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Similar figures
# Similar figures: side lengths and angle measures
Two polygons are similar if these two facts both must be true:
• Corresponding angles are equal.
• The ratios of pairs of corresponding sides must be equal.
(in other words, if they are proportional).
The symbol for "is similar to" is
If these two shapes are similar, what is the measure of the missing length b?
The corresponding sides of similar shapes are proportional.
The trapezoids' longer sides have a ratio of , which is .
The trapezoids' shorter sides have a ratio of .
Write a proportion:
Find the cross products and solve:
42 = 14b 42 ÷ 14 = 14b ÷ 14 3 = b
The missing length is 3 kilometers.
If these two shapes are similar, what is the measure of the missing length b?
21 m b 5 m 15 m
b = m | HuggingFaceTB/finemath | |
1,068 Answered Questions for the topic Math Equations
Math Equations
09/02/18
#### solve for x and explain 5x+7=10
solve for x and explain 5x+7=10
Math Equations
08/28/18
#### What is the next fraction in this sequence? Simplify your answer. 9 16 , 1 2 , 7 16 , 3 8 , ...
9 16 , 1 2 , 7 16 , 3 8 , ...
Math Equations
08/27/18
#### What is a word problem for the equation 5(x-1)=30
A word problem that uses 5(x-1)=30
08/27/18
#### child admission is $5.70 and adult admission is$9.20 .160 tickets were sold for a total sales of $1171.00 . How many child tickets were sold that day? At the city museum, child admission is$5.70 and adult admission is $9.20. On Friday, 160 tickets were sold for a total sales of$1171.00. How many child tickets were sold that day?
Math Equations Math Math Help
08/25/18
#### Find the equation in standard form of the circle with points (2,3), (3,4), (-1,2)
I can't keep up with the solution. I need it for my quiz on tuesday.
Math Equations
08/24/18
#### What is Z*+2, when *=2, and Z=2
Zx+2, when x=2, and Z=2
08/23/18
#### I need help solving word problem
define variables, write an equation, and then use the equation in the solving process. A building has a ramp to its front doors to accommodate the handicapped. If the distance from the building... more
Math Equations
08/23/18
#### How would I write a power function through the two points (1,9/4),(2,9/32)
Im pretty sure I know how to get the power but I don’t know how to write the equation.
Math Equations Math Algebra
08/22/18
#### How do I solve this math question?
If x and y are rational numbers and (3+4 (sqrt(3)) (x+y (sqrt(3)) =26, find the sum of x and y.
Math Equations
08/20/18
#### how do you know if you suppose to subtract or add in a variable equations ?
algebra equations with two variables how do i know to add or subtract?
Math Equations
08/19/18
#### What is the answer for Z/3 =6.5
The z/3 is a fraction and it is i dont know the same as 6.5?
Math Equations Math Question
08/18/18
#### if 1 day = 1000 years, then 1 min= ___ days
1 day equals 1000 years then 1 min is equals to how many days
Math Equations
08/16/18
#### 2 more than 3 times a number x is 17 as an equation
what is 2 more than 3 times a number x is 17 as an equation and solved
08/15/18
#### joe sold 15 Tshirts for a total of 82.50 what is the unit price
joe sold 15 shirts for a total of 82.50 what is the unit price
Math Equations
08/15/18
#### Four more than three times a number is equal to 27
How to express,Four more than three times a number is equal to 27
Math Equations Math College Algebra
08/14/18
#### Math Equations Math Problem Math Help
define variables, write an equation, and then use the equation in the solving process.Let C(x)=500+40x be the cost to manufacture x items. Find the average cost per item to produce 70 items.
08/14/18
#### College Algebra Word Problem
write an equation, and then use the equation in the solving process.An economist predicts that the buying power B(x) of a dollar x years from now will be given by the formula B(x) = 0.99x. How... more
#### Math question
define variables, write an equation, and then use the equation in the solving process. A lot is in the shape of a right triangle. The shorter leg measures 180 meters. The hypotenuse is 60 meters... more
Math Equations College Algebra Math Problem
08/14/18
#### Math Equation/ math word problem
define variables, write an equation, and then use the equation in the solving process. The length of a rectangular frame is 3 cm more than the width. The area inside the frame is 70... more
Math Equations College Math
08/14/18
#### Math problem
define variables, write an equation, and then use the equation in the solving process.projectile is thrown upward so that its distance, in feet, above the ground after t seconds is h= -10t2... more
Math Equations
08/08/18
#### How many times can I double x before reaching y?
I have no clue on how to do this eg. say x is 3 and y is 500 how many times can i double 3 (3, 6, 12 etc.) without exceeding 500?
