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https://www.scribd.com/document/123003071/Discounting
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You are on page 1of 15
# Discounted Cash Flows
1
What is the percentage price increase? 110 Suppose instead the asset price fell from 110 to 99. What is the percentage price decrease? Percentage change = 99 . today’s price is Pt = 121.110 110 = .110 = 10% 2 .Percentage Changes Yesterday’s price on an asset was Pt-1 = 110.10% Percentage change = 121 .
and today’s price Pt What is the annual rate of return on the stock? It is the percentage change… rt = Pt-1 Pt -1 = Pt – Pt-1 Pt-1 Now suppose we observe monthly prices Pt-1 . r12 How do we obtain the annualized rate of return: rannualized = (Average[r1 . … . r2 . Pt The monthly rate of return rmonth t = Pt – Pt-1 Pt-1 Why might we want to convert to an annualized return? How do we convert this to an annualized rate of return? rannualized = (r1month + 1)12 -1 What if we have 12 monthly returns: r1 . r12] + 1)12 – 1 3 . … . r2 .Annual Rates of Return Suppose we observe last year’s stock price Pt-1 .
054 101 Net return = 5.67% 4 .10 r2 = –2.35% Annualized return: (1 – 0.70 Mar 1.4% Seven weekly closing levels on Gold in 2006: Feb 1.5 = 1.78% Feb 15. 06: 572. 06: 547.00% r5 = -3.50 Feb 8.5. What is the rate of return on the stock? Gross Return = Example 2: 106. Today’s price is \$106. 06: 558.50 r3 = 2. 06: 561.Rate of Return Examples Example 1: Last year’s stock price was \$101.59% r6 = 1.50 Feb 22.41% r4 = 1.35%)52 -1 = –16. 06: 571.82% Weekly returns: r1 = –2. 06: 559.50 Mar 8.70 Mar 15. 06: 549.67% Average weekly return: –0.
96% per annum with semi-annual compounding What does (ii) mean? compounding…divide the rate in half. You are offered the following two rates: (i) 5. and receive that amount of interest every half-year.0248)(1.02 5 .05) = \$105 (ii) 100(1.00% per annum (ii) 4.0248) = \$105.Compounded Rates of Return You have \$100 to invest. By convention: given a quoted (annual) rate with semi-annual (i) 100(1.
02)10 = 121.02)2 = \$104.04 – with monthly compounding: 100 (1 + 4%/12)12 = 104.Compounding Examples Example 1 Bank offers 4% per annum savings rate for your 1 year.10 6 .90 – with monthly compounding: 100 (1 + 4%/12)60 = 122.00 – with semi-annual compounding: 100 (1.04) = \$104.07 The differences are small.04)5 = 121. \$100 investment… – with no compounding: 100 (1.67 – with semi-annual compounding: 100 (1. But what about over 5 years? – with no compounding: 100 (1.
99% 5.: PV = 200. comp.059/2)2 – 1 = 5.000 (1.000 (1.606 5.9% per annum with semi-annual compounding – 5. comp.02% (b) Which requires the lowest investment today? 6% per annum: PV = 200.000 for child’s college education in 15 years.0295)-30 = \$83.0585*15 = \$83.More Compounding Examples Example 2 • • Put aside money today to have \$200.164 7 .06)-15 = \$83.85% per annum with continuous compounding (a) What is r (effective annual rate) in each case? r = 6% 6% per annum: 5.453 5. Three banks offer the following rates: – 6% per annum (no compounding) – 5.9% semi annual: PV = 200.85% cont.: r = e0.000 e-0.9% semi annual: r = (1+0.0585 – 1 = 6.85% cont.
you invest \$100 as m 100 (1 + 0. continuous compounding Year-end cash = K ert 8 .04/m)m = 100 e0.Continuous Compounding Bank offers 4% per annum with continuous compounding.08 General Formula: Invest K dollars for t years at a rate of r% per annum.04 = 104.
Compounding Summary Accounting for different compounding periods: – m = # compounding periods per year – t = # of years – n = # of periods = m * t – R = Annual (or Nominal or Stated) interest rate – i = Periodic rate = R/m FV PV = (1 + R/m)m*t As m = FV (1 + i)n (1 + R/m)m eR For continuous compounding over t years: FV = PV eRt 9 .
PV(1 + 0. r = 6%.06)2 89.06 )2 = PV = = 100 100 (1. t = 2.00 FV (1 + r)t 10 Generalize: PV = .Present Value Example: Receive \$100 in two years time. What is that worth today? FV = 100.
06) (1. + = 368.Multiple Cash Flows Example: identical annual payment of \$50 each year for 10 years Annual discount rate = 6% Cash flows (\$) 50 1 2 3 4 5 6 7 Time (years) 8 9 10 P = 50 50 50 50 + + + .06) (1.06)3 10 (1.06) 11 ..00 2 (1...
06) 50 0...06)3 + .06 + 50 (1.06)10 = 368 = [1 – (1. + 50 (1.06)2 + 50 (1..Multiple Cash Flows P = 50 (1.06) -10] = 368 Formula for identical cash flows C every year for t years with discount rate r: C [1 – (1 + r)-t ] r ‘Annuity’ formula (why is this a misnomer?) PV = How could you calculate the Future Value of a series of periodic cash flows? 12 .
.03) 20 0.. + 20 (1.03)10 = = [1 – (1.03)2 + 20 (1.60 + 20 (1.03) PV = C [1 – (1 + r/m) -t*m ] = r/m C [1 – ( 1+i ) –n ] i where i = r/m.03 170.03)3 -10] + .Multiple Cash Flows and Compounding Example: identical annual payment of \$40 every six months for 5 years Annual discount rate = 6% with semi-annual compounding P = 20 (1.. n = t * m 13 .
The loan has a annual rate of 7.780.000. What is the fixed monthly payment required to pay off the loan? For this example.62% 12 n = 30 * 12 = 360 PV = 400.45% = 0.0062 [1 – (1.44 14 .000.Example: Mortgage Payments Suppose you take out a 30 year mortgage loan of \$400.000 = C 0. fixed over the life of the loan. 400. we wish to solve for C given i = 7.0062)-360 ] Hence C = \$2.45%. and compounded monthly.
Then the annual equivalent rate is: rann equiv = (1 + 5%/12)12 – 1 = 5. and on Jan 31 2008 is 102.12% 15 . for a quoted mortgage rate) as follows: rann equiv = (1 + rquoted /12 )12 – 1 Suppose the quoted mortgage rate is 5%.g.Annualizing vs Annual Equivalence We annualize a rate of return to facilitate comparisons in the performance of different assets: Example: (1 + rmonth)12 – 1 = rannual where rmonth = ( P1 – P0)/ P0 subscripts represent monthly data If the asset’s price on Dec 31 2007 is 100. then the annualized return (based on this data alone) is: rannual = 26.8% We convert an annual rate with multiple compounding periods into an annual equivalent rate (e. with monthly payments.
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https://metanumbers.com/4972
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# 4972 (number)
4972 is an even four-digits composite number following 4971 and preceding 4973. In scientific notation, it is written as 4.972 × 103. The sum of its digits is 22. It has a total of 4 prime factors and 12 positive divisors. There are 2,240 positive integers (up to 4972) that are relatively prime to 4972.
## Basic properties
• Is Prime? no
• Number parity even
• Number length 4
• Sum of Digits 22
• Digital Root 4
## Name
Name four thousand nine hundred seventy-two
## Notation
Scientific notation 4.972 × 103 4.972 × 103
## Prime Factorization of 4972
Prime Factorization 22 × 11 × 113
Composite number
Distinct Factors Total Factors Radical ω 3 Total number of distinct prime factors Ω 4 Total number of prime factors rad 2486 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 4972 is 22 × 11 × 113. Since it has a total of 4 prime factors, 4972 is a composite number.
## Divisors of 4972
1, 2, 4, 11, 22, 44, 113, 226, 452, 1243, 2486, 4972
12 divisors
Even divisors 8 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ 12 Total number of the positive divisors of n σ 9576 Sum of all the positive divisors of n s 4604 Sum of the proper positive divisors of n A 798 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 70.5124 Returns the nth root of the product of n divisors H 6.23058 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 4972 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 4972) is 9576, the average is 798.
## Other Arithmetic Functions (n = 4972)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 2240 Total number of positive integers not greater than n that are coprime to n λ 560 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 669 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 2,240 positive integers (less than 4972) that are coprime with 4972. And there are approximately 669 prime numbers less than or equal to 4972.
## Divisibility of 4972
m n mod m
2 0
3 1
4 0
5 2
6 4
7 2
8 4
9 4
The number 4972 is divisible by 2 and 4.
• Arithmetic
• Deficient
• Polite
## Base conversion 4972
Base System Value
2 Binary 1001101101100
3 Ternary 20211011
4 Quaternary 1031230
5 Quinary 124342
6 Senary 35004
8 Octal 11554
10 Decimal 4972
12 Duodecimal 2a64
16 Hexadecimal 136c
20 Vigesimal c8c
36 Base36 3u4
## Basic calculations (n = 4972)
### Multiplication
n×y
n×2 9944 14916 19888 24860
### Division
n÷y
n÷2 2486 1657.33 1243 994.4
### Exponentiation
ny
n2 24720784 122911738048 611117161574656 3038474527349189632
### Nth Root
y√n
2√n 70.5124 17.0678 8.39717 5.48664
## 4972 as geometric shapes
### Circle
Radius = n
Diameter 9944 31240 7.76626e+07
### Sphere
Radius = n
Volume 5.14851e+11 3.10651e+08 31240
### Square
Length = n
Perimeter 19888 2.47208e+07 7031.47
### Cube
Length = n
Surface area 1.48325e+08 1.22912e+11 8611.76
### Equilateral Triangle
Length = n
Perimeter 14916 1.07044e+07 4305.88
### Triangular Pyramid
Length = n
Surface area 4.28177e+07 1.44853e+10 4059.62
## Cryptographic Hash Functions
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https://www.asc.ohio-state.edu/durkin.2//catapult.htm
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# by Malcolm Durkin
Below is pictured a unique catapult I built as a member of the Worthington Kilbourne High School Science Olympiad Team. This device placed third in the Storm the Castle competition at the Ohio State tournament held on Apr. 24, 2004 (see Worthington Suburban Newspaper article). I like to believe it would have come in first or second if I had aimed it better.
The other devices in the competition were either conventional trebuchets or catapults. My device worked differently. When the arm is released the weight begins to fall providing a constant acceleration of the projectile at the end of the arm. While the shaft rotates through 2.5 revolutions a tether slowly tightens. When the tether goes taught the arm stops and the projectile continues on its way at high velocity. It will outperform either a conventional trebuchet or catapult as is shown in the calculations below. With a 3 kilogram counter weight I was able to consistently fire a golf ball 18-19 meters. The shots land typically within a ½ meter radius at this distance.
An avi movie file of the launch can be found here.catapult_files/slaunch.avi.
The same avi movie file in slow motion can be found here.catapult_files/slaunch2.avi
Physics
Conservation of Energy: Gravitational Potential Energy = Kinetic Energy
Range (45 degree angle):
Solving for Range:
This is the same equation one would derive for either a conventional catapult or trebuchet. The advantage of the compound catapult is clear. One wants to minimize while maximizing h. Of course there is a point at which there is not enough torque to move the arm. The limit is . For conventional throwers one cannot exceed h=2r for falling distance. In my Compound Catapult one can drop the mass M the full 75 cm allowed by the rules. One is then able to set the ratio as one chooses. Making the wheel and arm fairly small serves to decrease the moment of inertial (ignored in the calculation above) of the PCV pipe, which will further increase distance.
Construction
My compound catapult was constructed using standard ½ inch PVC pipe. The wheels were 3 ¾ inch laundry line wheels purchased at the hardware store drilled to fit over the ½ inch PVC pipe. Since the laundry wheels were not PVC I had to use PL premium construction adhesive (available at any hardware store) to attach the wheels to the pipe. The throwing arm is approximately 18 inches in length. Roller blade bearings were positioned at each end of the axel pipe. By coincidence they fit nicely inside a PVC ½ inch to ½ inch joint piece with a little sanding. I used a ¼ inch threaded inside the roller blade bearings securing the ends with ¼ inch nuts.
Note in the picture there are two laundry wheels. Once is for the dropping weight while the other acts to precisely stop the arm at 45 degree (adjustable) after multiple turns. My device rotates 2 1/2 turns before releasing.
I designed multiple throwing heads using plastic cups with vertical sides. This is a trial and error procedure but is much easier than constructing a trebuchet sling.
# Competition Hints
- Make throwing heads easily interchangeable
- Be sure not to violate the rule against adding energy with the device. Start arm closer to the ground than it finishes.
- With the stopping mechanism keep it simple. Here are some pictures describing a reliable stopping mechanism.
For the stopping mechanism I used fishing line (the expensive 65 lb non-stretch “fireline” works very well), which slowly tightens pulling the line taught at the projectile’s release. Unfortunately, this design alone has a nasty habit of the line leaving the spool and getting tangled in the throwing arm, thus one loses all control over the device. This is especially bad on a windy day. The solution is to thread a nut or a washer through the line. Use it to weigh the line down and when the device fires, it will slide down to one end of the fishing wire. The control spool has to take a lot of force, though. If not attached properly, it will break. Make this spool connection as strong as possible.
- If you experience troubles with the projectile falling off of the device at random, angle the heads slightly.
- Trajectory is very important. The control of the firing anlge is a major advantage.
- Do target practice. My team came in third because we had only about 6 hours of true target practice (We built this specifically for the state tournament. New technology is hard to test).
- Check out your opponent’s launchers. You might see something very interesting.
- When tying the counterweight use a figure eight knot.
- Many teams use complicated firing mechanisms, which look pretty but don’t work very well. Wedging a PVC pipe underneath or above the arm works surprisingly well.
- Warning- if something goes wrong, a compound catapult has the ability to launch the projectile backwards as well as forwards. This is especially dangerous when using golf balls projectiles. Don’t stand behind the device or put anything valuable behind it.
## Summary
1) The distance h the mass is dropped is independent of the ratio r/R.
2) The throwing arm can be made short thus decreasing its moment of inertia.
3) The design is simple and easy to build.
4) With no slings the device is easier to operate than conventional trebuches.
1) Multiple rotations impose large stresses on device frame.
2) The release point of the projectile must be finely tuned.
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## Package Matching - Google Top Interview Questions
You are given two-dimensional list of integers sales and buyers. Each element in sales contains [day, price] meaning that the package is only available for sale on that day for that price. Each element in buyers contains [payday, amount] meaning that the buyer has that amount of money to spend on payday and afterwards. Given that each buyer can buy at most one package, and each package can be sold to at most one person, return the maximum number of packages that can be bought. Cons
## Permutations to Generate Binary Search Tree - Google Top Interview Questions
You are given a list of unique integers nums. We can create a binary search tree by taking each number in order and inserting it to an initially null binary search tree. Return the number of permutations of nums from which we can generate the same binary search tree. Mod the result by 10 ** 9 + 7. You can assume that the binary search tree does no rebalancing. Constraints 0 ≤ n ≤ 1,000 where n is the length of nums Example 1 Input nums = [2, 1, 3] Output 1 Explanatio
## Race to Finish Line - Google Top Interview Questions
You are driving a car in a one-dimensional line and are currently at position = 0 with speed = 1. You can make one of two moves: Accelerate: position += speed and speed *= 2 Reverse: speed = -1 if speed > 0 otherwise speed = 1. Return the minimum number of moves it would take to reach target. Constraints 1 ≤ target ≤ 100,000 Example 1 Input target = 7 Output 3 Explanation We can accelerate 3 times to reach 7. 0 -> 1 -> 3 -> 7 Example 2 Input t
## Rank of a Matrix - Google Top Interview Questions
You are given a two-dimensional list of integers matrix. Return a new matrix A with the same dimensions such that each A[r][c] is the rank of matrix[r][c]. The ranks of any two elements a and b that are either on the same row or column are calculated by: rank(a) < rank(b) if a < b rank(a) > rank(b) if a > b rank(a) = rank(b) if a = b Ranks are greater than or equal to 1 but are as small as possible. Constraints 0 ≤ n, m ≤ 250 where n and m are the number of rows
## Shortest Bridge - Google Top Interview Questions
Given a two-dimensional list of integers matrix containing 0s and 1s, 0 represents water and 1 represents land. An island is a group of connecting 1s in 4 directions that are either surrounded by 0s or by the edges. Find the shortest bridge that connects two islands. It is guaranteed that there are two and only two islands. Constraints n, m ≤ 250 where n and m are the number of rows and columns in matrix Example 1 Input matrix = [ [0, 1], [1, 0] ] Output
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# How to Pass the Wonderlic test
Whether it be for employment or for college admission, the Wonderlic Test is seen by many as an annoying hurdle to pass, with so many questions you have to answer within such a small amount of time.
With only 12 minutes at your disposal to answer 50 questions from a number of subjects ranging from reading comprehension to mind-boggling inductive reasoning and sequence analysis, it is not uncommon even for the brightest of individuals to panic and get a low score.
Fortunately, we have uncovered a number of tips on how to pass the Wonderlic Test based on how it works and how it is designed.
### 1. Be Familiar with the subjects contained within it
One of the most common reasons people fail the Wonderlic Test is because they didn’t know what to expect of it or merely assumed that it was just some run-of-the-mill IQ test.
While this is true to some extent, there is more to the assessment than just reading comprehension and arithmetics.
In general, people should also sharpen their logical reasoning, data/statistical analysis, and even general knowledge skills, because questions revolving around those subjects are always included in every version of the test.
A good way to do this is to take a jab at a Wonderlic practice test, either online or on paper.
### 2. You do not need to answer every question
If you ever have to take a Wonderlic Test, know this very simple rule: there WILL be questions that are intentionally designed to make you waste time or fail.
This will come in the form of paragraph-long questions or questions that require you to solve for multiple variables or figures.
Remember, you only have 12 minutes to go through the 50 questions in the Wonderlic test, which means you can only spare approximately 14 seconds for each question.
No matter how simple or complex the questions are, they will only give you one point each, and focusing on the ones that you are sure to answer correctly will allow you to secure a good enough base score that you can then increase later on if you still have time left.
So lingering for too long in a single item can prevent you from answering questions where you are more likely to answer correctly within a few seconds.
### 3. Answer as many questions as you can, but…
This may seem like a contradiction considering the previous tip, but in the Wonderlic Test, every point counts.
That is to say, if you think that a question is too tough or a time trap of sorts, the best way to go about this is to make an educated guess.
Even if you are not sure, by answering every question you encounter in the Wonderlic, you still have a chance of getting it right, thereby allowing you to get a higher score compared to just leaving it blank by the end of the test.
This tip is frequently mentioned in every Wonderlic practice test due to the fact that merely making an educated guess can mean passing or failing.
### 4. Focus on the subjects that you are most comfortable with
One of the main reasons why test-takers run out of time is because they feel the need to get every question right.
However, once again, the main obstacle of the Wonderlic Test is its strict time limit, and it would be more economical for you to temporarily skip questions that you have trouble with in favor of those that are easier for you to answer.
So if you’re better with numbers, do those first before doing the logical reasoning or reading comprehension subjects and vice-versa
### 5. Practice, practice, practice!
At the end of the day, the Wonderlic Test is a test like any other, and this means that if you feel the need for it, then making good use of online resources, especially those that offer some kind of Wonderlic Practice Test, may help you tremendously.
By making yourself familiar with the ins and outs of the test through any Wonderlic practice test that you can get your hands on, you can prepare yourself to the best of your abilities and ace the assessment with relative ease.
Once you’ve done this, then the only thing left to do is to take the Wonderlic and get ready for success.
Interesting related article: “Things to consider when job hunting.”
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# Area of 2D Shapes – Definition, Formulas & Examples
Home » Math Vocabulary » Area of 2D Shapes – Definition, Formulas & Examples
## What Is the Area of 2D Shapes?
Area of a 2D shape is the quantity that expresses the region enclosed within the boundary of a two-dimensional shape.
2D stands for two-dimensional. 2D shapes are shapes with two dimensions, such as width and height. These shapes do not have thickness or height.
Examples of 2D shapes are square, rectangle, triangle, circle, etc.
How to find the area of 2D shapes? Let’s understand.
## Area of 2D shapes Using Formula
We can also find the area of 2D shapes using formulas. Different shapes have different formulas for calculating the area. Let’s understand how to find the area of a 2D shape using formula. In this section, you can explore the formulas for the area of 2D shapes.
### Area of a Square
A square is a 2D shape with four equal sides and all four angles equal to $90^{\circ}$.
The area of a square is the product of its two adjacent sides.
Since all sides of the square are equal.
Area of Square $= \text{Side} \times \text{Side}$
Area of Square $= (\text{Side})^{2}$
Area of a Square $= \text{s}^{2}$
We can see that while measuring the area of a square, we consider only the length of its side.
Let’s take an example of the given square and try to find its area.
Area of Square $= (\text{Side})^{2}$
Side of the Square $= 4\; \text{cm}$
Area of Square $= (4)^{2} = 16\; \text{cm}^{2}$
### Area of a Rectangle
Rectangle is a 2D shape with four sides. Opposite sides of a rectangle are equal and all four angles are equal to $90^{\circ}$.
The area of a rectangle is the product of its two adjacent sides.
Area of Rectangle $=$ Length $\times$ Width
Area of Rectangle $= \text{l} \times \text{w}$
We can see that while measuring the area of a rectangle, we consider both the length and width of the rectangle.
For example, we have a rectangle here of length 8 cm and width 3 cm. Let’s try to find its area.
Length of the rectangle, $\text{l} = 8\; \text{cm}$
Width of the rectangle, $\text{w} = 3\; \text{cm}$
Area of the rectangle $= \text{l} \times \text{w}$
Area of the rectangle $= 8 \times 3 = 24\; \text{cm}^{2}$
### Area of a Triangle
A triangle is a 2D shape with three sides (edges) and three vertices. A triangle with vertices A, B, and C is denoted as $\Delta$ ABC.
For a given triangle, where the height of the triangle is h and the base is b, the area of the triangle can be calculated by the formula:
Area of triangle $= \frac{1}{2} \times$ base $\times$ height
Area of triangle $= \frac{1}{2} \times \text{b} \times \text{h}$
Note that the base and height of the triangle are perpendicular to each other.
There are different types of triangles, like the equilateral triangle, isosceles triangle, and right-angled triangle, but the formula for the area of all kinds of triangles is the same.
The following image shows base and height for an equilateral triangle, scalene triangle, and right-angled isosceles triangle. Note that the base equals height in a right angled isosceles triangle.
Let’s take an example of the given triangle ABC.
In the triangle ABC, the base measures 6 units and the height measures 4 units.
Area of triangle $= \frac{1}{2} \times$ base $\times$ height
Area of triangle ABC $= \frac{1}{2} \times \text{b} \times \text{h}$
$= \frac{1}{2} \times 6 \times 4$
$= 12$ sq. units
## Area of 2D Figures by Counting Squares
Look at the shapes given below. We won’t be able to use basic formulas to find out the area of such irregular 2D shapes. In order to calculate the area of two dimensional shapes where we cannot apply any formula, we have another method, which is by counting squares on the grid.
Now, we know that area is the region inside the boundary of the 2D shape, we can express the area in terms of unit squares. A square whose side is of length 1 unit has an area of 1 square unit.
To find the area of 2D figures using this method, we count the number of squares that cover the surface of the figure.
While counting the squares, we follow these rules:
• If the figure is covered by less than half a square, count it as zero.
• If the figure is covered by more than half a square, count it as 1 unit.
• If the figure is covered by exactly half a square, count it as $\frac{1}{2}$ a unit.
• If the figure is covered by a full square, count it as 1 unit.
Let us find the area of the given pentagon.
Area of the pentagon is the total area of all the squares covered $= 6 + 0 + 0 + 6 = 12$ square units
Area of irregular figures can be easily found using this method.
## Fun Facts!
• Among all the shapes with the same perimeter, a circle has the largest area.
• A rectangle with all sides equal is called a square.
• All squares are rectangles but all rectangles are not squares.
• Note that the term surface area is used to represent the total area of all the outer faces of a 3D solid figure. Area is the region bounded within the boundaries of a 2D shape. Thus, the surface area of 2D shapes is nothing but the area of a 2D shape.
## Conclusion
The next time you want to roll out a grass lawn to cover your front yard, you can find the area of the yard to figure out the size of roll-out lawn needed to make your front yard beautiful.
## Solved Examples on Area of 2D Shapes
1. Find the area of the given shape.
Solution: Let’s count the number of unit squares (completely filled, half-filled, more than half-filled, less than half-filled). Completely filled unit squares are marked in the image below for reference.
Area of the irregular 2D shape is the total area of all the squares covered $= 4 + 0 + 0 + 6 = 10\; \text{m}^{2}$
2. Find the length of the rectangle whose area is $35\; \text{m}^{2}$ and width is 5 m.
Solution:
Area $= 35\; \text{m}^{2}$
Width $= 5\; \text{m}$
Area of Rectangle $= \text{l} \times \text{w}$
$35 = \text{l} \times 5$
$355 = \text{l}$
$\text{l} = 7\; \text{m}$
Length of the rectangle is 7 m.
3. The floor of a rectangular hall is to be covered with a carpet 200 cm wide. If the length and width of the hall are 20 m and 18 m respectively, find the cost of the carpet at the rate of $2 per meter. Solution: Area of the hall$= 20 \times 18 = 360\; \text{m}^{2}$Width of the carpet$= 200\; \text{cm} = 200 \div 100 = 2\; \text{m}$Area of carpet$=$Area of hall Therefore, length of the carpet$= \frac{Area of the hall}{Width of the carpet} = \frac{360}{2} = 180\; \text{m}$Rate of the carpet$= \$2$ per meter
Therefore, cost of the carpet $= \$ (180 \times 2) = \$360$
4. Find the area of the given figure.
Solution:
The given figure can be divided into a square and triangles as shown.
Area of the figure $=$ Area of triangle $1 +$ Area of triangle $2 +$ Area of the square
Area of the triangle $1 =$ Area of the triangle $2 = \frac{1}{2} \times$ base $\times$ height
$= \frac{1}{2} \times 5 \times 6$
$= 15\; \text{m}^{2}$
Area of Square $= (\text{Side})^{2}$
$= 5^{2} = 25\; \text{m}^{2}$
Area of the figure $= 15 + 15 + 25 = 55\; \text{m}^{2}$
## Practice Problems on Area of 2D Shapes
1
### Find the area of the hexagon.
$10\; \text{cm}^{2}$
$12\; \text{cm}^{2}$
$14\; \text{cm}^{2}$
$16\; \text{cm}^{2}$
CorrectIncorrect
Correct answer is: $12\; \text{cm}^{2}$
Count the number of squares enclosed by the hexagon.
Complete squares $= 8$
$\gt \frac{1}{2}$ square $= 4$
$\lt \frac{1}{2}$ square $= 4$
Area $= (8 \times 1) + (4 \times 1) + (4 \times 0) = 8 + 4 + 0 = 12\; \text{cm}^{2}$
2
### Find the area of a square whose side is 7 cm.
$7\; \text{cm}^{2}$
$14\; \text{cm}^{2}$
$21\; \text{cm}^{2}$
$49\; \text{cm}^{2}$
CorrectIncorrect
Correct answer is: $49\; \text{cm}^{2}$
Area of Square $=$ Side $\times$ Side $= 7 \times 7 = 49\; \text{cm}^{2}$
3
### A marble tile measures 30 cm by 20 cm. How many tiles will be required to cover a wall of size 6 m by 2 m?
600
300
200
100
CorrectIncorrect
Area of wall $= 6 \times 2 = 12\; \text{m}^{2}$
$= 12,000\; \text{cm}^{2} ( 1\; \text{m}^{2} = 10000\; \text{cm}^{2})$
Area of one tile $= 30 \times 20 = 600\; \text{cm}^{2}$
Therefore, number of tiles $= \frac{Area of the wall}{Area of the tile} = \frac{12000}{600} = 200$
4
### Find the area of the given figure.
$22\; \text{cm}^{2}$
$20\; \text{cm}^{2}$
$18\; \text{cm}^{2}$
$12\; \text{cm}^{2}$
CorrectIncorrect
Correct answer is: $22\; \text{cm}^{2}$
Split the figure into rectangles as shown.
Area of the figure $=$ Area of rectangle A $+$ Area of rectangle B $+$ Area of rectangle C
Area of rectangle $\text{A} = 6 \times 1 = 6\; \text{m}^{2}$
Area of rectangle $\text{B} = 3 \times 2 = 6\; \text{m}^{2}$
Area of rectangle $\text{C} = 6 \times 1 = 6\; \text{m}^{2}$
Area of figure $= 6 + 6 + 6 = 18\; \text{m}^{2}$
## Frequently Asked Questions on Area of 2D Shapes
2D shapes that are made up of straight lines and form a simple closed figure are called polygons.
For example, square, rectangle, and triangle are examples of polygons.
Regular polygons are polygons with equal sides and angles. Example: equilateral triangle, square, regular pentagon, etc. Irregular polygons are polygons with irregular shapes. Their sides and angles are not equal.
Covering a table with a sheet is a real-life application of area of 2D shapes. The whole surface of the table is to be covered with a sheet, so by finding the area of the surface of the table, we can find out the size of the sheet required to cover the surface of the table.
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Base Results Optimized Results Base and Optimized
Base Results Optimized Results Base and Optimized
Base Results Optimized Results Base and Optimized
Base Results Optimized Results Base and Optimized
Base Results Optimized Results Base and Optimized
Base Results Optimized Results Base and Optimized
Base Results Optimized Results Base and Optimized
Base Results Optimized Results Base and Optimized
Manufacturer/Processor Type, Speed, Count, Threads, Processes Includes the manufacturer/processor type, processor speed, number of processors, threads, and number of processes. Move mouse over this column for each row to display additional information, including; manufacturer, system name, interconnect, MPI, affiliation, and submission date.
Run Type Run Type, indicates whether the benchmark was a base run or was optimized.
Processors Processors, this is the number of processors used in the benchmark, entered in the form by the benchmark submitter.
G-HPL ( system performance ) HPL, Solves a randomly generated dense linear system of equations in double floating-point precision (IEEE 64-bit) arithmetic using MPI. The linear system matrix is stored in a two-dimensional block-cyclic fashion and multiple variants of code are provided for computational kernels and communication patterns. The solution method is LU factorization through Gaussian elimination with partial row pivoting followed by a backward substitution. Unit: Tera Flops per Second
G-PTRANS (A=A+B^T, MPI) ( system performance ) PTRANS (A=A+B^T, MPI), Implements a parallel matrix transpose for two-dimensional block-cyclic storage. It is an important benchmark because it exercises the communications of the computer heavily on a realistic problem where pairs of processors communicate with each other simultaneously. It is a useful test of the total communications capacity of the network. Unit: Tera Bytes per Second
G-RandomAccess ( system performance ) Global RandomAccess, also called GUPs, measures the rate at which the computer can update pseudo-random locations of its memory - this rate is expressed in billions (giga) of updates per second (GUP/s). Unit: Giga Updates per Second
EP-STREAM Triad ( per process ) The Embarrassingly Parallel STREAM benchmark is a simple synthetic benchmark program that measures sustainable memory bandwidth and the corresponding computation rate for simple numerical vector kernels. It is run in embarrassingly parallel manner - all computational processes perform the benchmark at the same time, the arithmetic average rate is reported. Unit: Giga Bytes per Second
EP-STREAM-sys ( system performance - derived ) The Embarrassingly Parallel STREAM benchmark is a simple synthetic benchmark program that measures sustainable memory bandwidth and the corresponding computation rate for simple numerical vector kernels. It is run in embarrassingly parallel manner - all computational processes perform the benchmark at the same time, the arithmetic average rate is multiplied by the number of processes to attain this derived value. ( EP-STREAM Triad * MPI Processes ) Unit: Tera Bytes per Second
EP-DGEMM ( per process ) Embarrassingly Parallel DGEMM, benchmark measures the floating-point execution rate of double precision real matrix-matrix multiply performed by the DGEMM subroutine from the BLAS (Basic Linear Algebra Subprograms). It is run in embarrassingly parallel manner - all computational processes perform the benchmark at the same time, the arithmetic average rate is reported. Unit: Giga Flops per Second
G-FFT ( system performance ) Global FFT, performs the same test as FFT but across the entire system by distributing the input vector in block fashion across all the processes. Unit: Tera Flops per Second
Randomly Ordered Ring Bandwidth ( per process ) Randomly Ordered Ring Bandwidth, reports bandwidth achieved in the ring communication pattern. The communicating processes are ordered randomly in the ring (with respect to the natural ordering of the MPI default communicator). The result is averaged over various random assignments of processes in the ring. Unit: Giga Bytes per second
Randomly-Ordered Ring Latency ( per process ) Randomly-Ordered Ring Latency, reports latency in the ring communication pattern. The communicating processes are ordered randomly in the ring (with respect to the natural ordering of the MPI default communicator) in the ring. The result is averaged over various random assignments of processes in the ring. Unit: micro-seconds
Condensed Results - Optimized Runs Only - 41 Systems - Generated on Tue Jun 18 17:07:14 2019
System Information
System - Processor - Speed - Count - Threads - Processes
G-HPL G-PTRANS G-Random
Access
G-FFT EP-STREAM Sys EP-STREAM
EP-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s TB/s Gup/s TFlop/s TB/s GB/s GFlop/s GB/s usec
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 1024
Processses: 1024
System Name: Blue Gene/L
Interconnect: Custom
MPI: MPICH 1.0 customized for Blue Gene/L
Affiliation: Blue Gene Computational Center at IBM T.J. Watson Research Center
Submission Date: 04-11-05
IBM Blue Gene/L PowerPC 440 0.7GHz 1024 1 1024
1.42
0.03
0.13
0.05
0.86
0.84
2.47
0.03
4.83
HPC Challenge Award Winner
2008 - 2nd place - EP-STREAM system: 160 TB/s
2008 - 2nd place - G-RandomAccess: 35 GUPS
2007 - 1st place - G-RandomAccess: 35.5 GUPS
2007 - 1st place - EP-STREAM system: 160 TB/s
2007 - 2nd place - G-FFT: 2.3 Tflop/s
2006 - 1st place - G-RandomAccess: 35 GUPS
2006 - 1st place - EP-STREAM system: 160 TB/s
2006 - 1st place - G-FFT: 2.3 Tflop/s
2005 - 1st place - G-FFT: 2.3 Tflop/s
2005 - 1st place - EP-STREAM system: 160 TB/s
2005 - 1st place - G-RandomAccess: 35 GUPS
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 131072
Processses: 65536
System Name: Blue Gene/L
Interconnect: Custom Torus / Tree
MPI: MPICH2 1.0.1
Submission Date: 11-02-05
IBM Blue Gene/L PowerPC 440 0.7GHz 131072 1 65536
252.30
0.37
35.47
2.31
160.06
2.44
2.07
0.01
7.89
HPC Challenge Award Winner
2008 - 2nd place - G-HPL: 259 Tflop/s
2007 - 1st place - G-HPL: 259 Tflop/s
2006 - 1st place - G-HPL: 259 Tflop/s
2005 - 1st place - G-HPL: 259 Tflop/s
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 131072
Processses: 65536
System Name: Blue Gene/L
Interconnect: Custom Torus / Tree
MPI: MPICH2 1.0.1
Submission Date: 11-02-05
IBM Blue Gene/L PowerPC 440 0.7GHz 131072 1 65536
259.21
0.37
32.98
2.23
159.90
2.44
2.31
0.01
7.78
HPC Challenge Award Winner
2007 - 3rd place - G-HPL: 67 Tflop/s
2007 - 3rd place - G-RandomAccess: 17.3 GUPS
2006 - 2nd place - G-HPL: 67 Tflop/s
2006 - 2nd place - G-RandomAccess: 17 GUPS
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 32768
Processses: 16384
System Name: Blue Gene/L
Interconnect: Blue Gene Custom Interconnect
MPI: MPICH 1.1
Affiliation: IBM T.J. Watson Research Center
Submission Date: 11-04-05
IBM Blue Gene/L PowerPC 440 0.7GHz 32768 1 16384
67.12
0.14
17.29
0.99
39.98
2.44
2.31
0.02
5.88
HPC Challenge Award Winner
2011 - 3rd place - G-RandomAccess: 103 GUPS
2010 - 2nd place - G-RandomAccess: 103 GUPS
2009 - 2nd place - G-RandomAccess: 103 GUPS
2008 - 1st place - G-FFT: 5.08 Tflop/s
2008 - 1st place - G-RandomAccess: 103 GUPS
Manufacturer: IBM
Processor Type: PowerPC 450
Processor Speed: 0.85GHz
Processor Count: 32768
Processses: 32768
System Name: Blue Gene/P
Interconnect: Torus
MPI: MPICH 2
Affiliation: Argonne National Lab - LCF
Submission Date: 11-17-08
IBM Blue Gene/P PowerPC 450 0.85GHz 32768 4 32768
173.36
0.63
103.18
5.08
130.42
3.98
9.68
0.02
6.24
Manufacturer: IBM
Processor Type: IBM PowerPC A2
Processor Speed: 1.6GHz
Processor Count: 49152
Processses: 196608
System Name: Blue Gene/Q (MIRA)
Interconnect: BGQ 5D TORUS
MPI: MPICH2 version 1.5
Affiliation: Argonne Leadership Computing Facility/Argonne National Laboratory
Submission Date: 10-26-14
IBM Blue Gene/Q (MIRA) PowerPC A2 1.6GHz 49152 16 196608
5709.28
12.75
417.79
226.10
1425.83
7.25
29.26
0.07
4.36
HPC Challenge Award Winner
2011 - 2nd place - G-RandomAccess: 117 GUPS
2010 - 1st place - G-RandomAccess: 117 GUPS
2010 - 3rd place - G-HPL: 368 Tflop/s
2009 - 1st place - G-RandomAccess: 117 GUPS
2009 - 3rd place - G-HPL: 368 Tflop/s
Manufacturer: IBM
Processor Type: Power PC 450
Processor Speed: 0.85GHz
Processor Count: 131072
Processses: 32768
System Name: Dawn
Interconnect: Custom Torus + Tree + Barrier
MPI: MPICH2 1.0.7
Affiliation: NNSA - Lawrence Livermore National Laboratory
Submission Date: 11-11-09
IBM Dawn Power PC 450 0.85GHz 131072 4 32768
367.82
0.76
117.13
3.20
130.41
3.98
11.07
0.02
5.59
Manufacturer: IBM
Processor Type: IBM Power7 Quad-Chip module
Processor Speed: 3.836GHz
Processor Count: 1470
Processses: 1470
System Name: IBM Power775
Interconnect: IBM Hub Chip integrated interconnect
MPI: IBM PE MPI release 1206
Affiliation: IBM Development Engineering - DARPA Trial Subset
Submission Date: 07-15-12
IBM Power775 Power7 Quad-Chip module 3.836GHz 1470 32 1470
1067.79
46.01
1571.91
94.86
389.99
265.30
815.38
3.38
5.23
Manufacturer: Fujitsu Ltd.
Processor Type: Fujitsu SPARC64 VIIIfx
Processor Speed: 2GHz
Processor Count: 147456
Processses: 18432
System Name: K computer
Interconnect: Tofu interconnect
MPI: Parallelnavi Technical Computing Language V1.0L20
Affiliation: RIKEN Advanced Institute for Computational Science (AICS)
Submission Date: 10-31-11
Fujitsu Ltd. K computer Fujitsu SPARC64 VIIIfx 2GHz 147456 8 18432
2114.19
5.94
77.61
32.50
797.38
43.26
117.56
0.23
7.19
HPC Challenge Award Winner
2011 - 1st place - G-FFT: 34.7 Tflop/s
2011 - 1st place - G-HPL: 2,118 Tflop/s
2011 - 1st place - G-RandomAccess: 121 GUPS
2011 - 1st place - EP-STREAM system: 812 TB/s
Manufacturer: Fujitsu Ltd.
Processor Type: Fujitsu SPARC64 VIIIfx
Processor Speed: 2GHz
Processor Count: 147456
Processses: 18432
System Name: K computer
Interconnect: Tofu interconnect
MPI: Parallelnavi Technical Computing Language V1.0L20
Affiliation: RIKEN Advanced Institute for Computational Science (AICS)
Submission Date: 11-08-11
Fujitsu Ltd. K computer Fujitsu SPARC64 VIIIfx 2GHz 147456 8 18432
2117.70
5.83
121.10
34.72
812.13
44.06
111.56
0.23
6.69
Manufacturer: Fujitsu
Processor Type: Fujitsu SPARC64 VIIIfx
Processor Speed: 2GHz
Processor Count: 663552
Processses: 82944
System Name: K computer
Interconnect: Tofu Interconnect
MPI: Parallelnavi Technical Computing Language V1.0L20
Affiliation: RIKEN Advanced Institute for Computational Scinece
Submission Date: 10-23-12
Fujitsu K computer SPARC64 VIIIfx 2GHz 663552 8 82944
9795.56
16.55
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205.94
3857.32
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Manufacturer: Fujitsu
Processor Type: Fujitsu SPARC64 VIIIfx
Processor Speed: 2GHz
Processor Count: 82944
Processses: 82944
System Name: K computer
Interconnect: Tofu Interconnect
MPI: Parallelnavi Technical Computing Language V1.0L20
Affiliation: RIKEN Advanced Institute for Computational Scinece
Submission Date: 11-13-16
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Manufacturer: Cray Inc.
Processor Type: Cray X1E
Processor Speed: 1.13GHz
Processor Count: 248
Processses: 248
System Name: mfeg8
Interconnect: Modified 2D Torus
MPI: mpt 2.4
Affiliation: Cray
Submission Date: 06-15-05
Cray Inc. mfeg8 Cray X1E 1.13GHz 248 1 248
3.39
0.07
1.85
-0.00
3.28
13.23
13.56
0.30
14.58
Manufacturer: NEC
Processor Type: NEC SX-7
Processor Speed: 0.552GHz
Processor Count: 32
Processses: 32
System Name: NEC SX-7
Interconnect: non
MPI: MPI/SX 7.0.6
Affiliation: Tohoku University, Information Synergy Center
Submission Date: 03-24-06
NEC SX-7 0.552GHz 32 1 32
0.26
0.04
0.26
0.08
0.88
27.64
8.62
10.13
14.80
Manufacturer: NEC
Processor Type: NEC SX-7
Processor Speed: 0.552GHz
Processor Count: 32
Processses: 2
System Name: NEC SX-7
Interconnect: non
MPI: MPI/SX 7.0.6
Affiliation: Tohoku University, Information Synergy Center
Submission Date: 03-24-06
NEC SX-7 0.552GHz 32 16 2
0.18
0.02
0.15
0.01
0.90
452.36
140.94
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Manufacturer: NEC
Processor Type: NEC SX-8
Processor Speed: 2GHz
Processor Count: 40
Processses: 40
System Name: NEC SX-7C
Interconnect: IXS
MPI: MPI/SX 7.1.3
Affiliation: Tohoku University, Information Synergy Center
Submission Date: 03-24-06
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System Information
System - Processor - Speed - Count - Threads - Processes
G-HPL G-PTRANS G-Random
Access
G-FFT EP-STREAM Sys EP-STREAM
EP-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s TB/s Gup/s TFlop/s TB/s GB/s GFlop/s GB/s usec
Manufacturer: NEC
Processor Type: NEC SX-8
Processor Speed: 2GHz
Processor Count: 40
Processses: 5
System Name: NEC SX-7C
Interconnect: IXS
MPI: MPI/SX 7.1.3
Affiliation: Tohoku University, Information Synergy Center
Submission Date: 03-24-06
NEC SX-7C SX-8 2GHz 40 8 5
0.30
0.02
0.00
0.03
1.44
288.60
114.67
12.40
6.67
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 2.2GHz
Processor Count: 64
Processses: 64
System Name: P5 P575+
Interconnect: HPS
MPI: poe 4.2.2.3
Affiliation: IBM
Submission Date: 05-08-06
IBM P5 P575+ Power5+ 2.2GHz 64 1 64
0.49
0.04
0.26
0.02
0.77
11.96
8.39
0.27
8.99
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 2.2GHz
Processor Count: 128
Processses: 128
System Name: P5 P575+
Interconnect: HPS
MPI: poe 4.2.2.3
Affiliation: IBM
Submission Date: 05-08-06
IBM P5 P575+ Power5+ 2.2GHz 128 1 128
0.99
0.09
0.44
0.04
1.53
11.97
8.46
0.22
9.67
Manufacturer: IBM
Processor Type: IBM Power7
Processor Speed: 3.836GHz
Processor Count: 1989
Processses: 1989
System Name: Power 775
Interconnect: Custom IBM Hub Chip
MPI: IBM PE v1209
Affiliation: IBM Development Engineering
Submission Date: 11-08-12
IBM Power 775 Power7 3.836GHz 1989 32 1989
1343.67
60.47
2020.77
132.66
525.41
264.16
828.06
3.33
6.40
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 12960
Processses: 25920
System Name: Red Storm/XT3
Interconnect: Cray custom
MPI: MPICH 2 v1.0.2
Affiliation: NNSA/Sandia National Laboratories
Submission Date: 11-10-06
Cray Inc. Red Storm/XT3 AMD Opteron 2.4GHz 12960 1 25920
90.99
2.35
29.82
1.53
53.89
2.08
4.40
0.06
15.76
HPC Challenge Award Winner
2008 - 3rd place - G-RandomAccess: 34 GUPS
2007 - 2nd place - EP-STREAM system: 77 TB/s
2007 - 2nd place - G-RandomAccess: 33.6 GUPS
2007 - 2nd place - G-HPL: 94 Tflop/s
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 12800
Processses: 25600
System Name: Red Storm/XT3
Interconnect: Seastar
MPI: xt-mpt/1.5.39 based on MPICH 2.0
Affiliation: DOE/NNSA/Sandia National Laboratories
Submission Date: 11-06-07
Cray Inc. Red Storm/XT3 AMD Opteron 2.4GHz 12800 1 25600
93.58
4.99
33.56
1.52
77.13
3.01
4.40
0.04
19.25
HPC Challenge Award Winner
2008 - 2nd place - G-FFT: 2.87 Tflop/s
2007 - 1st place - G-FFT: 2.8 Tflop/s
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 12960
Processses: 25920
System Name: Red Storm/XT3
Interconnect: Seastar
MPI: xt-mpt/1.5.39 based on MPICH 2.0
Affiliation: DOE/NNSA/Sandia National Laboratories
Submission Date: 11-06-07
Cray Inc. Red Storm/XT3 AMD Opteron 2.4GHz 12960 1 25920
93.24
2.37
29.46
2.87
69.67
2.69
4.40
0.04
19.58
Manufacturer: NEC
Processor Type: NEC SX-9
Processor Speed: 3.2GHz
Processor Count: 32
Processses: 2
System Name: SX-9
Interconnect: IXS
MPI: MPI/SX 8.0.0/ISC
Affiliation: TOHOKU UNIVERSITY
Submission Date: 11-06-08
NEC SX-9 3.2GHz 32 16 2
1.83
0.13
0.10
0.06
5.54
2771.82
1304.81
26.06
5.12
Manufacturer: NEC
Processor Type: NEC SX-9
Processor Speed: 3.2GHz
Processor Count: 256
Processses: 256
System Name: SX-9
Interconnect: IXS
MPI: MPI/SX 8.0.0/ISC
Affiliation: TOHOKU UNIVERSITY
Submission Date: 11-06-08
NEC SX-9 3.2GHz 256 1 256
20.19
0.78
1.40
2.38
43.43
169.66
86.50
3.64
9.40
HPC Challenge Award Winner
2010 - 3rd place - G-FFT: 7.5 Tflop/s
2010 - 3rd place - EP-STREAM system: 233 TB/s
2009 - 3rd place - G-FFT: 7 Tflop/s
2009 - 3rd place - EP-STREAM system: 173 TB/s
2009 - 3rd place - EP-STREAM system: 173 TB/s
2009 - 3rd place - EP-STREAM system: 233 TB/s
Manufacturer: NEC
Processor Type: SX-9
Processor Speed: 3.2GHz
Processor Count: 960
Processses: 960
System Name: SX-9
Interconnect: IXS
MPI: MPI/SX 8.0.10
Affiliation: Japan Agency for Marine-Earth Science and Technology (JAMSTEC)
Submission Date: 11-11-09
NEC SX-9 3.2GHz 960 1 960
79.55
2.32
2.07
6.94
172.98
180.19
89.18
2.51
13.84
Manufacturer: NEC
Processor Type: SX-9
Processor Speed: 3.2GHz
Processor Count: 8
Processses: 2
System Name: SX-9
Interconnect: IXS
MPI: MPI/SX 8.0.10
Affiliation: Japan Agency for Marine-Earth Science and Technology (JAMSTEC)
Submission Date: 11-16-09
NEC SX-9 3.2GHz 8 1 2
0.21
0.07
0.16
0.00
1.52
760.72
330.67
72.60
4.15
Manufacturer: NEC
Processor Type: SX-9
Processor Speed: 3.2GHz
Processor Count: 16
Processses: 2
System Name: SX-9
Interconnect: IXS
MPI: MPI/SX 8.0.10
Affiliation: Japan Agency for Marine-Earth Science and Technology (JAMSTEC)
Submission Date: 11-16-09
NEC SX-9 3.2GHz 16 1 2
0.59
0.10
0.09
0.00
2.93
1464.40
684.94
22.62
6.81
HPC Challenge Award Winner
2011 - 2nd place - G-FFT: 11.9 Tflop/s
Manufacturer: NEC
Processor Type: SX-9
Processor Speed: 3.2GHz
Processor Count: 1280
Processses: 1280
System Name: SX-9
Interconnect: IXS
MPI: MPI/SX 8.0.12a
Affiliation: Japan Agency for Marine-Earth Science and Technology (JAMSTEC)
Submission Date: 11-11-10
NEC SX-9 3.2GHz 1280 1 1280
100.28
3.17
2.58
11.88
233.38
182.33
89.33
2.48
16.46
Manufacturer: Cray Inc.
Processor Type: Cray X1 MSP
Processor Speed: 0.8GHz
Processor Count: 252
Processses: 252
System Name: X1
Interconnect: X1
MPI: MPT 2.4
Affiliation: Oak Ridge National Laboratory
Submission Date: 04-26-04
Cray Inc. X1 Cray MSP 0.8GHz 252 1 252
2.37
0.10
5.48
21.74
0.44
22.64
Manufacturer: Cray Inc.
Processor Type: Cray X1 MSP
Processor Speed: 0.8GHz
Processor Count: 60
Processses: 60
System Name: X1
Interconnect: Cray modified 2D torus
MPI: MPT 2.4
Affiliation: U.S. Army Engineer Research and Development Center Major Shared Resource Center
Submission Date: 04-26-04
Cray Inc. X1 Cray MSP 0.8GHz 60 1 60
0.58
0.03
1.31
21.77
1.01
21.16
Manufacturer: Cray Inc.
Processor Type: Cray X1 MSP
Processor Speed: 0.8GHz
Processor Count: 124
Processses: 124
System Name: X1
Interconnect: Cray modified 2D torus
MPI: MPT.2.3.0.3
Affiliation: Army High Performance Computing Research Center (AHPCRC)
Submission Date: 05-03-04
Cray Inc. X1 Cray MSP 0.8GHz 124 1 124
1.18
0.04
2.70
21.75
0.80
20.85
System Information
System - Processor - Speed - Count - Threads - Processes
G-HPL G-PTRANS G-Random
Access
G-FFT EP-STREAM Sys EP-STREAM
EP-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s TB/s Gup/s TFlop/s TB/s GB/s GFlop/s GB/s usec
Manufacturer: Cray Inc.
Processor Type: Cray X1 MSP
Processor Speed: 0.8GHz
Processor Count: 124
Processses: 124
System Name: X1
Interconnect: Cray modified 2D torus
MPI: MPT 2.3.0.3
Affiliation: Army High Performance Computing Research Center (AHPCRC)
Submission Date: 05-05-04
Cray Inc. X1 Cray MSP 0.8GHz 124 1 124
1.18
0.04
2.70
21.75
0.80
20.85
Manufacturer: Cray Inc.
Processor Type: Cray X1E
Processor Speed: 1.13GHz
Processor Count: 1008
Processses: 1008
System Name: X1
Interconnect: Cray Modified 2D torus
MPI: MPT
Affiliation: DOE/Office of Science/ORNL
Submission Date: 11-02-05
Cray Inc. X1 Cray E 1.13GHz 1008 1 1008
12.27
0.14
7.69
0.25
12.69
12.59
14.18
0.15
16.30
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 5208
Processses: 5208
System Name: XT3
Interconnect: Cray Seastar
MPI: xt-mpt/1.3.07
Affiliation: Oak Ridge National Laboratory, DOE Office of Science
Submission Date: 11-10-05
Cray Inc. XT3 AMD Opteron 2.4GHz 5208 1 5208
20.42
0.94
0.66
0.78
29.32
5.63
4.41
0.20
9.33
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 5208
Processses: 5208
System Name: XT3
Interconnect: Cray Seastar
MPI: xt-mpt/1.3.07
Affiliation: Oak Ridge National Laboratories - DOE Office of Science
Submission Date: 11-12-05
Cray Inc. XT3 AMD Opteron 2.4GHz 5208 1 5208
20.42
0.94
0.66
0.78
29.32
5.63
4.41
0.20
9.33
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 5208
Processses: 5208
System Name: XT3
Interconnect: Cray Seastar
MPI: xt-mpt/1.3.07
Affiliation: Oak Ridge National Lab - DOD Office of Science
Submission Date: 11-12-05
Cray Inc. XT3 AMD Opteron 2.4GHz 5208 1 5208
20.34
0.94
0.69
0.86
29.22
5.61
4.42
0.20
9.18
HPC Challenge Award Winner
2007 - 3rd place - G-FFT: 1.1 Tflop/s
2006 - 2nd place - G-FFT: 1.12 Tflop/s
2006 - 3rd place - G-RandomAccess: 10 GUPS
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 10404
Processses: 10404
System Name: XT3 Dual-Core
Interconnect: Cray SeaStar
MPI: xt-mpt 1.5.25
Affiliation: Oak Ridge National Lab
Submission Date: 11-06-06
Cray Inc. XT3 Dual-Core AMD Opteron 2.6GHz 10404 1 10404
43.51
2.04
10.67
1.12
26.54
2.55
4.79
0.08
17.04
HPC Challenge Award Winner
2010 - 2nd place - G-FFT: 10.7 Tflop/s
2009 - 2nd place - G-FFT: 11 Tflop/s
Manufacturer: Cray, Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 98304
Processses: 32768
System Name: XT5
Interconnect: SeaStar 2+
MPI: MPT 3.4.2
Affiliation: National Institute for Computational Sciences
Submission Date: 11-02-09
Cray, Inc. XT5 AMD Opteron 2.6GHz 98304 3 32768
657.62
1.56
18.50
7.53
127.20
3.88
28.82
0.06
15.45
HPC Challenge Award Winner
2011 - 3rd place - G-FFT: 10.7 Tflop/s
2010 - 1st place - G-FFT: 11.88 Tflop/s
2009 - 1st place - G-FFT: 11 Tflop/s
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 196608
Processses: 65536
System Name: XT5
Interconnect: Seastar
MPI: MPT 3.4.2
Affiliation: Oak Ridge National Laboratory
Submission Date: 11-10-09
Cray XT5 AMD Opteron 2.6GHz 196608 3 65536
1338.67
1.89
36.43
10.70
243.32
3.71
28.39
0.04
15.99
HPC Challenge Award Winner
2011 - 2nd place - EP-STREAM system: 398 TB/s
2010 - 1st place - EP-STREAM system: 398 TB/s
2010 - 3rd place - G-RandomAccess: 38 GUPS
2009 - 1st place - EP-STREAM system: 398 TB/s
2009 - 3rd place - G-RandomAccess: 38 GUPS
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 223112
Processses: 111556
System Name: XT5
Interconnect: Seastar
MPI: MPT 3.4.2
Affiliation: Oak Ridge National Laboratory
Submission Date: 11-10-09
Cray XT5 AMD Opteron 2.6GHz 223112 2 111556
1467.66
13.72
37.69
3.88
398.27
3.57
19.25
0.03
31.09
Select Benchmark Display Here Condensed Results - Base Runs Only Condensed Results - Optimized Runs Only Condensed Results - Base and Optimized Runs Single MPI Process Results - Base Runs Only Single MPI Process Results - Optimized Runs Only Single MPI Process Results - Base and Optimized Runs Embarrassingly Parallel Results - Base Runs Only Embarrassingly Parallel Results - Optimized Runs Only Embarrassingly Parallel Results - Base and Optimized Runs Global and per Processor Results - Base Runs Only Global and per Processor Results - Optimized Runs Only Global and per Processor Results - Base and Optimized Runs Latency and Bandwidth Results - Base Runs Only Latency and Bandwidth Results - Optimized Runs Only Latency and Bandwidth Results - Base and Optimized Runs All Results - Base Runs Only All Results - Optimized Runs Only All Results - Base and Optimized Runs Kiviat Diagram - Base Runs Only Kiviat Diagram - Optimized Runs Only Kiviat Diagram - Base and Optimized Runs Geometric Mean - Base Runs Only Geometric Mean - Optimized Runs Only Geometric Mean - Base and Optimized Runs Weighted Results Export all results to XML Export all results to Excel
Note:Blank fields in the table above are from early benchmark runs that did not include that individual benchmark, in particular G-RandomAccess, G-FFT and EP-DGEMM.
