029 0.0002325 0.000239 0.9997958 68 0.00143012 1029 0.0001128 0.000118 0.9999086 69 0.00066325 1029 0.0000523 0.000056 0.9999609 70 0.00029372 1029 0.0000232 0.000026 0.9999841 71 0.00012411 1029 0.0000098 0.000011 0.9999939 72 0.00004999 1029 0.0000039 0.000005 0.9999978 73 0.00001917 1029 0.0000015 0.000002 0.9999993 74 0.00000700 1029 0.0000006 0.000001 0.9999999 75 0.00000243 1029 0.0000002 0.000000 1.0000001 76 0.00000080 1029 0.0000001
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Statistics and Nonlinear Regression Allen R. Overman 75 1.0000002 77 0.00000025 1029 0.0000000 1.0000002 78 0.00000007 1029 79 0.00000002 1029 80 0.00000001 1029 C 12.67650601 1029 = 2100 Note the symmetry in the frequency distribution. x x F Z 14181 0 0907 7 2 1 2 1 2 erf 1 r = 0.9999959 000.50,0515.72 2 20515 7 00 50 exp 080010 0 2 exp 2 1 x x f ff00521.1000181.0 r = 0.999989
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Statistics and Nonlinear Regression Allen R. Overman 76 Case 8: 12 x 12 For this c ase n = 12 x 12 = 144 and x assumes values of 0, 1, 2, 3, 144 Corresponding values of xnc, are calculated from Eq. (1). Results are given in Table A24. Table A24 Frequency distribution for the square peg board with a 12 x 12 array. x c f F Z f 0.000000 0 1 0.000000 0.000000 1 144 0.000000 0.000000 0.000000 2 3 4 5 6 35 0.00000000 1042 0.000000 36 0.00000001 1042 0.000000 37 0.00000003 1042 0.000000 38 0.00000009 1042 0.00000000 39 0.00000025 1042 0.00000001 0.00000001 40 0.00000066 1042 0.00000003 0.00000004 41 0.00000168 1042 0.00000008 0.00000012 42 0.00000411 1042 0.00000018 0.000000 0.00000030 43 0.00000975 1042 0.00000044 0.000001 0.00000074 44 0.00002237 1042 0.00000100 0.000001 0.00000174 45 0.00004972 1042 0.00000223 0.000003
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Statistics and Nonlinear Regression Allen R. Overman 77 0.00000397 46 0.00010700 1042 0.00000480 0.000005 0.00000877 47 0.00022310 1042 0.00001000 0.000011 0.00001877 48 0.00045085 1042 0.00002022 0.000022 0.00003899 49 0.00088331 1042 0.00003961 0.000042 0.00007860 50 0.00167828 1042 0.00007526 0.000079 0.00015386 51 0.00309331 1042 0.00013871 0.000143 0.00029257 52 0.00553226 1042 0.00024808 0.000254 0.00054065 53 0.00960317 1042 0.00043062 0.000436 0.00097127 54 0.01618312 1042 0.00072568 0.000731 0.00169695 55 0.02648146 1042 0.00118747 0.001190 0.00288442 56 0.04208661 1042 0.00188723 0.001884 0.00477165 57 0.06497582 1042 0.00291362 0.002901 0.00768527 1.7166 58 0.09746373 1042 0.00437042 0.004344 0.01205569 1.5942 59 0.14206578 1042 0.00637045 0.006327 0.01842614 1.4738 60 0.20125985 1042 0.00902480 0.008962 0.02745094 1.3550 61 0.27714472 1042 0.01242760 0.012346 0.03987854 1.2371 62 0.37101632 1042 0.01663695 0.016540 0.05651549 1.1197 63 0.48291012 1042 0.02165444 0.021550 0.07816993 1.0023 64 0.61118313 1042 0.02740640 0.027306 0.10557633 0.8848 65 0.75222539 1042 0.03373095 0.033650 0.13930728 0.7668 66 0.90039099 1042 0.04037493 0.040329 0.17968221 0.6486 67 1.04821638 1042 0.04700365 0.047005 0.22668586 0.5302 68 1.18695090 1042 0.05322472 0.053282
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Statistics and Nonlinear Regression Allen R. Overman 78 0.27991058 0.4119 69 1.30736621 1042 0.05862433 0.058738 0.33853491 0.2941 70 1.40074951 1042 0.06281178 0.062973 0.40134669 0.1767 71 1.45993611 1042 0.06546580 0.065660 0.46681249 0.0593 72 1.48021300 1042 0.06637505 0.066581 0.53318754 +0.0593 73 1.45993611 1042 0.06546580 0.065660 0.59865334 +0.1767 74 1.40074951 1042 0.06281178 0.062973 0.66146512 +0.2941 75 1.30736621 1042 0.05862433 0.058738 0.72008945 +0.4119 76 1.18695090 1042 0.05322472 0.053282 0.77331417 +0.5302 77 1.04821638 1042 0.04700365 0.047005 0.82031782 +0.6486 78 0.90039099 1042 0.04037493 0.040329 0.86069275 +0.7668 79 0.75222539 1042 0.03373095 0.033650 0.89442370 +0.8848 80 0.61118313 1042 0.02740640 0.027306 0.92183010 +1.0023 81 0.48291012 1042 0.02165444 0.021550 0.94348454 +1.1197 82 0.37101632 1042 0.01663695 0.016540 0.96012149 +1.2371 83 0.27714472 1042 0.01242760 0.012346 0.97254909 +1.3550 84 0.20125985 1042 0.00902480 0.008962 0.98157389 +1.4738 85 0.14206578 1042 0.00637045 0.006327 0.98794434 +1.5942 86 0.09746373 1042 0.00437042 0.004344 0.99231476 +1.7166 87 0.06497582 1042 0.00291362 0.002901 0.99522838 88 0.04208661 1042 0.00188723 0.001884 0.99711561 89 0.02648146 1042 0.00118747 0.001190 0.99830308 90 0.01618312 1042 0.00072568 0.000731 0.99902876 91 0.00960317 1042 0.00043062 0.000436
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Statistics and Nonlinear Regression Allen R. Overman 79 0.99945938 92 0.00553226 1042 0.00024808 0.000254 0.99970746 93 0.00309331 1042 0.00013871 0.000143 0.99984617 94 0.00167828 1042 0.00007526 0.000079 0.99992143 95 0.00088331 1042 0.00003961 0.000042 0.99996104 96 0.00045085 1042 0.00002022 0.000022 0.99998126 97 0.00022310 1042 0.00001000 0.000011 0.99999126 98 0.00010700 1042 0.00000480 0.000005 0.99999606 99 0.00004972 1042 0.00000223 0.000003 0.99999829 100 0.00002237 1042 0.00000100 0.000001 0.99999929 101 0.00000975 1042 0.00000044 0.000001 0.99999973 102 0.00000411 1042 0.00000018 0.000000 0.99999991 103 0.00000168 1042 0.00000008 0.99999999 104 0.00000066 1042 0.00000003 1.00000002 105 0.00000025 1042 0.00000001 1.00000003 106 0.00000009 1042 0.00000000 107 0.00000003 1042 0.000000 108 0.00000001 1042 0.000000 109 0.00000000 1042 0.000000 143 144 144 1 C 22.30074520 1042 = 2144 ok Note the symmetry in the fre quency distribution.
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Statistics and Nonlinear Regression Allen R. Overman 80 xxFZ11801.049688.821212erf1 r = 0.9999985 00.72,47377.82 2 247377 8 00 72 exp 066581 0 2 exp 2 1 x x f ff00290.10000840.0 r = 0.9999960 Note that the total number of combinations is controlled by 2n. For n = 500 cells, this g ives C = 3.2733906 10150 total combinations !!! Is this a large number ? Compared to what ? Compared to molecules in a container of gas it may not be so big !! Remember Avogadros number from chemistry (6.0 1023) ? Or to the number of microorganisms in a living body. Statistics entered the field of physics when James Clerk Maxwell used it in his kinetic theory of gases in the mid 1800s, and was then developed further by Ludwig Boltzmann and Willard Gibbs. This led to the branch of physics known as statis tical mechanics These concepts were later incorporated in chemical kinetics by Henry Eyring in the absolute reaction rate theory.
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Statistics and Nonlinear Regression Allen R. Overman 81 Table A25 Spread of the distributions for peg boards with different number of holes ( n). Board n 2 2 2n Linear 2 0.9534 1.00 1.0489 0.9534 5 1.6286 2.50 1.5351 1.0300 6 1.8003 3.00 1.6664 1.0394 7 1.8892 3.50 1.8526 1.0098 8 2.0312 4.00 1.9692 1.0156 Triangular 3 1.2213 1.50 1.2282 0.9972 6 1.8003 3.00 1.6664 1.0394 10 2.2939 5.00 2.1797 1.0259 15 2.7711 7.50 2.7065 1.0119 21 3.2707 10.50 3.2103 1.0094 28 3.7810 14.00 3.7027 1.0105 Square 4 1.4391 2.00 1.3898 1.0176 9 2.1653 4.50 2.0782 1.0207 16 2.8703 8.00 2.7872 1.0148 25 3.5740 12.50 3.4975 1.0109 36 4.2786 18.00 4.2070 1.0085 49 4.9687 24.50 4.93087 1.0038 64 5.67576 32.00 5.63801 1.0033 81 6.38305 40.50 6.34493 1.0030 100 7.0907 50.00 7.05149 1.0028 144 8.49672 72.00 8.473858 1.00135 Note: 12n or n2 spread of distribution is approaching n 2 n center of distribution is equal to n/2 References Abramowitz, M. and I.A. Stegun. 1965. Handbook of Mathematical Functions. Dover Publications. New York, NY. Eigen, M. and R. Winkler. 1993. Laws of the Game: How the Principles of Nature Govern Chance. Princeton University Press. Princeton, NJ. Polster, B 2004. Q.E.D.: Beauty in Mathematical Proof. Walker & Co. New York, NY. Ruhla, C. 1992. The Physics of Chance: From Blaise Pascal to Niels Bohr. Oxford University Press. New York, NY. Watkins, M. 2000. Useful Mathematical and Physical Formulae. Walker & C o. New York, NY.
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Statistics and Nonlinear Regression Allen R. Overman 82 Table A26 Correlation between discrete ( f ) and continuous Gaussian ( f ) distributions. Pegboard Cells Regression Equation Correlation Linear 1 x 2 f f 2848 1 1045 0 1 1 x 3 ff1404.103885.0 1.0000 1 x 4 ff1052.10226.0 0.9971 1 x 5 f f 0882 1 01543 0 0.99723 1 x 6 f f 0809 1 01187 0 0.99730 1 x 7 f f 0689 1 01188 0 0.999198 1 x 8 ff0589.1008700.0 0.999084 Triangular n = 3 ff1432.103950.0 1.000000 n = 6 ff0809.101187.0 0.99730 n = 10 ff0383.100352.0 0.99894 n = 15 f f 02423 1 002046 0 0.999637 n = 21 ff0183.100135.0 0.99980 n = 28 f f 01674 1 000995 0 0.99985 Square 2 x 2 ff1053.102262.0 0.9971 3 x 3 ff0369.100373.0 0.99895 4 x 4 ff0248.100194.0 099962 5 x 5 ff017789.1001129.0 0.999827 6 x 6 f f 01356 1 000723 0 0.999909 7 x 7 f f 00727 1 000345 0 0.999965 8 x 8 f f 00622 1 000252 0 0.999978 9 x 9 f f 00595 1 000232 0 0.999985 10 x 10 ff00521.1000181.0 0.999989 12 x 12 ff00290.10000840.0 0.9999960 As number of cells increases, the intercept approaches 0 and slope approaches 1, which means that fit of the continuous Gaussian distribution to discrete distribution improves as the number of cells increases.
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Statistics and Nonlinear Regression Allen R. Overman 83 ABE 6933 Special Topics Mathematical and Statistical Characteristics of Nonlinear Regression Models A. R. Overman I. Elemen ts of Probability and Calculus A. Arithmetic the process of counting B. Natural numbers positive integers ( ,2,1,0 ) C. Rational numbers ratio of two integers ( ,3/2,3/1,2/1,,1/2,1/1 ) D. Irrational numbers (such as e, 2 etc.) E. Complex numbers z = x + i y with i = 1 F. Binomial theorem and Pascals triangle (a + b)0 = 1 (a + b)1 = a1 + b1 (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 (a + b)7 = a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab 6 + b7 (a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3 + 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8 (a + b)9 = a9 + 9a8b + 36a7b2 + 84a6b3 + 126a5b4 + 126a4b5 + 84a3b6 + 36a2b7 + 9ab8 + b9 Note symmetry in the distribution of coefficients for each expansion.
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Statistics and Nonlinear Regression Allen R. Overman 84 Pascals triangle for binomial coefficients 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 Note the pattern in the coefficients, including symmetry. G. Frequency distributions 1. Discrete distribution Consider the problem of a peg board. This is a two state system a cell (hole) is either filled or empty. Each cell holds one and only one object (peg), which can be viewed as a type of exclusion principle Define n as the total number of cells and x as the number of filled cells (pegs). Cells (hole s) are indistinguishable (all alike), as are the objects (pegs). Order of filling the cells is irrelevant. Note that a peg board can be linear, triangular, rectangular (Eigen and Winkler, 1993, p. 40; Polster, 2004, p. 33), or even 3dimensional. The numbe r of distinguishable combinations which are possible for each x, xnc, can be calculated from (Ruhla, 1992, p. 18; Watkins, 2000, p. 22) )!(!!,xnxnxnc (1)
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Statistics and Nonlinear Regression Allen R. Overman 85 where n n 3 2 1 and is called n factorial. Note that n can assume positive integers ( 3 2 1 n ) and x can also assume positive integers ( n x , 2 1 0 ). For small values of n it is easy to estimate c by intuition, but larger n calculations of c are best performed on a pocket calculato r or computer with the algorithm for computations (Eq. (1)) built in. The total number of combinations C for the system is defined as the sum of c values for all values of x and can be calculated from C = 2n. The frequency distribution of c values is then calculated from f = c/C. Cumulative frequency is calculated from the cumulative sum fF (2) so that F is normalized 10F It should be noted that F forms a discrete set of numbers for a particular case. 2. Cont inuous distribution The next step is to compare the discrete distribution to a continuous Gaussian distribution where x is considered a continuous variable and the cumulative distribution is described by 2 erf 1 2 1 x F (3) where and are the mean and spread of the distribution. The error function is defined by 202)exp(22erfxduux (4) where )exp(2u represents the Gaussian distribution (bell shaped curve). Values of the erf can be obtained from mathematical tables (cf. Abramowitz and Stegun, 1965, chp. 7). Some properties of the error function should be noted: erf (0) = 0, erf ( ) = 1, erf (x ) = erf (+ x ), erf ( ) = 1 Equation (3) can be rearranged to the linear form xFZ21212erf1 (5) where erf 1 is the inverse error function. For example, 00.1)8427.0(erf1 Linear regression of Z vs. x leads to values of the parameters and With these parameters now known the freq uency distribution for f vs. x can be calculated for the continuous distribution from
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Statistics and Nonlinear Regression Allen R. Overman 86 22exp21xf (6) The procedure can now be applied to a linear peg board, triangular peg board, and square peg board. It can even be applied to a 3dimensio nal system. This analysis falls within a branch of mathematics known as group theory Values of the error function can be calculated from the series approximation (Abramowitz and Stegun, 1965, p. 299) 4 4 3 2] 078108 0 000972 0 230389 0 278393 0 1 [ 1 1 erf x x x x x (7) for 8 1 0 x For the case where erf x is given, the inverse erf 1 and therefore x can be obtained on a scientific calculator or computer using the solver routine Note that for the case F < 0.5 and 2F 1 < 0 (negative) the procedure is to change the value from t o +, solve for the inverse by Eq. (7) and change the sign from + x to x Equation (7) does not work directly for x because the power series in Eq. (7) is not symmetric. H. Symmetry and conservation principle In all of the discrete and continuous Gauss ian distributions we note symmetry in the distributions around a mean point. A mathematical consequence of this property is that something is conserved (remains constant) in the system. Note that the number of filled cells is defined by x Since this is a two state (binary) system (cells are either empty of filled), it follows that the number of unfilled cells is n x The total capacity of the system is the sum of filled and unfilled cells so that total capacity is = x + n x = n While this is obvious f or our case, it illustrates the connection between symmetry and conservation. This property turns out to be very important in the various models of physics (including mechanics, electromagnetism, relativity, and quantum mechanics). It also shows up in chem istry and biology.
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Statistics and Nonlinear Regression Allen R. Overman 87 Gaussian Distribution The Gauss differential equation is given by kxydxdy with y = A at x = 0 (1) where y ) 0 ( y is a continuous function of x ( )x and k is the distribution coefficient. 1. Obtain the integral solution to Eq. (1). 2. Sketch the form of the solution y vs. x on linear graph paper. 3. Perform the 2nd derivative of y on x to obtain the inflection points at x Write the constant k in terms of 4. Evaluate the constant A by normalizing the integral 1 ydx (2) 5. Write the resulting solution y in terms of variable x and parameter 6. Obta in the cumulative probability distribution F 22 erf 1 2 1xx ydx F (3) where the error function is defined by zduuz02exp2erf (4) 7. Calculate and plot F vs. 2 / x on linear graph paper.
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Statistics and Nonlinear Regression Allen R. Overman 88 Solutions Note : at x = 0, dy/dx = 0 (maximum or minimum) 1. 2221exp21kxAydxkdxkxydy Solution is symmetric around x = 0 3. 2 2 2 2 21 0 2 1 exp 1 k kx kx kA dx y d 2 22 expx A y at x = 0, d2y/dx2 < 0 (maximum) 4. 2 1 1 2 exp 2 2 2 exp 2 2 exp2 2 2 2 A A du u A x d x A dx x A ydx A is chosen so that the distribut ion is normalized, hence the term normal distribution. 5. 22exp21xy 6. 2erf121exp2121exp21expexp1exp120220222020222xduuduuduuduuduuydxFxxxxx where 2 0 2exp 2 2 erfxdu u x This ties the cumulative frequency distribution to the error function of mathematical physics. Note the ch aracteristics of the error function: erf (0) = 0, erf ( ) = 1, erf ( x ) = erf (+ x ). It follows that F is bounded by 10F Note also that F is a well behaved function.
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Statistics and Nonlinear Regression Allen R. Overman 89 Solve Eq. (1) by the power series met hod. kxy dx dy with y = A at x = 0 (1) Assume that the solution is given by a power series 77665544332210xaxaxaxaxaxaxaay (2) The first derivative is given by 6 7 5 6 4 5 3 4 2 3 2 17 6 5 4 3 2 x a x a x a x a x a x a a dx dy (3) Substitution of Eqs. (2) and (3) into Eq. (1) leads to 7665544332210665544332210675645342321765432xkaxkaxkaxkaxkaxkaxkaxaxaxaxaxaxaakxxaxaxaxaxaxaa (4) Equating like coefficients in Eq. (4) gives the recursion relations 6426054240320034635022413021akkaakaaakkaakaakaaa It follows that the solution is given by
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Statistics and Nonlinear Regression Allen R. Overman 90 6 3 4 2 2 0 6 0 3 4 0 2 2 0 0 6 6 5 5 4 4 3 3 2 2 1 06 4 2 4 2 2 1 6 4 2 0 4 2 0 2 0 x k x k x k a x a k x a k x ka a x a x a x a x a x a x a a y (5) Now use the substitution 22xk (6) Then Eq. (5) becomes 2 0 0 3 2 0 3 2 02 exp exp 3 2 1 3 2 1 2 1 1 x k a a a a y (7) The constant 0a is evaluated from the boundary condition, which leads to 22 exp x k A y (8) This is the famous Gaussian distribution centered at x = 0. It remains to determine k in terms of the variance of the distribution and A to normalize the distribution. Check: kxy x k kxA dx dy 22 exp correct (9)
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Statistics and Nonlinear Regression Allen R. Overman 91 Probability distributions with dice. Consider a single die with 6 faces numbered 1 through 6. Assume that each number is equally likely. The frequency distribution can now be calculated. Table A27 Frequency distribution for a single die x S c f F Z Z f 0.00000000 1 1 1 0.16666667 0.093758 0.16666667 0.684656 0.669385 2 2 1 0.16666667 0.146760 0.33333334 0.304151 0.334693 3 3 1 0.16666667 0.183615 0.50000001 0.000000 0.000000 4 4 1 0.16666667 0.183615 0.66666668 + 0.304151 +0.334693 5 5 1 0.16666667 0.146760 0.83333335 +0.684656 +0.669385 6 6 1 0.16666667 0.093758 1.00000002 C 6 SSFZ334693.0171424.121212erf1 r = 0.998961 5000.3,9878.22 2 29878 2 5000 3 exp 18883 0 2 exp 2 1 S S f No correlation between f and f since f = constant = 1/6.
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Statistics and Nonlinear Regression Allen R. Overman 92 Table A28 Fre quency distribution for two di c e x S c f F Z Z f c 0.00000000 1,1 2 1 0.02777778 0.020242 0.73 0.02777778 1.3513 1.2999 1,2; 2,1 3 2 0.05555556 0.042894 1.54 0.08333334 0.9781 1.0110 1,3; 2,2; 3,1 4 3 0.08333333 0.076922 2.77 0.16666667 0.6847 0.7222 1,4; 2,3; 3,2; 4,1 5 4 0.11111111 0.116745 4.20 0.27777778 0.4164 0.4333 1,5; 2;4; 3,3; 4,2; 5,1 6 5 0.13888889 0.149951 5.40 0.41666667 0.1490 0.1444 1,6; 2,5; 3,4; 4,3; 5,2; 6,1 7 6 0.16666667 0.163000 5.87 0.58333334 +0.1490 +0.1444 2,6; 3,5; 4,4; 5,3; 6,2 8 5 0.13888889 0.149951 5.40 0.72222223 +0.4164 +0.4333 3,6; 4,5; 5,4; 6,3 9 4 0.11111111 0.116745 4.20 0.83333334 +0.6847 +0.7222 4,6; 5,5; 6,4 10 3 0.08333333 0.076922 2.77 0.91666667 +0.9781 +1.0110 5,6; 6,5 11 2 0.05555556 0.042894 1.54 0.97222223 +1.3513 +1.2999 6,6 12 1 0.02777778 0.020242 0.73 1.00000001 C 36 = 62 S S F Z 28885 0 02195 2 2 1 2 1 2 erf 1 r = 0.999211 0000.7,4620.32 224620.30000.7exp1630.02exp21SSf f f 1360 1 0145 0 r = 0.99319 fc36 The two dice problem is discussed by Speyer (1994, p. 62)
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Statistics and Nonlinear Regression Allen R. Overman 93 Table A29 Fre quency distribution for three di c e x S c f F Z Z f 0.000000 1,1,1 3 1 0.004630 0.005418 0.004630 1,1,2; 1,2,1; 2,1,1 4 3 0.013889 0.012058 0.018519 1.4726 1.4344 1,1,3; 1,2,2; 1,3,1; 2,1,2; 5 6 0.027778 0.023940 2,2,1; 3,1,1 0.046297 1.1878 1.1953 1,1,4; 1,2,3; 1,3,2; 1,4,1; 6 10 0.046296 0.042397 2,1,3; 2,2,2; 2,3,1; 3,1,2; 0.092593 0.9372 0.9563 3,2,1; 4,1,1 1,1,5; 1,2,4; 1,3,3; 1,4,2; 7 15 0.069444 0.066973 1,5,1; 2,1,4; 2,2,3; 2,3,2; 0.162037 0.6979 0.7172 2,4,1; 3,1,3; 3,2,2; 3,3,1; 4,1,2; 4,2,1; 5,1,1 1,1,6; 1,2,5; 1,3,4; 1,4,3; 8 21 0.097222 0.094367 1,5,2; 1,6,1; 2,1,5; 2,2,4; 0.259259 0.4563 0.4781 2,3,3; 2,4,2; 2,5,1; 3,1,4; 3,2,3; 3,3,2; 3,4,1; 4,1,3; 4,2,2; 4,3,1; 5,1,2; 5,2,1; 6,1,1 1,2,6; 1,3,5; 1,4,4; 1,5,3; 9 25 0.115741 0.118605 1,6,2; 2,1,6; 2,2,5; 2,3,4; 0.375000 0.2251 0.2391 2,4,3; 2,5,2; 2,6,1; 3,1,5; 3,2,4; 3,3,3; 3,4,2; 3,5,1; 4,1,4; 4,2,3; 4,3,2; 4,4,1; 5,1,3; 5,2,2; 5,3,1; 6,1,2; 6,2,1 1,3,6; 1,4,5; 1,5,4; 1,6,3; 10 27 0.125000 0.132967 2,2,6; 2,3,5; 2,4,4; 2,5,3; 0.500000 0.0000 0.0000 2,6,2; 3,1,6; 3,2,5; 3,3,4; 3,4,3; 3,5,2; 3,6,1; 4,1,5; 4,2,4; 4,3,3; 4,4,2; 4,5,1; 5,1,4; 5,2,3; 5,3,2; 5,4,1; 6,1,3; 6,2,2; 6,3,1 1,4,6; 1,5,5; 1,6,4; 2,3,6; 11 27 0.125000 0.132967 2,4,5; 2,5,4; 2,6,3; 3,2,6; 0.625000 +0.2251 +0.2391 3,3,5; 3,4,4; 3,5,3; 3,6,2;
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Statistics and Nonlinear Regression Allen R. Overman 94 4,1,6; 4,2,5; 4,3,4; 4,4,3; 4,5,2; 4,6,1; 5,1,5; 5,2,4; 5,3,3; 5,4,2; 5,5,1; 6,1,4; 6,2,3; 6,3,2; 6,4,1 1,5,6; 1,6,5; 2,4,6; 2,5,5; 12 25 0.115741 0.118605 2,6,4; 3,3,6; 3,4,5; 3,5,4; 0.740741 +0.4563 +0.4781 3,6,3; 4,2,6; 4,3,5; 4,4,4; 4,5,3; 4,6,2; 5,1,6; 5,2,5; 5,3,4; 5,4,3; 5,5,2; 5,6,1; 6,1,5; 6,2,4; 6,3,3; 6,4,2; 6,5,1 1,6,6; 2,5,6; 2,6,5; 3,4,6; 13 21 0.097222 0.094367 3,5,5; 3,6,4; 4,3,6; 4,4,5; 0.837963 +0.6979 +0.7172 4,5,4; 4,6,3; 5,2,6; 5,3,5; 5,4,4; 5,5,3; 5,6,2; 6,1,6; 6,2,5; 6,3,4; 6,4,3; 6,5,2; 6,6,1 2,6,6; 3,5,6; 3,6,5; 4,4,6; 14 15 0.069444 0.066973 4,5,5; 4,6,4; 5,3,6; 5,4,5; 0.907407 +0.9372 +0.9563 5,5,4; 5,6,3; 6,2,6; 6,3,5; 6,4,4; 6,5,3; 6,6,2 3,6,6; 4,5,6; 4,6,5; 5,4,6; 15 10 0.046296 0.042397 5,5,5; 5,6,4; 6,3,6; 6,4,5; 0.953703 +1.1878 +1.1953 6,5,4; 6,6,3; 4,6,6; 5,5,6; 5,6,5; 6,4,6; 16 6 0.027778 0.023940 6,5,5; 6,6,4 0.981481 +1.4726 +1.4344 5,6,6; 6,5,6; 6,6,5 17 3 0.013889 0.012058 0.995370 6,6,6 18 1 0.004630 0.005418 1.000000 C 216 = 63 SSFZ23906.05102.221212erf1 r = 0.999719 5001.10,1830.42
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Statistics and Nonlinear Regression Allen R. Overman 95 221830.4500.10exp13488.02exp21SSf ff0752.100590.0 r = 0.99800 For three dice t he discrete distribution is closely approximated by the continuous Gaussian distribution. I conclude that the peg board is a much simpler illustration of frequency distributions than dice.