Math Equations
07/31/18
#### need help please and thank you asap
Steven owns a model airplane manufacturing business. Steven wants to maximize his revenue. He starts with the linear function x = -40p + 1600, where x is the predicted number of model airplanes he... more
Math Equations Math Math Help
07/30/18
#### Explain to Larry how to represent his first 15 payments in sigma notation. Then explain how to find the sum of his first 15 payments, using complete sentences.
Larry has taken out a loan for college. He started paying off the loan with a first payment of \$150. Each month he pays, he wants to pay back 1.3 times as the amount he paid the month before.... more
Math Equations Math Maths
07/27/18
#### Is there a shortcut formula to use whenever one of the values inside a formula changes?
I have this formula.true=base×(1+A)×(1+B−C)×(1−D)×(1+E−F)×(1−G)If the E increases its value, is there a way to know the adjusted true without having to calculate the whole formula again using the... more
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
#### OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | HuggingFaceTB/finemath | |
# Physics Kinematics Review Game!
```TEAM JEOPARDY!
What is the definition of “distance.”
What is the definition of “displacement.”
If James has to get from the cafeteria to the IT
building and runs the 20m there, only to find
out class is in the media center 30m away,
what is the distance James traveled.
If I turn in the TV in the living room, walk
10m to the kitchen to get a snack, and then
walk 8m back to the living room and crash on
the couch, what is my displacement?
What is the definition of speed?
What is the formula for speed?
What is the definition of velocity?
Is the formula for speed and velocity the
same thing?
If Sarah can run 10 m in 2 seconds, what is
her speed?
If a car can drive 55 miles in 11 minutes, how
fast is the car traveling?
What is the definition of acceleration?
What is the formula for acceleration?
If Mary is at the track meet and she speeds up
from 6 m/s to 10m/s in 2s to win the race,
what was her acceleration?
What does negative acceleration mean?
What directions are negative for vector
values?
If a car is slowing down AND going west, is
the acceleration positive or negative?
What is centripetal acceleration? When do we
use it?
What is the formula for centripetal
acceleration?
If a horse is running around a ring with a 5m
radius, and the horse is trotting at 5m/s,
what is it’s centripetal acceleration?
What is freefall?
calculating how fast the object was falling?
What is the freefall formula to figure out how
far the object fell?
Tom wants to play a trick on Jerry and drop a
bucket of water on his head. If the bucket of
water has to fall in one second to hit Jerry
coming through the door, how high does the
bucket have to be above the door?
A rockslide happened in California last week
and a boulder hit a Toyota. If the rock took 3
seconds to fall and hit the car, what was the
rock’s final velocity?
What are the three types of projectile
motions? (the fancy word is trajectory)
What are the two types of velocities we have
to think about for projectile motion?
What is vertical acceleration ALWAYS going to
be?
What is vertical velocity when the object is
falling?
What is vertical velocity at the object’s peak.
What is horizonal velocity?
``` | HuggingFaceTB/finemath | |
Volume Of Hemisphere
Chapter 13 Class 9 Surface Areas and Volumes
Concept wise
This video is only available for Teachoo black users
This video is only available for Teachoo black users
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### Transcript
Ex 13.8,8 A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, find the inside surface area of the dome. We know that Area whitewashed × Cost of white wash = Total cost Area whitewashed × 2 = Rs 498.96 Area whitewashed = 498.96/2 Area whitewashed = 249.48 m2 Now, area whitewashed is curved surface area of hemisphere (as only walls are whitewashed, not floor) Therefore, inner surface area of dome = 249.48 m2 Ex 13.8,8 (ii) volume of the air inside the dome. Volume of air inside dome = Volume of hemisphere = 2/3 πr3 Let the radius of dome = r m First we find radius using surface area Surface area of dome = 249.48 m2 2πr2 = 249.48 2 × 22/7 × r2 = 249.48 r2 = (249.48 × 7)/(2 × 22) r2 = 39.69 r = √39.69 ∴ r = 6.3 m Volume of the air inside the dome = volume of hemisphere = 2/3 πr3 = 2/(3 ) ×22/7 × 6.3 × 6.3 ×6.3 m3 = 523.908 m3 (approx.) | HuggingFaceTB/finemath | |