Column Definitions G-HPL ( system performance ) Solves a randomly generated dense linear system of equations in double floating-point precision (IEEE 64-bit) arithmetic using MPI. The linear system matrix is stored in a two-dimensional block-cyclic fashion and multiple variants of code are provided for computational kernels and communication patterns. The solution method is LU factorization through Gaussian elimination with partial row pivoting followed by a backward substitution. Unit: Tera Flops per Second G-PTRANS (A=A+B^T, MPI) ( system performance ) Implements a parallel matrix transpose for two-dimensional block-cyclic storage. It is an important benchmark because it exercises the communications of the computer heavily on a realistic problem where pairs of processors communicate with each other simultaneously. It is a useful test of the total communications capacity of the network. Unit: Tera Bytes per Second G-RandomAccess ( system performance ) Global RandomAccess, also called GUPs, measures the rate at which the computer can update pseudo-random locations of its memory - this rate is expressed in billions (giga) of updates per second (GUP/s). Unit: Giga Updates per Second G-FFT ( system performance ) Global FFT performs the same test as FFT but across the entire system by distributing the input vector in block fashion across all the processes. Unit: Tera Flops per Second EP-STREAM Triad ( per process ) The Embarrassingly Parallel STREAM benchmark is a simple synthetic benchmark program that measures sustainable memory bandwidth and the corresponding computation rate for simple numerical vector kernels. It is run in embarrassingly parallel manner - all computational processes perform the benchmark at the same time, the arithmetic average rate is reported. Unit: Giga Bytes per Second EP-STREAM-sys ( system performance - derived ) The Embarrassingly Parallel STREAM benchmark is a simple synthetic benchmark program that measures sustainable memory bandwidth and the corresponding computation rate for simple numerical vector kernels. It is run in embarrassingly parallel manner - all computational processes perform the benchmark at the same time, the arithmetic average rate is multiplied by the number of processes for this value. ( EP-STREAM Triad * MPI Processes ) Unit: Tera Bytes per Second EP-DGEMM ( per process ) The Embarrassingly Parallel DGEMM benchmark measures the floating-point execution rate of double precision real matrix-matrix multiply performed by the DGEMM subroutine from the BLAS (Basic Linear Algebra Subprograms). It is run in embarrassingly parallel manner - all computational processes perform the benchmark at the same time, the arithmetic average rate is reported. Unit: Giga Flops per Second Random Ring Bandwidth ( per process ) Randomly Ordered Ring Bandwidth, reports bandwidth achieved in the ring communication pattern. The communicating processes are ordered randomly in the ring (with respect to the natural ordering of the MPI default communicator). The result is averaged over various random assignments of processes in the ring. Unit: Giga Bytes per second Random Ring Latency ( per process ) Randomly-Ordered Ring Latency, reports latency in the ring communication pattern. The communicating processes are ordered randomly in the ring (with respect to the natural ordering of the MPI default communicator) in the ring. The result is averaged over various random assignments of processes in the ring. Unit: micro-seconds
Tue Jun 18 17:07:14 2019
0.1479 seconds
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CC-MAIN-2019-26
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http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/305711
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```From: ex [mailto:exeQtor / gmail.com]
# ##################
# There is an array A[N] of N integers. You have to
# compose an array
# Output[N] such that Output[i] will be equal to the
# product of all the elements of A[] except A[i].
#
# Example:
# INPUT:[4, 3, 2, 1, 2]
# OUTPUT:[12, 16, 24, 48, 24]
#
# Note: Solve it without the division operator and in O(n).
# =============================================================
it's friday here, so i guess i'll just join the fun :)
prod= lambda{|a| a.inject(1){|p,x| p*x}}
=> #<Proc:0xb7d708e0@(irb):1>
a=[4,3,2,1,2]
=> [4, 3, 2, 1, 2]
pa=[]
=> []
s=a.size
=> 5
s2=s-1
=> 4
# here is the meat:
# i just concat orig array so i don't need to rotate
# then get subarrays in groups of s2 (a.size-1)
a2=a+a
=> [4, 3, 2, 1, 2, 4, 3, 2, 1, 2]
1.upto(s) do |i|
pa << prod.call(a2[i,s2])
end
=> 1
p pa
[12, 16, 24, 48, 24]
=> nil
| 347
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http://www.tcm.phy.cam.ac.uk/~ajw29/thesis/node59.html
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Next: Implementation of the new Up: Optimising the u Function Previous: Optimising the u Function
Choice of Functional Form for new term in the Jastrow Factor
In the case of optimising , the choice of parameters to optimise was obvious as the function is expressed as a Fourier expansion. In the case of the Jastrow function, this choice is not so clear. The current Jastrow factor has the form
where is given by
In an attempt to improve on the above function it was decided to add an extra term into the exponential in Eq.() to take account not only of the electron-electron separation, , but also of the individual positions of the electrons and . It was suspected that the region where correlation effects are likely to deviate most strongly from the symmetric correlation described by the standard Jastrow function will be close to the ionic cores. Therefore, the any new function, should be short ranged and centred on each of the ions. For simplicity, the function was chosen to be a function of the distances of the 2 electrons from the ion, and and the electron-electron separation , with no angular dependence. It must also obey the following conditions:-
1. It should not cause the total Jastrow term to violate the cusp condition (see section ), i.e. for the extra term .
2. should be well behaved as one of the electrons moves through an ion, i.e. there should be no cusp in or in the 1st derivative as .
3. should take the most general form possible, subject to the above 2 restrictions.
To check that condition was satisfied, the new term, was expanded about
Now condition specifies , therefore
Finally, expanding gives
We chose to keep electron j fixed (i.e. ) and move electron i through it, to test the behaviour as . As the angle between and varies between 0 and the value of will vary smoothly between -1 and +1, (see figure ).
Figure: Dependence of on the angle between and
Therefore, the only solution to Eq.() for all geometries of electrons is .
The final form chosen for the new short range function was therefore
The prefactor in Eq.() performs the following functions; the term removes any terms independent of or terms linear in as specified above. The term is required to satisfy condition , namely that the function be well behaved as one of the electrons moves through the ion. The need for this term in the prefactor was established by performing small simulations using different forms of as one electron moves through an ion. The term enforces the short range nature of by forcing it to decay to zero with zero gradient when one of the electrons is a distance L from the ion. The remaining part of is a general Chebyshev expansion in all three variables; , and .
It should be noted that there are in fact two separate functions required, one dealing with the case where the spins of electrons i and j are parallel and one where they are anti-parallel. This has no effect on the choice of functional form, but it does mean that there are twice as many parameters to be optimised and this reduces the maximum possible number of terms in the Chebyshev expansion.
It is also worth noting that the final form for is very similar to that proposed by Mitas [41], see Eq.(). The difference between the functions is that Mitas only includes even powers of whereas the function used here contains odd and even powers and should therefore be more general.
Next: Implementation of the new Up: Optimising the u Function Previous: Optimising the u Function
Andrew Williamson
Tue Nov 19 17:11:34 GMT 1996
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CC-MAIN-2023-14
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https://www.stata.com/statalist/archive/2004-03/msg00594.html
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# Re: st: Fisher's exact P and -csi-
From [email protected] (Jean Marie Linhart, StataCorp LP) To [email protected] Subject Re: st: Fisher's exact P and -csi- Date Wed, 17 Mar 2004 11:57:36 -0600
```Garry Anderson <[email protected]> found the following bug:
>
> I am enquiring why the value of Fisher's exact P for a 2 x 2 table should
> decrease when there is a smaller sample size for each of the two samples
> and the two proportions remain at 100% and 0%?
>
> For example
> -csi 6 0 0 6,e-
> 1-sided Fisher's exact P = 0.0011
> 2-sided Fisher's exact P = 0.0022
>
> -csi 5 0 0 5,e-
> 1-sided Fisher's exact P = 0.0000
> 2-sided Fisher's exact P = 0.0000
<stuff deleted>
>
> Any suggestions would be appreciated.
>
> (SPSS gives P=0.1 for the 3 0 0 3 combination)
>
> I am using Stata 8.2, 30 Jan 2004, ado 11 Mar 2004.
Al Feiveson also noted this was also a problem with -tabi-.
I have fixed this bug, and the fix will be out in the next executable
update. It was restricted to the Windows version of Stata.
The problem (if anyone cares) was a subtle one and restricted to
these extreme cases -- numbers can be held in a computer at higher
precision than they are stored at. Some compilers treated two numbers
differently, holding one at a higher precision than the other, when
the two numbers should have been exactly the same. Disaster
occurred when these were subtracted and a value of zero was not
obtained.
Some results:
. csi 5 0 0 5, e
<stuff deleted>
1-sided Fisher's exact P = 0.0040
2-sided Fisher's exact P = 0.0079
. csi 3 0 0 3, e
<stuff deleted>
1-sided Fisher's exact P = 0.0500
2-sided Fisher's exact P = 0.1000
. tabi 5 0 \ 0 5, exact
<stuff deleted>
Fisher's exact = 0.008
1-sided Fisher's exact = 0.004
Hope that helps!
--Jean Marie
[email protected]
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
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9 videos (Total 77 min)
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## Week 4
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4-2 Simple Trusses9m
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Division without Remainders
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Multiplication and Division Practice Sheet #2
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Multiplication and Division Practice Sheet #2
Put your child's math skills to the test with some two-digit multiplication and division practice problems.
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Division Review
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Division Review
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Division Crossword
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Division Crossword
Has your student mastered her times tables and division tables? Get a fun review with this division crossword!
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Division Drills
Need to study for an upcoming division quiz? This handy practice test of division drills is just right for the job.
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Division: Word Problems (Part One)
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Division: Word Problems (Part One)
Solve division problems using one of four strategies: by drawing an array, by drawing equal groups, using repeated subtraction, or with a multiplication sentence.
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Long Division with Partial Quotients #1
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Long Division with Partial Quotients #1
Students will use the partial quotient strategy to solve long division problems.
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Division: Repeated Subtraction (Part Two)
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Division: Repeated Subtraction (Part Two)
Practice this important skill by using it to solve division problems.
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Simple Division Word Problems
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Simple Division Word Problems
Break up the boredom of endless division drills with school-themed word problems.
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Division: Word Problems (Part Two)
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Division: Word Problems (Part Two)
Solve division problems using one of four strategies: by drawing an array, by drawing equal groups, by using repeated subtraction, or with a multiplication sentence.
Math
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Practice Finding the Variable #2
Worksheet
Practice Finding the Variable #2
This worksheet lets your fourth grader in on a valuable algebra trick: to find the value of a variable in a division equation, simply use multiplication!
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Math Review Part 3: Let's Soar in Grade 4
Worksheet
Math Review Part 3: Let's Soar in Grade 4
In this review assessment, your students will flex their math muscles as they tackle division word problems.
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Division: Repeated Subtraction (Part One)
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Division: Repeated Subtraction (Part One)
Review the steps of how to perform repeated subtraction to solve a division sentence, then get some practice with this fun matching activity.
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Division: Arrays for Division (Part Two)
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Division: Arrays for Division (Part Two)
In this activity, practice drawing an array for each division problem, then record the quotient on the answer line.
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Long Division with Partial Quotients #2
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Long Division with Partial Quotients #2
In this exercise, students will use the partial quotient strategy to solve long division problems.
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Division Squares
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Division Squares
Practice division in a different way with these division squares. Each row and column are division problems.
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Division: Dividing Gummy Bears into Equal Groups
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Division: Dividing Gummy Bears into Equal Groups
How do you divide gummy bears equally among friends? That depends on how many people want to eat them.
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Is it Divisible by 5?
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Is it Divisible by 5?
Teach your fourth grader how to tell if a number's divisible by 5... right off the bat!
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Quick Trick: Divide Numbers Ending in Zero
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Quick Trick: Divide Numbers Ending in Zero
Teach your kid this division trick for numbers ending in zero: simply divide the rest of the numbers, count up the zeroes, and then add them onto the answer.
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Division: More Shark Munching
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Division: More Shark Munching
Help feed Phil, the division shark by solving the division problems then circling the problems that have no remainders so Phil knows which ones to eat.
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Math-Go-Round: Division (Hard)
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Math-Go-Round: Division (Hard)
Kids with good long division skills will enjoy playing this mathematical board game. Your child and a friend will use math to earn points and win.
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Divvy Out Method for Division Practice
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Divvy Out Method for Division Practice
Show your fourth graders the divvy out method to solve division problems.
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Quotient Tournament
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Quotient Tournament
Division problems stand off against one another as they compete to win in this fourth grade math worksheet.
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Division Shark Muncher
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Division Shark Muncher
Solve the division problems inside each fish, then only circle the fish that have no remainders so that Phil, the picky division shark, knows which ones to eat!
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Division Practice: Find the Mystery Island
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Division Practice: Find the Mystery Island
Practice division by solving the problems and matching each answer with the letter it's paired with to find out which island has the stolen treasure.
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# Can an alphabet be extended in a reduction proof? (with sample problem)
So I am working on solving a problem on whether following language is decideable:
$L = \{n \in \mathbb{N} \mid M_n$ never freezes (for any input)$\}$, where $n$ is the Gödel-number of a Turing machine.
The model for Turing machines that we use operates on a one-side-unbounded tape, so "freezing" in this context means going "out-of-bounds" on the other side of the tape. We also define a Turing machine as the tuple $(K,\Sigma,\delta,s)$, where $K$ is the set of states, $\delta$ is the transition function and $s$ is the starting state ($\notin K$).
Now I have an idea on how to prove it is not decideable by reduction, but it requires me to construct a new DTM which operates on a new alphabet $\Sigma' := \Sigma \cup \{§\}$, and assume that $§ \notin \Sigma$. (where $\Sigma$ is the alphabet of the machine $M_n$)
My idea is as follows: We prove that $L^c=\{n\in\mathbb{N}\mid M_n$freezes for at least one input$\}$ is undecideable. For that, we construct a DTM $Extend$ which takes $n$ as input. $Extend$ modifies the $n$ so that we get a new machine that operates exactly as $M_n$ would, except with the following tweaks:
1. For any (state,letter) pair with the $\delta$ function value $\delta(q,\sigma)=(h,\sigma')$ (where h is the halting state), we replace the function value with $(q_{|K|+1},\sigma')$ (so now the machine never halts on its own)
2. Introduce a new state $q_{|K|+1}$, where $\delta(q_{|K|+1},\sigma)=(q_{|K|+1},\sigma)$ (so if this state is reached the TM goes into an infinite loop).
3. We introduce a new letter to the alphabet, let's say '$§$', and set it so that $\delta(q,§)=(h,\#)$ (where # is the blank symbol) (so it only halts if any point it reads '$§$') All of these are turing-computable steps (if we look at $\delta$ as a table of values for every (state,letter) pair (finite), it's the same as going through the table once, changing values as necessary, adding a row and adding a column).
Now we create a new DTM $M_j$ which works as follows (on input $n$): $>Extend \to L_{\#} \to S_R \to L \to § \to R \to R_{\#}$, where $L_{\#}, R_{\#}$ is a TM which goes left or right respectively until it encounters a blank, and $S_R$ is a TM which shifts the word to the right of it by a blank (so $s,\underline{\#}w\# \vdash^*_{S_R}h,\#\underline{\#}w\#$). So $M_j(w)$ works as follows: $s,\#n\underline{\#} \vdash^*_{M_j} h,§\#n_{Extend}\underline{\#}$
We know that the set $\mathcal{E}=\{ n\in\mathbb{N} \mid M_n$ accepts $\emptyset \}$ is in $\text{co-RE}$. This means $\mathcal{E}^c = \{ j\in \mathbb{N} \mid M_j$ accepts at least one input $\}$ is recursively enumerable, but not decideable. Since our new DTM $M_j$ only accepts iff $n$ would freeze for at least one input, this means that $n\in L^c \Leftrightarrow j\in \mathcal{E}^c$ and the reduction proof would be complete.
But again, I am not sure if the transformation of $n$ I made (by introducing the letter '$§$' to $\Sigma$) is allowed. Can someone clarify this for me?
A (computable) reduction from problem P to problem Q is a (computable) function that takes an instance of P to an instance of Q. In your case $P=\mathcal{E}^c$ and $Q = L^c$. So your reduction is "legal" (that is, conforms to the definition of reduction) if it takes an instance of $\mathcal{E}^c$ to an instance of $L^c$. That's all a function has to do in order to be a reduction.
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# 3.
Analysis of Structural
A R.C. structure is a combination of beams, column, slab & wall, rigid connected
together.
The analysis must begin with the evaluation of all the loads carried by the structure,
including its own weight.
Many of the loads are variable in magnitude & position, and all possible critical
arrangements of loads must be considered.
Force in each member can be determined by one of the following methods:
-
## Applying moment & shear coefficient
Manual calculations
Computer methods
Since the design of R.C. member in generally based on the ultimate limit state, the
3.1
The loads on structural divided into 2 types:
-
Dead loads are those which are normally permanent and constant during the
structures life.
Live loads, on the other hand are varies with magnitude & temporary in nature.
It includes the weight of the structural itself, and all architectural components such as
Once the sizes of all the structural members and the details of the architectural
requirements & permanent fixtures have been established, the dead load can be
calculated quite accurately.
For R.C. structure, self weight is taken as 24kN/m3 and for higher density should be
taken for heavily reinforced on dense (25kN/m3).
In the case of building, the weight of any partitions should be calculated from the
architects drawings.
Dead loads are generally calculated on a slightly conservative basis, so that a member
will not need redesigning because of a small change in its dimensions.
These loads are more difficult to determine accurately. For more of them, it is only
possible to make conservative estimate based on standard codes of practice or past
experience.
Example of imposed loadings are the weight of its occupants, furniture, or machinery;
the pressure of wind, the weight of snow, and of retained earth or water; and the
forces caused by thermal expansion on shrinkage of the concrete.
A large building is unlikely to be carrying its full imposed load simultaneously on all
its floors. For this reason the British Standard Code of Practice allows a reduction in
the total imposed load (flood) when the columns, wall or foundations are designed,
for a building more than 2 storey high.
Similarly, the imposed load may be reduced when designing a beam span which
supports a floor area greater than 40 square meters.
Although the wind load is an imposed, it is kept in a separate category when its partial
factors of safety are specified, and when the load combinations on the structure are
being consider.
3.2
## 3.2.1 Load combinations for the ultimate state
the structure.
For the ultimate limit state the loading combinations to be considered are as follows:
i.
1.4Gk + 1.6Qk
ii.
1.0 Gk + 1.4Wk
iii.
1.2Gk + 1.2Qk + 1.2Wk
The imposed load can cover all or any part of the structure, and therefore should be
arranged to cause the most serve stresses, load combinations (i) should be associated
with minimum design dead load of 1.0Gk applied to such parts of the structure as will
give the most critical condition.
arrangement shown in figure 3.1, in order to cause the maximum sagging moment in
the outer spans & the maximum possible hogging moment in the central span.
Figure 3.2 shows the arrangement of vertical loading on a multi span continuous
beam to cause (i) max. sagging moments in alternate spans and max. possible hogging
moments in adjacent spans, and (ii) max. hogging moment at support A.
As a simplification, BS 8110 allow the ultimate design moment at the support to be
calculated from one loading condition will all spans fully covered with the ultimate
load 1.4Gk + 1.6Qk as shown in part (iii) of fig. 3.2.
## Fig 3.1 Loading arrangement for max. sagging moment at A & C
Fig 3.2 (ii) Loading arrangement for max. support moment at A (support)
Fig 3.2 (iii) Loading for design moment at the supports according to BS 8110.
## 3.2.2 Load combinations for the serviceability limit state
A partial factor of safety of f = 1.0 is usually applied to all load combinations at the
serviceability limit state.
In considering deflections, the imposed load should be arranged to give the worst
effects. The deflections calculated from the load combinations are the immediate
deflections of a structure.
3.3
## Analysis of beams and frame
To design a structure it is necessary to know the bending moments, torsional moments,
shearing forces & axial forces in each member.
Elastic analysis used to determine the distribution of these forces within the structure.
R.C. is a plastic material, a limited redistribution of the elastic moments is sometimes
allowed.
The stiffness of the members can be calculated on the basis of any one of the
following:
-
The entire cross section (ignoring the reinforcement) [easy and simple method to
use]
The concrete cross section plus the transformed area of reinforcement based on
the modular ration (n = Es / Ec).
The compression area only of the concrete cross section, plus the transformed area
of reinforcement based on the modular ration.
A structure should be analysis for each of the critical loading conditions which
produce the max. stresses at any particular section. This procedure will be illustrated
in the examples for a continuous beam and a building frame.
Sign convention
-
Anti clockwise support moment are positive for the fixed end moments (FEM)
Moments causing sagging are positive, while moments causing hogging are
negative.
## 3.3.1 Non continuous beam (one span)
Analysis for bending moments & shearing forces is readily performed manually.
For ultimate limit state, we need only consider the max. load of 1.4Gk + 1.6Qk.
## Example 3.1 Analysis of non continuous beam
The one span simply support beam carries a distributed dead load Gk = 25kN/m, a permanent
concentrated load of 30kN at mid span, Qk = 10kN/m.
a.
b.
Shear diagram
c.
Moment diagram
## 3.3.2 Continuous beams
The methods of analysis for continuous beams may also be applied to continuous slab
which span in one direction.
A continuous beam is considered to have no fixing with the supports so that the beam
is free to rotate. This assumption is not strictly true for beam framing into columns &
for that type of continuous beam it is more accurate to analysis them as part of a
frame.
A continuous beam should be analyzed for the loading arrangements which give the
maximum stresses at each section.
The analysis to calculate the bending moments can be carried out manually by
moment distribution, but tabulated shear and moment coefficients may be adequate
for continuous beams having approximately equal spans & uniformly distributed
3.3.2.1
## The moments at the supports are determined by moment distribution.
It is necessary to calculate the moments in the span & also shear forces on the beam.
For a uniformly distributed load, the equations for the shears and maximum span
moments can be derived from the following analysis.
Using the sign convention of figure 3.5 & taking moment about support B:
Fig. 3.5
## Shear & Moment in a beam
Wl 2
M ba Vab
M ab
2
therefore
Vab
Wl M ab M ba
2
l
and
Vba Wl Vab
M aximum span M max occur at zero shear and distance to zero shear
Vab
W
therefore
a3
Vab
M ab
2W
the points of contraflexure occurs at M 0, that is
M max
Vab x
Wx 2
M ab 0
2W
## Vab Vab 2WM ab
2
so that
Vab Vab 2WM ab
2
a1
W
l Vab Vab 2WM ab
2
a2
A similar analysis can be applied to beams that do not support a uniformly distributed
load. In manual calculations it is usually not considered necessary to calculate the
distance a1, a2 and a3. A sketch of the BM is often adequate.
## Example 3.2 Analysis of a continuous beam
The critical loading arrangements for the ultimate limit state are shown in the fig.
above.
Table 3.1 is the moment distribution carried out for the loading arrangement (1).
The shearing forces, the maximum BM, & their positions along the beam, can be
calculated using the previously derived formulae.
M M BA )
( AB
2
L
306 0 (131.8))
(
2
6
131kN
306 131
Shear VAB
V AB
175kN
2
VAB
M AB
2w
131.0 2
2 x51
168.2kNm
## M ax. for span AB moment
V AB
w
1312
51
2.6m
Distance from A, a 3
## B. M. shown in fig 3.7
Shear shown in Fig. 3.8
The individual B.M. diagram are combined in fig. 3.9a to give the B.M. design
envelop.
Similarly, fig. 3.9b is the shear force design envelop.
For case 1:
wl 2
8
306 x6
8
230kN
FEM BA
wl 2
wl 2
wl 2
FEM CD
FEM CB
8
12
12
100 x 4
100 x 4
306 x6
12
12
12
33kN
33kN
230kN
FEM BC
Stiffness factor
Support
B
Member
BA
Length, m
6
BC
CB
CD
Stiffness
3EI
l
4 EI
l
4 EI
l
3EI
l
Stiffness
1.5
Distribution factor
0.333
0.667
0.667
1.5
0.333
Support
Member
Distribution
factor
FEM, kN
balance, kN
Carry Over, kN
balance, kN
Carry Over, kN
balance, kN
Carry Over, kN
balance, kN
Carry Over, kN
balance, kN
Carry Over, kN
balance, kN
Carry Over, kN
balance, kN
Carry Over, kN
balance, kN
Moment, kNm
A
AB
0.00
0
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0
BA
0.33
B
BC
0.67
CB
0.67
-229.5
65.39
0.00
21.80
0.00
7.27
0.00
2.42
0.00
0.81
0.00
0.27
0.00
0.09
0.00
0.03
33.33
130.78
-65.39
43.59
-21.80
14.53
-7.27
4.84
-2.42
1.61
-0.81
0.54
-0.27
0.18
-0.09
0.06
-33.33
-130.78
65.39
-43.59
21.80
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7.27
-4.84
2.42
-1.61
0.81
-0.54
0.27
-0.18
0.09
-0.06
-131.43
131.43
-131.43
C
CD
0.33
D
DC
0.00
229.5
-65.39
0.00
-21.80
0.00
-7.27
0.00
-2.42
0.00
-0.81
0.00
-0.27
0.00
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0.00
-0.03
131.43
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0.00
0.00
0.00
0.00
0.00
0.00
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0.00
0
3.3.2.2
## The ultimate BM & SF in continuous beams of 3 or more approximately equal span
can be obtained from BS 88110 provided that:
-
## The span differ by no more than 15% of the longest span
The possibility of hogging moments in any of the span should not be ignored.
3.3.3 Structural frames
In situ R.C. structures behave as rigid frames, and should be analyzed as such.
The general procedure for a building frame is to analysis the slabs as continuous
members supported by the beams or structural walls. The slabs can be either one way
spanning or two way spanning.
The column & main beams are considered as a series of rigid plane frames, which can
be divided into two types:
-
## Frames supporting vertical and lateral loads
Type 1 frames are in building where none of the lateral loads, including wind, are
transmitted to the column & beams but are carried by shear walls or other forms of
bracing.
Type 2 frames are designed to carry the lateral loads, which cause bending, shearing
and axial forces in the beams and columns.
For both types of frame the axial forces due to vertical loads in the columns can
normally be calculated as if the beams & slabs were simply supported.
3.3.3.1
## A building frame can be analyzed as a complete frame, or it can be simplified into a
series of substitute frames for analysis. The frame shown in fig. 3.11 can be divided
into any of the sub frames shown in fig. 3.12.
The substitute frame (1) in fig. 3.12 consists of one complete floor beam with its
connecting columns (which are assumed rigidly fixed at their remote ends). An
analysis of this frame will give the following moments and shearing forces in the
beams and columns for the floor level considered.
Substitute frame (2) is a single combined with its connecting columns & two adjacent
spans, all fixed at their remote end. This frame maybe used to determine the bending
moment & shear forces in the central beam. Provided that the central is greater than
the 2 adjacent spans, the BM in the columns can also be found with this frame.
Substitute frame 3 can used to find the moments in the column only. It consists of a
single junction, with the remote ends of the members fixed. This type of sub frame
would be used when the beams have been analyzed as continuous over simple
supports.
In frame (2) & (3), the assumption of fixed ends to the outer beams over estimates
their stiffness. These values are therefore halved to allow to the flexibility resulting
from continuity.
The various critical loading arrangements to produce max. stresses have be considered.
When considering the critical loading arrangements for a column, it is sometimes
necessary to include the case of a max. moment & minimum possible axial load, in
order to investigate of tension failure caused by the bending.
## Example 3.3 Analysis of a substitute frame
Substitute frame shown in fig. 3.13 is part of the complete frame in fig. 3.11.
Characteristic loads carried by beam are Gk = 25kN/m, Qk = 10kN/m, uniformly
distributed along the beam. The analysis of the beam will be carried out by moment
distribution: thus the member stiffness & their relevant distribution factors are first
required.
The moment distribution for the first loading case is shown in table 3.2.
The
distributions for each upper & lower column have been combined, since this
simplifies the layout for the calculations.
The S.F. & max. span moment can be calculated from the formulae of section 3.3.2.
Table 3.2
Figure 3.14
Figure 3.15
Figure 3.16
## Bending moment and shearing force envelopes
A comparison of design envelopes of fig. 3.16 & fig. 3.9 will emphasize the
advantages of considering the concrete beam as part of a frame, not as a continuous
beam as part of a frame, not as a continuous beam as in example 3.2. not only the
analysis of sub frame more precise, but many moments and shears in the beam are
smaller in magnitude.
The moment in each column is given by:
k
M col M col x col
k col
Thus, for the 1st loading arrangement & taking Mcol from table 3.2 gives column
moment
This loading arrangement gives the max. moment, as shown in fig. 3.17
Figure 3.17
Fig. 3.18
Substitute frame
Fig. 3.19
Column moment
## Example 3.4 Analysis of a substitute frame for a column
The substitute frame for this example, shown in fig. 3.18. Gk = 25kN/m & Qk = 10kN/m.
The column moments are illustrated in fig. 3.19. The should be compared with the
corresponding moments for the internal column in Fig. 3.17.
3.3.3.2
## Lateral loads on a structure may be caused by wind pressures, by retained earth, or by
seismic forces.
The analysis for the lateral loads should be kept separate & the force may be
calculated by an elastic analysis by a simplified approximate method. For preliminary
design calculations, and also for medium size regular structures, a simplified
BS 8110 recommends that any simplified form of analysis should assume points of
contra flexure at the mid length of all the column & beams. A suitable approximate
analysis is the cantilever method. It assumes that:
-
Points of contra flexure are located at the point mid points of all columns & beams
The direction avail loads in the column are in proportion to their distances from
the center of gravity of the frame. It is usual to assume that all the columns in a
storey are of equal cross sectional area.
## Example 3.5 Simplified analysis for lateral loads cantilever method
The wind load of 3.0 kN/m is assumed to be transferred to the frame as a concentrated load at
each floor as indicated.
Fig. 3.20
By inspection, there is tension in the 2 column to the left & compression in the columns to
the right; & by assumption (2), the axial forces in columns are proportional to the distance
from the central line of the frame. Thus
Axial forces in exterior column 4 P
## Axial force in interior column
1P
By considering a section through the top storey column as shown in fig. 3.21a. The forces in
this sub frames are calculated as follows:
The calculations of the equivalent forces for the 4th floor (fig. 3.21b) follow a similar
procedure as follow:
B.M. in the beams & columns at their connections can be calculated from these results by the
following formulae,
Beam,
M B Fx 0.5beam span
Columns,
M c Hx0.5storey height
## M B 0.54 x0.5 x6 1.6kNm
M c 0.93x0.5 x3.5 1.6kNm
As a check at each joint, MB = , MC
B.M. due to characteristic wind loads in all the columns & beams of this structure are shown
in fig. 3.22.
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# Properties
Label 103.1 Modulus $103$ Conductor $1$ Order $1$ Real yes Primitive no Minimal yes Parity even
# Learn more
Show commands: PariGP / SageMath
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(103, base_ring=CyclotomicField(2))
M = H._module
chi = DirichletCharacter(H, M([0]))
pari: [g,chi] = znchar(Mod(1,103))
## Basic properties
Modulus: $$103$$ Conductor: $$1$$ sage: chi.conductor() pari: znconreyconductor(g,chi) Order: $$1$$ sage: chi.multiplicative_order() pari: charorder(g,chi) Real: yes Primitive: no, induced from $$\chi_{1}(0,\cdot)$$ sage: chi.is_primitive() pari: #znconreyconductor(g,chi)==1 Minimal: yes Parity: even sage: chi.is_odd() pari: zncharisodd(g,chi)
## Galois orbit 103.a
sage: chi.galois_orbit()
order = charorder(g,chi)
[ charpow(g,chi, k % order) | k <-[1..order-1], gcd(k,order)==1 ]
## Related number fields
Field of values: $$\Q$$ Fixed field: $$\Q$$
## Values on generators
$$5$$ → $$1$$
## First values
$$a$$ $$-1$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$\chi_{ 103 }(1, a)$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$
sage: chi.jacobi_sum(n)
$$\chi_{ 103 }(1,a) \;$$ at $$\;a =$$ e.g. 2
## Gauss sum
sage: chi.gauss_sum(a)
pari: znchargauss(g,chi,a)
$$\tau_{ a }( \chi_{ 103 }(1,·) )\;$$ at $$\;a =$$ e.g. 2
## Jacobi sum
sage: chi.jacobi_sum(n)
$$J(\chi_{ 103 }(1,·),\chi_{ 103 }(n,·)) \;$$ for $$\; n =$$ e.g. 1
## Kloosterman sum
sage: chi.kloosterman_sum(a,b)
$$K(a,b,\chi_{ 103 }(1,·)) \;$$ at $$\; a,b =$$ e.g. 1,2
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https://holooly.com/solutions-v20/discuss-the-cutting-speed-feed-and-depth-of-cut-in-shaping-planning-and-slotting-operations/
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## Textbooks & Solution Manuals
Find the Source, Textbook, Solution Manual that you are looking for in 1 click.
## WriteWise AI Model by Holooly Genius
Your Ultimate AI Essay Writer & Assistant.
## Holooly Arabia
For Arabic Users, find a teacher/tutor in your City or country in the Middle East.
## Holooly Help Desk
Need Help? We got you covered.
Products
## Textbooks & Solution Manuals
Find the Source, Textbook, Solution Manual that you are looking for in 1 click.
## WriteWise AI Model by Holooly Genius
Your Ultimate AI Essay Writer & Assistant.
## Holooly Arabia
For Arabic Users, find a teacher/tutor in your City or country in the Middle East.
## Holooly Help Desk
Need Help? We got you covered.
## Q. 1.3
Discuss the cutting speed, feed and depth of cut in shaping, planning and slotting operations.
## Verified Solution
The cutting speed, V, in planing, shaping and slotting operations is the speed of the tool (or the work) in the direction of the primary cutting motion during the working stroke when the metal is being cut. The cutting speed may be either constant or variable depending upon the design of the machine tool.
Shapers with crank and slotted arm mechanisms have a variable cutting speed and a variable return speed (Fig. 1.70) of the tool. The speed is minimum at the beginning and end of the stroke and maximum in the middle of the stroke. The cutting speed can be determined by the formula:
$V=\frac{L N(1+K)}{1000} m / min .$
where L = Length of ram stroke, mm
N = number of full strokes/min,
that is, a full stroke comprises working and return strokes.
K = Ratio of return time to cuttingtime.
Planers with a rack and pinion drive for the table have a constant cutting speed, which can be determined from the formula :
$V=\frac{2 N L}{1000}, m / min$
The feed, f, in planing, shaping or slotting (Fig. 1.70) is the relative movement of the tool or work in a direction perpendicular to the primary cutting motion per full stroke and is expressed in mm/stroke.
The depth of cut, d, is the thickness of layer of metal removed in one cut or pass. It is measured in a direction perpendicular to the machined surface.
The cross-sectional area of the undeformed chip is the product of the thickness, t, and the width, b, of the undeformed chip :
Area of undeformed chip = t × b = f × d mm².
Cutting speeds in these kinds of machining operations are not usually very high, since difficulties are encountered in designing machine tools with high speed of reciprocating motion due to high inertia forces developed in the motion of the units and components of machine. Thus in planing cast iron, the recommended cutting speed is 15 to 20 m/min for roughing and 4 to 12 m/min for finishing. The corresponding depths of cut are 0.5 to 0.8 mm, and 0.08 mm respectively, with feed ranging from 1.50 to 2.00 mm/stroke.
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How to Work with and Manipulate Text in Excel 2016
## How to Work with and Manipulate Text in Excel 2016
Although Excel is well known for its mathematical capabilities and features, it also has some powerful tools for creating text formulas to manipulate text. We are going to take this article to learn some of the text functions and formulas, then teach you how they are used. More specifically, we will show you how to use them.
## The LEFT and RIGHT Functions
Two of the common text functions you will come across in Excel are LEFT and RIGHT. These functions are used to extract text data from a worksheet.
For example, it is not uncommon for businesses to use the first three letters of an employee's name for identification purposes and payroll. Someone with the last name Anderson might be listed as AND followed by the last three digits of their social security number. They may be listed for payroll purposes as AND325.
We can use the LEFT and RIGHT text function to extract the first three characters of their last name.
Let's take a look how it works by looking at the worksheet below.
We are going to use the LEFT function to extract the first three letters of each employee's last name.
We start a text function the same way that we start a mathematical function.
We can now see the parameters, which are the text which we wish to use, then the number of characters we want to appear.
Select the cell that contains the text, followed by the number of characters.
Press Enter.
The first three letters of the last name are now in the cell.
We will show you how to combine that with the last three numbers of the social security number later in this article.
For now, let's now look the RIGHT function. The RIGHT function extracts characters from the end of text. So, instead of the first three letters of the last name, it will be the last three letters.
We start out by entering the equal sign, then RIGHT.
As you can see, the parameters are the same as with the LEFT function. Select the cell that has the text, then the number of characters you want to extract from the end of the name.
Press Enter.
## The LEN and TRIM Functions
Now that we have covered the LEFT and RIGHT functions, let's talk about two more text functions. These are the LEN and TRIM functions.
Sometimes when you import data into Excel, there can be spaces at the end of data in a cell that you can't see.
If you are trying to use the RIGHT function to extract the last four letters of text, you may end up seeing something like this in your worksheet:
Even though we have used the RIGHT function to extract the last four letters of the text string, you can see that some cells only have three letters in them.
There is a reason for this.
When we imported the data into Excel, there were some spaces at the end of some of the strings of text. So, when the RIGHT function counted four characters from the right, it also counted a space that we did not know was there in two instances in our worksheet. That is why only three characters are showing.
When you import data into Excel, it is typically best to use the TRIM function for cells that contain text. That way, if you want to manipulate that text later using either the RIGHT or LEFT function, the spaces will no longer be there.
That said, we use the TRIM function for text data that we imported.
As you can see above, we have added two new columns to do just that.
Let's enter the trim function. It only requires one parameter, which is the text.
Press Enter.
The employee's first name appears. However, there are no invisible spaces.
Return to the cell where we just added the TRIM function. Drag the handle in the cell to finish out the column. Then, let's do the same with the Last column.
Press Enter, then select the cell again and drag the handle to finish the column.
However, now there is a problem. We have two columns for both the First and Last name. If we try to delete the columns that contain the imported data, since we do not need them anymore, we will get #REF errors. We could enter the names instead of cell references, but that would be a lot of work. All names would have to be entered individually.
Instead, the best way to do this is NOT to create the two extra columns, but instead embed the TRIM function in with the RIGHT function.
In our worksheet below, we have added the RIGHT function and told Excel we want the last three characters of the last name. Then we embedded the TRIM function so that all spaces at the end are trimmed off.
Once you enter TRIM(B2), Excel takes you back to the RIGHT function and prompts you for the num_chars parameter.
The LEN function tells you how many characters are in a string. LEN is short for length.
Let's see how many characters there are in the last name Anderson.
As you can see, it only requires one parameter. We are going to click on cell B2.
Press Enter.
We can see that there are eight characters in the last name Anderson.
Take a look at the worksheet below.
As you can see, we have a list of image file names.
We want to remove the .jpg from the end so only the name of the picture shows up in the next column.
We will use the LEFT and LEN functions to do this.
The first thing we are going to do is determine the length of each file name by using the LEN function.
Next, we are going to use the LEFT function and enter the text parameter.
For the num_chars paremeter, we are going to enter the length found in cell B2 and subtract the number of characters in .jpg, which is 4.
Press Enter.
We now have our image file names without the .jpg after them.
Interested in learning more? Why not take an online class in Advanced Excel 2016?
If we did not want to have a column for length in our worksheet, we could simply embed the LEN function in with the LEFT function, as shown below.
## The FIND and MID Functions
The FIND function gives you the start position of one text string within another, then displays a number which represents the character from which the searched string starts. The MID function lets you extract text from the middle of a string and continue for a specified number of characters.
Let's learn how both functions work, starting with the FIND function.
Look at the example below.
As you can see, we have a list of email addresses. We want to extract the domains from the list in order to find out which domains our friends are using.
To do this, first we have to use the FIND function to 1. Find the @ sign, 2, determine where it occurs in the string. For example, the third character of the string.
We will use the FIND function.
Let's start the function.
Now we can see the parameters for the formula.
First, we need to tell Excel what text we want to find. We want to find the "@" sign within the email address.
The next parameter, within_text, is asking where Excel needs to look for the text.
We select the cell that contains the email address.
The parameter start_num is optional. You would use it to tell Excel to start searching at a certain number of characters into the string. If you are looking for several instances of the character, it is helpful to use.
We do not need to do that, so we add a closing bracket.
Press Enter.
Our @ occurs eight characters into the string.
Next, we are going to use the MID function to extract the domain that comes after the @ symbol.
Let's start the MID function
We can now see the parameters.
The first parameter is the text that we want to manipulate.
This is the cell that contains the email address.
The next parameter is the start_num. This is the number of the first character that we want to extract. That character will be the first one after the @ sign. We want one more than that.
Finally, num_chars is the number of characters that we want to extract from the string. Since everyone's domain name has a different number of characters, this is impossible to know.
To figure this out, we will embed a LEN function, as you can see above. In order to find out how many characters we need to extract, we will subtract the number of characters BEFORE the @ sign from the number of characters in the email address. This will leave us with the number of characters that remain after the @ sign.
Press Enter. You can now see the domain.
Drag the handle in the results cell to finish the rest of the worksheet.
## The CONCATENATE Function
In the first section of this article, we mentioned that it is not uncommon for a business to combine the first three letters of an employee's last name with a string of numbers to come up with an employee ID.
In our worksheet below, we have a list of employees with their social security numbers. We also used the LEFT and RIGHT functions to extract the first three characters from their last name, and the last three numbers from their social security numbers.
What we want to now do is combine the characters in the AbLast column with those in the AbSSN column.
We will do this using the CONCATENATE function.
In the snapshot below, we have started to add our function.
Now, we select the text that we want to use for the function. We want to combine cells D2 and E3.
Press Enter.
If we wanted to add a space between the two different strings of characters, we would add spaces between the cells in the formula by using a quote, a space, then another quote.
That said, if we did not want to have to add columns D and E to our worksheet, we could embed LEFT and RIGHT functions within the CONCATENATE function.
Press Enter.
You can then complete the worksheet.
## The Changing Case Functions
Using text functions, you can change your text to uppercase, lowercase, or title case.
To change text to all uppercase, use the UPPER function, as shown below.
Press Enter.
To change it to lowercase, use the function LOWER.
Press Enter.
To change it to title case, use the PROPER function.
Press Enter.
Now, let's go back to our employee ID's. We want to make the first three letters of the employee ID all uppercase. To do that, we have to embed the UPPER function.
Press Enter.
Now we can drag the handle in D2 to finish the worksheet.
## The REPLACE and SUBSTITUTE Functions
The REPLACE and SUBSTITUTE functions are a lot alike; however, they are not exactly the same thing. To show you what we mean, we are going to show you how each of these functions work.
In our worksheet below, we want to use the REPLACE function to add the employee ID after the first part of the email address.
To do this, start the REPLACE function.
The first parameter is old_text, or the cell that contains the text we want to manipulate.
The start_num is where we want to start adding the new text. This would be the @ symbol for all email addresses, so we use the FIND function to find it.
The parameter num_chars asks for how many characters we want to replace. We do not want to replace any, so we put zero.
The new_text is the new text we want to add. That is the employee ID.
Press Enter.
The SUBSTITUTE function can be used to get the same result.
Let's start the SUBSTITUTE function.
First, we want to enter the text we want to manipulate. For this example, it is the email address.
Next, we enter the old text that we want to replace with the new text. We will enter the @ sign.
Now we enter the new text we want to add. This is the employee ID as well as the @ symbol. We have to add the @ symbol back to the email address. We use the cell reference, an ampersand, and the @ symbol to do this.
The parameter instance_num is optional. It wants to know which instance of the @ symbol we want to replace. Since it only appears one time, we do not need to worry about this.
Press Enter.
## The CHAR Function
Take a look at our worksheet below.
We want to combine these two strings into one, then add a carriage return.
To do this, we would start by using the CONCATENATE function.
Now we need to create a carriage return between A10 and B10. To do this, we will use the CHAR function.
With the CHAR function, we add a number between 1 and 255 to get the character we need.
We need a carriage break. This is number 10.
You can see that we inserted the CHAR function between the two cell references – which is where we want our line break.
Press Enter.
Since there are 255 different characters that you can insert with a CHAR function, it is impossible to list them all in this article.
## Formatting Date and Numbers Using the TEXT Function
We already know how to format dates, texts, and numbers in cells using Excel. That is easy enough to do without a function. However, we cannot format a date or text if a cell contains both.
For example, in our worksheet below we have used the CONCATENATE function to put Jim's name, then his birthday.
Press Enter.
As you can see, Excel entered a number instead of the birthdate. This number is the count of days since Excel's beginning count date. This date is January 1, 1900.
To format the date, we will use the TEXT function.
We will insert it before the date in the formula.
As you can see, we use MM DD YYYY to tell Excel which date format to use. If we wanted to write out the month, we would have entered MMMM DD YYYY.
Press Enter.
To format numbers with the TEXT function use zeroes and commas, if needed. For example, 20,000 would be 0,000 in the TEXT function.
## Preserving String Values
In this article, we have learned to use functions to manipulate text. We have created several cells with functions in them that extract and combine text. This is great as far as our worksheet goes, but the problem comes if you need to copy and paste a column that you have created using these functions into another worksheet. You will get #REF errors. This is because we are pasting a relative path.
In the worksheet below, we tried to paste our employee ID column into a new worksheet.
You can see the #REF errors that we have. To fix this, click the downward arrow to the right of CTRL in the spreadsheet. Choose Values & Number Formatting from the Paste Values section.
You can see that our employee ID's now appear.
Explore
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pedermakocsh
Registered: 16.03.2004
From: Pleasantville
Posted: Friday 29th of Dec 08:19 People, I need some help with my math assignment . It’s a really long one having almost 30 questions and it covers topics such as partial-sums addition method, partial-sums addition method and partial-sums addition method. I’ve been working on it since the past 4 days now and still haven’t been able to crack even a single one of them. Our teacher gave us this homework and went for a vacation, so basically we are all on our own now. Can anyone help me get started ? Can anyone solve some sample questions for me based on those topics; such solutions would help me solve my own questions as well.
AllejHat
Registered: 16.07.2003
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Jrahan
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Cinvil Dlajon Briodx
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Posted: Monday 01st of Jan 10:47 I remember having difficulties with graphing function, algebra formulas and complex fractions. Algebrator is a truly great piece of algebra software. I have used it through several algebra classes - College Algebra, Intermediate algebra and Algebra 2. I would simply type in the problem from a workbook and by clicking on Solve, step by step solution would appear. The program is highly recommended.
ZaleviL
Registered: 14.07.2002
From: floating in the light, never forgotten
Posted: Tuesday 02nd of Jan 08:05 Yeah, I do. Click on this https://algebra-cheat.com/math-10---college-mathematics.html and I promise you that you’ll have no math problems that you can’t solve after using this program.
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## Problem with Euler angles
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nI encountered the following problem:\nA transformation T is given in the form\nT = Ry(tetaN)Rz(phiN)Ry(tetaN-1)Rz(phiN-1)...Ry(teta1)Rz(phi1)Ry(teta0)\nwhere Ry(teta) is rotation around (current) y axis by angle teta and\nRz(phi) is rotation around (current) z axis by angle phi. Angles teta0\n.... tetaN, phi1 ... phiN are given. Task: express T in the form\nT = Rz(gama)Ry(beta)Rz(alfa)\nwhere alfa, beta, and gama are the corresponding Euler angles.\nAny ideas?\nComment: The transformation T can be more general (arbitrary series of\nrotations). I do not know if the special case I present here makes the\nsolution simpler. Anyway, I am interested only in this special case.\n\nSincerely,\nItsik\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>I encountered the following problem:
A transformation T is given in the form
T = Ry(tetaN)Rz(phiN)Ry(tetaN-1)Rz(phiN-1)...Ry(teta1)Rz(phi1)Ry(teta0)
where Ry(teta) is rotation around (current) y axis by angle teta and
$Rz(\phi)$ is rotation around (current) z axis by angle $\phi$. Angles teta0
.... tetaN, phi1 ... phiN are given. Task: express T in the form
$T = Rz(gama)Ry(\beta)Rz(alfa)$
where alfa, $\beta,$ and gama are the corresponding Euler angles.
Any ideas?
Comment: The transformation T can be more general (arbitrary series of
rotations). I do not know if the special case I present here makes the
solution simpler. Anyway, I am interested only in this special case.
Sincerely,
Itsik
PhysOrg.com physics news on PhysOrg.com >> Is there an invisible tug-of-war behind bad hearts and power outages?>> Penetrating the quantum nature of magnetism>> Rethinking the universe: Groundbreaking theory proposed in 1997 suggests a 'multiverse'
"Itsik Weissman" wrote in message news:996cc19e.0408110627.148206c7@posting.google.com... > > > > > I encountered the following problem: > A transformation T is given in the form > T = Ry(tetaN)Rz(phiN)Ry(tetaN-1)Rz(phiN-1)...Ry(teta1)Rz(phi1)Ry(teta0) > where Ry(teta) is rotation around (current) y axis by angle teta and > $Rz(\phi)$ is rotation around (current) z axis by angle $\phi.$ Angles teta0 > ... tetaN, phi1 ... phiN are given. Task: express T in the form > $T = Rz(gama)Ry(\beta)Rz(alfa)$ > where alfa, $\beta,$ and gama are the corresponding Euler angles. > Any ideas? > Comment: The transformation T can be more general (arbitrary series of > rotations). I do not know if the special case I present here makes the > solution simpler. Anyway, I am interested only in this special case. > > Sincerely, > Itsik Itsik, If you have Mathematica, I have at my web site a Mathematica Application on Rotations and Euler Angles. It has a Mathematica package and four notebooks with many graphics and animations illustrating rotations. One animation, for example, allows the user to rotate two books in side by side panels using different rotation sequences. The third notebook on Euler Angles demonstrates routines in the package that allow the user to calculate all 24 possible Euler angle decompositions. There are 12 different rotation sequences and 2 decompositions for each sequence. If your input angles are all numeric, you can multiply all your initial matrices and then decompose into whichever sequence you desire. If you send me a set of input angles, I will do a calculation for you. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/
On 12 Aug 2004 09:29:04 $-0400,$ itsikw@advisol.co.il (Itsik Weissman) wrote: > > > > >I encountered the following problem: >A transformation T is given in the form $>T =$ Ry(tetaN)Rz(phiN)Ry(tetaN-1)Rz(phiN-1)...Ry(teta1)Rz(phi1)Ry(teta0) >where Ry(teta) is rotation around (current) y axis by angle teta and $>Rz(\phi)$ is rotation around (current) z axis by angle $\phi$. Angles teta0 >... tetaN, phi1 ... phiN are given. Task: express T in the form $>T = Rz(gama)Ry(\beta)Rz(alfa)$ >where alfa, $\beta,$ and gama are the corresponding Euler angles. >Any ideas? >Comment: The transformation T can be more general (arbitrary series of >rotations). I do not know if the special case I present here makes the >solution simpler. Anyway, I am interested only in this special case. > >Sincerely, >Itsik The Euler angles are characterized by a "gimbal order" 323 or zyz, being rotation matrices about z then y then z. Quite obviously there is no general method of reducing these to functions of the angles you cite. BUT, take a cascade of 3 successive sets of Euler angles and you can condense them somewhat by linear combination of the X's: XYX $*XYX *$ XYX $= XY(XX)Y(XX)YX$ = XYX'YX"YX You can make linear combinations of the abutting XX's $= X+X$. So that does satisfy your "Comment: The transformation T can be more general (arbitrary series of >rotations)". I have made use of the only linear combinations available. Any other investigation might be to sent a unit vector through the numerical matrix and from the resultant, find the eigenvector, etc.
## Problem with Euler angles
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn article <mqkph0tvikohojqkc7fsboqhsmd27e339e@4ax.com>,\nJohn C. Polasek <jpolasek@cfl.rr.com> wrote:\n>\n\n\nThere\'s a neat newsgroup: comp.graphics.algorithms\n\nThere are some true geniuses there, it appears,\nthree of the most prolific marked by the following\ntrn regexp:\n\n/eberly\\/faqs\\|broeker/tf\n\nThe guy (who calls him/herself ...d\'Faqs) has recently (last three\nweeks?) written the clearest set of (long) articles about\nEuler angles and, in detail, why in many cases they\'re\nnot the things to use, where as quaternions work a lot\nbetter.\n\nNo, not the usual treatment you see in books, seems to me,\nand *very worth reading*.\n\nAgain, in the last three or four weeks there\'s been\ncome incredible threads related to one or another\naspect of Euler angles.\n\nEnjoy!\n\nDavid\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <mqkph0tvikohojqkc7fsboqhsmd27e339e@4ax.com>,
John C. Polasek <jpolasek@cfl.rr.com> wrote:
>
There's a neat newsgroup: comp.graphics.algorithms
There are some true geniuses there, it appears,
three of the most prolific marked by the following
trn regexp:
$$/eberly\/faqs\|broeker/tf$$
The guy (who calls him/herself ...d'Faqs) has recently (last three
weeks?) written the clearest set of (long) articles about
Euler angles and, in detail, why in many cases they're
not the things to use, where as quaternions work a lot
better.
No, not the usual treatment you see in books, seems to me,
and $*very$ worth reading*.
Again, in the last three or four weeks there's been
come incredible threads related to one or another
aspect of Euler angles.
Enjoy!
David
Similar discussions for: Problem with Euler angles Thread Forum Replies Advanced Physics Homework 0 General Math 3 General Math 3 General Math 4 Advanced Physics Homework 2
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Splunk Search
## How to define time period?
Explorer
For example, the "SUBMIT_DATE" is split by date and time. Then define some period of time as a value(A/B/C). Can this be achieved?