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Statistics and Nonlinear Regression Allen R. Overman 96 Table A30 Fre quency distribution for four di c e x S c f 1,1,1,1 4 1 0.000772 1,1,1,2; 1,1,2,1; 1,2,1,1; 5 4 0.003086 2,1,1,1 1,1,1,3; 1,1,2,2; 1,1,3,1; 6 10 0.007716 1,2,1,2; 1,2,2,1; 1,3,1,1; 2,1,1,2; 2,1,2,1; 2,2,1,1; 3,1,1,1 1,1,1,4; 1,1,2,3; 1,1,3,2; 7 20 0.015432 1,1,4,1; 1,2,1,3; 1,2,2,2; 1,2,3,1; 1,3,1,2; 1,3,2,1; 1,4,1,1; 2,1,1,3; 2,1,2,2; 2,1,3,1; 2,2,1,2; 2,2,2,1; 2,3,1,1; 3,1,1,2; 3,1,2,1; 3,2,1,1; 4,1,1,1 1,1,1,5; 1,1,2,4; 1,1,3,3; 8 35 0.027006 1,1,4,2; 1,1,5,1; 1,2,1,4; 1,2,2,3; 1,2,3,2; 1,2,4,1; 1,3,1,3; 1,3,2,2; 1,3,3,1; 1,4,1,2; 1,4,2,1; 1,5,1,1; 2,1,1,4; 2,1,2,3; 2,1,3,2; 2,1,4,1; 2,2,1,3; 2,2,2,2; 2,2,3,1; 2,3,1,2; 2,3,2,1; 2,4,1,1; 3,1,1,3; 3,1,2,2; 3,1,3,1; 3,2,1,2; 3,2,2,1; 3,3,1,1; 4,1,1,2; 4,1,2,1; 4,2,1,1; 5,1,1,1 1,1,1,6; 1,1,2,5; 1,1,3,4; 9 56 0.043210 1,1,4,3; 1,1,5,2; 1,1,6,1; 1,2,1,5; 1,2,2,4; 1,2,3,3; 1,2,4,2; 1,2,5,1; 1,3,1,4; 1,3,2,3; 1,3,3,2; 1,3,4,1; 1,4,1,3; 1,4,2,2; 1,4,3,1; 1,5,1,2; 1,5,2,1; 1,6,1,1; 2,1,1,5; 2,1,2,4; 2,1,3,3; 2,1,4,2; 2,1,5,1; 2,2,1,4; 2,2,2,3; 2,2,3,2; 2,2,4,1; 2,3,1,3; 2,3,2,2; 2,3,3,1;
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Statistics and Nonlinear Regression Allen R. Overman 97 2,4,1,2; 2,4,2,1; 2,5,1,1; 3,1,1,4; 3,1,2,3; 3,1,3,2; 3,1,4,1; 3,2,1,3; 3,2,2,2; 3,2,3,1; 3,3,1,2; 3,3,2,1; 3,4,1,1; 4,1,1,3; 4,1,2,2; 4,1,3,1; 4,2,1,2; 4,2,2,1; 4,3,1,1; 5,1,1,2; 5,1,2,1; 5,2,1,1; 6,1,1,1 1,1,2,6; 1,1,3,5; 1,1,4,4; 10 80 0.061728 1,1,5,3; 1,1,6,2; 1,2,1,6; 1,2,2,5; 1,2,3,4; 1,2,4,3; 1,2,5,2; 1,2,6,1; 1,3,1,5; 1,3,2,4; 1,3,3,3; 1,3,4,2; 1,3,5,1; 1,4,1,4; 1,4,2,3; 1,4,3,2; 1,4,4,1; 1,5,1,3; 1,5,2,2; 1,5,3,1; 1,6,1,2; 1,6,2,1; 2,1,1,6; 2,1,2,5; 2,1,3,4; 2,1,4,3; 2,1,5,2; 2,1,6,1; 2,2,1,5; 2,2,2,4; 2,2,3,3; 2,2,4,2; 2,2,5,1; 2,3,1,4; 2,3,2,3; 2,3,3,2; 2,3,4,1; 2,4,1,3; 2,4,2,2; 2,4,3,1; 2,5,1,2; 2,5,2,1; 2,6,1,1; 3,1,1,5; 3,1,2,4; 3,1,3,3; 3,1,4,2; 3,1,5,1; 3,2,1,4; 3,2,2,3; 3,2,3,2; 3,2,4,1; 3,3,1,3; 3,3,2,2; 3,3,3,1; 3,4,1,2; 3,4,2,1; 3,5,1,1; 4,1,1,4; 4,1,2,3; 4,1,3,2; 4,1,4,1; 4,2,1,3; 4,2,2,2; 4,2,3,1; 4,3,1,2; 4,3,2,1; 4,4,1,1; 5,1,1,3; 5,1,2,2; 5,1,3,1; 5,2,1,2; 5,2,2,1; 5,3,1,1; 6,1,1,2; 6,1,2,1; 6,2,1,1 1,1,3,6; 1,1,4,5; 1,1,5,4; 11 104 0.080247 1,1,6,3; 1,2,2,6; 1,2,3,5; 1,2,4,4; 1,2,5,3; 1,2,6,2; 1,3,1,6; 1,3,2,5; 1,3,3,4; 1,3,4,3; 1,3,5,2; 1,3,6,1; 1,4,1,5; 1,4,2,4; 1,4,3,3; 1,4,4,2; 1,4,5,1; 1,5,1,4; 1,5,2,3; 1,5,3,2; 1,5,4,1; 1,6,1,3; 1,6,2,2; 1,6,3,1;
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Statistics and Nonlinear Regression Allen R. Overman 98 2,1,2,6; 2,1,3,5; 2,1,4,4; 2,1,5,3; 2,1,6,2; 2,2,1,6; 2,2,2,5; 2,2,3,4; 2,2,4,3; 2,2,5,2; 2,2,6,1; 2,3,1,5; 2,3,2,4; 2,3,3,3; 2,3,4,2; 2,3,5,1; 2,4,1,4; 2,4,2,3; 2,4,3,2; 2,4,4,1; 2,5,1,3; 2,5,2,2; 2,5,3,1; 2,6,1,2; 2,6,2,1; 3,1,1,6; 3,1,2,5; 3,1,3,4; 3,1,4,3; 3,1,5,2; 3,1,6,1; 3,2,1,5; 3,2,2,4; 3,2,3,3; 3,2,4,2; 3,2,5,1; 3,3,1,4; 3,3,2,3; 3,3,3,2; 3,3,4,1; 3,4,1,3; 3,4,2,2; 3,4,3,1; 3,5,1,2; 3,5,2,1; 3,6,1,1; 4,1,1,5; 4,1,2,4; 4,1,3,3; 4,1,4,2; 4,1,5,1; 4,2,1,4; 4,2,2,3; 4,2,3,2; 4,2,4,1; 4,3,1,3; 4,3,2,2; 4,3,3,1; 4,4,1,2; 4,4,2,1; 4,5,1,1; 5,1,1,4; 5,1,2,3; 5,1,3,2; 5,1,4,1; 5,2,1,3; 5,2,2,2; 5,2,3,1; 5,3,1,2; 5,3,2,1; 5,4,1,1; 6,1,1,3; 6,1,2,2; 6,1,3,1; 6,2,1,2; 6,2,2,1; 6,3,1,1 1,1,4,6; 1,1,5,5; 1,1,6,4; 12 125 0.096451 1,2,3,6; 1,2,4,5; 1,2,5,4; 1,2,6,3; 1,3,2,6; 1,3,3,5; 1,3,4,4; 1,3,5,3; 1,3,6,2; 1,4,1,6; 1,4,2,5; 1,4,3,4; 1,4,4,3; 1,4,5,2; 1,4,6,1; 1,5,1,5; 1,5,2,4; 1,5,3,3; 1,5,4,2; 1,5,5,1; 1,6,1,4; 1,6,2,3; 1,6,3,2; 1,6,4,1; 2,1,3,6; 2,1,4,5; 2,1,5,4; 2,1,6,3; 2,2,2,6; 2,2,3,5; 2,2,4,4; 2,2,5,3; 2,2,6,2; 2,3,1,6; 2,3,2,5; 2,3,3,4; 2,3,4,3; 2,3,5,2; 2,3,6,1; 2,4,1,5; 2,4,2,4; 2,4,3,3; 2,4,4,2; 2,4,5,1; 2,5,1,4; 2,5,2,3; 2,5,3,2; 2,5,4,1; 2,6,1,3; 2,6,2,2; 2,6,3,1; 3,1,2,6; 3,1,3,5; 3,1,4,4;
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Statistics and Nonlinear Regression Allen R. Overman 99 3,1,5,3; 3,1,6,2; 3,2,1,6; 3,2,2,5; 3,2,3,4; 3,2,4,3; 3,2,5,2; 3,2,6,1; 3,3,1,5; 3,3,2,4; 3,3,3,3; 3,3,4,2; 3,3,5,1; 3,4,1,4; 3,4,2,3; 3,4,3,2; 3,4,4,1; 3,5,1,3; 3,5,2,2; 3,5,3,1; 3,6,1,2; 3,6,2,1; 4,1,1,6; 4,1,2,5; 4,1,3,4; 4,1,4,3; 4,1,5,2; 4,1,6,1; 4,2,1,5; 4,2,2,4; 4,2,3,3; 4,2,4,2; 4,2,5,1; 4,3,1,4; 4,3,2,3; 4,3,3,2; 4,3,4,1; 4,4,1,3; 4,4,2,2; 4,4,3,1; 4,5,1,2; 4,5,2,1; 4,6,1,1; 5,1,1,5; 5,1,2,4; 5,1,3,3; 5,1,4,2; 5,1,5,1; 5,2,1,4; 5,2,2,3; 5,2,3,2; 5,2,4,1; 5,3,1,3; 5,3,2,2; 5,3,3,1; 5,4,1,2; 5,4,2,1; 5,5,1,1; 6,1,1,4; 6,1,2,3; 6,1,3,2; 6,1,4,1; 6,2,1,3; 6,2,2,2; 6,2,3,1; 6,3,1,2; 6,3,2,1; 6,4,1,1 1,1,5,6; 1,1,6,5; 1,2,4,6; 13 140 0.108025 1,2,5,5; 1,2,6,4; 1,3,3,6; 1,3,4,5; 1,3,5,4; 1,3,6,3; 1,4,2,6; 1,4,3,5; 1,4,4,4; 1,4,5,3; 1,4,6,2; 1,5,1,6; 1,5,2,5; 1,5,3,4; 1,5,4,3; 1,5,5,2; 1,5,6,1; 1,6,1,5; 1,6,2,4; 1,6,3,3; 1,6,4,2; 1,6,5,1; 2,1,4,6; 2,1,5,5; 2,1,6,4; 2,2,3,6; 2,2,4,5; 2,2,5,4; 2,2,6,3; 2,3,2,6; 2,3,3,5; 2,3,4,4; 2,3,5,3; 2,3,6,2; 2,4,1,6; 2,4,2,5; 2,4,3,4; 2,4,4,3; 2,4,5,2; 2,4,6,1; 2,5,1,5; 2,5,2,4; 2,5,3,3; 2,5,4,2; 2,5,5,1; 2,6,1,4; 2,6,2,3; 2,6,3,2; 2,6,4,1; 3,1,3,6; 3,1,4,5; 3,1,5,4; 3,1,6,3; 3,2,2,6; 3,2,3,5; 3,2,4,4; 3,2,5,3; 3,2,6,2; 3,3,1,6; 3,3,2,5; 3,3,3,4; 3,3,4,3; 3,3,5,2;
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Statistics and Nonlinear Regression Allen R. Overman 100 3,3,6,1; 3,4,1,5; 3,4,2,4; 3,4,3,3; 3,4,4,2; 3,4,5,1; 3,5,1,4; 3,5,2,3; 3,5,3,2; 3,5,4,1; 3,6,1,3; 3,6,2,2; 3,6,3,1; 4,1,2,6; 4,1,3,5; 4,1,4,4; 4,1,5,3; 4,1,6,2; 4,2,1,6; 4,2,2,5; 4,2,3,4; 4,2,4,3; 4,2,5,2; 4,2,6,1; 4,3,1,5; 4,3,2,4; 4,3,3,3; 4,3,4,2; 4,3,5,1; 4,4,1,4; 4,4,2,3; 4,4,3,2; 4,4,4,1; 4,5,1,3; 4,5,2,2; 4,5,3,1; 4,6,1,2; 4,6,2,1; 5,1,1,6; 5,1,2,5; 5,1,3,4; 5,1,4,3; 5,1,5,2; 5,1,6,1; 5,2,1,5; 5,2,2,4; 5,2,3,3; 5,2,4,2; 5,2,5,1; 5,3,1,4; 5,3,2,3; 5,3,3,2; 5,3,4,1; 5,4,1,3; 5,4,2,2; 5,4,3,1; 5,5,1,2; 5,5,2,1; 5,6,1,1; 6,1,1,5; 6,1,2,4; 6,1,3,3; 6,1,4,2; 6,1,5,1; 6,2,1,4; 6,2,2,3; 6,2,3,2; 6,2,4,1; 6,3,1,3; 6,3,2,2; 6,3,3,1; 6,4,1,2; 6,4,2,1; 6,5,1,1 1,1,6,6; 1,2,5,6; 1,2,6,5; 14 146 0.112654 1,3,4,6; 1,3,5,5; 1,3,6,4; 1,4,3,6; 1,4,4,5; 1,4,5,4; 1,4,6,3; 1,5,2,6; 1,5,3,5; 1,5,4,4; 1,5,5,3; 1,5,6,2; 1,6,1,6; 1,6,2,5; 1,6,3,4; 1,6,4,3; 1,6,5,2; 1,6,6,1; 2,1,5,6; 2,1,6,5; 2,2,4,6; 2,2,5,5; 2,2,6,4; 2,3,3,6; 2,3,4,5; 2,3,5,4; 2,3,6,3; 2,4,2,6; 2,4,3,5; 2,4,4,4; 2,4,5,3; 2,4,6,2; 2,5,1,6; 2,5,2,5; 2,5,3,4; 2,5,4,3; 2,5,5,2; 2,5,6,1; 2,6,1,5; 2,6,2,4; 2,6,3,3; 2,6,4,2; 2,6,5,1; 3,1,4,6; 3,1,5,5; 3,1,6,4; 3,2,3,6; 3,2,4,5; 3,2,5,4; 3,2,6,3; 3,3,2,6; 3,3,3,5; 3,3,4,4; 3,3,5,3; 3,3,6,2; 3,4,1,6; 3,4,2,5;
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Statistics and Nonlinear Regression Allen R. Overman 101 3,4,3,4; 3,4,4,3; 3,4,5,2; 3,4,6,1; 3,5,1,5; 3,5,2,4; 3,5,3,3; 3,5,4,2; 3,5,5,1; 3,6,1,4; 3,6,2,3; 3,6,3,2; 3,6,4,1; 4,1,3,6; 4,1,4,5; 4,1,5,4; 4,1,6,3; 4,2,2,6; 4,2,3,5; 4,2,4,4; 4,2,5,3; 4,2,6,2; 4,3,1,6; 4,3,2,5; 4,3,3,4; 4,3,4,3; 4,3,5,2; 4,3,6,1; 4,4,1,5; 4,4,2,4; 4,4,3,3; 4,4,4,2; 4,4,5,1; 4,5,1,4; 4,5,2,3; 4,5,3,2; 4,5,4,1; 4,6,1,3; 4,6,2,2; 4,6,3,1; 5,1,2,6; 5,1,3,5; 5,1,4,4; 5,1,5,3; 5,1,6,2; 5,2,1,6; 5,2,2,5; 5,2,3,4; 5,2,4,3; 5,2,5,2; 5,2,6,1; 5,3,1,5; 5,3,2,4; 5,3,3,3; 5,3,4,2; 5,3,5,1; 5,4,1,4; 5,4,2,3; 5,4,3,2; 5,4,4,1; 5,5,1,3; 5,5,2,2; 5,5,3,1; 5,6,1,2; 5,6,2,1; 6,1,1,6; 6,1,2,5; 6,1,3,4; 6,1,4,3; 6,1,5,2; 6,1,6,1; 6,2,1,5; 6,2,2,4; 6,2,3,3; 6,2,4,2; 6,2,5,1; 6,3,1,4; 6,3,2,3; 6,3,3,2; 6,3,4,1; 6,4,1,3; 6,4,2,2; 6,4,3,1; 6,5,1,2; 6,5,2,1; 6,6,1,1 1,2,6,6; 1,3,5,6; 1,3,6,5; 15 140 0.108025 1,4,4,6; 1,4,5,5; 1,4,6,4; 1,5,3,6; 1,5,4,5; 1,5,5,4; 1,5,6,3; 1,6,2,6; 1,6,3,5; 1,6,4,4; 1,6,5,3; 1,6,6,2; 2,1,6,6; 2,2,5,6; 2,2,6,5; 2,3,4,6; 2,3,5,5; 2,3,6,4; 2,4,3,6; 2,4,4,5; 2,4,5,4; 2,4,6,3; 2,5,2,6; 2,5,3,5; 2,5,4,4; 2,5,5,3; 2,5,6,2; 2,6,1,6; 2,6,2,5; 2,6,3,4; 2,6,4,3; 2,6,5,2; 2,6,6,1; 3,1,5,6; 3,1,6,5; 3,2,4,6; 3,2,5,5; 3,2,6,4; 3,3,3,6; 3,3 4,5; 3,3,5,4; 3,3,6,3; 3,4,2,6; 3,4,3,5; 3,4,4,4;
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Statistics and Nonlinear Regression Allen R. Overman 102 3,4,5,3; 3,4,6,2; 3,5,1,6; 3,5,2,5; 3,5,3,4; 3,5,4,3; 3,5,5,2; 3,5,6,1; 3,6,1,5; 3,6,2,4; 3,6,3,3; 3,6,4,2; 3,6,5,1; 4,1,4,6; 4,1,5,5; 4,1,6,4; 4,2,3,6; 4,2,4,5; 4,2,5,4; 4,2,6,3; 4,3,2,6; 4,3,3,5; 4,3,4,4; 4,3,5,3; 4,3,6,2; 4,4,1,6; 4,4,2,5; 4,4,3,4; 4,4,4,3; 4,4,5,2; 4,4,6,1; 4,5,1,5; 4,5,2,4; 4,5,3,3; 4,5,4,2; 4,5,5,1; 4,6,1,4; 4,6,2,3; 4,6,3,2; 4,6,4,1; 5,1,3,6; 5,1,4,5; 5,1,5,4; 5,1,6,3; 5,2,2,6; 5,2,3,5; 5,2,4,4; 5,2,5,3; 5,2,6,2; 5,3,1,6; 5,3,2,5; 5,3,3,4; 5,3,4,3; 5,3,5,2; 5,3,6,1; 5,4,1,5; 5,4,2,4; 5,4,3,3; 5,4,4,2; 5,4,5,1; 5,5,1,4; 5,5,2,3; 5,5,3,2; 5,5,4,1; 5,6,1,3; 5,6,2,2; 5,6,3,1; 6,1,2,6; 6,1,3,5; 6,1,4,4; 6,1,5,3; 6,1,6,2; 6,2,1,6; 6,2,2,5; 6,2,3,4; 6,2,4,3; 6,2,5,2; 6,2,6,1; 6,3,1,5; 6,3,2,4; 6,3,3,3; 6,3,4,2; 6,3,5,1; 6,4,1,4; 6,4,2,3; 6,4,3,2; 6,4,4,1; 6,5,1,3; 6,5,2,2; 6,5,3,1; 6,6,1,2; 6,6,2,1 1,3,6,6; 1,4,5,6; 1,4,6,5; 16 125 0.096451 1,5,4,6; 1,5,5,5; 1,5,6,4; 1,6,3,6; 1,6,4,5; 1,6,5,4; 1,6,6,3; 2,2,6,6; 2,3,5,6; 2,3,6,5; 2,4,4,6; 2,4,5,5; 2,4,6,4; 2,5,3,6; 2,5,4,5; 2,5,5,4; 2,5,6,3; 2,6,2,6; 2,6,3,5; 2,6,4,4; 2,6,5,3; 2,6,6,2; 3,1,6,6; 3,2,5,6; 3,2,6,5; 3,3,4,6; 3,3,5,5; 3,3,6,4; 3,4,3,6; 3,4,4,5; 3,4,5,4; 3,4,6,3; 3,5,2,6; 3,5,3,5; 3,5,4,4; 3,5,5,3; 3,5,6,2; 3,6,1,6; 3,6,2,5;
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Statistics and Nonlinear Regression Allen R. Overman 103 3,6,3,4; 3,6,4,3; 3,6,5,2; 3,6,6,1; 4,1,5,6; 4,1,6,5; 4,2,4,6; 4,2,5,5; 4,2,6,4; 4,3,3,6; 4,3,4,5; 4,3,5,4; 4,3,6,3; 4,4,2,6; 4,4,3,5; 4,4,4,4; 4,4,5,3; 4,4,6,2; 4,5,1,6; 4,5,2,5; 4,5,3,4; 4,5,4,3; 4,5,5,2; 4,5,6,1; 4,6,1,5; 4,6,2,4; 4,6,3,3; 4,6,4,2; 4,6,5,1; 5,1,4,6; 5,1,5,5; 5,1,6,4; 5,2,3,6; 5,2,4,5; 5,2,5,4; 5,2,6,3; 5,3,2,6; 5,3,3,5; 5,3,4,4; 5,3,5,3; 5,3,6,2; 5,4,1,6; 5,4,2,5; 5,4,3,4; 5,4,4,3; 5,4,5,2; 5,4,6,1; 5,5,1,5; 5,5,2,4; 5,5,3,3; 5,5,4,2; 5,5,5,1; 5,6,1,4; 5,6,2,3; 5,6,3,2; 5,6,4,1; 6,1,3,6; 6,1,4,5; 6,1,5,4; 6,1,6,3; 6,2,2,6; 6,2,3,5; 6,2,4,4; 6,2,5,3; 6,2,6,2; 6,3,1,6; 6,3,2,5; 6,3,3,4; 6,3,4,3; 6,3,5,2; 6,3,6,1; 6,4,1,5; 6,4,2,4; 6,4,3,3; 6,4,4,2; 6,4,5,1; 6,5,1,4; 6,5,2,3; 6,5,3,2; 6,5,4,1; 6,6,1,3; 6,6,2,2; 6,6,3,1 1,4,6,6; 1,5,5,6; 1,5,6,5; 17 104 0.080247 1,6,4,6; 1,6,5,5; 1,6,6,4; 2,3,6,6; 2,4,5,6; 2,4,6,5; 2,5,4,6; 2,5,5,5; 2,5,6,4; 2,6,3,6; 2,6,4,5; 2,6,5,4; 2,6,6,3; 3,2,6,6; 3,3,5,6; 3,3,6,5; 3,4,4,6; 3,4,5,5; 3,4,6,4; 3,5,3,6; 3,5,4,5; 3,5,5,4; 3,5,6,3; 3,6,2,6; 3,6,3,5; 3,6,4,4; 3,6,5,3; 3,6,6,2; 4,1,6,6; 4,2,5,6; 4,2,6,5; 4,3,4,6; 4,3,5,5; 4,3,6,4; 4,4,3,6; 4,4,4,5; 4,4,5,4; 4,4,6,3; 4,5,2,6; 4,5,3,5; 4,5,4,4; 4,5,5,3; 4,5,6,2; 4,6,1,6; 4,6,2,5; 4,6,3,4; 4,6,4,3; 4,6,5,2;
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Statistics and Nonlinear Regression Allen R. Overman 104 4,6,6,1; 5,1,5,6; 5,1,6,5; 5,2,4,6; 5,2,5,5; 5,2,6,4; 5,3,3,6; 5,3,4,5; 5,3,5,4; 5,3,6,3; 5,4,2,6; 5,4,3,5; 5,4,4,4; 5,4,5,3; 5,4,6,2; 5,5,1,6; 5,5,2,5; 5,5,3,4; 5,5,4,3; 5,5,5,2; 5,5,6,1; 5,6,1,5; 5,6,2,4; 5,6,3,3; 5,6,4,2; 5,6,5,1; 6,1,4,6; 6,1,5,5; 6,1,6,4; 6,2,3,6; 6,2,4,5; 6,2,5,4; 6,2,6,3; 6,3,2,6; 6,3,3,5; 6,3,4,4; 6,3,5,3; 6,3,6,2; 6,4,1,6; 6,4,2,5; 6,4,3,4; 6,4,4,3; 6,4,5,2; 6,4,6,1; 6,5,1,5; 6,5,2,4; 6,5,3,3; 6,5,4,2; 6,5,5,1; 6,6,1,4; 6,6,2,3; 6,6,3,2; 6,6,4,1 6,6,1,5; 6,6,2,4; 6,6,3,3; 18 80 0.061728 6,6,4,2; 6,6,5,1; 6,5,1,6; 6,5,2,5; 6,5,3,4; 6,5,4,3; 6,5,5,2; 6,5,6,1; 6,4,2,6; 6,4,3,5; 6,4,4,4; 6,4,5,3; 6,4,6,2; 6,3,3,6; 6,3,4,5; 6,3,5,4; 6,3,6,3; 6,2,4,6; 6,2,5,5; 6,2,6,4; 6,1,5,6; 6,1,6,5; 5,6,1,6; 5,6,2,5; 5,6,3,4; 5,6,4,3; 5,6,5,2; 5,6,6,1; 5,5,2,6; 5,5,3,5; 5,5,4,4; 5,5,5,3; 5,5,6,2; 5,4,3,6; 5,4,4,5; 5,4,5,4; 5,4,6,3; 5,3,4,6; 5,3,5,5; 5,3,6,4; 5,2,5,6; 5,2,6,5; 5,1,6,6; 4,6,2,6; 4,6,3,5; 4,6,4,4; 4,6,5,3; 4,6,6,2; 4,5,3,6; 4,5,4,5; 4,5,5,4; 4,5,6,3; 4,4,4,6; 4,4,5,5; 4,4,6,4; 4,3,5,6; 4,3,6,5; 4,2,6,6; 3,6,3,6; 3,6,4,5; 3,6,5,4; 3,6,6,3; 3,5,4,6; 3,5,5,5; 3,5,6,4; 3,4,5,6; 3,4,6,5; 3,3,6,6; 2,6,4,6; 2,6,5,5; 2,6,6,4; 2,5,5,6; 2,5,6,5; 2,4,6,6; 1,6,5,6; 1,6,6,5; 1,5,6,6
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Statistics and Nonlinear Regression Allen R. Overman 105 6,6,1,6; 6,6,2,5; 6,6,3,4; 19 56 0.043210 6,6,4,3; 6,6,5,2; 6,6,6,1; 6,5,2,6; 6,5,3,5; 6,5,4,4; 6,5,5,3; 6,5,6,2; 6,4,3,6; 6,4,4,5; 6,4,5,4; 6,4,6,3; 6,3,4,6; 6,3,5,5; 6,3,6,4; 6,2,5,6; 6,2,6,5; 6,1,6,6; 5,6,2,6; 5,6,3,5; 5,6,4,4; 5,6,5,3; 5,6,6,2; 5,5,3,6; 5,5,4,5; 5,5,5,4; 5,5,6,3; 5,4,4,6; 5,4,5,5; 5,4,6,4; 5,3,5,6; 5,3,6,5; 5,2,6,6; 4,6,3,6; 4,6,4,5; 4,6,5,4; 4,6,6,3; 4,5,4,6; 4,5,5,5; 4,5,6,4; 4,4,5,6; 4,4,6,5; 4,3,6,6; 3,6,4,6; 3,6,5,5; 3,6,6,4; 3,5,5,6; 3,5,6,5; 3,4,6,6; 2,6,5,6; 2,6,6,5; 2,5,6,6; 1,6,6,6 6,6,2,6; 6,6,3,5; 6,6,4,4; 20 35 0.027006 6,6,5,3; 6,6,6,2; 6,5,3,6; 6,5,4,5; 6,5,5,4; 6,5,6,3; 6,4,4,6; 6,4,5,5; 6,4,6,4; 6,3,5,6; 6,3,6,5; 6,2,6,6; 5,6,3,6; 5,6,4,5; 5,6,5,4; 5,6,6,3; 5,5,4,6; 5,5,5,5; 5,5,6,4; 5,4,5,6; 5,4,6,5; 5,3,6,6; 4,6,4,6; 4,6,5,5; 4,6,6,4; 4,5,5,6; 4,5,6,5; 4,4,6,6; 3,6,5,6; 3,6,6,5; 3,5,6,6; 2,6,6,6; 6,6,3,6; 6,6,4,5; 6,6,5,4; 21 20 0.015432 6,6,6,3; 6,5,4,6; 6,5,5,5; 6,5,6,4; 6,4,5,6; 6,4,6,5; 6,3,6,6; 5,6,4,6; 5,6,5,5; 5,6,6,4; 5,5,5,6; 5,5,6,5; 5,4,6,6; 4,6,5,6; 4,6,6,5; 4,5,6,6; 3,6,6,6 6,6,4,6; 6,6,5,5; 6,6,6,4; 22 10 0.007716 6,5,5,6; 6,5,6,5; 6,4,6,6; 5,6,5,6; 5,6,6,5; 5,5,6,6;
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Statistics and Nonlinear Regression Allen R. Overman 106 4,6,6,6 6,6,6,5; 6,6,5,6; 6,5,6,6; 23 4 0.003086 5,6,6,6 6,6,6,6 24 1 0.000772
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Statistics and Nonlinear Regression Allen R. Overman 107 Tab le A31 Summary of frequency distribution for four dice. S c f F Z Z f 0.000000 4 1 0.000772 0.001510 0.000772 5 4 0.003086 0.003455 0.003858 6 10 0.007716 0.007246 0.011574 1.605467 1.565423 7 20 0.015432 0.013929 0.027006 1.359960 1.356700 8 35 0.027006 0.024540 0.054012 1.135382 1.147977 9 56 0.043210 0.039627 0.097222 0.917975 0.939254 10 80 0.061728 0.058651 0.158950 0.706891 0.730531 11 104 0.080247 0.079564 0.239197 0.501122 0.521808 12 125 0.096451 0.098928 0.335648 0.299664 0.313085 13 140 0.108025 0.112740 0.443673 0.100551 0.104362 14 146 0.112654 0.117760 0556327 +0.100551 +0.104362 15 140 0.108025 0.112740 0.664352 +0.299664 +0.313085 16 125 0.096451 0.098928 0.760803 +0.501122 +0.521808 17 104 0.080247 0.079564 0.841050 +0.706891 +0.521808 18 80 0.061728 0.058651 0.902778 +0.917975 +0.730531 19 56 0.043210 0.039627 0.945988 +1.135382 +0.939254 20 35 0.027006 0.024540 0.972994 +1.359960 +1.147977 21 20 0.015432 0.013929 0.988426 +1.605467 +1.565423 22 10 0.007716 0.007246 0.996142 23 4 0.003086 0.003455 0.999228 24 1 0.000772 0.001510 1.000000
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Statistics and Nonlinear Regression Allen R. Overman 108 C 1296 = 64 SSFZ208723.09221.221212erf1 r = 0.999770 0000 14 79104 4 2 2279104.4000.14exp11776.02exp21SSf f f 07548 1 00512 0 r = 0.99883 The frequency distribution for a set of dice more closel y conforms to the continuous Gaussian distribution as the number of dice increases. Complexity of computing the discrete distribution increases dramatically with the number of dice, and becomes unwieldy beyond four dice. For five dice the total number of c ombinations is 65 = 7776! The two dice problem illustrates how well the continuous Gaussian distribution approximates the discrete distribution (triangular). Even though the approximation is not exact, it does bring in an analytic function which we have used in the model for plant growth. The peg board offers a simpler model of the frequency distribution than does a set of dice. Reference: Speyer, E. 1994. Six Roads from Newton: Great Discoveries in Physics. John Wiley & Sons. New York. NY.