# Thread: Integration by parts problem?!?!
1. ## Integration by parts problem?!?!
I know how to do integration by parts. ∫ u dv= uv-∫ v du
as you may know for some functions like ∫ e^x cosx dx when you use integration by parts it repeats it's self. if you don't know what I mean, see the 3rd example on this page http://www.math.hmc.edu/calculus/tutoria...
what I need to know is a function that repeats when you use itegration by parts like ∫ e^x cosx dx
but instead of 2 steps it has 3.
if anyone knows of one let me know
2. The following example is very artificial and therefore not quite satisfying, but it answers you question anyway.
Consider $f(x)=e^{-\frac{x}{2}}\cos\left(\frac{\sqrt{3}}{2}x\right)$. You can check that $f'''(x)=f(x)$, so that you may compute $\int f(x)e^x\,dx$ using 3 integrations by parts, which will end up with the initial integral.
In fact, if you know complex numbers, $f(x)$ is the real part of $e^{jx}$ where $j=e^{\frac{2i\pi}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ is a cubic root of the unity: $j^3=1$, so what I claimed about $f$ is a simple consequence of this property.
Of course however, since $e^xf(x)=e^{x/2}\cos\left(\frac{\sqrt{3}}{2}x\right)$, two integrations by parts would suffice if you split $e^xf(x)$ into $u(x)=e^{x/2}$ and $v'(x)=\cos\left(\frac{\sqrt{3}}{2}x\right)$. I don't know of a genuine example where after 3 integrations by parts, and not possibly less, we recover the initial integral, in a way enabling to find its value.
3. Thanks, Laurent.
If anyone can find a specific example of, as Laurent put it, where after 3 integrations by parts, and not possibly less, we recover the initial integral, that'd be great. | HuggingFaceTB/finemath | |
# Question: What Is Absolute Value Function?
## What are absolute functions?
An absolute value function is a function that contains an algebraic expression within absolute value symbols. Recall that the absolute value of a number is its distance from 0 on the number line. The absolute value parent function, written as f(x)=| x |, is defined as. f(x)={x if x>00 if x=0−x if x<0.
## What is the equation of the absolute value function?
The general form of an absolute value function is f(x)=a|x-h|+k.
## What is an absolute value in math?
Absolute value describes the distance from zero that a number is on the number line, without considering direction. The absolute value of a number is never negative. Take a look at some examples. The absolute value of 5 is 5.
## What are absolute value functions used for?
The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems.
## How do you do absolute value?
Absolute value equations are equations where the variable is within an absolute value operator, like |x-5|=9. The challenge is that the absolute value of a number depends on the number’s sign: if it’s positive, it’s equal to the number: |9|=9. If the number is negative, then the absolute value is its opposite: |-9|=9.
## What is the absolute value of 8?
Absolute value is always nonnegative, since distance is always nonnegative. For example, the absolute value of 8 is 8, since 8 is 8 units from 0 on the number line. The absolute value of − 8 is also 8, since − 8 is also 8 units from 0 on the number line.
## What is the greatest integer function?
The Greatest Integer Function is also known as the Floor Function. It is written as f(x)=⌊x⌋. The value of ⌊x⌋ is the largest integer that is less than or equal to x.
## Can two numbers have the same absolute value?
Hence if domain is real number for each absolute value there are two different numbers one can have with same absolute value. However, if domain is Complex numbers, absolute value of a number a+bi is √a2+b2 and for each absolute value there could be infinite different numbers with same absolute value.