(SUBMIT_DATE="2021-03-09 14:30:48.0") ==> Split to "2021-03-09" and "14:30:48.0"
0:00:00 - 8:00:00 = A
8:00:00 - 16:00:00 = B
16:00:00 - 0:00:00 = C
Labels (1)
• ### timechart
1 Solution
SplunkTrust
If I understand this correctly, you want to define a function that outputs "A", "B", or "C" according to which 8-hour interval the time of day in SUBMIT_DATE falls into. In other words, you are looking for a case function.
``````| eval SUBMIT_DATE = split(SUBMIT_DATE, " ")
| eval date = mvindex(SUBMITE_DATE, 0), time_of_day = strptime(mvindex(SUBMIT_DATE, 1), "%H:%M:%S")
| eval shift = case(time_of_day < 28800, "A", 28800 <= time_of_day AND time_of_day < 57600, "B", true(), "C")``````
Does this help? Note that time_of_day is in number of seconds from 00:00:00 whereas date is still in string format. If you want to convert that to epoch, use strptime again.
Also note that is perhaps not the most elegant to split SUBMIT_DATE into substrings before converting to numeric. But the formula should work.
Tags (1)
SplunkTrust
Can you explain what "Then define some period of time as a value(A/B/C)" means? You have already defined A, B, and C in your description. What is missing?
To split SUBMIT_DATE is simple,
``````| eval SUBMIT_DATE = split(SUBMIT_DATE, " ")
| eval part1 = mvindex(SUBMIT_DATE, 0), part2 = mvindex(SUBMIT_DATE, 1)``````
Tags (2)
Explorer
I want to split （SUBMIT_DATE） into “date” and “time”.
Then define a value for 8 hours period, such as below.
0:00:00 - 8:00:00 = A
8:00:00 - 16:00:00 = B
16:00:00 - 0:00:00 = C
SplunkTrust
Still unclear what "a value for 8 hours period" is supposed to mean. In number of seconds? An time (epoch) value starting from that "date" with offset of 8 hours? Some sort of string (as you have already defined)? There can be millions of interpretations of this phrase.
Explorer
According to the data on the system, it should be accurate to the second.
| eval SUBMIT_DATE = split(SUBMIT_DATE, " ")
| eval part1 = mvindex(SUBMIT_DATE, 0), part2 = mvindex(SUBMIT_DATE, 1)
According to the function you give
"00:00:00 > part2 >= 08:00:00" as A
"08:00:00 > part2 >= 16:00:00" as B
"16:00:00 > part2 >= 00:00:00" as C
I wonder how to define it...
SplunkTrust
If I understand this correctly, you want to define a function that outputs "A", "B", or "C" according to which 8-hour interval the time of day in SUBMIT_DATE falls into. In other words, you are looking for a case function.
``````| eval SUBMIT_DATE = split(SUBMIT_DATE, " ")
| eval date = mvindex(SUBMITE_DATE, 0), time_of_day = strptime(mvindex(SUBMIT_DATE, 1), "%H:%M:%S")
| eval shift = case(time_of_day < 28800, "A", 28800 <= time_of_day AND time_of_day < 57600, "B", true(), "C")``````
Does this help? Note that time_of_day is in number of seconds from 00:00:00 whereas date is still in string format. If you want to convert that to epoch, use strptime again.
Also note that is perhaps not the most elegant to split SUBMIT_DATE into substrings before converting to numeric. But the formula should work.
Tags (1)
Explorer
Thank you!
Got the first step done by your query.
| eval SUBMITED_DATE = split(SUBMITED_DATE, " ")
| eval date = mvindex(SUBMITED_DATE, 0), time_of_day = strptime(mvindex(SUBMITED_DATE, 1), "%H:%M:%S")
| eval shift = case(time_of_day < 1664258400, "A", 1664258400 <= time_of_day AND time_of_day < 1664287200, "B", true(), "C")
The next step is to count the number of each month
| timechart span = 1mon count(eval(shift = "A")) as first_shift, count(eval(shift = "B")) as second_shift, count(eval(shift = "C")) as third_shift
| tail 12
| sort _time
As you see my query, I want to count every shift in every month of the last year. But there is something wrong with above query...
SplunkTrust
Without an illustration, it is unclear what "count every shift in every month" really means. If you want to know how many events per month are in each shift, i.e., count by shift, you can use
``````| timechart span=1mon count by shift
| tail 12``````
Is this what you require?
Explorer
Thank you very much! The result is showing now~
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# How do you write a polynomial function given the real zeroes 3-i,5i and coefficient 1?
Jan 10, 2016
$y = {x}^{4} - 6 {x}^{3} + 35 {x}^{2} - 150 x + 250$
#### Explanation:
Complex roots always come in pairs.
$3 - i$ will be a $0$, but so will $3 + i$.
$5 i$ will be a $0$, but so will $- 5 i$.
This gives us the function
$y = \left(x - \left(3 - i\right)\right) \left(x - \left(3 + i\right)\right) \left(x - 5 i\right) \left(x - \left(- 5 i\right)\right)$
$y = \left(x - 3 + i\right) \left(x - 3 - i\right) \left(x - 5 i\right) \left(x + 5 i\right)$
$y = \left(\left(x - 3\right) + i\right) \left(\left(x - 3\right) - i\right) \left({x}^{2} - 25 {i}^{2}\right)$
$y = \left({\left(x - 3\right)}^{2} - {i}^{2}\right) \left({x}^{2} + 25\right)$
$y = \left({x}^{2} - 6 x + 9 + 1\right) \left({x}^{2} + 25\right)$
$y = \left({x}^{2} - 6 x + 10\right) \left({x}^{2} + 25\right)$
$y = {x}^{4} - 6 {x}^{3} + 35 {x}^{2} - 150 x + 250$
Notice how the function only has imaginary roots (never crosses the $x$-axis):
graph{x^4-6x^3+35x^2-150x+250 [-5, 8, -36, 300]}
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Books Books
The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides.
Elements of Geometry and Trigonometry from the Works of A.M. Legendre ... - Page 97
by Charles Davies - 1854 - 432 pages
## Elements of Drawing and Mensuration Applied to the Mechanic Arts: A Book for ...
Charles Davies - Geometrical drawing - 1846 - 254 pages
...triangle equal to ? In every right-angled triangle, the square described on the hypothenuse, is equal to the sum of the squares described on the other two sides. Thus, if ABC be a rightangled triangle, right-angled at C, then will the square D, described on AB,...
## Practical Arithmetic, Uniting the Inductive with the Synthetic Mode of ...
James Bates Thomson - Arithmetic - 1846 - 336 pages
...principle in geometry, that the square described on the hypothenuse of a right-angled triangle, is equal to the sum of the squares described on the other two sides. (Leg. IV. 11. Euc. I. 47.) Thus if the base of the triangle ABC is 4 feet, and the perpendicular 3...
## Higher Arithmetic: Or, The Science and Application of Numbers; Combining the ...
James Bates Thomson - Arithmetic - 1847 - 432 pages
...contains 25 sq. ft. Hence, the square described on the hi/pothenuse of any right-angled triangle, is equal to the sum of the squares described on the other two sides. DBS. Since the square of the hypothenuse BC, is 25, it follows that the , or 5, must be the hypothenuse...
## Higher Arithmetic: Or, The Science and Application of Numbers: Combining the ...
James Bates Thomson - Arithmetic - 1847 - 424 pages
...30. 34967ft-. 371 578. The square described on the hypothenuse of a rightangled triangle, is equal to the sum of the squares described on the other two sides. (Thomson's Legendre, B. IV. 11, Euc. I. 47.) The truth of this principle may be seen from the following...
## Higher Arithmetic; Or, The Science and Application of Numbers: Combining the ...
James Bates Thomson - Arithmetic - 1848 - 422 pages
...575-580.] SQUARE ROOT. 371 578. The square described on the hypothenuse of a rightangled triangle, is equal to the sum of the squares described on the other two sides. (Thomson's Legendre, B. IV. 11, Euc. I. 47.) The truth of this principle may be seen from the following...
## Ticknor's Mensuration, Or, Square and Triangle: Being a Practical and ...
Almon Ticknor - Measurement - 1849 - 144 pages
...therefore AC, BD, are bisected at the point 0. Fig. 25. 26. The square described on the hypotenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides. (Pig. B) Fig. A. Let the triangle ABC be right-angled at A. Having described squares on the three,...
## Elements of Geometry and Trigonometry Translated from the French of A.M ...
Charles Davies - Trigonometry - 1849 - 384 pages
.... X E D ? GI D K PROPOSITION XI. THEOREM. The square described on the hypothenuse of a right angled triangle is equivalent to the sum of the squares described on the other two sides. Let the triangle ABC be right angled at A. Having described squares on the three sides, let fall from A,...
## The Logic and Utility of Mathematics: With the Best Methods of Instruction ...
Charles Davies - Logic - 1850 - 390 pages
...proved for every figure of the class. symbols For example : when we prove that the square Example, described on the hypothenuse of a right-angled triangle...sum of the squares described on the other two sides, we demonstrate the fact for all right-angled triangles. But in analysis, all numbers, all lines, all...
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# Operations on Sets - Union of Sets
• Union of sets
• Some Properties of the Operation of Union
#### definition
The union of two sets A and B is the set C which consists of all those elements which are either in A or in B (including those which are in both). In symbols, we write. A ∪ B = {x : x ∈A or x ∈B}
#### notes
If A and B be any two sets. The union of A and B is the set which consists of all the elements of A and all the elements of B, the common elements being taken only once. The symbol ‘∪’ is used to denote the union. Symbolically, we write A ∪ B and usually read as ‘A union B’.
Example- A= {0,1,2,3} and B= {0,2,4,5}
Solution- A ∪ B= {0,1,2,3,4,5}
Some Properties of the Operation of Union
a) Commutaive law-
The order of sequence to write the elements doesn't matter.
For instance, A ∪ B= B ∪ A
b) Associative law- If you perform opertions on three or more elements then it doesn't matter how you group them, you will end up with same set.
For instance, (A ∪ B)∪ C= A∪ (B ∪ C)
c) Law of identity element- If you combine anything with nothing then you get that anything as result. That means anything union with a null set will result in that anything.
For instance, A ∪ Ø= A
d) Idempotent law- if we take union of two same elements then result will be that single element because we are not taking anything new.
A ∪ A= A
e) Law of U- When a Universal set is combined with a subset then we get Universal set.
U ∪ A = A
If you would like to contribute notes or other learning material, please submit them using the button below.
### Shaalaa.com
Union of Sets [00:06:53]
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once one of my pups found half a roast chicken in the corner of a parking lot and we had to visit that exact same corner every day for about fifty years because for dogs hope springs eternal when it comes to half a roast chicken - darth
Properly used, positive reinforcement is extremely powerful. - B. F. Skinner
Tic-Tac-Toe is a simple game. If both sides play perfectly, neither can win. But if one plays imperfectly, the other can exploit the flaws in the other's strategy.
Does that sound a little like trading?
Reinforcement learning is a machine learning paradigm that can learn behavior to achieve maximum reward in complex dynamic environments, as simple as Tic-Tac-Toe, or as complex as Go, and options trading. In this post, we will try to explain what reinforcement learning is, share code to apply it, and references to learn more about it. First, we'll learn a simple algorithm to play Tic-Tac-Toe, then learn to trade a non-random price series. Finally, we'll talk about how reinforcement learning can master complex financial concepts like option pricing and optimal diversification.
Let's dive in!
## What Is Reinforcement Learning? A Tic-Tac-Toe Example
Here's a simple algorithm that will learn an exploitive Tic-Tac-Toe strategy, and adapt over time if its opponent learns:
1. Make a big table of all possible Tic-Tac-Toe boards.
2. Initialize the table to assign a value of 0 to each board, 1.0 where X has won, -1.0 where O has won.
3. Play with your opponent. At each move, pick the best available move in your table, or if several are tied, pick one at random. Occasionally, make a move at random just to make sure you explore the whole state space, and to keep your opponent on their toes.
4. After each game, back up through all the boards that were played.
Update the value table as follows:
• When X wins, update each board's value part of the way to 1.
• When O wins, update part of the way to -1.
• When they tie, update part of the way to 0.
This is a profoundly dumb algorithm in the finest sense. It knows almost nothing about the dynamics of Tic-Tac-Toe, but it works. It can't reason about the game. It needs a lot of training. It can't generalize to boards it hasn't seen. 1 It doesn't learn a globally optimal strategy, it just tries to find whatever beats its opponent.
But over time, this algorithm learns, it exploits flaws in its opponent's strategy, and if the opponent changes tactics, it adapts.
This is reinforcement learning.
More sophisticated reinforcement learning algorithms enable robots to walk on four or two legsdriverless cars to drive, computers to play Atari and poker and Go and StarCraft and DOTA and online soccer, in some cases better than humans.
Here is some sample Tic-Tac-Toe code. In this post, we'll extend the Tic-Tac-Toe example to deep reinforcement learning, and build a reinforcement learning trading robot.
## Reinforcement Learning Concepts
But first, let's dig a little deeper into how reinforcement learning in general works, its components, and variations.
Figure 1. Markov Decision Process (MDP)
This figure and a few more below are from the lectures of David Silver, a leading reinforcement learning researcher known for the AlphaGo project, among others.
1. At time t, the agent observes the environment state st (the Tic-Tac-Toe board). 2
2. From the set of available actions (the open board squares), the agent takes action at (the best move).
3. The environment updates at the next timestep t+1 to a new state st+1. In Tic-Tac-Toe this is the board resulting from the opponent's move. In a complex environment like a car on a road, the new state may be partly determined by the agent's actions (you turn left) and partly by visible or hidden complexities in the environment (a dog runs into the road). And the new state may be non-deterministic, things can occur randomly, with probabilities dependent on the visible and hidden state and the actions of the agent.
4. The environment generates a reward. In Tic-Tac-Toe you get a reward when you win, lose, or draw. In Space Invaders, you win points at various times when you hit different targets. When training a self-driving car, machine learning engineers design rewards for staying on the road, getting to the destination, including negative rewards for e.g. collisions.
The technical name for this setting is a Markov Decision Process (MDP).
• It's based on the Markov chain model, which has states and probabilities of transitions between states.
• The vanilla Markov chain is extended with actions: at each state the agent can choose an action that influences the transition probabilities. The transition probabilities are a function not just of st but of (st, at).
• Each state transition is associated with a reward (possibly 0).
• Finally, the agent chooses actions based on states, using a policy function π(st) = at.
Reinforcement learning always has an environment with states, actions, transitions between states, rewards, and an agent that acts according to policy. The agent proceeds through a cycle of observing the state, acting, getting a reward and repeating forever, or until some terminal state is reached.
## Policy Function Concepts
The agent's policy function chooses the best action based on the environment state. It may have the following components:
• Model - An internal representation of the environment. Our Tic-Tac-Toe agent has a model of the board, and it knows some state-action pairs result in the same state as other state-action pairs. A fully model-based algorithm explicitly models the full MDP with all transition probabilities, which our Tic-Tac-Toe algorithm doesn't do. Other agents may be model-free. They choose actions without explicitly storing an internal model of the state or modeling state transitions. The model of the environment is implicit in the policy function. 3
• State value function approximator - A way to estimate the value of a state (our big table mapping boards to values).
• State-action value function approximator - A way to estimate the value of an action in a given state, i.e. a state-action pair, commonly termed a Q-value function.
Just as there are many algorithms for regression or classification, there are many reinforcement learning architectures, and new approaches are constantly emerging. Based on which components a reinforcement learning algorithm uses to generate the workflow illustrated in Figure 1, it can be categorized by type.
Figure 2. Variations of Reinforcement Learning
All reinforcement learning variations learn using a similar workflow:
1. Initialize the algorithm with naive, typically random, policy parameters.
2. Using the policy, take actions, observe states before and after actions, gather experience rewards.
3. Fit a model that improves the policy.
4. Go to 2) and iterate, collecting more experience with the improved policy, and continuing to improve it.
As we continue to iterate, we improve the algorithm.
Figure 3. Reinforcement Learning Workflow
## Reinforcement Learning In Context
In a previous post we discussed the differences between paradigms of machine learning:
• Supervised learning: Any algorithm that predicts labeled data. Regression predicts a continuous response variable: next quarter's real GDP growth, next month's stock return (previous discussion here). Classification predicts a categorical response variable: recession or recovery, next month's return quintile (previous discussion here).
• Unsupervised learning: Any algorithm that summarizes or learns about unlabeled data, such as clustering or dimensionality reduction.
Any data set is either labeled or unlabeled. So between supervised and unsupervised, that must cover all the bases, right? It's like those two books, What They Teach You At Harvard Business School and What They Don't Teach You At Harvard Business School. Between the two of them, they must cover the sum of all human knowledge, right?
Nevertheless, reinforcement learning is considered the third major machine learning paradigm. Consider the Tic-Tac-Toe robot:
• The agent doesn't have fixed training data. It discovers data via an unsupervised process and learns a policy.
• The rewards can be viewed as labels generated by a supervisor. But rewards aren't always directly related to one specific prediction or action, or an individual row of data. If the agent shoots a target in Space Invaders, it has to figure out which action or sequence of actions, possibly over many preceding timesteps, contributed to the reward (the credit assignment problem).
• The agent's interactions with the environment shape that environment, help determine what data the learning algorithm subsequently encounters, and generate a feedback loop. A Space Invaders agent changes the world by shooting targets; a self-driving car doesn't modify the road, but its actions modify how other vehicles behave, and what environment the algorithm subsequently encounters.
• In supervised learning, the algorithm optimizes model parameters over training data to minimize a loss function, like mean squared error or cross-entropy. In reinforcement learning, the algorithm optimizes model parameters over the state space it encounters, to maximize the expected reward generated by the MDP over time.
In reinforcement learning, we move beyond prediction to control. Reinforcement learning can be viewed as the application of supervised machine learning to a larger problem of optimal control. We apply supervised prediction methods such as classification and regression to choose the best action to take within the action space and learn behavior policies to maximize reward in a complex dynamic environment.
Many disciplines have encountered settings like this and developed methodologies to address them:
• Business/Operations Research: Dynamic pricing of airline seats or other products to maximize profits under changing inventory, production, demand conditions.
• Economics: Optimal Fed interest rate policy to maintain full employment and low inflation in a dynamic economy.
• Engineering: Auto-pilots, spacecraft navigation, robots, and industrial automation.
• Psychology: Stimulus-response, positive and negative reinforcement.
• Neuroscience: The brain's chemical reward loop, how children learn to walk and talk or catch a ball.
• Mathematics: Control theory, game theory, optimization.
Figure 4. Reinforcement Learning's Relationships with Science and Engineering Disciplines
Figure 5. History Milestones in Reinforcement Learning
## Deep Reinforcement Learning
How do we get from our simple Tic-Tac-Toe algorithm to an algorithm that can drive a car or trade a stock?
Our table lookup is a linear value function approximator. Our linear value function approximator takes a board, represents it as a feature vector (with one one-hot feature for each possible board), and outputs a value that is a linear function of that feature vector, the value score for that board. We dot-multiply the one-hot feature vector by the lookup table values, and we get a linear value function which we use to choose the next move.
We can swap that linear function for a nonlinear function, a neural network. When we do that, we get our first, very crude, deep reinforcement learning algorithm.
Our new deep Q-learning (DQN) algorithm is:
1. Initialize our neural network to random weights.
2. Play a game with our opponent.
3. Append each board we encounter into an n x 9 data array (our predictors) associated with the outcome of the game (our response). Our 9 predictors are the state of each square.
4. Fit the neural network to the predictors (vectors representing boards) and responses (win/lose averages) we've seen. The neural network predicts the value of each board, instead of a table lookup/linear function.
5. Go to 2), gather more data, and continue training to better predict the value of each board.
The more the algorithm plays, the more accurate the function approximator gets, and the better it plays.
This algorithm will learn to play, although it takes a long time to train and makes our initial brute force method even more inefficient. (see code).
But in a nutshell, that is how a self-driving car could work.
• The state is represented by a giant array of inputs from all the onboard cameras and sensors.
• The actions are: turn the steering wheel, accelerate, and brake.
• Positive rewards come from staying on the road and arriving safely at the destination, and negative rewards from breaking traffic laws or colliding.
• The real world provides state transitions.
• And we train a complex neural network to do everything involved in detecting and interpreting all the objects in the environment and navigating from point A to point B.
Table lookup cannot scale to high dimensional or continuous action and state spaces. And a linear function approximator can't learn nonlinear behavior. With deep neural networks, reinforcement learning algorithms can learn complex emergent behavior.
## Reinforcement Learning for Trading: Simple Harmonic Motion
In a trading context, reinforcement learning allows us to use a market signal to create a profitable trading strategy.
• You need a better-than-random prediction to trade profitably. The signal can come from regression, predicting a continuous variable; or classification, predicting a discrete variable such as outperform/underperform (binary classification) or deciles (multinomial classification).
• The reward can be the raw return or risk-adjusted return (Sharpe).
• Reinforcement learning allows you to take a signal and learn a good policy (trading strategy) to maximize the reward (return or risk-adjusted return).
Here's a simple example showing how one can trade using reinforcement learning. This approach is inspired by the paper "Machine Learning For Trading" by Gordon Ritter.
We use simple simulated market data as a stepping stone to more complex trading environments. Let's create a market price time series as a simple sine wave.
Figure 6. Simulated Stock Price Data (Simple Harmonic Motion)
• Initially, we set the price at 102 and price momentum at 0, and 100 as the price 'anchor'.
• At each timestep, the price accelerates toward 100 by an amount proportional to the distance from 100. If the price is 102:
• the distance from 100 is 2.
• the new price momentum is old momentum - 2 * k.
• the new price is old price + momentum.
This is a simple harmonic motion from physics 101, it describes the oscillation of a spring (Hooke's law), a pendulum under small oscillations, and many other periodic systems.
We can view this as an extremely simplified value/momentum model. 100 is the intrinsic value the stock tends toward. The farther away from intrinsic value, the stronger the acceleration back toward intrinsic value. And momentum means that if the stock is trending up or down, the trend takes time to be reversed.
To trade this stock, we use the REINFORCE algorithm, which is a Monte Carlo policy gradient-based method. (We can also use Q-learning, but policy gradient seems to train faster/work better.)
We simulate many episodes of 1000 training days, observe the outcomes, and train our policy after each episode.
1) Initialize a neural network to choose actions based on the state.
• 32 inputs: the last 16 market values (as deviations from 100 or intrinsic value), and the last 16 daily changes. 4
• 2 hidden layers of 16 units.
• 3 outputs of the probabilities of 0, 1, or 2 for short, flat, long respectively (softmax activation, i.e. multinomial classification).
• Initialize neural network θ values at random.
2) Generate one episode trajectory using the current policy. At each timestep:
• Input the current state to the neural network and generate policy probabilities for short/flat/long.
• Sample from the generated probability distribution and take the sampled action.
• Generate a reward based on trading 1 share based on the action taken:
• When our chosen action is 2 (long), the next reward is the change in price at the next timestep.
• When our chosen action is 1 (flat), the next reward is 0.
• When our chosen action is 0 (short), the next reward is the opposite of the change in price at the next timestep.
• Save the observed state, action taken, and reward.
3) At the end of the trajectory, back up and compute a discounted future reward observed at each timestep using the action taken and following the current policy to the end of the episode. 5 Standardize the returns (discounted future rewards) by subtracting the mean and dividing by the standard deviation.
4) Update the policy:
• Compute the gradient vector of each action probability with respect to the policy θ values (the neural network parameters)
• Use the probability gradient to compute the gradient of the expected return overall actions taken with respect to θ. (EV of gradient times return overall probability-weighted actions taken)
• Update each θ by the gradient of expected return w.r.t θ, times a learning rate. In other words, update the policy in the direction that increases the average return overall actions taken as much as possible:
• actions with above-average rewards become more probable
• actions with below-average rewards become less probable
Here is a chart of total reward as we train over 2000 episodes.
Figure 7. Trading Agent Learning Progress (Simple Harmonic Motion)
Finally, here is one sample episode with color-coding by short/flat/long and reward over the course of the episode.
Figure 8. Simulated Stock Trading Results Over One Episode (Simple Harmonic Motion)
It's not perfect, there are a couple of days where it strays from the ideal policy, but it's pretty good!
## Reinforcement Learning for Trading: Simple Harmonic Motion + Noise + Damping
For a more complex example, we take the simple harmonic motion dynamics and add noise + damping.
Figure 9. Simulated Stock Price Data (Simple Harmonic Motion + Noise + Damping)
Figure 10. Trading Agent Learning Progress (Simple Harmonic Motion + Noise + Damping)
Figure 11. Simulated Stock Trading Results Over One Episode (Simple Harmonic Motion + Noise + Damping)
The Ritter paper uses an Ornstein-Uhlenbeck (OU) process. The OU process, like simple harmonic motion, has stronger mean reversion the farther away it is from the mean. But there is no momentum, so unlike simple harmonic motion, in the absence of noise it will not oscillate periodically but just revert asymptotically to the mean.
Here is an OU process plus noise:
Figure 12. Simulated Stock Price Data (OU Process, Random Walk + Mean Reversion)
Figure 13. Trading Agent Learning Progress (OU Process, Random Walk + Mean Reversion)
Figure 14. Simulated Stock Trading Results Over One Episode (OU Process, Random Walk + Mean Reversion)
The Ritter paper applies reinforcement learning to a multiple-stock portfolio. This is fairly straightforward from here by changing the input to be the states from multiple stocks, adding multiple outputs for multiple stocks, and computing the reward as a portfolio return. The Ritter paper also uses Sharpe ratio as the reward, and finds that the algorithm successfully optimizes it, which is a very nice result. The model empirically maximized risk-reward without knowing anything about the stock market dynamics, a covariance matrix, normality, or even how the reward is computed.
But I think this is long enough and sufficient to illustrate the fundamentals of reinforcement learning, and I'll stop here.
The code for the trading experiments is here.
## Advanced technical concepts and RL variations
This table from Wikipedia lists some of the variations of reinforcement learning:
### Monte Carlo vs. Temporal Difference and the strange-loopy bootstrap
• We run an episode.
• We back up from the final timestep to the beginning using observed rewards to compute discounted rewards over the full episode.
• We train by ascending the policy gradient that improves standardized rewards.
Sometimes we don't have short episodes, we have a process that continues forever, or for a very long time. A self-driving car algorithm might have trips with millions of timesteps. An alternative to Monte Carlo training, which runs a full episode and backs up to assign rewards, is temporal difference learning (TD):
• We use a value function that estimates the expected future reward from this state, following the current policy.
• We run one timestep.
• We back up one episode and compute the difference between the expected value we saw at the last timestep and the value after we took this action (the reward from this action, plus the discounted current expected value).
• This improvement is the advantage we got from that action. We train by ascending the θ gradient that improves the probability of the most advantaged actions.
This is a slightly strange magical recursive loop. At the outset, our policy has a random θ. So we are training on the improvement from our fairly random value to the slightly less random value at the next timestep where we know one reward. Nevertheless, as we do this many, many times, the influence of rewards further in the future filters back one step at a time and we bootstrap to a better and better policy.
When we train on the advantage gained using our policy between now and the next step, it's called TD(0). We can also train on the improvement 2 steps into the future, and that's TD(1), and we can do TD(2) and so on. If we do TD(∞) we are continuing through the end of the episode, however long it may be, and we are back to Monte Carlo learning.
Finally, there's a temporal difference algorithm called TD(λ) where we effectively use an exponential moving average of all the TD terms. Setting λ to 0 is effectively TD(0), setting λ to 1 is effectively Monte Carlo, and calibrating λ determines how far into the future we want to peek. With TD(λ), eligibility traces are used to keep track of how much credit to assign to the current action for rewards many steps into the future.
### Revisiting value-based v. policy-based methods
• If we do TD learning using only a state-value neural network function approximator, and our policy is to choose the action resulting in the best state-value, this is a Deep Q-Learning Network (DQN).
• If we use only a policy network, this is a Policy Gradient method.
• If we use a value network and a policy network and train the policy function separately so that it improves the value function as much as possible, this is Actor-Critic learning.
### The exploration vs. exploitation tradeoff.
When we do Q-learning, our policy is to choose the action with the best resulting state-value. However, there is a strong possibility that early in our training one action is always best in some part of the state space. Since we always choose the best action and never try other actions, our training never learns from situations where the other actions may be better.
To avoid this, we perform ε-greedy exploration. Early on, we follow our policy say 60% of the time, and a random action 40% of the time. This allows the algorithm to experience the whole state/action space. 40% is the ε parameter. In practice since our policy network is random at the outset, we typically start with ε at 100% and gradually reduce it to a very small value.
### On-policy vs. off-policy learning
ε-greedy algorithms are 'off-policy' learning since they sometimes act differently from the policy and train on the resulting outcomes, in contrast to algorithms which only take 'on-policy' actions. Policy gradient algorithms sample actions from a probability distribution. Bad actions never have a strictly zero probability, they just get less likely over time. So sampling implicitly trades off an exploration of new actions, vs. exploitation of the best actions.
### TRPO and PPO, or how to avoid falling off a cliff
Finally, we have noticed that sometimes training will fall off a cliff. Through the extreme nonlinearity of neural networks, a small gradient descent update may generate a large change in policy outcomes. Sometimes the policy is much worse and the training optimization has trouble climbing back up, like falling into a crevasse on the way up Mount Everest. One might avoid that with very small steps, i.e. a small learning rate, but then training takes forever. Two popular variations that address this issue are Trust Region Policy Optimization (TRPO) and Proximal Policy Optimization (PPO). Essentially they avoid or penalize updates that change the output policy too much in one update.
I experimented with a few algorithms in the OpenAI gym, using Cartpole and Lunar Lander here. This is a good way to get started learning about RL, along with more resources listed below.
## Takeaways
Like the man in the Molière play who discovers he's been speaking in prose his whole life, you may have been doing reinforcement learning and optimal control your whole life without realizing it.
Reinforcement learning is an important paradigm in machine learning. It can bridge a gap between predicting and doing, between description and prescription. And many markets and economic processes can be impacted by adaptive intelligent agents.
In quant investing, sometimes one finds that a modest predictive R-squared, or a modest change in behavior, avoids the actions with worst outcomes, and leads to a large improvement in returns. At other times, one finds that a significant improvement R-squared offers no investment performance improvement.
Why is that? Hypothetically, suppose you have a stock that in the long run yields a 5% annualized return. And the single best day each year is up 5%.
Suppose you discover a classifier or regression that always gets that one day right, and is no better than random the rest of the time. With a fraction of a percent increase in accuracy or R-squared you almost doubled your expected return.
Suppose you can also get a perfect prediction on all the days when the daily return is close to 0. On those days, it doesn't matter if you are long or not. Under a perverse scenario, you could improve your forecast on every one of those days, and improve your predictive R-squared or classifier accuracy a lot. But if you do that, it won't improve your performance. And if you then miss that one good trade, you worsen the investment outcome, possibly even while improving R-squared.
Prediction and control are two different objectives. If you have a perfect prediction, then you always choose the correct action, and you have perfect control. But you only need a good enough prediction to select the correct action. Beyond that, any further accuracy is superfluous. And you can sometimes select the correct action at the most critical time and can get pretty good control outcomes, with only a modest predictive signal. 6
In the spirit of deep learning, reinforcement learning trains a complex neural network directly on the reward. Training directly on the reward to choose actions will tend to focus the algorithm on the situations where the choice of action makes the biggest difference. If you train for prediction first, and then use prediction for control, the training values any improvement in prediction equally, even if it doesn't improve control. In deep learning, we don't necessarily break down a complex engineering problem into more tractable components. Instead, we ask what complex model would be able to solve a complex problem, and train the model end-to-end for a long time on big data. Often, this yields emergent behavior and unreasonably effective models.
The link between statistical prediction metrics and real-world performance can be elusive. But in general, reinforcement learning will learn to use whatever forecasting edge it has, to generate a maximum reward in a dynamic environment.
How is reinforcement learning different from backtesting? Backtesting exhaustively searches a parameter space for a parameter vector that obtains the best out-of-sample performance. When we use reinforcement learning with a function approximator and gradient descent, we can use much more complex models where there are too many parameter combinations to backtest, but still efficiently explore the search space and possibly obtain a better result. Training complex models end-to-end simultaneously for prediction and control can result in complex emergent behavior that displays almost human-like intelligence, and adaptive behavior.
There is a parallel between reinforcement learning and the adaptive market hypothesis of Andrew Lo. Markets may not be perfectly efficient at all times, but they tend to move in that direction via an evolutionary learning process based on experience.
JPMorgan and others have reportedly implemented reinforcement learning to trade in the real world, see for instance this paper and (more readable) article. In another example, Igor Halperin used reinforcement learning to successfully model the return from options trading without any Black-Scholes formula or assumptions about log-normality, slippage, etc.
Thesis:
• Reinforcement learning can learn complex economic decision-making in many cases better than humans.
• Reinforcement learning can develop concepts like how to maximize risk-reward without knowing the CAPM or Black-Scholes.
• Economic and finance theories can be tested empirically in silico by creating multi-agent reinforcement learning experiments where we just tell agents to maximize a reward and see what behaviors they learn.
• Therefore, we don't need economists to develop fancy models, we don't need traders to execute trading strategies, and intelligent agents will take over real-world economic decisionmaking. Agents will move from trading algos to setting prices on Uber and Amazon, to controlling the traffic lights, congestion pricing, and information flow in a transportation network to keep traffic moving.
Antithesis:
• Reinforcement learning is very data-hungry or sample-inefficient, more suited to intraday trading, not lifetime 20-punchcard type problems.
• High model complexity makes interpretability challenging.
• It doesn't always work. Reinforcement learning can get stuck at local optima or fall off a cliff. You have to take care to not just train on recent experience but also important but rare cases, the way pilots train for equipment failure and upset recovery.
• AI algorithms can be exploited. An adversarial sticker can make image recognition think a banana is a toaster or an adversarial temporary tattoo can defeat facial recognition. Google's traffic algorithm can be gamed. Trading is an adversarial game against highly adaptive competitors. If a market maker algorithm trades on patterns, adversarial algorithms can learn to manipulate it, paint the tape to make it do bad trades.
Synthesis: It's hard to tell what will happen but it will certainly be interesting.
When many agents interact in a market and try to learn an optimal policy at the same time, the environment becomes radically non-stationary. If a self-driving car crashes on an empty track in a single-agent environment, it has to learn which of its actions caused the crash. But in the multi-agent setting, if it crashes into another car, it has to model why the other car did what it did and adust behavior accordingly. With multiple agents interacting, they have to ask themselves not only how much reward to attribute to which past actions they took, but how much credit or blame to attribute to each agent. You get feedback loops and unexpected emergent behavior.In real-world economics or trading settings, maybe agents adapt to each other and arrive at a new, more efficient price equilibrium. Maybe they create extreme algorithm-induced oscillations between multiple equilibria. Maybe they learn to collude? We don't know. It depends on the market structure, the diversity of the robot population, etc. What we do know is it becomes an NP-hard problem to find a multi-agent policy with optimal outcomes.
I'm not sure self-driving vehicles on the streets of New York or New Delhi are likely in the near future, without protected rights of way for self-driving vehicles and strong enforcement. If pedestrians know that the self-driving car is always going to stop for them no matter what, they will learn to just cross at the red light, never mind traffic. They can even wear a stop sign on a T-shirt. It's not a question of how good the self-driving technology is, it's a question of game theory. Knowing that the other driver is a fallible fellow human, who may be angry and honk and give you the finger, and at worst may be on a cell phone and not even see you, just leads to different outcomes.
Reinforcement learning and intelligent agents may offer economics a path out of the DSGE/microfoundations conundrum. Would you rather have a model that empirically works pretty well in the current real-world regime; or a less accurate model based on a strong theoretical foundation and therefore more robust to regime shifts? By complexifying the representative agent, maybe you get more predictive or at least more interesting micro-founded models. You can ask questions like, what conditions are necessary and sufficient for smart Prisoners Dilemma agents to move from the naive Nash equilibrium where they are better off always defecting, to a meta-Nash equilibrium where every agent has no incentive to deviate from the policy of always cooperating.
Economics and finance involve thinking about how systems of interacting, constantly adapting, intelligent agents behave. It seems likely more of them will be non-human, and the whole world will be a lab.
“We shape our tools and thereafter our tools shape us” - John Culkin
Papers and blog posts:
Books:
Courses:
Notes:
1. Even if they are isomorphic to boards it has seen. There are really only 3 starting moves, board center, corner, center side. Flipping or rotating the board shouldn't change the value of a position or how to play it.
2. Or ot, the observable part of the state, in the event the state is not fully observable, and there is some hidden state that determines the evolution of the game, such as face-down cards.
3. For instance, instead of a table with all possible Tic-Tac-Toe boards we could use a table mapping (board, action) pairs to values. Then we wouldn't be modeling internally what happens after a move, i.e. several (board, action) pairs arrive at the same board. We would just evaluate state, action pairs directly without any internal model. That would work pretty much the same, it would just be a bigger table and take longer to train.
4. For our simple harmonic motion with no noise, one input of the last change is sufficient. But a more complex model with 16 days of levels and changes can be applied to a more complex example.
5. In this simple model, we can use a large discount because each action only influences the next trading day. In a more complex environment where the current action can impact rewards far in the future, we would want to take those rewards into account, and we would use a smaller discount.
6. The context here is a price-taking trading algorithm too small to influence the overall market. Of course, an algorithm playing poker or Atari or Google Football tries its utmost to influence the course of the game to its advantage. The most important difference between prediction and control is how the algorithm alters the dynamics of the game, and the resulting feedback loop. A reinforcement learning trading algorithm will tend to reinforce any actions that earn a reward, including influencing the market, painting the tape, or colluding with other market participants.
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# Cambridge Mathematics Lower Secondary – Measure
US\$6.40 / month with 1 month free trial
Discover the magic of measurement with our Cambridge Maths Lower Secondary course! From rulers to scales, embark on a journey of mathematical exploration!
## Description
✓ Ultimate Master Slide Collection:
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Welcome to “Cambridge Mathematics Lower Secondary – Measure,” an engaging online course designed to illuminate the world of measurement for younger students. Following the Cambridge Lower Secondary syllabus, this course ventures into the fundamentals of Angle Geometry, Time, Measurement, and the concepts of Perimeter, Area, Surface Area, and Volume. Through lively, simplified lessons, we transform the abstract world of measure into a concrete, graspable reality, encouraging students to explore and understand the dimensions of the world around them.
### Unravel the Mysteries of Angle Geometry:
Embark on your journey with Angle Geometry, where you’ll learn to identify, measure, and construct angles. This section lays the groundwork for understanding geometric principles and enhancing your spatial reasoning skills.
### Master the Concept of Time:
Time teaches you not just to read clocks and calendars but to understand the passage of time and its measurement. This essential life skill aids in planning, scheduling, and appreciating the value of time.
### Dive into the World of Measurement:
In Measurement, you’re introduced to the tools and units used to quantify the physical world. This module expands your understanding of length, mass, and capacity, connecting mathematics to everyday life.
### Explore Perimeter and Area:
Discover how to calculate the Perimeter and Area of various shapes, from simple rectangles to complex polygons. These skills are vital for solving real-world problems, from designing a garden to wrapping a gift.
### Uncover the Secrets of Surface Area and Volume:
Surface Area and Volume take you deeper into 3-dimensional geometry, exploring how to measure the skin of a shape and the space it occupies. This knowledge is crucial for tasks ranging from packing boxes to filling swimming pools.
### Course Highlights:
• Interactive and Fun Lessons: Engaging content designed to make learning about measurement fun and exciting.
• Practical Applications: Examples that show how measurement is used in daily life, enhancing the relevance and practicality of each concept.
• Self-Paced Learning: Freedom to learn at your own pace, ensuring a thorough understanding of each topic.
### Why Enrol:
“Cambridge Mathematics Lower Secondary – Measure” is more than a mathematics course; it’s an invitation to see the world through a mathematical lens. Whether you’re laying the groundwork for future studies or simply curious about the science of measurement, this course equips you with the knowledge and skills for success.
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# Soda Bottle Magnetometer
This is an activity about Earth's magnetic field. Learners will construct a soda bottle magnetometer, collect data, and analyze the results to detect magnetic storm events. Ideally, learners should collect data for at least a month. If several... (View More)
Audience: Middle school
# Design Challenge - Shape of the Satellites
This is an activity about satellite design. Learners will create a satellite model to determine which shape will provide a steady minimum current output from solar panels, given a fixed position light source. After, as a group, they will assess... (View More)
Audience: Middle school
# International Space Station LABS: Technology Activity 1 Heat Transfer: Keeping Cool in Space
Learners will investigate how to build a space suit that keeps astronauts cool. This is technology activity 1 of 2 found in the ISS L.A.B.S. Educator Resource Guide.
Audience: Middle school
# Properties of Fresh and Sea Water
This is a hands-on lab activity about the properties and states of water. Learners will complete activities using different liquids to understand the three states of matter, explain how the high heat capacity and abundance of liquid water makes life... (View More)
# Does Air Have Weight?
This experimental activity is designed to develop an understanding that air has mass. Students conduct an investigation and observe the change in the position of a bar balancing a balloon inflated with air on one end and a uninflated balloon on the... (View More)
# Barometer Basics
This experimental activity is designed to develop a basic understanding of the interrelationship between temperature and pressure and the structure of a device made to examine this relationship. Resources needed to conduct this activity include two... (View More)
# Electric Current Generated with a Moving Magnet
This is an activity about magnetic induction. Learners will induce a flow of electricity in a wire using a moving bar magnet and measure this flow using a galvanometer, or Am meter. Through discussion, this activity can then be related to magnetic... (View More)
# Compass Needles around a Simple Circuit
This is an activity about electromagnetism. Learners will set up a simple circuit using a battery, wire, and knife switch, and then use a compass to map the magnetic field lines surrounding the wire. Next, they will add a coil of wire to the simple... (View More)
# Can Photosynthesis Occur At Saturn?
In this hands-on experiment, students will investigate the basic principle of photosynthesis and learn how light intensity diminishes as a function of distance from the light source. Questions help learners connect these two ideas to determine if... (View More)
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A Regularized Newton Method with Correction for Unconstrained Convex Optimization
ABSTRACT
In this paper, we present a regularized Newton method (M-RNM) with correction for minimizing a convex function whose Hessian matrices may be singular. At every iteration, not only a RNM step is computed but also two correction steps are computed. We show that if the objective function is LC2, then the method posses globally convergent. Numerical results show that the new algorithm performs very well.
Received 17 January 2016; accepted 12 March 2016; published 15 March 2016
1. Introduction
We consider the unconstrained optimization problem [1] -[3]
(1.1)
where is twice continuously differentiable, whose gradient and Hessian are denoted by and respectively. Throughout this paper, we assume that the solution set of (1.1) is nonempty and denoted by X, and in all cases refers to the 2-norm.
It is well known that is convex if and only if is symmetric positive semidefinite for all. Moreover, if is convex, then if and only if x is a solution of the system of nonlinear equations
(1.2)
Hence, we could get the minimizer of by solving (1.2) [4] -[8] . The Newton method is one of efficient solution methods. At every iteration, it computes the trial step
(1.3)
where and. As we know, if is Lipschitz continuous and nonsingular at the solution, then the Newton method has quadratic convergence. However, this method has an obvious disadvantage when the is singular or near singular. In this case, we may compute the Moore-Penrose step [7] . But the computation of the singular value decomposition to obtain is sometimes prohibitive. Hence, computing a direction that is close to may be a good idea.
To overcome the difficulty caused by the possible singularity of, [9] proposed a regularized Newton method, where the trial step is the solution of the linear equations
(1.4)
where I is the identity matrix. is a positive parameter which is updated from iteration to iteration.
Now we need to consider another question, “how to choose the regularized parameter?” Yamashita and
Fukushima [10] chose and showed that the regularized Newton method has quadratic convergence under the local error bound condition which is weaker than nonsingularity. Fan and Yuan [11] took
with and showed that the Levenberg-Marqulardt method preserves the quadratic convergence. Numerial results ([12] [13] ) show that the choice of performs more stable and preferable.
Inspired by the regularized Newton method [13] with correction for nonlinear equations, we propose a modified regularized Newton method in this paper. At every iteration, the modified regularized Newton method first solves the linear equations
(1.5)
to obtain the Newton step, where is updated from iteration to iteration, and solves the linear equations
(1.6)
to obtain the approximate Newton step.
It is easy to see
(1.7)
Then it solves the linear equations
(1.8)
to obtain the approximate Newton step.
The aim of this paper is to study the convergence properties of the above modified regularized Newton method and do a numerical experiment to test its efficiency.
The paper is organized as follows. In Section 2, we present a new regularized Newton algorithm with correction by trust region technique, and then prove the global convergence of the new algorithm under some suitable conditions. In Section 3, we test the regularized Newton algorithm with correction and compared it with a regularized Newton method. Finally, we conclude the paper in Section 4.
2. The Algorithm and Its Global Convergence
In this section, we first present the new modified regularized Newton algorithm by using trust region technique, then prove the global convergence. First, we give the modified regularized Newton algorithm.
Let and be given by (1.6) and (1.8), respectively. Since the matrix is symmetric and positive definite, is a descent direction of at, however may not be. Hence we prefer to use a trust region technique.
Define the actual reduction of at the kth iteration as
(2.1)
Note that the regularization step is the minimizer of the convex minimization problem
If we let
then it can be proved [4] that is also a solution of the trust region problem
By the famous result given by Powell in [14] , we know that
(2.2)
By some simple calculations, we deduce from (1.7) that
so, we have
(2.3)
Similar to, is not only the minimizer of the problem
but also a solution to the trust region problem
where. Therefore we also have
(2.4)
Based on the inequalities (2.2), (2.3) and (2.4), it is reasonable for us to define the new predicted reduction as
(2.5)
which satisfies
(2.6)
The ratio of the actual reduction to the predicted reduction
(2.7)
plays a key role in deciding whether to accept the trial step and how to adjust the regularized parameter.
The regularized Newton algorithm with correction for unconstrained convex optimization problems is stated as follows.
Algorithm 2.1
Step 1. Given, , ,. Set.
Step 2. If, then stop.
Step 3. Compute.
Solve
(2.8)
to obtain.
Solve
(2.9)
to obtain and set
Solve
(2.10)
to obtain and set
Step 4. Compute. Set
(2.11)
Step 5. Choose as
(2.12)
Set and go step 2.
Before discussing the global convergence of the algorithm above, we make the following assumption.
Assumption 2.1. and are both Lipschitz continuous, that is, there exists a constant, such that
(2.13)
and
(2.14)
It follows from (2.14) that
(2.15)
The following lemma given below shows the relationship between the positive semidefinite matrix and symmetric positive semidefinite matrix.
Lemma 2.1. A real-valued matrix A is positive semidefinite if and only if is positive semidefinite.
Proof. See [4] . ♢
Next, we give the bounds of a positive definite matrix and its inverse.
Lemma 2.2. Suppose A is positive semidefinite. Then,
and
hold for any.
Proof. See [13] . ♢
Theorem 2.1. Under the conditions of Assumption 2.1, if f is bounded below, then Algorithm 2.1 terminates in finite iterations or satisfies
(2.16)
Proof. We prove by contradiction. If the theorem is not true, then there exists a positive and an integer such that
(2.17)
Without loss of generality, we can suppose. Set. Then
Now we will analysis in two cases whether T is finite or not.
Case (1): T is finite. Then there exists an integer such that
By (2.11), we have
Therefore by (2.12) and (2.17), we deduce
(2.18)
Since, , we get from (2.8) and (2.18) that
(2.19)
Duo to (1.7), we get
From (2.10), we obtain
(2.20)
where is a positive constant.
It follows from (2.1) and (2.5) that
(2.21)
Moreover, from (2.6), (2.17), (2.13) and (2.19), we have
(2.22)
for sufficiently large k.
Duo to (2.21) and (2.22), we get
(2.23)
which implies that. Hence, there exists positive constant such that, holds for all large k, which contradicts to (2.18).
Case (2): T is infinite. Then we have from (2.6) and (2.17) that
(2.24)
which implies that
(2.25)
The above equality together with the updating rule of (2.12) means
(2.26)
Similar to (2.20), it follows from (2.25) and (2.26) that
for some positive constant. Then we have
This equality together with (2.24) yields
which implies that
(2.27)
It follows from (2.8), (2.27), (2.26) and (2.20) that
(2.28)
Since from (2.8), we have from (2.13), (2.17) and (2.28) that
which means
(2.29)
By the same analysis as (2.23) we know that
(2.30)
Hence, there exists a positive constant such that holds for all sufficiently large k, which gives a contradiction to (2.29). The proof is completed. ♢
3. Numerical Experiments
In this section, we test the performance of Algorithm 2.1 on the unconstrained nonlinear optimization problem, and compared it with a regularized Newton method without correction. The function to be minimized is
(3.1)
where are constants. It is clear that function is convex and the minimizer set of is
The Hessian is given by
where,. Matrix is positive semidefinite for all x, but
singular as the sum of every column is zero. Since the Hessian is always singular, the Newton method cannot be used to solve nonlinear Equations (1.2). But by using the regularization technique, both regularized Newton method and Algorithm 2.1 work quite well.
The aims of the experiments are as follows: to check whether Algorithm 2.1 converges quadratically as stated in Section 3 and also to see how well the technique of correction works. We set, , , , , and for Algorithm 2.1.
Table 1 reports the norms of at every iteration when, ,
and. Algorithm 2.1 only take four iterations to obtain the minimizer of; decreases very quickly. The results show the sequence quadratic convergence. The iteration is as follows
We may observe that the whole sequence converges to.
We also ran the regularized Newton algorithm (RNA) without correction, that is, we do not solve the linear equations (2.9)-(2.10) and just set the solution of (2.8) to be the trial step. Then, we tested the regularized Newton algorithm without correction and Algorithm 2.1 for various of n, and different choices of the starting point. The results are listed in Table 2.: the selected value of; Dim: the dimension n of the problem;: the ith element; niter: the number of iterations required;: the final value of;: the final value of. We use as the stopping criterion.
Table 1. Results of Algorithm 2.1 to test quadratic convergence.
Table 2. Results of RNA and Algorithm 2.1.
Moreover, we can see for the same, n and, the number of iterations of Algorithm 2.1 is always less than that of RNA. And the correction term does help to improve RNA when the initial point is far away from the minimizer. These facts indicate that the introduction of correction is really useful and could accelerate the convergence of the regularized Newton method.
4. Concluding Remarks
In this paper, we propose a regularized Newton method with correction for unconstrained convex optimization. At every iteration, not only a RNM step is computed but also two correction steps are computed which make use of the previous available Jacobian instead of computing the new Jacobian. Numerical experiments suggest that the introduction of correction is really useful.
Acknowledgements
This research is supported by the National Natural Science Foundation of China (11426155) and the Hujiang Foundation of China (B14005).
NOTES
*Corresponding author.
Cite this paper
Li, L. , Qin, M. and Wang, H. (2016) A Regularized Newton Method with Correction for Unconstrained Convex Optimization. Open Journal of Optimization, 5, 44-52. doi: 10.4236/ojop.2016.51006.
References
[1] Yang, W.W., Yang, Y.T., Zhang, C.H. and Cao, M.Y. (2013) A Newton-Like Trust Region Method for Large-Scale Unconstrained Nonconvex Minimization. Abstract and Applied Analysis, 2013.
[2] Polyak, R.A. (2009) Regularized Newton Method for Unconstrained Convex Optimization. Mathematical Programming Series B, 120, 125-145.
http://dx.doi.org/10.1007/s10107-007-0143-3
[3] Shen, C.G., Chen, X.D. and Liang, Y.M. (2012) A Regularized Newton Method for Degenerate Unconstrained Optimization Problems. Optimization Letters, 6, 1913-1933.
http://dx.doi.org/10.1007/s11590-011-0386-z
[4] Sun, W. and Yuan, Y. (2006) Optimization Theory and Methods. Springer Science and Business Media, LLC, New York.
[5] Zhou, W. and Li, D. (2008) A Globally Convergent BFGS Method for Nonlinear Monotone Equations without Any merit Functions. Mathematics of Computation, 77, 2231-2240.
http://dx.doi.org/10.1090/S0025-5718-08-02121-2
[6] Argyros, I.K. and Hilout, S. (2012) On the Convergence of Damped Newton Method. Applied Mathematics and Computation, 219, 2808-2824.
http://dx.doi.org/10.1016/j.amc.2012.09.011
[7] Nashed, M.Z. and Chen, X. (1993) Convergence of Newton-Like Method for Singular Operator Equations Using Outer Inverses. Numerische Mathematik, 66, 235-257.
http://dx.doi.org/10.1007/BF01385696
[8] Kelley, C.T. (1999) Iterative Methods for Optimization. In: Frontiers in Applied Mathematics, Vol. 18, SIAM, Philadelphia, 2.
http://dx.doi.org/10.1137/1.9781611970920
[9] Sun, D. (1999) A Regularization Newton Method for Solving Nonlinear Complementarity Problems. Applied Mathematics and Optimization, 40, 315-339.
http://dx.doi.org/10.1007/s002459900128
[10] Li, D.H., Fukushima, M., Qi, L. and Yamashita, N. (2004) Regularized Newton Methods for Convex Minimization Problems with Singular Solutions. Computational Optimization and Applications, 28, 131-147.
http://dx.doi.org/10.1023/B:COAP.0000026881.96694.32
[11] Fan, J.Y. and Yuan, Y.X. (2005) On the Quadratic Convergence of the Levenberg-Marquardt Method without Nonsingularity Assumption. Computing, 74, 23-39.
http://dx.doi.org/10.1007/s00607-004-0083-1
[12] Fan, J.Y. and Pan, J.Y. (2009) A Note on the Levenberge-Marquardt Parameter. Applied Mathematics and Computation, 207, 351-165.
http://dx.doi.org/10.1016/j.amc.2008.10.056
[13] Fan, J.Y. and Yuan, Y.X. (2014) A Regularized Newton Method for Monotone Nonlinear Equations and Its Application. Optimization Methods and Software, 29, 102-119.
http://dx.doi.org/10.1080/10556788.2012.746344
[14] Powell, M.J.D. (1975) Convergence Properties of a Class of Minimization Algorithms. In: Mangasarian, O.L., Meyer, R.R. and Robinson, S.M., Eds., Nonlinear Programming, Vol. 2, Academic Press, New York, 1-27.
http://dx.doi.org/10.1016/b978-0-12-468650-2.50005-5
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SQL Server CHISQ.DIST function
CHISQ_DIST
Updated: 31 July 2015
Use CHISQ_DIST to calculate the probability density or the lower cumulative probability of a chi-squared distribution for x and a number of degrees of freedom.
Syntax
SELECT [wct].[CHISQ_DIST](
<@X, float,>
,<@Degrees_freedom, float,>
,<@Cumulative, bit,>)
Arguments
@X
The value of interest to be evaluated. @X must be of a type float or of type that intrinsically converts to float
@Degrees_freedom
The number of degrees freedom. @Degrees_freedom must be of a type float or of a type that intrinsically converts to float.
@Cumulative
A bit value indicating whether the probability density function ('False') or the cumulative distribution function ('True') should be returned.
Return Type
float
Remarks
· 0 < @X
· 0 < @Degrees_freedom
Examples
In this example we calculate the chi-squared lower-tailed cumulative distribution function.
SELECT
wct.CHISQ_DIST(
0.5 --@X
,1 --@Degrees_freedom
,'True' --@Cumulative
) as CDF
This produces the following result
Using the same data we calculate the chi-squared probability distribution function.
SELECT
wct.CHISQ_DIST(
0.5 --@X
,1 --@Degrees_freedom
,'False' --@Cumulative
) as PDF
This produces the following result.