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Statistics and Nonlinear Regression Allen R. Overman 109 Derivatives for power functions (as defined by Cauchy) Function Derivative 0xy 011)(00xxxy 00xxy 0lim0 x y dx dyx 1xy x x x x x x x y 1 1) ( 1 x x x y 1lim0 x y dx dyx 2x y 2 2 2 2 2 2) ( 2 ) ( 2 ) ( x x x x x x x x x x x y x x x x x x x y 2 ) ( 22 x x y dx dyx2lim0 3x y 3 2 2 3 3 2 2 3 3 3) ( ) ( 3 3 ) ( ) ( 3 3 ) ( x x x x x x x x x x x x x x x y 22)(33xxxxxy 203limxxydxdyx 4xy 322343223444)(4)(64)(4)(64)(xxxxxxxxxxxxxxxxxy 223)(464xxxxxxy 304limxxydxdyx nxy 1 nnx dx dy
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Statistics and Nonlinear Regression Allen R. Overman 110 The reader may note that the binomial expansion with n = 0, 1, 2, ,3 leads to finite power series. Newton was the first to prove this. Then he showed that for n either negative or a fraction the expansion leads to an infinite power series (Berlinski, 2000, p. 30). This led to Newtons first memoir in 1669 entitled On Analysis by Infinite Series which preceded his development of calculus. It is common to write the general solution as an infinite power series 044332210iiixaxaxaxaxaay where ai are the expansion coefficients. Now the coefficients can be evaluated using the derivatives fro m calculus and the boundary values at x = 0. This leads to the following relationships 0 0) 0 ( a y x y 1 0a dx dyx 0 2 2 2 2 0 2 2! 2 1 1 2 x xdx y d a a dx y d 03333033!31123xxdxydaadxyd 04444044!411234xxdxydaadxyd 00!1!xnnnnxnndxydnaandxyd This leads finally to the infinite series nxnnxxxdxydnxdxydxdxdyyy0202200!1!21!11
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Statistics and Nonlinear Regression Allen R. Overman 111 may be recognized as the Taylor series, which was discovered in 1715 by Brook Taylor. This approach assumes that the derivatives exist. It also assumes that the series converges to a finite value for any value of x or at least for a limited domain of x Physics usually enters into the process by way of a differential equation with initial conditions at x = 0. Berlinski, D. 2000. Newtons Gift: How Sir Isaac Newton Unlocked the System of the World. Simon & Schuster. New York, NY. Example: Consider the first order differential equation k dx dy with the initial condition 0 at0 x y y where k is a constant. Solution: kdxdyx0 0022xdxyd 00044033xnnxxdxyddxyddxyd The solution becomes kx y x dx y d n x dx y d x dx dy y yn x n n x x 0 0 2 0 2 2 0 0! 1 2 1 1 1 which is the equation of a straight line.
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Statistics and Nonlinear Regression Allen R. Overman 112 Example: Consider the first order differential equation kydxdy with the initial condition 0 at0 x y y where k is a constant. Solution: 0 0ky dx dyx 0 2 0 0 2 2y k dx dy k dx y dx x 0 3 0 3 3y k dx y dx 0 0y k dx y dn x n n The solution becomes n x n n x xx dx y d n x dx y d x dx dy y y0 2 0 2 2 0 0! 1 2 1 1 1 n nx y k n x y k x ky y0 2 0 2 0 0! 1 2 1 nkxnkxkxy!1!21!111210 kxyexp0 where nkxnkxkxkx!1!21!111exp2 as defined by Euler.
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Similar Variable Names
I have variables $X1,...,X10$ and $X1hat,..., X10hat$ in a system of equations. How can I refer to these variables in a summation? I am looking for something like:
Sum[Xi' Xi'hat,{i,10}]
• Instead of x1, use x[1]. Then you can use x[i]. Commented May 7, 2017 at 7:37
• Thanks! It solved my problem for variables. But is there also a way to do something similar for an equation to avoid writing it multiple times? For example instead of writing eq1=X[1]^2,...,eq10=X[10]^2, write something like: {eq[i] = X[i]^2, {i, 10}} in a system of equations? Commented May 7, 2017 at 7:51
• Use Table to make a list of equations. Commented May 7, 2017 at 8:55
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# 2 D Shape Pattern Worksheet
In Free Printable Worksheets216 views
4.17 / 5 ( 180votes )
Top Suggestions 2 D Shape Pattern Worksheet :
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Lie algebra, exponential
Lie algebra of type $( E)$
A finite-dimensional real Lie algebra $\mathfrak g$ for any element $X$ of which the operator of adjoint representation (cf. Adjoint representation of a Lie group) $\mathop{\rm ad} X$ does not have purely imaginary eigen values. The exponential mapping $\mathop{\rm exp} : \mathfrak g \rightarrow G$ of the algebra $\mathfrak g$ into the corresponding simply-connected Lie group $G$ is a diffeomorphism, and $G$ is an exponential Lie group (cf. Lie group, exponential).
Every exponential Lie algebra is solvable (cf. Lie algebra, solvable). A nilpotent Lie algebra (cf. Lie algebra, nilpotent) over $\mathbf R$ is an exponential Lie algebra. The class of exponential Lie algebras is intermediate between the classes of all solvable and all supersolvable Lie algebras (cf. Lie algebra, supersolvable); it is closed with respect to transition to subalgebras, quotient algebras and finite direct sums, but it is not closed with respect to extensions.
The simplest example of an exponential Lie algebra that is not a supersolvable Lie algebra is the three-dimensional Lie algebra with basis $X , Y , Z$ and multiplication specified by the formulas
$$[ X , Y ] = 0 ,\ \ [ Z , X ] = a _ {11} X + a _ {12} Y ,\ \ [ Z , Y ] = a _ {21} X + a _ {22} Y ,$$
where $[ a _ {ij} ]$ is a real matrix that has complex but not purely imaginary eigen values. The three-dimensional Lie algebra $\mathfrak g _ {0}$ with basis $X , Y , Z$ and defining relations
$$[ X , Y ] = 0 ,\ \ [ Z , X ] = Y ,\ \ [ Z , Y ] = - X$$
is a solvable, but not an exponential Lie algebra.
A Lie algebra $\mathfrak g$ is exponential if and only if all roots of $\mathfrak g$( cf. Root system) have the form $\alpha + i \beta$, where $\alpha$ and $\beta$ are real linear forms on $\mathfrak g$ and $\beta$ is proportional to $\alpha$( see ), or if $\mathfrak g$ has no quotient algebra containing a subalgebra isomorphic to $\mathfrak g _ {0}$.
For references see Lie group, exponential.
How to Cite This Entry:
Lie algebra, exponential. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Lie_algebra,_exponential&oldid=47623
This article was adapted from an original article by V.V. Gorbatsevich (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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• Unity uses the PhysX system developed by Nvidia for 3D physics.
• High level functions are available for creating rigid bodies, adding forces, and coupling bodies with joints and springs.
• Inherently discrete, so errors will occur, and it is not always deterministic.
## Manual spring/damper calculations
• A spring/damper system is an example of PD controller, and are used to correct for some error (difference between desired and actual position). The spring applies a force proportional to the error, and the damper applies a force proportional to the derivative of the error (velocity), relative to a desired velocity. Constants (`Kp` and `Kd`) are used to control this force. Usually you want `Kp` to be positive (to correct the error) and `Kd` to be negative (to slow the system down - simulating drag or friction)
• If you want to manually calculate a drag force using a Rigidbody's velocity, you will always be one frame behind any forces you apply, since velocity only gets updated when the physics simulation is updated (i.e. during `FixedUpdate`. Try to use the built-in drag parameters where you can.
• For rotational/torsion springs, the Rigidbody Inertia Tensor property describes how the mass of the object is distributed in 3D space. It is conceptually equivalent to the Moment of inertia. Along with the Inertia Tensor Rotation property, these can be use to change how much torque is required to rotate an object. These are approximated by Unity if left unchanged, but you can manually assign these to make an object more resistant to rotation in a particular direction.
• Classical calculations for PD controllers can cause issues such as overshooting and oscillations in discrete physics calculation. A more stable Spring/Damper system has been helpfully typed up in this article by Digital Opus. With this you can create critically damped systems which smoothly correct for error with no oscillation.
## Configurable Joints
• When an axis is set to limited, all external forces/torques applied to the axis seem to be ignored.
## Unity's drag calculation
• The drag value assigned in the inspector is quite opaque, and it is not clear what the units are. It seems that the drag is simply a multiplier for the velocity, which asymptotes the velocity to zero as drag increases. To simulate it yourself:
``````body.velocity *= Mathf.Clamp01(1f - drag * Time.fixedDeltaTime);
``````
Published 2022-01-04
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# How do you FOIL (4x^5-9)(x^6-7)?
$\left(4 {x}^{5} - 9\right) \left({x}^{6} - 7\right)$
First $\implies 4 {x}^{11}$
Outer $\implies - 28 {x}^{5}$
Inner $\implies - 9 {x}^{6}$
Last $\implies 63$
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Term Rewriting System R:
[X]
af(X) -> g(h(f(X)))
af(X) -> f(X)
mark(f(X)) -> af(mark(X))
mark(g(X)) -> g(X)
mark(h(X)) -> h(mark(X))
Innermost Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
MARK(f(X)) -> AF(mark(X))
MARK(f(X)) -> MARK(X)
MARK(h(X)) -> MARK(X)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Argument Filtering and Ordering`
Dependency Pairs:
MARK(h(X)) -> MARK(X)
MARK(f(X)) -> MARK(X)
Rules:
af(X) -> g(h(f(X)))
af(X) -> f(X)
mark(f(X)) -> af(mark(X))
mark(g(X)) -> g(X)
mark(h(X)) -> h(mark(X))
Strategy:
innermost
The following dependency pair can be strictly oriented:
MARK(f(X)) -> MARK(X)
There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(MARK(x1)) = x1 POL(h(x1)) = x1 POL(f(x1)) = 1 + x1
resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
f(x1) -> f(x1)
h(x1) -> h(x1)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳Argument Filtering and Ordering`
Dependency Pair:
MARK(h(X)) -> MARK(X)
Rules:
af(X) -> g(h(f(X)))
af(X) -> f(X)
mark(f(X)) -> af(mark(X))
mark(g(X)) -> g(X)
mark(h(X)) -> h(mark(X))
Strategy:
innermost
The following dependency pair can be strictly oriented:
MARK(h(X)) -> MARK(X)
There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(MARK(x1)) = x1 POL(h(x1)) = 1 + x1
resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
h(x1) -> h(x1)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳AFS`
` ...`
` →DP Problem 3`
` ↳Dependency Graph`
Dependency Pair:
Rules:
af(X) -> g(h(f(X)))
af(X) -> f(X)
mark(f(X)) -> af(mark(X))
mark(g(X)) -> g(X)
mark(h(X)) -> h(mark(X))
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes
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A home loan calculator is an automated tool that enables you to calculate and determine any financial implications is financing your house. Every home loan calculator is able to calculate differing variable in a home loan and they can successfully help you in finding out what you need to pay in certain periods of time.
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If you have time to read on the formulas then use the knowledge you will gain to your advantage. However, if you really have no time to check these things then you can have someone check it for you perhaps.
1. #### Makes Use of Updated Formulas
Apart from using the right formulas yielding to the right results, another thing you must consider is if the formulas being used are indeed the latest ones.
Math and financing grows through time and new ways of solving things emerge. New formulas are developed which can make computations much easier. Furthermore, these new formulas can perhaps calculate more variables also which a good thing is so you wouldn’t be caught by surprise.
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# Sta641 GDB discussion
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Correlation:
Correlation is the concept of linear relationship between two variables. Please note that I emphasized the word linear. It is linear relationship nor any other relationship.Correlation is the process of studying the cause and effect relationship that exists between two variables.
Explanation:
Co relationship tells you're insidious or amusing in your companions group and its coefficient tells that how much more insidious or clever are you companions gather in case compared to best clever friend.
In measurement, reliance is any measurable relationship between two random Variable and two sets of information. Relationship alludes to any of a wide lesson of factual Relationships including reliance. Commonplace cases of subordinate marvels Include the relationship between the physical statures of guardians and their offspring and the relationship between the request for a numerically it is given by
# STA641 GDB Solution # 1 Fall 2020 - Statistical Packages and its Applications
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# 统计代写|网络分析代写Network Analysis代考|Discovering complexes by finding dense regions
## 统计代写|网络分析代写Network Analysis代考|Discovering complexes by finding dense regions
The molecular complex detection (MCODE) algorithm, described in the early work of Bader et al., [8], tries to find complexes by building clusters. The rationale of MCODE that takes in as input the interaction network, is the separation of dense regions based on an ad hoc defined local density. MCODE has three main steps: (i) node weighting, (ii) complexes prediction, (iii) post processing. In its first stage, MCODE weights all vertices based on their local network density. The local area, in which density is calculated, is delimited by an ad hoc defined subgraph structure called $k$-core. A $k$-core of a graph is the central most densely connected subgraph with minimal degree $k$. Thus the core-clustering coefficient of a vertex $v$ is the density of the highest $k$-core of the immediate neighborhood of $v$. Finally, the weight of a vertex is the product of the vertex core-clustering coefficient and the highest $k$-core level, $k m a x$, of the immediate neighborhood of the vertex.
The resulting weighted graph is given as input of the second stage. The algorithm, hence, starting from the highest weighted vertex, tries to span a region visiting vertices, whose weight is above a certain threshold, called vertex weight percentage (WWP). This step stops when no more vertices can be added to the complex, and it is repeated considering the next highest weighted network not already considered. Finally, the third step has to filter the complexes, which do not contain at least a $k$-core with $k=2$.
The third stage is the post-processing. Complexes are filtered if they do not contain at least a 2-core (graph of minimum degree equal to 2). The algorithm has two main options that determine the characteristics of this phase: fluff and haircut.
The algorithm has two modes of execution: a direct mode (in which the search starts from a given node), and an undirect mode (in which the seed is randomly selected). MCODE is freely available on the author’s website, ${ }^5$ and there also exists a version that runs as a plugin for the Cytoscape software.
## 统计代写|网络分析代写Network Analysis代考|Complex prediction via clustering
The paper of Ul-Amin et al. [2] presents another approach based on clustering an interaction network to find complexes. The rationale of the algorithm is the building of a cluster as a dense region embedded into a sparse region. The algorithm is logically organized in five major steps: (i) initialization, (ii) termination check, (iii) selection of starting node, (iv) cluster growth and (v) output.
In the first step, the algorithm takes as input an undirected graph and initializes its main variables: cluster density, cluster property, cluster ID. The algorithm calculates the minimum value of density for each generated cluster, i.e., the ratio of the number of edges present in the cluster and the maximum possible number of edges in the cluster. The cluster property $c p_{n, k}$ of any node $\mathrm{n}$, with respect to any cluster $k$ of density $d_k$ and size $\left|N_k\right|$, is the ratio between the total number of edges between the node $\mathrm{n}$ and each of the nodes of cluster and the product between the density and the size of the cluster $d_k$. The cluster ID, $k$ is initialized to 1.
In the second step, the algorithm verifies the termination conditions, and if the graph has no edges, the algorithm will end. Conversely, if the termination check fails, the algorithm in its third step, namely selection of starting node, will select a node as a starting point to build a new cluster. Hence, in the fourth step, namely cluster growth, the algorithm adds nodes to the cluster chosen from the neighbors of starting node. Neighbors are labeled with a priority in order to guide the cluster formation. Finally, when a cluster is generated, it is removed from the graph, and the clusterID, $k$ is incremented.
The algorithm is polynomial, and its complexity in the worst case is $O\left(N^3\right)$, where $N$ is the number of nodes. This complexity is due to the cost of sorting clusters.
# 网络分析代考
## 统计代写|网络分析代写Network Analysis代考|Discovering complexes by finding dense regions
Bader 等人 [8] 早期工作中描述的分子复合物检测 (MCODE) 算法试图通过构建簇来寻找复合物。将交互网络作为输入的 MCODE 的基本原理是基于临时定义的局部密度分离密集区域。MCODE 具有三个主要步骤：(i) 节点加权，(ii) 复合物预测，(iii) 后处理。在其第一阶段，MCODE 根据本地网络密度对所有顶点进行加权。计算密度的局部区域由一个名为k-核。Ak-图的核心是具有最小度的中心最密集连接的子图k. 因此顶点的核心聚类系数在是密度最高的k- 紧邻的核心在. 最后，顶点的权重是顶点核心聚类系数与最高点的乘积k-核心水平，k米AX, 顶点的直接邻域。
## 统计代写|网络分析代写Network Analysis代考|Complex prediction via clustering
Ul-Amin 等人的论文。[2] 提出了另一种基于聚类交互网络来寻找复合物的方法。该算法的基本原理是将集群构建为嵌入到稀疏区域中的密集区域。该算法在逻辑上分为五个主要步骤：（i）初始化，（ii）终止检查，（iii）起始节点的选择，（iv）集群增长和（v）输出。
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# MV-1025: Model Point-to-Curve (PTCV) Higher-Pair Constraint
In this tutorial you will learn how to model a PTCV (point-to-curve) joint.
A PTCV (point-to-curve) joint is a higher pair constraint. This constraint restricts a specified point on a body to move along a specified curve on another body. The curve may be open or closed, planar or in 3-d space. The point may belong to a rigid, flexible or point body. This constraint can help avoid modeling contact in some systems. It may prove advantageous since proper contact modeling (refer to MV-1010: 3D Mesh to Mesh Contact Simulation) in many cases involves fine-tuning of contact parameters. One good example for such a system is a knife-edge cam follower mechanism. You can avoid modeling the contact between the cam and follower by defining a PTCV joint: the curve being the cam profile and the point being the tip of the follower. In this tutorial, you will model a knife-edge cam follower mechanism with the help of a PTCV joint.
## Create Points
In this step you will create points for the model.
Copy all the files from the mbd_modeling\interactive folder to your <working directory>.
1. Start a new MotionView session.
We will work with the default units (kg, mm, s, N).
• From the Project Browser right-click on Model and select Add Reference Entity > Point.
• On the Model-Reference toolbar, right-click the (Point) icon.
3. For Label, enter PivotPoint. Accept the default Variable name.
4. Click OK.
5. Click the Properties tab and specify the X, Y, and Z coordinates as 0.0.
6. Repeat steps 2 through 5 to create the points specified in Table 1.
Table 1.
Point X Y Z
FollowerCM 0.0 65.557 0.0
FollowerPoint 0.0 25.0 0.0
FollowerJoint 0.0 85.0 0.0
CamCM 0.0 -14.1604 0.0
## Create Bodies
In this step you will create the Cam and Follower bodies.
You will use pre-specified inertia properties to define the bodies.
• From the Project Browser right-click on Model and select Add Reference Entity > Body.
• On the Model-Reference toolbar, right-click on the (Body) icon.
2. Add the Cam and Follower bodies.
1. For the first Label, enter Cam.
2. Click Apply.
3. For the second Label, enter Follower.
4. Click OK to close the Add Body or BodyPair dialog.
The bodies you created will appear in the Model Tree under the Bodies folder.
3. In the Properties panel for each body, specify the information given in Table 2.
Table 2.
Body Mass Ixx Iyy Izz Ixy Iyz Izx
Cam 0.174526 60.3623 63.699 123.276 0.0 0.0 0.0
Follower 0.0228149 7.10381 0.219116 7.22026 0.0 0.0 0.0
4. Specify the CM Coordinate tab settings for the Cam body.
1. Check the Use center of mass coordinate system box.
2. Double-click .
3. In the Select a Point dialog, choose CamCM.
4. Click OK.
5. Accept the defaults for axes orientation properties.
5. Specify the CM Coordinate tab settings for the Follower body.
1. Check the Use CM Coordsys box.
2. Double-click .
3. In the Select a Point dialog, choose FollowerCM.
4. Click OK.
5. Accept the defaults for axes orientation properties.
## Create Joints
In this step you will define the revolute joint and the translational joint needed for the model.
The revolute joint is between the Cam and Ground Body. The translational joint is between the Follower and the Ground Body. You will define the PTCV joint in a different step.