## What has an absolute value of 3?
Absolute Value means The absolute value of −9 is 9. The absolute value of 3 is 3.
## What has an absolute value of 4?
Absolute (denoted by the vertical bars) means that everything between them is converted to non-negative. So |− 4 |= 4 and so is | 4 |= 4.
You might be interested: How To Do Absolute Value In Excel?
## Who uses absolute value in real life?
A geophysicist uses absolute value to look at the total amount of energy used. In an energy wave, there are both negative and positive directions of movement. Another example is when scuba divers discuss their location in regards to sea level. “50 feet below sea level” doesn’t have to be represented as -50 feet.
## Where do we use absolute value in real life?
The absolute value is used in the real world to define the DIFFERENCE or change from one point to another. A good example I found was that if the everybody is going 55 mph and you are going 70 or 40 mph you will most likely get a ticket.
## What is an absolute value in real life?
The absolute value of a number can be thought of as the value of the number without regard to its sign. For example |3| = 3 and |-5| = 5. When plotted on a number line, it’s the distance from zero. We use the absolute value when subtracting a positive number and a negative number. | HuggingFaceTB/finemath | |
# Greater Than Lesssheets For First Grade 1st Math Comparing Tosheet Numbers
By Silke Klein at November 17 2018 06:35:00
What do you mean by that? How will you know if you've arrived? A better goal statement is "I am going to lose 10 pounds, be able to do 50 push-ups without a break, and run 3 miles in under 25 minutes by my next birthday." No wiggle room there! You will know if you've succeeded or failed. And, assuming the targets are also appropriately challenging or significant, you will have a strong goal statement. Why do I want to achieve this goal? What are the benefits I'm seeking. This could be a very long list. Referring back to the fitness goal, you may want to look better at the beach, beat a friend in a race, improve your heart health or any number of other possibilities. The purpose of this step is to identify your deepest motivations, get them on paper, and refer to them as you progress towards your goal.
Realize that children who are having difficulty with math dread math worksheets, which is reason why they procrastinate and do their homework at the last minute or after several reminds. Generating Questions - Students need to be aware of their own background schema and how it relates to the current reading selection. Students determine what they already know about the topic, what they need to know, then what they learned. By developing their own questions, students increase their active processing of text which results in increased comprehension.
## Gallery of Greater Than Less Than Worksheets For First Grade
In math bingo, each student is given a bingo card (also known as a "bingo worksheet" or "bingo board") printed with numbers. These aren't necessarily the standard bingo numbers, but rather are the answers to a number of different math problems. The game is then played exactly like a normal game of bingo, with the teacher playing the part of the bingo caller, but instead of the teacher calling out the numbers printed on the cards, the teacher instead calls out math problems (the teacher may also write the problem on the blackboard). The students' task is to solve each problem, and then look for the number on their bingo card. If you are looking for an article that describes the basics of Excel and introduces the interface and concepts for beginners, you have come to the right place. Microsoft Excel is a powerful business application that is organized into a structural hierarchy of Workbooks, Worksheets, and Cells.
Math and games if combined together can improve your child's learning habit. Games can make learning math fun, exciting and engaging. They can offer a lot of benefits not only to improve but also to make the learning not very stressful and tedious. Lost in their favorite gizmos, today's kids are devoid of the fun learning aspect offered by preschool worksheets. For generations, worksheets for kids have been used by educators to develop logical, lingual, analytical, and problem-solving capabilities. It is a proven fact that children learn quickly in their formative years than at any time in their life. | HuggingFaceTB/finemath | |
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Get in touch with us
# Measurement of Distances
Jun 7, 2023
Introduction:
An object is said to be in motion if its position changes with time. An object is said to be at rest if its position does not change with time. When an object is in motion for some time, it covers some distance. This distance can be measured in various ways using various units.
Explanation:
## Distance:
An object changes its position while in motion. This makes it cover some distance. The distance traveled by a body is the length of the path that it has traveled.
Suppose a person starts from the black location and reaches the red location by the road shown in the picture below. The person takes the road shown in yellow.
Here, the distance traveled by this person from the black location to the red location is the length of the path ABCDEF.