The chi-squared distribution is a special case of the gamma distribution
SELECT
wct.CHISQ_DIST(x,df,'True') as [Chi-squared]
,wct.GAMMAP(0.5*df,0.5*x) as GammaP
FROM (VALUES (42,57))n(x,df)
This produces the following result.
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# Why couldn't use the mathematical induction?
We can use mathematical induction which is deduced from Peano axioms and illustrated on Terence Tao's Real Analysis(here it is)
• Axiom 2.1 $$0$$ is a natural number.
• Axiom 2.2 If $$n$$ is a natural number, then $$n++$$ is also a natural number.
• Axiom 2.3 $$0$$ is not the successor of any natural number; i.e., we have $$n++\neq 0$$ for every natural number $$n$$.
• Axiom 2.4 Different natural numbers must have different successors; i.e., if $$n,m$$ are natural numbers and $$n\neq m$$, then $$n++\neq m++$$. Equivalently, if $$n++=m++$$, then we must have $$n=m$$.
• Axiom 2.5 (Principle of mathematical induction). Let $$P(n)$$ be any property pertaining to a natural numbers $$n$$. Suppose that $$P(0)$$ is true, and suppose that whenever $$P(n)$$ is true, $$P(n++)$$ is also true. Then $$P(n)$$ is true for every natural number $$n$$.
to prove a proposition such as $$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$$ Here is an exercise about limitation $$\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)=1\neq 0.$$ so that we couldn't use mathematical induction here. Notice that I don't mean you can exchange the order of the limits operation and addition operation.
An other example from Royden and Fitzpatrick's Real Analysis:
Proposition 5 The union of a finite collection of measurable sets is measurable.
Proposition 7 The union of a countable collection of measurable sets is measurable.
The authors use the definition to prove Proposition 7. However, why don't we just use the technology above with Proposition 5 to obtain Proposition 7?
• Let $a_1,a_2,\dots$ be a sequence. Sometimes one uses induction to prove that $a_n$ satisfies certain inequalities. These may be useful in finding $\lim_{n\to\infty}a_n$. – André Nicolas Aug 12 '13 at 4:21
• Another "induction limit proof": Proof of 1=0 by mathematical induction? – Martin Sleziak Aug 12 '13 at 8:18
Just because $P(n)$ is true for every natural number $n$ doesn't mean that $P(\infty)$ is true.
If $P$ is a proposition, then mathematical induction can show that $P(n)$ is true for every $n$ in the natural numbers. That is, it shows that a statement is true for every finite case - but a countable union doesn't have to be a finite union, so the usual form of induction won't give us a strong enough result. That is, we can prove the result for infinitely many finite cases, but no infinite ones.
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TOPICS
# Orthic Triangle
Given a triangle , the triangle whose vertices are endpoints of the altitudes from each of the vertices of is called the orthic triangle, or sometimes the altitude triangle. The three lines , , and are concurrent at the orthocenter of .
The orthic triangle is therefore both the pedal triangle and Cevian triangle with respect to (Kimberling 1998, p. 156). It is also the cyclocevian triangle of the triangle centroid .
(1)
The area of the orthic triangle is given by
(2)
where is the circumradius of .
The orthic triangle has the minimum perimeter of any triangle inscribed in a given acute triangle (Johnson 1929, pp. 161-165). The lengths of the legs of the orthic triangle are given by
(3) (4) (5)
The inradius of the orthic triangle is
(6)
where is the circumradius of the reference triangle (Johnson 1929, p. 191), and the circumradius is
(7)
For an obtuse triangle or right triangle, the semiperimeter is
(8)
which simplifies in the case of an acute triangle to
(9)
where is the triangle area of and(Johnson 1929, p. 191).
Given a triangle , construct the orthic triangle and determine the symmedian points , , and of , , and , respectively. Then the -symmedian of the corner triangle is the -median of triangle , and similarly for the (Honsberger 1995, p. 75). In addition, the -median of the corner triangle is the -symmedian of triangle , and similarly for the other two corner triangles.
Finally, the Euler lines of the three corner triangles , and pass through the Euler points, and concur at a point on the nine-point circle of triangle such that one of the following holds
(10) (11) (12)
(Thébault 1947, 1949; Thébault et al. 1951).
The sides of the orthic triangle are parallel to the tangents to the circumcircle at the vertices (Johnson 1929, p. 172). This is equivalent to the statement that each line from a triangle's circumcenter to a vertex is always perpendicular to the corresponding side of the orthic triangle (Honsberger 1995, p. 22), and to the fact that the orthic and tangential triangles are homothetic at Kimberling center .
The triangle centroid of the orthic triangle has triangle center function
(13)
(Casey 1893, Kimberling 1994), which is Kimberling center . The symmedian point of the orthic triangle has triangle center function
(14)
(Casey 1893, Kimberling 1994), which is Kimberling center .
The following table gives the centers of the orthic triangle in terms of the centers of the reference triangle that correspond to Kimberling centers .
center of orthic triangle center of reference triangle incenter radical center of (circumcircle, Parry circle, Bevan circle) triangle centroid triangle centroid of orthic triangle circumcenter nine-point center orthocenter orthocenter of orthic triangle nine-point center nine-point center of orthic triangle symmedian point symmedian point of orthic triangle Euler infinity point isogonal conjugate of -of-orthic-triangle Tarry point -of-orthic-triangle Steiner point| -of-orthic-triangle psi(symmedian point, orthocenter) -of-orthic-triangle focus of Kiepert parabola -of-orthic-triangle Parry point -of-orthic-triangle psi(orthocenter, symmedian point) -of-orthic-triangle Parry reflection point isogonal conjugate of isogonal conjugate of Napoleon crossdifference
Altitude, Fagnano's Problem, Orthic Inconic, Orthocenter, Pedal Triangle, Symmedian Point
## Explore with Wolfram|Alpha
More things to try:
## References
Casey, J. A Treatise on the Analytical Geometry of the Point, Line, Circle, and Conic Sections, Containing an Account of Its Most Recent Extensions, with Numerous Examples, 2nd ed., rev. enl. Dublin: Hodges, Figgis, & Co., p. 9, 1893.Coxeter, H. S. M. and Greitzer, S. L. "The Orthic Triangle." §1.6 in Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 9 and 16-18, 1967.Honsberger, R. "The Orthic Triangle." §2.3 in Episodes in Nineteenth and Twentieth Century Euclidean Geometry. Washington, DC: Math. Assoc. Amer., pp. 21-25, 1995.Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.Kimberling, C. "Central Points and Central Lines in the Plane of a Triangle." Math. Mag. 67, 163-187, 1994.Kimberling, C. "Triangle Centers and Central Triangles." Congr. Numer. 129, 1-295, 1998.Thébault, V. "Concerning the Euler Line of a Triangle." Amer. Math. Monthly 54, 447-453, 1947.Thébault, V. "Problem 4328." Amer. Math. Monthly 56, 39-40, 1949.Thébault, V.; Ramler, O. J.; and Goormaghtigh, R. "Solution to Problem 4328: Euler Lines." Amer. Math. Monthly 58, 45, 1951.
Orthic Triangle
## Cite this as:
Weisstein, Eric W. "Orthic Triangle." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/OrthicTriangle.html
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Find the number of Islands
Count number of islands where every island is row-wise and column-wise separated
Count total number of islands present in given matrix where every island is separated row-wise and column-wise.
Given a rectangular matrix which has only two values ‘1’ and ‘0’.
1 represents land.
0 represents water.
Islands are all rectangular in shape that is values ‘1’ will always appear in form of rectangular islands and these islands are always row-wise and column-wise separated by at least one line of ‘0’s.
If there is any diagonal island then it will be considered adjacent islands and will be separate.
Count total number of islands in the given matrix.
Lets understand what is the input and the expected output.
Sample 1:
Input:
int[][] island = new int[][] {
{ 0, 0, 0 },
{ 1, 1, 0 },
{ 1, 1, 0 },
{ 0, 0, 1 },
{ 0, 0, 1 },
{ 1, 1, 0 }
};
Output: 3
Sample 2:
Input:
int[][] island = new int[][] {
{ 1, 0, 0 },
{ 0, 1, 0 },
{ 0, 0, 1 },
{ 0, 1, 0 },
{ 1, 0, 0 },
{ 0, 1, 0 }
};
Output:6
Sample 3:
Input:
int[][] island = new int[][] {
{ 1, 1, 0 },
{ 1, 1, 0 },
{ 1, 1, 0 },
{ 0, 1, 0 },
{ 0, 0, 1 },
{ 1, 1, 1 }
};
Output:3
Algorithm
We can solve the problem by using BFS and DFS approach but the problem is even simple and can be done in O(M*N) and constant extra space.
As all islands are perfect rectangular and separated by at least one line of ‘0’s, It is sure that first element of islands(rectangle) will definitely have LEFT coordinate and TOP coordinate '0' and all other elements of same island will definitely have either Left or Top coordinate.
So just count all the first element of island and we are done.
Java Program to find the number of Islands.
```public class CountNumberOfIsland {
public static void main(String[] args) {
int[][] island = new int[][] {
{1, 0, 0},
{0, 1, 0},
{0, 0, 1},
{0, 1, 0},
{1, 0, 0},
{0, 1, 0}
};
System.out.println(countIsland(island));
}
public static int countIsland(int[][] island){
int count=0;
for (int i = 0; i < island.length; i++) {
for (int j = 0; j < island[i].length; j++) {
if(island[i][j] == 1){
//check top and left coordinate
if( (i-1 < 0 || island[i-1][j] == 0) && (j-1 < 0 || island[i][j-1] == 0) ){
count++;
}
}
}
}
return count;
}
}
```
You may also like to see
Find the number of Islands using DFS.
Enjoy !!!!
If you find any issue in post or face any error while implementing, Please comment.
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https://byjus.com/question-answer/an-anther-has-1200-pollen-grains-how-many-pollen-mother-cells-must-have-been-there/
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Question
# An anther has 1200 pollen grains.How many pollen mother cells must have been there to produce them? 1200 300 150 2400 I need the explanation too:)
Solution
## 300 One pollen mother cell produces 4 pollen grains in plants. So if there are 1200 total pollen grains then there would have been originally 300 pollen mother cells.
Suggest corrections
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https://math.stackexchange.com/questions/2435888/showing-improper-integrals-are-divergent-or-convergent
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# Showing improper integrals are divergent or convergent
We are given $$\int^{\infty}_{3} \frac{x}{(x-3)^{\frac{3}{2}}}\, \mathrm dx = I_1 + I_2,$$ $where$ $I_1 = \int^{4}_{3} \frac{x}{(x-3)^{\frac{3}{2}}}\, \mathrm dx$ and $I_2 = \int^{\infty}_{4} \frac{x}{(x-3)^{\frac{3}{2}}}\, \mathrm dx$
State whether $I_1$ and $I_2$ are convergent or divergent.
I tried comparing to $\frac{1}{(x-3)^{\frac{3}{2}}}$ and other similar integrals but I couldn't get a solution. All help is appreciated, thanks in advance.
Set $u=x-3$. The integral becomes $$\displaystyle\int^{\infty}_{3} \frac{x}{(x-3)^{3/2}}\, \mathrm dx = \int^{\infty}_{0} \frac{u+3}{u^{3/2}}\, \mathrm du = \underbrace{\int^{1}_{0} \frac{u+3}{u^{3/2}}\, \mathrm du}_{\mathstrut J_1}+\underbrace{\int^{\infty}_{10} \frac{u+3}{u^{3/2}}\, \mathrm du}_{\mathstrut J_2}$$
• Now near $0$, $\dfrac{u+3}{u^{3/2}}\sim\dfrac{3}{u^{3/2}}$, so $J_1$, hence $I_1$, diverges.
• Near $\infty$, $\dfrac{u+3}{u^{3/2}}\sim\dfrac{u}{u^{3/2}}=\dfrac{1}{\sqrt u}$, so $J_2$, hence $I_2$, diverges.
Hint. Note that for $x\in (3,4]$ then $$\frac{x}{(x-3)^{\frac{3}{2}}}\geq \frac{1}{(x-3)^{\frac{3}{2}}}\implies I_1\geq \int_3^4\frac{dx}{(x-3)^{\frac{3}{2}}}$$ Moreover for $x\geq 4$, $$\frac{x}{(x-3)^{\frac{3}{2}}}\geq \frac{x}{(x)^{\frac{3}{2}}}=\frac{1}{x^{\frac{1}{2}}}\implies I_2\geq \int_4^{\infty}\frac{dx}{x^{\frac{1}{2}}}.$$ Can you take it from here?
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http://www.tricksgroup.co.in/average/
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# AVERAGE
### Average
is the sum of observations over a number of observation .In this chapter, we are providing the both normal method and tricks method.
QUES: Average age of 20 students is 16. When the teacher joins the class than the average increases by 1. Then what is the teacher’s age?
ANS: Normal Solutions:
[No. of students × Average Age] – [No. of Students (including teacher) × Average Age (Including Teacher)]= Answer.
Here,
20×16= 320
21×17= 357,
357-320= 37.
Shortcut Method:
No. of Students + Average Age + 1 (denotes the teacher) = Answer
= 20+16+1
QUES: Average age of 15 students is 12. When the teacher joins the class than the average increases by 1. Then what is the teacher’s age?
ANS:
Normal Solutions:
15×12= 180
16×13= 208
Then, 208-180= 28.
Shortcut Trick:
15+12+1= 28.
QUES: Average age of 32 students is 25. When the teacher joins the class than the average increases by 1. Then what is the teacher’s age?
ANS:
Normal Solutions:
32×25= 800
33×26= 858
Then, 800-858= 58.
Shortcut Trick:
32+25+1= 58.
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Today in Precalculus - PowerPoint PPT Presentation
Today in Precalculus. Go over homework Notes: Parabolas Completing the square Writing the equation given the graph Applications Homework. Example 1. Prove that the graph of y 2 + 2y – 8x – 7 = 0 is a parabola y 2 + 2y = 8x + 7 y 2 + 2y + 1 = 8x + 7 + 1 y 2 + 2y + 1 = 8x + 8
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Presentation Transcript
• Go over homework
• Notes: Parabolas
• Completing the square
• Writing the equation given the graph
• Applications
• Homework
Prove that the graph of y2 + 2y – 8x – 7 = 0 is a parabola
y2 + 2y = 8x + 7
y2 + 2y + 1 = 8x + 7 + 1
y2 + 2y + 1 = 8x + 8
(y+1)2 = 8 (x + 1)
Standard form for a parabola
(y+1)2 = 8 (x + 1)
Graph:
Opens to the right
vertex (-1, -1)
4p = 8, p = 2
focal length =2
focal width = 8
focus: (1, -1)
directrix: x = -3
axis: y = -1
Prove that the graph of x2 – 2x + 3y + 7 = 0 is a parabola
x2 – 2x = -3y – 7
x2 – 2x + 1 = -3y – 7 + 1
x2 – 2x + 1 = -3y – 6
(x – 1)2 = -3(y + 2)
Standard form for a parabola
(x – 1)2 = -3(y + 2)
Graph:
Opens downward
vertex (1, - 2)
p = -3/4
focal length = -3/4
focal width = 3
focus: (1, -11/4)
directrix: y = -5/4
axis: x = 1
Find vertex, substitute into general form vertex: (-2, -2)
(y+2)2=4p(x+2)
Find another point and substitute for x and y (1,-5)
(-5 + 2)2 = 4p(1 + 2)
Solve for p
9 = 12p
p = ¾
(y+2)2=4(¾)(x+2)
(y+2)2=3(x+2)
vertex: (-4, -1)
(x + 4)2=4p(y+1)
point (2,-4)
(2 + 4)2 = 4p(-4 + 1)
36 = -12p
p = -3
(x + 4)2=4(-3)(y + 1)
(x+4)2=-12(y+1)
vertex: (2, 1)
(y – 1)2=4p(x – 2)
point (-1,4)
(4 – 1)2 = 4p(-1 – 2)
9 = -12p
p = -3/4
(y – 1)2=4(-3/4)(x – 2)
(y – 1)2=-3(x – 2)
Let the focus F have coordinates (0, p) and the vertex be at (0,0)
equation: x2 = 4py
Because the reflector is 3ft across and 1 ft deep, the points (1.5, 1) and (-1.5, 1) must lie on the parabola.
(1.5)2 = 4p(1)
2.25 = 4p
p = .5625ft
= 6.75in
So the microphone should be placed inside the reflector along its axis and 6.75 inches from its vertex.
Let the roadway be the x-axis.
Then the vertex is at (500,25)
So the equation is:
(x – 500)2 = 4p(y – 25)
Points (0, 150) and (1000, 150) also on the parabola.
So the equation is:
(0 – 500)2 = 4p(150 – 25)
250,000 = 500p
p = 500
(x – 500)2 = 2000(y – 25)
Equation for the shape of the main cable.
To find the length of the support cables, solve equation for y
Support cables are every 100ft, so starting with x = 100, then 200, 300, etc.
Cables are 105ft, 70ft, 45ft, 30ft, 25ft, 30ft, 45ft, 70ft, 105ft.
Page 641: 51-56, 59-63
Quiz:
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# Operations & Algebraic Thinking
January 9, 2020
## Puzzles: Mobile Puzzles
Mobile puzzles are pictorial representations of systems of equations; each beam that balances two sets of objects (representing variables as unknown “weights”) represents an equation. Children […]
December 3, 2019
## The Number Line – subtraction, and measurement
The number line is not just a school object. It is as much a mathematical idea as functions. Unlike the Number Line Hotel, hundreds charts, Cuisenaire rods, and base ten blocks, the number line is not just a pedagogical aid used only to help children learn; mathematicians refer to it, too.
December 3, 2019
## Difference of Squares
This page describes one silent teaching activity that serves, at the superficial level, to give students useful practice with multiplication facts, but that also builds deeper […]
December 3, 2019
Addition and subtraction of whole numbers can be developed together, systematically, starting with recognition of quantities and combinations of quantities, the 10-based structure of the language […]
November 21, 2019
## Hundreds Chart and the Number Line Hotel
Hundreds chart A hundreds chart in which “higher numbers” are higher, and “the 30s” are all on the same row. Unlike the number line, which is […]
November 21, 2019
## Multiplication
This article deals with the meaning of multiplication, how it is introduced in Think Math!, and the early acquisition of facts.
November 21, 2019
## Puzzles: KenKen Puzzles
KenKen(R) puzzles were not built into Think Math! but are a wonderful material to make regularly available to children. These puzzles give excellent arithmetic practice while […]
November 21, 2019
## Guess My Rule
How to Play Guess My Rule games are games in which one person thinks up and gives examples of some “rule,” and the player(s) try to […]
November 21, 2019
## Puzzles: Who Am I Puzzles
Who-Am-I? puzzles give clues about some mathematical object — usually a number or a shape — and you are to figure out what the object is. […]
November 21, 2019
## Measurement: Discovering formulas for area
Area formulas Students who have the informal notion that area is the “amount of 2-D ‘stuff'” contained inside a region can invent for themselves most of […]
November 20, 2019
## Algebraic Thinking
Number tricks are fun for children. The fun, all by itself, is valuable, but is not mathematics. But understanding how the trick works is good mathematical, often algebraic, learning.
November 20, 2019
## Puzzles: Introducing KenKen Puzzles
This page focuses on a way to introduce KenKen puzzles so that students see, focus on, and learn the logic, not just guessing. These puzzles give […]
November 4, 2019
## Square Number
Informally: When you multiply an integer (a “whole” number, positive, negative or zero) times itself, the resulting product is called a square number, or a perfect […]
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# How to stop Robot's arm from falling down
Basically let’s say I press a button to bring the arm half way up then let go of said button, Gravity will force the arm down. Is the any way to “lock” the motors in place? I’ve thought of having the motors at a resting speed of 5, but this would just over heat the motors, and wouldn’t work if a picked up a cube/star, any ideas?
Rubber Bands. Put enough on both sides so that the robot comes up at about the same speed as it comes down. More rubber bands also helps you hold more.
Our team uses a 6 bar lift, and we have the screws loose enough that the motors can still lift without burning out, but tight enough that it puts up lots of resistance to falling.
To be honest, any extra friction is really just an eventual loss in energy. I personally would much rather put enough elastics to compensate for weight because adding friction tends to wear down on the parts over time and will only slow you down. Even if you have the motor running at 5-10% ish to keep your arm up, I would still argue that it is more efficient than adding unnecessary friction. Just my opinion.
Our team put encoders on the arm and a pi code that went into effect whenever the driver wasn’t manipulating the arm to hold the arm’s position. A much easier but less effective way is to program a button that gives the motor say 30 power… just enough to hold the arm in place when carrying something… of course, this fails when your load varies too much and if you have a particularly well charged or badly charged battery… or a number of other things…
I would usually agree with this, but we were able to add 4 extra rubber bands to the lift by adding the friction, and this friction was able to keep the tray on the ground when the lift was down.
Instead of adding friction to the lift you could use a simple while loop like this:
``````while(VexRT[Btn5U] != 1)//while button 5U isn't pressed...
{
motor[liftMotor] = 15;//You need to add all of your lift motors here
}
``````
You would then post this into your user control task so that while button 5U isn’t pressed, there would be a small amount of power that gets supplied to the lift motor(s). You would also need to either create another while loop or say “&&” after the button and add an additional button after it for multiple buttons.
This is what we do and it doesn’t overheat the motors, if it does overheat your motors then it may just be because your trying to give it a power value that’s too low.
We have a button that applies 12% power, or 15 power, to the motors and it doesn’t over heat unless you have all the game peaces on your lift.
Correction – 15 power, not 15%. It’s about 12%.
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# Dihedral in a regular spherical polygon
Planes rotate around a central symmetry axis pass through a sphere centre to intersect on the sphere forming a regular spherical polygon of $$n$$ sides.
A small circle forms as base of cone semi-vertical angle $$\alpha$$ circumscribing the vertices of the polygon.
If $$\delta$$ is the dihedral angle between successive planes then show that $$\tan \dfrac {\delta}{2} =\tan \dfrac {\pi}{n} \;\cos \alpha$$
Special case in Euclidean geometry when $$\alpha\rightarrow 0$$ sum of external angles of a regular polygon: $$n \delta = {2\pi}.$$
The planes carve a regular spherical polygon in the surface of the (unit) sphere, such that the dihedral angles $$\delta$$ between the planes match the exterior angles of the polygon. The semi-vertical angle $$\alpha$$ of the cone gives the (spherical) circumradius of the polygon.
If $$P$$ is the center of the polygon, $$Q$$ one of its vertices, and $$A$$ the midpoint of a side adjacent to $$Q$$, then $$\triangle PQA$$ is a spherical right triangle with hypotenuse $$\alpha$$ and acute angles $$\pi/n$$ and $$(\pi-\delta)/2$$.
\begin{align} \cos A &= -\cos P \cos Q + \sin P \sin Q \cos\alpha \\[4pt] 0 &= -\cos\frac{\pi}{n}\sin\frac\delta2+\sin\frac{\pi}{n}\cos\frac\delta2\cos\alpha \\[4pt] \tan\frac{\delta}{2} &= \tan\frac{\pi}{n}\;\cos\alpha \end{align} as desired. $$\square$$
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## linear first order differential equations calculator
The differential equation in the picture above is a first order linear differential equation, with $$P(x) = 1$$ and $$Q(x) = 6x^2$$. + . Find more Mathematics widgets in Wolfram|Alpha. Linear Equations – In this section we solve linear first order differential equations, i.e. Sturm–Liouville theory is a theory of a special type of second order linear ordinary differential equation. Solve Differential Equation. While general solutions to ordinary differential equations involve arbitrary constants, general solutions to partial differential equations involve arbitrary functions. To eliminate constants, see Solve Differential Equations with Conditions. A linear first order ordinary differential equation is that of the following form, where we consider that y = y(x), and y and its derivative are both of the first degree. The number of grid vectors in state-space diagram can be set in the numeric field for the grid points. Homogeneous Differential Equations Calculator. A linear first-order equation takes the following form: To use this method, follow these steps: Calculate the integrating factor. And that should be true for all x's, in order for this to be a solution to this differential equation. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. In … Solve the first-order differential equation dy dt = ay. •The general form of a linear first-order ODE is . This video explains how to find the general solutions to linear first order differential equations. Differential Equations Calculator. To solve a system of differential equations, see Solve a System of Differential Equations. Restate […] However, we would suggest that you do not memorize the formula itself. The method for solving such equations is similar to the one used to solve nonexact equations. Get the free "General Differential Equation Solver" widget for your website, blog, Wordpress, Blogger, or iGoogle. dy / dx + y = 2x + 52. dy / dx + y = x4Answers to Above Exercises1. First, the long, tedious cumbersome method, and then a short-cut method using "integrating factors". The differential equation in first-order … A first‐order differential equation is said to be linear if it can be expressed in the form where P and Q are functions of x. In a previous post, we talked about a brief overview of ODEs. Multiply the DE by this integrating factor. A first-order differential equation is defined by an equation: dy/dx =f (x,y) of two variables x and y with its function f(x,y) defined on a region in the xy-plane.It has only the first derivative dy/dx so that the equation is of the first order and no higher-order derivatives exist. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'=AX. In this case, unlike most of the first order cases that we will look at, we can actually derive a formula for the general solution. There are two methods which can be used to solve 1st order differential equations. In other words, we confine ourdiscussion to first-order equations with or withoutdiscontinuities. But, the solution to the first order partial differential equations with as many arbitrary constants as the number of independent variables is called the complete integral.The following n-parameter family of solutions which can be written in matrix form as X'=AX, where A is the coefficients matrix. The initial values y 01 and y 02 can be varied with the sliders on the vertical axis at x 0 in the first chart. The value for x 0 can be set in the numeric input field. They are Separation of Variables. You can check this for yourselves. Linear Equations: TI-84 Plus and TI-83 Plus graphing calculator program for performing calculations related to linear equations including intercepts, distance, midpoint and gradient. To solve it there is a special method: We invent two new functions of x, call them u and v, and say that y=uv. Solve a differential equation analytically by using the dsolve function, with or without initial conditions. Delta functions are covered in Section 6.4, and convolution is discussed in Section 6.5. Differential Equation, Equations, Linear Equations. DSolve labels these arbi-trary functions as C@iD. What is Meant by Second Order Differential Equation? Solve Differential Equation with Condition. :) https://www.patreon.com/patrickjmt !! GRADE CALCULATOR: Course Evaluations: WolframAlpha: Problems: Tests: Weeks: Dates ... First-order linear differential equations: V1 ... and why we cannot solve very many differential equations: 3-11 S1, S2, S3; SLD PR: 3: Sep 8, 10 You da real mvps! The procedure to use the second-order differential equation solver calculator is as follows: Step 1: Enter the ordinary differential equation in the input field Step 2: Now click the button “Calculate” to get the ODEs classification Step 3: Finally, the classification of the ODEs will be displayed in the new window. Second Order Differential Equations. We give an in depth overview of the process used to solve this type of differential equation as well as a derivation of the formula needed for the integrating factor used in the solution process. Home » Elementary Differential Equations » Differential Equations of Order One. The solution of the differential equations is calculated numerically. Linear Equations | Equations of Order One . A clever method for solving differential equations (DEs) is in the form of a linear first-order equation. in the equation u(x) y = ò Returns the biggest integer n with n < x. One can see that this equation is not linear with respect to the function $$y\left( x \right).$$ However, we can try to find the solution for the inverse function $$x\left( y \right).$$ We write the given equation in terms of differentials and make some transformations: This method involves multiplying the entire equation by an integrating factor. First Order Linear Differential Equations How do we solve 1st order differential equations? Integrating Factor Tutorials. The Demonstration explains the "variation of parameters" method of solving a linear first-order differential equation. So in order for this to satisfy this differential equation, it needs to … The general solution is derived below. We'll talk about two methods for solving these beasties. Learn the First Order Differential Equations and know the formulas for Linear Equation, Separable Equation, Homogeneous Equation and a lot more. The columns can be normal, stacked, or by percent. It also outputs slope and intercept parameters and displays line on a graph. \$1 per month helps!! Then, solve the equation by using dsolve. differential equations in the form $$y' + p(t) y = g(t)$$. = ( ) •In this equation, if 1 =0, it is no longer an differential equation and so 1 cannot be 0; and if 0 =0, it is a variable separated ODE and can easily be solved by integration, thus in this chapter 0 cannot be 0. The function equation_solver can solve first order linear differential equations online, to solve the following differential equation : y'+y=0, you must enter equation_solver(y'+y=0;x). Solving first order linear differential equation. Three Runge-Kutta methods are available: Heun, Euler and Runge-Kutta 4.Order. Here is the general solution to a linear first-order PDE. Remember, the solution to a differential equation is not a value or a set of values. It is helpful, as a matter of notation first, to consider differentiation as an abstract operation, accepting a function and returning another (in the style of a higher-order function in computer science). We then solve to find u, and then find v, and tidy up and we are done! We have now reached... ¡Únete a 100 millones de usuarios felices! If P(x) or Q(x) is equal to 0, the differential equation can be reduced to a variables separable form which can be easily solved. A linear first order equation is one that can be reduced to a general form – $${\frac{dy}{dx} + P(x)y = Q(x)}$$ where P(x) and Q(x) are continuous functions in the domain of validity of the differential equation. A calculator for solving differential equations. General solution and complete integral. First you have to transform the second order ode in a system of two first order equations and then you can use one of the functions included in the package. There are several different formulas for the equation of a line. There, the nonexact equation was multiplied by an integrating factor, which then made it easy to solve (because the equation became exact). The first special case of first order differential equations that we will look at is the linear first order differential equation. Contributed by: Izidor Hafner (March … Before tackling second order differential equations, make sure you are familiar with the various methods for solving first order differential equations. Free linear first order differential equations calculator - solve ordinary linear first order differential equations step-by-step This website uses cookies to ensure you get the best experience. ar. Solving second order differential equation . Use * for multiplication a^2 is a 2 First-Order Linear ODE. Learn the First Order Differential Equations and know the formulas for Linear Equation, Separable Equation, Homogeneous Equation and a lot more. Get the free "1st order lineardifferential equation solver" widget for your website, blog, Wordpress, Blogger, or iGoogle. It is a function or a set of functions. Thanks to all of you who support me on Patreon. Their solutions are based on eigenvalues and corresponding eigenfunctions of linear operators defined via second-order homogeneous linear equations.The problems are identified as Sturm-Liouville Problems (SLP) and are named after J.C.F. Section 6.3 extends the discussion to second-orderequations. Description. In this post, we will focus on a specific type of ODE, linear first order differential equations. The general solution to the first order partial differential equation is a solution which contains an arbitrary function. syms y(t) a eqn = diff(y,t) == a*y; S = dsolve(eqn) S = C 1 e a t C1*exp((a*t)) The solution includes a constant. Advanced Math Solutions – Ordinary Differential Equations Calculator, Linear ODE Ordinary differential equations can be a little tricky. Specify the first-order derivative by using diff and the equation by using ==. A differential operator is an operator defined as a function of the differentiation operator. A first order differential equation is linear when it can be made to look like this: dy dx + P(x)y = Q(x) Where P(x) and Q(x) are functions of x. Toggle Nav. A linear equation or polynomial, with one or more terms, consisting of the derivatives of the dependent variable with respect to one or more independent variables is known as a linear differential equation. The used method can be selected. A general first-order differential equation is given by the expression: dy/dx + Py = Q where y is a function and dy/dx is a derivative. Find more Mathematics widgets in Wolfram|Alpha. Is not a value or a set of functions first-order differential equation analytically by diff... And tidy up and we are done focus on a graph + =. One used to solve nonexact equations matrix form as X'=AX, where a the... A linear first-order PDE with conditions a function of the differentiation operator linear! Eliminate constants, see solve differential equations is calculated numerically which contains an arbitrary.... Method for solving differential equations ( DEs ) is in the form \ ( y ' + (... First-Order equation specify the first-order differential equation to solve nonexact equations ( March general! The columns can be written in matrix form as X'=AX, where a the! True for all x 's, in order for this to be a little tricky form of a line such. Widget for your website, blog, Wordpress, Blogger, or by percent is coefficients. The coefficients matrix 1st order differential equations we solve linear first order differential equations involve arbitrary functions type second! » differential equations of order One are familiar with the various methods for solving such equations is to. C @ iD Returns the biggest integer n with n < x we will focus on a graph memorize! The nature of the differentiation operator be written in matrix form as X'=AX where..., blog, Wordpress, Blogger, or iGoogle line on a specific type of order. X'=Ax+By y'=cx+dy focus on a graph point ( when ) and solutions of the operator! The columns can be normal, stacked, or iGoogle solving differential equations type! = g ( t ) y = g ( t ) y = g ( t \! Short-Cut method using integrating factors '' Section 6.4, and convolution is discussed in Section 6.5 such is... Little tricky first-order equation with conditions the formulas for the grid points and solutions the! Solutions of the critical point ( when ) and solutions of the differentiation operator partial! An operator defined as a function or a set of values and integral! Separable equation, Separable equation, Separable equation, Separable equation, Separable equation Separable. By an integrating factor similar to the One used to solve nonexact equations are different. 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Is the coefficients matrix Runge-Kutta 4.Order + y = ò Returns the biggest integer n with n x! Differentiation operator previous post, we confine ourdiscussion to first-order equations with or withoutdiscontinuities linear system X'=AX or set! And then a short-cut method using integrating factors '' methods are available: Heun, Euler and 4.Order! The numeric input field are two methods which can be used to solve nonexact equations equation and a more! Available: Heun, Euler and Runge-Kutta 4.Order we solve linear first order differential and... Nature of the linear system X'=AX involve arbitrary functions in order for this to a. To eliminate constants, general solutions to linear first order differential equations involve arbitrary constants, general solutions to first!
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# An intuitive explanation of Adagrad, its purpose and its formula
It (Adagrad) adapts the learning rate to the parameters, performing smaller updates (i.e. low learning rates) for parameters associated with frequently occurring features, and larger updates (i.e. high learning rates) for parameters associated with infrequent features.
If a parameter is associated with an infrequent feature then yes, it is more important to focus on properly adjusting that parameter since it is more decisive in classification problems. But how does making the learning rate higher in this situation help?
If it only changes the size of the movement in the dimension of the parameter (makes it larger) wouldn't that make things even more imprecise? Since the network depends more on those infrequent features, shouldn't adjusting those parameters be done more precisely instead of just faster? The more decisive parameters should have a higher "slope", thus why should they also have high learning rates? I must be missing something, but what is it?
Further, in the article, the formula for parameter adjustments with Adagrad is given. Where exactly in that formula do you find the information about the frequency of a parameter? There must be a relationship between the gradients of a parameter and the frequency of features associated with it because it's the gradients that play an important role in the formula. What is that relationship?
TLDR: I don't understand both the purpose and formula behind Adagrad. What is an intuitive explanation of it that also provides an answer to the questions above, or shows why they are irrelevant?
• I'm fairly sure lecture 3 of cs231n on youtube provides a good explanation at some point. I'm just not sure if this is the specific lecture or not. – Recessive Aug 19 '19 at 3:17
## 1 Answer
I found a somewhat more accessible introduction here:
https://medium.com/konvergen/an-introduction-to-adagrad-f130ae871827
Let me start from the last part of your question. The frequency of a parameter is in G_t, which is the accumulated sum of squared gradients from all the time steps up to step t. If the gradient vanishes in many of the previous steps, then you divide the learning rate with a smaller number for that parameter.
And for the first part, you want the parameter that is more frequent to have a smaller learning rate as it is updated on more iterations compared to a parameter which is updated only a small number of times.
• For the first part: Maybe its just the fact that SGD doesnt update the neurons associated with the infrequent features much because they provide relatively small improvement (low gradient) and thus that must be compensated for with a higher learning rate? So basically, how does the frequency of feature influence frequency of adjustment (and not only the gradient and thus adjustment itself) if adjustments don't occur after every example, but after a mini-batch of examples where an example with the infrequent feature (for the sake of simplicity) has occured? – DaddyMike Aug 20 '19 at 12:08
• A good point for which I have no answer. I could not find anything about this in the original paper by Duchi et al, or the deep learning book of Goodfellow. I suppose the parameters have to be sparse enough (compared to mini batch size) for the algorithm to work effectively. – serali 9 mins ago Delete – serali Aug 20 '19 at 13:23
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# In this project, you will demonstrate your mastery of the following competencies:
Competencies
In this project, you will demonstrate your mastery of the following competencies:
Interpret psychological data using quantitative and qualitative methods
Apply the principles of statistical methods to inform a research problem
Scenario
You work for a market research company that supplies information to non-profit organizations throughout the nation. Your supervisor has asked you to provide an objective description of the data that will provide information regarding how to target different audiences in ways that bring about empathy. This will help your business support non-profit organizations in obtaining donations. You will create a short memorandum that includes graphical representations of your data in order to communicate this information.
Directions:
For this project you will submit a memorandum as a Word file. You will complete your memo using the templates on this page. Your memorandum must be a minimum of 1–2 pages (not including graphs). For more details on how you’ll be graded, refer to the Project One Guidelines and Rubric page in Brightspace.
Introduction: Describe the purpose of your memo and the plan to address the scenario in 1 to 3 sentences.
Conclusions: Describe your findings in an executive summary of 4 to 6 sentences. Include the following in your conclusions:
• The main points you want to convey to your audience
• Rationale for your points in the form of data summaries
Main Analysis: Describe the summary statistics and frequency distributions, taking into account the scale of measurement for your data. Refer to the graphs you created. Your main analysis section should be about 2 to 5 sentences.
Graph One: Create a graphical representation of the qualitative (nominal and often ordinal) data to support your main analysis and upload it here as a JPG or PNG file. Ensure your graph meets the following criteria:
• You include a narrative to introduce your graph into your memo.
• Your graph is accurate and objective.
• Your graph appropriately represents the data.
• You use the appropriate type of graph for the data.
• Your graph is labeled appropriately.
Graph Two: Create a graphical representation of the quantitative (interval and/or ratio) data to support your main analysis and upload it here as a JPG or PNG file. Ensure your graph meets the following criteria:
• You include a narrative to introduce your graph into your memo.
• Your graph is accurate and objective.
• Your graph appropriately represents the data.
• You use the appropriate type of graph for the data.
• Your graph is labeled appropriately.
Recommendations: Describe the actions you believe your audience should take in 2 to 5 sentences.
Limitations: Describe the limitations of both your data and your summaries in 1 to 3 sentences.
## ")96% of our customers have reported a 90% and above score. You might want to place an order with us."
##### Affordable prices
You might be focused on looking for a cheap essay writing service instead of searching for the perfect combination of quality and affordable rates. You need to be aware that a cheap essay does not mean a good essay, as qualified authors estimate their knowledge realistically. At the same time, it is all about balance. We are proud to offer rates among the best on the market and believe every student must have access to effective writing assistance for a cost that he or she finds affordable.
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# [Vm-dev] Tangential: A Lisp Interpreter Implemented in Conway's Game of Life
Vanessa Freudenberg vanessa at codefrau.net
Sun Jan 22 22:55:46 UTC 2023
```On Sat, Jan 21, 2023 at 00:48 Stéphane Rollandin <lecteur at zogotounga.net>
wrote:
>
> > I can not map
> > that implementation to the rules of the game... Can someone explain to
> > me or point me to a paper/description of why that implementation works?
>
> There is the Byte paper, around
>
> https://archive.org/details/byte-magazine-1981-08/page/n205/mode/2up?view=theater
>
> The figure there shows how ancilliary forms are used to count, in
> parallel, the number of neighbors for each point in the pattern.
>
> For each direction in the Moore neighborood, the pattern is shifted and
> these eight shifted grids are combined together.
>
> This is an addition in base two, and we need to count up to eight, so
> the algorithm uses three forms (nbr1, nbr2 and nbr4) to store the
> corresponding bits, and two other forms (carry2 and carry4) to store the
> bits carried from one bit to the next (so here from nbr1 to nbr2, and
> from nbr2 to nbr4), just as you need to do when performing an addition
> by hand.
>
> The last step (the last four lines in #nextLifeGeneration from the Byte
> paper) generates the next pattern from nbr1, nbr2 and nbr4. The rules of
> the Game of Life are encoded here in the successive combination rules,
> in a nifty and quite obfuscated way that I did not even tried to
> comprehend.
>
> Stef
Right. You basically need to understand how binary counting works at a
logic-gate level, to be able to comprehend the neighbor counting code.
Typically that logic is invisible. I love this mechanical version:
https://youtu.be/zELAfmp3fXY
You can see the bits (the big 0s and 1s). The carry is the little latch
sticking out to the left, and it gets set whenever a bit flips from 1 to 0,
which means the next bit has to flip.
The “bit + carry” logic is called a “half adder”, you can google that, but
basically it generates the next bit and carry from the previous bit and
carry using two logic operations, an XOR for the bit and an AND for the
carry.
With that understanding, Dan’s explanation in the Byte article might make
more sense – it does the XOR and AND for all pixels at the same time.
The 3 bit result count is then used to determine if a cell should be born,
survive, or die. Again, logic ops do that for all pixels at once.
The Life rule is that cells with 2 or 3 neighbors survive, any with fewer
or more neighbors die, and any empty cell with 3 neighbors is born.
To implement that rule with minimal logic, we restructure it a bit
A live cell will be there in the next generation if
(1) the cell is alive AND has 2 neighbors
(2) OR has 3 neighbors (independent of itself being alive or not)
(3) BUT it must not have 4 or more neighbors
(1) is implemented as “self AND 2s” because the 2s bit is only set for 2, 6
and 7 neighbors (010, 110 and 111 in binary), and we deal with the >= 4
neighbors case in (3)
(2) is implemented as “self OR (1s AND 2s)” because those bits are only set
together for 3 and 7 neighbors (011 and 111 in binary), and we deal with
the 7 neighbors case in (3)
(3) is implemented as “self AND NOT 4s” which kills any remaining 6 and 7
neighbor cells (4, 5, and 8 neighbor cells wouldn’t have made it to this
step anyways because their lower bits look like 0 and 1 neighbors: 100,
101, and 000)
This also explains why we don’t need to count to 8. With 3 bits we can only
count to 7, if we add one more to 7 we get 0. But in Life this is fine
because both cells with 8 and 0 neighbors have the same result.
Hope that helps,
Vanessa
>
>
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## ››Convert minute to Gregorian year
minutes year
Did you typical to convert minutes to Gregorian yearJulian yeartropical yearsidereal yearanomalistic yeardraconic yearlunar yearGaussian year
You are watching: How many minutes in a year
How many minutes in 1 year?The price is 525949.2.We assume you space converting in between minute and also Gregorian year.You can view much more details on each measurement unit:minutes oryearThe SI base unit because that time is the second.1 second is same to 0.016666666666667 minutes, or 3.1688738506811E-8 year.Note that round off errors might occur, so constantly check the results.Use this page to learn exactly how to convert in between minutes and years.Type in your own numbers in the type to convert the units!
## ››Date difference in between calendar days
You may also want to uncover out how plenty of daysare in between two dates on the calendar. Use thedate calculatorto acquire your age in work or measure up the term of one event.
## ››Want other units?
You have the right to do the turning back unit switch fromyear come minutes, or enter any type of two devices below:
## Enter 2 units to convert
From: To:
## ››Common time conversions
minutes to decademinutes to hourminutes to quarterminutes to centuryminutes to microsecondminutes to weekminutes come millisecondminutes come nanosecondminutes to dayminutes come fortnight
## ››Definition: Minute
A minute is: * a unit that time same to 1/60th of one hour and to 60 seconds. (Some rare minutes have 59 or 61 seconds; check out leap second.)
## ››Definition: Year
An typical Gregorian year is 365.2425 job (52.1775 weeks, 8765.82 hours, 525949.2 minute or 31556952 seconds). For this calendar, a common year is 365 job (8760 hours, 525600 minute or 31536000 seconds), and also a leap year is 366 work (8784 hours, 527040 minute or 31622400 seconds). The 400-year cycle of the Gregorian calendar has actually 146097 days and hence precisely 20871 weeks.ag = a_g = 365.2425 days for the typical Gregorian year
## ››Metric conversions and more
jajalger2018.org provides an onlineconversion calculator for all varieties of measure up units.You can uncover metric switch tables for SI units, as wellas English units, currency, and also other data. Kind in unitsymbols, abbreviations, or full names for units of length,area, mass, pressure, and also other types. Examples include mm,inch, 100 kg, US fluid ounce, 6"3", 10 stone 4, cubic cm,metres squared, grams, moles, feet per second, and many more!
See more: Which Conversion Strategy Converts Each Location To The New System At A Separate Time?
Convert ·Time ·Dates ·Salary ·Chemistry ·Forum ·Search ·Privacy ·Bibliography ·Contact© 2021 jajalger2018.org
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# Police Constable Quiz Test 11
Selling a book in 720 makes 20% profit. If you want to make 10% profit on it, how much do you have to sell it for?
660
What is 90% of Rs.315 = __?
350
If the sum of age of parents and two sons was 40 years before 2 years, what will be the sum of their ages after 3 years?
60
What is the probability of a specific event?
A. Always 0 B. Always between 1 C. -1, D. 1
What is the sum of the digits from 1 to 50?
1275
[ays_quiz id=’251′]
Dividing a number by 56 increases the remainder by 29. What is the remainder of that number divided by 8?
5
What is the GSA of 425 and 525?
25
1 inch = how many cm?
2.54 cm
1089 and 1090 of G.S.A.
1
6,12,20,30,42 _
A.60 B.56 C.58 D.50
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# Solve question one the drawing is right beside it and the key answer I want explaining...
###### Question:
Solve question one the drawing is right beside it and the key answer I want explaining step by step for the drawing and the forces found please
Problem 1: Find the support reactions and draw the moment diagram for the beam: (CLO 2) 15kN/m Find support moment: 20x3-(15x3/2)+M-Mp-7.5kN-m. Static Moment: 15x3/8-16.875kN-m Ok MB Ra Rcv 15kN Rehx3-5x2-5x4-5x6 0 RG.hz 60/3-20kN =-Rah Ro.b 3 in Problem 2 Solve the truss forces at node A using (a) The method of Sections and (b) the method of Nodal Force Equilibrium n N P From equilibrium of Node G in y: FAO 15 (T) S kN From equilibrium of Node G In x: For 20(T) (b) Method of Joints (Nodes), NodeA Take part BDF: Moments about F FaBx3+5x2+5x4-0FAB (30/3)-10 (C) Take moments about B: FaFx2sin56.3+Forx3-5x2-5x40 FAF (10+20-20x3)/2/0.83-18kN (C) 20F40 For FBA Forsin56.3+150 For-15/sin56.3-18kN (C) 20-18cos56.3-FRA 0FRA 20-18cos56.3 10kN(C)
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Question
# QUESTION 9 Nu-Tek is expanding rapidly. As a result, the company expects to pay annual dividends...
QUESTION 9
1. Nu-Tek is expanding rapidly. As a result, the company expects to pay annual dividends of \$0.5, \$0.65, and \$0.75 per share over the next three years, respectively. After that, the dividend is projected to increase by 2 percent annually. What is the current value of this stock if the required return is 12 percent?
\$8.94 \$9.94 \$5.94 \$6.94 \$7.94
The current value of the stock is computed as shown below:
= Dividend in year 1 / (1 + required rate of return)1 + Dividend in year 2 / (1 + required rate of return)2 + Dividend in year 3 / (1 + required rate of return)3 + 1 / (1 + required rate of return)3[ ( Dividend in year 3 (1 + growth rate) / ( required rate of return - growth rate) ]
= \$ 0.50 / 1.12 + \$ 0.65 / 1.122 + \$ 0.75 / 1.123 + 1 / 1.123 [ ( \$ 0.75 x 1.02) / ( 0.12 - 0.02) ]
= \$ 0.50 / 1.12 + \$ 0.65 / 1.122 + \$ 0.75 / 1.123 + \$ 7.65 / 1.123
= \$ 6.94 Approximately
Feel free to ask in case of any query relating to this question
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## FANDOM
34,204 Pages
In statistics, a censored variable is a variable that is not fully measured or observed because of floor or ceiling effects providing a cut off in the range, beyond which data cannot be quantified.
For example, suppose a study is conducted to measure the impact of a drug on mortality. In such a study, it may be known that an individual's age at death is at least 75 years. Such a situation could occur if the individual withdrew from the study at age 75, or if the individual is currently alive at the age of 75.
Censoring also occurs when a value occurs outside the range of a measuring instrument. For example, a bathroom scale might only measure up to 300 lbs. If a 350 lb individual is weighed using the scale, the observer would only know that the individual's weight is at least 300 lbs.
## Types Edit
• Left censoring – a data point is below a certain value but it is unknown by how much
• Interval censoring – a data point is somewhere on an interval between two values
• Right censoring – a data point is above a certain value but it is unknown by how much
• Type I censoring occurs if an experiment has a set number of subjects or items and stops the experiment at a predetermined time, at which point any subjects remaining are right-censored.
• Type II censoring occurs if an experiment has a set number of subjects or items and stops the experiment when a predetermined number are observed to have failed; the remaining subjects are then right-censored.
• Random (or non-informative) censoring is when each subject has a censoring time that is statistically independent of their failure time. The observed value is the minimum of the censoring and failure times; subjects whose failure time is greater than their censoring time are right-censored.
Censoring should not be confused with the related idea truncation. With censoring, observations result either in knowing the exact value that applies, or in knowing that the value lies within an interval. With truncation, observations never result in values outside a given range — values in the population outside the range are never seen or never recorded if they are seen. Note that in statistics, truncation is not the same as rounding.
The problem of censored data, in which the observed value of some variable is partially known, is related to the problem of missing data, where the observed value of some variable is unknown.
Interval censoring can occur when observing a value requires follow-ups or inspections. Left and right censoring are special cases of interval censoring, with the beginning of the interval at zero or the end at infinity, respectively.
Left-censored data, is observed, for example, in environmental analytical data where trace concentrations of chemicals may indeed be present in an environmental sample (e.g., groundwater, soil) but are "non-detectable," i.e., below the detection limit of the analytical instrument or laboratory method. Estimation methods for using left-censored data vary, and not all methods of estimation may be applicable to, or the most reliable, for all data sets.[1]
## EpidemiologyEdit
One of the earliest attempts to analyse a statistical problem involving censored data was Daniel Bernoulli's 1766 analysis of smallpox morbidity and mortality data to demonstrate the efficacy of vaccination.[2]
## AnalysisEdit
Special techniques may be used to handle censored data.
## ReferencesEdit
1. Helsel, D. Much ado about next to Nothing: Incorporating Nondetects in Science, Ann. Occup. Hyg., Vol. 54, No. 3, pp. 257-262, 2010
2. Bernoulli D. (1766) "Essai d’une nouvelle analyse de la mortalité causée par la petite vérole. Mem. Math. Phy. Acad. Roy. Sci. Paris, reprinted in Bradley (1971) 21 and Blower (2004)
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# If $E/F$ is algebraic and every $f\in F[X]$ has a root in $E$, why is $E$ algebraically closed? [duplicate]
Suppose $E/F$ is an algebraic extension, where every polynomial over $F$ has a root in $E$. It's not clear to me why $E$ is actually algebraically closed.
I attempted the following, but I don't think it's correct:
I let $f$ be an irreducible polynomial in $E[X]$. I let $\alpha$ be a root in some extension, so $f=m_{\alpha,E}$. Since $\alpha$ is algebraic over $E$, it is also algebraic over $F$, let $m_{\alpha,F}$ be it's minimal polynomial. I now let $K$ be a splitting field of $m_{\alpha,F}$, which is a finite extension since each root has finite degree over $F$.
If $m_{\alpha,F}$ is separable, then $K/F$ is also separable, so as a finite, separable extension, we can write $K=F(\beta)$ for some primitive element $\beta$. By assumption, $m_{\alpha,F}$ has a root in $E$, call it $r$. Then we can embed $F(\beta)$ into $r$ by mapping $\beta$ to $r$. It follows that $m_{\alpha,F}$ splits in $E$. Since $f\mid m_{\alpha,F}$, we must also have the $f$ is split in $E$.
But what happens if $m_{\alpha,F}$ is not separable? In such case, $F$ must have characteristic $p$. I know we can express $m_{\alpha,F}=g(X^{p^k})$ for some irreducible, separable polynomial $g(X)\in F[X]$. But I'm not sure what follows after that.
NB: I say $E$ is algebraically closed if every nonconstant polynomial in $E[X]$ has a root in $E$.
## marked as duplicate by Brahadeesh, YuiTo Cheng, Shogun, José Carlos Santos, Yanior WegJul 8 at 16:20
• There are (unfortunately) two slightly different definitions of algebraically closed. Which one are you using? – Eric Towers Mar 21 '14 at 22:43
• @EricTowers I say $E$ is algebraically closed if every nonconstant polynomial in $E[X]$ has a root in $E$. – Nastassja Mar 21 '14 at 22:44
• You cannot map $\beta$ to $r$ in general: a primitive element for the splitting field of a polynomial need not be a root of the polynomial. – Jan Ladislav Dussek Mar 21 '14 at 23:04
• Your title and first sentence don't match. Does every polynomial in $F[x]$ have a root in $F$ or in $E$? – Greg Martin Mar 21 '14 at 23:08
• @GregMartin I'm sorry, that was a typo, every polynomial in $F[X]$ has a root in $E$. – Nastassja Mar 21 '14 at 23:09
This is a recent qual question at my school. :)
If $F$ is not perfect, say it has characteristic $p$. Let $F_{perf}$ be the extension of $F$ obtained by adjoining a root for $x^{p^n}-a$ for every $n \in \mathbb N, a \in F$. Since these polynomials are purely inseparable, $F_{perf}$ embeds (via a unique isomorphism) into $E$, and we may identify $F_{perf}$ with its isomorphic copy in $E$.
Two facts for you to verify: $F_{perf}$ is a perfect field, and every element of $F_{perf}$ is a root of $x^{p^n}-a$ for some $n \in \mathbb N$ and $a \in F$.
Now let $f$ be a polynomial with coefficients in $F_{perf}$, say $f(x) = x^k + a_{k-1} x^{k-1} + \cdots + a_0$. Then for sufficiently large $n$, $a_i^{p^n} \in F$ for all $i$. So $$f^{p^n}(x) = x^{k p^n} + a_{k-1}^{p^n} x^{(k-1)p^n} + \cdots + a_0^{p^n} \in F[x],$$ which has a root in $E$ by assumption. But $f$ and $f^{p^n}$ have the same roots, hence $f$ has a root in $E$. So without loss of generality, we may assume $F = F_{perf}$, and Bruno's answer gets us the rest of the way.
• Ah, so you want to take a perfect closure of $F$. May I ask, why is it that each $x^{p^n}-a$ is purely inseparable, and why is that needed to embed $F_{perf}$ into $E$? – Nastassja Mar 22 '14 at 1:05
• Purely inseparable means that the polynomial has only one root. If $\alpha$ is a root of $x^{p^n}-a$, then $x^{p^n}-a = x^{p^n}-\alpha^{p^n} = (x-\alpha)^{p^n}$. Since each of these polynomials has only one root, they all split in $E$, which can't be said, a priori, for a polynomial with more than one root. – Dustan Levenstein Mar 22 '14 at 1:08
• Thanks, I had not heard that terminology for polynomials before. – Nastassja Mar 22 '14 at 1:10
• Done! Thanks both for your help. – Nastassja Mar 22 '14 at 1:55
• lol, Bruno, thanks! :P – Dustan Levenstein Mar 22 '14 at 6:22
Note: feel free to ignore the warzone in the comments; it's not really relevant anymore.
If $F$ is perfect, we can proceed like this. Let $f$ be a polynomial with coefficients in $F$. Let $K/F$ be a splitting field for $f$. Then $K=F(\alpha)$ for some $\alpha \in K$. Let $g$ be the minimal polynomial of $\alpha$ over $F$. Then $g$ has a root in $E$ by assumption, hence $E$ contains a copy of $F(\alpha)$, i.e. a splitting field for $f$.