• From the Project Browser, right-click on Model and select Add > Constraint > Joint.
• On the Model-Constraints toolbar, click the (Joints) icon.
2. Create the CamPivot joint.
1. In the Add Joint or JointPair, for Label enter CamPivot.
2. For Type, select Revolute Joint.
3. Click OK.
4. In the Connectivity tab, double click on and resolve it to Cam.
5. Resolve to Ground Body and click OK.
6. In the Connectivity tab, double click and resolve it to PivotPoint.
7. Change the Alignment Axis to . Resolve Vector to Global Z.
3. Create the FollowerJoint.
1. In the Add Joint or JointPair, for Label enter FollowerJoint.
2. For Type, select Translational Joint.
3. Click OK.
4. In the Connectivity tab, double click on and resolve it to Follower.
5. Resolve to Ground Body and click OK.
6. In the Connectivity tab, double click and resolve it to FollowerJoint.
7. Change the Alignment Axis to . Resolve Vector to Global Y and click OK.
## Create Markers
In this step you will define markers required for defining the PTCV joint.
You need a marker associated with the Cam (for the curve) and one associated with the Follower (for the point).
• From the Project Browser, right-click on Model and select Add > Reference Entity > Marker.
• On the Model-Reference toolbar, click the (Marker) icon.
2. Create the CamMarker.
1. In the Add Marker or MarkerPair dialog, for Label enter CamMarker.
2. Click OK.
3. In the Properties tab, double click on and resolve it to Cam and click OK.
4. In the Properties tab, double-click and resolve it to PivotPoint.
5. Click OK. Accept the defaults for axes orientation.
3. Create the FollowerMarker.
1. In the Add Marker or MarkerPair dialog, for Label enter FollowerMarker.
2. Click OK.
3. In the Properties tab, double click on and resolve it to Follower and click OK.
4. In the Properties tab, double-click and resolve it to FollowerPoint.
5. Click OK. Accept the defaults for axes orientation.
## Create Graphics
In this step you will load graphics for the Cam and Follower and create graphics for the joints in the model.
Graphics for the cam and follower have been provided as h3d files. We need to associate the h3ds with bodies defined in our model.
• From the Project Browser, right-click on Model and select Add > Reference Entity > Graphics.
• On the Model-Reference toolbar, click the (Graphics) icon.
2. From the dialog, for Label, enter Cam.
3. From the drop-down menu, click File.
4. Click the (File Browser) icon and open the CamProfile.h3d from the model folder.
5. Click OK.
6. In the Connectivity tab, double-click on and resolve the graphic to the Cam body.
7. Repeat step 1 to create another graphic. For Label, enter Follower.
8. From the drop-down menu, click File.
9. Click the (File Browser) icon and open the FollowerProfile.h3d from the model folder.
10. Click OK.
11. In the Connectivity tab, double-click on and resolve the graphic to the Follower body.
12. Create cylinder graphics for the Cam pivot in your model using the specifications listed in Table 3.
Table 3.
Label Type Direction ()
PivotGraphicOne Cylinder Ground Body PivotPoint Global Z
PivotGraphicTwo Cylinder Cam PivotPoint Global Z
13. In the Project Browser, click PivotGraphicOne.
14. In the Properties tab, specify the values listed in Table 4:
Table 4.
Property Value
Length 7.5
Offset -3.75
15. For Cap properties, choose Cap Both Ends.
16. In the Project Browser, click PivotGraphicTwo.
17. In the Properties tab, specify the values listed in Table 5:
Table 5.
Property Value
Length 7.6
Offset -3.8
18. For Cap properties, choose Cap Both Ends.
19. Create a box graphic for the Follower translational joint with the specifications in Table 6 and the Properties listen in Table 7.
Table 6.
Label Type Z-Axis () XZ Plane ()
FollowerJointGraphic Center (choose this option in the Connectivity tab) Ground Body FollowerJoint Global Z Global X
Table 7.
Property Value
Length X 15
Length Y 10
Length Z 10
Your model should look like the example in Figure 3.
## Create the Curve
In this step you will create the curve that defines the profile of the Cam.
You will use data provided in .csv format to define the curve.
1. Open the Add Curve dialog by doing one of the following:
• From the Project Browser, right-click on Model and select Add > Reference Entity > Curve.
• On the Model-Reference toolbar, click the (Curves) icon.
2. For Label, enter CamProfile and click OK.
3. In the Properties tab, click the first drop-down menu and change the curve from 2D Cartesian to 3D Cartesian.
4. Click the fourth drop-down menu and set the curve type to Closed curve.
5. Click the x radio button.
6. Click and open CamProfile.csv.
7. Choose the properties of the curve given in Figure 4.
8. Click the y radio button. Change the Component to Column 2.
9. Click the z radio button. Change the Component to Column 3.
## Create the PTCV Joint
In this step, you will create the PTCV joint.
1. Open the Add AdvJoint dialog by doing one of the following:
• From the Project Browser, right-click on Model and select Add > Constraint > Advanced Joint.
• On the Model-Reference toolbar, click the (Advanced Joint) icon.
2. In the dialog, enter the Label PTCV.
3. From the drop-down menu, choose PointToCurveJoint and click OK.
4. Configure the Connectivity tab for the PTCV joint.
1. Double click . In the dialog, choose Follower and click OK.
2. Double click . In the dialog, choose FollowerPoint and click OK.
3. Double click . In the dialog, choose CamProfile and click OK.
4. Double click . In the dialog, choose CamMarker and click OK.
## Specify the Cam Motion
In this step, you will specify a motion for the cam using an expression.
1. Open the Add Motion or MotionPair dialog by doing one of the following:
• From the Project Browser, right-click on Model and select Add > Constraint > Motion.
• On the Model-Reference toolbar, click the (Motion) icon.
2. In the dialog, enter the Label CamMotion and click OK.
3. In the Connectivity tab, double-click . Choose CamPivot and click OK.
4. in the Properties tab, in the drop-down menu define the motion by Expression.
5. Enter 10*TIME in the Expression field.
## Specify Gravity
In this step you will specify the gravity in the negative Y direction.
1. In the Project Browser, click to expand Misc. > Forms.
2. In the Forms folder, click Gravity. In the panel, specify the following values:
• X Component = 0
• Y Component = -9810
• Z Component = 0
## Specify Output Requests
In this step you will specify the output requests in order to monitor the reaction on the PTCV joint.
1. Open the Add Output dialog by doing one of the following:
• From the Project Browser, right-click on Model and select Add > General MDL Entity > Output.
• On the General toolbar, click the (Outputs) icon.
2. In the dialog, for Label enter PTCV Reaction and click OK.
3. In the Properties tab, from the drop-down menu define the output by Expression.
4. Click in the F2 field to activate the button.
5. Click the button.
6. In the Expression Builder, populate the expression as 'PTCV({aj_0.idstring},0,2,0)'.
7. Click OK.
8. Repeat steps 4 through 6 for F3, F4, F6, F7, and F8 by changing the third parameter in the expression to 3, 4, 6, 7, and 8 accordingly.
The PTCV (id, jflag, comp, ref_marker) function returns the reaction on the PTCV joint:
• id ID of the PTCV joint
• jflag 0 gives reaction on the I-marker and 1 on J-marker
• comp component of the reaction
• ref_marker reference marker (0 implies Global Frame)
## Run the Model
In this step you will run the knife-edge cam follower model.
The model is now defined completely and is ready to run.
1. On the toolbar, click (Run).
2. In the Run panel, specify the values shown in Figure 9.
3. Click the Save and run current model radio button.
4. Click the (browser icon) and specify a name for the solver file.
5. Click Save.
6. Click the (Check Model) button to check the model for errors.
7. To invoke the solver, click the Run button.
8. Once the solver has finished, click the (Start/Pause Animation) button to view the animation.
## View the Results
In this step you will learn how to view the animation and plot the Y displacement of the follower.
1. Once the solver has finished, the Animate button will be active. Click on Animate.
Click the (Start/Pause Animation) button to view the animation.
One would also like to inspect the displacement profile of the follower in this mechanism. For this, you will plot the Y position of the center of mass of the follower.
2. From the Page Controls toolbar, click > . Click in the window on the lower right to make it the active window.
3. In the Select application drop-down menu, change the client from MotionSolve to HyperGraph 2D .
4. On the Curves toolbar, click (Build Plots).
5. Click the (file browser) and open the results.abf file.
6. Configure the Plots panel as show in Figure 10.
7. Click Apply. This will plot the Y profile of the center of mass of the follower.
The profile for the Y-displacement of the follower should look like the one in Figure 11.
If you set the x-axis properties to zoom in on the cycle, the profile looks as shown in Figure 12.
## Check the Model for Potential Lift-Off
In this step, you will check the knife-edge cam follower mechanism for potential lift-off by plotting the Y-reaction on the follower.
In some cases, the dynamics of the system may cause the follower to lose contact with the cam. This is called ‘lift-off’. In such cases, modeling the system with a PTCV will give you incorrect results because the joint constrains the follower point to always be on the curve (and hence cannot model lift-offs). For such cases, you must use contact modeling (refer to MV-1010: 3D Mesh to Mesh Contact Simulation). However, you will want to start with a PTCV model since it is a lot easier than modeling contact. Given this scenario, model the system using a PTCV joint and monitor the PTCV joint reaction. If the reaction on the follower is a ‘pulling’ reaction, this indicates lift-off would have occurred and you must switch to a contact model. Otherwise, the contact model is unnecessary. Now, you will check the model you used in the tutorial. The follower is moving along the Y-axis, so any negative reaction along the Y-axis is a 'pulling' reaction.
1. Click (Add Page) to add a new page to the session.
2. Switch the client to HyperGraph 2D .
3. Click on (Build Plots).
4. Click on (browser icon) and load the results.abf file.
5. Configure the Plot panel as shown in Figure 13.
6. Click Apply.
The profile should look like the one shown in Figure 14.
7. Scale the x-axis to view one cycle on the profile.
As shown in Figure 15, the Y component of the PTCV reaction on the follower is always positive. There is no pulling reaction, so the PTCV model is acceptable for this mechanism.
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My Math Forum if light cannot bend then how refraction & reflection of light works?
Physics Physics Forum
December 14th, 2014, 06:02 PM #1 Senior Member Joined: Aug 2014 From: India Posts: 343 Thanks: 1 if light cannot bend then how refraction & reflection of light works? if light cannot bend then how refraction & reflection of light are working? is it still mystery or any other theory is there for this?
December 14th, 2014, 06:48 PM #2 Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 Math Focus: Mainly analysis and algebra Light can bend! This was one of the predictions that validated relativity. Well, strictly speaking it's the space that curves rather than the light, but the effect is the same. Reflection is basically the same as bouncing balls off a wall, although here the balls would be photons. Refraction is caused by the different velocity of waves in two media.
December 14th, 2014, 07:16 PM #3
Senior Member
Joined: Aug 2014
From: India
Posts: 343
Thanks: 1
Quote:
Originally Posted by v8archie the space that curves rather than the light,
sorry My question is regarding light bend, so i don't know why you mention "space" in your reply, i am not asking about space bends or not.
December 14th, 2014, 07:21 PM #4
Senior Member
Joined: Jul 2014
From: भारत
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Quote:
Originally Posted by Ganesh Ujwal sorry My question is regarding light bend, so i don't know why you mention "space" in your reply, i am not asking about space bends or not.
December 14th, 2014, 07:25 PM #5 Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 Reflection can happen in two quite different ways. If you have a smooth, highly polished surface and you shine a narrow beam of light at it, you get a narrow beam of light reflected back off it. This is called specular reflection and it's what happens if you shine a flashlight or laser into a mirror: you get a well-defined beam of light bouncing back towards you. Most objects aren't smooth and highly polished: they're quite rough. So, when you shine light onto them, it's scattered all over the place. This is called diffuse reflection and it's how we see most objects around us as they scatter the light falling on them. If you can see your face in something, it's specular reflection; if you can't see your face, it's diffuse reflection. Polish up a teaspoon and you can see your face quite clearly. But if the spoon is dirty, all the bits of dirt and dust are scattering light in all directions and your face disappears. Refraction : Light waves travel in straight lines through empty space (a vacuum), but more interesting things happen to them when they travel through other materials—especially when they move from one material to another. That's not unusual: we do the same thing ourselves. Have you noticed how your body slows down when you try to walk through water? You go racing down the beach at top speed but, as soon as you hit the sea, you slow right down. No matter how hard you try, you cannot run as quickly through water as through air. The dense liquid is harder to push out of the way, so it slows you down. Exactly the same thing happens to light if you shine it into water, glass, plastic or another more dense material: it slows down quite dramatically. This tends to make light waves bend—something we usually call refraction. How refraction works : Refraction of laser beams inside crystals Photo: Laser beams bending (refracting) through a crystal. Photo by Warren Gretz courtesy of US Department of Energy/National Renewable Energy Laboratory (DOE/NREL). You've probably noticed that water can bend light. You can see this for yourself by putting a straw in a glass of water. Notice how the straw appears to kink at the point where the water meets the air above it. The bending happens not in the water itself but at the junction of the air and the water. You can see the same thing happening in this photo of laser light beams shining between two crystals. As the beams cross the junction, they bend quite noticeably. Artwork explaining how refraction (the bending of light) happens when light rays slow down Why does this happen? You may have learned that the speed of light is always the same, but that's only true when light travels in a vacuum. In fact, light travels more slowly in some materials than others. It goes more slowly in water than in air. Or, to put it another way, light slows down when it moves from air to water and it speeds up when it moves from water to air. This is what causes the straw to look bent. Let's look into this a bit more closely. Imagine a light ray zooming along through the air at an angle to some water. Now imagine that the light ray is actually a line of people swimming along in formation, side-by-side, through the air. The swimmers on one side are going to enter the water more quickly than the swimmers on the other side and, as they do so, they are going to slow down—because people move more slowly in water than in air. That means the whole line is going to start slowing down, beginning with the swimmers at one side and ending with the swimmers on the other side some time later. That's going to cause the entire line to bend at an angle. This is exactly how light behaves when it enters water—and why water makes a straw look bent. Last edited by Prakhar; December 14th, 2014 at 07:28 PM.
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• 100% Satisfaction Guarantee
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# Can I received non-qualified plan benefit from my ex-husband
### Resolved Question:
Can I received non-qualified plan benefit from my ex-husband company when I divorce him.
He retired and now all pensions are pay in status.
We must divide his pension by time rule.
But I'd like to know how much I can get Non-qualified benefit such as supplemental retirement plan.
Submitted: 2 years ago.
Category: Legal
Expert: Dimitry K., Esq. replied 2 years ago.
In terms of a 'non-qualified' benefit, the benefit is generally calculated the same way as a 'qualified' benefit, meaning based on length of time of marriage based on how long the benefit was earned by the spouse and what percentage of that benefit was earned while married to you. To use an example, if the benefit grew 60% while you were married, that 60% growth is split in half, and you would be entitled to about 30% growth of the benefit going forward. But if this benefit was scheduled and obtained after divorce was granted and s a separate benefit, you would not be naturally entitled to it since it was not earned or obtained while married. If you can provide more specific information I can better break it down, but in general this is how this is reviewed and calculated.
Hope that helps clarify.
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Customer: replied 2 years ago.
Dear Mr. Dimitry K Esq.
I heard that qualified plan and non-qualified will be divide in different way.
qualified will divide in time rule and non-qualified will not.
Can you check one more time?
Also I want to know about time rule.
my ex worked for the company for 36 years and we had married for 12 years during his working.
That means I can get pension half of 1/3(period) we married?
It means that I will get half of the value of 12 years(last 12 years) ?
Which is the right way to calculate as follows?
1/6(=half of 1/3married period) of the total amount of the pension?
Or
(Total amount of pension - the value of we married) divide in half.
Thank you.
Expert: Dimitry K., Esq. replied 2 years ago.
Thank you for your follow-up, Mai Mura.
Qualified and non-qualified splits are not based on the funds but on when they were earned. For example if your spouse earned a non-qualified bonus during the time you were married, such a bonus is considered communal and marital and is generally split 50/50. But at the same time a non-qualified pension plan that was added to by the employer and your spouse and gradually grew in value much like a qualified plan is likewise split like a qualified plan, base don time and proportion of value. Hence, there may be a difference in the split but it would be based on how the funds were earned, not on the fact that the funds are potentially qualified or non-qualified in nature.
The formula for proper calculation is the first one you listed, as the split is not a straight 1/6 of the pension, it is based on evaluation from time of marriage to divorce. For example if pension was worth \$50,000 at time of marriage, and is now worth \$80,000, the growth in value, the \$30,000, is the marital growth--that is what wold be split in half with half going to the other spouse.
Hope that helps.
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mersenneforum.org Gaps between maximal prime gaps
Register FAQ Search Today's Posts Mark Forums Read
2018-05-22, 22:43 #1 Bobby Jacobs May 2018 23×5×7 Posts Gaps between maximal prime gaps Consider the sequence of record prime gaps. 1, 2, 4, 6, 8, 14, 18, 20, 22, 34, ... Now, take the differences between consecutive numbers in this sequence. 1, 2, 2, 2, 6, 4, 2, 2, 12, 2, ... These are the gaps between record prime gaps. Is there a pattern here?
2018-05-22, 23:07 #2
rudy235
Jun 2015
Vallejo, CA/.
3·383 Posts
Quote:
Originally Posted by Bobby Jacobs Consider the sequence of record prime gaps. 1, 2, 4, 6, 8, 14, 18, 20, 22, 34, ... Now, take the differences between consecutive numbers in this sequence. 1, 2, 2, 2, 6, 4, 2, 2, 12, 2, ... These are the gaps between record prime gaps. Is there a pattern here?
Try this for size
2018-05-24, 14:40 #3
ldesnogu
Jan 2008
France
22·149 Posts
Quote:
Originally Posted by Bobby Jacobs Consider the sequence of record prime gaps. 1, 2, 4, 6, 8, 14, 18, 20, 22, 34, ... Now, take the differences between consecutive numbers in this sequence. 1, 2, 2, 2, 6, 4, 2, 2, 12, 2, ... These are the gaps between record prime gaps. Is there a pattern here?
http://oeis.org/A053695
2018-05-24, 16:38 #4
robert44444uk
Jun 2003
Suva, Fiji
23×3×5×17 Posts
Quote:
Originally Posted by ldesnogu http://oeis.org/A053695
This is an interesting OEIS entry. Worded rather poorly though - "maximal gaps" was invented to describe the "record gaps"
2018-05-24, 20:03 #5
rudy235
Jun 2015
Vallejo, CA/.
21758 Posts
Quote:
Originally Posted by robert44444uk This is an interesting OEIS entry. Worded rather poorly though - "maximal gaps" was invented to describe the "record gaps"
Maximal gaps is a clear-cut term. Irreversible.
"Record gaps" is rather vacuous. As far a I understand a "record gap" can still be reverted up to the time it becomes a CFC. And even when a "record gap" becomes definitive as a CFC, it does not, as a rule, become a 'Maximal Gap".
2018-05-25, 01:20 #6 Bobby Jacobs May 2018 23·5·7 Posts The biggest known number in the sequence is 208. That is the gap between the maximal gaps of sizes 924 and 1132. That is very big!
2018-05-26, 22:25 #7
Bobby Jacobs
May 2018
23·5·7 Posts
Quote:
Originally Posted by rudy235 Maximal gaps is a clear-cut term. Irreversible. "Record gaps" is rather vacuous. As far a I understand a "record gap" can still be reverted up to the time it becomes a CFC. And even when a "record gap" becomes definitive as a CFC, it does not, as a rule, become a 'Maximal Gap".
Sorry about the confusion. I say, "record gaps" when I am talking about maximal prime gaps. I know that "record gaps" mean something different on this site. I will try to say, "maximal gaps."
Last fiddled with by Bobby Jacobs on 2018-05-26 at 22:27
2018-05-27, 00:36 #8
rudy235
Jun 2015
Vallejo, CA/.
100011111012 Posts
Quote:
Originally Posted by Bobby Jacobs Sorry about the confusion. I say, "record gaps" when I am talking about maximal prime gaps. I know that "record gaps" mean something different on this site. I will try to say, "maximal gaps."
Don't feel like this had to do with you. It was the posting of the http://oeis.org/A053695 what made both Robert and myself make that comment.
If you think about it, all and everyone of the 1000 gaps in the Dr. Nicely 's table of gaps from 1 to 1998 are, in some way or another, "record gaps". Some are first occurrences, some are first known occurrences, and some are maximal gaps.
All maximal gaps are first ocurrences but not the other way around.
2018-07-03, 23:45 #9 Bobby Jacobs May 2018 23×5×7 Posts Record gaps between maximal prime gaps Here are the known record gaps between maximal prime gaps. 1, 2, 6, 12, 20, 26, 30, 32, 62, 100, 208 The number 208 is more than twice the previous record of 100. In fact, 208 is only 3 terms after 100 in the sequence of gaps between maximal prime gaps. That is surprisingly large.
2018-07-04, 08:15 #10
robert44444uk
Jun 2003
Suva, Fiji
7F816 Posts
Quote:
Originally Posted by Bobby Jacobs Here are the known record gaps between maximal prime gaps. 1, 2, 6, 12, 20, 26, 30, 32, 62, 100, 208 The number 208 is more than twice the previous record of 100. In fact, 208 is only 3 terms after 100 in the sequence of gaps between maximal prime gaps. That is surprisingly large.
i.e. this list:
https://oeis.org/A270878
2018-07-04, 16:56 #11 rudy235 Jun 2015 Vallejo, CA/. 100011111012 Posts MERITS OF MAXIMAL GAPS 1 1 1 0 0 1 0 0 1 1 0 0 1 1 1 1 1 0 0 1 1 0 1 1 0 0 0 1 0 1 0 0 1 1 1 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 1 1 1 0 1 1 0 0 0 1 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 In this sequence 0 represents that the merit is not the highest until this point, while 1 means that that merit is higher than all previous merits. There are 80 data points: (gap 1 to gap 1550) Point 75 in Red represents TOES 1476 gap and it is a 1 because there is no previous merit higher than its own) Point 76 is Cyan and belongs to Axn gap of 1488 (it is represented by "0" because TOES gap is larger) Point 77 is Orange and it represents Danaj gap of 1510 (it is represented by "0" because TOES gap is larger) Point 78 is Plum and it represents Steve Coles gap of 1526 (it is represented by "0" because TOES gap is larger) Points 79 & 80 in Black are probable gaps of 1530 and 1550 by the late Be.Nyman (they are represented by "0" because TOES gap is larger) Last fiddled with by rudy235 on 2018-07-04 at 16:59
Similar Threads Thread Thread Starter Forum Replies Last Post Bobby Jacobs Prime Gap Searches 45 2021-09-02 15:47 Terence Schraut Miscellaneous Math 10 2020-09-01 23:49 Bobby Jacobs Prime Gap Searches 11 2018-07-02 00:28 HellGauss Computer Science & Computational Number Theory 18 2015-11-16 14:21 gd_barnes Riesel Prime Search 11 2007-06-27 04:12
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Energy Activity Answers A model glider has a mass of 1 kg. How much potential energy does it have 2 meters off the ground? Ep = mgh = (1)(9.8)(2) = 19.6 j The same model has a velocity of 2.2 m/s. How much kinetic energy does it have? Ek = 1/2(m)V2 = (1/2)(1)(2.2)2 = 2.42 j If the same model descends 2 meters and all it's potential energy is converted to kinetic energy, what is the glider's change in velocity? Ek = 1/2(m)U2 A full-sized glider has a weight of 4,900 N, while it's pilot has a weight of 825 N. If it is 1,000 meters off the ground, how much potential energy do the plane and pilot have? Ep = mgh or Fwh = (4,900 + 825)(1,000) = 5,725,000 j The same glider from Problem 4 has a velocity of 35 m/s. How much kinetic energy does it have? Ek = 1/2(m)V2 = 1/2 [(4.900 + 825)/9.8](352) = 357,813 j The same glider from Problem 4 has a velocity of 35 m/s. The glider descends 900 meters. What is it's new velocity? Ep = Fwh = (4,900 + 825)(900) = 3.63825 * 109 j New Vel. = Old + Change = 35 + 132 = 167 m/s Compare the velocity you calculated in Problem 6 to the speed of sound. Is this answer reasonable? Why or why not? 167/346 = .48 Approximately 1/2 the speed of sound. (Note: This speed is faster than a B-17 Flying Fortress.)
Related Pages:
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## Loloo 3 years ago HELPP! Solve -8(y+4)=10y+4
1. Loloo
How?
2. Loloo
Hehee?
3. michael0195
luis has his ways ^_-
4. Loloo
Ahhh
5. Sandeez
-8(y+4)=10y+4 -8y-32=10y+4 -32-4=10y+8y
6. Directrix
-8(y+4)=10y+4 -8y -32 = 10y + 4 -8y - 10y = 32 + 4 -18y = 36 y = -2
7. Loloo
Huh? I got y=.2..
8. jerwyn_gayo
it's -2
9. Loloo
....
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LSU
Mathematics
# Calendar
Time interval: Events:
Thursday, August 28, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted August 26, 2014
3:00 pm - 4:00 pm Lockett 232
Jacob Matherne, Department of Mathematics, LSU
Representation Theory of the Symmetric Group
Thursday, September 11, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted September 8, 2014
3:00 pm - 4:00 pm Lockett 113
Richard Frnka, Department of Mathematics, LSU Graduate Student
Farey Sequences, Ford Circles, and Their Application in Rademacher's Theorem for the Partition Function
Abstract:The Farey Sequence of order n on an interval is the complete ordered sequence of reduced fractions whose denominator does not exceed n in the interval. These fractions can be used to generate Ford Circles, which have some very nice properties including a relation to modular forms. For two consecutive fractions in the sequence of order n (called Farey neighbors), the Ford Circles generated by them are tangent at only one point. By taking the arc on a circle between the two tangent points from both of its Farey neighbors for every fraction in the sequence, we can form a periodic, continous path. Rademacher used this path to integrate the generating function for partitions to come up with an exact formula for the partition number, which had only been approximated before. This talk does not require any background, and will be accessible to any graduate/undergraduate students with a basic knowledge of Euclidean geometry.
Thursday, September 18, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted September 13, 2014
3:30 pm - 4:30 pm Lockett 113
Lucius Schoenbaum, LSU
Tropical Geometry I
Tropical geometry is a relatively new subject in mathematics which draws connections between algebraic geometry and discrete mathematics and applies them, for example, to enumerative geometry and areas of theoretical physics. In this talk I will introduce the subject and present the graph-theoretic proof of the Riemann-Roch theorem for tropical curves due to Gathmann (2006), based on work of Baker and Norine. This talk should be accessible to any graduate student, or motivated undergraduate familiar with basic abstract algebra and discrete math.