Thus, the distance covered = AB + BC + CD + DE + EF
### Measuring Lengths:
In ancient times people used various units of measurement of length which were easily accessible to everyone such as the handspan, the cubit, the arm span, the pace, and the foot span.
However, everyone’s body parts are of slightly different sizes. Therefore, in 1790, the French created a standard unit of measurement called the metric system. Thus, for the sake of uniformity, scientists all over the world have accepted a set of standard units of measurement, which is known as the International System of Units or SI units.
Units of measurement of distances:
The SI unit of length is a meter. A meter rule is a device that is used to measure the length of various objects. It is of length 1 meter (m). There are smaller and larger units of measurement of length other than meters.
The units of measurement of length which are smaller than 1 meter (m) are centimeters (cm) and millimeters (mm).
1 m = 100 cm
1 cm = 10 mm
1 m = 1000 mm
A meter rule has the cm and mm marked on it.
It has 1 m divided into 100 equal parts, each part of length 1 cm, and each 1 cm part is divided into 10 equal parts, each of length 1 mm.
The units of measurement of length which are greater than 1 meter (m) are kilometer (km) and mile (mi).
1 km = 1000 m
1 mi = 1.609 km
1 mi = 1609 m
Large distances such as the distance between two cities are measured in the units of km or mi.
However, smaller distances such as the length of a table or the dimensions of a room are measured in the units of meter.
1. A person travels from city A to city F on the road shown. Find the distance traveled by the person.
Total distance traveled
= AB + BC + CD + DE + EF
= 25 + 17 + 5 + 25 + 28
= 100 miles
Thus, the distance traveled by the person is 100 miles.
2. Which objects are in motion and at rest in the picture below? Fill in the blanks.
1. 7 km = —- m
2. 2 mi = —- m
3. 12 m = —– cm
4. 24 cm = —– mm
5. 2 m = —- mm
1. 7 km = 7 x 1000 = 7000 m
2. 2 mi = 2 x 1609 = 3218 m
3. 12 m = 12 x 100 = 1200 cm
4. 24 cm = 24 x 10 = 240 mm
5. 2 m = 2 x 1000 = 2000 mm
3. The length of a sofa is measured using a meter rule. The length of the sofa turned out to be 2 and a half-length of the meter rule. What is its length?
The length of a meter rule is 1 m.
The length of the sofa is 2 and a half of the meter rule.
Therefore, the length of the sofa
= 2.5 x 1 = 2.5 m
#### Different Types of Waves and Their Examples
Introduction: We can’t directly observe many waves like light waves and sound waves. The mechanical waves on a rope, waves on the surface of the water, and a slinky are visible to us. So, these mechanical waves can serve as a model to understand the wave phenomenon. Explanation: Types of Waves: Fig:1 Types of waves […]
#### Dispersion of Light and the Formation of Rainbow
Introduction: Visible Light: Visible light from the Sun comes to Earth as white light traveling through space in the form of waves. Visible light contains a mixture of wavelengths that the human eye can detect. Visible light has wavelengths between 0.7 and 0.4 millionths of a meter. The different colors you see are electromagnetic waves […]
#### Force: Balanced and Unbalanced Forces
Introduction: In a tug of war, the one applying more force wins the game. In this session, we will calculate this force that makes one team win and one team lose. We will learn about it in terms of balanced force and unbalanced force. Explanation: Force Force is an external effort that may move a […] | HuggingFaceTB/finemath | |
# Proof $\sqrt2$ is supremum of $x^2 <2$
1,159
The wording is a bit imprecise, but the reasoning is correct.
You start with: "Let $$u$$ be another upper bound for $$M$$.
Assumption: $$\sqrt{2}>u$$"
Right now I could set $$u=10$$ and the proof ends with "and so $$u$$ can't be another upper bound of $$M$$", which is obviously wrong.
So start the proof with "Let $$u$$ be another upper bound for $$M$$ that fulfills $$\sqrt{2}>u$$." and the rest is fine.
Share:
1,159
Author by
### franz3
Updated on November 16, 2020
• franz3 almost 3 years
Let $$M = \{x \in \mathbb{R}| x^2 <2\}$$. Prove that $$\sqrt2$$ is the supremum of M.