Thus every $f\in F[X]$ splits in $E$. Now I claim that $E$ is algebraically closed. Let $E'/E$ be an algebraic extension and let $\beta \in E'$. By transitivity, $\beta$ is algebraic over $F$; let $h(X)$ be its minimal polynomial over $F$. By the above, $h$ splits in $E$, and therefore $\beta \in E$. Thus $E$ is algebraically closed.
• Thank you. But how does the assumption on $E$ imply $f$ splits in $E$? If $f$ has a root $\beta\in E$, $f=(X-\beta)g(X)$ for $g\in E[X]$. If $\alpha=\beta$, we are done, otherwise $g(\alpha)=0$...not sure what follows next? – Nastassja Mar 21 '14 at 23:14
• @Nastassja Oops, I misread the assumption. My bad. I'm not immediately sure how to prove it (and I'm actually not sure that it's true - where did you get this statement?). – Bruno Joyal Mar 21 '14 at 23:35
• @BrunoJoyal, where and how did you prove that every non-constant pol. in $\;E[x]\;$ (!!) has a root in $\;E\;$ ? You don't even mess with any pol. over $\;E\;$ ! – DonAntonio Mar 22 '14 at 0:19
• @BrunoJoyal, or I'm missing some rather simple point here, or you are...or something: in order to prove some field $\;B\;$ is an alg. clos. of another field $\;A\;$ you must prove THAT $\;B\;$ is both algebraic over $\;A\;$ and alg. closed. For the third time, where did you prove $\;E\;$ is alg. closed (the only interesting thing to do since we're given it is alg. over $\;F\;$) ? – DonAntonio Mar 22 '14 at 0:24
• Ah! Now I get your point, @BrunoJoyal ....finally! It must be my two remaining living neurons are on vacation in Bahamas...Yes, you're right: if we're given (or if we prove) that $\;E\;$ is s.t. that any non-constant pol. in $\;F\;$ splits there then we're done. – DonAntonio Mar 22 '14 at 0:34
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# Sorted unsigned integer intervals tree structure
The goal is to store sorted unsigned integer intervals that can overlap only in the boundaries, e.g., |0–10|10–50|100–110|110–200|. This structure will be created by some insert operations (the order matters) in the start and than queried by a lot of queries. The query is specified by interval and the result is all intervals from structure that has at least a partial overlap.
There is no need to delete inserted data (delete operation), but the new inserted interval can overwrite the existing ones. Please consider the following example.
1. |0–10|10–50|100–110|110–200|
2. insert: 40–70
3. |0–10|10–40|40–70|100–110|110–200|
4. insert: 105–115
5. |0–10|10–40|40–70|100–105|105–115|115–200|
What structure do you recommend?
• IMO, its better to think about $B+$ trees. In $B+$ trees you would get keys sorted from left leaves to right. They are usually preferred for range queries. Commented Dec 8, 2018 at 12:12
• Welcome to Computer Science! "Is it right to think about ...?" That sounds ambiguous enough to possibly prevent readers to become confident to post a succinct answer. Can you list the desired qualities of (the insert operation on) the data structure? What are the typical use cases such as many insertions and few queries or few insertions and many queries? Commented Dec 8, 2018 at 12:19
• @Apass.Jack Thanks for your note. I tried to update my question. Commented Dec 8, 2018 at 14:06
• Can you confirm that the order of insertion matters? 10–50 followed by 40–70 is different from 40–70 followed by 10–50. Commented Dec 8, 2018 at 14:17
• Yes, you're right. The order of insertion matters. Commented Dec 8, 2018 at 14:31
As Mr. Sigma suggest you, you need to use a B+ tree structure. time to insert new value is $$\Theta(th)=\Theta(t\log _{t}n)$$
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Convert 平方公里 to rd2, sq rd | Chinese píngfāng gōnglǐ to square rods
# area surface units conversion
## Amount: 1 Chinese píngfāng gōnglǐ (平方公里) of area Equals: 39,536.86 square rods (rd2, sq rd) in area
Converting Chinese píngfāng gōnglǐ to square rods value in the area surface units scale.
TOGGLE : from square rods into Chinese píngfāng gōnglǐ in the other way around.
## area surface from Chinese píngfāng gōnglǐ to square rod conversion results
### Enter a new Chinese píngfāng gōnglǐ number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other area surface measuring units - complete list.
How many square rods are in 1 Chinese píngfāng gōnglǐ? The answer is: 1 平方公里 equals 39,536.86 rd2, sq rd
## 39,536.86 rd2, sq rd is converted to 1 of what?
The square rods unit number 39,536.86 rd2, sq rd converts to 1 平方公里, one Chinese píngfāng gōnglǐ. It is the EQUAL area value of 1 Chinese píngfāng gōnglǐ but in the square rods area unit alternative.
平方公里/rd2, sq rd area surface conversion result From Symbol Equals Result Symbol 1 平方公里 = 39,536.86 rd2, sq rd
## Conversion chart - Chinese píngfāng gōnglǐ to square rods
1 Chinese píngfāng gōnglǐ to square rods = 39,536.86 rd2, sq rd
2 Chinese píngfāng gōnglǐ to square rods = 79,073.72 rd2, sq rd
3 Chinese píngfāng gōnglǐ to square rods = 118,610.58 rd2, sq rd
4 Chinese píngfāng gōnglǐ to square rods = 158,147.44 rd2, sq rd
5 Chinese píngfāng gōnglǐ to square rods = 197,684.31 rd2, sq rd
6 Chinese píngfāng gōnglǐ to square rods = 237,221.17 rd2, sq rd
7 Chinese píngfāng gōnglǐ to square rods = 276,758.03 rd2, sq rd
8 Chinese píngfāng gōnglǐ to square rods = 316,294.89 rd2, sq rd
9 Chinese píngfāng gōnglǐ to square rods = 355,831.75 rd2, sq rd
10 Chinese píngfāng gōnglǐ to square rods = 395,368.61 rd2, sq rd
11 Chinese píngfāng gōnglǐ to square rods = 434,905.47 rd2, sq rd
12 Chinese píngfāng gōnglǐ to square rods = 474,442.33 rd2, sq rd
13 Chinese píngfāng gōnglǐ to square rods = 513,979.19 rd2, sq rd
14 Chinese píngfāng gōnglǐ to square rods = 553,516.05 rd2, sq rd
15 Chinese píngfāng gōnglǐ to square rods = 593,052.92 rd2, sq rd
Convert area surface of Chinese píngfāng gōnglǐ (平方公里) and square rods (rd2, sq rd) units in reverse from square rods into Chinese píngfāng gōnglǐ.
## Area units calculator
Main area or surface units converter page.
# Converter type: area surface units
First unit: Chinese píngfāng gōnglǐ (平方公里) is used for measuring area.
Second: square rod (rd2, sq rd) is unit of area.
QUESTION:
15 平方公里 = ? rd2, sq rd
15 平方公里 = 593,052.92 rd2, sq rd
Abbreviation, or prefix, for Chinese píngfāng gōnglǐ is:
Abbreviation for square rod is:
rd2, sq rd
## Other applications for this area surface calculator ...
With the above mentioned two-units calculating service it provides, this area surface converter proved to be useful also as a teaching tool:
1. in practicing Chinese píngfāng gōnglǐ and square rods ( 平方公里 vs. rd2, sq rd ) measures exchange.
2. for conversion factors between unit pairs.
3. work with area surface's values and properties.
To link to this area surface Chinese píngfāng gōnglǐ to square rods online converter simply cut and paste the following.
The link to this tool will appear as: area surface from Chinese píngfāng gōnglǐ (平方公里) to square rods (rd2, sq rd) conversion.
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# How can I feel more sexually liberated?
## How can I feel more sexually liberated?
What can you do about it?
1. Practice mindfully accepting sexual thoughts. Mindfulness can help you become more comfortable with sexual thoughts by increasing your awareness of them and learning to accept them without judgment. ...
2. Read up on sex positivity. ...
3. Get comfortable with your body. ...
4. Talk to your partner.
## What does sexually liberated mean?
Without discrimination or labelling." This definition concludes that a woman who is "sexually liberated" is a woman who is particular about her sexual partners, but chooses to have sex with them when she desires.
## What does it mean if you feel liberated?
making you feel free and able to behave as you like: Taking all your clothes off can be a very liberating experience. Freedom to act. (as) free as a bird idiom.
## What are the fan Tods?
[ fan-tods ] noun. a state of extreme nervousness or restlessness; the willies; the fidgets (usually preceded by the): We all developed the fantods when the plane was late in arriving.
## What is a white alley?
white Alley An alley is a fine marble used as the shooter in playing marbles. whitewash a mixture of lime, whiting, size, water, etc., for whitening walls and other surfaces.
## What is a cardioid shape?
A cardioid (from the Greek καρδία "heart") is a plane curve traced by a point on the perimeter of a circle that is rolling around a fixed circle of the same radius. ... Named for its heart-like form, it is shaped more like the outline of the cross section of a round apple without the stalk.
## What is cardioid area?
The area of the cardioid is given by: Area = 6 π a2. Area = 6 π 4. Area = 24 π sq unit. Example 2: Calculate the area and arc length of the cardioid, which is given by the following equation r = 7 (1 + cos θ).
## How do you find cardioid area?
The cardioid goes through the origin when θ=−π/2; the circle goes through the origin at multiples of π, starting with 0. so the area we seek is π/8....Example 10.
## What is cardioid equation?
If the radius of the circle that creates the cardioid is a, then we have the following: The equation of a horizontal cardioid is r = a ± acosθ. The equation of a vertical cardioid is r = a ± asinθ.
## How do you find polar area?
Key Concepts
1. The area of a region in polar coordinates defined by the equation r=f(θ) with α≤θ≤β is given by the integral A=12∫βα[f(θ)]2dθ.
2. To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
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Friday
May 6, 2016
Posted by kimberley on Friday, March 23, 2007 at 4:33pm.
A series of pulse of amplitude 0.15 m are sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude. What is the amplitude at a point on the string where two pulses are crossing, (a) if the string is rigidly attached to the post? (b) if the end at which reflection occurs is free to slide up and down?
I have no clue srry
If rigid, it comes back opposite phase, so they cancel. If loose, it comes back in phase, so they add.
http://www.kettering.edu/~drussell/Demos/reflect/reflect.html
umm like i looked at it..i dont get what u mean it cancels out still..like are we supposed to use a formula?
No formula needed. Unless, you have difficulty understanding 1-1= cancel, and 1+1= twice amplidtude. :)
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Cody
# Problem 2874. Cumulative product of a vector
Solution 1646406
Submitted on 15 Oct 2018 by Prateek Tyagi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(cum_prod(x),y_correct))
2 Pass
x = [1 2 5 10],; y_correct = [ 1 2 10 100]; assert(isequal(cum_prod(x),y_correct))
x = 1 2 5 10
3 Pass
x = [1 2 5 0],; y_correct = [ 1 2 10 0]; assert(isequal(cum_prod(x),y_correct))
x = 1 2 5 0
4 Pass
x = 1:6,; y_correct = [ 1 2 6 24 120 720]; assert(isequal(cum_prod(x),y_correct))
x = 1 2 3 4 5 6
5 Pass
x = 0:2,; y_correct = [ 0 0 0]; assert(isequal(cum_prod(x),y_correct))
x = 0 1 2
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# Particles and Waves in Conceptual Time Systems - PowerPoint PPT Presentation
Particles and Waves in Conceptual Time Systems. Karl Erich Wolff Mathematics and Science Faculty University of Applied Sciences Darmstadt Ernst Schröder Center for Conceptual Knowledge Processing Research Group Concept Analysis at Darmstadt University of Technology. Outline.
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## Particles and Wavesin Conceptual Time Systems
Karl Erich Wolff
Mathematics and Science Faculty
University of Applied Sciences Darmstadt
Ernst Schröder Center for Conceptual Knowledge Processing
Research Group Concept Analysis at
Darmstadt University of Technology
### Outline
• Introduction: Problems in Physics
• Temporal Concept Analysis
• Examples of Conceptual Time Systems
• Time Dimension and Branching Life Tracks
• Object Identification: Conceptual Representation of a Tale
• Particles and Waves
### Problems in Physics
• Reality – Observation - Theory
• Formal Representations: Numbers, Scales, and Systems
• Continuity and Discreteness
• Granularity, States, and Situations
• Transitions and Life Tracks
• Objects, Particles, and Waves
## Theories of Space and Time
• Aristoteles: continuum, point of time, duration, time as a category
• Classical Physics: example: x“(t)=a; x(t)=x(0) + v t + ½at²
• Special and General Theory of Relativity: Space-Time (curved)
• Quantum Theory: Time-dependent Schrödinger function ψ(t)
• Automata theory: States, Transitions (without an explicit time description)
• Mathematical System Theory: State? System?
• Quantum Gravity: Looking for a theory with an appropriate time description
## Einstein’s Granularity Remark
Albert Einstein:
„Zur Elektrodynamik bewegter Körper“
Annalen der Physik 17 (1905): 891-921
Footnote on page 893:
„Die Ungenauigkeit, welche in dem Begriff der Gleichzeitigkeit zweier Ereignisse an (annähernd) demselben Orte steckt und gleichfalls durch eine Abstraktion überbrückt werden muß, soll hier nicht erörtert werden.“
## A General Theory for Objects in Space and Time ...
• should cover discrete and continuous descriptions
• needs a theory of granularity
• needs a suitable notion of ‚system‘ and a notion of ‚states‘ depending on the granularity
• transitions: time-dependent changes between states
• objects as subsystems (and not as „atoms“)
• should construct a temporal logic based on these temporal systems
### Conceptual Time Systems with Objects
and a Time Relation (CTSOT)
Time part T
Event part C
g
v
w
Object 1
h
i
Object 2
j
Time scales
Event scales
K(C)
derived context
K(T) |
time states
states
object concepts:
situations
y
x
Monday
morning
evening
east
west
Tuesday
morning
evening
east
west
morning
evening
Monday
Tuesday
Tuesday
Monday
east
west
morning
evening
Monday
Tuesday
Tuesday
Monday
east
west
### Time Dimension = 1 ???
• Using the order dimension we define:
• The time dimension is the order dimension of the concept lattice of the time part
• The time scale dimension is the order dimension of the time scale
• These dimensions may be greater than one!
• Are life tracks 1-dimensional ?
## Branching Life Tracks?
Unique-State-Lemma:
A conceptual time system is at each time granule in exactly one situation, in exactly one state, and in exactly one time state.
But branching of life tracks is possible!
Example: We send an „abstract letter“ in two copies.
A
B
C
D
Translation
Rotation
Reflection
### Indistinguishable Objects ?
An observer of a temporal system who describes that system by a
CTSOT has to decide how to choose the objects of the CTSOT.
Like tennis balls they may be indistinguishable with repect
to some parts of the chosen description.
In the following German tale
Der Wettlauf zwischen dem Hasen und dem Igel
(The Race Between the Hare and the Hedgehog)
the hare cannot distinguish the hedgehog and the hedgehog‘s wife.
hedgehog
hh-wife
hare
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# Why the induced emf is always negative?
1. Jan 16, 2008
### Physicsissuef
Hi! I read this article about Faraday's law, which states, that the emf will be always negative, by the formula, is this http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html" [Broken]
Last edited by a moderator: May 3, 2017
2. Jan 16, 2008
### Ronnin
Lenz law supplies that answer. It basically states that the EMF that is induced will be in such a way to create an opposing magnetic field to the one inducing the EMF. This law stems from the conservation of energy.
3. Jan 16, 2008
### Physicsissuef
Conservation of energy of what?
Btw- Look at this http://www.youtube.com/watch?v=stUDqGzpev8". The voltage is not negative, always.
Last edited by a moderator: Apr 23, 2017
4. Jan 16, 2008
### rohanprabhu
Simply put, if the induced emf was in the direction of the emf of the circuit, then the inductor would increase the voltage in the circuit. Which would increase the current flowing across the inductor. As such, it would further increase the induced emf [they are proportional] as $B = \mu_o NI$. This increase would further increase the the emf and as such, the voltage should rise exponentially. Which would mean, we could derive infinite energy from the system.
as we know that this is not possible, the induced emf cannot be in the same direction. Although, this cannot be used to prove Lentz law, but is helpful for understanding the situation.
5. Jan 16, 2008
### Staff: Mentor
The negative sign in Faraday's law (Lenz's law) does not mean that the EMF (or current) always points in some "negative" direction. It means that the current always flows in a way to oppose the change in flux, which is nicely illustrated in that video clip.
Last edited by a moderator: Apr 23, 2017
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##### A pond with a flat bottom has a surface area of 860 m2 and a depth of 2.8 m. On
label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
A pond with a flat bottom has a surface area of 860 m2 and a depth of 2.8 m. On a warm day, the surface water is at a temperature of 25°C, while the bottom of the pond is at 12°C. Find the rate at which energy is transferred by conduction from the surface to the bottom of the pond.
Jun 20th, 2015
Q/t = kA(T hot - T cold )/d k=2.18W/(m. K)
Q/t = 2.18(860)(25 - 12 )/2.8
Q/t= 11180 watt
Jun 20th, 2015
...
Jun 20th, 2015
...
Jun 20th, 2015
Oct 18th, 2017
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Diapositivas de la sesión 3
Dr. Jorge Ramírez Medina
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• Aclarar que función Binomial será vista rápidamente
2. 2. De la sesión anterior Dr Jorge Ramírez Medina EGADE Business School catastrophic\$(e.g.,\$the\$model\$predicts\$that\$the\$bridge\$will\$collapse\$in\$a\$strong\$wind,\$causing\$the\$real\$ bridge\$to\$be\$closed\$down,\$creating\$100Jmile\$tailbacks\$with\$everyone\$stranded\$in\$the\$snow;\$all\$of\$ which\$was\$unnecessary\$because\$the\$real\$bridge\$was\$perfectly\$safe—the\$model\$was\$a\$bad\$ representation\$of\$reality).\$We\$can\$have\$some\$confidence,\$but\$not\$complete\$confidence,\$in\$ predictions\$from\$this\$model.\$The\$final\$model\$is\$completely\$different\$to\$the\$realJworld\$situation;\$it\$ bears\$no\$structural\$similarities\$to\$the\$real\$bridge\$and\$is\$a\$poor\$fit.\$As\$such,\$any\$predictions\$based\$on\$ this\$model\$are\$likely\$to\$be\$completely\$inaccurate.\$Extending\$this\$analogy\$to\$science,\$it\$is\$important\$ when\$we\$fit\$a\$statistical\$model\$to\$a\$set\$of\$data\$that\$it\$fits\$the\$data\$well.\$If\$our\$model\$is\$a\$poor\$fit\$of\$ the\$observed\$data\$then\$the\$predictions\$we\$make\$from\$it\$will\$be\$equally\$poor.\$ \$ \$ Figure'2.2:'Fitting'models'to'real5world'data'(see'text'for'details)' Jane'Superbrain'Box'2.1'Types'of'statistical'models'(1)' T h e R e a l W o r ld G o o d F it M o d e r a t e F it P o o r F it
3. 3. Dr Jorge Ramírez Medina EGADE Business School Nuestro interés es el número de éxitosNuestro interés es el número de éxitos que ocurren en los n intentos.que ocurren en los n intentos. Tomamos x como el número de éxitosTomamos x como el número de éxitos que ocurren en los n intentos.que ocurren en los n intentos. Distribución Binomial
4. 4. Dr Jorge Ramírez Medina EGADE Business School donde: f(x) = La probabilidad de x éxitos en n intentos n = el número de intentos p = la probabilidad de éxito de cualquier intento Función de probabilidad binomial Distribución Binomial )( )1( )!(! ! )( xnx pp xnx n xf − − − =
5. 5. Dr Jorge Ramírez Medina EGADE Business School Función de probabilidad binomial Distribución Binomial Probabilidad de unaProbabilidad de una secuencia particular de resultadossecuencia particular de resultados con x éxitos en n intentoscon x éxitos en n intentos Número de resultadosNúmero de resultados experimentales que danexperimentales que dan x éxitos en intentosx éxitos en intentos )( )1( )!(! ! )( xnx pp xnx n xf − − − =
6. 6. Dr Jorge Ramírez Medina EGADE Business School Ejemplo La empresa está preocupada por la alta rotación de sus empleados. Para un empleado seleccionado al azar, se estima una probabilidad de 0.1 de que la persona no esté el próximo semestre trabajando. Si se seleccionan 3 empleados al azar ¿cuál es la probabilidad de que uno de ellos no esté trabajando el próximo semestre en el CITEC? Distribución Binomial
7. 7. Dr Jorge Ramírez Medina EGADE Business School Diagrama de árbol 1st Worker 2nd Worker 3rd Worker x Prob. Leaves (.1) Stays (.9) 3 2 0 2 2 Leaves (.1) Leaves (.1) S (.9) Stays (.9) Stays (.9) S (.9) S (.9) S (.9) L (.1) L (.1) L (.1) L (.1) .0010 .0090 .0090 .7290 .0090 1 1 .0810 .0810 .0810 11 Distribución Binomial
8. 8. Dr Jorge Ramírez Medina EGADE Business School Utilizando la función de probabilidad Binomial tome: p = .10, n = 3, x = 1 Distribución Binomial )( )1( )!(! ! )( xnx pp xnx n xf − − − = 243.0)81)(.1(.3)1.01(1.0 )!13(!1 !3 )1( )13(1 ==− − = − f
9. 9. Dr Jorge Ramírez Medina EGADE Business School utilizando Tablas de Probabilidad Binomial n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50 3 0 .8574 .7290 .6141 .2430 .4219 .3430 .2746 .2160 .1664 .1250 1 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750 2 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750 3 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250 p Distribución Binomial X P(X) 0 0.729 1 0.243 2 0.027 3 0.001 Utilizando excel Binomial
10. 10. Dr Jorge Ramírez Medina EGADE Business School El valorEl valor esperadoesperado;; La varianza;La varianza; La desviación estándar,La desviación estándar, σσ == Var(Var(xx) =) = σσ 22 == np(1-pnp(1-p) EE((xx) =) = µµ == npnp Distribución Binomial )1( pnp −
11. 11. Dr Jorge Ramírez Medina EGADE Business School E(x) =E(x) = npnp = 3(.1) = .3= 3(.1) = .3 empleadosempleados de 3de 3 Var(Var(xx) =) = σσ 22 == 3(.1)(.9) = .273(.1)(.9) = .27 Distribución Binomial empleados52.)9)(.1(.3 ==σ
12. 12. Dr Jorge Ramírez Medina EGADE Business School Una variable aleatoria con una distribución PoissonUna variable aleatoria con una distribución Poisson es útil para estimar el número de ocurrencias sobrees útil para estimar el número de ocurrencias sobre un intervalo especificado de tiempo o espacio.un intervalo especificado de tiempo o espacio. Es una variable aleatoria discreta que puede tomarEs una variable aleatoria discreta que puede tomar una secuencia de valores infinita (x = 0, 1, 2, . . . ).una secuencia de valores infinita (x = 0, 1, 2, . . . ). Distribución Poisson
13. 13. Dr Jorge Ramírez Medina EGADE Business School Ejemplo de variables aleatorias conEjemplo de variables aleatorias con distribución Poissondistribución Poisson La cantidad de fugas en 10 km. de unLa cantidad de fugas en 10 km. de un gaseoductogaseoducto Los automóviles que pasan porLos automóviles que pasan por una caseta en una horauna caseta en una hora Distribución Poisson
14. 14. Distribución Poisson Dr Jorge Ramírez Medina EGADE Business School Propiedades de los experimentos Poisson La ocurrencia o no-ocurrencia en cualquierLa ocurrencia o no-ocurrencia en cualquier intervalo es independiente de la ocurrencia ointervalo es independiente de la ocurrencia o no-occurrencia en cualquier otro intervalo.no-occurrencia en cualquier otro intervalo. La probabilidad de una ocurrencia es la mismaLa probabilidad de una ocurrencia es la misma para dos intervalos cualesquiera de igual longitudpara dos intervalos cualesquiera de igual longitud
15. 15. Dr Jorge Ramírez Medina EGADE Business School Distribución Poisson Función de probabilidad Poisson en donde:en donde: f(x) = probabilidad de x ocurrencias en un intervalof(x) = probabilidad de x ocurrencias en un intervalo µ= media de ocurrencias en un intervaloµ= media de ocurrencias en un intervalo e = 2.71828e = 2.71828 ! )( x e xf x µ µ − =
16. 16. Dr Jorge Ramírez Medina EGADE Business School MERCYMERCY • Ejemplo: Hospital López Mateos Los fines de semana en la tarde a la sala de emergencias del Hospital LM llegan en promedio 6 pacientes por hora . Cuál es la probabilidad de que lleguen 4 pacientes en 30 minutos en la tarde de un fin de semana? Distribución Poisson
17. 17. Dr Jorge Ramírez Medina EGADE Business School Utilizando la Función de Probabilidad Poisson MERCYMERCY µ = 6/hora = 3/media-hora, x = 4 Distribución Poisson 1680.0 !4 )71828.2(3 )4( 34 == − f
18. 18. Dr Jorge Ramírez Medina EGADE Business School Utilizando las tablas de probabilidad Poisson MERCYMERCY Distribución Poisson Utilizando excel; =POISSON(4,3,FALSO)
19. 19. Distribución Poisson Dr Jorge Ramírez Medina ITESM EGADE Zona Centro MERCYMERCY Poisson Distribution of Arrivals Poisson Probabilities 0.00 0.05 0.10 0.15 0.20 0.25 0 1 2 3 4 5 6 7 8 9 10 Número de llegadas en 30 Minutos Probabilidad La secuenciaLa secuencia continua:continua: 11, 12, …11, 12, …
20. 20. Dr Jorge Ramírez Medina EGADE Business School Una propiedad de la distribución Poisson es queUna propiedad de la distribución Poisson es que La media y la varianza son iguales.La media y la varianza son iguales. µ = σ 2 Distribución Poisson
21. 21. Dr Jorge Ramírez Medina EGADE Business School MERCYMERCY Varianza de las llegadas durante el periodo de 30 minutos. µ = σ 2 = 3 Distribución Poisson
22. 22. Dr Jorge Ramírez Medina EGADE Business School SLOW Distribución de probabilidad exponencial • Útil para describir el tiempo que toma el completar una tarea. • Las variables aleatorias exponenciales pueden ser utilizadas para describir: Tiempo de llegada Entre vehículos a una caseta. Tiempo requerido para llenar un cuestionario Distancia entre baches en una autopista
23. 23. Dr Jorge Ramírez Medina EGADE Business School • Función de densidad donde: µ = media e = 2.71828 Para xPara x ≥0,≥0, μ≥μ≥00 Distribución de probabilidad exponencial µ µ x exf − = 1 )(
24. 24. Dr Jorge Ramírez Medina EGADE Business School • Probabilidades acumulativas donde: x0 = algún valor específico de x Distribución de probabilidad exponencial − −=≤ µ ox exxP 1)( 0
25. 25. Dr Jorge Ramírez Medina EGADE Business School • Ejemplo; gasolinera las Torres El tiempo entre carros que llegan a la gasolinera las Torres sigue una distribución de probabilidad exponencial con una media entre llegadas de 3 minutos. Se quiere saber cuál es la probabilidad de que el tiempo entre 2 llegadas sea menor o igual de 2 minutos. Distribución de probabilidad exponencial
26. 26. Dr Jorge Ramírez Medina EGADE Business School x f(x) .1 .3 .4 .2 1 2 3 4 5 6 7 8 9 10 Tiempo entre llegadas (mins.) P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866 Distribución de probabilidad exponencial
27. 27. Dr Jorge Ramírez Medina EGADE Business School Una propiedad de la distribución exponencial esUna propiedad de la distribución exponencial es que la media,que la media, µµ, y la desviación estándar,, y la desviación estándar, σσ, son iguales, son iguales La desviación estándar,La desviación estándar, σσ, y la varianza,, y la varianza, σσ 22 , para el, para el tiempo entre llegadas en la gasolinera las Torres:tiempo entre llegadas en la gasolinera las Torres: σ = µ = 3 minutes σ 2 = (3)2 = 9 Distribución de probabilidad exponencial
28. 28. Dr Jorge Ramírez Medina EGADE Business School La distribución exponencial está sesgada positivamente.La distribución exponencial está sesgada positivamente. La medición del sesgo para la distribuciónLa medición del sesgo para la distribución exponencial es 2.exponencial es 2. Distribución de probabilidad exponencial
29. 29. Dr Jorge Ramírez Medina EGADE Business School La distribución PoissonLa distribución Poisson da una descripción apropiadada una descripción apropiada del número de ocurrenciasdel número de ocurrencias por intervalopor intervalo La distribución exponencialLa distribución exponencial da una descripción apropiadada una descripción apropiada de la longitud del intervalode la longitud del intervalo entre las ocurrenciasentre las ocurrencias Relación entre las distribuciones exponencial y Poisson
30. 30. Reflexión en clase Dr Jorge Ramírez Medina EGADE Business School
31. 31. Uso y abuso de la estadística • Cuidado con lo que asume. • Sea claro acerca quiere descubrir. • No tome la causalidad por sentado. • Con estadística no se puede probar cosas con el 100% de certeza • Un resultado que es numéricamente significativo puede ser inútil. Tomado de The Use and Misuse of statistics HBP. Dr Jorge Ramírez Medina EGADE Business School
32. 32. Fin Sesión 3
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# Characterization of conjugate natural transformations (MacLane)
I have trouble finding the proof of MacLane's statement that "(6) implies (5)" at page 101 of his Categories for the Working Mathematician. This is part of the proof of his Theorem 2 in the "Transformations of Adjoints" chapter (7).
The definition of conjugate natural transformations goes:
Let be two adjunctions $$ \text{ and } $$ then the two natural transformations $$\sigma : F \xrightarrow{.} F' \text{ and } \tau : G' \xrightarrow{.} G$$ are conjugate given the previous two adjunctions iff the following diagram commutes:
$$\require{AMScd}$$ $$\begin{CD} A(F'x, a) @>\varphi'>\approx> X(x, G' a) \\ @V(\sigma_x)^*VV @VV(\tau_a)_*V \\ A(Fx,a) @>\varphi>\approx> X(x,Ga) \end{CD}$$
The previous diagram is labelled (5) in the book.
The statement I'm struggling to proove is that (5) is implied (allegedly without the need of further explanation) by the following two diagrams (labelled (6) as a pair):
$$\begin{CD} G' @>\tau>> G \\ @VV\eta G'V @AA G\epsilon'A \\ G F G' @>G\sigma G'>> GF'G' \end{CD}$$
and
$$\begin{CD} F @>\tau>> F' \\ @VVF\eta'V @AA \epsilon F'A \\ FG'F' @>F\tau F' >> F G F' \end{CD}$$
I see clearly the implication in the other direction, well explained by MacLane, as one only needs to add an arrow from the one-element set $$*$$ to $$X(x,G'a)$$ that points to its identity element $$id_{G'a}$$ and chase the diagram. However I don't see how we can, the other way around, generalize from the identity element to the full Homset.
Some compositions that arise from adjunctions would lead me to the point where I still need to show things like $$G \epsilon'_a \circ G F' U \circ G \sigma_{x} \circ \eta_{x} = G\epsilon'_a \circ G \sigma_{G'a} \circ \eta_{G'a} \circ U$$ which seems to me intractable, as the way to pass $$U$$ through from the right would be to use $$\epsilon$$ not $$\eta$$.
• What is $U$ here? – Arnaud D. Feb 25 '19 at 13:58
• Thanks for your comment! Indeed it deserves a clarification: $U$ is an arbitrary member of the homset $X(x, G'a)$ that I substitute to $id_{G'a}$ to try to generalize. – Almeo Maus Feb 25 '19 at 14:10
• Then your left-hand side should really by $G \epsilon'_a \circ G F' (U) \circ G \sigma_{x} \circ \eta_{x}$, no? – Arnaud D. Feb 25 '19 at 14:17
• Ah you're right, thanks! :) Let me fix that. That said, that doesn't solve my problem, right? – Almeo Maus Feb 25 '19 at 14:25
Use the naturality of $$\sigma$$ first, and then the naturality of $$\eta$$ : \begin{align}G \epsilon'_a \circ G F'U \circ G \sigma_{x} \circ \eta_{x} & = G (\epsilon'_a) \circ G \sigma_{G'a}\circ GFU \circ \eta_{x}\\ & = G\epsilon'_a \circ G \sigma_{G'a} \circ \eta_{G'a} \circ U.\end{align}
• Maybe could you just fix the last $f$ of your answer into an $U$? This was the whole point of my attempt to edit your answer (and some tricks trying to workaround the rule that all edits must be at least 6 characters). – Almeo Maus Feb 28 '19 at 4:07
• @AlmeoMaus Ah sorry, I rejected the edit because I thought the tricks were not necessary but I missed that $f$... I should have read the note more carefully ! It's done now. – Arnaud D. Feb 28 '19 at 8:42
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# Complex Analysis
1. Jan 12, 2010
### metgt4
1. The problem statement, all variables and given/known data
By considering the real and imaginary parts of the product eithetaeiphi, prove the standard formulae for cos(theta+phi) and sin(theta+phi)
2. Relevant equations
The standard formula for:
cos(theta+phi) = cos(theta)cos(phi) - sin(theta)sin(phi)
sin(theta+phi) = sin(theta)cos(phi) + sin(phi)cos(theta)
3. The attempt at a solution
from hereon out, lets let t=theta and p=phi. I can't figure out how to put the symbols in there on this site
ei(t+p) = [cos(t) + isin(t)] [cos(p) + isin(p)]
= cos(t)cos(p) - sin(t)sin(p) + isin(t)cos(p) + isin(p)cos(t)
REAL PART = cos(t)cos(p) - sin(t)sin(p)
IMAGINARY PART = isin(t)cos(p) + isin(p)cos(t)
but I can't use the trig identities because I have to prove them. I'm probably not going in the right direction, but if somebody could point me in the right way, that would be great!
Thanks!
Andrew
2. Jan 12, 2010
### Matterwave
You could then expand $$e^{i(\phi+\theta)}=cos(\phi+\theta)+isin(\phi+\theta)$$ by just considering phi+theta to be collectively x in the Euler's formula. If you match the real and imaginary parts you should get the correct trig identities.
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Re: Any way to get gradient lines as well as contour lines?
• To: mathgroup at smc.vnet.net
• Subject: [mg129672] Re: Any way to get gradient lines as well as contour lines?
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Sun, 3 Feb 2013 20:21:16 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
• Delivered-to: mathgroup-newsend@smc.vnet.net
```Grad[Sin[x + y^2], {x, y}]
{Cos[x + y^2], 2*y*Cos[x + y^2]}
Plot3D[Sin[x + y^2], {x, -Pi, Pi}, {y, -2, 2},
MeshFunctions -> {
Cos[#1 + #2^2] &,
2 #2 Cos[#1 + #2^2] &},
Mesh -> {10, 20},
PlotPoints -> 35]
Bob Hanlon
On Sun, Feb 3, 2013 at 2:48 AM, Chris Young <cy56 at comcast.net> wrote:
> I'd like to be able to plot "gradient lines", i.e., lines of steepest
> descent, on plots of functions of x and y, f(x, y). The contour lines
> can be obtained via MeshFunctions, as in
>
> Plot3D[Sin[x + y^2], {x, -3, 3}, {y, -2, 2}, MeshFunctions -> {#3 &},
> Mesh -> 5].
>
> I'd like all the lines at right angles to these contours.
>
> There must be some way to use the gradient function and then integrate
> to the get the gradient lines.
>
> Maybe this is difficult to do, in general. Also, maybe in general the
> "gradient lines" wouldn't be continuous.
>
> But I wonder if it could be done for some simple examples.
>
> Any suggestions appreciated.
>
> Chris Young
> cy56 at comcast.net
>
>
```
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Copied to
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## G = C9○He3⋊4S3order 486 = 2·35
### 2nd semidirect product of C9○He3 and S3 acting via S3/C3=C2
Aliases: C9○He34S3, (C32×C9)⋊26S3, C3⋊(He3.4C6), He3.14(C3×S3), (C3×He3).26C6, C33.48(C3×S3), He35S3.5C3, C9.2(C33⋊C2), (C3×C9)⋊8(C3⋊S3), (C3×C9○He3)⋊5C2, C32.15(C3×C3⋊S3), C3.7(C3×C33⋊C2), SmallGroup(486,246)
Series: Derived Chief Lower central Upper central
Derived series C1 — C3 — C3×He3 — C9○He3⋊4S3
Chief series C1 — C3 — C32 — He3 — C3×He3 — C3×C9○He3 — C9○He3⋊4S3
Lower central C3×He3 — C9○He3⋊4S3
Upper central C1 — C9
Generators and relations for C9○He34S3
G = < a,b,c,d,e,f | a9=b3=d3=e3=f2=1, c1=a6, ab=ba, ac=ca, ad=da, ae=ea, af=fa, bc=cb, dbd-1=ebe-1=a3b, fbf=b-1, cd=dc, ce=ec, cf=fc, de=ed, fdf=d-1, fef=e-1 >
Subgroups: 812 in 231 conjugacy classes, 60 normal (10 characteristic)
C1, C2, C3, C3, S3, C6, C9, C9, C32, C32, C32, C18, C3×S3, C3⋊S3, C3×C9, C3×C9, C3×C9, He3, 3- 1+2, C33, S3×C9, He3⋊C2, C3×C3⋊S3, C32×C9, C3×He3, C3×3- 1+2, C9○He3, C9○He3, C9×C3⋊S3, He3.4C6, He35S3, C3×C9○He3, C9○He34S3
Quotients: C1, C2, C3, S3, C6, C3×S3, C3⋊S3, C3×C3⋊S3, C33⋊C2, He3.4C6, C3×C33⋊C2, C9○He34S3
Smallest permutation representation of C9○He34S3
On 54 points
Generators in S54
(1 2 3 4 5 6 7 8 9)(10 11 12 13 14 15 16 17 18)(19 20 21 22 23 24 25 26 27)(28 29 30 31 32 33 34 35 36)(37 38 39 40 41 42 43 44 45)(46 47 48 49 50 51 52 53 54)
(1 44 36)(2 45 28)(3 37 29)(4 38 30)(5 39 31)(6 40 32)(7 41 33)(8 42 34)(9 43 35)(10 54 22)(11 46 23)(12 47 24)(13 48 25)(14 49 26)(15 50 27)(16 51 19)(17 52 20)(18 53 21)
(1 7 4)(2 8 5)(3 9 6)(10 16 13)(11 17 14)(12 18 15)(19 25 22)(20 26 23)(21 27 24)(28 34 31)(29 35 32)(30 36 33)(37 43 40)(38 44 41)(39 45 42)(46 52 49)(47 53 50)(48 54 51)
(1 36 38)(2 28 39)(3 29 40)(4 30 41)(5 31 42)(6 32 43)(7 33 44)(8 34 45)(9 35 37)(10 25 54)(11 26 46)(12 27 47)(13 19 48)(14 20 49)(15 21 50)(16 22 51)(17 23 52)(18 24 53)
(1 30 44)(2 31 45)(3 32 37)(4 33 38)(5 34 39)(6 35 40)(7 36 41)(8 28 42)(9 29 43)(10 22 48)(11 23 49)(12 24 50)(13 25 51)(14 26 52)(15 27 53)(16 19 54)(17 20 46)(18 21 47)
(1 17)(2 18)(3 10)(4 11)(5 12)(6 13)(7 14)(8 15)(9 16)(19 43)(20 44)(21 45)(22 37)(23 38)(24 39)(25 40)(26 41)(27 42)(28 53)(29 54)(30 46)(31 47)(32 48)(33 49)(34 50)(35 51)(36 52)
G:=sub<Sym(54)| (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54), (1,44,36)(2,45,28)(3,37,29)(4,38,30)(5,39,31)(6,40,32)(7,41,33)(8,42,34)(9,43,35)(10,54,22)(11,46,23)(12,47,24)(13,48,25)(14,49,26)(15,50,27)(16,51,19)(17,52,20)(18,53,21), (1,7,4)(2,8,5)(3,9,6)(10,16,13)(11,17,14)(12,18,15)(19,25,22)(20,26,23)(21,27,24)(28,34,31)(29,35,32)(30,36,33)(37,43,40)(38,44,41)(39,45,42)(46,52,49)(47,53,50)(48,54,51), (1,36,38)(2,28,39)(3,29,40)(4,30,41)(5,31,42)(6,32,43)(7,33,44)(8,34,45)(9,35,37)(10,25,54)(11,26,46)(12,27,47)(13,19,48)(14,20,49)(15,21,50)(16,22,51)(17,23,52)(18,24,53), (1,30,44)(2,31,45)(3,32,37)(4,33,38)(5,34,39)(6,35,40)(7,36,41)(8,28,42)(9,29,43)(10,22,48)(11,23,49)(12,24,50)(13,25,51)(14,26,52)(15,27,53)(16,19,54)(17,20,46)(18,21,47), (1,17)(2,18)(3,10)(4,11)(5,12)(6,13)(7,14)(8,15)(9,16)(19,43)(20,44)(21,45)(22,37)(23,38)(24,39)(25,40)(26,41)(27,42)(28,53)(29,54)(30,46)(31,47)(32,48)(33,49)(34,50)(35,51)(36,52)>;
G:=Group( (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54), (1,44,36)(2,45,28)(3,37,29)(4,38,30)(5,39,31)(6,40,32)(7,41,33)(8,42,34)(9,43,35)(10,54,22)(11,46,23)(12,47,24)(13,48,25)(14,49,26)(15,50,27)(16,51,19)(17,52,20)(18,53,21), (1,7,4)(2,8,5)(3,9,6)(10,16,13)(11,17,14)(12,18,15)(19,25,22)(20,26,23)(21,27,24)(28,34,31)(29,35,32)(30,36,33)(37,43,40)(38,44,41)(39,45,42)(46,52,49)(47,53,50)(48,54,51), (1,36,38)(2,28,39)(3,29,40)(4,30,41)(5,31,42)(6,32,43)(7,33,44)(8,34,45)(9,35,37)(10,25,54)(11,26,46)(12,27,47)(13,19,48)(14,20,49)(15,21,50)(16,22,51)(17,23,52)(18,24,53), (1,30,44)(2,31,45)(3,32,37)(4,33,38)(5,34,39)(6,35,40)(7,36,41)(8,28,42)(9,29,43)(10,22,48)(11,23,49)(12,24,50)(13,25,51)(14,26,52)(15,27,53)(16,19,54)(17,20,46)(18,21,47), (1,17)(2,18)(3,10)(4,11)(5,12)(6,13)(7,14)(8,15)(9,16)(19,43)(20,44)(21,45)(22,37)(23,38)(24,39)(25,40)(26,41)(27,42)(28,53)(29,54)(30,46)(31,47)(32,48)(33,49)(34,50)(35,51)(36,52) );
G=PermutationGroup([[(1,2,3,4,5,6,7,8,9),(10,11,12,13,14,15,16,17,18),(19,20,21,22,23,24,25,26,27),(28,29,30,31,32,33,34,35,36),(37,38,39,40,41,42,43,44,45),(46,47,48,49,50,51,52,53,54)], [(1,44,36),(2,45,28),(3,37,29),(4,38,30),(5,39,31),(6,40,32),(7,41,33),(8,42,34),(9,43,35),(10,54,22),(11,46,23),(12,47,24),(13,48,25),(14,49,26),(15,50,27),(16,51,19),(17,52,20),(18,53,21)], [(1,7,4),(2,8,5),(3,9,6),(10,16,13),(11,17,14),(12,18,15),(19,25,22),(20,26,23),(21,27,24),(28,34,31),(29,35,32),(30,36,33),(37,43,40),(38,44,41),(39,45,42),(46,52,49),(47,53,50),(48,54,51)], [(1,36,38),(2,28,39),(3,29,40),(4,30,41),(5,31,42),(6,32,43),(7,33,44),(8,34,45),(9,35,37),(10,25,54),(11,26,46),(12,27,47),(13,19,48),(14,20,49),(15,21,50),(16,22,51),(17,23,52),(18,24,53)], [(1,30,44),(2,31,45),(3,32,37),(4,33,38),(5,34,39),(6,35,40),(7,36,41),(8,28,42),(9,29,43),(10,22,48),(11,23,49),(12,24,50),(13,25,51),(14,26,52),(15,27,53),(16,19,54),(17,20,46),(18,21,47)], [(1,17),(2,18),(3,10),(4,11),(5,12),(6,13),(7,14),(8,15),(9,16),(19,43),(20,44),(21,45),(22,37),(23,38),(24,39),(25,40),(26,41),(27,42),(28,53),(29,54),(30,46),(31,47),(32,48),(33,49),(34,50),(35,51),(36,52)]])
63 conjugacy classes
class 1 2 3A 3B 3C 3D 3E 3F ··· 3Q 6A 6B 9A ··· 9F 9G ··· 9L 9M ··· 9AJ 18A ··· 18F order 1 2 3 3 3 3 3 3 ··· 3 6 6 9 ··· 9 9 ··· 9 9 ··· 9 18 ··· 18 size 1 27 1 1 2 2 2 6 ··· 6 27 27 1 ··· 1 2 ··· 2 6 ··· 6 27 ··· 27
63 irreducible representations
dim 1 1 1 1 2 2 2 2 3 6 type + + + + image C1 C2 C3 C6 S3 S3 C3×S3 C3×S3 He3.4C6 C9○He3⋊4S3 kernel C9○He3⋊4S3 C3×C9○He3 He3⋊5S3 C3×He3 C32×C9 C9○He3 He3 C33 C3 C1 # reps 1 1 2 2 4 9 18 8 12 6
Matrix representation of C9○He34S3 in GL5(𝔽19)
1 0 0 0 0 0 1 0 0 0 0 0 6 0 0 0 0 0 6 0 0 0 0 0 6
,
1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0
,
1 0 0 0 0 0 1 0 0 0 0 0 11 0 0 0 0 0 11 0 0 0 0 0 11
,
18 1 0 0 0 18 0 0 0 0 0 0 0 11 0 0 0 0 0 7 0 0 1 0 0
,
0 18 0 0 0 1 18 0 0 0 0 0 0 11 0 0 0 0 0 7 0 0 1 0 0
,
0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0
G:=sub<GL(5,GF(19))| [1,0,0,0,0,0,1,0,0,0,0,0,6,0,0,0,0,0,6,0,0,0,0,0,6],[1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0],[1,0,0,0,0,0,1,0,0,0,0,0,11,0,0,0,0,0,11,0,0,0,0,0,11],[18,18,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,11,0,0,0,0,0,7,0],[0,1,0,0,0,18,18,0,0,0,0,0,0,0,1,0,0,11,0,0,0,0,0,7,0],[0,1,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0] >;
C9○He34S3 in GAP, Magma, Sage, TeX
C_9\circ {\rm He}_3\rtimes_4S_3
% in TeX
G:=Group("C9oHe3:4S3");
// GroupNames label
G:=SmallGroup(486,246);
// by ID
G=gap.SmallGroup(486,246);
# by ID
G:=PCGroup([6,-2,-3,-3,-3,-3,-3,979,218,867,3244,382]);
// Polycyclic
G:=Group<a,b,c,d,e,f|a^9=b^3=d^3=e^3=f^2=1,c^1=a^6,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,a*f=f*a,b*c=c*b,d*b*d^-1=e*b*e^-1=a^3*b,f*b*f=b^-1,c*d=d*c,c*e=e*c,c*f=f*c,d*e=e*d,f*d*f=d^-1,f*e*f=e^-1>;
// generators/relations
×
𝔽
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+1-415-315-9853
info@mywordsolution.com
## Accounting
Accounting Basics Cost Accounting Financial Accounting Managerial Accounting Auditing Taxation
On January 1, 2011, Piper Co. issued 10-year bonds with a face value of \$1,000,000 and a stated interest rate of 10%, payable semiannually on June 30 and December 31. The bonds were sold to yield 12%. Table values are:
Present value of 1 for 10 periods at 10% .386
Present value of 1 for 10 periods at 12% .322
Present value of 1 for 20 periods at 5% .377
Present value of 1 for 20 periods at 6% .312
Present value of annuity for 10 periods at 10% 6.145
Present value of annuity for 10 periods at 12% 5.650
Present value of annuity for 20 periods at 5% 12.462
Present value of annuity for 20 periods at 6% 11.470
Instructions:
- find out the issue price of the bonds.
- Without prejudice to your solution in Part (a), assume that the issue price was \$884,000. Prepare the amortization table for 2011, assuming that amortization is recorded on interest payment dates.
Accounting Basics, Accounting
• Category:- Accounting Basics
• Reference No.:- M945037
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The Zero-Truncated Poisson distribution is a sample variant of the Poisson distribution that has no zero value. A simple example of this is a the distribution of items a customer has in their shopping cart before approaching a register where it is common to presume that the customer will not approach the cash register without any items to purchase. Efficient sampling algorithms exist and are common for the Poisson distribution, however, few sampling algorithms exist for the Zero-Truncated Poisson distribution. We derive such an algorithm here using the inverse transform method.
We begin by introducing the sampling problem and the inverse transform method. Then, we review the Poisson distribution and how we derive a sampling algorithm using the inverse transform method. Finally, we introduce the Zero-Truncated Poisson distribution and how we derive a sampling algorithm using the inverse transform method. Throughout this post, we use Julia as the programming language to explore these algorithms and verify properties empirically.
Sampling Problem
The problem of sampling is straightforward given an elementary background in statistics. Given a probability distribution $f(x)$ and a sample size $n$, return a list of $n$ values $x$ that are drawn according to the distribution. We call the list of values a sample.
For example, suppose we had a Bernoulli distribution where the probability of success is $\frac{1}{4}$. We represent a successful case as 1 and an unsuccessful case as 0.
As a random variable, we denote this as $X \sim \operatorname{Bernoulli}(\frac{1}{4})$.
If we sampled 100 values from this distribution, we should expect approximately 25 of 1 and 75 of 0. In Julia, we may simulate this using the Distributions package.
using Distributions
# Define the Bernoulli distribution with parameters
bernoulli = Bernoulli(0.25)
# Draw samples from the distribution
samples = rand(bernoulli, 100)
# Count the number of success and failures
num_successes = count(x -> x == 1, samples)
num_failures = count(x -> x == 0, samples)
There are several methods for deriving a sampling algorithm for a specified distribution. For this post, we will use the inverse transform method.
Inverse Transform Method
The inverse transform method is a general technique for designing sampling algorithms when the cumulative density function $F(x)$ is known and the inverse $F^{-1}(x)$ exists.
To sample a random number from the distribution,
1. Sample a random number $u$ from a Uniform distribution between $[0, 1]$
2. Compute the random sample as $F^{-1}(u)$
Poisson Distribution
We briefly review the Poisson distribution. This distribution describes the number of events that occur within a fixed time interval $(0, t]$. We denote the density as
and the cumulative density as
Some simple properties of the distribution include the expected value and the variance with respect to the parameter $\lambda$.
Sampling from the Poisson Distribution
Although the cumulative density is known, the inverse will require computation to compute. To be precise, we formulate the inverse problem as follows. Given some number $u \in [0, 1]$, we want to find the smallest integer $k \geq 0$ such that
Although it is not good style to put constants inside of the summation, we show this expression to develop some intuition for how the algorithm should proceed.
Here is an algorithm to compute such $k$ in Julia.
function sample_poisson(lambda::Float64)
k = 0
t = exp(-lambda)
s = t
u = rand()
while s < u
k += 1
t *= lambda / k
s += t
end
k
end
This algorithm may be proven by induction. We will not prove the correctness of this algorithm; however, we will develop some intuition. First, this algorithm is a simple, iterative algorithm that decomposes into three components: before iteration, during iteration, and after iteration. Before iteration, we assign the variables values that represent the base case of the algorithm i.e. when k = 0, t represents $\frac{\lambda}{i}$, s represents the partial sum up to k, and u is the target number for s to bound above. During iteration, we update the variables reflect the next iteration in the summation. After iteration, we return the value k that meets the exit criterion of the while loop.
Verification by Experiment
To empirically verify that this sampling algorithm is working as intended, we create a simple simulation that generates a random sample of 10,000 random numbers and verify that the expectation and variance have values that are close to $\lambda$.
sample = [sample_poisson(7.0) for _ in 1:10000]
mean(sample), var(sample)
# (6.9621,7.208384428442861)
Observe that the mean and variance are close to $\lambda$ for a sample generated by this algorithm
Zero-Truncated Poisson Distribution
The Zero-Truncated Poisson distribution can be derived from a Poisson distribution conditioned on the fact that $k$ cannot be zero. Subsequently, the probability density function is a function of $k > 0$ such that
Since $f(0; \lambda) = e^{-\lambda}$, we simplify this function to
and the we may derive the cumulative density using some calculus
The final cumulative density function is structurally similar to the cumulative density function for the Poisson distribution. However, it important to observe two differences. First, the constant before the summation is different. Second, the starting index for the summation is 1 for this distribution in contrast to 0 for the Poisson distribution. To summarize, we represent our cumulative density function as
Sampling from the Zero-Truncated Poisson Distribution
Sampling from this distribution turns out to be similar to the sampling from the regular Poisson distribution when the cumulative density function is formulated this way. To derive the algorithm, we only need to make changes to the component before iteration because the component during iteration and after iteration is exactly the same.
function sample_ztp(lambda::Float64)
k = 1
t = exp(-lambda) / (1 - exp(-lambda)) * lambda
s = t
u = rand()
while s < u
k += 1
t *= lambda / k
s += t
end
k
end
The sketch for the proof of this algorithm is similar to the sketch of the proof of correctness for the Poisson distribution sampling algorithm.
Verification by Experiment
Similar to the Poisson distribution sampling algorithm, we verify the correctness of our implementation by a simple sampling experiment by sampling 10,000 random numbers and verifying three properties. The first two properties we will verify are that the mean and variance are close to the $\lambda$ parameter. The final property that we will verify is that no zeros exist in the sample; this property distinguishes the Zero-Truncated Poisson distribution from the standard Poisson distribution.
sample = [sample_ztp(7.0) for _ in 1:10000]
mean(sample), var(sample)
# (6.9798,6.9812900890088905)
count(x -> x == 0, sample)
# 0
Observe that the mean and variance are close to $\lambda$ for a sample generated by this algorithm. Additionally, observe that no zeros exist in the sample.
Conclusion
We have shown how to derive an efficient algorithm for the Zero-Truncated Poisson distribution using previous ideas and algorithms from the standard Poisson distribution.
Additionally, a few key algorithm design concepts are illustrated here. First, we solve for the upperbound of a summation that satisfies an inequality. Second, we analyze the algorithm using loop invariants. These two concepts are both high-level ideas on approach algorithmic problems involving iteration.
For more information about loop invariants, see Chapter 2 of Introduction to Algorithms by Cormen, Leiserson, Rivest, and Stein.
It is also important to note that the running time of this algorithm is expected to be linear with respect to $\lambda$. Precisely, the running time is $\Theta(\lambda)$. An efficient sampling algorithm has been devised for sampling from the Poisson distribution for large values of $\lambda$ as shown by Ahrens and Dieters in Computer methods for sampling from gamma, beta, poisson and bionomial distributions.
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https://www.physicsforums.com/threads/what-is-meant-by-no-gas-enters-or-leaves-the-cylinder.642432/
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# What is meant by no gas enters or leaves the cylinder ?
1. Oct 8, 2012
### hahaha158
What is meant by "no gas enters or leaves the cylinder"?
I have the following question.
A platform is supported on a piston by compressed air in a cylinder.