Thursday, September 25, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted September 19, 2014
3:30 pm - 4:30 pm Lockett 113
Bach Nguyen, LSU
Derived Categories
Thursday, October 16, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted October 9, 2014
3:30 pm - 4:30 pm Lockett 113
Lucius Schoenbaum, LSU
Tropical Geometry II
Thursday, October 30, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted October 18, 2014
3:30 pm - 4:30 pm Lockett 113
Sean Taylor, LSU
Introduction to Theory of Sheaf, Part I
Sheaves are an important tool in modern mathematics that were introduced by Jean Leray during the 1940's. They have since become an integral part of algebraic geometry due to the work of many, not the least of which are Jean-Pierre Serre and Alexander Grothendieck. However, they are fruitful not only to algebraic geometry, but also to such areas as algebraic topology and representation theory. In this talk I will present the basic definition of sheaves, the category of sheaves on a topological space, and the functors that can be associated with these categories. I will also present a theorem relating certain types of sheaves on a topological space and the fundamental group of that space.
Thursday, November 6, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted October 30, 2014
3:30 pm - 4:30 pm Lockett 113
Sean Taylor, LSU
Introduction to Theory of Sheaf, Part II
Thursday, November 13, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted November 9, 2014
3:30 pm - 4:30 pm Lockett 113
Dun Liang, LSU
Hodge Theory
Tuesday, November 25, 2014
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted November 19, 2014
3:30 pm - 4:30 pm Lockett 113
Dun Liang, LSU
Curves, Jacobian, and Their Moduli
Thursday, February 19, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted February 6, 2015
4:30 pm - 5:30 pm Lockett 233
Eric Bucher, LSU
Maximal Green Sequences
Thursday, March 12, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted March 5, 2015
3:40 pm - 4:30 pm Lockett 233
Joseph Timmer, Louisiana State University
An Introduction to Hopf Algebras
Hopf algebras have grown in usefulness since their introduction and seem to be a pervasive element in many areas of mathematics. They appear in Topology, Representations of Groups, Lie Theory, Category Theory and even Applied Mathematics. In this talk, we introduce the definitions, structures and "well known" theory. We will focus on examples and ideas of the field. The talk will be accessible to first year graduate students. The only assumed knowledge will be very basic ring theory and some linear algebra.
Thursday, March 19, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted March 15, 2015
3:30 pm - 4:30 pm Lockett 233
Ian Runnels, LSU
Higher Polytopal Relations for Wigner's 6j-symbol
In the quantum theory of angular momentum, Eugene Wigner introduced a gadget called the 6j-symbol to calculate recoupling coefficients for interactions between several particles. Mathematically, these gadgets found uses in the representation theory of what Wigner called "Simply Reducible Groups", classical groups with the property that each irreducible representation is self-dual and multiplicity free; one such example is the Lie group SU(2,C). Over the next few decades, both physicists and mathematicians worked out many symmetries and relations for 6j-symbols, the most famous of which is called the Elliot-Biedenharn identity (secretly this is just the pentagon identity for tensor categories). In this talk, I will develop the definition of the 6j-symbol through the representation theory of SU(2) and introduce some new identities stemming from the combinatorics of these gadgets.
Thursday, August 27, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted August 25, 2015
3:15 pm - 4:15 pm Lockett 233
Jacob Matherne, Department of Mathematics, LSU
The Hilbert scheme of points in the plane
Abstract: We will begin with a few brief notions in algebraic geometry needed to introduce the Hilbert scheme of points in the plane. The Hilbert scheme is a compactification of a configuration space of n distinct particles moving around on a plane. Our main tool for studying it will be the combinatorics of Young diagrams. Time permitting, we may discuss torus actions on the Hilbert scheme and the computation of its cohomology. The talk should be accessible to first-year graduate students (even if the words seem scary at first), so come by!
Thursday, September 3, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted August 27, 2015
3:15 pm - 4:15 pm Lockett 233
Kyle Istvan, LSU
Quantum Invariant Theory
This informal discussion will focus on motivating the use of quantum groups to creating topological invariants, following the perspective of Manin. We will begin with a brief discussion of SL(2,C), its action on C^2, and why this particular group is of interest in geometry, topology, (and vaguely, number theory.) We will then define a new "geometric" object, the quantum complex plane C_q^{2}, and proceed to derive the necessary deformation of SL(2,C) in order to have a useful action on the quantum plane. If time permits, we will see the Kauffman relations (from the study of links and 3-manifolds) appear very naturally in this setting as the quantum analogue of the Cayley-Hamilton Identity, and hopefully motivate the further study of deformations of classical groups. This talk is based on a series of lectures given by Roland van der Veen at Gazi University in August 2015.
Thursday, September 24, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted September 21, 2015
3:15 pm - 4:15 pm Lockett 233
Neal Livesay, LSU
An excursion into the mathematics of Jacques Tits
Abstract: The theory of buildings provides a beautiful combinatorial and geometric viewpoint to the theory of algebraic groups. It can be shown that every group G with an algebraic condition (having a BN-pair) corresponds to a simplicial complex (a building) endowed with a compatible action by G. We will discover this correspondence by working out the details for G=GL_2(k) and GL_3(k).
Thursday, October 15, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted October 12, 2015
3:15 pm - 4:15 pm Lockett 233
Bach Nguyen, LSU
A New Look into the Center of the Quantized Enveloping Algebra of a Complex Semi-simple Lie Algebra
Abstract: In the paper \textit{Local Finiteness of the Adjoint Action for Quantized Enveloping Algebras''} by Anthony Joseph and Gail Letzter, they show that the center of $U_{q}(\mathfrak{g})$ ($\mathfrak{g}$ is the Kac-Moody algbra) is isomorphic to the $W$-invariants in the ring $k(q)[T^0]$, where $W$, the Weyl group, acting by traslation, and $T^{0}=T_{<}^{-1}T_{<}$, where $T_<=-R^+$, and $R^+$ is the intersection of four times the dominant weight with the extended root lattice. Recently, in \textit{Generalized Joseph's Decompositions,''} Arkady Berenstein and Jacob Greenstein give a new construction for the basis of $\mathcal{Z}(U_q)$ which allows us to identify the center with the ring of symmetric functions. In this talk, we'll be discussing the construction that lead to this new basis of $\mathcal{Z}(U_q)$.
Thursday, October 22, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted October 17, 2015
3:15 pm - 4:15 pm Lockett 232
Richard Frnka, Department of Mathematics, LSU Graduate Student
Partitions, Unimodal Sequences, and some Congruence Relations
Abstract: The partition number p(n) is an important mathematical idea that has applications in Combinatorics, Number Theory, and even Representation Theory. In this talk, we will discuss the generating function for p(n), some important relations linking infinite products to theta functions, and will consider different restrictions on the parts that make up the partitions. The only requirement for this talk will be a basic knowledge of geometric series. The plan is to make this talk very accessible and basic, so we will be able to go deeper into the asymptotics of the partition function and unimodal sequences in a later talk.
Thursday, November 12, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted November 9, 2015
3:15 pm - 4:15 pm Lockett 233
Cris Negron, Mathematics Department, LSU
Cohomology for the Young and Restless
Thursday, December 3, 2015
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted November 30, 2015
3:15 pm - 4:15 pm Lockett 233
Jesse Levitt, LSU
Nearly Commutative Rings and Algebras: Properties preserved by controlled non-commutativity
Thursday, February 18, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted February 17, 2016
4:30 pm - 5:30 pm Lockett 233
Bach Nguyen, LSU
Poisson Structures arising from Noncommutative Algebras I
Thursday, February 25, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted February 23, 2016
3:40 pm - 4:40 pm Lockett 233
Bach Nguyen, LSU
Poisson Structures arising from Noncommutative Algebras II
We will continue to discuss examples of Poisson Structures coming from noncommutative algebras such as Lie algebra, algebra of polynomial differential, and algebra of quantum matrices.
Wednesday, March 9, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted March 7, 2016
3:40 pm - 4:10 pm Lockett 233
Neal Livesay, LSU
Regular filtrations on the symplectic loop algebra
Filtrations on the symplectic loop algebra can give explicit normal forms for formal flat $Sp$-bundles with toral'' singularities. Conjecturally, these filtrations will be useful for constructing well-behaved moduli spaces, generalizing the work of Bremer and Sage on flat $GL$-bundles. I will discuss what is meant by a toral'' singularity and illustrate the theory for some small rank examples.
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted March 7, 2016
4:10 pm - 4:40 pm Lockett 233
Lucius Schoenbaum, LSU
Introduction to Topos Theory
What does left exactness of a left R-module over a commutative ring have to do with the existence of atoms in a Boolean algebra? We will learn about this and more in a leisurely, very short introduction to topos theory.
Thursday, April 14, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted April 13, 2016
3:40 pm - 4:40 pm Lockett 233
Cris Negron, Mathematics Department, LSU
A Suprised Talk
Thursday, April 21, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted April 20, 2016
3:40 pm - 4:40 pm Lockett 233
Maitreyee Kulkarni, LSU
Introduction to Nakajima's quiver varieties
Kashiwara's crystals give a combinatorial description of characters of irreducible finite dimensional U_q(g)-modules. The situation gets much more complicated in the case of the quantum loop algebras U_q(Lg), as the characters are now polynomials in terms of q. Nakajima has given a geometric description of the irreducible q-characters of U_q(Lg) using his graded quiver varieties. In this talk, we will define these quiver varieties, compute some examples and see how they are related to the q-characters.
Thursday, April 28, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted April 27, 2016
1:30 pm - 2:20 pm Lockett 233
Sean Taylor, LSU
An Introduction to Algebraic Stacks
Algebraic stacks constitute an often over-looked area of modern algebraic geometry, because of the seeming abstractness of the subject. In this talk, I will attempt to give a (very) brief and hopefully gentle introduction to stacks and what makes a stack "algebraic." This will hopefully be approachable even for those who do not know what a scheme is, since the audience will be encouraged to think about the geometric objects in their favorite categories (manifolds, complex analytic spaces, etc.). In short, these are powerful geometric (!) objects that are becoming more and more fundamental to algebraic geometry and representation theory among other subjects.
Tuesday, May 3, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted May 2, 2016
1:30 pm - 2:30 pm Lockett 233
Sean Taylor, LSU
Introduction to Algebraic Stacks Part 2
After introducing what a Grothendieck topology was last week, we will be able to proceed to actually define stacks - which are higher analogues of sheaves - and then algebraic stacks. If time provides, we will look at common examples of these beautiful and powerful geometric objects.
Thursday, July 21, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted July 11, 2016
2:00 pm - 3:00 pm Lockett 233
Cris Negron, Mathematics Department, LSU
Gauge invariants from the antipode for a Chevalley Hopf algebras
The antipode of a given Hopf algebra is an easily overlooked, yet mysteriously informative, part of the Hopf algebra structure. (For the uninitiated, the antipode of a Hopf algebra is an algebra anti-automorphism which acts like the inversion operator of a group.) For example, a result of of Larson and Radford states that a finite dimensional Hopf algebra in characteristic 0 is a semisimple ring if and only if the square of the antipode is the identity. For a finite dimensional non-semisimple Hopf algebra we only know that the order of the antipode is some positive even integer. This number can be seen as a measure of non-semisimplicity. In this talk I will discuss gauge invariance of the order of the antipode for a certain class of finite dimensional Hopf algebras, and some other related invariants. Rather, I will discuss how the order of the antipode for a given Hopf algebra can, in some cases, be extricated from its associated tensor category of representations.
Thursday, August 25, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted August 23, 2016
3:35 pm - 4:35 pm Lockett 233
Emily Cowie, LSU
Introduction to Lie algebras and their representations
This talk is intended to give an introduction to the basics of Lie algebra representation theory. This talk will provide the necessary background and definitions before describing the combinatorics of the most fundamental Lie algebra, sl(2, C).
Thursday, September 29, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted September 27, 2016
3:30 pm - 4:30 pm Lockett 233
Bach Nguyen, LSU
Orders: "You are out of order!"
Abstract: We will define orders and discuss examples and some of their important properties.
Thursday, October 13, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted October 6, 2016
3:30 pm - 4:30 pm Lockett 233
Lucius Schoenbaum, LSU
Cartesian Closed Categories and Lambda Calculus
During the 1960's and 1970's, connections between logic and category theory were discovered through the work of Grothendieck, Lawvere, Lambek, Benabou, and others. In the 1980's, these developments made an impact on areas of computer science, such as functional programming and the design of many functional programming languages. In this talk, I will focus on cartesian closed categories and the (simply-typed) lambda calculus, which are related via the Curry-Howard-Lambek correspondence (I will explain what this is). Prerequisites: No category theory other than a basic idea of what categories and functors are.
Thursday, November 10, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted November 9, 2016
3:30 pm - 4:30 pm Lockett 233
Sean Taylor, LSU
Algebraic Geometry over Fields of Characteristic p
Over fields of characteristic p there exist new phenomena that arise. One important example of this is the action of the Galois group, and in particular the Frobenius element, on the geometric points of a scheme and on the etale cohomology groups. In this talk we will discuss some of these structures and more.
Wednesday, November 16, 2016
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted November 16, 2016
4:30 pm - 5:30 pm Lockett 233
Shotaro Makisumi, Stanford University
Introduction to Soergel Bimodules
Thursday, January 12, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted January 11, 2017
3:00 pm - 4:00 pm Lockett 233
Jacob Matherne, University of Massachusetts at Amherst
Combinatorial Fourier transform for quiver representation varieties in type A
Thursday, January 19, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted January 17, 2017
10:00 am - 11:00 am Lockett 233
Sean Taylor, LSU
Perverse Sheaves on Toric Varieties
Toric varieties (or "torus embeddings" as they were originally known) are defined as algebraic varieties that have an algebraic torus as an open dense subset such that the natural action of the torus on itself extends smoothly to the whole variety. They have slowly become of more and more interest to both algebraic geometers and combinatorists. The reason for this is that a large class of toric varieties can be functorially associated to several combinatorial objects, the most well known of which are called combinatorial fans. They are also of interest to algebraic geometers for their own sake - much in the way that algebraic curves or surfaces are - and because, though they obey a whole host of interesting and powerful geometric, topological, and combinatorial properties, they paradoxically turn out to be a fantastic place to test new theorems in algebraic geometry. In this talk, we will discuss recent research associated with the speaker's thesis with Pramod Achar. The ultimate goal is to finish creating a mixed category of perverse sheaves on toric varieties in the sense of Beilinson, Ginzburg, and Soergel. Along the way, however, it has become necessary to make an extended study of perverse sheaves on toric varieties and the special properties that they possess. We will begin with some basic definitions and arrive at some fascinating decompositions of Ext groups on toric varieties.
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted January 17, 2017
3:00 pm - 4:00 pm Lockett 233
Sean Taylor, LSU
Representations of SO($\infty$)
One of the primary goals of representation theory and harmonic analysis is to decompose "natural" representations into irreducible pieces. In the 20th century, the regular representation of semisimple Lie groups on symmetric spaces stood out as both an example of extraordinary success of this task and as a model of beauty. In the study of symmetric spaces and their connections to representation theory, we find an interaction between algebra, geometry, and analysis. Since the introduction of Kac-Moody Lie algebras and Kac-Moody Lie groups, infinite-dimensional Lie theory has been an important area of exploration for representation theorists. It is a vista that is still very open, however, and it turns out to be important to consider even "simple" examples of infinite-dimensional Lie groups. In this talk, we will explore recent research of the speaker along with Matthew Dawson, Stephan Merignon, and Gestur Olafsson on a "replacement" for the regular representation of SO($\infty$) and a construction of new representations on these spaces.
Thursday, March 2, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted February 28, 2017
3:00 pm - 4:00 pm Lockett 233
Bach Nguyen, LSU
Quantum Groups: Definitions, Motivations/ History and Applications
The term quantum groups is often used in a loosed way to describe objects in mathematics which may or may not related to each other. To clear up some confusion, we will discuss the definitions of quantum groups in various settings such as: universal enveloping algebra of semisimple lie algebras/ Kac-Moody algebras, coordinate ring of simple algebraic groups, C^*- algebras, and infinite dimension quantum groups. If time permits, the history/ motivation and applications of these objects will also be mentioned.
Wednesday, March 15, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted March 14, 2017
4:30 pm - 5:30 pm Lockett 233
Trey Trampel, LSU
Computing Noncommutative Discriminants via Poisson Primes
We will present a general method for computing discriminants of noncommutative algebras obtained from specialization at roots of unity. This method builds a connection with Poisson geometry and will express the discriminants as products of Poisson primes. The method will be used to compute the discriminants of specializations at roots of unity of algebras of quantum square matrices. We will also evaluate the more general case of specialization of any quantum Schubert cell algebra.
Thursday, August 31, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted August 31, 2017
3:30 pm - 4:30 pm Lockett 233
Andrew Alaniz, LSU
Hecke Operators and Ramanujan's tau function
Abstract: The purpose of this talk is, primarily, to show how applying algebraic methods to situations involving analytic objects can often simplify the situation. Often, fourier coefficients encode interesting arithmetic or combinatorial information about a particular generating series considered as a holomorphic function, this is a common theme in number theory. Hecke's revolutionary insight was understanding the universial meaning of constructions like these. We will consider a certain class of operators, the so-called Hecke operators, and Ramanujan's tau function via the theory of modular forms.
Thursday, September 7, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted September 6, 2017
3:15 pm - 4:15 pm Lockett 233
William Hardesty, Louisiana State University
Baby Verma modules for p-restricted Lie algebras
Abstract: Will give a brief introduction to the representation of theory of (semi-simple) restricted Lie algebras with an emphasis on an important class of representations called "Baby Verma modules". Various results concerning the structure of these modules will be presented and, if time permits, we will briefly discuss some relevant ongoing joint work between the speaker and V. Nandakumar.
Monday, September 11, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted September 6, 2017
3:30 pm - 4:30 pm Lockett 233
Matt Lee, UC Riverside
Demazure Flags and q-Hypergeometric Series
Abstract: Since the current algebra of sl_2 is not semisimple, we need to understand more than just the irreducible representations. We will look at the filtration for a family of modules parametrized by a partition. Attempts to generalize this filtration to other modules leads to an interesting connection to q-Hypergeometric series and related concepts. This talk is accessible to anyone with basic knowledge of sl_2.
Tuesday, November 21, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted November 19, 2017
3:30 pm - 4:20 pm Lockett 233
Sean Taylor, LSU
Mixed Categories of Sheaves on Toric Varities
Thursday, December 7, 2017
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted December 7, 2017
3:30 pm
Bach Nguyen, Louisiana State University
Hecke algebra and its interesting applications
Thursday, March 8, 2018
Student Algebra Seminar Graduate Student Algebra and Number Theory Seminar
Posted March 5, 2018
3:15 pm - 4:15 pm Lockett 233
Bach Nguyen, LSU
Geometric and Algebraic Approaches to Representation Theory of Algebras
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KG - 4th
15 Q
4th
Multiplication strategies
11 Q
4th - 5th
Unit 5 Math Review (2 x 2 digit) Learning Checkpoint
15 Q
4th
Division strategies
17 Q
4th
15 Q
3rd - 4th
Multiplication Strategies
15 Q
3rd - 4th
Subtraction
15 Q
4th - 7th
Multiplication Strategies!
15 Q
4th - 5th
Multiplication Strategies
10 Q
4th
10 Q
4th
Test Taking Strategies
19 Q
4th - 6th
Multiplication Strategies
20 Q
3rd - 4th
Multiplication Strategies
10 Q
4th
15 Q
3rd - 4th
Multiplication Strategies
20 Q
3rd - 4th
## Explore printable Subtraction Strategies worksheets for 4th Grade
Subtraction Strategies worksheets for Grade 4 are an essential tool for teachers looking to help their students master the concept of subtraction in mathematics. These worksheets provide a variety of subtraction problems, ranging from simple single-digit subtractions to more complex multi-digit subtractions with regrouping. By incorporating these worksheets into their lesson plans, teachers can ensure that their Grade 4 students have a solid understanding of subtraction and are well-prepared for more advanced math topics. Additionally, these worksheets can be used as a formative assessment tool, allowing teachers to gauge their students' progress and identify any areas where additional support may be needed. In conclusion, Subtraction Strategies worksheets for Grade 4 are an invaluable resource for teachers seeking to enhance their students' math skills and overall understanding of subtraction.
Quizizz is an excellent platform that offers a wide range of educational resources, including Subtraction Strategies worksheets for Grade 4, to help teachers create engaging and interactive learning experiences for their students. In addition to these worksheets, Quizizz also provides teachers with access to thousands of quizzes and games that cover various math topics, making it easy for educators to find the perfect activity to complement their lesson plans. Teachers can also create their own quizzes and games, allowing them to tailor the content to their students' specific needs and learning objectives. Furthermore, Quizizz offers real-time feedback and analytics, enabling teachers to monitor their students' progress and make data-driven decisions about their instruction. Overall, Quizizz is an invaluable tool for teachers looking to enhance their Grade 4 students' math skills and understanding of subtraction through the use of engaging worksheets and other interactive learning resources.
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# [Numpy-discussion] How to compare an array of arrays elementwise to None in Numpy 1.13 (was easy before)?
Eric Wieser wieser.eric+numpy at gmail.com
Mon Jul 17 13:52:30 EDT 2017
```Here’s a hack that lets you keep using ==:
class IsCompare:
__array_priority__ = 999999 # needed to make it work on either side of `==`
def __init__(self, val): self._val = val
def __eq__(self, other): return other is self._val
def __neq__(self, other): return other is not self._val
a == IsCompare(None) # a is None
a == np.array(IsCompare(None)) # broadcasted a is None
Eric
On Mon, 17 Jul 2017 at 17:45 Robert Kern <robert.kern at gmail.com> wrote:
> On Mon, Jul 17, 2017 at 2:13 AM, <Martin.Gfeller at swisscom.com> wrote:
> >
> > Dear all
> >
> > I have object array of arrays, which I compare element-wise to None in
> various places:
> >
> > >>> a =
> numpy.array([numpy.arange(5),None,numpy.nan,numpy.arange(6),None],dtype=numpy.object)
> > >>> a
> > array([array([0, 1, 2, 3, 4]), None, nan, array([0, 1, 2, 3, 4, 5]),
> None], dtype=object)
> > >>> numpy.equal(a,None)
> > FutureWarning: comparison to `None` will result in an elementwise object
> comparison in the future.
> >
> >
> > So far, I always ignored the warning, for lack of an idea how to resolve
> it.
> >
> > Now, with Numpy 1.13, I have to resolve the issue, because it fails with:
> >
> > ValueError: The truth value of an array with more than one element is
> ambiguous. Use a.any() or a.all()
> >
> > It seem that the numpy.equal is applied to each inner array, returning a
> Boolean array for each element, which cannot be coerced to a single Boolean.
> >
> > The expression
> >
> > >>> numpy.vectorize(operator.is_)(a,None)
> >
> > gives the desired result, but feels a bit clumsy.
>
> Wrap the clumsiness up in a documented, tested utility function with a
> descriptive name and use that function everywhere instead.
>
> --
> Robert Kern
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion at python.org
> https://mail.python.org/mailman/listinfo/numpy-discussion
>
-------------- next part --------------
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```
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volume of a parallelogram [duplicate]
Possible Duplicate:
Determinants and volume of parallelotopes
Can you give me a direction about how to prove that $|det(UVW)|$ is the 3D volume of the parallelogram that defined by $U, V$ and $W$, for $U, V, W \in \mathbb{R}^{3}$ ?
Thanks.
-
marked as duplicate by Phira, t.b., David Speyer, Asaf Karagila, J. M.Nov 25 '11 at 0:02
Not a duplicate as far as I can see. The earlier question asks whether there's a generalization to dimension $>3$; this one seeks a proof for the 3-dimensional case. An answer to one is not the same as an answer to the other. – Henning Makholm Nov 21 '11 at 2:03
First note that $a=\frac{V\times W}{\|V\times W)\|}$ is an unit vector orthogonal to the plane spanned by $V$ and $W$. If $\theta$ is the angle between $U$ and $a$, then $U\cdot a=\|U\|\|a\|\cos\theta=\|U\|\cos\theta$ since $a$ is an unit vector. Notice that $a\cdot U$ is the height $h$ of the parallelepiped formed by $U$, $V$ and $W$ with the parallelogram spanned by $V$ and $W$ as the base (It would be easier to understand if you draw a picture), i.e. $$h=a\cdot U.$$ On the other hand, the area $A$ of the base is equal to the area the parallelogram spanned by $V$ and $W$, that is $$A=\|V\times W\|.$$ Therefore, the volume of the parallelepiped is given by $$hA=(U\cdot a)\|V\times W\|=U\cdot (V\times W)$$ since $a=\frac{V\times W}{\|V\times W)\|}$. Now the result follows from $$\det(U,V,W)=U\cdot(V\times W).$$
Is there a elegant way to expand this prove to $\mathbb{R}^n$? – ofer Nov 20 '11 at 10:55
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mersenneforum.org Mersenne Primes Help!!
Register FAQ Search Today's Posts Mark Forums Read
2021-10-29, 10:51 #12
Dobri
"刀-比-日"
May 2018
2×7×17 Posts
Quote:
Originally Posted by Batalov That's not good, calling a double function, and casting? why (long int)? Maybe (long long int)? How about simply Code: Mp=(1ULL<63) {perror("Can only handle 64-bit values, hence p must be < 64\n");exit(-1);}
Indeed, this is unprotected demo code for the OP to get some ideas and choose their own way of coding the task.
Here "long long int" is needed only when computing s*s for the prime exponent p = 19, and just "long int" is sufficient for p = 2, 3, 5, 7, 13 and 17.
The "pow" function could be avoided by using bit-by-bit operations to compute Mp by simply setting multiple bits to '1'.
2021-10-29, 13:47 #13
Dr Sardonicus
Feb 2017
Nowhere
3×29×59 Posts
Quote:
Originally Posted by Dobri Here "long long int" is needed only when computing s*s for the prime exponent p = 19, and just "long int" is sufficient for p = 2, 3, 5, 7, 13 and 17.
Correct me if I'm wrong, but AFAIK a C long multiply will cause an integer overflow if the product is greater than 231 - 1. So if s > 215.5 then s*s will overflow. Since 17 > 15.5 I found your statement suspect.
I told Pari-GP to do the LL calculations for p = 17, flagging with a * any residues large enough to cause trouble:
EDIT: I realized that my script was a bit too mindless. I rewrote it only to flag residues r such that both r and 217 - 1 - r were large enough to cause trouble.