Proof:
$$\forall x \in M: x^2 <2 \Leftrightarrow |x| <\sqrt 2 \Leftrightarrow -\sqrt2 < x < \sqrt 2 \Rightarrow \sqrt 2$$ is an upper bound for $$M$$.
Hypothesis: $$\sup(M) = \sqrt 2$$.
Let $$u$$ be another upper bound for $$M$$.
Assumption: $$\sqrt2 >u$$.
Let $$\epsilon:= \sqrt2 - u >0$$. By the theorem of Eudoxos: $$\forall \epsilon > 0$$ $$\exists m \in \mathbb{N}: \dfrac{1}{m} < \epsilon$$, that means: $$\dfrac{1}{m} < \sqrt2 - u \Leftrightarrow u< \sqrt2 - \dfrac{1}{m}$$. Since $$m>0 \Rightarrow \dfrac{1}{m} >0 \Rightarrow \sqrt2 - \dfrac{1}{m} \in M$$ and so $$u$$ can't be another upper bound of $$M$$. Then $$\sqrt2 \leq u \Rightarrow \sqrt2 = \sup(M)$$.
Is this proof correct?
• TonyK almost 4 years
I think your instances of $2>\sqrt u$ should be $\sqrt 2>u$ (or $2>u^2$).
• Nurator almost 4 years
Oh yeah sorry, I copied incorrectly. Thank you! | HuggingFaceTB/finemath | |
# The Efficient Brewer
### Help Support Homebrew Talk - Beer, Wine, Mead, & Cider Brewing Discussion Forum:
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Why is my efficiency so low? Can someone explain my efficiency? I thought I understood efficiency. These statements are just some examples of the many threads relating to efficiency that I have seen on HomeBrewTalk. It would seem the forum is replete with individuals scratching their noodles in an effort to solve a conundrum their brewing data has presented. Fortunately, there are many helpful members who are ready, willing, and able to crunch the numbers, point out possible errors, and in many instances solve the riddle.
The purpose of this article is not to explore brewing-science's formulae and mathematics, but to illustrate some simple steps involved in the collection and interpretation of useful brewing data. Many homebrewers, myself included, use brewing software to crunch the numbers and calculate efficiencies. Through accurate data collection and analysis, brewers can more easily localize problems with their equipment or methods and find effective solutions. Lets look at two important numbers, mash efficiency and brewhouse efficiency, and explore what each of them means.
Mash Efficiency
This is a measure of how effective your methods are at extracting the starches from the grains, converting those starches to sugars, and getting the resultant sugars out of the mash-tun and into your boil-kettle. In other words, how effective you are at getting the available sugars into your kettle. Its no more complicated than that. To calculate mash efficiency we need to know three things.
• The estimated maximum potential gravity points in the grain-bill.
• The specific-gravity of the sweet-wort
• The pre-boil volume of sweet-wort in the boil-kettle
How Much Sugar Do You Get From Your Grain
Maximum Potential Gravity
Every grain has a variety of descriptive numbers associated with it. One of these is its potential gravity and is measured in points per pound per gallon (PPG). This is the theoretical specific gravity of a gallon of wort containing 100% of the potential sugars in one pound of the grain. A grain-bills potential gravity is the sum of the gravity points available from each grain-type. Unless you perform your own grain analysis, this number is dependent on the accuracy of two things; the malsters grain data and weight of the grain-bill. Accurately weighing each grain-type is therefore very important.
Specific Gravity
A representative sample of the sweet-wort is taken from the kettle, cooled to the calibration temperature of the measuring device, and its specific-gravity measured. Hydrometers or refractometers can be used. I favor narrow-range hydrometers for their ease of use. Both the sample and hydrometer are cooled prior to taking a reading. If only the sample is cooled, adding a room-temperature hydrometer will induce a temperature change and small measurement error.
Potential Gravity Measured In Points Per Pound Per Gallon
Volume
The volume of sweet-wort prior to the boil can be measured in a variety of ways. A calibrated dowel, ruler or sight-glass are popular methods. Another approach is to etch volume-markings directly onto the kettle. This is the method I chose. Owing to the high temperatures (~170F) of the sweet-wort, pre-boil measurements must be corrected to account for volumetric shrinkage to the hydrometers calibration point. This correction, approximately a 2.5% reduction in volume from pre-boil temperatures is not automatically made by the popular brewing software I use.