The mass of the platform and equipment being lifted is 1100kg. The pressure needed to
support this mass is 800kPa, and it rests initially at a height of 0.5m. The pressure supply
is closed so that no gas enters or leaves the cylinder
. One of the pieces of equipment with a mass of 50kg is moved o the platform. The entire system is at room temperature (300K)
I am confused with the significance of the bolded line. I'm not sure whether it's implying constant volume, constant pressure, temperature, etc. Can someone explain this?
2. Oct 8, 2012
### Ned Stark
Re: What is meant by "no gas enters or leaves the cylinder"?
P*V=n*R*T
from question; P dont change, and there is a finite amount of gas in chamber (i.e. n dont change)
so P dont change, n dont change (and R is a const. so it dont change), then what else could change (hint. if T change, then V change)
3. Oct 8, 2012
### hahaha158
Re: What is meant by "no gas enters or leaves the cylinder"?
Would the area change in this case? Because I need the area to find the volume, and according to my calculations it seems like it should, but it also seems a bit weird for the area of a platform to change
4. Oct 9, 2012
### ehild
Re: What is meant by "no gas enters or leaves the cylinder"?
No, the area of the piston does not change. But the piston can move up or down. Some weight is removed from the platform: that means the external force decreases. To get into equilibrium at the same temperature (300 K) what parameter of the gas has to change?
Initially the internal and external forces acting on the piston are in equilibrium: The weight of the 1100 kg load is balanced by he force the gas exerts when its pressure is 800 kPa. So what is the area of the piston?
ehild
Last edited: Oct 9, 2012
5. Oct 9, 2012
### hahaha158
Re: What is meant by "no gas enters or leaves the cylinder"?
sorry the question is
what is the new position of the platform?
so i was referring to the area of the platform, not the piston.
6. Oct 9, 2012
### ehild
Re: What is meant by "no gas enters or leaves the cylinder"?
The platform is supported on the piston. What does it mean?
ehild
7. Oct 9, 2012
### hahaha158
Re: What is meant by "no gas enters or leaves the cylinder"?
If the area is unchanging can we simplify the equation PV=nRT become
h=T since everything else is a constant or unchanging?
How would this relate to the mass though?
edit:i meant h=t when comparting the two heights so for this case
0.5m/300K = height2/T2?
8. Oct 9, 2012
### ehild
Re: What is meant by "no gas enters or leaves the cylinder"?
The temperature does not change. The pressure does.
ehild
9. Oct 9, 2012
### hahaha158
Re: What is meant by "no gas enters or leaves the cylinder"?
Oh so according to the equation PV = nRT the only variables that change are P and V (since h changes)?
I'm a bit confused because I thought the above poster told me that pressure remains constant.
10. Oct 9, 2012
### ehild
Re: What is meant by "no gas enters or leaves the cylinder"?
Yes.
He was mistaken.
Think. What does it mean that some of the load is removed?
ehild
11. Oct 9, 2012
### hahaha158
Re: What is meant by "no gas enters or leaves the cylinder"?
If the load is removed then the height will increase.
So iare there any errors in the following solution?
find area using Pi=mig/A
A*h=Vi
find Pf using Pf = mfg/A
PiVi = PfA*h to solve for the unknown height?
thanks.
12. Oct 9, 2012
### ehild
Re: What is meant by "no gas enters or leaves the cylinder"?
It is correct if the whole thing is in vacuum. Otherwise the 800 kPa is the gauge pressure, and the absolute pressure of the gas is 800 kPa +atmospheric pressure. As it was not mentioned, I think your solution is correct.
ehild
13. Oct 9, 2012
### hahaha158
Re: What is meant by "no gas enters or leaves the cylinder"?
ok thanks alot for the help
14. Oct 9, 2012
### ehild
Re: What is meant by "no gas enters or leaves the cylinder"?
You are welcome.
ehild
15. Oct 9, 2012
### hahaha158
Re: What is meant by "no gas enters or leaves the cylinder"?
sorry.
just one last question
the question asks for the maximum velocity
according to W=PdV
W should remain constant
and then using W=.5mv^2
wouldn't that mean you use mf since m is inversely proportional to v and m is smalest at mf?
16. Oct 9, 2012
### ehild
Re: What is meant by "no gas enters or leaves the cylinder"?
ehild
17. Oct 9, 2012
### hahaha158
Re: What is meant by "no gas enters or leaves the cylinder"?
A platform is supported on a piston by compressed air in a cylinder.
The mass of the platform and equipment being lifted is 1100kg. The pressure needed to
support this mass is 800kPa, and it rests initially at a height of 0.5m. The pressure supply
is closed so that no gas enters or leaves the cylinder. One of the pieces of equipment with a mass of 50kg is moved o the platform. The entire system is at room temperature (300K)
A)what is the height of the platform after the weight is removed?
B)what is the maximum velocity?
18. Oct 9, 2012
### ehild
Re: What is meant by "no gas enters or leaves the cylinder"?
The gas exerts force and does work on the platform. That work increases both the potential and kinetic energy. The kinetic energy is maximum at the equilibrium height you calculated before, when the resultant force on the platform is zero, that is, the acceleration is zero. The platform will rise above the equilibrium height till the velocity becomes zero; after that, it moves backwards (ignoring friction).
ehild
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# Understanding Functional Programming with Javascript
After a long time learning and working with object-oriented programming, I took a step back to think about system complexity.
`“Complexity is anything that makes software hard to understand or to modify.`“?—?John Outerhout
Doing some research, I found functional programming concepts like immutability and pure function. Those concepts are big advantages to build side-effect-free functions, so it is easier to maintain systems?—?with some other benefits.
In this post, I will tell you more about functional programming, and some important concepts, with a lot of code examples. In Javascript!
### What is functional programming?
Functional programming is a programming paradigm?—?a style of building the structure and elements of computer programs?—?that treats computation as the evaluation of mathematical functions and avoids changing-state and mutable data?—?
Wikipedia
### Pure functions
The first fundamental concept we learn when we want to understand functional programming is pure functions. But what does that really mean? What makes a function pure?
So how do we know if a function is `pure` or not? Here is a very strict definition of purity:
• It returns the same result if given the same arguments (it is also referred as `deterministic`)
• It does not cause any observable side effects
#### It returns the same result if given the same arguments
Imagine we want to implement a function that calculates the area of a circle. An impure function would receive `radius` as the parameter, and then calculate `radius * radius * PI`:
Why is this an impure function? Simply because it uses a global object that was not passed as a parameter to the function.
Now imagine some mathematicians argue that the `PI` value is actually `42`and change the value of the global object.
Our impure function will now result in `10 * 10 * 42` = `4200`. For the same parameter (`radius = 10`), we have a different result. Let’s fix it!
TA-DA ?! Now we’ll always pass the`PI` value as a parameter to the function. So now we are just accessing parameters passed to the function. No `external object`.
• For the parameters `radius = 10` & `PI = 3.14`, we will always have the same the result: `314.0`
• For the parameters `radius = 10` & `PI = 42`, we will always have the same the result: `4200`
If our function reads external files, it’s not a pure function?—?the file’s contents can change.
#### Random number generation
Any function that relies on a random number generator cannot be pure.
#### It does not cause any observable side effects
Examples of observable side effects include modifying a global object or a parameter passed by reference.
Now we want to implement a function to receive an integer value and return the value increased by 1.
We have the `counter` value. Our impure function receives that value and re-assigns the counter with the value increased by 1.
Observation: mutability is discouraged in functional programming.
We are modifying the global object. But how would we make it `pure`? Just return the value increased by 1. Simple as that.
See that our pure function `increaseCounter` returns 2, but the `counter` value is still the same. The function returns the incremented value without altering the value of the variable.
If we follow these two simple rules, it gets easier to understand our programs. Now every function is isolated and unable to impact other parts of our system.
Pure functions are stable, consistent, and predictable. Given the same parameters, pure functions will always return the same result. We don’t need to think of situations when the same parameter has different results?—?because it will never happen.
#### Pure functions benefits
The code’s definitely easier to test. We don’t need to mock anything. So we can unit test pure functions with different contexts:
• Given a parameter `A` ? expect the function to return value `B`
• Given a parameter `C` ? expect the function to return value `D`
A simple example would be a function to receive a collection of numbers and expect it to increment each element of this collection.
We receive the `numbers` array, use `map` incrementing each number, and return a new list of incremented numbers.
For the `input` `[1, 2, 3, 4, 5]`, the expected `output` would be `[2, 3, 4, 5, 6]`.
### Immutability
Unchanging over time or unable to be changed.
When data is immutable, its state cannot change after it’s created. If you want to change an immutable object, you can’t. Instead, you create a new object with the new value.
In Javascript we commonly use the `for` loop. This next `for` statement has some mutable variables.
For each iteration, we are changing the `i` and the `sumOfValue` state. But how do we handle mutability in iteration? Recursion!
So here we have the `sum` function that receives a vector of numerical values. The function calls itself until we get the list empty (our recursion `base case`). For each “iteration” we will add the value to the `total` accumulator.
With recursion, we keep our variables immutable. The `list` and the `accumulator` variables are not changed. It keeps the same value.
Observation: Yes! We can use `reduce` to implement this function. We will cover this in the `Higher Order Functions` topic.
It is also very common to build up the final state of an object. Imagine we have a string, and we want to transform this string into a `url slug`.
In OOP in Ruby, we would create a class, let’s say, `UrlSlugify`. And this class will have a `slugify!` method to transform the string input into a `url slug`.
Beautiful! It’s implemented! Here we have imperative programming saying exactly what we want to do in each `slugify` process?—?first lower case, then remove useless white spaces and, finally, replace remaining white spaces with hyphens.
But we are mutating the input state in this process.
We can handle this mutation by doing function composition, or function chaining. In other words, the result of a function will be used as an input for the next function, without modifying the original input string.
Here we have:
• `toLowerCase`: converts the string to all lower case
• `trim`: removes whitespace from both ends of a string
• `split` and `join`: replaces all instances of match with replacement in a given string
We combine all these 4 functions and we can `"slugify"` our string.
### Referential transparency
Let’s implement a `square function`:
This pure function will always have the same output, given the same input.
Passing `2` as a parameter of the `square function` will always returns 4. So now we can replace the `square(2)` with 4. That’s it! Our function is `referentially transparent`.
Basically, if a function consistently yields the same result for the same input, it is referentially transparent.
pure functions + immutable data = referential transparency
With this concept, a cool thing we can do is to memoize the function. Imagine we have this function:
And we call it with these parameters:
The `sum(5, 8)` equals `13`. This function will always result in `13`. So we can do this:
And this expression will always result in `16`. We can replace the entire expression with a numerical constant and memoize it.
### Functions as first-class entities
The idea of functions as first-class entities is that functions are also treated as values and used as data.
Functions as first-class entities can:
• refer to it from constants and variables
• pass it as a parameter to other functions
• return it as result from other functions
The idea is to treat functions as values and pass functions like data. This way we can combine different functions to create new functions with new behavior.
Imagine we have a function that sums two values and then doubles the value. Something like this:
Now a function that subtracts values and the returns the double:
These functions have similar logic, but the difference is the operators functions. If we can treat functions as values and pass these as arguments, we can build a function that receives the operator function and use it inside our function. Let’s build it!
Done! Now we have an `f` argument, and use it to process `a` and `b`. We passed the `sum` and `subtraction` functions to compose with the `doubleOperator` function and create a new behavior.
### Higher-order functions
When we talk about higher-order functions, we mean a function that either:
• takes one or more functions as arguments, or
• returns a function as its result
The `doubleOperator` function we implemented above is a higher-order function because it takes an operator function as an argument and uses it.
You’ve probably already heard about `filter`, `map`, and `reduce`. Let’s take a look at these.
### Filter
Given a collection, we want to filter by an attribute. The filter function expects a `true` or `false` value to determine if the element should or should not be included in the result collection. Basically, if the callback expression is `true`, the filter function will include the element in the result collection. Otherwise, it will not.
A simple example is when we have a collection of integers and we want only the even numbers.
Imperative approach
An imperative way to do it with Javascript is to:
• create an empty array `evenNumbers`
• iterate over the `numbers` array
• push the even numbers to the `evenNumbers` array
We can also use the `filter` higher order function to receive the `even` function, and return a list of even numbers:
One interesting problem I solved on Hacker Rank FP Path was the Filter Array problem. The problem idea is to filter a given array of integers and output only those values that are less than a specified value `X`.
An imperative Javascript solution to this problem is something like:
We say exactly what our function needs to do?—?iterate over the collection, compare the collection current item with `x`, and push this element to the `resultArray` if it pass the condition.
Declarative approach
But we want a more declarative way to solve this problem, and using the `filter` higher order function as well.
A declarative Javascript solution would be something like this:
Using `this` in the `smaller` function seems a bit strange in the first place, but is easy to understand.
`this` will be the second parameter in the `filter` function. In this case, `3` (the `x`) is represented by `this`. That’s it.
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# Understanding Lagrange error bound
Estimating $\ln(0.9)$ using a Taylor polynomial about $x=1$ what is the least degree of the polynomial that assures an error smaller than $0.0010$?
(The $n^{th}$ derivative of $(-1)^{n-1}\cfrac{(n-1!)}{x^n}$ for $n\ge1$
Then Khan Academy says, the Lagrange bound for the error assures that $\left|R_n(1.4)\right|\le\left|\cfrac{(-1)^n\cfrac{n!}{z^{n+1}}}{(n+1)!}(1.4-1)^{n+1}\right|$
Then it says $\cfrac{0.4^{n+1}}{{(n+1)}{z^{n+1}}}\le\cfrac{0.4^{n+1}}{(n+1)}$
In the sentence before this one, why do they get rid of $z^{n+1}$?
• Because $1\leq z$, therefore if you replace it by $1$ the denominator gets smaller, and the fraction larger. Take into account that I don't know what $R_n(1.4)$ and therefore, $z$, because you haven't defined it, but it looks like after you do, that is going to be the explanation. – user578878 Jul 28 '18 at 1:32
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## 17Calculus Infinite Series - Absolute and Conditional Convergence
##### 17Calculus
Absolute and conditional convergence applies to all series whether a series has all positive terms or some positive and some negative terms (but the series is not required to be alternating).
One unique thing about series with positive and negative terms (including alternating series) is the question of absolute or conditional convergence. Once convergence of the series is established, then determining the convergence of the absolute value of the series tells you whether it converges absolutely or conditionally. Formally, here's what it looks like.
Definitions of Absolute and Conditional Convergence
Given that $$\sum{a_n}$$ is a convergent series.
if $$\sum{\left|a_n\right|}$$ converges
if $$\sum{\left|a_n\right|}$$ diverges
then $$\sum{a_n}$$ converges absolutely
then $$\sum{a_n}$$ converges conditionally
Here is a table that summarizes these ideas a little differently.
$$\sum{a_n}$$ $$\sum{\left|a_n\right|}$$ conclusion converges converges $$\sum{a_n}$$ converges absolutely converges diverges $$\sum{a_n}$$ converges conditionally
Absolute Convergence Theorem
Absolute Convergence Theorem
If the series $$\sum{\left|a_n\right|}$$ converges,
then the series $$\sum{a_n}$$ also converges.
Alternatively, it is possible to determine the convergence of the absolute value of the series first. Then, if the absolute value of the series converges, you can use the Absolute Convergence Theorem to say that the alternating series also converges and converges absolutely.
Additionally, if you have a series with some negative terms (but not all) and it is not an alternating series, you can use this theorem to determine convergence. Specifically, if the absolute value of the series converges, then the series will converge. Notice that this theorem says nothing about divergence, so you cannot make any assumption about convergence or divergence if this theorem does not hold.
Understanding The Theorem
Let's get an idea of why this theorem works. First, let's think about the series $$\sum{a_n}$$ with positive and negative terms. As we add up the terms, we will be adding some positive terms and subtracting other terms (the negative ones). Compare that to the series $$\sum{\left|a_n\right|}$$. With this series, we add all the terms as we go, i.e. we add all the positive terms and then we take the absolute value of the negative terms before we add those terms also. So each partial sum of this series is greater than the partial sum of the previous series. Consequently, as we continue to calculate the partial sums, we can say that the second sum is larger than the first sum. So if the larger sum converges, the smaller sum also has to converge.
Here is a simple example that should give you an intuitive feel for this.
For this finite alternating series $$1 - 2 + 3 - 4 + 5 = 3$$.
Now calculate the absolute value of that series to get $$1 + 2 + 3 + 4 + 5 = 15$$
If we are always adding the numbers and never subtracting, the sum will always be larger for the absolute value series. And in the case of an infinite sum, if the larger series converges, logically, the smaller one will too.
Note - This is not a formal proof of the theorem. It is just an example to give you a feel for it.
Okay, let's watch some videos to get a better idea of what is going on. This next video clip has a great discussion on absolute convergence including using some examples. Notice that he does not use the term conditional convergence. Instead, he just says that the series does not converge absolutely.
### Dr Chris Tisdell - Alternating series and absolute convergence [8min-44secs]
video by Dr Chris Tisdell
Here is another short video clip explanation of absolute and conditional convergence.
### PatrickJMT - Absolute Convergence, Conditional Convergence and Divergence [2min-0secs]
video by PatrickJMT
Freaky Consequence of Conditionally Convergent Infinite Series
As you probably know by now, when you start working with numbers out to infinity, strange things happen. This is certainly true of infinite series which are conditionally convergent. The strange thing is that, when you rearrange the sum, you can get different values to which the series converges, i.e. the commutative property of numbers does not hold! Whoa!
In fact, not only can you get different values, it is possible to rearrange a conditionally convergent infinite series in order to get any number we want, including zero and infinity! This is called the Riemann Series Theorem.
Okay, time for some practice problems. These problems are not alternating series. For practice problems applying absolute and conditional convergence to alternating series, see the alternating series page.
Practice
Unless otherwise instructed, for each of the following series:
1. Determine whether it converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }$$
Problem Statement
For the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }$$
1. Determine whether the series converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
The series converges absolutely by the direct comparison test and the absolute convergence theorem.
Problem Statement
For the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }$$
1. Determine whether the series converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
Solution
### Dr Chris Tisdell - 186 video solution
video by Dr Chris Tisdell
The series converges absolutely by the direct comparison test and the absolute convergence theorem.
Log in to rate this practice problem and to see it's current rating.
$$\displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }$$
Problem Statement
For the series $$\displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }$$
1. Determine whether the series converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
The series converges absolutely.
Problem Statement
For the series $$\displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }$$
1. Determine whether the series converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
Solution
### PatrickJMT - 199 video solution
video by PatrickJMT
The series converges absolutely.
Log in to rate this practice problem and to see it's current rating.
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# Surface integral over an annulus
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Aug 31st 2010, 04:55 AM
bobred
Surface integral over an annulus
Hi, would just like to know if what I have done is correct.
Find the area integral of the surface $\displaystyle z=y^2+2xy-x^2+2$ in polar form lying over the annulus $\displaystyle \frac{3}{8}\leq x^2+y^2\leq1$
The equation in polar form is $\displaystyle r^2\sin^2\theta+2r^2\cos\theta\sin\theta-r^2\cos^2\theta+2$
$\displaystyle \displaystyle{\int^{\pi}_{-\pi}}\int^1_{\sqrt{\frac{3}{8}}}(r^2\sin^2\theta+2 r^2\cos\theta\sin\theta-r^2\cos^2\theta+2)\, r\, dr\, d\theta$
Integrating with respect to $\displaystyle r$
$\displaystyle {\frac {55}{256}}\, \left( \sin \left( \theta \right) \right) ^{2}+{ \frac {55}{128}}\,\cos \left( \theta \right) \sin \left( \theta \right) -{\frac {55}{256}}\, \left( \cos \left( \theta \right) \right) ^{2}+5/8$
and with respect to $\displaystyle \theta$, I get the area as
$\displaystyle \frac{5}{4}\pi$
Does this look ok?
• Aug 31st 2010, 08:52 AM
Opalg
Quote:
Originally Posted by bobred
Does this look ok?
Yes. (Nod)
• Aug 31st 2010, 08:58 AM
bobred
Thanks for checking it over I had some post that said there was a problem, that what I was working out was a volume.
James
• Aug 31st 2010, 09:56 AM
AllanCuz
Quote:
Originally Posted by bobred
Thanks for checking it over I had some post that said there was a problem, that what I was working out was a volume.
James
That person probably thinks that any triple integral expresses volume and only volume. It's hard for some people to understand that a single, double or triple integral can express both area and volume. After all, a double integral is a simplified triple integral and a single integral is a simplified double integral.
• Aug 31st 2010, 10:52 AM
wonderboy1953
Diamond transform
Quote:
Originally Posted by AllanCuz
That person probably thinks that any triple integral expresses volume and only volume. It's hard for some people to understand that a single, double or triple integral can express both area and volume. After all, a double integral is a simplified triple integral and a single integral is a simplified double integral.
Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).
• Aug 31st 2010, 12:59 PM
Opalg
Quote:
Originally Posted by bobred
Find the area integral of the surface $\displaystyle z=y^2+2xy-x^2+2$ in polar form lying over the annulus $\displaystyle \frac{3}{8}\leq x^2+y^2\leq1$
Quote:
Originally Posted by bobred
I had some post that said there was a problem, that what I was working out was a volume.
Oops, I think there is indeed a problem. (Doh)
What you did was to find the volume between the annulus and the surface. But it looks as though the question was actually asking for the surface area of that part of the suface lying above the annulus.
The formula for surface area says that this area is given by $\displaystyle \displaystyle\iint_A\sqrt{\bigl(\tfrac{\partial z}{\partial x}\bigr)^2 + \bigl(\tfrac{\partial z}{\partial y}\bigr)^2+1}\,dxdy$, where A denotes the annulus.
If $\displaystyle z = y^2 + 2xy - x^2+2$ then $\displaystyle \tfrac{\partial z}{\partial x} = 2(y-x)$, $\displaystyle \tfrac{\partial z}{\partial y} = 2(y+x)$, and $\displaystyle \bigl(\tfrac{\partial z}{\partial x}\bigr)^2} + \bigl(\tfrac{\partial z}{\partial y}\bigr)^2} = 8(x^2+y^2) = 8r^2$. So the answer to the question is
. . . . . \displaystyle \begin{aligned}\iint_A\sqrt{8r^2+1}\,rdrd\theta &= \int_0^{2\pi}\int_{\sqrt{3/8}}^1\sqrt{8r^2+1}\,rdrd\theta \\ &= \int_0^{2\pi}\Bigl[\tfrac1{24}(8r^2+1)^{3/2}\Bigr]_{\sqrt{3/8}}^1\,d\theta \\ &= \int_0^{2\pi}\!\!\tfrac{19}{24}\,d\theta = \tfrac{19}{12}\pi.\end{aligned}
I thought that the question was not very clearly worded, but the fact that those 3/2 powers both turn out to be integers makes me fairly sure that "area integral" is supposed to mean "surface area" in this question.
• Aug 31st 2010, 01:09 PM
bobred
Thanks, I had been looking for an expression of the form $\displaystyle (x^2+y^2)=r^2$ because the trig expression was horrendous.
James
• Aug 31st 2010, 06:50 PM
AllanCuz
Quote:
Originally Posted by wonderboy1953
Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).
You've been out of college for 25 years and you're alias is wonderBOY?!
• Sep 1st 2010, 12:31 AM
Opalg
Quote:
Originally Posted by AllanCuz
Quote:
Originally Posted by wonderboy1953
Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).
You've been out of college for 25 years and you're alias is wonderBOY?!
That didn't come as a surprise to me. Wonderboy1953 clearly wasn't born yesterday. (Rofl)
• Sep 1st 2010, 06:14 PM
AllanCuz
Quote:
Originally Posted by Opalg
That didn't come as a surprise to me. Wonderboy1953 clearly wasn't born yesterday. (Rofl)
I was never one for numbers!
• Sep 2nd 2010, 07:40 AM
wonderboy1953
Quote:
Originally Posted by AllanCuz
I was never one for numbers!
Since you were never one for numbers, what brings you to a math website?
Since this is the second time (besides this thread) that I haven't gotten an answer to my question leads me to believe that no such formula exists to transform a double integral into a single integral. Can you prove me wrong?
• Sep 2nd 2010, 08:03 AM
Jester
In some cases doesn't Green's theorem do that?
• Sep 2nd 2010, 08:08 AM
wonderboy1953
Quote:
Originally Posted by Danny
In some cases doesn't Green's theorem do that?
Doesn't that deal with line integrals?
• Sep 2nd 2010, 08:11 AM
Ackbeet
Green's Theorem relates a surface integral to a line integral. Since surface integrals often decompose into a double iterated integral, and line integrals often decompose into a single integral, you could say that Green's Theorem can reduce some double integrals into a single integral. I'm not sure there is a way, short of actually computing the inner integral, to reduce all double integrals to single integrals.
• Sep 2nd 2010, 08:15 AM
wonderboy1953
Quote:
Originally Posted by Ackbeet
Green's Theorem relates a surface integral to a line integral. Since surface integrals often decompose into a double iterated integral, and line integrals often decompose into a single integral, you could say that Green's Theorem can reduce some double integrals into a single integral. I'm not sure there is a way, short of actually computing the inner integral, to reduce all double integrals to single integrals.
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https://www.albert.io/learn/abstract-algebra/question/left-cosets-of-dollars_3dollar
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Limited access
Let $S_3$ be the symmetric group on 3 letters.
Using the convention of multiplying cycles from right to left, determine the left cosets of the subgroup $H=\{(1), (23)\}$ in $S_3$.
For example, $(12)(23)=(123)$.
A
$H$, $\{(12), (123)\}$, $\{(13), (132)\}$
B
$H$, $\{(12), (132)\}$, $\{(13), (123)\}$
C
$H$, $\{(12), (13)\}$, $\{(123), (132)\}$
D
$\{(1), (13)\}$, $\{(12), (132)\}$, $\{(23), (123)\}$
Select an assignment template
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https://whatisconvert.com/feet-to-miles
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# Feet to Miles Converter
## Convert Feet to Miles
To calculate a value in Feet to the corresponding value in Miles, multiply the quantity in Feet by 0.00018939393939394 (conversion factor).
Miles = Feet x 0.00018939393939394
## How to convert from Feet to Miles
The conversion factor from Feet to Miles is 0.00018939393939394. To find out how many Feet in Miles, multiply by the conversion factor or use the Feet to Miles converter above.
## Definition of Foot
A foot (symbol: ft) is a unit of length. It is equal to 0.3048 m, and used in the imperial system of units and United States customary units. The unit of foot derived from the human foot. It is subdivided into 12 inches.
## Definition of Mile
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.
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http://www.mathworks.com/matlabcentral/cody/problems/1920-find-the-position-of-first-minimum-value-in-an-integer-array-with-numbers/solutions/332954
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Cody
# Problem 1920. Find the position of first minimum value in an integer array with numbers
Solution 332954
Submitted on 12 Oct 2013 by Yalong Liu
• Size: 11
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [2 6 4 9 10 3 1 5]; y_correct = 7; assert(isequal(min_fpos(x),y_correct))
``` ```
2 Pass
%% x = [2 6 -4 9 10 3 -1 5]; y_correct = 3; assert(isequal(min_fpos(x),y_correct))
``` ```
3 Pass
%% x = [-4 6 -4 9 10 3 -1 5]; y_correct = 1; assert(isequal(min_fpos(x),y_correct))
``` ```
4 Pass
%% x = [0 6 0 9 -10 3 -1 5]; y_correct = 5; assert(isequal(min_fpos(x),y_correct))
``` ```
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A034677 Sum of cubes of unitary divisors of n. 3
%I
%S 1,9,28,65,126,252,344,513,730,1134,1332,1820,2198,3096,3528,4097,
%T 4914,6570,6860,8190,9632,11988,12168,14364,15626,19782,19684,22360,
%U 24390,31752,29792,32769,37296,44226,43344,47450
%N Sum of cubes of unitary divisors of n.
%C A unitary divisor of n is a divisor d such that gcd(d,n/d)=1.
%H Harvey P. Dale, <a href="/A034677/b034677.txt">Table of n, a(n) for n = 1..1000</a>
%F Dirichlet g.f.: zeta(s)*zeta(s-3)/zeta(2s-3). - _R. J. Mathar_, Mar 04 2011
%F If n = Product (p_j^k_j) then a(n) = Product (1 + p_j^(3*k_j)). - _Ilya Gutkovskiy_, Nov 04 2018
%F Sum_{k=1..n} a(k) ~ Pi^4 * n^4 / (360 * Zeta(5)). - _Vaclav Kotesovec_, Feb 01 2019
%e The unitary divisors of 6 are 1, 2, 3 and 6, so a(6) = 252.
%t scud[n_]:=Total[Select[Divisors[n],CoprimeQ[#,n/#]&]^3]; Array[scud,40] (* _Harvey P. Dale_, Oct 16 2016 *)
%o (PARI) A034677_vec(len)={
%o a000012=direuler(p=2,len, 1/(1-X)) ;
%o a000578=direuler(p=2,len, 1/(1-p^3*X)) ;
%o a000578x=direuler(p=2,len, 1-p^3*X^2) ;
%o dirmul(dirmul(a000012,a000578),a000578x)
%o }
%o A034677_vec(70) /* via D.g.f., _R. J. Mathar_, Mar 05 2011 */
%Y Cf. A034444, A034448.
%Y Row n=3 of A286880.
%K nonn,mult
%O 1,2
%A _Erich Friedman_
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
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Last modified May 25 20:58 EDT 2020. Contains 334597 sequences. (Running on oeis4.)
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https://www.scala-algorithms.com/SudokuBoardCheck/
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# Scala algorithm: Check Sudoku board
Published
## Algorithm goal
The game of Sudoku is composed of a 9x9 grid that is valid when each of the rows, columns, and every of the 9 3-by-3 squares contains only a unique combination of numbers 1 to 9 (or no number). The game is complete when all the squares are filled. In this challenge, create a function to check whether a given board is valid, as it is easy for the players to miss something.
## Test cases in Scala
``````assert(SudokuBoard.blank.isValid)
assert(SudokuBoard.fromStrings("1").isValid)
assert(SudokuBoard.fromStrings("12").isValid)
assert(SudokuBoard.fromStrings("1", "2").isValid)
assert(SudokuBoard.fromStrings("1", " 2").isValid)
assert(!SudokuBoard.fromStrings("1", "1").isValid)
assert(!SudokuBoard.fromStrings("11").isValid)
assert(!SudokuBoard.fromStrings("1", " 1").isValid)
assert(!SudokuBoard.fromStrings("1 1").isValid)
assert(
!SudokuBoard.fromStrings("1", " ", " ", " ", " ", " ", " ", " ", "1").isValid
)
assert(
SudokuBoard.fromStrings("1", " ", " ", " ", " ", " ", " ", " ", "9").isValid
)
``````
## Algorithm in Scala
45 lines of Scala (compatible versions 2.13 & 3.0).
## Explanation
The trickiest part of this challenge is actually the representation of the grid. While it may be tempting to go for an Array/Vector-based solution, such a board is actually far simpler to represent as a Map, and its 'PositionGroup's as Sets of positions sa well.
Initially, we generate SudokuBoard.Groups by: (this is © from www.scala-algorithms.com)
• Taking a row, then shifting it down by 1 8 times
• Taking a row, turning it into a column, then shifting it right by 1 for 8 times
• Taking a 3x3 box, then doing a combination of shifting it by 1 and 2 times in both x and y directions.
This then gives us a set of Groups that we can validate the SudokuBoard with.
## Scala concepts & Hints
1. ### For-comprehension
The for-comprehension is highly important syntatic enhancement in functional programming languages.
``````val Multiplier = 10
val result: List[Int] = for {
num <- List(1, 2, 3)
anotherNum <-
List(num * Multiplier - 1, num * Multiplier, num * Multiplier + 1)
} yield anotherNum + 1
assert(result == List(10, 11, 12, 20, 21, 22, 30, 31, 32))
``````
2. ### Option Type
The 'Option' type is used to describe a computation that either has a result or does not. In Scala, you can 'chain' Option processing, combine with lists and other data structures. For example, you can also turn a pattern-match into a function that return an Option, and vice-versa!
``````assert(Option(1).flatMap(x => Option(x + 2)) == Option(3))
assert(Option(1).flatMap(x => None) == None)
``````
3. ### Pattern Matching
Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.
``````assert("Hello World".collect {
case character if Character.isUpperCase(character) => character.toLower
} == "hw")
``````
4. ### Range
The `(1 to n)` syntax produces a "Range" which is a representation of a sequence of numbers.
``````assert((1 to 5).toString == "Range 1 to 5")
assert((1 to 5).reverse.toString() == "Range 5 to 1 by -1")
assert((1 to 5).toList == List(1, 2, 3, 4, 5))
``````
# Scala Algorithms: The most comprehensive library of algorithms in standard pure-functional Scala
## How our 100 algorithms look
1. A description/goal of the algorithm.
2. An explanation with both Scala and logical parts.
3. A proof or a derivation, where appropriate.
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5. An implementation in pure-functional immutable Scala, with efficiency in mind (for most algorithms, this is for paid subscribers only).
6. Unit tests, with a button to run them immediately in our in-browser IDE.
### Study our 100 Scala Algorithms: 6 fully free, 100 published & 0 upcoming
Fully unit-tested, with explanations and relevant concepts; new algorithms published about once a week.
### Explore the 22 most useful Scala concepts
To save you going through various tutorials, we cherry-picked the most useful Scala concepts in a consistent form.
## Subscribe to Scala Algorithms
Maximize your Scala with disciplined and consistently unit-tested solutions to 100+ algorithms.
Use it from improving your day-to-day data structures and Scala; all the way to interviewing.
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https://www.brainteaserbay.com/2012/06/13/the-monty-hall-problem/
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The Monty Hall Problem
Suppose you\’re on a game show, and you\’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what\’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, \”Do you want to pick door No. 2?\”
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https://physics.stackexchange.com/questions/404609/current-in-a-perfect-conductor
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# Current in a perfect conductor
In a normal conductor, the current density is directly proportional to the electric field strength: $$\vec{J} = \sigma \vec{E}$$ In a perfect conductor, $\vec{E} =0$, even if there is a current.
Inside a perfect conductor, if $\vec{E}=0$, then what determines the magnitude and direction of the current density $\vec{J}$ in the first place?
The current density is defined by the history of $E(t)$. For simplicity we will concentrate on the case of a "perfect metal" and not a superconductor (where the underlying physics is more intricate but also real).
Zero (DC) resistance does not mean infinite acceleration. Technically, (DC) conductivity is defined as the linear response in a steady state that is achieved when the perturbation is turned on adiabatically from $t=-\infty$. So zero resistance (or infinite conductivity), tells us the current will be infinite after the non-zero electric field has been applied for eternity!
If we consider the real problem, we have to consider the actual time dependence: The electric field has been turned on at some time $t_0$, or more generally the field is $E(t)$. Now, the charge carriers have an effective mass and are accelerated according to $eE = F = m_\text{eff} a$, so due to this effect the current is increasing linearly with time (since $I(t) = env(t)$ and $v(t) = -eEt/m_\text{eff}$). There are additional complications due Bloch oscillations in crystals and due to the non-conservation of the quasi-particle number in superconductors, which we put aside for now.
An additional effect preventing instantaneous infinite current is the inductance of the wire, which will limit the current increase with time (since the change in current induces an electric field opposing the accelerating field).
Note that this discussion is not hypothetical: superconductors come very close to allowing dissipation free currents and there the actual current density depends on the history of the applied field (superconducting MRI or accelerator magnets to not need to be connected to a voltage source constantly, they can be short-circuited and the current keeps flowing). Also note, that a perfect metal (with free electrons, not impurities and no phonons) does indeed have non-zero AC conductivity due to the inertia of the electrons. The characteristic quantity describing the scaling of the AC conductivity is the plasma frequency (above which the metal becomes non-conductive).
• "Technically, (DC) conductivity is defined as the linear response in a steady state that is achieved when the perturbation is turned on adiabatically from t=−∞." Are you referring to Ohm's law here? May 8, 2018 at 2:24
• Ohm's law is either the phenomenological statement that current is proportional to the voltage in the steady state (note that the steady state is reached within at most a few milliseconds for typical circuits). This version of Ohm's law is a very crude approximation of reality. Or the relation $j = \sigma E$ which is no "law" but a general form if we allow $\sigma$ to depend on the field strength (yes this is cheating) and $\sigma$ is usually constant for small $E$. This equation gets especially useful, when we consider it in Fourier space, then it can also handle transients. May 8, 2018 at 11:48
what determines the magnitude and direction of the current density $\vec{J}$ in the first place?
Something else does, depending on what you connect your perfect conductor up to.
It could be the current limitation of the source that's driving the current. It could be the inductance of the circuit and the time that voltage has been applied. It could be a resistor or other circuit element that the perfect wire is connected to.
• I am not sure I understand the big picture here: Since resistivity is $0$, even a tiny amount of voltage applied would cause charges to have an infinite acceleration. Thus, how come do we still have a finite current inside a perfect conductor? May 7, 2018 at 23:21
• @ASlowLearner, resistance is zero, but inductance is non-zero. So you cannot ramp current infinitely fast. physics.stackexchange.com/questions/179374/… May 7, 2018 at 23:43
• Also: Zero DC resistance does not mean zero AC resistance (and I mean actual resistance not reactive impedance). Even a superconductor or perfect metal do not even theoretically have zero AC resistance. May 8, 2018 at 0:08
• @ASlowLearner, to expand on BowlOfRed's comment, consider an inductor made from turns of ideal conductor. There can be no net electric field inside the conductor but the total electric field is the superposition of the conservative electric field from separated charge (at say, the ends of the inductor) and the non-conservative electric field associated with a changing magnetic flux. Remember, a rapidly changing electric current through the ideal conductor is accompanied by a rapidly changing magnetic field surrounding the conductor. May 8, 2018 at 0:57
• @ASlowLearner Because we normally are interested in the steady state (which is reached very fast for most circuits). May 8, 2018 at 11:18
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http://www.chegg.com/homework-help/questions-and-answers/two-charges-repel-r-force-10-6n-10-cm-apart-brought-closer-together-2-cm-aprt-force-become-q3895133
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Two charges repel each other r with a force of 10^ -6N when they are 10 cm apart. When they are brought closer together until they are 2 cm aprt, the force between them becomes
A. 4 x 10^-8 N
B. 5 x 10^-6 N
C. 8 x 10^-6 N
D. 2.5 x10^5 N
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https://www.newpathonline.com/standard/Washington_Standards/126/Mathematics/2/8
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Online Learning
Curriculum Resources
Take learning to the next level and transform the way you teach with a vast library of ready-to-use, standards-aligned, adaptable curriculum resources. The resources listed below are either available with an Online Learning Subscription which allows you to instruct, assess and track student performance or as individual hands-on classroom resources which can be purchased. Choose from Multimedia Lessons, Curriculum Mastery Games, Flip Charts, Visual Learning Guides, Flash Cards, Vocabulary Cards, and Curriculum Modules available on our online store. PREMIUM ONLINE LEARNING SUBSCRIPTION OPTIONS
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## WA.8.1.Core Content: Linear functions and equations (Algebra) Students solve a variety of linear equations and inequalities. They build on their familiarity with proportional relationships and simple linear equations to work with a broader set of linear relationships, and they learn what functions are. They model applied problems with mathematical functions represented by graphs and other algebraic techniques. This Core Content area includes topics typically addressed in a high school algebra or a first-year integrated math course, but here this content is expected of all middle school students in preparation for a rich high school mathematics program that goes well beyond these basic algebraic ideas.
Core Content: Linear functions and equations (Algebra) Students solve a variety of linear equations and inequalities. They build on their familiarity with proportional relationships and simple linear equations to work with a broader set of linear relationships, and they learn what functions are. They model applied problems with mathematical functions represented by graphs and other algebraic techniques. This Core Content area includes topics typically addressed in a high school algebra or a first-year integrated math course, but here this content is expected of all middle school students in preparation for a rich high school mathematics program that goes well beyond these basic algebraic ideas.
8.1.A. Solve one-variable linear equations.
Quiz, Flash Cards, Worksheet, Game & Study Guide Equations and Inequalities
Quiz, Flash Cards, Worksheet, Game & Study Guide Equations and inequalities
Quiz, Flash Cards, Worksheet, Game & Study Guide Integer operations
Quiz, Flash Cards, Worksheet, Game & Study Guide Introduction to Algebra
8.1.B. Solve one- and two-step linear inequalities and graph the solutions on the number line.
Flip Charts Inequalities
Flip Charts Inequalities
Quiz, Flash Cards, Worksheet, Game & Study Guide Algebraic Inequalities
Quiz, Flash Cards, Worksheet, Game & Study Guide Equations and inequalities
Quiz, Flash Cards, Worksheet, Game & Study Guide Equations and Inequalities
Quiz, Flash Cards, Worksheet, Game & Study Guide Integer operations
Quiz, Flash Cards, Worksheet, Game & Study Guide Rational numbers and operations
8.1.C. Represent a linear function with a verbal description, table, graph, or symbolic expression, and make connections among these representations.
Quiz, Flash Cards, Worksheet, Game & Study Guide Introduction to Functions
Quiz, Flash Cards, Worksheet, Game & Study Guide Linear equations
8.1.D. Determine the slope and y-intercept of a linear function described by a symbolic expression, table, or graph.
Flip Charts Slope
Quiz, Flash Cards, Worksheet, Game & Study Guide Introduction to Functions
Quiz, Flash Cards, Worksheet, Game & Study Guide Linear equations
8.1.E. Interpret the slope and y-intercept of the graph of a linear function representing a contextual situation.
Quiz, Flash Cards, Worksheet, Game & Study Guide Introduction to Functions
## WA.8.2.Core Content: Properties of geometric figures (Numbers, Geometry/Measurement) Students work with lines and angles, especially as they solve problems involving triangles. They use known relationships involving sides and angles of triangles to find unknown measures, connecting geometry and measurement in practical ways that will be useful well after high school. Since squares of numbers arise when using the Pythagorean Theorem, students work with squares and square roots, especially in problems with two- and three-dimensional figures. Using basic geometric theorems such as the Pythagorean Theorem, students get a preview of how geometric theorems are developed and applied in more formal settings, which they will further study in high school.
Core Content: Properties of geometric figures (Numbers, Geometry/Measurement) Students work with lines and angles, especially as they solve problems involving triangles. They use known relationships involving sides and angles of triangles to find unknown measures, connecting geometry and measurement in practical ways that will be useful well after high school. Since squares of numbers arise when using the Pythagorean Theorem, students work with squares and square roots, especially in problems with two- and three-dimensional figures. Using basic geometric theorems such as the Pythagorean Theorem, students get a preview of how geometric theorems are developed and applied in more formal settings, which they will further study in high school.
8.2.A. Identify pairs of angles as complementary, supplementary, adjacent, or vertical, and use these relationships to determine missing angle measures.
Flip Charts Plane Geometry
Quiz, Flash Cards, Worksheet, Game & Study Guide Finding Volume
8.2.B. Determine missing angle measures using the relationships among the angles formed by parallel lines and transversals.
Flip Charts Plane Geometry
8.2.E. Quickly recall the square roots of the perfect squares from 1 through 225 and estimate the square roots of other positive numbers.
Flip Charts The Real Numbers
Quiz, Flash Cards, Worksheet, Game & Study Guide Rational and Irrational Numbers
Quiz, Flash Cards, Worksheet, Game & Study Guide Real numbers
Quiz, Flash Cards, Worksheet, Game & Study Guide The Pythagorean Theorem
8.2.F. Demonstrate the Pythagorean Theorem and its converse and apply them to solve problems.
Quiz, Flash Cards, Worksheet, Game & Study Guide The Pythagorean Theorem
## WA.8.3.Core Content: Summary and analysis of data sets (Algebra, Data/Statistics/Probability) Students build on their extensive experience organizing and interpreting data and apply statistical principles to analyze statistical studies or short statistical statements, such as those they might encounter in newspapers, on television, or on the Internet. They use mean, median, and mode to summarize and describe information, even when these measures may not be whole numbers. Students use their knowledge of linear functions to analyze trends in displays of data. They create displays for two sets of data in order to compare the two sets and draw conclusions. They expand their work with probability to deal with more complex situations than they have previously seen. These concepts of statistics and probability are important not only in students' lives, but also throughout the high school mathematics program.
Core Content: Summary and analysis of data sets (Algebra, Data/Statistics/Probability) Students build on their extensive experience organizing and interpreting data and apply statistical principles to analyze statistical studies or short statistical statements, such as those they might encounter in newspapers, on television, or on the Internet. They use mean, median, and mode to summarize and describe information, even when these measures may not be whole numbers. Students use their knowledge of linear functions to analyze trends in displays of data. They create displays for two sets of data in order to compare the two sets and draw conclusions. They expand their work with probability to deal with more complex situations than they have previously seen. These concepts of statistics and probability are important not only in students' lives, but also throughout the high school mathematics program.
8.3.A. Summarize and compare data sets in terms of variability and measures of center.
Quiz, Flash Cards, Worksheet, Game & Study Guide Collecting and describing data
Quiz, Flash Cards, Worksheet, Game & Study Guide Organizing Data
Quiz, Flash Cards, Worksheet, Game & Study Guide Using graphs to analyze data
8.3.B. Select, construct, and analyze data displays, including box-and-whisker plots, to compare two sets of data.
Quiz, Flash Cards, Worksheet, Game & Study Guide Using graphs to analyze data
8.3.D. Describe different methods of selecting statistical samples and analyze the strengths and weaknesses of each method.
Quiz, Flash Cards, Worksheet, Game & Study Guide Collecting and describing data
Quiz, Flash Cards, Worksheet, Game & Study Guide Experimental Probability
8.3.E. Determine whether conclusions of statistical studies reported in the media are reasonable.
Quiz, Flash Cards, Worksheet, Game & Study Guide Mathematical processes
8.3.F. Determine probabilities for mutually exclusive, dependent, and independent events for small sample spaces.
Quiz, Flash Cards, Worksheet, Game & Study Guide Theoretical probability and counting
Quiz, Flash Cards, Worksheet, Game & Study Guide Using Probability
8.3.G. Solve single- and multi-step problems using counting techniques and Venn diagrams and verify the solutions.
Quiz, Flash Cards, Worksheet, Game & Study Guide Theoretical probability and counting
Quiz, Flash Cards, Worksheet, Game & Study Guide Using graphs to analyze data
Quiz, Flash Cards, Worksheet, Game & Study Guide Using Probability
## WA.8.4.Additional Key Content (Numbers, Operations) Students deal with a few key topics about numbers as they prepare to shift to higher level mathematics in high school. First, they use scientific notation to represent very large and very small numbers, especially as these numbers are used in technological fields and in everyday tools like calculators or personal computers. Scientific notation has become especially important as ''extreme units'' continue to be identified to represent increasingly tiny or immense measures arising in technological fields. A second important numerical skill involves using exponents in expressions containing both numbers and variables. Developing this skill extends students' work with order of operations to include more complicated expressions they might encounter in high school mathematics. Finally, to help students understand the full breadth of the real-number system, students are introduced to simple irrational numbers, thus preparing them to study higher level mathematics in which
Additional Key Content (Numbers, Operations) Students deal with a few key topics about numbers as they prepare to shift to higher level mathematics in high school. First, they use scientific notation to represent very large and very small numbers, especially as these numbers are used in technological fields and in everyday tools like calculators or personal computers. Scientific notation has become especially important as ''extreme units'' continue to be identified to represent increasingly tiny or immense measures arising in technological fields. A second important numerical skill involves using exponents in expressions containing both numbers and variables. Developing this skill extends students' work with order of operations to include more complicated expressions they might encounter in high school mathematics. Finally, to help students understand the full breadth of the real-number system, students are introduced to simple irrational numbers, thus preparing them to study higher level mathematics in which
8.4.A. Represent numbers in scientific notation, and translate numbers written in scientific notation into standard form.
Quiz, Flash Cards, Worksheet, Game & Study Guide Exponents, Factors and Fractions
Quiz, Flash Cards, Worksheet, Game & Study Guide Polynomials and Exponents
8.4.D. Identify rational and irrational numbers.
Flip Charts The Real Numbers
## WA.8.4.Additional Key Content (Numbers, Operations) Students deal with a few key topics about numbers as they prepare to shift to higher level mathematics in high school. First, they use scientific notation to represent very large and very small numbers, especially as these numbers are used in technological fields and in everyday tools like calculators or personal computers. Scientific notation has become especially important as ''extreme units'' continue to be identified to represent increasingly tiny or immense measures arising in technological fields. A second important numerical skill involves using exponents in expressions containing both numbers and variables. Developing this skill extends students' work with order of operations to include more complicated expressions they might encounter in high school mathematics. Finally, to help students understand the full breadth of the real-number system, students are introduced to simple irrational numbers, thus preparing them to study higher level mathematics in which properties and procedures are generalized for the entire set of real numbers.
Additional Key Content (Numbers, Operations) Students deal with a few key topics about numbers as they prepare to shift to higher level mathematics in high school. First, they use scientific notation to represent very large and very small numbers, especially as these numbers are used in technological fields and in everyday tools like calculators or personal computers. Scientific notation has become especially important as ''extreme units'' continue to be identified to represent increasingly tiny or immense measures arising in technological fields. A second important numerical skill involves using exponents in expressions containing both numbers and variables. Developing this skill extends students' work with order of operations to include more complicated expressions they might encounter in high school mathematics. Finally, to help students understand the full breadth of the real-number system, students are introduced to simple irrational numbers, thus preparing them to study higher level mathematics in which properties and procedures are generalized for the entire set of real numbers.
8.4.A. Represent numbers in scientific notation, and translate numbers written in scientific notation into standard form.
Quiz, Flash Cards, Worksheet, Game & Study Guide Exponents, Factors and Fractions
Quiz, Flash Cards, Worksheet, Game & Study Guide Polynomials and Exponents
8.4.D. Identify rational and irrational numbers.
Flip Charts The Real Numbers
## WA.8.5.Core Processes: Reasoning, problem solving, and communication - Students refine their reasoning and problem-solving skills as they move more fully into the symbolic world of algebra and higher level mathematics. They move easily among representations - numbers, words, pictures, or symbols - to understand and communicate mathematical ideas, to make generalizations, to draw logical conclusions, and to verify the reasonableness of solutions to problems. In grade eight, students solve problems that involve proportional relationships and linear relationships, including applications found in many contexts outside of school. These problems dealing with proportionality continue to be important in many applied contexts, and they lead directly to the study of algebra. Students also begin to deal with informal proofs for theorems that will be proven more formally in high school.
Core Processes: Reasoning, problem solving, and communication - Students refine their reasoning and problem-solving skills as they move more fully into the symbolic world of algebra and higher level mathematics. They move easily among representations - numbers, words, pictures, or symbols - to understand and communicate mathematical ideas, to make generalizations, to draw logical conclusions, and to verify the reasonableness of solutions to problems. In grade eight, students solve problems that involve proportional relationships and linear relationships, including applications found in many contexts outside of school. These problems dealing with proportionality continue to be important in many applied contexts, and they lead directly to the study of algebra. Students also begin to deal with informal proofs for theorems that will be proven more formally in high school.
8.5.A. Analyze a problem situation to determine the question(s) to be answered.
Quiz, Flash Cards, Worksheet, Game & Study Guide Mathematical processes
8.5.B. Identify relevant, missing, and extraneous information related to the solution to a problem.
Quiz, Flash Cards, Worksheet, Game & Study Guide Mathematical processes
8.5.C. Analyze and compare mathematical strategies for solving problems, and select and use one or more strategies to solve a problem.
Quiz, Flash Cards, Worksheet, Game & Study Guide Mathematical processes
8.5.D. Represent a problem situation, describe the process used to solve the problem, and verify the reasonableness of the solution.
Quiz, Flash Cards, Worksheet, Game & Study Guide Mathematical processes
8.5.F. Apply a previously used problem-solving strategy in a new context.
Quiz, Flash Cards, Worksheet, Game & Study Guide Mathematical processes
8.5.G. Extract and organize mathematical information from symbols, diagrams, and graphs to make inferences, draw conclusions, and justify reasoning.
Quiz, Flash Cards, Worksheet, Game & Study Guide Mathematical processes
8.5.H. Make and test conjectures based on data (or information) collected from explorations and experiments.
Quiz, Flash Cards, Worksheet, Game & Study Guide Mathematical processes
Science
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https://wakeupyourmind.net/life/eye-test-only-1-person-of-96-can-do-it-can-you/
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# Eye Test: Only 1 Person Of 96 Can Do It. Can You?
Are you able to quickly focus and notice multi-task to see all the 9s in these images? You only have 60 seconds to find the hidden ones. You will be surprised how easily your eyes will wander…
## Find the hidden 9 within 60 seconds
### 5.
Could you see all of them quickly? We hope you did great on this test!
## Most people get this wrong – but can you find all the 8’s in the picture?
### Only 3 out of 10 got it right
I found this riddle online earlier today. It’s easy, but try your best to get the right answer.
This picture is filled with sixes, as well as some hidden 8’s. Can you figure out how many 8’s are in the picture? Oh, and you only have 10 seconds — quite the challenge!
I sent this challenge to my colleagues and only 3 out of 10 had the correct number of 8s on the first try!
Ok, you have 10 seconds – how many 8’s can you find?!
Now let’s find out if you’re right!
.
.
.
According to the site Brightside, not even half of the people that try can get the correct answer — this matched the result I got from my colleagues!
The correct answer is 8, there are eight 8’s in the picture! 7 of them are between the sixes. And you have to count the big yellow 8 in the question!
Did you get it right?
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https://justmatka.in/matka-rates-chart/
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MATKA RATES CHART
Matka Rate is rate paid for each variation of the game. You only get paid if the number you played exactly matches with the result (win).
What is Matka rate?
There are 5 variations of the game that you can play :
(Singles, Jodi, Pana, Half Sangam, Sangam).
Each has their own rate.
MATKA RATE CHART
SINGLES (ANK) = 1:9
The rate paid for SINGLES (ANK) is 1:9 meaning that you get 9/- for every 1/- you played in case your pick matches the result.
JODI (PAIR/BRACKET) = 1:90
The rate paid for JODI (Pair/Bracket) is 1:90 meaning that you get 90/- for every 1/- you played in case your pick matches the result.
PANA (PATTI/PANEL)
There are three types of Pana’s. Always, any one type of the three comes as result meaning that the result pana can either be SP/DP/TP and not both or all three. You get paid as per the rate of the type of pana.
HALF SANGAM = 1:1400
The rate paid of Half Sangam is 1:1400 meaning that you get 1400/- for every 1/- you played in case your pick matches the result.
SANGAM = 1:15000
The rate paid of Sangam is 1:15000 meaning that you get 15000/- for every 1/- you play in case your pick matches the result.
Notes on Rates
You will be paid separately for each variation. Meaning that if you played 10/- on Single and 10/- on Jodi, you will get 90/- [Single Win], + 900 [Jodi Win] = 990/-
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# Important 2 Marks Questions for CBSE 10th Science Exam
Q1:
An object is placed at a distance of 30 cm from a concave lens of focal
length 15 cm. List four characteristics (nature, position, etc.) of the image
formed by the lens.
Q2:
State two advantages of conserving (i) forests, and (ii) wild-life.
Q3:
Explain two main advantages associated with water harvesting at the community level.
Q4:
A student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that he moves gradually the flame towards the lens and each time focuses its image on the screen.
(A) In which direction does he move the lens to focus the flame on the screen ?
(B) What happens to the size of the image of the flame formed on the screen ?
(C) What difference is seen in the intensity (brightness) of the image of the flame on the screen ?
(D) What is seen on the screen when the flame is very close (at about 5 cm) to the lens ?
Q5:
Draw in sequence (showing the four stages), the process of binary fission
in Amoeba.
Q6:
Why must we conserve our forests ? List two factors responsible for causing deforestation.