Code:
? b=sqrtint(2^31-1);print(b)
46340
? n=2^17-1;s=Mod(4,n);for(i=1,15,s=s^2-2;r=lift(s);if(n-b>r&&r>b,c="*",c="");print(i" "r" "c))
1 14
2 194
3 37634
4 95799
5 119121
6 66179 *
7 53645 *
8 122218
9 126220
10 70490 *
11 69559 *
12 99585
13 78221 *
14 130559
15 0
Last fiddled with by Dr Sardonicus on 2021-10-29 at 14:10 Reason: Improved script
2021-10-29, 13:57 #14
Dobri
"刀-比-日"
May 2018
2·7·17 Posts
Quote:
Originally Posted by Dobri Here "long long int" is needed only when computing s*s for the prime exponent p = 19.
To be precise, "long long int" is also needed for the prime exponent p = 17.
2021-10-29, 14:25 #15 Dobri "刀-比-日" May 2018 3568 Posts Here is a hint how the LL (and PRP) computations could be made much faster for small prime exponents. The modulo operation "%" (which in essence is a division) can be replaced by multiplication and shift operations. Said multiplication+shift approach requires roughly four times less computing cycles than division. For large prime exponents, more memory would be needed. I couldn't find a discussion in the forum comparing the computation time of the FFT approach used by the prime95 app, standard modulo division, and the multiplication+shift approach. Please share a link if there is such a thread.
2021-10-29, 23:54 #16
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·23·137 Posts
Quote:
Originally Posted by Dobri The modulo operation "%" (which in essence is a division) can be replaced by multiplication and shift operations.
It can be even simpler with just shift and add.
But with the exponents used here all this extra code is overkill. Simple TF and done in a few (milli?)seconds.
Last fiddled with by retina on 2021-10-29 at 23:55
Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Miscellaneous Math 3 2017-08-10 13:47 emily Math 34 2017-07-16 18:44 siegert81 Math 2 2011-09-19 17:36 Unregistered Information & Answers 0 2011-01-31 15:41 optim PrimeNet 13 2004-07-09 13:51
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Sun Dec 5 05:12:51 UTC 2021 up 134 days, 23:41, 0 users, load averages: 1.86, 1.72, 1.56
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# Quant Quiz on “COORDINATE GEOMETRY” for SSC CGL 2016
For the first time SSC 2016 examination will be held in Computer Based pattern and managing time will be one of the important factor while solving the questions. So in order to make students familiar with such a situations we are providing questions in a time based manner , which will help students to manage the time properly.
1. What is the distance of point of intersection of straight lines 2x+3y=6 and y=x+7 from origin ?
7
3
4
5
Solution:
2. The length intercepted by the straight line 12x-9y=108 between the coordinate axes is
12 unit
18 unit
15 unit
9 unit
Solution:
3. Area of triangle formed by straight lines 4x-3y+4=0,4x+3y-20=0 and x- axis is
3 sq. unit
6 sq. unit
12 sq. unit
24 sq. unit
Solution:
Area =1/2*(difference between x-intercept)*(y-coordinate of point of intersection)
Required Area =1/2 |5-(-1) |×4=12 square unit
4. Area of triangle formed by straight lines 4x-y=4,3x+2y=14 and y-axis is
11/2 sq. unit
11/4 sq. unit
22 sq. unit
11 sq. unit
Solution:
Required Area =1/2 |-4-7|×2=11 square unit
5. Ratio of area of triangle formed by straight lines 2x+3y=4 and 3x-y+5=0 with x-axis and y-axis is
1 : 2
2 : 1
4 : 1
None of these
Solution:
6. Area of quadrilateral formed by straight lines 2x=-5,2y=3,x+1=0 and y+2=0 is
21/2 sq. unit
21/4 sq. unit
21/8 sq. unit
21/16 sq. unit
Solution:
7. Area enclosed by equation |x|+|y|=4 is
16
32
24
48
Solution:
8. For what value of k system of equations 3x+4y=19, y-x=3 and 2x+3y=k has a solution ?
11
-11
14
-14
Solution:
Solving 3x+4y=19 and y-x=3
We get x=1,y=4
Putting (x,y)=(1,4) in 2x+3y=k
We have 2×1+3×4=k
⇒k=14
9. Which of the following pair represent equation of parallel straight lines.
2x+3y=4,4x+6y=9
x+2y=4,2x+y=4
y=3x+5,x=3y+5
None of these
Solution:
In option (a) a1/a2 =b1/b2 ≠c1/c2 . Hence lines given in alternative (a) shows parallel lines.
10. For what value of Ksystem of equation x+3y=K and 2x+6y=2K has infinitely many solution ?
K=1
K=2
for all real values of K
for no real value of K
Solution:
Here a1/a2 =b1/b2 =c1/c2 is always true. It has infinitely many solution for all real values of K.
11. Values of a andb so that system of equations 2x+3y=7 and 2ax+(a+b)y=28 has infinitely many solutions are
a=4,b=8
a=8,b=4
a=-4,b=-8
a=-8,b=-4
Solution:
Required condition is 2a/2=(a+b)/3=28/7
or, a=(a+b)/3=4
∴a=4,a+b=12
or,a=4,b=8
12. Value of k for which system of equations kx+2y=5, 3x+y=1 has unique solution is
k=1
k=2
k=3
All are true
Solution:
System has unique solution if a1/a2 ≠b1/b2
⇒k/3≠2/1
⇒k≠6
so options (a), (b), (c) are correct
13.
a
b
c
d
Solution:
14. Area of triangle formed by the straight line 8x-3y+24=0 and coordinate axes is
24 sq. unit
12 sq. unit
6 sq. unit
18 sq. unit
Solution:
15. Area of triangle formed by straight line y=mx+c with coordinate axes is
a
b
c
d
Solution:
×
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SLC Kiki Term 3
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kiki Huang
on 8 September 2010
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Transcript of SLC Kiki Term 3
KiKi's SLC UOI Presentation
Central Idea : Human understand and use simple
machines to make live easier. How I under stand the Central Idea? What make our live easier are
the "Seven Simple Machines".[continue
on the next one] The Seven Simple Machines: That talk about the seven simple
machines,we'll talk about "Wheel"
first. 1. Wheel: Wheel make our live easier because
we travel with car and if the car did
not have any wheel then the car couldn't
move around.[The wheel make the car to
travel around.Go to 2.lever.]
The Seven Simple
Machines list: 1.Wheel and axel
2.lever
3.pulley
4.screw
5.inclinded plane
6.wedge
7.gears 2. Lever Lever make our live easier because
the lever could lift the things up
as you do not have to use much effot.
[Go to the 3.Pulley.] 3. Pulley Pulley make our live easier because
it help us to lift heavy thing without
much forces.If there are no pulley you
use a lot of forces to lift the heavy
thing.[Go to 4.Screw.] 4. Screw Screw make live easier because
screw something to stay together
and stable.But if there are no
screw,everything will fall off
into pieces.So it is very useful.
[Go to 5.Inclinded plane.] 5. Inclinded plane Inclined plane make our lives easier because
if you have a inclined plane you could push
the heavy thing up somewhere with less forces.
If there are no inclined plane, you will have
to use more forces to push/bring anything up
somewhere.[Go to 6.Wedge.] 6. Wedge There are wedges I think because
it help the door don't move and to
stay open, if there are no wedge
the door wouldn't stay open.
Full transcript
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https://www.coursehero.com/file/p7d7fj5/public-SignedNumber-long-left-long-mid-long-right-super-Mathabs-left-mid-right/
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public SignedNumber long left long mid long right super Mathabs left mid right
# Public signednumber long left long mid long right
• 40
This preview shows page 23 - 25 out of 40 pages.
public SignedNumber (long left, long mid, long right) { super (Math.abs (left), mid, right); isNegative = left < 0; } //... lots more is required here in the same vein Exercise 11.26 Modify the VeryLong constructor to create the equivalent of ZERO when any one of the parameters is negative or has more than eighteen digits. Exercise 11.27 (harder) Write the doubleValue method for the VeryLong class. Exercise 11.28 (harder) Write the equals method for the VeryLong class. Exercise 11.29 (harder) Write a simplified valueOf method for the VeryLong class, for which you have a precondition that the parameter is a string of 1 to 54 digits. Exercise 11.30* Write the full valueOf method for the VeryLong class. Allow the input to contain commas among the digits. Use the preceding exercise to get started. Exercise 11.31* Write a Real subclass of Numeric for ordinary numbers with one double instance variable. This lets the clients mix in ordinary numbers with the special ones. Exercise 11.32* Write the subtract method for the VeryLong class. Exercise 11.33* The Complex instance variables are final but the Fraction instance variables are not, even though both are immutable classes. Explain why. Exercise 11.34** Write the compareTo method for the VeryLong class. Exercise 11.35** The VeryLong toString method produces leading zeros when the leftmost one or more components of itsItem are zero. Revise it to fix this problem. 11.8 Implementing The NumericArray Class With Null-Terminated Arrays These mathematicians often deal with numbers in big bunches. They may read in a bunch of numbers from a file, then calculate the average of the whole bunch, find the smallest and the largest, insert a value in order, etc. For this, you decide to store a lot of Numeric objects in an array of Numeric values. Call it the NumericArray class. Once you define an array, you cannot change its size. So you need to make it big enough for the largest number of values you expect. But then the array is generally only partially filled. So you have to have some way of noting the end of the array. One way is to keep track of the size. Another way is to put the null value in all the components after the end of the actual Numeric values in the array, as shown in Figure 11.5. Figure 11.5 picture of a null-terminated array with six Complex values For instance, if the array has size 1000 and currently contains only 73 values, then those 73 values will be stored in components 0 through 72 and the null value will be stored in each of components 73...1000. We do not allow a full array. A precondition for these null-terminated arrays is that they contain at least one instance of null. Precondition for NumericArrays For the array parameter item , there is some integer n such that 0 <= n < item.length and item[k] is not null when k < n and item[k] is null otherwise. The non-null values are the values on the conceptual list in order, with the first at index 0.
Java Au Naturel by William C. Jones 11-24 11-24 The find method To illustrate how to work with such a null-terminated array, consider the problem of searching through the array to find whether a particular non-null value is there.
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# Distance between Excel AB and Abbey SK
The distance from Excel to Abbey is 200 kilometers by road. Road takes approximately 2 hours and 15 minutes and goes through Oyen, Eatonia, Leader and Lancer.
Shortest distance by air 146 km Car route length 200 km Driving time 2 h 15 min Fuel amount 16 L Fuel cost 19.1 CAD
City
Town
Village
Hamlet
Motorway
Primary
Secondary
Unclassified
Point Distance Time Fuel Excel 0 km 0:00 00 min 0.0 L 4 km, 02 min Cereal 4 km 0:02 02 min 0.1 L 9 16 km, 10 min Acadia Valley 20 km 0:13 13 min 0.6 L 41 570 53 km, 35 min Mantario 73 km 0:49 49 min 5.8 L 44 35 km, 23 min Eatonia 108 km 1:13 1 h 13 min 7.3 L 21 28 km, 18 min Leader 136 km 1:31 1 h 31 min 10.4 L 21 32 29 km, 19 min Sceptre 165 km 1:51 1 h 51 min 12.1 L 32 23 km, 15 min Lancer 188 km 2:07 2 h 07 min 14.3 L 32 12 km, 08 min Abbey 200 km 2:15 2 h 15 min 15.8 L
## Book a Hotel for your trip to Abbey
### How much does it cost to drive from Excel to Abbey?
This fuel cost is calculated as: (Route length 200 km / 100 km) * (Fuel consumption 8 L/100 km) * (Fuel price 1.19 CAD / L)
You can adjust fuel consumption and fuel price here.
### How long is a car ride from Excel to Abbey?
Driving time: 2 h 15 min
This time is calculated for driving at the maximum permitted speed, taking into account traffic rules restrictions.
• 194 km with a maximum speed 90 km/h = 2 h 9 min
• 2 km with a maximum speed 80 km/h = 1 min
• 2 km with a maximum speed 50 km/h = 2 min
• 1 km with a maximum speed 40 km/h = 1 min
The calculated driving time does not take into account intermediate stops and traffic jams.
### How far is Excel to Abbey by land?
The distance between Excel and Abbey is 200 km by road.
Precise satellite coordinates of highways were used for this calculation. The start and finish points are the centers of Excel and Abbey respectively.
### How far is Excel to Abbey by plane?
The shortest distance (air line, as the crow flies) between Excel and Abbey is 146 km.
This distance is calculated using the Haversine formula as a great-circle distance between two points on the surface of a sphere. The start and finish points are the centers of Excel and Abbey respectively. Actual distance between airports may be different.
### How many hours is Excel from Abbey by plane?
Boeing 737 airliner needs 10 min to cover the distance of 146 km at a cruising speed of 800 km/h.
Small plane "Cessna 172" needs 39 min to flight this distance at average speed of 220 km/h.
This time is approximate and do not take into account takeoff and landing times, airport location and other real world factors.
### How long is a helicopter ride from Excel to Abbey?
Fast helicopter "Eurocopter AS350" or "Hughes OH-6 Cayuse" need 36 min to cover the distance of 146 km at a cruising speed of 240 km/h.
Popular "Robinson R44" needs 41 min to flight this distance at average speed of 210 km/h.
This time is approximate and do not take into account takeoff and landing times, aerodrome location and other real world factors.
### What city is halfway between Excel and Abbey?
The halfway point between Excel and Abbey is Eatonia. It is located about 8 km from the exact midpoint by road.
The distance from Eatonia to Excel is 108 km and driving will take about 1 h 13 min. The road between Eatonia and Abbey has length 92 km and will take approximately 1 h 2 min.
The other cities located close to halfway point:
• Mantario is in 73 km from Excel and 127 km from Abbey
### Where is Excel in relation to Abbey?
Excel is located 146 km north-west of Abbey.
Excel has geographic coordinates: latitude 51.38339, longitude -110.56669.
Abbey has geographic coordinates: latitude 50.73455, longitude -108.75799.
The distance between Excel and Abbey is ranked 29,156th in the ranking popularity.
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# The greatest and least resultant of two forces acting at a point are 16N and 4N. If they are acting at an angle 120 degrees with one another their resultant is
√76N
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What are you looking for?
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LFD Book Forum (http://book.caltech.edu/bookforum/index.php)
- Chapter 4 - Overfitting (http://book.caltech.edu/bookforum/forumdisplay.php?f=111)
ghbcode 10-10-2017 09:49 AM
Exercise 4.3
The only posting for Chapter 4 that touched on this topic is listed below though it did not explicitly cover exercise 4.3, it is somewhat touchy feely and not that exact.
Deterministic noise depends on H, as some models approximate f better
than others.
(a) Assume H is fixed and we increase the complexity of f. Will deter*ministic noise in general go up or down? Is there a higher or lower tendency to overfit?
(b) Assume f is fixed and we decrease the complexity of H. Will deter*ministic noise in general go up or down? Is there a higher or lower tendency to overfit? [Hint: There is a race between two factors that affect overfitting in opposite ways, but one wins.]
The hint to me implies the the response to a and b would move in different directions. This is what I have for an answer:
a) By increasing the target function complexity deterministic noise will increase since H remains fixed and f becomes more complex. There will be lower overfitting in the out-of-sample data. As a matter of fact this goes counter to the summary table in page 124, however, it does not make sense to me that, keeping all else constant, by increasing the target complexity we are increasing the overfit. If anything by increasing the target complexity, your fixed H would underfit.
b) By lowering the target function complexity deterministic noise would increase and there would be a tendency to lower overfit.
I'm not sure that my answer is correct so if you could enlighten me that would be most helpful. In the thread that I posted above there was discussion and use of the formula:
$E_{out}=\sigma^2+bias+var$
Question 1:
In this exercise do we assume that E_out remains fixed so that the expression is tweaked by changing H or f?
Question 2:
In part a) if you keep H fixed and increase f, what happens to deterministic noise and what happens to overfit?
Question 3:
In part b) if you keep f fixed and decrease H, what happens to deterministic noise and what happens to overfit?
All times are GMT -7. The time now is 12:56 AM.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2012-09-10 21:29:24
n1corponic
Member
Registered: 2012-09-10
Posts: 7
### Natural logarithms problem (ln)
Hello all! I am new here and I am a student having a hard time in math. I have the following natural logarithms problem and it is giving a hard time. Can someone please help with the solution? And also if possible evaluate the difficulty of this problem?
Solve for x: 4^x-4^(x-1)=3^(x+1)-3^x
Last edited by n1corponic (2012-09-10 21:36:19)
Offline
## #2 2012-09-10 21:50:36
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 106,389
### Re: Natural logarithms problem (ln)
Hi n1corponic;
I might be missing something but I would say very difficult. I do not see a way using algebra so you have to use the methods of numerical analysis. This will require a computer or a programmable calculator.
A little background:
1) Where does the problem come from? This will determine what methods can be used.
2) Are you just interested in a solution?
3) What type of solution, real or complex?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.
Offline
## #3 2012-09-10 22:01:21
n1corponic
Member
Registered: 2012-09-10
Posts: 7
### Re: Natural logarithms problem (ln)
Hi Bobbym! I don't think i have to find a rational number. Usually the answer to these logarithmic problems is smth like x=ln3/ln5 .To answer your great questions:
1)It comes from the book called Essential Mathematics for Economic Analysis 4th edition on page 123 problem 3.(a)
2)I am mostly interested in the logical thinking process in these kind of problems.
Offline
## #4 2012-09-10 22:05:08
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 106,389
### Re: Natural logarithms problem (ln)
Hi;
There is a simple answer according to Mathematica.
Seems like I spoke to soon. You can derive it like this:
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.
Offline
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## Pure Sequencing PT4 & PT6
Prepare for the LSAT or discuss it with others in this forum.
epokei
Posts: 21
Joined: Thu Dec 23, 2010 8:22 pm
### Pure Sequencing PT4 & PT6
PT4 1992 Feb. LG Game#1 Q4. (or if you are working on the "PowerScore LSAT Game Type Training", pg.76)
How do you do this? Who should I look for for the same salary as Nassar's or I don't even have to know that?
This is how I diagram:
K-I-F-M-G-J-H
K-L-N
PT6 1992 Oct. LG Game#2 Q.10 and Q12 (or "PowerScore LSAT Game Type Training", pg.77)
Q10. I simply don't understand what the question means....
Q12. Why not (A)two? L and P is the only numbers that I can determine since I don't know the N or J's relation....
This is how I diagram:
L-P-N
L-P-J-O-K-M
Thank you.
TMC116
Posts: 284
Joined: Mon Jun 06, 2011 6:08 pm
### Re: Pure Sequencing PT4 & PT6
PT4 1992 Feb. LG Game#1 Q4. (or if you are working on the "PowerScore LSAT Game Type Training", pg.76)
How do you do this? Who should I look for for the same salary as Nassar's or I don't even have to know that?
This is how I diagram:
K-I-F-M-G-J-H
K-L-N
PT6 1992 Oct. LG Game#2 Q.10 and Q12 (or "PowerScore LSAT Game Type Training", pg.77)
Q10. I simply don't understand what the question means....
Q12. Why not (A)two? L and P is the only numbers that I can determine since I don't know the N or J's relation....
This is how I diagram:
L-P-N
L-P-J-O-K-M
Thank you.
I did this game today actually. Weird.
To answer your first question, it doesn't matter who has the same salary as Nassar. Pick any one you like and it will work out fine. (i remember the answer, so it doesn't make any difference). Give it a shot and then you'll see when you go through the answer choices
I don't have the other games in front of me so i'd have to look over it before getting back to you.
epokei
Posts: 21
Joined: Thu Dec 23, 2010 8:22 pm
### Re: Pure Sequencing PT4 & PT6
Thank you! I guess I have forgotten the 2nd rule:"Lopez's salary is greater than Nassar's." and yes I did get it.
I still need the help for PT6 if anyone could explain.
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# GME +GLOBAL MINING MACHINE
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Math (cummulative review)
posted by .
Factor the given polynomial: 3y2+5y-8
the 2 is an exponent...
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i also have theses to solve Factor the polynomial x2 + 5x + 6 completely Solve the given equation by using the quadratic formula. x2 – 7x – 4 = 0 What value is under the square root (or radical) in your answer?
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factor each polynomial completely. to begin, state which method should be applied to the first step, given the guidelines of this section. Then continue the exercise and factor each polynomial completely. 2p - 6q + pq - 3q^2 Please …
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Suppose you divide a polynomial by a binomial. How do you know if the binomial is a factor of the polynomial?
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Part 1: Suppose you divide a polynomial by a binomial. How do you know if the binomial is a factor of the polynomial?
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identify the conic sections below (circle,hyperbola,parabola,ellipse). a)3x2+3y2-2y=4 b)3x2-9y2+2x-4y=7 c)2x2+5y2-7x+3y-4=0 d)3y2-4x+17y=-10
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# Fuzzy subset
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To be Completed !!
## Introduction
The fuzzy subset of "small numbers" and the fuzzy subset of "numbers close to 6".
In the everyday activity it is usual to adopt vague properties as "to be small", "to be close to 6" and so on. Now, in set theory given a set S and a "well defined" property P, the axiom of comprehension reads that a subset B of S exists whose members are precisely those objects in S satisfying P. Such a set is called "the extension of P in S". For example if S is the set of natural numbers and P is the property "to be prime", then the subset B of prime numbers is defined. Assume that P is a vague property, then the question arises whether there is a way to define notions as "the subset of small numbers", "the subset of numbers close to 6". An answer to such a question was proposed in 1965 by Lotfi Zadeh and, at the same time, by Dieter Klaua in the framework of multi-valued logic. The idea is to extend the notion of characteristic function.
Definition. Let S be a nonempty set, then a fuzzy subset of S is a map s from S into the real interval [0,1]. If S1,...Sn are nonempty sets, then a fuzzy subset of S1×. . .×Sn is called an n-ary fuzzy relation.
The elements in [0,1] are interpreted as truth degree and, in accordance, given x in S, the number s(x) is interpreted as the membership degree of x to s. We say that a fuzzy subset s is crisp if s(x) is in {0,1} for every x in S. By associating every classical subsets of S with its characteristic function, we can identify the subsets of S with the crisp fuzzy subsets. In particular we call "empty subset" of S the fuzzy subset of S constantly equal to 0. Notice that in such a way different sets have different empty subsets and therefore there is not a unique empty subset as in set theory.
## Some set-theoretical notions
The intersection of the "fuzzy subset of small numbers" with the "fuzzy subset of numbers close to 6" (obtained by the minimum and the product).
In classical mathematics the definitions of union, intersection and complement are related with the interpretation of the basic logical connectives ${\displaystyle \vee ,\wedge ,\neg }$. In order to define the same operations for the fuzzy subsets of a given set, we have to fix suitable operations ${\displaystyle \oplus ,\otimes }$ and ${\displaystyle \backsim }$ in [0,1] to interpret these connectives. Once this was done, we can define these operations by the equations
${\displaystyle (s\cup t)(x)=s(x)\oplus t(x)}$,
${\displaystyle (s\cap t)(x)=s(x)\otimes t(x)}$,
${\displaystyle (-s)(x)=\backsim s(x)}$.
If we denote by [0,1]S the class of all the fuzzy subsets of S, then an algebraic structure ${\displaystyle ([0,1]^{S},\cup ,\cap ,-,\emptyset ,S)}$ is defined. This structure is the direct power of the structure ${\displaystyle ([0,1],\oplus ,\otimes ,\backsim ,0,1)}$ with index set S. In Zadeh's original papers the operations ${\displaystyle \oplus ,\otimes ,\backsim }$ are defined by setting for every x and y in [0,1]:
${\displaystyle x\otimes y=min(x,y)}$ ; ${\displaystyle x\oplus y=max(x,y)}$ ; ${\displaystyle \backsim x=1-x}$.
In such a case ${\displaystyle ([0,1]^{S},\cup ,\cap ,-,\emptyset ,S)}$ is a complete, completely distributive lattice with an involution. Usually one assumes that ${\displaystyle \otimes }$ is a triangular norm in [0,1] and that ${\displaystyle \oplus }$ is the corresponding triangular co-norm defined by setting ${\displaystyle x\oplus y=\backsim ((\backsim x)\otimes (\backsim y))}$. For example, the picture represents the intersection of the fuzzy subset of small number with the fuzzy subset of numbers close to 6 obtained by the minimum and the product. In all the cases the interpretation of a logical connective is conservative in the sense that its restriction to {0,1} coincides with the classical one. This entails that the map associating any subset X of a set S with the related characteristic function is an embedding of the Boolean algebra ${\displaystyle (\{0,1\}^{S},\cup ,\cap ,-,\emptyset ,S)}$ into the algebra ${\displaystyle (L^{S},\cup ,\cap ,-,\emptyset ,S)}$.
## L-subsets
The notion of fuzzy subset can be extended by substituting the interval [0,1] by any bounded lattice L. Indeed, we define an L-subsetas a map s from a set S into the lattice L. Again one assumes that in L suitable operations are defined to interpret the logical connectives and therefore to extend the set theoretical operations. This extension was done mainly in the framework of fuzzy logic.
## Bibliography
• Cox E., The Fuzzy Systems Handbook (1994), ISBN 0-12-194270-8
• Gerla G., Fuzzy logic: Mathematical Tools for Approximate Reasoning, Kluwer, 2001.
• Gottwald S., A treatase on Multi-Valued Logics, Research Studies Press LTD, Baldock 2001.
• Hájek P., Metamathematics of fuzzy logic. Kluwer 1998.
• Klaua D., Über einen Ansatz zur mehrwertigen Mengenlehre, Monatsberichte der Deutschen Akademie der Wissenschaften Berlin, vol 7 (1965), pp 859-867.
• Klir G. and Folger T., Fuzzy Sets, Uncertainty, and Information (1988), ISBN 0-13-345984-5.
• Klir G. and Bo Yuan, Fuzzy Sets and Fuzzy Logic (1995) ISBN 0-13-101171-5
• Kosko B., Fuzzy Thinking: The New Science of Fuzzy Logic (1993), Hyperion. ISBN 0-7868-8021-X
• Novák V., Perfilieva I, Mockor J., Mathematical Principles of Fuzzy Logic, Kluwer Academic Publishers, Dordrecht, (1999).
• Yager R. and Filev D., Essentials of Fuzzy Modeling and Control (1994), ISBN 0-471-01761-2
• Zimmermann H., Fuzzy Set Theory and its Applications (2001), ISBN 0-7923-7435-5.
• Zadeh L.A., Fuzzy Sets, Information and Control, 8 (1965) 338-353.
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# prime
(redirected from prime number)
Also found in: Dictionary, Thesaurus, Medical, Encyclopedia, Wikipedia.