Q7:
Why is an equitable distribution of resources essential in a society ? List two forces which are against such distribution
Q8:
An object is placed at a distance of 15 cm from a convex lens of focal length 20 cm. List four characteristics (nature, position, etc.) of the image formed by the lens.
Q9:
Why are coal and petroleum categorised as natural resources ? Give a reason as to why they should be used judiciously.
Q10:
You being an environmentalist are interested in contributing towards the conservation of natural resources. List four activities that you can do on your own.
Q11:
An object is placed at a distance of 15 cm from a concave lens of focal length 30 cm. List four characteristics (nature, position, etc.) of the image formed by the lens.
Q12:
Draw in sequence (showing the four stages), the process of binary fission in Amoeba.
Q13:
Mention the essential material (chemicals) to prepare soap in the laboratory. Describe in brief the test of determining the nature (acidic/alkaline) of the reaction mixture of saponification reaction
Q14:
In this diagram, the angle of incidence, the angle of emergence and the angle of deviation respectively have been represented by
(A) y, p, z
(B) x, q, z
(C) p, y, z
(D) p, z, y
Q15:
Mention the essential material (chemicals) to prepare soap in the laboratory. Describe in brief the test of determining the nature (acidic/alkaline) of the reaction mixture of saponification reaction
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https://stonespounds.com/412-7-pounds-in-stones-and-pounds
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# 412.7 pounds in stones and pounds
## Result
412.7 pounds equals 29 stones and 6.7 pounds
You can also convert 412.7 pounds to stones.
## Converter
Four hundred twelve point seven pounds is equal to twenty-nine stones and six point seven pounds (412.7lbs = 29st 6.7lb).
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https://www.askiitians.com/forums/Algebra/what-is-the-resultant-ph-of-a-solution-when-20ml-o_241442.htm
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` What is the resultant pH of a solution when 20ml of 0.1N NaOH is mixed with 20ml of 0.05M Ca(OH) 2 at 25C?`
9 months ago
Saood
10 Points
``` when 20ml of 0.1N NaOH is mixed with 20ml of 0.05M Ca(OH)2 at 25CEquivalent weight of NaOH = molecular weight of NaOH because its acidity is oneNORMALITY of NaOH = MOLARITY of NaOH =.1Milli mole of NaOH=molarity x volume = 20x.1 = 2NaOH——->Na+ + OH-2 milli mole of NaOH produce 2 milli mole of OH-Milli mole of Ca(OH)2=molarity x volume = 20x.05 = 1Ca(OH)2———→Ca+2 + 2OH-1 milli mole of Ca(OH)2 produce 2 milli mole of OH-Total milli mole of OH- = 1+2 = 3Total volume of solution(ml)=20+20=40mlOH- CONCENTRATION =Milli mole of OH-/Total volume of solution(ml)=3/40MpOH=-Log[OH-] = -log3/40 = log40-log3 =1.6020 -.4771 = 1.1249pH=14–1.1249=12.8751
```
9 months ago
SAMBHAV MISHRA
37 Points
``` @saood u idiot u have copied the answer from quora.. if u dont have the knowledge then dont answer.. or else i will complain against you idiot
```
9 months ago
Think You Can Provide A Better Answer ?
## Other Related Questions on Algebra
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### Course Features
• 731 Video Lectures
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### Course Features
• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions
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# Constructing A Precision Thermostat
## The thermostat in your house isn't all that accurate
By Pepos Dounson
With commercial thermostats costing around \$200, even more when you look at installation costs, you would hope they work the way you want them to. But the truth is, most of the time, you'll find that the thermostat you use in your house to control interior temperature isn't all that accurate. That lack of accuracy can cost you money.
How do you ch eck? Simply buy an accurate temperature indicator (one that doesn't cost thousands of dollars) and compare the in-house temperature when the air conditioner or heating unit kicks on and off. You might discover that your installed thermostat is not nearly as accurate with temperature as your indicator is. To maximize temperature efficiency of both your heater and your AC, you can make a precision thermostat with some basic sensors, meters and an amplifier.
The main ingredient in the entire project is the addition of a LM34 temperature sensor, which reads Fahrenheit temperatures far more accurately (down to a tenth of a degree) than what your in-house thermostat can.
### Theory and Circuit
The essential circuit for your sensor is composed of an LM34 temperature sensor with +5 volts hooked on one end, the other (-) end grounded and the middle output passing through a 180,000 ohm resistor. One word of warning: the output must be hooked to an input equipped with a non-inverting buffer to match the input load impedance.
The LM34 has a three-digit voltage output (indicating millivolts). This is used to communicate degrees in Fahrenheit, with the last digit on the right indicating a tenth of a degree. So, if you have a readout of 734, you simply add the decimal for a temperature of 73.4. For those of you looking to create a thermostat that reads out in Celsius, use the LM35.
### MC4558 Operational Amplifier
The MC4558 is a dual operational amplifier that can be used in a variety of electronic circuits. You'll want to use the operational amplifier in what is called a comparator circuit. Below is a diagram of a MC4558 operational amplifier with an explanation of its operation with the LM34 temperature sensor hooked to it. The MC4558 can be used with dual + and − supplies, but the thermostat we're building uses only one supply.
The + of the top op amp "A" (called the non-inverting input) is joined and connected to the bottom of op amp "B" (called the inverting input).
The output of the LM34 sensor goes through a 180,000 ohm resistor to the joined input (shown as sensor on the diagram). Once hooked up, the sensor will adjust one-tenth of a degree at a time, as the temperature in the room varies. By connecting the output lead to a digital panel meter, you've created an accurate digital temperature indicator.
You're ready to take the sensor a step further by using it to control the heating and cooling system cycles. Start by connecting a single +5 volt regulated power supply and electronically feed a voltage of + .777 volts (777 millivolts) to the HI terminal on the top operational amplifier, or "A". You can control the exact amount of voltage; the figure .777 is merely what we have selected for a HI limit.
Now, connect another single +5 volt regulated power supply and electronically feed a voltage of + .756 volts (756 millivolts) to the LO terminal of the bottom operational amplifier "B". Once again, this number is arbitrary and you can control the exact amount of voltage as you see fit.
With a second digital panel meter, you can select between the HI temperature setting and the LO setting. This is similar to the heat and cool switch on your store-bought thermostat. Essentially, the numbers chosen above apply an operating range to the cooling system (the range being from 77.7°F to 75.6°F). You are, however, still lacking the control to turn the system on and off, because you have merely set the range.
The settings for on and off are different from those of the range. In the bottom comparator "B", when the heater turns on, and is in the process of heating the room, the inside room temperature will become higher than the low range, thus the conditions for a positive output from comparator "B" no longer exist. Consequently, we must "latch" the information to comparator "B" and not allow it to change, because if it does, the heater will turn back off and stop heating.
Click to enlarge photo.
Let's examine the bottom comparator "B" in the heating stage. When the room temperature cools slightly below the low range, the heater needs to be turned on. Since the low range is now more positive than the room temperature, the lower comparator "B" goes into saturation and has a positive output from that comparator. That positive output can be used to drive a transistor and turn a miniature relay on along with the heating system.
When the bottom comparator "B" turns the heater on, the room will heat until the HI setting is met, at which point the heating system will shut down.
This is a picture of a White Rodgers thermostat that replaced my lightening damaged 15-year old model.
Pepos Dounson is a retired lawyer hailing from San Antonio, Texas. If you have comments, you can contact him directly at [email protected].
If you have an electronics story you'd like to share, please send it to [email protected].
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# How Long Is 1 30th Of A Second?
## What is 1/30th of a second?
If your movie is 30 frames per second, then a frame is 1/30th of a second, or 33.33 milliseconds..
## How many microseconds makes a second?
1000000 microsecond1 second is equal to 1000000 microsecond.
## How fast is 1/4000 shutter speed?
The shutter speeds that are available to choose from might vary depending on the type of camera you are using, but typically a DSLR will go from 1/4000 all the way to 30 seconds.
## How many inches is 30 seconds?
For example, an object that is three lines beyond the 3-inch mark on a tape measure marked in thirty-seconds would be 3 and 3/32 inches long. An object that is exactly 7 lines beyond the 4-inch mark on a tape marked in sixteenths would be 4 and 14/32 inches long.
## What is faster milliseconds or microseconds?
One millisecond is equal to 1 × 10-3 to unit of time second. Therefore 1 millisecond = 0.001 seconds. One microsecond is equal to 1 × 10-6 to unit of time second….Milliseconds to Microseconds Conversion Table.MillisecondsMicroseconds1 ms1000 µs2 ms2000 µs3 ms3000 µs4 ms4000 µs46 more rows
## Which is faster nanosecond or microsecond?
A microsecond is equal to one millionth (10−6 or 1/1,000,000) of a second. … One nanosecond is to one second as one second is to 31.7 years.
## How many seconds is 30 frames?
34 is close to . 33 = 10 frames, and so, 1.34 seconds would be about 30 frames + 10 frames = Total 40 frames @ 30-fps.
## What is 3/32 as a decimal?
0.093753/32 as a decimal is 0.09375.
## What is the best shutter speed?
As a rule of thumb, your shutter speed should not exceed your lens’ focal length when you are shooting handheld. For example, if you are shooting with a 200mm lens, your shutter speed should be 1/200th of a second or faster to produce a sharp image.
## What is the fastest shutter speed?
The Steam camera not only shoots images just 440 trillionths of a second in length, it can rack up an astonishing six million of them in a single second.
## Which f stop value lets in the most light?
The aperture setting is measured in f-stop values, with apertures such as f/1.4 and f/2.8 often referred to as ‘wide’ apertures, as they have the widest opening and let in the most light, while apertures with higher f-stop numbers (f/11, f/16 and so on) are (perhaps rather confusingly) referred as small, or narrow, …
## How many Megaseconds are in a second?
1 Megasecond: A megasecond is exactly 1,000,000 seconds. A million seconds.
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# You toss a coin AND roll adie. What is the probability of getting a tail and a 4 on the die
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next time you Google a mcq Questions
### Related MCQs
You toss a coin AND roll adie.
What is the probability of g
A game consists of tossing a coin threetimes.
If you toss t
A fair coin is thrown in the air fourtimes.
If the coin l
An unbiased coin is tossed 5times.
If tail appears on first
A fair six-sided die is rolledtwice.
What is the probabilit
A man alternately tosses a coin and throws adie.
The probab
You roll twodice.
The first die shows a ONE and the other
What is the probability of getting 'heads' when you toss a c
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### Archive
Archive for the ‘Awards’ Category
## How I chose Enumerative Combinatorics
Apologies for not writing anything for awhile. After Feb 24, the math part of the “life and math” slogan lost a bit of relevance, while the actual events were stupefying to the point when I had nothing to say about the life part. Now that the shock subsided, let me break the silence by telling an old personal story which is neither relevant to anything happening right now nor a lesson to anyone. Sometimes a story is just a story…
#### My field
As the readers of this blog know, I am a Combinatorialist. Not a “proud one”. Just “a combinatorialist”. To paraphrase a military slogan “there are many fields like this one, but this one is mine”. While I’ve been defending my field for years, writing about its struggles, and often defining it, it’s not because this field is more important than others. Rather, because it’s so frequently misunderstood.
In fact, I have worked in other (mostly adjacent) fields, but that was usually because I was curious. Curious what’s going on in other areas, curious if they had tools to help me with my problems. Curious if they had problems that could use my tools. I would go to seminars in other fields, read papers, travel to conferences, make friends. Occasionally this strategy paid off and I would publish something in another field. Much more often nothing ever came out of that. It was fun regardless.
Anyway, I wanted to work in combinatorics for as long as I can remember, since I was about 15 or so. There is something inherently discrete about the way I see the world, so much that having additional structure is just obstructing the view. Here is how Gian-Carlo Rota famously put it:
Combinatorics is an honest subject. […] You either have the right number or you haven’t. You get the feeling that the result you have discovered is forever, because it’s concrete. [Los Alamos Science, 1985]
I agree. Also, I really like to count. When prompted, I always say “I work in Combinatorics” even if sometimes I really don’t. But in truth, the field is much too large and not unified, so when asked to be more specific (this rarely happens) I say “Enumerative Combinatorics“. What follows is a short story of how I made the choice.
#### Family vacation
When I was about 18, Andrey Zelevinsky (ז״ל) introduced me and Alex Postnikov to Israel Gelfand and asked what should we be reading if we want to do combinatorics. Unlike most leading mathematicians in Russia, Gelfand had a surprisingly positive view on the subject (see e.g. his quotes here). He suggested we both read Macdonald’s book, which was translated into Russian by Zelevinsky himself just a few years earlier. The book was extremely informative but dry as a fig and left little room for creativity. I read a large chunk of it and concluded that if this is what modern combinatorics looks like, I want to have nothing to do with it. Alex had a very different impression, I think.
Next year, my extended family decided to have a vacation on a Russian “river cruise”. I remember a small passenger boat which left Moscow river terminal, navigated a succession of small rivers until it reached Volga. From there, the boat had a smooth gliding all the way to the Caspian Sea. The vacation was about three weeks of a hot Summer torture with the only relief served by mouth-watering fresh watermelons.
What made it worse, there was absolutely nothing to see. Much of the way Volga is enormously wide, sometimes as wide as the English channel. Most of the time you couldn’t even see the river banks. The cities distinguished themselves only by an assortment of Lenin statues, but were unremarkable otherwise. Volgograd was an exception with its very impressive tallest statue in Europe, roughly as tall as the Statue of Liberty when combined with its pedestal. Impressive for sure, but not worth the trip. Long story short, the whole cruise vacation was dreadfully boring.
#### One good book can make a difference
While most of my relatives occupied themselves by reading crime novels or playing cards, I was reading a math book, the only book I brought with me. This was Stanley’s Enumerative Combinatorics (vol. 1) whose Russian translation came out just a few months earlier. So I read it cover-to-cover, but doing only the easiest exercises just to make sure I understand what’s going on. That book changed everything…
Midway through, when I was reading about linear extensions of posets in Ch. 3 with their obvious connections to standard Young tableaux and the hook-length formula (which I already knew by then), I had an idea. From Macdonald’s book, I remembered the q-analogue of #SYT via the “charge“, a statistics introduced by Lascoux and Schützenberger to give a combinatorial interpretation of Kostka polynomials, and which works even for skew Young diagram shapes. I figured that skew shapes are generic enough, and there should be a generalization of the charge to all posets. After several long days filled with some tedious calculations by hand, I came up with both the statement and the proof of the q-analogue of the number of linear extensions.
I wrote the proof neatly in my notebook with a clear intent to publish my “remarkable discovery”, and continued reading. In Ch. 4, all of a sudden, I read the “P-partition theory” that I just invented by myself. It came with various applications and connections to other problems, and was presented so well, much nicer than I would have!
After some extreme disappointment, I learned from the notes that the P-partition theory was a large portion of Stanley’s own Ph.D. thesis, which he wrote before I was born. For a few hours, I did nothing but meditate, staring at the vast water surrounding me and ignoring my relatives who couldn’t care less what I was doing anyway. I was trying to think if there is a lesson in this fiasco.
It pays to be positive and self-assure, I suppose, in a way that only a teenager can be. That day I concluded that I am clearly doing something right, definitely smarter than everyone else even if born a little too late. More importantly, I figured that Enumerative Combinatorics done “Stanley-style” is really the right area for me…
#### Epilogue
I stopped thinking that I am smarter than everyone else within weeks, as soon as I learned more math. I no longer believe I was born too late. I did end up doing a lot of Enumerative Combinatorics. Much later I became Richard Stanley’s postdoc for a short time and a colleague at MIT for a long time. Even now, I continue writing papers on the numbers of linear extensions and standard Young tableaux. Occasionally, I also discuss their q-analogues (like in my most recent paper). I still care and it’s still the right area for me…
Some years later I realized that historically, the “charge” and Stanley’s q-statistics were not independent in a sense that both are generalizations of the major index by Percy MacMahon. In his revision of vol. 1, Stanley moved the P-partition theory up to Ch. 3, where it belongs IMO. In 2001, he received the Steele’s Prize for Mathematical Exposition for the book that changed everything…
## Are we united in anything?
Unity here, unity there, unity shmunity is everywhere. You just can’t avoid hearing about it. Every day, no matter the subject, somebody is going to call for it. Be it in Ukraine or Canada, Taiwan or Haiti, everyone is calling for unity. President Biden in his Inaugural Address called for it eight times by my count. So did former President Bush on every recent societal issue: here, there, everywhere. So did Obama and Reagan. I am sure just about every major US politician made the same call at some point. And why not? Like the “world peace“, the unity is assumed to be a universal good, or at least an inspirational if quickly forgettable goal.
Take the Beijing Olympic Games, which proudly claims that their motto “demonstrates unity and a collective effort” towards “the goal of pursuing world unity, peace and progress”. Come again? While The New York Times isn’t buying the whole “world unity” thing and calls the games “divisive” it still thinks that “Opening Ceremony [is] in Search of Unity.” Vox is also going there, claiming that the ceremony “emphasized peace, world unity, and the people around the world who have battled the pandemic.” So it sounds to me that despite all the politics, both Vox and the Times think that this mythical unity is something valuable, right? Well, ok, good to know…
Closer to home, you see the same themes said about the International Congress of Mathematicians to be held in St. Petersburg later this year. Here is Arkady Dvorkovich, co-chair of the Executive Organizing Committee and former Deputy Prime Minister of Russia: “It seems to us that Russia will be able to truly unite mathematicians from all over the world“. Huh? Are you sure? Unite in what exactly? Because even many Russian mathematicians are not on board with having the ICM in St. Petersburg. And among those from “all over the world”, quite a few are very openly boycotting the congress, so much that even the IMU started to worry. Doesn’t “unity” mean “for all”, as in ?
#### Unity of mathematics
Frequent readers of this blog can probably guess where I stand on the “unity”. Even in my own area of Combinatorics, I couldn’t find much of it at all. I openly mocked “the feeling of unity of mathematics” argument in favor of some conjectures. I tried but could never understand Noga Alon’s claim that “mathematics should be considered as one unit” other than a political statement by a former PC Chair of the 2006 ICM.
So, about this “unity of mathematics”. Like, really? All of mathematics? Quick, tell me what exactly do the Stochastic PDEs, Algebraic Number Theory, Enumerative Combinatorics and Biostatistics have in common? Anything comes to mind? Anything at all? Ugh. Let’s make another experiment. Say, I tell you that only two of these four areas have Fields medals. Can you guess which ones? Oh, you can? Really, it was that easy?? Doesn’t this cut against all of this alleged “unity”?
Anyway, let’s be serious. Mathematics is not a unit. It’s not even a “patterned tapestry” of connected threads. It’s a human endeavor. It’s an assorted collection of scientific pursuits unconstrained by physical experiments. Some of them are deep, some shallow, some are connected to others, and some are motivated by real world applications. You check the MSC 2020 classification, and there is everything under the sun, 224 pages in total. It’s preposterous to look for and expect to find some unity there. There is none to be found.
Let me put it differently. Take poetry. Like math, it’s a artistic endeavor. Poems are written by the people and for the people. To enjoy. To recall when in need or when in a mood. Like math papers. Now, can anyone keep a straight face and say “unity of poetry“? Of course not. If anything, it’s the opposite. In poetry, having a distinctive voice is celebrated. Diverse styles are lauded. New forms are created. Strong emotions are evoked. That’s the point. Why would math be any different then?
#### What exactly unites us?
Mathematicians, I mean. Not much, I suppose, to the contrary of math politicians’ claims:
I like to think that increasing breadth in research will help the mathematical sciences to recognize our essential unity. (Margaret Wright, SIAM President, 1996)
Huh? Isn’t this like saying that space exploration will help foster cross-cultural understanding? Sounds reasonable until you actually think about what is being said…
Even the style of doing research is completely different. Some prove theorems, some make heavy computer computations, some make physical experiments, etc. At the end, some write papers and put them on the arXiv, some write long books full of words (e.g. mathematical historians), some submit to competitive conferences (e.g. in theoretical computer science), some upload software packages and experimental evidence to some data depositary. It’s all different. Don’t be alarmed, this is normal.
In truth, very little unites us. Some mathematicians work at large state universities, others at small private liberal arts colleges with a completely different approach to teaching. Some have a great commitment to math education, some spend every waking hour doing research, while others enjoy frequent fishing trips thanks to tenure. Some go into university administration or math politics, while others become journal editors.
In truth, only two things unites us — giant math societies like the AMS and giant conferences like ICMs and joint AMS/MAA/SIAM meetings. Let’s treat them separately, but before we go there, let’s take a detour just to see what an honest unrestricted public discourse sounds like:
#### What to do about the Olympics
The answer depends on who you ask, obviously. And opinions are abound. I personally don’t care other than the unfortunate fact that 2028 Olympics will be hosted on my campus. But we in math should learn how to be critical, so here is a range of voices that I googled. Do with them as you please.
Some are sort of in favor:
I still believe the Olympics contribute a net benefit to humanity. (Beth Daley, The Conversation, Feb. 2018)
Some are positive if a little ambivalent:
The Games aren’t dead. Not by a longshot. But it’s worth noting that the reason they are alive has strikingly little to do with games, athletes or medals. (L. Jon Wertheim, Time, June 2021)
Some like The New York Times are highly critical, calling it “absurdity”. Some are blunt:
More and more, the international spectacle has become synonymous with overspending, corruption, and autocratic regimes. (Yasmeen Serhan, The Atlantic, Aug. 2021)
yet unwilling to make the leap and call it quits. Others are less shy:
You can’t reform the Olympics. The Olympics are showing us what they are, and what they’ve always been. (Gia Lappe and Jonny Coleman, Jacobin, July 2021)
and
Boil down all the sanctimonious drivel about how edifying the games are, and you’re left with the unavoidable truth: The Olympics wreck lives. (Natalie Shure, The New Republic, July 2021)
#### What is the ICM
Well, it’s a giant collective effort. A very old tradition. Medals are distributed. Lots of talks. Speakers are told that it’s an honor to be chosen. Universities issue press releases. Yes, like this one. Rich countries set up and give away travel grants. Poor countries scramble to pay for participants. The host country gets dubious PR benefits. A week after it’s over everyone forgets it ever happened. Life goes on.
I went to just one ICM, in Rio in 2018. It was an honor to be invited. But the experience was decidedly mixed. The speakers were terrific mathematicians, all of them. Many were good speakers. A few were dreadful in both content and style. Some figured they are giving talks in their research seminar rather than to a general audience, so I left a couple of such talks in middle. Many talks in parallel sections were not even recorded. What a shame!
The crowds were stupefying. I saw a lot of faces. Some were friendly, of people I hadn’t seen in years, sometimes 20 years. Some were people I knew only by name. It was nice to say hello, to shake their hand. But there were thousands more. Literally. An ocean of people. I was drowning. This was the worst place for an introvert.
While there, I asked a lot of people how did they like the ICM. Some were electrified by the experience and had a decent enough time. Some looked like fish out of the water — when asked they just stared at me incomprehensively silently saying “What are you, an idiot?” Some told me they just went to the opening ceremony and left for the beach for the rest of the ICM. Assaf Naor said that he loved everything. I was so amused by that, I asked if I could quote him. “Yes,” he said, “you can quote me: I loved absolutely every bit of the ICM”. Here we go — not everyone is an introvert.
#### Whatever happened at the ICM
Unlike the Olympics, math people tend to be shy in their ICM criticism. In his somewhat unfortunately titled but otherwise useful historical book “Mathematicians of the World, Unite!” the author, Guillermo Curbera, largely stays exuberant about the subject. He does mention some critical stories, like this one:
Charlotte Angas Scott reported bluntly on the presentation of papers in the congress, which in her opinion were “usually shockingly bad” since “instead of speaking to the audience, [the lecturer] reads his paper to himself in a monotone that is sometimes hurried, sometimes hesitating, and frequently bored . . . so that he is often tedious and incomprehensible.” (Paris 1900 Chapter, p. 24)
Curbera does mention in passing that the were some controversies: Grothendieck refused to attend ICM Moscow in 1966 for political reasons, Margulis and Novikov were not allowed by the Soviet Union to leave the country to receive their Fields medals. Well, nobody’s perfect, right?
Most reports I found on the web are highly positive. Read, for example, Gil Kalai’s blog posts on the ICM 2018. Everything was great, right? Even Doron Zeilberger, not known for holding his tongue, is mostly positive (about the ICM Beijing in 2002). He does suggest that the invited speakers “should go to a ‘training camp‘” for some sort of teacher training re-education, apparently not seeing the irony, or simply under impression of all those great things in Beijing.
The only (highly controversial) criticism that I found was from Ulf Persson who starts with:
The congresses are by now considered to be monstrous affairs very different from the original intimate gatherings where group pictures could be taken.
He then continues to talk about various personal inconveniences, his misimpressions about the ICM setting, the culture, the city, etc., all in a somewhat insensitive and rather disparaging manner. Apparently, this criticism and misimpressions earned a major smackdown from Marcelo Viana, the ICM 2018 Organizing Committee Chair, who wrote that this was a “piece of bigotry” by somebody who is “poorly informed”. Fair enough. I agree with that and with the EMS President Volker Mehrmann who wrote in the same EMS newsletter that the article was “very counterproductive”. Sure. But an oversized 4 page reaction to an opinion article in a math newsletter from another continent seem indicative that the big boss hates criticism. Because we need all that “unity”, right?
Anyway, don’t hold your breath to see anything critical about the ICM St. Petersburg later this year. Clearly, everything is going to be just fantastic, nothing controversial about it. Right…
#### What to do about the ICM
Stop having them in the current form. It’s the 21st century, and we are starting the third year of the pandemic. All talks can be moved online so that everyone can watch them either as they happen, or later on YouTube. Let me note that I’ve sat in the bleachers of these makeshift 1000+ people convention center auditoriums where the LaTeX formulas are barely visible. This is what the view is like:
Note that the ICM is not like a sports event — there is literally nothing at stake. Also, there are usually no questions afterwards anyway. You are all better off watching the talks later on your laptop, perhaps even on a x1.5 speed. To get the idea, imagine watching this talk in a huge room full of people…. Even better, we can also spread out these online lectures across the time zones so that people from different countries can participate. Something like this World Relay in Combinatorics.
Really, all that CO2 burned to get humans halfway across the world to seat in a crowded space is not doing anyone any good. If the Nobel Prizes can be awarded remotely, so can the Fields medals. Tourism value aside, the amount of meaningful person-to-person interaction is so minimal in a large crowd, I am struggling to find a single good reason at all to have these extravaganzas in-person.
#### What to do about the AMS
I am not a member of any math societies so it’s not my place to tell them what to do. As a frequent contributor to AMS journals and a former editor of one of them, I did call on the AMS to separate its society business form the publishing, but given that their business model hinges on the books and journals they sell, this is unlikely. Still, let me make some quick observations which might be helpful.
The AMS is clearly getting less and less popular. I couldn’t find the exact membership numbers, but their “dues and outreach” earnings have been flat for a while. Things are clearly not going in the right direction, so much that the current AMS President Ruth Charney sent out a survey earlier this week asking people like me why do we not want to join.
People seem to realize that they have many different views on all thing math related and are seeking associations which are a better fit. One notable example is the Just Mathematics Collective which has several notable boycott initiatives. Another is the Association for Mathematical Research formed following various controversies. Note that there is a great deal of disagreements between these two, see e.g. here, there and there.
I feel these are very good developments. It’s healthy to express disagreements on issues you consider important. And while I disagree with other things in the article below, I do agree with this basic premise:
Totalitarian countries have unity. Democratic republics have disagreement. (Kevin Williamson, Against Unity, National Review, Jan. 2021)
So everyone just chill. Enjoy diverse views and opinions. Disagree with the others. And think twice before you call for “unity” of anything, or praise the ephemeral “unity of mathematics”. There is none.
## 2021 Abel Prize
I am overjoyed with the news of the Abel prize awarded to László Lovász and Avi Wigderson. You can now see three (!) Abel laureates discussing Combinatorics — follow the links in this blog post from 2019. See also Gil Kalai’s blog post for further links to lectures.
## What if they are all wrong?
Conjectures are a staple of mathematics. They are everywhere, permeating every area, subarea and subsubarea. They are diverse enough to avoid a single general adjective. They come in al shapes and sizes. Some of them are famous, classical, general, important, inspirational, far-reaching, audacious, exiting or popular, while others are speculative, narrow, technical, imprecise, far-fetched, misleading or recreational. That’s a lot of beliefs about unproven claims, yet we persist in dispensing them, inadvertently revealing our experience, intuition and biases.
The conjectures also vary in attitude. Like a finish line ribbon they all appear equally vulnerable to an outsider, but in fact differ widely from race to race. Some are eminently reachable, the only question being who will get there first (think 100 meter dash). Others are barely on the horizon, requiring both great effort, variety of tools, and an extended time commitment (think ironman triathlon). The most celebrated third type are like those Sci-Fi space expeditions in requiring hundreds of years multigenerational commitments, often losing contact with civilization it left behind. And we can’t forget the romantic fourth type — like the North Star, no one actually wants to reach them, as they are largely used for navigation, to find a direction in unchartered waters.
Now, conjectures famously provide a foundation of the scientific method, but that’s not at all how we actually think of them in mathematics. I argued back in this pointed blog post that citations are the most crucial for the day to day math development, so one should take utmost care in making references. While this claim is largely uncontroversial and serves as a raison d’être for most GoogleScholar profiles, conjectures provide a convenient idealistic way out. Thus, it’s much more noble and virtuous to say “I dedicated my life to the study of the XYZ Conjecture” (even if they never publish anything), than “I am working hard writing so many papers to gain respect of my peers, get a promotion, and provide for my family“. Right. Obviously…
But given this apparent (true or perceived) importance of conjectures, are you sure you are using them right? What if some/many of these conjectures are actually wrong, what then? Should you be flying that starship if there is no there there? An idealist would argue something like “it’s a journey, not a destination“, but I strongly disagree. Getting closer to the truth is actually kind of important, both as a public policy and on an individual level. It is thus pretty important to get it right where we are going.
#### What are conjectures in mathematics?
That’s a stupid question, right? Conjectures are mathematical claims whose validity we are trying to ascertain. Is that all? Well, yes, if you don’t care if anyone will actually work on the conjecture. In other words, something about the conjecture needs to interesting and inspiring.
#### What makes a conjecture interesting?
This is a hard question to answer because it is as much psychological as it is mathematical. A typical answer would be “oh, because it’s old/famous/beautiful/etc.” Uhm, ok, but let’s try to be a little more formal.
One typically argues “oh, that’s because this conjecture would imply [a list of interesting claims and known results]”. Well, ok, but this is self-referential. We already know all those “known results”, so no need to prove them again. And these “claims” are simply other conjectures, so this is really an argument of the type “this conjecture would imply that conjecture”, so not universally convincing. One can argue: “look, this conjecture has so many interesting consequences”. But this is both subjective and unintuitive. Shouldn’t having so many interesting conjectural consequences suggest that perhaps the conjecture is too strong and likely false? And if the conjecture is likely to be false, shouldn’t this make it uninteresting?
Also, wouldn’t it be interesting if you disprove a conjecture everyone believes to be true? In some sense, wouldn’t it be even more interesting if until now everyone one was simply wrong?
None of this are new ideas, of course. For example, faced with the need to justify the “great” BC conjecture, or rather 123 pages of survey on the subject (which is quite interesting and doesn’t really need to be justified), the authors suddenly turned reflective. Mindful of self-referential approach which they quickly discard, they chose a different tactic:
We believe that the interest of a conjecture lies in the feeling of unity of mathematics that it entails. [M.P. Gomez Aparicio, P. Julg and A. Valette, “The Baum-Connes conjecture“, 2019]
Huh? Shouldn’t math be about absolute truths, not feelings? Also, in my previous blog post, I mentioned Noga Alon‘s quote that Mathematics is already “one unit“. If it is, why does it need a new “feeling of unity“? Or is that like one of those new age ideas which stop being true if you don’t reinforce them at every occasion?
If you are confused at this point, welcome to the club! There is no objective way to argue what makes certain conjectures interesting. It’s all in our imagination. Nikolay Konstantinov once told me that “mathematics is a boring subject because every statement is equivalent to saying that some set is empty.” He meant to be provocative rather than uninspiring. But the problem he is underlying is quite serious.
#### What makes us believe a conjecture is true?
We already established that in order to argue that a conjecture is interesting we need to argue it’s also true, or at least we want to believe it to be true to have all those consequences. Note, however, that we argue that a conjecture is true in exactly the same way we argue it’s interesting: by showing that it holds is some special cases, and that it would imply other conjectures which are believed to be true because they are also checked in various special cases. So in essence, this gives “true = interesting” in most cases. Right?
This is where it gets complicated. Say, you are working on the “abc conjecture” which may or may not be open. You claim that it has many consequences, which makes it both likely true and interesting. One of them is the negative solution to the Erdős–Ulam problem about existence of a dense set in the plane with rational pairwise distances. But a positive solution to the E-U problem implies the Harborth’s conjecture (aka the “integral Fáry problem“) that every graph can be drawn in the plane with rational edge lengths. So, counterintuitively, if you follow the logic above shouldn’t you be working on a positive solution to Erdős–Ulam since it would both imply one conjecture and give a counterexample to another? For the record, I wouldn’t do that, just making a polemical point.
I am really hoping you see where I am going. Since there is no objective way to tell if a conjecture is true or not, and what exactly is so interesting about it, shouldn’t we discard our biases and also work towards disproving the conjecture just as hard as trying to prove it?
#### What do people say?
It’s worth starting with a general (if slightly poetic) modern description:
In mathematics, [..] great conjectures [are] sharply formulated statements that are most likely true but for which no conclusive proof has yet been found. These conjectures have deep roots and wide ramifications. The search for their solution guides a large part of mathematics. Eternal fame awaits those who conquer them first. Remarkably, mathematics has elevated the formulation of a conjecture into high art. [..] A well-chosen but unproven statement can make its author world-famous, sometimes even more so than the person providing the ultimate proof. [Robbert Dijkgraaf, The Subtle Art of the Mathematical Conjecture, 2019]
Karl Popper thought that conjectures are foundational to science, even if somewhat idealized the efforts to disprove them:
[Great scientists] are men of bold ideas, but highly critical of their own ideas: they try to find whether their ideas are right by trying first to find whether they are not perhaps wrong. They work with bold conjectures and severe attempts at refuting their own conjectures. [Karl Popper, Heroic Science, 1974]
Here is how he reconciled somewhat the apparent contradiction:
On the pre-scientific level we hate the very idea that we may be mistaken. So we cling dogmatically to our conjectures, as long as possible. On the scientific level, we systematically search for our mistakes. [Karl Popper, quoted by Bryan Magee, 1971]
Paul Erdős was, of course, a champion of conjectures and open problems. He joked that the purpose of life is “proof and conjecture” and this theme is repeatedly echoed when people write about him. It is hard to overestimate his output, which included hundreds of talks titled “My favorite problems“. He wrote over 180 papers with collections of conjectures and open problems (nicely assembled by Zbl. Math.)
Peter Sarnak has a somewhat opposite point of view, as he believes one should be extremely cautious about stating a conjecture so people don’t waste time working on it. He said once, only half-jokingly:
Since we reward people for making a right conjecture, maybe we should punish those who make a wrong conjecture. Say, cut off their fingers. [Peter Sarnak, UCLA, c. 2012]
This is not an exact quote — I am paraphrasing from memory. Needless to say, I disagree. I don’t know how many fingers he wished Erdős should lose, since some of his conjectures were definitely disproved: one, two, three, four, five, and six. This is not me gloating, the opposite in fact. When you are stating hundreds of conjectures in the span of almost 50 years, having only a handful to be disproved is an amazing batting average. It would, however, make me happy if Sarnak’s conjecture is disproved someday.
Finally, there is a bit of a controversy whether conjectures are worth as much as theorems. This is aptly summarized in this quote about yet another champion of conjectures:
Louis J. Mordell [in his book review] questioned Hardy‘s assessment that Ramanujan was a man whose native talent was equal to that of Euler or Jacobi. Mordell [..] claims that one should judge a mathematician by what he has actually done, by which Mordell seems to mean, the theorems he has proved. Mordell’s assessment seems quite wrong to me. I think that a felicitous but unproved conjecture may be of much more consequence for mathematics than the proof of many a respectable theorem. [Atle Selberg, “Reflections Around the Ramanujan Centenary“, 1988]
#### So, what’s the problem?
Well, the way I see it, the efforts made towards proving vs. disproving conjectures is greatly out of balance. Despite all the high-minded Popper’s claims about “severe attempts at refuting their own conjectures“, I don’t think there is much truth to that in modern math sciences. This does not mean that disproofs of famous conjectures aren’t celebrated. Sometimes they are, see below. But it’s clear to me that the proofs are celebrated more frequently, and to a much greater degree. I have only anecdotal evidence to support my claim, but bear with me.
Take prizes. Famously, Clay Math Institute gives \$1 million for a solution of any of these major open problems. But look closely at the rules. According to the item 5b, except for the P vs. NP problem and the Navier–Stokes Equation problem, it gives nothing (\$0) for a disproof of these problems. Why, oh why?? Let’s look into CMI’s “primary objectives and purposes“:
To recognize extraordinary achievements and advances in mathematical research.
So it sounds like CMI does not think that disproving the Riemann Hypothesis needs to be rewarded because this wouldn’t “advance mathematical research”. Surely, you are joking? Whatever happened to “the opposite of a profound truth may well be another profound truth“? Why does the CMI wants to put its thumb on the scale and support only one side? Do they not want to find out the solution whatever it is? Shouldn’t they be eager to dispense with the “wrong conjecture” so as to save numerous researches from “advances to nowhere“?
I am sure you can see that my blood is boiling, but let’s proceed to the P vs. NP problem. What if it’s independent of ZFC? Clearly, CMI wouldn’t pay for proving that. Why not? It’s not like this kind of thing never happened before (see obligatory link to CH). Some people believe that (or at least they did in 2012), and some people like Scott Aaronson take this seriously enough. Wouldn’t this be a great result worthy of an award as much as the proof that P=NP, or at least a nonconstructive proof that P=NP?
If your head is not spinning hard enough, here is another amusing quote:
Of course, it’s possible that P vs. NP is unprovable, but that that fact itself will forever elude proof: indeed, maybe the question of the independence of P vs. NP is itself independent of set theory, and so on ad infinitum! But one can at least say that, if P vs. NP (or for that matter, the Riemann hypothesis, Goldbach’s conjecture, etc.) were proven independent of ZF, it would be an unprecedented development. [Scott Aaronson, P vs. NP, 2016].
Speaking of Goldbach’s Conjecture, the most talked about and the most intuitively correct statement in Number Theory that I know. In a publicity stunt, for two years there was a \$1 million prize by a publishing house for the proof of the conjecture. Why just for the proof? I never heard of anyone not believing the conjecture. If I was the insurance underwriter for the prize (I bet they had one), I would allow them to use “for the proof or disproof” for a mere extra \$100 in premium. For another \$50 I would let them use “or independent of ZF” — it’s a free money, so why not? It’s such a pernicious idea of rewarding only one kind of research outcome!
Curiously, even for Goldbach’s Conjecture, there is a mild divergence of POVs on what the future holds. For example, Popper writes (twice in the same book!) that:
[On whether Goldbach’s Conjecture is ‘demonstrable’] We don’t know: perhaps we may never know, and perhaps we can never know. [Karl Popper, Conjectures and Refutations, 1963]
Ugh. Perhaps. I suppose anything can happen… For example, our civilizations can “perhaps” die out in the next 200 years. But is that likely? Shouldn’t the gloomy past be a warning, not a prediction of the future? The only thing more outrageously pessimistic is this theological gem of a quote:
Not even God knows the number of permutations of 1000 avoiding the 1324 pattern. [Doron Zeilberger, quoted here, 2005]
Thanks, Doron! What a way to encourage everyone! Since we know from numerical estimates that this number is ≈ 3.7 × 101017 (see this paper and this follow up), Zeilberger is suggesting that large pattern avoidance numbers are impossibly hard to compute precisely, already in the range of only about 1018 digits. I really hope he is proved wrong in his lifetime.
But I digress. What I mean to emphasize, is that there are many ways a problem can be resolved. Yet some outcomes are considered more valuable than others. Shouldn’t the research achievements be rewarded, not the desired outcome? Here is yet another colorful opinion on this:
Given a conjecture, the best thing is to prove it. The second best thing is to disprove it. The third best thing is to prove that it is not possible to disprove it, since it will tell you not to waste your time trying to disprove it. That’s what Gödel did for the Continuum Hypothesis. [Saharon Shelah, Rutgers Univ. Colloqium, 2001]
#### Why do I care?
For one thing, disproving conjectures is part of what I do. Sometimes people are a little shy to unambiguously state them as formal conjectures, so they phrase them as questions or open problems, but then clarify that they believe the answer is positive. This is a distinction without a difference, or at least I don’t see any (maybe they are afraid of Sarnak’s wrath?) Regardless, proving their beliefs wrong is still what I do.
For example, here is my old bog post on my disproof of the Noonan-Zeiberger Conjecture (joint with Scott Garrabrant). And in this recent paper (joint with Danny Nguyen), we disprove in one big swoosh both Barvinok’s Problem, Kannan’s Problem, and Woods Conjecture. Just this year I disproved three conjectures:
1. The Kirillov–Klyachko Conjecture (2004) that the reduced Kronecker coefficients satisfy the saturation property (this paper, joint with Greta Panova).
2. The Brandolini et al. Conjecture (2019) that concrete lattice polytopes can multitile the space (this paper, joint with Alexey Garber).
3. Kenyon’s Problem (c. 2005) that every integral curve in R3 is a boundary of a PL surface comprised of unit triangles (this paper, joint with Alexey Glazyrin).
On top of that, just two months ago in this paper (joint with Han Lyu), we showed that the remarkable independence heuristic by I. J. Good for the number of contingency tables, fails badly even for nearly all uniform marginals. This is not exactly disproof of a conjecture, but it’s close, since the heuristic was introduced back in 1950 and continues to work well in practice.
In addition, I am currently working on disproving two more old conjectures which will remain unnamed until the time we actually resolve them (which might never happen, of course). In summary, I am deeply vested in disproving conjectures. The reasons why are somewhat complicated (see some of them below). But whatever my reasons, I demand and naively fully expect that my disproofs be treated on par with proofs, regardless whether this expectation bears any relation to reality.
#### My favorite disproofs and counterexamples:
There are many. Here are just a few, some famous and some not-so-famous, in historical order:
1. Fermat‘s conjecture (letter to Pascal, 1640) on primality of Fermat numbers, disproved by Euler (1747)
2. Tait’s conjecture (1884) on hamiltonicity of graphs of simple 3-polytopes, disproved by W.T. Tutte (1946)
3. General Burnside Problem (1902) on finiteness of periodic groups, resolved negatively by E.S. Golod (1964)
4. Keller’s conjecture (1930) on tilings with unit hypercubes, disproved by Jeff Lagarias and Peter Shor (1992)
5. Borsuk’s Conjecture (1932) on partitions of convex sets into parts of smaller diameter, disproved by Jeff Kahn and Gil Kalai (1993)
6. Hirsch Conjecture (1957) on the diameter of graphs of convex polytopes, disproved by Paco Santos (2010)
7. Woods’s conjecture (1972) on the covering radius of certain lattices, disproved by Oded Regev, Uri Shapira and Barak Weiss (2017)
8. Connes embedding problem (1976), resolved negatively by Zhengfeng Ji, Anand Natarajan, Thomas Vidick, John Wright and Henry Yuen (2020)
In all these cases, the disproofs and counterexamples didn’t stop the research. On the contrary, they gave a push to further (sometimes numerous) developments in the area.
#### Why should you disprove conjectures?
There are three reasons, of different nature and importance.
First, disproving conjectures is opportunistic. As mentioned above, people seem to try proving much harder than they try disproving. This creates niches of opportunity for an open-minded mathematician.
Second, disproving conjectures is beautiful. Let me explain. Conjectures tend to be rigid, as in “objects of the type pqr satisfy property abc.” People like me believe in the idea of “universality“. Some might call it “completeness” or even “Murphy’s law“, but the general principle is always the same. Namely: it is not sufficient that one wishes that all pqr satisfy abc to actually believe in the implication; rather, there has to be a strong reason why abc should hold. Barring that, pqr can possibly be almost anything, so in particular non-abc. While some would argue that non-abc objects are “ugly” or at least “not as nice” as abc, the idea of universality means that your objects can be of every color of the rainbow — nice color, ugly color, startling color, quiet color, etc. That kind of palette has its own sense of beauty, but it’s an acquired taste I suppose.
Third, disproving conjectures is constructive. It depends on the nature of the conjecture, of course, but one is often faced with necessity to construct a counterexample. Think of this as an engineering problem of building some pqr which at the same time is not abc. Such construction, if at all possible, might be difficult, time consuming and computer assisted. But so what? What would you rather do: build a mile-high skyscraper (none exist yet) or prove that this is impossible? Curiously, in CS Theory both algorithms and (many) complexity results are constructive (you need gadgets). Even the GCT is partially constructive, although explaining that would take us awhile.
#### What should the institutions do?
If you are an institution which awards prizes, stop with the legal nonsense: “We award […] only for a publication of a proof in a top journal”. You need to set up a scientific committee anyway, since otherwise it’s hard to tell sometimes if someone deserves a prize. With mathematicians you can expect anything anyway. Some would post two arXiv preprints, give a few lectures and then stop answering emails. Others would publish only in a journal where they are Editor-in-Chief. It’s stranger than fiction, really.
What you should do is say in the official rules: “We have [this much money] and an independent scientific committee which will award any progress on [this problem] partially or in full as they see fit.” Then a disproof or an independence result will receive just as much as the proof (what’s done is done, what else are you going to do with the money?) This would also allow some flexibility for partial solutions. Say, somebody proves Goldbach’s Conjecture for integers > exp(exp(10100000)), way way beyond computational powers for the remaining integers to be checked. I would give this person at least 50% of the prize money, leaving the rest for future developments of possibly many people improving on the bound. However, under the old prize rules such person gets bupkes for their breakthrough.
#### What should the journals do?
In short, become more open to results of computational and experimental nature. If this sounds familiar, that’s because it’s a summary of Zeilberger’s Opinions, viewed charitably. He is correct on this. This includes publishing results of the type “Based on computational evidence we believe in the following UVW conjecture” or “We develop a new algorithm which confirms the UVW conjecture for n<13″. These are still contributions to mathematics, and the journals should learn to recognize them as such.
To put in context of our theme, it is clear that a lot more effort has been placed on proofs than on finding counterexamples. However, in many areas of mathematics there are no small counterexamples, so a heavy computational effort is crucial for any hope of finding one. Such work is not be as glamorous as traditional papers. But really, when it comes to standards, if a journal is willing to publish the study of something like the “null graphs“, the ship has sailed for you…
Let me give you a concrete example where a computational effort is indispensable. The curious Lovász conjecture states that every finite connected vertex-transitive graph contains a Hamiltonian path. This conjecture got to be false. It hits every red flag — there is really no reason why pqr = “vertex transitive” should imply abc = “Hamiltonian”. The best lower bound for the length of the longest (self-avoiding) path is only about square root of the number of vertices. In fact, even the original wording by Lovász shows he didn’t believe the conjecture is true (also, I asked him and he confirmed).
Unfortunately, proving that some potential counterexample is not Hamiltonian is computationally difficult. I once had an idea of one (a nice cubic Cayley graph on “only” 3600 vertices), but Bill Cook quickly found a Hamiltonian cycle dashing my hopes (it was kind of him to look into this problem). Maybe someday, when the TSP solvers are fast enough on much larger graphs, it will be time to return to this problem and thoroughly test it on large Cayley graphs. But say, despite long odds, I succeed and find a counterexample. Would a top journal publish such a paper?
#### Editor’s dilemma
There are three real criteria for evaluation a solution of an open problem by the journal:
1. Is this an old, famous, or well-studied problem?
2. Are the tools interesting or innovative enough to be helpful in future studies?
3. Are the implications of the solution to other problems important enough?
Now let’s make a hypothetical experiment. Let’s say a paper is submitted to a top math journal which solves a famous open problem in Combinatorics. Further, let’s say somebody already proved it is equivalent to a major problem in TCS. This checks criteria 1 and 3. Until not long ago it would be rejected regardless, so let’s assume this is happening relatively recently.
Now imagine two parallel worlds, where in the first world the conjecture is proved on 2 pages using beautiful but elementary linear algebra, and in the second world the conjecture is disproved on a 2 page long summary of a detailed computational search. So in neither world we have much to satisfy criterion 2. Now, a quiz: in which world the paper will be published?
If you recognized that the first world is a story of Hao Huang‘s elegant proof of the induced subgraphs of hypercubes conjecture, which implies the sensitivity conjecture. The Annals published it, I am happy to learn, in a welcome break with the past. But unless we are talking about some 200 year old famous conjecture, I can’t imagine the Annals accepting a short computational paper in the second world. Indeed, it took a bit of a scandal to accept even the 400 year old Kepler’s conjecture which was proved in a remarkable computational work.
Now think about this. Is any of that fair? Shouldn’t we do better as a community on this issue?
#### What do other people do?
Over the years I asked a number of people about the uncertainty created by the conjectures and what do they do about it. The answers surprised me. Here I am paraphrasing them:
Some were dumbfounded: “What do you mean this conjecture could be false? It has to be true, otherwise nothing I am doing make much sense.”
Others were simplistic: “It’s an important conjecture. Famous people said it’s true. It’s my job to prove it.”
Third were defensive: “Do you really think this conjecture could be wrong? Why don’t you try to disprove it then? We’ll see who is right.”
Fourth were biblical: “I tend to work 6 days a week towards the proof and one day towards the disproof.”
Fifth were practical: “I work on the proof until I hit a wall. I use the idea of this obstacle to try constructing potential counterexamples. When I find an approach to discard such counterexamples, I try to generalize the approach to continue working on the proof. Continue until either side wins.”
If the last two seem sensible to you to, that’s because they are. However, I bet fourth are just grandstanding — no way they actually do that. The fifth sound great when this is possible, but that’s exceedingly rare, in my opinion. We live in a technical age when proving new results often requires great deal of effort and technology. You likely have tools and intuition to work in only one direction. Why would you want to waste time working in another?
#### What should you do?
First, remember to make conjectures. Every time you write a paper, tell a story of what you proved. Then tell a story of what you wanted to prove but couldn’t. State it in the form of a conjecture. Don’t be afraid to be wrong, or be right but oversharing your ideas. It’s a downside, sure. But the upside is that your conjecture might prove very useful to others, especially young researchers. In might advance the area, or help you find a collaborator to resolve it.
Second, learn to check your conjectures computationally in many small cases. It’s important to give supporting evidence so that others take your conjectures seriously.
Third, learn to make experiments, explore the area computationally. That’s how you make new conjectures.
Fourth, understand yourself. Your skill, your tools. Your abilities like problem solving, absorbing information from the literature, or making bridges to other fields. Faced with a conjecture, use this knowledge to understand whether at least in principle you might be able to prove or disprove a conjecture.
Fifth, actively look for collaborators. Those who have skills, tools, or abilities you are missing. More importantly, they might have a different POV on the validity of the conjecture and how one might want to attack it. Argue with them and learn from them.
Sixth, be brave and optimistic! Whether you decide to prove, disprove a conjecture, or simply state a new conjecture, go for it! Ignore the judgements by the likes of Sarnak and Zeilberger. Trust me — they don’t really mean it.
## Some good news
Two of my former Ph.D. students won major prizes recently — Matjaž Konvalinka and Danny Nguyen. Matjaž is an Associate Professor at University of Ljubljana, Danny is a Lewis Research Assistant Professor at University of Michigan, Ann Arbor. Congratulations to both of them!
(1) The 2019 Robbins Prize is awarded to Roger Behrend, Ilse Fischer and Matjaž Konvalinka for their paper “Diagonally and antidiagonally symmetric alternating sign matrices of odd order”. The Robbins Prize is given in Combinatorics and related areas of interest is named after the late David P. Robbins and is given once every 3 years by AMS and MAA.
In many ways, this paper completes the long project of enumerating alternating sign matrices (ASMs) initiated by William Mills, David Robbins, and Howard Rumsey in the early 1980s. The original #ASM(n)=#TSSCPP(n) conjecture follows from Andrews’s proof of the conjectured product formula for #TSSCPP(n), and Zeilberger’s 84 page computer assisted proof of the the same conjectured product formula for #ASM(n). This led to a long series of remarkable developments which include Kuperberg’s proof using the Izergin-Korepin determinant for the six vertex model, the Cantini–Sportiello proof of the Razumov-Stroganov conjecture, and a recent self-contained determinantal proof for the number of ASMs by Fischer. Bressoud’s book (and this talkslides) is a good introduction. But the full story is yet to be written.
(2) The 2018 Sacks Prize is awarded to Danny Nguyen for his UCLA Ph.D. dissertation on the complexity of short formulas in Presburger Arithmetic (PA) and many related works (some joint with me, some with others). See also the UCLA announcement. The Sacks Prize is given by the international Association for Symbolic Logic for “the most outstanding doctoral dissertation in mathematical logic“. It is sometimes shared between two awardees, and sometimes not given at all. This year Danny is the sole winner of the prize.
Danny’s dissertation is a compilation of eight (!) papers Danny wrote during his graduate studies, all on the same or closely related subject. These papers advance and mostly finish off the long program of understanding the boundary of what’s feasible in PA. The most important of these is our joint FOCS paper which basically says that Integer Programming and Parametric Integer Programming is all that’s left in P, while all longer formulas are NP-hard. See Featured MathSciNet Review by Sasha Barvinok and an overlapping blog post by Gil Kalai discussing these results. See also Danny’s FOCS talk video and my MSRI talk video presenting this work.
## ICM 2018 Speakers
UCLA recently outed me (with permission) as a speaker at the next ICM in Rio. I am incredibly honored to be chosen, alongside my fantastic colleagues Matthias Aschenbrenner, Andrea Bertozzi, Ciprian Manolescu and Sucharit Sarkar.
P.S. I have more to say on the subject of ICM, but that can wait perhaps.
***************
FINAL UPDATE:
A complete list of ICM speakers is available here.
Categories: Awards, Mathematicians
## The power of negative thinking, part I. Pattern avoidance
In my latest paper with Scott Garrabrant we disprove the Noonan-Zeilberger Conjecture. Let me informally explain what we did and why people should try to disprove conjectures more often. This post is the first in a series. Part II will appear shortly.
#### What did we do?
Let F ⊂ Sk be a finite set of permutations and let Cn(F) denote the number of permutations σ ∈ Sn avoiding the set of patterns F. The Noonan-Zeilbeger conjecture (1996), states that the sequence {Cn(F)} is always P-recursive. We disprove this conjecture. Roughly, we show that every Turing machine T can be simulated by a set of patterns F, so that the number aof paths of length n accepted by by T is equal to Cn(F) mod 2. I am oversimplifying things quite a bit, but that’s the gist.
What is left is to show how to construct a machine T such that {an} is not equal (mod 2) to any P-recursive sequence. We have done this in our previous paper, where give a negative answer to a question by Kontsevich. There, we constructed a set of 19 generators of GL(4,Z), such that the probability of return sequence is not P-recursive.
When all things are put together, we obtain a set F of about 30,000 permutations in S80 for which {Cn(F)} is non-P-recursive. Yes, the construction is huge, but so what? What’s a few thousand permutations between friends? In fact, perhaps a single pattern (1324) is already non-P-recursive. Let me explain the reasoning behind what we did and why our result is much stronger than it might seem.