## PRIME
Stands for Prescribed Right to Income and Maximum Equity, a certificate that entitles the owner to the dividend/income from an underlying security, but not to the capital appreciation of that security.
## Prime
1. In commercial banking, the best available interest rate under most circumstances. Generally speaking, only the most creditworthy customers receive the prime, but this is not always true. In any case, a prime serves as a benchmark against which other interest rates are compared. In this sense, it is also called the prime rate.
2. Describing the highest possible credit rating on a bond, either Aaa (for Moody's) or AAA (for S&P and Fitch).
## prime
1. Of or relating to a debt security rated AAA or Aaa.
2. See prime rate.
## prime
To come before another creditor in terms of priority of payment should there be insufficient assets to pay all creditors.A first mortgage holder primes a second mortgage holder,who primes a later judgment creditor, who primes a general unsecured creditor. Filing for bankruptcy reshuffles the deck,as the bankruptcy trustee primes large categories of creditors.
The Complete Real Estate Encyclopedia by Denise L. Evans, JD & O. William Evans, JD. Copyright © 2007 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
The rational functions GA([lambda], p) for four different prime numbers (p = 3, 5, 7,11) are represented by red lines in Figure 5.3.
In the following example, the sequence has the next prime number added to its list of distinct prime factors at each step: {2, 6, 30, 210, 2310, 30 030, 510 510 ...}.
If [J.sub.[phi]] or [J.sub.[kappa]] < [[mu].sub.p] + [b.sub.q] +2, then suppose that [J.sub.v] is the smallest odd prime number which is not smaller than [[mu].sub.p] + [b.sub.q] +2.
The search for ever longer prime numbers will continue, in Missouri and elsewhere.
Although there are an infinite amount of prime numbers, the hunt for the largest has in recent years centred on rare Mersenne primes, named after Marin Mersenne, a 17th-century French monk and mathematician.
for the generalized twin prime numbers p = 2a - D, p' = 2a + D.
This theorem assures us that if n is a prime number then [b.sup.n-1] [equivalent to ] 1 mod n for every integer b co-prime to n.
"One important discovery that children can make with Chartworld is the difference between a prime number and a composite number.
Joe Fanning, who rode Prime Number, completed a double on Trimaran in the following maiden.
That's a prime number!) who falls for a younger man David (Bryan Greenberg, who's only 23 - woo!
PRIME NUMBER (4.10), who bounced back to winning form at Windsor on Monday, can make a quick follow-up at Pontefract this afternoon.
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# Analyzing Mixed Costs Using High-Low Method, Preparing a Contribution Margin Income Statement,...
Analyzing Mixed Costs Using High-Low Method, Preparing a Contribution Margin Income Statement, Analyzing Break-Even Point, and Setting Target Profit Tina Sutton delivers flowers for several local flower stores. She charges clients $0.85 per mile driven. Tina has determined that if she drives 1,200 miles in a month, her average operating cost is$0.70 per mile. If she drives 2,000 miles in a month, her average operating cost is $0.60 per mile. Required: 1. Using the high-low method, determine Tina’s variable and fixed operating cost components. Show this as a cost formula. 2. Determine how many miles Tina will need to drive to break even. 3. Assume Tina drove 900 miles last month. Without making any additional calculations, determine whether she earned a profit or a loss last month. 4. Prepare a contribution margin income statement for Tina’s business last month. 5. If Tina wants to earn$800 a month, determine how many miles she must drive.
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# Group Theory in Finance
Written By
Written By
Dan Buckley
Dan Buckley is an US-based trader, consultant, and part-time writer with a background in macroeconomics and mathematical finance. He trades and writes about a variety of asset classes, including equities, fixed income, commodities, currencies, and interest rates. As a writer, his goal is to explain trading and finance concepts in levels of detail that could appeal to a range of audiences, from novice traders to those with more experienced backgrounds.
Updated
Group Theory, a branch of abstract algebra, is fundamentally about the study of symmetry and structure.
While it might not seem immediately relevant to finance, trading, and investing, Group Theory’s principles and concepts find application in these fields, particularly in quantitative finance, risk management, and complex financial modeling.
## Key Takeaways – Group Theory in Finance
• Group theory is the study of mathematical groups, which are sets of elements combined with an operation that satisfies certain conditions, like closure, associativity, identity, and invertibility.
• The application of Group Theory in finance enables the modeling of complex financial systems.
• Group Theory could aid in the design of new financial instruments and strategies.
• It provides a framework for understanding the underlying structure of financial products and assets, and help in creating innovative derivatives, optimizing portfolio construction, and developing algorithmic trading strategies.
## Group Theory – A Non-Technical Example
Imagine a Rubik’s Cube, a well-known puzzle consisting of a cube with colored squares that can be rotated in various directions.
The goal is to get all sides of the cube to be a uniform color.
### Basic Concepts of Group Theory
#### Elements and Operations
In Group Theory, a “group” consists of “elements” and an “operation” that combines any two elements.
In the Rubik’s Cube analogy, each possible state of the cube is an “element,” and the “operation” is the act of rotating a part of the cube.
#### Closure
Closure means if you combine two elements of a group using the group’s operation, the result will always be another element of the same group.
In the Rubik’s Cube, no matter how you twist it (the operation), you always end up with another valid state of the cube (another element of the group).
#### Associativity
This means that when combining multiple elements, the order in which you combine them doesn’t change the final result.
With the Rubik’s Cube, if you plan a series of rotations, the final state of the cube will be the same no matter how you group these rotations.
#### Identity Element
There’s always one element in a group that, when combined with any other element, leaves the other element unchanged.
In the Rubik’s Cube, this is the solved state.
If you don’t make any moves (the identity operation), the cube stays solved.
#### Inverse Elements
For every element in the group, there is another element that, when combined, results in the identity element.
On the Rubik’s Cube, for every sequence of moves, there’s a reverse sequence that brings the cube back to its solved state.
## Group Theory in Portfolio Construction
Here are some examples of how group theory might be used in portfolio construction:
### Hierarchical risk parity
This is a type of risk parity that uses group theory to construct a hierarchy of asset classes.
This hierarchy is used to allocate capital to different asset classes in a way that balances risk and return.
### Group-based diversification
This is a type of diversification that uses group theory to identify groups of assets that have low correlations with each other.
These groups are then used to construct portfolios that are more diversified than traditional portfolios.
## Group Theory in Other Applications of Finance
### Risk Management and Portfolio Theory
In finance, risk and return profiles of assets often exhibit certain symmetrical properties.
Group Theory can help in understanding these symmetries.
For instance, the concept of diversification in portfolio theory can be examined through the lens of symmetry – how different investments behave in relation to each other under various market conditions.
Financial markets can exhibit patterns that repeat over time or under certain conditions.
Group Theory can be used to model these patterns and their transformations.
This can be extremely valuable in algorithmic trading strategies where recognizing and exploiting patterns is the name of the game.
### Pricing Models
Derivatives pricing, particularly for complex instruments, can involve understanding symmetries and invariances in market behaviors and asset relationships.
Group Theory provides a framework for analyzing these relationships for more accurate and robust pricing models.
### Financial Cryptography
Group Theory forms the mathematical basis of many cryptographic techniques used in secure financial transactions.
Public-key cryptography, used for secure online transactions, often relies on the properties of mathematical groups.
### Econophysics and Market Analysis
Group Theory can be applied in the analysis of large-scale market behaviors (drawing parallels from physics).
This interdisciplinary approach, known as econophysics, uses concepts from physics, like statistical mechanics (which often employs Group Theory), to understand complex financial systems.
### Statistical Arbitrage
This involves using statistical methods to identify and exploit market inefficiencies.
Group Theory can assist in identifying patterns and relationships between different financial instruments, which can be important in developing arbitrage strategies.
## Python Code for Group Theory in Finance – Example: Risk Parity Portfolio
Given we talked about risk parity as an example application of Group Theory in finance, let’s use it in our coding example.
Risk parity is a portfolio allocation strategy that balances the risk contributed by each asset in the portfolio.
In a Group Theory context, we can think of different asset allocations as elements of a group.
And the operation of the group could be the rebalancing process that aims to achieve risk parity.
For simplicity, we’ll generate synthetic data. In a real-world application, you would replace this with actual financial data, typically obtained from a financial data API or database.
Let’s write the Python code for a basic risk parity model:
```import numpy as np
import pandas as pd
# Generate synthetic asset return data
np.random.seed(42)
dates = pd.date_range('2020-01-01', periods=100)
assets = ['Asset1', 'Asset2', 'Asset3']
data = np.random.randn(100, 3) # synthetic returns for 3 assets
returns = pd.DataFrame(data, index=dates, columns=assets)
# Function to calculate portfolio risk
def portfolio_risk(weights, cov_matrix):
return np.sqrt(np.dot(weights.T, np.dot(cov_matrix, weights)))
# Function to calculate risk contribution of each asset
def risk_contribution(weights, cov_matrix):
total_portfolio_risk = portfolio_risk(weights, cov_matrix)
marginal_risk_contribution = np.dot(cov_matrix, weights)
risk_contribution = np.multiply(marginal_risk_contribution, weights) / total_portfolio_risk
return risk_contribution / total_portfolio_risk
# Risk Parity Optimization
def risk_parity_portfolio(returns, target_risk_contribution=None):
if target_risk_contribution is None:
target_risk_contribution = np.ones(len(returns.columns)) / len(returns.columns)
cov_matrix = returns.cov()
# Initial guess for weights (equal weights)
weights = np.ones(len(returns.columns)) / len(returns.columns)
constraints = ({'type': 'eq', 'fun': lambda x: np.sum(x) - 1}) # sum of weights is 1
bounds = tuple((0, 1) for asset in range(len(returns.columns)))
# Objective function
def objective(weights):
return np.sum((risk_contribution(weights, cov_matrix) - target_risk_contribution)**2)
# Optimize
result = minimize(objective, weights, method='SLSQP', bounds=bounds, constraints=constraints)
return result.x
# Compute the risk parity portfolio
optimal_weights = risk_parity_portfolio(returns)
print("Optimal weights for risk parity portfolio:", optimal_weights)```
In this example, the risk_parity_portfolio function calculates the optimal asset weights for a risk parity portfolio.
(Be sure to indent the code where appropriate like in the image below.)
It uses the squared difference between actual and target risk contributions as the objective function to minimize.
Note that this example is a simplified version of risk parity.
In practice, you would need to consider additional factors like transaction costs, liquidity constraints, and more sophisticated risk models.
Also, the integration of group theory into this model is more conceptual, focusing on the idea of balancing elements (assets) in a group (portfolio) to achieve a desired state (equal risk contribution)
## Conclusion
In finance, Group Theory translates to identifying patterns, correlations, and structures that can be mathematically described and used for various financial strategies and analyses.
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# economy homework
The human resource manager of the XYZ Company makes the following claim: “Our workers make an average of $500 per week. We produce$6000 worth of output each week using only 10 workers. That averages out to $600 per worker per week. We should therefore hire more people as long as the wage is$500 per week.” Assess this claim - is the reasoning solid or do you disagree? (2 points) 2. Consider a firm that uses both labor and capital in production. The price of capital is $20 per unit and the wage rate is$10 per hour. • Draw the firm’s isocost line assuming a total production cost of $100. How steep is this line (that is, what is its slope)? Be sure to clearly label the axes. (3 points) • Suppose the wage drops to$5 per unit. In which direction does the substitution effect change the firm’s demand for labor and capital? In which direction does the scale effect change the firm’s demand for labor and capital? (2 points) • If the firm chooses its labor and capital combination to minimize its production costs, will the marginal product of labor be higher than, lower than, or equal to the marginal product of capital? Why? (Assume that the price of labor and capital are those given in part b.) (2 points) 3. Consider a firm that uses two inputs: skilled workers and computers. Explain what it means if skilled workers and computers are complements in production. If the price of computers falls, and skilled workers and computers are complements, will the firm want to hire more or fewer skilled workers? (2 points) 4. Suppose the hourly wage is $20, the price of each unit of capital is$25, and the price of output is $50 per unit. Assume that the firm cannot affect any of these prices. The production function of the firm is , so that the marginal product of labor is . • If the current capital stock is fixed at 1,600 units, how many hours of labor should the firm hire in the short run (i.e., what should E be)? How much profit will the firm earn (hint: remember that profit is just price×output – wage×E – price of capital×K)? (3 points total: 2 for the value of E and 1 for the profit). • Now let’s think about the long run, in which the firm can freely choose both labor and capital. Based on the production function, the marginal product of capital can be written as (the fact that we’re telling you what MPK is in this part is a big hint). If the firm is maximizing profits, what must the ratio of MPE to MPK be? What must the ratio of E to K be? (2 points) Attachments: ## 1 Approved Answer 1. Output per worker per week =$ 600 per week Worker's wage = $500 per week Productivity = 600/500*100 = 120% Based on productivity we can't decide to hire more people. Hiring more people depends on the demand and based on that demand the labor requirement. Hence, the reasoning is not solid 2. Let amount of capital be X and amount of labor be Y rX + wY = C where r is the capital cost per unit and w is the wage per hour C is the total production cost Equation of isocost line: 20X + 10Y = 100 20X + 10Y = 100 can be written as Y = 10 - 2X Slope of line = y/x = -2 Hence, slope of the line is -2 3. With substitution effect, the consumers replace more expensive items with less expensive items. Hence, demand for labor will increase and that of capital will decrease as cost of labor has decreased With scale effect, the demand for capital will increase in order to reduce the demand for labor and thus reduce the labor costs. 4. The marginal cost of both inputs i.e. labor and capital will be equal because: THe cost of capitalwill be$20 per unit while cost of labor will be $10 per unit. Marginal cost of labor will be defined as increase in output by increasing one unit of labor. Similarly we can define marginal cost of Capital. Lets suppose: Marginal cost labor > Marginal cost of Capital => In this case, decreasinglabor by one dollarand increasing capital by one dollarwill reduce costs by marginal labor cost and will increase by marginalcapital cost. Hence there will... Related Questions in Inventory Management and Control • Operations Management Assignment #3 June 07, 2011 Looking for assistance on this operations assignment. I am submitting the assignment for help to get a price quote. I am not willing to pay more than$45 for this assignment. Please...
• Operations Management Questions (Solved) August 21, 2012
will then have to pay $250 per person- hour to hire contract consultants on short notice to make -up for the remaining work requirements. a) ( 4 points)Mr. Grisham has asked you for... Solution Preview : After baking the vases move to a vase glazing station where workers use glaze to hand-paint designs on the pieces. Each worker does one piece at a time, and it takes an average of 30 minutes... • 1. Inventory Management Warm-up Questions (7 points) a. (3 points) A... November 14, 2017 unit 12%, an order quantity 500 units , and a cost per unit of inventory of$ 100 ? B . EOQ Model) Assuming no safety stock , what is the re-order point (ROP) given an average daily...
• Operation management (Solved) April 26, 2014
. Usage rate is 25 stones per day, and ordering costs are $48. If annual carrying costs are 30 percent of unit cost , what is the optimal order size? 2 ) A chemical firm produces sodium... Solution Preview : Solution: 1. This is a problem of identifying optimal order size when the price per unit varies according to the quantity ordered. Order Quantity Price Per UNit >600$ 8 400-599 $9$ 10 EOQ...
• My exam is scheduled for next Friday 1/20 starting at 1pm. (Solved) January 12, 2012
There will be 50 multiple choice questions similar to the sample I've attached. I only have that same Friday to complete the exam. This is for an MBA Operations Management course. I would...
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MBA 626 Winter 2012 Test 1 Submission Requirements January 20, 2012 You must submit 3 copies of your answer as following: Submit one copy of this file and supporting Email all your files...
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9.26 The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour.
(a) At ± = .05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 indicate that the manufacturer's claim is overstated?
(b) Why might the assumption of a normal population be doubtful? (See Aviation Week and Space Technology 162, no. 4 (January 24, 2005), p. 42.)
#### Solution Preview
9.26)
The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour.
(a) At ± = .05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 indicate that the manufacturer's claim is overstated?
(b) Why might the ...
#### Solution Summary
Test of hypothesis about mean of an airport baggage scanning machine claims.
H0: mu=> 530bags/Hr (Claim is not overstated)
Ha: mu < 530 bags/Hr (Claim is overstated)
where mu is the population mean number of bags per hour
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# Einstein velocity transformations problem
So I made this problem up to visualize the einstein velocity transformations between inertial frames.
## Homework Statement
I throw a frisbee due north. It goes north at a constant velocity of .7c. At the same time I throw it, a bird flies in a straight line at a constant velocity of .5c at such an angle that its northward component is .3c and its eastward component is .4c, relative to the frisbee. It is going toward a birdfeeder located 1 light-minute north and .4 light-minutes east of where I stand. Is the bird really going toward the birdfeeder in both the frisbee's inertial frame and my inertial frame?
## Homework Equations
vx = (vx' + β)/(1+vx'β)
vy = (vy'(√1 - β2)/(1-vx'β)
Apostrophied velocities are measured in the frisbee's frame, which moves at velocity "beta" relative to my frame.
## The Attempt at a Solution
In the frisbee's frame, the birdfeeder is heading south at .7c. In one minute, it will be .7 light-minutes south of where it was before. The bird moves relative to the frisbee up .3 light-minutes and east .4 light-minutes, so it should meet the birdfeeder in one minute.
In the home frame,
The northward component of the bird is .3+.7 / 1+(.3)(.7) = .82645
The eastward component of the bird is .4(sqrt(1-.49))/1+(.3)(.7) = .23608
Since .82645/.23608 does not equal 1/.4, the bird is not heading toward the birdfeeder.
I definitely did something wrong to get this contradiction. Would anyone like to try it?
Last edited:
Related Introductory Physics Homework Help News on Phys.org
Simon Bridge
Homework Helper
Is the bird really going toward the birdfeeder in both the frisbee's inertial frame and my inertial frame?
... to word "really" does not belong here.
You can check your setup by sketching the space-time diagrams for each observer.
You forgot about space contraction which changes the angle towards the location of the bird feeder.
1 person
Dauto,
Thanks for the reply, but I'm not sure I get it. Wouldn't the Einstein velocity transformations already account for the space contraction between the points of view of me and the frisbee?
Okay, I figured it out. From my point of view, the bird flew north at .82645c, but if I were to use a simple Lorentz contraction and multiply this velocity by sqrt(1-(.7)^2), you get a velocity of .5902c north, which, coupled with the .23608c component East, will get the bird to the birdfeeder.
Simon Bridge
Homework Helper
Well done - it's easier with the space-time diagrams though.
1 person
I'm afraid I don't know space-time diagrams very well. Do you mean, a graph with perpendicular axes "time" and "distance" with a second "time" axis at slope 1/.7 and a second "distance" axis at slope .7, all from the origin?
Simon Bridge
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# . A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h
Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
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• Level: GCSE
• Subject: Maths
• Word count: 5131
# Handling Data - Mayfield High School
Extracts from this document...
Introduction
GCSE Mathematics Coursework Handling Data - Mayfield High School Key Stage 2 results at Mayfield School The Plan I am investigating the factors that affect Key Stage 2 results. Factors that may affect Key Stage 2 results are: IQ If someone is intelligent they are more likely to get good Key Stage 2 results Primary school The standard of education at different primary schools varies so the understanding of subjects of different pupils is likely to be affected by their primary schools'. Age The older pupils in a year, those whose birthday is in September, may have acquired a better understanding in their subjects due to their brains being more fully developed whereas younger students in a year, those whose birthday is in August, will be at more of a disadvantage concerning Key Stage 2 results as their minds will not be as developed as those older than them. Year group New teaching methods, such as the Numeracy Hour, have been introduced since older years, such as years 10 and 11, have been educated and so the younger years, for example years 7 and 8, may have benefited from these new teaching methods and have a better understanding of their subjects therefore are more likely to get better Key Stage 2 results. Gender It is said that in general girls cope better in exam conditions than boys and so are more likely to get better Key Stage 2 results. I cannot investigate how different primary schools and how new teaching methods, such as the Numeracy Hour, affect the students' Key Stage 2 results as I do not have the data to support this. I am going to investigate how IQ and gender affect Key Stage 2 results. Hypothesis 1: There is a strong positive correlation between IQ and Key Stage 2 results. Hypothesis 2: (a) IQ of boys and girls is normally distributed, in line with national ....figures. ...read more.
Middle
49 20 102 49 122 109 50 35 96 50 53 100 I will now group my data for the boys' IQ and use this to draw a histogram. Group Frequency Class width Frequency Density 70 - 85 2 15 0.1 86 - 95 4 10 0.4 96 - 100 26 5 5.2 101 - 105 8 5 1.6 106 - 115 4 10 0.4 116 - 125 3 10 0.3 126 - 135 3 10 0.3 Histogram to show the IQ of boys in year 8 at Mayfield School The histogram I have drawn to show the IQ of boys in year 8 at Mayfield School has a curve with a 'bell-shape' which suggests that the IQ is normally distributed but to provide further evidence of this I am going to now find the mean and standard deviation of the data. If 68% of the data lies within 1 standard deviation of the mean, and 95% of the data lies within 2 standard deviations of the mean, this means it is normally distributed. I found the mean of the boy's IQ from my sample to be 101. To find 1 standard deviation of my data I will use the following formula: f is the symbol I'm using for frequency x is the symbol I'm using for the midpoint of the group ? is the symbol I'm using for standard deviation ? is the symbol I'm using for 'the sum of' is the symbol I'm using for the mean Group x f x- (x-)2 f(x-)2 70 - 85 77.5 2 -23.5 552.25 1104.5 86 - 95 90.5 4 -10.5 110.25 441.0 96 - 100 98.0 26 -3.0 9.0 234.0 101 - 105 103.0 8 2.0 4.0 32.0 106 - 115 110.5 4 9.5 90.25 361.0 116 - 125 120.5 3 19.5 380.25 1140.75 126 - 135 130.5 3 29.5 870.25 2610.75 ?f 50 ? ...read more.
Conclusion
10 girls 89 95 3 3 3 3 66 90 3 3 3 3 25 105 5 5 4 5 18 100 4 4 4 4 Year 11 girls 37 100 4 4 4 4 2 108 5 5 5 5 39 97 3 3 3 3 31 104 4 4 4 4 Scatter graph showing relationship between Boys IQ and Key Stage 2 results Scatter graph showing relationship between Girls' IQ and Key Stage 2 results Comparing these two scatter graphs I've drawn I can see that they are very similar. Neither of the graphs show either boys or girls to have gotten higher grades than the other. I think this is quite significant as it provides good evidence to support my hypothesis that gender makes no difference to Key Stage 2 results. I am confident that my results are reliable because I have used the same stratified sample as I used in Hypothesis 1 and I believe this to be a reliable sample as it avoided bias because I used the random number function on my calculator to obtain it. I am confident that my results are valid for the whole school because the stratified sample I used was representative of the whole school. I managed to find evidence to support all three hypotheses. I did not come across any suspect data throughout my investigation so as far as I know all the data I used was reliable but I really cannot be sure as I wasn't the person who collected the data. I could extend my investigation by testing whether there is still strong positive correlation between IQ and just the English Key Stage 2 result as opposed to an average of all three, English, Maths and Science, results. I could also extend my investigation by testing whether Age influences Key Stage 2 results. For example, those whose birthday is in September may be at an advantage compared to those whose birthday is in July or August. Jess Blair 10.7 1 ...read more.
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# Related GCSE Height and Weight of Pupils and other Mayfield High School investigations essays
1. ## mayfield high statistics coursework
Right -Handed 11 6 16 Left -Handed 1 3 4 Total 11 9 20 Probability Tree Diagram My variables are Male and Female and Left and Right Handed. M= MALE R= RIGHT F=FEMALE L= LEFT R = RM M L = ML R = FR F L = FL Calculations
2. ## Conduct an investigation comparing height and weight from pupils in Mayfield School.
-6 36 36 -10 100 45 -1 1 51 5 25 43 -3 9 40 -6 36 47 1 1 44 -2 4 38 -8 64 47 1 1 Sum of standard deviations 753 Variance = 753 = 30.32 25 Standard Deviation = V30.32 = 5.5 (to 1 d.p)
1. ## Liquid chromatography is a technique used to separate components of a mixture to isolate ...
There is then a ~10% decrease between 200�l and 350�l, followed by another decrease of ~4% between 350�l and 500�l and a ~2% decrease in efficiency between 500�l and 1000�l. This suggests that the column becomes overloaded with caffeine between 200�l and 350�l.
2. ## The aim of the statistics coursework is to compare and contrast 2 sets of ...
I did not pick another person because I wanted to see what the results were like. Likely the in the results there was not disputing what the overall favourite eye colour was. If I had to do the project again I would probably pick a different person in they had
1. ## I would like to know whether there is a link between ability in Maths ...
Females 0.96 0.68 [Table 5: Male / Female Product-Moment Correlation Coefficients] The calculation of the product-moment correlation coefficient is such that it will lie between 1 and -1, with 1 meaning that there is perfect positive correlation, 0 no correlation and -1 perfect negative correlation.
2. ## Maths Data Handling
I also predict that most boys in my sample will have a greater height and weight than girls. This means that I predict that my results will show that the relationship between height and weight is affected by gender. Plan As stated in the introduction, my line of enquiry is: 'The relationship between height and weight'.
1. ## Maths data investgation
I have found there only 2 female students who has Attendence (%) between 50>Mth<60. The majoraty of female students have a Attendence (%) beween 80>Mth<90. Tally chart for Male data GCSE maths (%) GCSE maths (%) 10>Mth<20 Tally Frequency 2 20>Mth<30 3 30>Mth<40 0 40>Mth<50 0 50>Mth<60 4 60>Mth<70 2 70>Mth<80 4 Attendence (%)
2. ## Mayfield High School
I will now go in depth to show that the next hypothesis is correct. I will draw more graphs so that I could have more to support my second hypothesis. Hypothesis Two- "Year 10's have a better Ks2 Results and higher IQ than Year" Mean, Median & Mode of Grouped
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These Operations with Rational Numbers notes and practice are differentiated based on some common needs found in the middle school math classroom. Modifications are considered for both struggling learners and high flyers. In these activities
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In these activities, students practice graphing linear relationships by making a table. First students are given equations in standard form and must rearrange the equations to be in slope intercept form (solve for y). Then students practice
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These Solving Equations with Special Solutions notes and practice are differentiated based on some common needs found in the middle school math classroom. Modifications are considered for both struggling learners and high flyers. In these
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These Solving Equations with Rational Coefficients notes and practice are differentiated based on some common needs found in the middle school math classroom. Modifications are considered for both struggling learners and high flyers. In these
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This Bivariate Data Bundle contains differentiated notes and practice for each lesson related to scatter plots and two-way tables. This resource is great for distance learning because each file can be uploaded separately as a PDF to a
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\$11.95
15
Bundle
These Calculating Relative Frequency notes and practice are differentiated based on some common needs found in the middle school math classroom. Modifications are considered for both struggling learners and high flyers. In these activities
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In these activities, students practice using the Pythagorean Theorem to find the distance between two points. The first example has students find the distance between vertical and horizontal points so they can begin to count units and consider
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8
In these activities, students practice writing linear equations given two points. Students will be able to write an equation in slope intercept form given two points and will be able to write an equation in slope intercept form to model a real
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In these activities, students practice finding rate of change of linear functions given a table or a verbal description. Given a table, students may be able to use change in y over change in x and in other cases they may need to select two points
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In these activities, students practice identifying functions represented as a set of ordered pairs, table, mapping, and graph. Students will use the definition of a function as well as the Vertical Line Test to identify whether a relation is a
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7
These Translating Algebraic Equations notes and practice are differentiated based on some common needs found in the middle school math classroom. Modifications are considered for both struggling learners and high flyers. In these activities,
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CCSS:
6.EE.A.2
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7
This Solving Linear Equations Bundle contains differentiated notes and practice for each lesson. Now also includes two quizzes, an end-of-unit study guide and a unit test. This resource is great for distance learning because each file can be
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7
Bundle
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# Tuning fork clanging mode: boundary conditions?
The equation of motion of a tuning fork tine is given by: $$EI\frac{\partial^4z}{\partial x^4}+m\frac{\partial^2z}{\partial t^2}=0$$ or in shorthand: $$EIz_{xxxx}+mz_{tt}=0$$ Boundary conditions: $$z(0,t)=0$$ $$z_x(0,t)=0$$ $$z_{xx}(L,t)=0$$ $$z_{xxx}(L,t)=0$$ Initial condition: $$z_t(x,0)=0$$ Using the Ansatz: $$z(x,t)=X(x)T(t)$$ ... the PDE can be separated to: $$\frac{X''''}{X}=-\frac{m}{EI}\frac{T''}{T}=k^4$$ So: $$X''''-k^4X=0$$ ... which has four roots: $+k$, $-k$, $+ki$ and $-ki$ and the general solution: $$X=A\cosh kx+B\sinh kx+C\cos kx+D\sin kx$$ Using the boundary conditions we get: $$B=\frac{\sinh kL-\sin kL}{\cosh kL+\cos kL}A=\gamma A$$ $$C=-A$$ $$D=-B$$ $$\cos kL=-\frac{1}{\cosh kL}$$ The latter with first root: $$kL=0.597\pi$$ Solving the DE for $T(t)$ then gives the angular velocity and fundamental frequency (not shown).
For $X(x)$ we then have: $$X(x)=A(\cosh kx+\gamma \sinh kx-\cos kx-\gamma \sin kx)$$ A dimensionless plot for that first root gave me:
But there are several modes of oscillation available to a tuning fork. Two important ones are:
Left: fundamental mode (derived above), right: clanging mode. The latter is said to be obtained by hitting the tine "hard" and has a frequency roughly 6.26 times higher than the fundamental.
I suspect that this hitting "hard" changes the boundary condition(s), leading to a new set of coefficients for $X(x)$. I suspect some boundary condition may have to be applied to $x=L/2$ rather than $x=L$ but cannot see how hitting the tine hard can have that effect.
Can anyone see it?
• I don't think your drawing on the left shows the fundamental mode. – M. Enns Jul 3 '16 at 16:39
• @Gert's drawing does match the fundamental mode as defined in T. D. Rossing, D. A. Russell, and D. E. Brown, ‘‘On the acoustics of tuning forks,’’ Am. J. Phys. 60 , 620–626 ~ 1992 here Looking over the paper it seems the clang mode is a higher order in plane normal mode of treating the fork as a cantilevered beam, i.e same boundary conditions as the in plane fundamental, but a higher order mode. – paisanco Jul 3 '16 at 16:45
• A better picture has been inserted. And great animations can be found here: acs.psu.edu/drussell/Demos/TuningFork/fork-modes.html – Gert Jul 3 '16 at 16:49
• @paisanco: not sure what you mean by higher order. A harmonic? The first harmonic is for $kL=1.492\pi$ and gives the same shape for $X(x)$ as the fundamental mode. But the frequency would be $(1.492/0.597)^2\approx 6.3$ times higher, so that would fit. – Gert Jul 3 '16 at 17:04
• Yes that's what I mean, fundamental the first normal mode, next higher order mode (first harmonic ) the "clang" mode. Are you sure the X(x) is the same shape though? I'd expect the wavelength of the clang mode to be shorter. – paisanco Jul 3 '16 at 17:09
Based on T. D. Rossing, D. A. Russell, and D. E. Brown, ‘‘On the acoustics of tuning forks,’’ Am. J. Phys. 60 , 620–626 ~ 1992, I'd conclude that the fundamental and "clang" modes are both in plane modes (X plane in the questions' notation ) and have the same boundary conditions.
So I don't think the "clang" mode has a different boundary condition from the fundamental mode. They are simply different normal modes of vibration , the "fundamental" mode being the first normal mode, and the "clang" mode being the second normal mode (first harmonic is really a misnomer here as the frequencies aren't harmonics). Both of these modes have boundary conditions of treating the fork as a cantilevered beam (ie. one end, the stem end, fixed).
I suspect the comment about striking the fork hard to excite the "clang" mode really refers to exciting the motion of the "clang" mode by hitting the whole fork at a hard surface. This would excite motion at the antinodes of the "clang" mode along the fork , since it would have shorter wavelength than the fundamental mode. The fundamental mode would best be excited by tapping the open end of the fork.
It's similar in idea to exciting natural harmonics on a guitar- you fix the string at the 12th fret with your finger and pluck along the string, exciting it at at the antinode.
The treatment by @Gert in the question is using the Euler-Bernoulli (E-B) beam theory. There is a decent Wikipedia article on this in which the in plane transverse normal modes of vibration of a cantilevered bar (free at one end, clamped at the other end in this case $x=0$) are computed and plotted.
Here is the plot of the shape of the bar displacement due to vibration, for the first few normal modes (normalized to the length of the bar being 1.0) from the Wikipedia article:
Mode 1 is the fundamental mode and has no node or antinode along the bar.
Mode 2 is the next normal mode (the "clang" mode) and shows a node (zero displacement) at just below 0.8 times the length of the bar, and an antinode (local maximum of displacement) at a shorter distance along the bar.
The higher order normal modes also show nodes and antinodes along the bar.
The statement that the clang mode displacement does not show optima or roots for $x<L$ isn't consistent with the second drawing of the tuning fork in the question showing the clang mode having a maximum and minimum of displacement along the length of the fork. Each term in the E-B equation's solution should indeed have a different constant.
Of course both analyzing a tuning fork in terms of a analytical solution using Euler-Bernoulli theory, and in analogy to a cantilevered bar, are approximations. A more accurate treatment would use Timoshenko beam theory and finite element analysis to find a solution.
• $X(x)=A(\cosh kx+\gamma \sinh kx-\cos kx-\gamma \sin kx)$ shows no optima, it's not wavy at all. But the first harmonic's frequency does fit the 6.26 ratio to the fundamental frequency. – Gert Jul 3 '16 at 17:49
• X_m(x) should depend on wavenumber of the m-th normal mode. k should be a wavenumber depending on the mode. There should be units of frequency in the initial equation's second term. It's not simple harmonic (sinusoidal) wave motion, but it's definitely "wavy" -the drawings of the modes from the papers show nodes and antinodes. – paisanco Jul 3 '16 at 18:32
• $X(x)$ does depend on order: both $\gamma$ and $k$ depend on it. But for $kL=1.494\pi$ (first harmonic) the function does not become wavy, it continues to grow for $x$, without maxima or minima. – Gert Jul 3 '16 at 21:34
• As regards the Am. J. Phys. 60 paper, I've had it for a while (the second graphic in my question was lifted from it) and while interesting it doesn't contain any ab initio derivations at all. – Gert Jul 3 '16 at 21:49
• For the second normal mode $kL = 1.49\pi$ yes, it will grow without bound for $x>L$. But there should be a zero of $X(x)$ for a position on the fork $x<L$ (a node) and somewhere below the position of the node, the displacement $X(x)$ should reach a local maximum (antinode). – paisanco Jul 4 '16 at 1:14
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# Explain The Procedure Of Finding Specific Heat Of Solid Experimentally?
Specific heat is the quantity of heat required to raise the temperature of any substance by 1 degree Celsius.
Specific Heat formula is
C= $$\frac{\Delta Q}{m \Delta T}$$
Where,
m is the mass
Δ Q is the heat gained or lost
Δ T is the temperature difference
The procedure to determine the specific heat capacity of a solid:
• A calorimeter measures the specific heat capacity of a solid.
• Arrange all the apparatus as shown in the above diagram.
• Place the thermometers A and B in a beaker containing water and make a note of their reading.
• Allow thermometer A to be a standard to find the correction that is to be applied to thermometer B
• Place the thermometer B into the copper tube of a hypsometer containing the powder of the given solid.
• Now place the hypsometer on the burner by adding a sufficient amount of water.
• Record the weight of calorimeter along with stirrer and lid over it
• Make a note of the temperature. Add water when the temperature is between 5℃ to 8℃ and up to the half-length of the calorimeter
• Record the weight of calorimeter along with stirrer and lid over it
• Heat the hypsometer till the temperature of the solid is steady.
• Note the temperature of water in calorimetry.
• Add solid powder from the hypsometer into the calorimeter
• Record the final temperature of the mixture.
• Remove the thermometer from the calorimeter and weigh the calorimeter with the contents and lid.
• Now, as per the obtained values, calculate the specific heat capacity of a solid.
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International
Tables for
Crystallography
Volume A
Space-group symmetry
Edited by M. I. Aroyo
International Tables for Crystallography (2015). Vol. A, ch. 1.4, pp. 59-61
## Section 1.4.3. Generation of space groups
H. Wondratscheke
### 1.4.3. Generation of space groups
| top | pdf |
In group theory, a set of generators of a group is a set of group elements such that each group element may be obtained as a finite ordered product of the generators. For space groups of one, two and three dimensions, generators may always be chosen and ordered in such a way that each symmetry operation can be written as the product of powers of h generators (j = ). Thus,where the powers are positive or negative integers (including zero). The description of a group by means of generators has the advantage of compactness. For instance, the 48 symmetry operations in point group can be described by two generators. Different choices of generators are possible. For the space-group tables, generators and generating procedures have been chosen such as to make the entries in the blocks General position' (cf. Section 2.1.3.11 ) and Symmetry operations' (cf. Section 2.1.3.9 ) as transparent as possible. Space groups of the same crystal class are generated in the same way (see Table 1.4.3.1 for the sequences that have been chosen), and the aim has been to accentuate important subgroups of space groups as much as possible. Accordingly, a process of generation in the form of a composition series has been adopted, see Ledermann (1976). The generator is defined as the identity operation, represented by (1) . The generators , , and are the translations with translation vectors a, b and c, respectively. Thus, the coefficients , and may have any integral value. If centring translations exist, they are generated by translations (and in the case of an F lattice) with translation vectors d (and e). For a C lattice, for example, d is given by . The exponents (and ) are restricted to the following values:
• Lattice letter A, B, C, I: or 1.
Table 1.4.3.1| top | pdf | Sequence of generators for the crystal classes
The space-group generators differ from those listed here by their glide or screw components. The generator 1 is omitted, except for crystal class 1. The generators are represented by the corresponding Seitz symbols (cf. Tables 1.4.2.1 –1.4.2.3 ). Following the conventions, the subscript of a symbol denotes the characteristic direction of that operation, where necessary. For example, the subscripts 001, 010, 110 etc. refer to the directions [001], [010], [110] etc. For mirror reflections m, the `direction of m' refers to the normal of the mirror plane.
Hermann–Mauguin symbol of crystal classGenerators (sequence left to right)
1 1
2 2
m m
222
mm2
mmm
4
422
4mm
3
(rhombohedral coordinates )
(rhombohedral coordinates
321
(rhombohedral coordinates )
312
3m1
(rhombohedral coordinates )
31m
(rhombohedral coordinates )
6
622
6mm
23
432
• Lattice letter R (hexagonal axes): , 1 or 2.
• Lattice letter : or 1; or 1.
As a consequence, any translation of with translation vectorcan be obtained as a productwhere are integers determined by . The generators and are enclosed between parentheses because they are effective only in centred lattices.
The remaining generators generate those symmetry operations that are not translations. They are chosen in such a way that only terms or occur. For further specific rules, see below.
The process of generating the entries of the space-group tables may be demonstrated by the example in Table 1.4.3.2, where denotes the group generated by . For , the next generator is introduced when , because in this case no new symmetry operation would be generated by . The generating process is terminated when there is no further generator. In the present example, completes the generation: (178).
Table 1.4.3.2| top | pdf | Generation of the space group (178)
The entries in the second column designated by the numbers (1)–(12) correspond to the coordinate triplets of the general position of .
Coordinate tripletsSymmetry operations
(1) ; Identity
The group of all translations of P6122 has been generated
Threefold screw rotation
Threefold screw rotation
Now the space group has been generated
Twofold screw rotation
Sixfold screw rotation
Sixfold screw rotation
Now the space group has been generated
Twofold rotation, direction of axis [110]
Twofold rotation, axis [100]
Twofold rotation, axis [010]
Twofold rotation, axis
Twofold rotation, axis [120]
Twofold rotation, axis [210]
#### 1.4.3.1. Selected order for non-translational generators
| top | pdf |
For the non-translational generators, the following sequence has been adopted:
• (a) In all centrosymmetric space groups, an inversion (if possible at the origin O) has been selected as the last generator.
• (b) Rotations precede symmetry operations of the second kind. In crystal classes and and and , as an exception, and are generated first in order to take into account the conventional choice of origin in the fixed points of and .
• (c) The non-translational generators of space groups with C, A, B, F, I or R symbols are those of the corresponding space group with a P symbol, if possible. For instance, the generators of (24) are those of (19) and the generators of Ibca (73) are those of Pbca (61), apart from the centring translations.
Exceptions: I4cm (108) and I4/mcm (140) are generated via P4cc (103) and P4/mcc (124), because P4cm and P4/mcm do not exist. In space groups with d glides (except , No. 122) and also in (88), the corresponding rotation subgroup has been generated first. The generators of this subgroup are the same as those of the corresponding space group with a lattice symbol P.
#### Example
(227):
.
• (d) In some cases, rule (c) could not be followed without breaking rule (a), e.g. in Cmme (67). In such cases, the generators are chosen to correspond to the Hermann–Mauguin symbol as far as possible. For instance, the generators (apart from centring) of Cmme and Imma (74) are those of Pmmb, which is a non-standard setting of Pmma (51). (A combination of the generators of Pmma with the C- or I-centring translation results in non-standard settings of Cmme and Imma.)
For the space groups with lattice symbol P, the generation procedure has given the same triplets (except for their sequence) as in IT (1952). In non-P space groups, the triplets listed sometimes differ from those of IT (1952) by a centring translation.
### References
International Tables for X-ray Crystallography (1952). Vol. I, Symmetry Groups, edited by N. F. M. Henry & K. Lonsdale. Birmingham: Kynoch Press. [Abbreviated as IT (1952).]
Ledermann, W. (1976). Introduction to Group Theory. London: Longman.
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# Thread: 2 Boolean Algebra expressions
1. ## 2 Boolean Algebra expressions
Hi
I need help solving these two boolean algebra expressions. I need to simplify them using the rules of boolean algebra. The two expressions in question are:
Just to clarify, I need help simplifying both expressions.
Thank you in advance if you can help.
2. Originally Posted by rushhour
Hi
I need help solving these two boolean algebra expressions. I need to simplify them using the rules of boolean algebra. The two expressions in question are:
Just to clarify, I need help simplifying both expressions.
Thank you in advance if you can help.
Well I will follow different notation and I apologize for that...
NOTATIONS I shall use:
$\displaystyle \overline{P}$ for NOT P.
A+B for OR of A and B.
A.B for AND of A and B.
$\displaystyle (P.Q) + (\overline{P} . Q) = (P + \overline{P}) . Q = 1.Q = Q$
$\displaystyle P \implies Q$ is equivalent to $\displaystyle \overline{P} + Q$
So $\displaystyle (P \implies \overline{Q}) + P.Q = \overline{P} + \overline{Q} + P.Q = \overline{P} + \overline{Q} + P = 1$
Here I have used the Boolean relation $\displaystyle X + \overline{X}.Y = X + Y$
Hence $\displaystyle (P \implies \overline{Q}) + P.Q$ is a tautology.
3. Thanks, but could you possibly convert that into my version, simply because I really have no idea how to follow what you have done! The type of way I do it is the following:
I hope you can help me format it in this type of way.
4. You must supply the reasons according to your textbook/notes.
$\displaystyle \begin{gathered} \left( {P \Rightarrow \neg Q} \right) \vee \left( {P \wedge Q} \right) \hfill \\ \left( {\neg P \vee \neg Q} \right) \vee \left( {P \wedge Q} \right) \hfill \\ \neg P \vee \left[ {\neg Q \vee \left( {P \wedge Q} \right)} \right] \hfill \\ \neg P \vee \left( {\neg Q \vee P} \right) \wedge \underbrace {\left( {\neg Q \vee Q} \right)}_{TRUE} \hfill \\ \end{gathered}$
$\displaystyle \begin{gathered} \neg P \vee \left( {\neg Q \vee P} \right) \hfill \\ \neg Q \vee \underbrace {\left( {\neg P \vee P} \right)}_{TRUE} \hfill \\ TRUE \hfill \\ \end{gathered}$
5. May I ask if this is the solution to both or just the bottom one?
6. Just wish to clarify that there are 2 problems I need help solving, the first is this:
the second one is this:
Thanks if you can help me simplify them.
7. $\displaystyle \begin{gathered} \left( {P \wedge Q} \right) \vee \left( {\neg P \wedge Q} \right) \hfill \\ \underbrace {\left( {P \vee \neg P} \right)}_{TRUE} \wedge Q \hfill \\ Q \hfill \\ \end{gathered}$
8. Originally Posted by Plato
$\displaystyle \begin{gathered} \left( {P \wedge Q} \right) \vee \left( {\neg P \wedge Q} \right) \hfill \\ \underbrace {\left( {P \vee \neg P} \right)}_{TRUE} \wedge Q \hfill \\ Q \hfill \\ \end{gathered}$
Thanks this helped alot, and I included the rules that you used, I am just having trouble understanding what you wrote for this:
Originally Posted by Plato
$\displaystyle \begin{gathered} \left( {P \Rightarrow \neg Q} \right) \vee \left( {P \wedge Q} \right) \hfill \\ \left( {\neg P \vee \neg Q} \right) \vee \left( {P \wedge Q} \right) \hfill \\ \neg P \vee \left[ {\neg Q \vee \left( {P \wedge Q} \right)} \right] \hfill \\ \neg P \vee \left( {\neg Q \vee P} \right) \wedge \underbrace {\left( {\neg Q \vee Q} \right)}_{TRUE} \hfill \\ \end{gathered}$
$\displaystyle \begin{gathered} \neg P \vee \left( {\neg Q \vee P} \right) \hfill \\ \neg Q \vee \underbrace {\left( {\neg P \vee P} \right)}_{TRUE} \hfill \\ TRUE \hfill \\ \end{gathered}$
Could you please state the rules you used for this if possible. Thanks.
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https://fr.maplesoft.com/support/help/maplesim/view.aspx?path=student(deprecated)%2FLineint
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student(deprecated)/Lineint - Maple Help
student
Lineint
inert Form of Line Integral
Calling Sequence Lineint(f(x,y), x, y) Lineint(f(x,y), x=x(t), y=y(t)) Lineint(f(x,y), x, y =a..b) Lineint(f(x,y,z), x, y, z)
Parameters
f(x, y) - expression in x and y a, b - (optional) lower and upper bounds
Description
• Important: The student package has been deprecated. Use the superseding command Student[VectorCalculus][LineInt] instead.
• This two-dimensional version of the procedure constructs a line integral'' of f(x,y) expressed respect to y. In general, the first variables are regarded as parameters of the last.
• Lineint uses an unevaluated'' form of Maple's int function, and only minor simplifications are performed.
• A range can be specified for the parameter, as in (z = a..b).
• Use value to force Lineint to evaluate like int.
• The command with(student,Lineint) allows the use of the abbreviated form of this command.
Examples
Important: The student package has been deprecated. Use the superseding command Student[VectorCalculus][LineInt] instead.
> $\mathrm{with}\left(\mathrm{student}\right):$
> $\mathrm{Lineint}\left(f\left(x,y\right),x,y\right)$
${\int }{f}{}\left({x}{}\left({y}\right){,}{y}\right){}\sqrt{{\left(\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{x}{}\left({y}\right)\right)}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}$ (1)
> $\mathrm{Lineint}\left(f\left(x,y\right),x,y=2..4\right)$
${{\int }}_{{2}}^{{4}}{f}{}\left({x}{}\left({y}\right){,}{y}\right){}\sqrt{{\left(\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{x}{}\left({y}\right)\right)}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}$ (2)
> $\mathrm{Lineint}\left(f\left(x,y\right),x=t,y={t}^{2}\right)$
${\int }{f}{}\left({t}{,}{{t}}^{{2}}\right){}\sqrt{{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{t}\right)}^{{2}}{+}{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{t}}^{{2}}\right)\right)}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}$ (3)
> $\mathrm{Lineint}\left(f\left(x,y\right),x,y,t=a..b\right)$
${{\int }}_{{a}}^{{b}}{f}{}\left({x}{}\left({t}\right){,}{y}{}\left({t}\right)\right){}\sqrt{{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{x}{}\left({t}\right)\right)}^{{2}}{+}{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right)}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}$ (4)
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https://query.libretexts.org/Kiswahili/Chuo_Kikuu_Fizikia_II_-_Thermodynamics%2C_Umeme%2C_na_Magnetism_(OpenStax)/06%3A_Sheria_ya_Gauss/6.0S%3A_Sheria_ya_Gauss_(muhtasari)
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# 6.S: Sheria ya Gauss (muhtasari)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
## Masharti muhimu
eneo vector vector na ukubwa sawa na eneo la uso na mwelekeo perpendicular kwa uso ulinganifu cylindrical mfumo tu inatofautiana na umbali kutoka mhimili, si mwelekeo flux ya umeme dot bidhaa ya uwanja wa umeme na eneo kwa njia ambayo ni kupita flux wingi wa kitu kinachopita katika eneo fulani elektroni huru pia huitwa elektroni za upitishaji, hizi ni elektroni katika kondakta ambazo hazifungwa na atomu yoyote fulani, na hivyo ni huru kuzunguka Gaussia uso yoyote iliyoambatanishwa (kawaida imaginary) uso ulinganifu wa planar mfumo tu inatofautiana na umbali kutoka ndege ulinganifu wa spherical mfumo tu inatofautiana na umbali kutoka asili, si katika mwelekeo
## Mlinganyo muhimu
Ufafanuzi wa umeme wa umeme, kwa shamba la umeme sare $$\displaystyle Φ=\vec{E}⋅\vec{A}→EAcosθ$$ Umeme wa umeme kupitia uso wazi $$\displaystyle Φ=∫_S\vec{E}⋅\hat{n}dA=∫_S\vec{E}⋅d\vec{A}$$ Umeme wa umeme kupitia uso uliofungwa $$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=∮_S\vec{E}⋅d\vec{A}$$ Sheria ya Gauss $$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=\frac{q_{enc}}{ε_0}$$ Sheria ya Gauss kwa mifumo yenye ulinganifu $$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=E∮_SdA=EA=\frac{q_{enc}}{ε_0}$$ Ukubwa wa uwanja wa umeme nje ya uso wa conductor $$\displaystyle E=\frac{σ}{ε_0}$$
## 6.2 Umeme Flux
• Flux ya umeme kupitia uso ni sawa na idadi ya mistari ya shamba inayovuka uso huo. Kumbuka kwamba hii inamaanisha ukubwa ni sawa na sehemu ya shamba perpendicular kwa eneo hilo.
• Flux ya umeme inapatikana kwa kutathmini muhimu ya uso
$$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=∮_S\vec{E}⋅d\vec{A}$$,
ambapo notation kutumika hapa ni kwa ajili ya uso kufungwa S.
## 6.3 Akifafanua Sheria ya Gauss
• Sheria ya Gauss inahusiana na mtiririko wa umeme kupitia uso uliofungwa kwa malipo ya wavu ndani ya uso huo,
$$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=\frac{q_{enc}}{ε_0}$$,
• ambapo qencqenc ni malipo ya jumla ndani ya uso wa Gaussia S.
• Nyuso zote zinazojumuisha kiasi sawa cha malipo zina idadi sawa ya mistari ya shamba inayovuka, bila kujali sura au ukubwa wa uso, kwa muda mrefu kama nyuso zinazingatia kiasi sawa cha malipo.
## 6.4 Kutumia Sheria ya Gauss
• Kwa usambazaji wa malipo na ulinganifu fulani wa anga (spherical, cylindrical, na planar), tunaweza kupata uso wa Gaussia juu ya ambayo$$\displaystyle \vec{E}⋅\hat{n}=E$$, ambapo E ni mara kwa mara juu ya uso. Shamba la umeme linaamua na sheria ya Gauss.
• Kwa ulinganifu wa spherical, uso wa Gaussia pia ni nyanja, na sheria ya Gauss inafungua$$\displaystyle 4πr^2E=\frac{q_{enc}}{ε_0}$$.
• Kwa ulinganifu wa cylindrical, tunatumia uso wa Gaussia wa cylindrical, na kupata kwamba sheria ya Gauss inaeleza$$\displaystyle 2πrLE=\frac{q_{enc}}{ε_0}$$.
• Kwa ulinganifu wa mpango, uso rahisi wa Gaussia ni sanduku linalopenya ndege, na nyuso mbili zinazofanana na ndege na perpendicular iliyobaki, na kusababisha sheria ya Gauss kuwa$$\displaystyle 2AE=\frac{q_{enc}}{ε_0}$$.
## 6.5 Wafanyabiashara katika Msawazo wa umeme
• Shamba la umeme ndani ya conductor hupotea.
• Malipo yoyote ya ziada yaliyowekwa kwenye conductor inakaa kabisa juu ya uso wa conductor.
• Sehemu ya umeme ni perpendicular kwa uso wa conductor kila mahali juu ya uso huo.
• Ukubwa wa uwanja wa umeme tu juu ya uso wa conductor hutolewa na$$\displaystyle E=\frac{σ}{ε_0}$$.
## Wachangiaji na Majina
Template:ContribOpenStaxUni
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