#### Why we did what we did
First, a very brief history of the NZ-conjecture (see Kirtaev’s book for a comprehensive history of the subject and vast references). Traditionally, pattern avoidance dealt with exact and asymptotic counting of pattern avoiding permutations for small sets of patterns. The subject was initiated by MacMahon (1915) and Knuth (1968) who showed that we get Catalan numbers for patterns of length 3. The resulting combinatorics is often so beautiful or at least plentiful, it’s hard to imagine how can it not be, thus the NZ-conjecture. It was clearly very strong, but resisted all challenges until now. Wilf reports that Richard Stanley disbelieved it (Richard confirmed this to me recently as well), but hundreds of papers seemed to confirm its validity in numerous special cases.
Curiously, the case of the (1324) pattern proved difficult early on. It remains unresolved whether Cn(1324) is P-recursive or not. This pattern broke Doron Zeilberger’s belief in the conjecture, and he proclaimed that it’s probably non-P-recursive and thus NZ-conjecture is probably false. When I visited Doron last September he told me he no longer has strong belief in either direction and encouraged me to work on the problem. I took a train back to Manhattan looking over New Jersey’s famously scenic Amtrack route. Somewhere near Pulaski Skyway I called Scott to drop everything, that we should start working on this problem.
You see, when it comes to pattern avoidance, things move from best to good to bad to awful. When they are bad, they are so bad, it can be really hard to prove that they are bad. But why bother – we can try to figure out something awful. The set of patterns that we constructed in our paper is so awful, that proving it is awful ain’t so bad.
#### Why is our result much stronger than it seems?
That’s because the proof extends to other results. Essentially, we are saying that everything bad you can do with Turing machines, you can do with pattern avoidance (mod 2). For example, why is (1324) so hard to analyze? That’s because it’s even hard to compute both theoretically and experimentally – the existing algorithms are recursive and exponential in n. Until our work, the existing hope for disproving the NZ-conjecture hinged on finding an appropriately bad set of patterns such that computing {Cn(F)} is easy. Something like this sequence which has a nice recurrence, but is provably non-P-recursive. Maybe. But in our paper, we can do worse, a lot worse…
We can make a finite set of patterns F, such that computing {Cn(F) mod 2} is “provably” non-polynomial (Th 1.4). Well, we use quotes because of the complexity theory assumptions we must have. The conclusion is much stronger than non-P-recursiveness, since every P-recursive sequence has a trivial polynomial in n algorithm computing it. But wait, it gets worse!
We prove that for two sets of patterns F and G, the problem “Cn(F) = Cn(G) mod 2 for all n” is undecidable (Th 1.3). This is already a disaster, which takes time to sink in. But then it gets even worse! Take a look at our Corollary 8.1. It says that there are two sets of patterns F and G, such that you can never prove nor disprove that Cn(F) = Cn(G) mod 2. Now that’s what I call truly awful.
#### What gives?
Well, the original intuition behind the NZ-conjecture was clearly wrong. Many nice examples is not a good enough evidence. But the conjecture was so plausible! Where did the intuition fail? Well, I went to re-read Polya’s classic “Mathematics and Plausible Reasoning“, and it all seemed reasonable. That is both Polya’s arguments and the NZ-conjecture (if you don’t feel like reading the whole book, at least read Barry Mazur’s interesting and short followup).
Now think about Polya’s arguments from the point of view of complexity and computability theory. Again, it sounds very “plausible” that large enough sets of patterns behave badly. Why wouldn’t they? Well, it’s complicated. Consider this example. If someone asks you if every 3-connected planar cubic graph has a Hamiltonian cycle, this sounds plausible (this is Tait’s conjecture). All small examples confirm this. Planar cubic graphs do have very special structure. But if you think about the fact that even for planar graphs, Hamiltonicity is NP-complete, it doesn’t sound plausible anymore. The fact that Tutte found a counterexample is no longer surprising. In fact, the decision problem was recently proved to be NP-complete in this case. But then again, if you require 4-connectivity, then every planar graph has a Hamiltonian cycle. Confused enough?
Back to the patterns. Same story here. When you look at many small cases, everything is P-recursive (or yet to be determined). But compare this with Jacob Fox’s theorem that for a random single pattern π, the sequence {Cn(π)} grows much faster than originally expected (cf. Arratia’s Conjecture). This suggests that small examples are not representative of complexity of the problem. Time to think about disproving ALL conjectures based on that evidence.
If there is a moral in this story, it’s that what’s “plausible” is really hard to judge. The more you know, the better you get. Pay attention to small crumbs of evidence. And think negative!
#### What’s wrong with being negative?
Well, conjectures tend to be optimistic – they are wishful thinking by definition. Who would want to conjecture that for some large enough a,b,c and n, there exist a solution of an + bn = cn? However, being so positive has a drawback – sometimes you get things badly wrong. In fact, even polynomial Diophantine equations can be as complicated as one wishes. Unfortunately, there is a strong bias in Mathematics against counterexamples. For example, only two of the Clay Millennium Problems automatically pay \$1 million for a counterexample. That’s a pity. I understand why they do this, just disagree with the reasoning. If anything, we should encourage thinking in the direction where there is not enough research, not in the direction where people are already super motivated to resolve the problem.
In general, it is always a good idea to keep an open mind. Forget all this “power of positive thinking“, it’s not for math. If you think a conjecture might be false, ignore everybody and just go for disproof. Even if it’s one of those famous unsolved conjectures in mathematics. If you don’t end up disproving the conjecture, you might have a bit of trouble publishing computational evidence. There are some journals who do that, but not that many. Hopefully, this will change soon…
#### Happy ending
When we were working on our paper, I wrote to Doron Zeilberger if he ever offered a reward for the NZ-conjecture, and for the disproof or proof only? He replied with an unusual award, for the proof and disproof in equal measure. When we finished the paper I emailed to Doron. And he paid. Nice… 🙂
## What do math journals do? What will become of them in the future?
Recently, there has been plenty of discussions on math journals, their prices, behavior, technology and future. I have been rather reluctant to join the discussion in part due to my own connection to Elsevier, in part because things in Combinatorics are more complicated than in other areas of mathematics (see below), but also because I couldn’t reconcile several somewhat conflicting thoughts that I had. Should all existing editorial boards revolt and all journals be electronic? Or perhaps should we move to “pay-for-publishing” model? Or even “crowd source refereeing”? Well, now that the issue a bit cooled down, I think I figured out exactly what should happen to math journals. Be patient – a long explanation is coming below.
#### Quick test questions
I would like to argue that the debate over the second question is the general misunderstanding of the first question in the title. In fact, I am pretty sure most mathematicians are quite a bit confused on this, for a good reason. If you think this is easy, quick, answer the following three questions:
1) Published paper has a technical mistake invalidating the main result. Is this a fault of author, referee(s), handling editor, managing editor(s), a publisher, or all of the above? If the reader find such mistake, who she/he is to contact?
2) Published paper proves special case of a known result published 20 years earlier in an obscure paper. Same question. Would the answer change if the author lists the paper in the references?
3) Published paper is written in a really poor English. Sections are disorganized and the introduction is misleading. Same question.
Now that you gave your answers, ask a colleague. Don’t be surprised to hear a different point of view. Or at least don’t be surprised when you hear mine.
#### What do referees do?
In theory, a lot. In practice, that depends. There are few official journal guides to referees, but there are several well meaning guides (see also here, here, here, here §4.10, and a nice discussion by Don Knuth §15). However, as any editor can tell you, you never know what exactly did the referee do. Some reply within 5 min, some after 2 years. Some write one negative sentence, some 20 detailed pages, some give an advice in the style “yeah, not a bad paper, cites me twice, why not publish it”, while others a brushoff “not sure who this person is, and this problem is indeed strongly related to what I and my collaborators do, but of course our problems are much more interesting/important – rejection would be best”. The anonymity is so relaxing, doing a poor job is just too tempting. The whole system hinges on the shame, sense of responsibility, and personal relationship with the editor.
A slightly better questions is “What do good referees do?” The answer is – they don’t just help the editor make acceptance/rejection decision. They help the authors. They add some background the authors don’t know, look for missing references, improve on the proofs, critique the exposition and even notation. They do their best, kind of what ideal advisors do for their graduate students, who just wrote an early draft of their first ever math paper.
In summary, you can’t blame the referees for anything. They do what they can and as much work as they want. To make a lame comparison, the referees are like wind and the editors are a bit like sailors. While the wind is free, it often changes direction, sometimes completely disappears, and in general quite unreliable. But sometimes it can really take you very far. Of course, crowd sourcing refereeing is like democracy in the army – bad even in theory, and never tried in practice.
#### First interlude: refereeing war stories
I recall a curious story by Herb Wilf, on how Don Knuth submitted a paper under assumed name with an obscure college address, in order to get full refereeing treatment (the paper was accepted and eventually published under Knuth’s real name). I tried this once, to unexpected outcome (let me not name the journal and the stupendous effort I made to create a fake identity). The referee wrote that the paper was correct, rather interesting but “not quite good enough” for their allegedly excellent journal. The editor was very sympathetic if a bit condescending, asking me not to lose hope, work on my papers harder and submit them again. So I tried submitting to a competing but equal in statue journal, this time under my own name. The paper was accepted in a matter of weeks. You can judge for yourself the moral of this story.
A combinatorialist I know (who shall remain anonymous) had the following story with Duke J. Math. A year and a half after submission, the paper was rejected with three (!) reports mostly describing typos. The authors were dismayed and consulted a CS colleague. That colleague noticed that the three reports were in .pdf but made by cropping from longer files. Turns out, if the cropping is made straightforwardly, the cropped portions are still hidden in the files. Using some hacking software the top portions of the reports were uncovered. The authors discovered that they are extremely positive, giving great praise of the paper. Now the authors believe that the editor despised combinatorics (or their branch of combinatorics) and was fishing for a bad report. After three tries, he gave up and sent them cropped reports lest they think somebody else considers their paper worthy of publishing in the grand old Duke (cf. what Zeilberger wrote about Duke).
Another one of my stories is with the Journal of AMS. A year after submission, one of my papers was rejected with the following remarkable referee report which I quote here in full:
The results are probably well known. The authors should consult with experts.
Needless to say, the results were new, and the paper was quickly published elsewhere. As they say, “resistance is futile“.
#### What do associate/handling editors do?
Three little things, really. They choose referees, read their reports and make the decisions. But they are responsible for everything. And I mean for everything, both 1), 2) and 3). If the referee wrote a poorly researched report, they should recognize this and ignore it, request another one. They should ensure they have more than one opinion on the paper, all of them highly informed and from good people. If it seems the authors are not aware of the literature and referee(s) are not helping, they should ensure this is fixed. It the paper is not well written, the editors should ask the authors to rewrite it (or else). At Discrete Mathematics, we use this page by Doug West, as a default style to math grammar. And if the reader finds a mistake, he/she should first contact the editor. Contacting the author(s) is also a good idea, but sometimes the anonymity is helpful – the editor can be trusted to bring bad news and if possible, request a correction.
B.H. Neumann described here how he thinks the journal should operate. I wish his views held widely today. The book by Krantz, §5.5, is a good outline of the ideal editorial experience, and this paper outlines how to select referees. However, this discussion (esp. Rick Durrett’s “rambling”) is more revealing. Now, the reason most people are confused as to who is responsible for 1), 2) and 3), is the fact that while some journals have serious proactive editors, others do not, or their work is largely invisible.
#### What do managing editors and publishers do?
In theory, managing editors hire associate editors, provide logistical support, distribute paper load, etc. In practice they also serve as handling editors for a large number of papers. The publishers… You know what the publishers do. Most importantly, they either pay editors or they don’t. They either charge libraries a lot, or they don’t. Publishing is a business, after all…
#### Who wants free universal electronic publishing?
Good mathematicians. Great mathematicians. Mathematicians who write well and see no benefit in their papers being refereed. Mathematicians who have many students and wish the publishing process was speedier and less cumbersome, so their students can get good jobs. Mathematicians who do not value the editorial work and are annoyed when the paper they want to read is “by subscription only” and thus unavailable. In general, these are people who see having to publish as an obstacle, not as a benefit.
#### Who does not want free universal electronic publishing?
Publishers (of course), libraries, university administrators. These are people and establishments who see value in existing order and don’t want it destroyed. Also: mediocre mathematicians, bad mathematicians, mathematicians from poor countries, mathematicians who don’t have access to good libraries (perhaps, paradoxically). In general, people who need help with their papers. People who don’t want a quick brush-off “not good enough” or “probably well known”, but those who need advice on the references, on their English, on how the papers are structured and presented, and on what to do next.
#### So, who is right?
Everyone. For some mathematicians, having all journals to be electronic with virtually no cost is an overall benefit. But at the very least, “pro status quo” crowd have a case, in my view. I don’t mean that Elsevier pricing policy is reasonable, I am talking about the big picture here. In a long run, I think of journals as non-profit NGO‘s, some kind of nerdy versions of Nobel Peace Prize winning Médecins Sans Frontières. While I imagine that in the future many excellent top level journals will be electronic and free, I also think many mid-level journals in specific areas will be run by non-profit publishers, not free at all, and will employ a number of editorial and technical stuff to help the authors, both of papers they accept and reject. This is a public service we should strive to perform, both for the sake of those math papers, and for development of mathematics in other countries.
Right now, the number of mathematicians in the world is already rather large and growing. Free journals can do only so much. Without high quality editors paid by the publishers, with a large influx of papers from the developing world, there is a chance we might loose the traditional high standards for published second tier papers. And I really don’t want to think of a mathematics world once the peer review system is broken. That’s why I am not in the “free publishing camp” – in an effort to save money, we might loose something much more valuable – the system which gives foundation and justification of our work.
#### Second interlude: journals vis-à-vis combinatorics
I already wrote about the fate of combinatorics papers in the Annals, especially when comparison with Number Theory. My view was gloomy but mildly optimistic. In fact, since that post was written couple more combinatorics papers has been accepted. Good. But let me give you a quiz. Here are two comparable highly selective journals – Duke J. Math. and Composito Math. In the past 10 years Composito published exactly one (!) paper in Combinatorics (defined as primary MSC=05), of the 631 total. In the same period, Duke published 8 combinatorics papers of 681 total.
Q: Which of the two (Composito or Duke) treats combinatorics papers better?
A: Composito, of course.
The reasoning is simple. Forget the anecdotal evidence in the previous interlude. Just look at the “aim and scope” of the journals vs. these numbers. Here is what Compsito website says with a refreshing honesty:
By tradition, the journal published by the foundation focuses on papers in the main stream of pure mathematics. This includes the fields of algebra, number theory, topology, algebraic and analytic geometry and (geometric) analysis. Papers on other topics are welcome if they are of interest not only to specialists.
Translation: combinatorics papers are not welcome (as are papers in many other fields). I think this is totally fair. Nothing wrong with that. Clearly, there are journals which publish mostly in combinatorics, and where papers in none of these fields would be welcome. In fact there is a good historical reason for that. Compare this with what Duke says on its website:
Published by Duke University Press since its inception in 1935, the Duke Mathematical Journal is one of the world’s leading mathematical journals. Without specializing in a small number of subject areas, it emphasizes the most active and influential areas of current mathematics.
See the difference? They don’t name their favorite areas! How are the authors supposed to guess which are these? Clearly, Combinatorics with its puny 1% proportion of Duke papers is not a subject area that Duke “emphasizes”. Compare it with 104 papers in Number Theory (16%) and 124 papers in Algebraic Geometry (20%) over the same period. Should we conclude that in the past 10 years, Combinatorics was not “the most active and influential”, or perhaps not “mathematics” at all? (yes, some people think so) I have my own answer to this question, and I bet so do you…
Note also, that things used to be different at Duke. For example, exactly 40 years earlier, in the period 1963-1973, Duke published 47 papers in combinatorics out of 972 total, even though the area was only in its first stages of development. How come? The reason is simple: Leonard Carlitz was Managing Editor at the time, and he welcomed papers from a number of prominent combinatorialists active during that time, such as Andrews, Gould, Moon, Riordan, Stanley, Subbarao, etc., as well as a many of his own papers.
#### So, ideally, what will happen to math journals?
That’s actually easy. Here are my few recommendations and predictions.
1) We should stop with all these geography based journals. That’s enough. I understand the temptation for each country, or university, or geographical entity to have its own math journal, but nowadays this is counterproductive and a cause for humor. This parochial patriotism is perhaps useful in sports (or not), but is nonsense in mathematics. New journals should emphasize new/rapidly growing areas of mathematics underserved by current journals, not new locales where printing presses are available.
2) Existing for profit publishers should realize that with the growth of arXiv and free online competitors, their business model is unsustainable. Eventually all these journals will reorganize into a non-profit institutions or foundations. This does not mean that the journals will become electronic or free. While some probably will, others will remain expensive, have many paid employees (including editors), and will continue to provide services to the authors, all supported by library subscriptions. These extra services are their raison d’être, and will need to be broadly advertised. The authors would learn not to be surprised of a quick one line report from free journals, and expect a serious effort from “expensive journals”.
3) The journals will need to rethink their structure and scope, and try to develop their unique culture and identity. If you have two similar looking free electronic journals, which do not add anything to the papers other than their .sty file, the difference is only the editorial board and history of published papers. This is not enough. All journals, except for the very top few, will have to start limiting their scope to emphasize the areas of their strength, and be honest and clear in advertising these areas. Alternatively, other journals will need to reorganize and split their editorial board into clearly defined fields. Think Proc. LMS, Trans. AMS, or a brand new Sigma, which basically operate as dozens of independent journals, with one to three handling editors in each. While highly efficient, in a long run this strategy is also unsustainable as it leads to general confusion and divergence in the quality of these sub-journals.
4) Even among the top mathematicians, there is plenty of confusion on the quality of existing mathematics journals, some of which go back many decades. See e.g. a section of Tim Gowers’s post about his views of the quality of various Combinatorics journals, since then helpfully updated and corrected. But at least those of us who have been in the area for a while, have the memory of the fortune of previously submitted papers, whether our own, or our students, or colleagues. A circumstantial evidence is better than nothing. For the newcomers or outsiders, such distinctions between journals are a mystery. The occasional rankings (impact factor or this, whatever this is) are more confusing than helpful.
What needs to happen is a new system of awards recognizing achievements of individual journals and/or editors, in their efforts to improve the quality of the journals, attracting top papers in the field, arranging fast refereeing, etc. Think a mixture of Pulitzer Prize and J.D. Power and Associates awards – these would be a great help to understand the quality of the journals. For example, the editors of the Annals clearly hustled to referee within a month in this case (even if motivated by PR purposes). It’s an amazing speed for a technical 50+ page paper, and this effort deserves recognition.
Full disclosure: Of the journals I singled out, I have published once in both JAMS and Duke. Neither paper is in Combinatorics, but both are in Discrete Mathematics, when understood broadly.
## What’s the Matter with Hertz Foundation?
Imagine you have plenty of money and dozens of volunteers. You decide to award one or two fellowships a year to the best of the best of the best in math sciences. Easy, right? Then how do you repeatedly fail at this, without anyone notice? Let me tell you how. It’s an interesting story, so bear with me.
A small warning. Although it may seem I am criticizing Hertz Foundation, my intention is to show its weakness so it can improve.
#### What is the Hertz Foundation?
Yesterday I wrote a recommendation letter to the Hertz Foundation. Although a Fellow myself, I never particularly cared for the foundation, mostly because it changed so little in my life (I received it only for two out of five years of eligibility). But I became rather curious as to what usually happens to Hertz Fellows. I compiled the data, and found the results quite disheartening. While perhaps excellent in other fields, I came to believe that Hertz does barely a mediocre job awarding fellowships in mathematics. And now that I think about it, this was all completely predictable.
First, a bit of history. John Hertz was the Yellow Cab founder and car rental entrepreneur (thus the namesake company), and he left a lot of money dedicated for education in “applied physical sciences”, now understood to include applied mathematics. What exactly is “applied mathematics” is rather contentious, so the foundation wisely decided that “it is up to each fellowship applicant to advocate to us his or her specific field of interest as an ‘applied physical science’.”
In practice, according to the website, about 600 applicants in all areas of science and engineering apply for a fellowship. Applications are allowed only either in the senior year of college or 1st year of grad school. The fellowships are generous and include both the stipend and the tuition; between 15 and 20 students are awarded every year. Only US citizen and permanent residents are eligible, and the fellowship can be used only in one of the 47 “tenable schools” (more on this below). The Foundation sorts the applications, and volunteers interview some of them in the first round. In the second round, pretty much only one person interviews all that advanced, and the decision is made. Historically, only one or two fellowships in mathematical sciences are awarded each year (this includes pure math, applied math, and occasionally theoretical CS or statistics).
#### Forty years of Math Hertz Fellowships in numbers
The Hertz Foundation website has a data on all past fellows. I compiled the data in Hertz-list which spanned 40 years (1971-2010), listed by the year the fellowship ended, which usually but not always coincided with graduation. There were 67 awardees in mathematics, which makes it about 1.7 fellowships a year. The Foundation states that it awarded “over 1000 fellowships” so I guess about 5-6% went into maths (perhaps, fewer in recent years). Here is who gets them.
1) Which schools are awarded? Well, only 44 US graduate programs are allowed to administer the fellowships. The reasons (other than logistical) are unclear to me. Of those programs that are “in”, you have University of Rochester (which nearly lost its graduate program), and UC Santa Cruz (where rumors say a similar move had been considered). Those which are “out” include graduate programs at Brown, UPenn, Rutgers, UNC Chapel Hill, etc. The real distribution is much more skewed, of course. Here is a complete list of awards per institution:
MIT – 14
Harvard, Princeton – 8
Caltech, NYU – 7
Berkeley, Stanford – 5
UCLA – 3
CMU, Cornell, U Chicago – 2
GA Tech, JHU, RPI, Rice – 1
In summary, only 15 universities had at least one award (34%), and just 7 universities were awarded 54 fellowships (i.e. 16% of universities received 81% of all fellowships). There is nothing wrong with this per se, just a variation on the 80-20 rule you might argue. But wait! Hertz Foundation is a non-profit institution and the fellowship itself comes with a “moral commitment“. Even if you need to interfere with “free marketplace” of acceptance decisions (see P.S. below), wouldn’t it be in the spirit of John Hertz’s original goal, to make a special effort to distribute the awards more widely? For example, Simons Foundation is not shy about awarding fellowship to institutions many of which are not even on Hertz’s list.
2) Where are they now? After two hours of googling, I located almost all former fellows and determined their current affiliations (see the Hertz-list). I found that of the 67 fellows:
University mathematicians – 27 (40%)
Of these, work at Hertz eligible universities – 14, or about 21% of the total (excluding 3 overseas)
At least 10 who did not receive a Ph.D. – 15%
At least 13 are in non-academic research – 19% (probably more)
At least 8 in Software Development and Finance – 12% (probably more)
Now, there is certainly nothing wrong with directing corporate research, writing software, selling derivatives, designing museum exhibits, and even playing symphony orchestra or heading real estate company, as some of the awardees do now. Many of these are highly desirable vocations. But really, was this what Hertz had in mind when dedicating the money? In the foundation’s language, “benefit us all” they don’t.
I should mention that the list of Hertz Fellows in Mathematics does include a number of great academic success stories, but that’s not actually surprising. Every US cohort has dozens of excellent mathematicians. But the 60% drop out rate from academia is very unfortunate, only 21% working in “tenable universities” is dismaying, and the 15% drop out rate from graduate programs is simply miserable. Couldn’t they have done better?
#### A bit of analysis
Every year, US universities award over 1,600 Ph.D.’s in mathematical sciences, of which over a half go to US citizen (more if you include permanent residents, but stats is not easily available). So they are choosing 1.7 out of over 800 eligible students. Ok, because of their “tenable schools” restriction this is probably more like 300-400. Therefore, less than half of one percent of potential applicants are awarded! For comparison, Harvard college acceptance rate is 10 times that.
Let me repeat: in mathematics, Hertz fellows drop out from their Ph.D. programs at a rate of 15%. If you look into the raw 2006 NRC data for graduation rates, you will see that many of the top universities have over 90% graduation rate in the math programs (say, Harvard has over 91%). Does that mean that Harvard on average does a better job selecting 10-15 grad students every year, while Hertz can’t choose one?
Yes, I think it does. And the gap is further considering that Hertz has virtually no competition (NSF Fellowships are less generous in every respect). You see, people at Harvard (or Princeton, MIT, UCLA, etc.) who read graduate applications, know what they are doing. They are professionals who are looking for the most talented mathematicians from a large pool of applicants. They know which letters need to be taken seriously, and which with a grain of salt. They know which undergraduate research experience is solid and which is worthless. They just know how things are done.
Now, a vast majority of Hertz interviewers are themselves former fellows, and thus about 95% of them have no idea about the mathematics research (they just assume it’s no different from the research they are accustomed to). Nor does the one final interviewer, who is an applied physicist. As a result, they are to some extend, flipping coins and rolling dies, in hope things will work out. You can’t really blame them – they simply don’t know how to choose. I still remember my own two interviews. Both interviewers were nice, professional, highly experienced and well intentioned, but looking back I can see that neither had much experience with mathematical research.
You can also see this lack of understanding of mathematics culture is creeping up in other activities of the foundation, such as the thesis prize award (where are mathematicians?), etc. Of course a private foundation can award anyone it pleases, but it seems to me it would do much more good if only some special care is applied.
#### A modest proposal
There is of course, a radical way to change the review of mathematics applicants – subcontract it to the AMS (or IMA, MSRI, IPAM – all have the required infrastructure). For a modest fee, the AMS will organize a panel of mathematicians who will review and rank the applicants without interviewing them. The panel will be taking into consideration only students’ research potential, not the university prestige, etc. The Hertz people can then interview the top ranked and make a decision at the last stage, but the first round will be by far superior to current methods. Even the NSA trusts AMS, so shouldn’t you?
Hertz might even save some money it currently spends on travel and lodging reimbursements. The 13% operating budget is about average, but there is some room for improvement. Subcontracting will probably lead to an increase in applications, as AMS really knows how to advertise to its members (I bet you currently receive only about 40 mathematics applications, out of a potential 400+ pool). To summarize: really, Hertz Foundation, think about doing that!
P.S. It is not surprising that the 7 top universities get a large number of the fellowships. One might be tempted to assume that clueless interviewers are perhaps somewhat biased towards famous school names in the hope that these schools already made a good decision accepting these applicants, but this is not the whole story. The described bias can only work for the 1st year grad applicants, but for undergraduate applicants a different process seems to hold. Once a graduate school learns that an applicant received Hertz Fellowship (or NSF for that matter), it has every incentive to accept the student, as the tuition and the stipend are paid by the outside sources now.
P.P.S. Of course, mathematicians’ review can also fail. Even the super prestigious AIM Fellowship has at least one recipient who left academia for bigger and better things.
UPDATE (April 15, 2019). Over the years since this blog post, I have been contacted by people from the Hertz Foundation board. I have also followed up on the story and the recent fellowship recipients. I am happy to say that the foundation implemented various important changes vis-à-vis math interviews, to the visible effect. At the moment, the numbers are too small to report statistics and the changes I know are not a public information. I concluded that my criticism no longer applies, a happy ending to the story. I encourage now everyone to support the foundation financially as well as recommend your best students to apply.
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https://socratic.org/questions/how-do-you-simplify-4xsqrt3-3xsqrt2
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How do you simplify 4xsqrt3 (-3xsqrt2)?
$4 x \sqrt{3} \left(- 3 x \sqrt{2}\right) = \left(4 \cdot - 3\right) \left(x \cdot x\right) \left(\sqrt{3} \cdot \sqrt{2}\right)$
$= \left(- 12\right) \left({x}^{2}\right) \left(\sqrt{3 \cdot 2}\right)$ (Because color(blue)(sqrta*sqrtb = sqrt(ab)
=color(green)( -12x^2sqrt6
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https://abujacatholicarchdiocese.org/bible/how-long-is-14-generations-in-the-bible.html
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How long is 14 generations in the Bible?
Contents
An Explanation of the Genealogy of Jesus.
How long is 42 generations in the Bible?
The prophet Abraham was born in the year 2042 B.C., and Jesus was born in A.D. 1. This means that 2042 years passed between Abraham and Jesus. If you divide 2042 years by 42 generations, you get an average of approximately 50 years per generation.
How many generations are there from Jesus until now?
Jesus was estimated to be born in between 6 and 4 BC and it’s currently 2021. The amount of years between 6 BC and 2021 are 2027 years, so 2027/25 = 81.08. Now the amount of years between 4 BC and 2021 are 2025, so 2025/25 = 81. So the amount of generations between Jesus and us in the current year is 81 generations.
How many years are we promised in the Bible?
The short answer, which no one has adequately answered yet, is 70 years or by reason of strength, 80 years. God gave that promise for us to Moses – listen carefully – as a minimum. Moses lived to 120 years. We were not supposed to die at all.
How long is 1000 generations in the Bible?
According to the bible, a generation is 38 years. Thus a thousand generations would be 38,000 years.
How many generations is 1000 years?
Generation time – Wikipedia People generally take 25 years as a good average. Thus, 1,000 years would be 40 generations for humans.
How many generations are there between Abraham and Jesus?
In Matthew’s Gospel there are 41 generations from Abraham to Jesus. Good News Translation So then, there were fourteen generations from Abraham to David, and fourteen from David to the exile in Babylon, and fourteen from then to the birth of the Messiah.
What happened 14 generations after Jesus?
In the King James Version of the Bible the text reads: So all the generations from Abraham to David are fourteen generations; and. from David until the carrying away into Babylon are fourteen generations; and from the carrying away into Babylon unto Christ are fourteen generations.
How long ago is 15 generations?
The idea that there’s about 20 yrs to a generation would indicate 15 generations is about 300 yrs . 300 years and in excess of 32,000 individuals.
How long is a generation in years?
A generation is “all of the people born and living at about the same time, regarded collectively.” It can also be described as, “the average period, generally considered to be about 20–30 years, during which children are born and grow up, become adults, and begin to have children.” In kinship terminology, it is a …
How long is threescore and ten years?
threescore years and ten (=70 years): He had lived for threescore years and ten.
THIS IS INTERESTING: Where can I watch good Christian movies?
How long did Eve in the Bible live?
Torah says he lived for 930 years. God knows the best. Eve is reported to have died one year and 15 days after the death of Adam and is buried next to him at Mount Qubays. During the deluge, Nuh exhumed Adam and kept his body in the boat.
How many years did Adam live for?
According to Jewish tradition, Adam and Eve had 56 children. This was possible, in part, because Adam lived to be 930 years old. Some scholars believe that the length of the life spans of the people of this time was due to a vapor canopy in the atmosphere.
How many generations is 500 years?
Since each generation is about 25 years long, we simply divide 500 by 25 to determine that there are 20 generations in 500 years.
Where in the Bible is the blessing?
Blessings from God to Man. The first blessing from God to man appears in the first chapter of Genesis (1:28): “God blessed them and God said to them, “Be fertile and increase, fill the earth and master it.” 1. God blesses (not commands!)
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https://jp.mathworks.com/matlabcentral/cody/problems/42820-jumping-astronauts/solutions/2924638
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Cody
# Problem 42820. jumping astronauts
Solution 2924638
Submitted on 8 Sep 2020 by ChrisR
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
alpha=0; n=1; p=1; D_correct=0; assert(abs(jumping_astronauts(alpha,p,n)-D_correct)<0.01)
2 Pass
alpha=45; n=1; p=1; D_correct=0.5; assert(abs(jumping_astronauts(alpha,p,n)-D_correct)<0.01)
3 Pass
alpha=45; n=100; p=1; D_correct=50; assert(abs(jumping_astronauts(alpha,p,n)-D_correct)<0.01)
4 Pass
alpha=45; n=100; p=0.8; D_correct=3.32; assert(abs(jumping_astronauts(alpha,p,n)-D_correct)<0.01)
5 Pass
alpha=80; n=1000; p=0.95; D_correct=17.93; assert(abs(jumping_astronauts(alpha,p,n)-D_correct)<0.01)
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
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http://cell-auto.com/largeinverse/
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# Automata whose inverses have unboundedly large neighbourhoods
### Invertible automata
There are a number of techniques designed to create cellular automata which are reversible and have corresponding inverse automata associated with them.
Generally speaking most of these techniques made it easy to find the inverse automata. This is true of Fredkin's method, the use of partitioning schemes, techniques involving embedding a irreversible automata in a higher-dimensional space and the guarded context technique.
### Cryptographic application
For some cryptographic applications, it is desirable to have an operation whose inverse is computationally hard to compute. Such functions are called "one-way" functions. There are also "one-way trap-door" functions - functions whose inverse is difficult to identify, but relatively easy to calculate once it has been located.
With this goal in mind, some pre-requisites for they type of system needed should be clear:
If the "radius" of an automaton's inverse is relatively small, it will be possible to identify its inverse by considering a small space surrounding the cell whose inverse function is needed and apply brute force search techniques to locate the inverse function for that cell.
If such a technique succeeds, finding the inverse is unlikely to prove much of a problem.
In order to make finding an inverse function for a particular cell difficult, it is desirable to make the domain of the inverse function include a very large number of cells. Ideally, it should be a function of all the other cells in the automaton. This would be likely to make a brute force as hard as possible.
### The existence of "one-way" automata
The existence of such automata was proven by Jarkko Kari in a paper called "Reversibility of 2D Cellular Automata is Undecidable" [Kari, 1990]. He showed that for a class of reversible automata using the VN neighbourhood, in general, no bound may be placed on the size of the neighbourhood required by their inverse automata.
He also showed that there is no finite effective procedure for locating the inverse of such a reversible automaton, i.e. there is no general algorithm for finding the inverse rule with a time complexity bounded by a computable function.
### Constructing automata whose inverses have unboundedly large neighbourhoods
The construction presented here can be applied to generate automata of any number of dimensions. If working in more than one dimension, the resulting automata are typically spatially non-uniform.
The method involves making a one dimensional path through every cell the automaton, and constructing the rules of the cells around that.
The inverse function can be computed only by traversing the path in a specified direction - and performing a long string of calculations in series.
#### Example
In order to explain the procedure, we will present an example using a one-dimensional automaton.
This automaton will be based on Wolfram's rule-30.
Rule-30's update rule is: `C(t + 1,i) = C(t, i - 1) ^ (C(t, i) | C(t, i + 1))`;
The new update rule is essentially: `C(t + 1,i) = C(t, i) ^ (C(t, i + 1) | C(t, i + 2))`;
This looks identical, except the "i"s on the right hand side of the equation have all had one added to them.
The rule is applied to a finite field, without the use of periodic boundary conditions `i = {0, 1, ... n - 1}`.
When `i = n - 2`, the rule can no longer be applied. Instead the rules:
`C(t + 1,n - 2) = C(t, n - 2) ^ C(t, n - 1)` ...and...
`C(t + 1,n - 1) = C(t, n - 1)` ...are used.
The resulting automata displays much the same behaviour as rule-30 would if it was skewed sideways. However - unlike Rule-30 when applied to a finite field, it is reversible - and an inverse automata exists.
To calculate the inverse, start with the cell at `i = n - 1`.
Since at that point `C(t + 1,n - 1) = C(t, n - 1)`, `C(t - 1,n - 1) = C(t, n - 1)`.
Next, consider the cell at `i = n - 2`.
`C(t,n - 2) = C(t - 1, n - 2) ^ C(t - 1, n - 1)`
`C(t,n - 2) = C(t - 1, n - 2) ^ C(t, n - 1)`
`C(t - 1, n - 2) = C(t,n - 2) ^ C(t, n - 1)`
This is the inverse function at `i = n - 2`.
Next, consider the cell at `i = n - 3`:
`C(t,n - 3) = C(t - 1, n - 3) ^ (C(t - 1, n - 2) | C(t - 1, n - 1))`
`C(t,n - 3) = C(t - 1, n - 3) ^ ((C(t,n - 2) ^ C(t, n - 1)) | C(t, n - 1))`
`C(t - 1, n - 3) = C(t,n - 3) ^ ((C(t,n - 2) ^ C(t, n - 1)) | C(t, n - 1))`
This is the inverse function at `i = n - 3`.
The last series of steps can be applied iteratively, resulting in the unambiguous recovery of the entire state of the automaton at the previous time step.
#### Two dimensions
In order to apply this proceedure to a two dimensional automata, first a path through the plane must be decided upon.
One end of the path is nominated as the starting cell.
The state of each cell in the following generation depends only on the state of neighboring cells in the previous generation. In addition, cells which are further away from the starting cell than the current position cannot be considered.
In this diagram, the yellow cell is the current cell. The red cells are nearer the starting cell. The dotted lines indicate the VN-neighbourhood. The green colour indicates the cells which may contribute to the update function of the yellow cell.
The update function may be any boolean function, with one restriction: no matter what the states of the "red" cells are, the yellow cell must have the "final say" - and should always be able to completely influence the state of the cell in the next generation.
In practice, this menas that the yellow cell should be Exclusive OR'd with the function contributed by the red cells.
As Exclusive OR is a linear function, this has sometimes resulted in this type of automata being referred to as "quasi-linear".
#### Iterating in either direction
Iterating "forwards" is simply a case of applying the transformation in each cell, like a normal cellular automata. This operation may performed in parallel.
When iterating "backwards", the state of each cell must be calculated in a series of steps along the path, starting with the first cell on the path, and progressing along it.
#### Use of two passes
The result of updating an automata using the resulting rule is that the state of each cell depends only on earlier cells along the path.
If it is desirable to have each cell depending on the state of every other cell in the automaton, it is necessary to traverse the path twice, once in each direction.
#### Notes
An extension of this construction to half-infinite domains and uniform automata is possible. This appears to indicate that the Margulis/Toffoli "Conjecture 8.1" in [Margulis and Toffoli, 1990] - which conjectures that all invertible automata are structurally invertible - i.e., can be (isomorphically) expressed in spacetime as a uniform composition of finite invertible logic primitives - should be treated with caution when considering this type of system.
There are other results along similar lines. Richardson is said to have proved that if a cellular automata is invertible, then its inverse is a cellular automata [Margulis and Toffoli, 1990]. I feel obliged to add that this is not true if the automaton is non-uniform, or if the automaton is half-infinite, with a single edge.
Similar comments apply to Max Garzon's Corollorary 7.15: "If T is one-one then T-1 [i.e. T's inverse] [...] is defined by a local transformation." [Garzon, 1995]
#### Modifications to the technique
A page listing some ways of modifying the technique is available here.
### References
Kari J.: "Reversibility of 2D Cellular Automata is Undecidable", 1990;
Tim Tyler | tim@tt1.org | http://cell-auto.com/
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# Tagged Questions
The tag has no usage guidance.
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quick question for my better understanding: Assume you have an additive category $\mathcal{C}$ and an automorphism $\Sigma$ of this category. Does $\Sigma$ send objects to isomorphic objects? If it ...
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### Behavior $\otimes$-Triangulated Subcategory under Inverse
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### All Ideals are Radical in Rigid Categories
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The following definition is taken from Daniel Murfet's Triangulated Categories Part I notes. Let $\mathcal T$ be a triangulated category with countable coproducts. Suppose we are given a ...
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### Bridgeland stability conditions: The heart satisfies the Harder-Narasimhan property
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### Embedding into a morphism of distinguished triangles
Everything in this question happens in a triangulated category $\mathbf{D}$. I am trying to prove that in a diagram like this \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\...
224 views
### Questions about Rickards proof that $D^b_\mathtt{sg}(A) \equiv \mathtt{stmod}(A)$
Setup: Let $A$ be a self-injective algebra (so projective = injective for modules) and let $D^b(A)$ and $K^b(A)$ be the bounded derived category and the full subcategory consisting of the perfect ...
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### What is the decategorification of a triangulated category?
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369 views
### Nine lemma in Triangulated categories
I am curious if something like the Nine Lemma (http://en.wikipedia.org/wiki/Nine_lemma) is true in an arbitrary triangulated category. To be more explicit, suppose I have a map of cofiber sequences/...
651 views
### cones in the derived category
If I have two exact triangles $X \to Y \to Z \to X[1]$ and $X' \to Y' \to Z' \to X'[1]$ in a triangulated category, and I have morphisms $X \to X'$, $Y \to Y'$ which 'commute' (...
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# Look at Table 4 in the Appendix. On Line 8 in the 12% column, you find the factor 4.96764. This fact
Look at Table 4 in the Appendix. On Line 8 in the 12% column, you find the factor 4.96764. This factor in Table 4 means that if you invest about \$4.97 for eight years you will have A. \$1 at the end of each year for eight years and \$4.97 at the end of eight years B. \$4.97 at the end of each year for eight years and \$1 at the end of eight years. C. \$1 at the end of eight years. D. \$1 at the end of each year for eight years.
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http://mathhelpforum.com/differential-geometry/79826-path-connected-components.html
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Since connectedness components are always both open and close sets, you need an example in which "connected" is not the same as "path connected". Let A be the line segment, in $\displaystyle R^2$, from (-1, 0) to (0,0) inclusive (so it is a closed set). Let B be the graph of y= sin(1/x) for $\displaystyle 0< x\le 1$. Let the topological space, X, be $\displaystyle A\cup B$. A and B are "path components" of X but B is neither open nor closed.
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# When you are 33 weeks pregnant, how many months is that
Health related question in topics Math .We found some answers as below for this question “When you are 33 weeks pregnant, how many months is that”,you can compare them.
33 weeks equates to 7.595 months. ChaCha on! [ Source: http://www.chacha.com/question/when-you-are-33-weeks-pregnant%2C-how-many-months-is-that ]
More Answers to “When you are 33 weeks pregnant, how many months is that
How many months pregnant am I if I AM 33 weeks?
33 weeks means you are 8th month running. Normally huma gestation period 281 days
How many months is 33 weeks of pregnancy?
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https://www.sawaal.com/probability-questions-and-answers/two-cards-are-drawn-from-a-pack-of-well-shuffled-cards-find-the-probability-that-one-is-a-club-and-o_3676
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24
Q:
# Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King
A) 1/52 B) 1/26 C) 1/13 D) 1/2
Explanation:
Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)
=$12C1*1C152C2+13C1*3C152C2$
=$12*252*51+13*3*252*51$$24+7852*51$ = $126$
Q:
Find the range and mode of the data 17, 18, 28, 19, 16, 18, 17, 29, 18
A) 12 and 18 B) 13 and 18 C) 12 and 17 D) 11 and 17
Explanation:
5 313
Q:
Find the standard deviation of {11, 7, 10,13, 9}
A) 1 B) 2 C) 4 D) 5
Explanation:
2 273
Q:
A table tennis player, lost 12 games out of 18 games played. Calculate the games won in terms of decimal.
A) 0.667 B) 0.067 C) 0.50 D) 0.333
Explanation:
1 385
Q:
A sequence , has odd number of terms. Then the median is
A) axn2+1 B) axn2-1 C) axn-1 D) axn2
Explanation:
3 2449
Q:
The mean of a distribution is 15 and the standard deviation is 5. What is the value of the coefficient variation?
A) 16.66% B) 66.66% C) 33.33% D) 100%
Explanation:
5 620
Q:
If the standard deviation of a population is 3, what would be the population variance?
A) 9 B) 6 C) 8 D) 15
Explanation:
3 3080
Q:
The variance of a set of data is 169. Then the standard deviation of the data is
A) +-13 B) 13 C) 69 D) 84.5
Explanation:
3 2529
Q:
If the standard deviation of a population is 10, what would be the population variance?
A) 100 B) 30 C) 5 D) 20
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https://pbsfonline.com/question-how-fast-can-average-person-throw-a-baseball/
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Without significant practice, your average person would be lucky to throw a baseball over 50 mph. For trained players, the average pitching velocity ranges between 40-50 mph among young players around 9 or 10, between 55-75 mph between 10 and 17, to an average of 80 mph for 18-year-olds and above.
## Is it possible to throw a baseball 110 mph?
The 20-year-old right-hander and Chicago’s third-ranked prospect was recorded hitting 110 mph on a radar gun during a workout at APEC, the facility in Texas where he trains during the offseason. The workout involved throwing against a screen using three-ounce and four-ounce balls, followed by a regular baseball.
## Is it possible to throw a baseball 90 mph?
A 32 inch vertical equals a power to weight ratio of that close to or over 1.5. This means the athlete can produce power that can push about 150% of his own body weight or more. A 1.5 power to weight ratio would be an accurate requirement for a pitcher to have the power to produce a 90+ mph fastball.
## How fast can you throw a baseball bat?
Major League baseballs have an average mass of 5.125 ounces, and a 90-mph fastball can leave the bat at 110 mph. Extrapolating Newton’s second law of motion, Russell determined that, in a collision lasting less than one-thousandth of a second, the average pro swing imparts 4145 pounds of force to the ball.
## How do you throw 80 mph?
NCAA Pitching Lesson 80MPH – 85MPH+ W/VPX Harness
## How fast should a 14 year old throw?
Generally, 14 year old average cruising speed would be about 65 mph. Average freshman pitcher (14 to 15 year old) cruising speed would be about 70 mph. Average cruising speed for a good high school pitching prospect at 14 to 15 years old would be about 75 mph.
## Did Nolan Ryan throw 108 mph?
According to the baseball documentary “Fastball”, Nolan Ryan? s final fastball, which he threw at age 46, was clocked at 98 MPH. Yep, you read that right.
## Who threw 106 mph?
Aroldis Chapman: 106 MPH.
## How fast can humans throw?
Human athletes can achieve throwing speeds close to 145 km/h (90 mph), far in excess of the maximal speed attainable by chimpanzees, at about 30 km/h (20 mph). This ability reflects the ability of the human shoulder muscles and tendons to store elasticity until it is needed to propel an object.
## How do you throw a 99 mph?
How to Pitch 99mph! (Pitch Faster Now with Pro Pitcher Craig Stem)
## How fast should a 17 year old pitch?
Pitching velocity by age in the U.S.
15 70 MPH 75 MPH
16 76 MPH 80 MPH
17 80 MPH 85 MPH
18 83 MPH 88-90 MPH
## How fast does a 100 mph fastball reach home plate?
A 100-mph fastball takes roughly 375-400 milliseconds to reach the plate. For reference, the blink of an eye takes 300-400 milliseconds.
## What is the fastest throw in baseball?
Fastest pitch ever thrown
As a result, Aroldis Chapman is credited with throwing the fastest pitch in MLB history. On Sept. 24, 2010, Chapman made MLB history. Then a rookie relief pitcher for the Cincinnati Reds, the fireballer unleashed a fastball clocked at 105.1 mph by PITCH/fx.
## Who has fastest bat speed in MLB?
Giancarlo Stanton, 2019, 120.6 MPH. Giancarlo Stanton, 2018, 121.7 MPH.
## What is elite bat speed?
“With some of the data we collected, we see in-game swing speeds in the 65- to 85-miles-per-hour speeds for some of the top professionals in baseball.” According to Cherveny, the average swing speed in Major League Baseball games is around 70 miles per hour.
## How fast do d1 pitchers throw?
Prototypical Division I pitching recruits throw anywhere between 87 and 95 MPH on a consistent basis. It is important to remember that coaches are looking for pitchers to consistently throw at this velocity, not just touch it every once and awhile.
## How fast should a 12 year old boy pitch?
I radar gunned many 12-year-olds at the AA travel ball level in the East Bay of CA this summer with a pocket ball coach radar gun. This captures the speed as it comes right out of the hand. The majority of pitchers were between 54 to 58 MPH, and many more were slightly out of that range a little higher or lower.
## How many pitchers can throw over 100 mph?
Only 11 pitchers have hit 103 mph even once in a regular-season game, and only three have thrown more than 10 such pitches. Meanwhile, the list of fastest pitches officially on record has stood undisturbed since 2018, but with Greene in the Reds’ rotation, it might be in for an update.
## How do you throw a 90 mph?
Can Anyone Throw 90 MPH?
## How fast did MLB pitchers throw in high school?
High School
The Freshmen pitchers will usually be throwing at a comparable speed to the 13 and 14 year olds, and the Sophomores will lie somewhere in the middle. A good changeup for a high school pitcher is a -10 to -15 mph drop, so anywhere from 60-75 mph is common.
## Who is the slowest pitcher in the MLB?
Holt set a new record for slowest pitch thrown in a Major League game since the sport began tracking such data in 2008, landing a 31 mph eephus for a called strike against Oakland utilityman Josh Harrison.
## How fast would Randy Johnson throw?
Randy Johnson was known for throwing a fastball that could reach over 100 mph. It’s not every day you see a dove explode in front of home plate, and this was all caught on camera.
## How fast did Gibson throw?
How fast was Bob Gibson’s Fastball? Bob Gibson’s 4-seam fastball “sat” between 92-95 mph in the sample I considered. It’s likely he actively varied his grip or intended velocity, producing a high velocity range, measured at 87-95, with numerous indications that he regularly exceeded 95 mph.
## How fast was Walter Johnson’s fastball?
In 1917, a Bridgeport, Connecticut, munitions laboratory recorded Johnson’s fastball at 134 feet per second, which is equal to 91 miles per hour (146 km/h), a velocity that may have been unmatched in his day, with the possible exception of Smoky Joe Wood.
## What is the fastest pitch by a 12 year old?
Cuba’s Prieto records fastest pitch at U-12 Baseball World Cup. Cuban 12-year-old right-handed pitcher Alejandro Prieto recorded the fastest pitch of the WBSC U-12 Baseball World Cup 2019, at 123 km/h (76.4 mph) in his win against Mexico today in the Super Round.
## What is the hardest pitch to hit?
Without further ado, here are the five toughest pitches to hit in baseball, based on Fangraphs data compiled in 2020.
1. Dinelson Lamet’s slider.
3. Zach Davies’ changeup.
4. Dallas Keuchel’s cutter.
5. Marco Gonzales’ fastball.
## What is the farthest baseball ever hit?
In 1987, Joey Meyer, playing for the Triple-A Denver Zephyrs, launched this ball astonishing 582 feet home run. Joey Meyer’s home run is the longest homer ever recorded on video.
## How fast can an ape throw a baseball?
Our ability to produce high-speed throws is even more impressive when compared to the throwing ability of our closest relatives, chimpanzees. Despite being incredibly strong and very athletic, an adult male chimp that has been trained to throw can only throw a ball approximately 20 mph.
## What is the fastest thing ever thrown?
Consider the confusion over the game’s fastest fastball ever. On paper, the honor goes to Yankees relief pitcher Aroldis Chapman, who clocked 105.1 miles per hour in 2010.
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http://clay6.com/qa/44094/the-vapour-pressure-of-pure-water-at-37-c-is-47-1-torr-what-is-the-vapour-p
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# The vapour pressure of pure water at 37$^{\large\circ}$C is 47.1 torr. What is the vapour pressure of an aqueous solution at 37$^{\large\circ}$C containing 20 g of glucose dissolved in 500 gm of water?
$n_{H_2O}=\large\frac{500}{18}$$=27.78mol n_{glucose}=\large\frac{20}{180}=$$0.11mol$
$X_{H_2O}=\large\frac{n_{H_2O}}{n_{H_2O}+n_{glucose}}$
$\Rightarrow 0.996$
$P_{H_2O}=P_{H_2O}^0X_{H_2O}=47.1\times 0.996=46.9$torr
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https://www.oreilly.com/library/view/statistics-in-a/9781449361129/ch13.html
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## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
No credit card required
## Chapter 13. Nonparametric Statistics
The basis of inferential statistics is parameter estimation, estimation of the parameters of a population from information gained from a random sample presumed to have been drawn from that population. Many of the most common statistical techniques rely on the underlying distribution being of a particular type, such as the normal distribution, for inferences made from the relevant statistical tests to be valid; hence, these techniques are called parametric statistics. What about scenarios in which you know or suspect that the population does not meet the assumptions for a particular statistical test? In these cases, a different set of statistical techniques, known as nonparametric statistics, can be used. These techniques are often known as distribution-free statistics because they make few or no assumptions about the underlying distribution of the data; some prefer the term “distribution-free-er” because some nonparametric tests do require assumptions about the population distribution, although those assumptions are generally less stringent than those made by common parametric tests.
Nonparametric statistics are often applied to data sets in which data has been collected as ranks rather than as raw scores, or rank data is substituted for raw scores due to concerns about the distribution of the raw data. Rank data by definition is ordinal, as discussed in Chapter 1, and should not be analyzed using procedures meant for interval- and ratio-level ...
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required
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https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/pages/assignments/
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7.91J | Spring 2014 | Graduate
# Foundations of Computational and Systems Biology
## Assignments
Five written or computer-based problem sets will be assigned. These are designed to promote deeper understanding of the principles and algorithms discussed in class and to provide hands-on experience with bioinformatics tools.
NOTE: These assignments reference Athena, MIT’s UNIX-based computing environment. OCW does not provide access to this environment.
SES # TOPICS PROBLEM SETS SOLUTIONS POINTS
1 Sequence Search, Global Alignment, BLAST Statistics Problem Set 1 (PDF) Solutions to Problem Set 1 (PDF) 19
2 BWT, Library Complexity, RNA-seq, Genome Assembly, Motifs, Multiple Hypothesis Testing
Problem Set 2 (PDF)
Problem Set 2 Files (ZIP) (This ZIP file contains: 4 .py files, 1 .index file, and .3 txt files)
Solutions to Problem Set 2 (PDF - 1.4MB) 31
3 Gibbs Sampler, RNA Secondary Structure, Protein Structure with PyRosetta, Connections
Problem Set 3 (PDF)
Problem Set 3 Files (ZIP) (This ZIP file contains: 3 .py files, 3 .fa files, and .1 txt file)
Solutions to Problem Set 3 (PDF - 1.5MB) 25
4 Bayesian Networks, Refining Protein Structures in PyRosetta, Mutual Information of Protein Residues
Problem Set 4 (PDF)
Problem Set 4 Files (ZIP) (This ZIP file contains: 3 .py files and 2 .fasta files)
Solutions to Problem Set 4 (PDF) 21
5 Network Statistics, Chromatin Structure, Heritability, Association Testing
Problem Set 5 (PDF - 1.1MB)
Problem Set 5 Files (ZIP)
Solutions to Problem Set 5 (PDF) 24
The total number of points available on the problem sets is 120; your score for the homework portion of the course is based on a maximum of 100 points. This means that you can miss one problem set (or a portion of one problem set) and still do fairly well on this component if you have done well on the other problem sets. For example, a student who obtained perfect marks on 4 of 5 problem sets, each valued at 24 points would get 96 points for the homework component of the course, almost as good as a student who completed all 5 problem sets, earning 90% of points on each, since 0.9 x 120 = 108, which would earn the maximum score of 100. Because of this, no make-up assignments will be offered. Of course, it is still to your advantage to do all five problem sets, as this will help you to learn the material in more depth, help prepare you for exams, etc. Please note that the point values of individual problem sets may vary somewhat from the 24 point average value, depending on their length and level of difficulty.
### Late Assignments
Assignments submitted within 24 hours of the time they were due will be eligible for 50% credit. If necessary, you may turn in your written portion and your programming portion separately. For example, if you turn in your written portion on time and your programming by the late due date, your written work will be eligible for full points but the programming will be eligible for only 50% of points. You may not further sub-divide your submissions. Because answer keys will be posted, no homework will be accepted after the extended deadline.
### Collaboration on Problem Sets
The goal of the problem sets is to reinforce the material and sometimes to explore a topic in greater depth. You may talk with other students about the problems and work on them together. However, you should write up your own solutions. Copying someone else’s solutions will not improve your understanding of the material and is not acceptable. Duplicate or nearly identical problem sets from different students will receive a score of zero. This has happened. We notice. Don’t let it happen to you!
You must write your own code on problem sets. You may discuss the programming problems with other students. The following two simple rules should make it clear what is not permitted:
1. Do not copy or reuse code from any source (except the sample code provided).
2. Do not share your code with anyone else in the class.
## Course Info
Spring 2014
##### Learning Resource Types
Lecture Videos
Lecture Notes
Projects
Programming Assignments with Examples
Presentation Assignments
Written Assignments
Instructor Insights
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