Why is it important to measure mash efficiency?
Multiple enzyme mediated chemical reactions occur in a mash. These reactions produce the fermentable sugars and other compounds responsible for giving a beer its signature characteristics. Knowing how good a job we did at mashing (getting the sugars into our boil-kettle), can highlight procedural problems before and during the mash. Factors negatively impacting the efficacy and efficiency of a mash include:
• Poorly crushed grain
• An overly thick or gelatinous mash consistency
• Excessively low or high mash temperatures
• Ineffective sparging methods
• Poor management of mash pH*
*Mash pH has greater impact on flavor but does effect efficiency, albeit to a much lesser extent.
After mashing and lautering, the wort is heated and boiling ensues with or without corrective measures. If mash efficiency is not in-line with the brewers objective, such corrections can include an addition of malt extract or an adjustment to the planned boil duration. This is a completely different, albeit related topic, beyond the scope of this article. When the boil is complete, the wort is cooled and transferred to the fermentor. Brewhouse efficiency can now be calculated.
Grain Crush, Mash Thickness And Temperature Contribute To Mash Efficiency
Brewhouse Efficiency
This is a measure of how effective your entire brewing process is. What portion of the total potential sugars made it into the fermentor. It can never be greater than your mash efficiency and is affected by volume losses during and after the mash. To calculate brewhouse efficiency we again, need to know three things.
• The estimated maximum potential gravity points in the grain-bill.
• The original-gravity of the wort (OG)
• The volume of wort in the fermentor
Grain Potential, Wort Gravity And Fermentor Volume Are Used To Calculate Brewhouse Efficiency
Maximum Potential Gravity
This is the same measurement as previously calculated. It is based on the malsters data and accurate weighings of the various grain-types
Original Gravity
After the boil a representative sample of the wort is taken from the kettle or the fermentor. The sample is prepared in the same manner as before, allowing accurate measurement of the original gravity.
Volume
The volume of wort transferred to the fermentor is easily measured using calibrated markings on the fermentor. I ferment my beer in glass carboys and added etched volume-markings to them. Accurate markings eliminate any guesswork.
Really Get To Know Your Volumes
Why is it important to measure brewhouse efficiency?
Monitoring and maintaining a high brewhouse efficiency is extremely important for commercial breweries as it significantly impacts their overhead costs and bottom line. Efficiency related cost savings are minimal on the homebrewing scale. I would argue however, that there are important, non-cost related benefits to developing more efficient brewing practices.
Knowing how good a job was done at getting the sugars into our fermentor enables the brewer to more accurately formulate recipes and water volume requirements at each stage of the brewing process. Assuming mash efficiency is known, the value can also highlight volume losses after the mash and areas where ones methods could be improved. Factors negatively impacting brewhouse efficiency include:
• Mash efficiency (brewhouse efficiency can never exceed this number)
• Spilled wort
• Dead-space in the kettle, chiller, and hoses
• Wort absorbed by hops
• Kettle-trub
With accurate data collection, and carefully directed refinement of the approach one takes to brewing, a greater understanding and control of its fundamental processes can be gleaned. Improved control of the home-brewery will inevitably lead to increased efficiency and a welcome additional byproduct; consistency. This is all-too-often overlooked when discussing efficiency numbers.
Take And Record Accurate Measurements
Unlike the astronauts in The Simpsons you dont need to be "a mathematician, a different kind of mathematician [or] a statistician", but you do need to take accurate measurements. Consistent and predictable efficiency better allows us to craft beers with intended characteristics, not ones dictated by inaccuracies inherent in our homebrewery. Isn't that a worthy goal?
Brewhouse Efficiency Calculations Aid In Brewing Consistency
The calculations involved in assessing brewing efficiency are not complex. Nonetheless, there are many beneficial reasons for using some form of brewing-software. There are lots of great options from which to choose. Many, like BrewersFriend.com are available for free.
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Fantastic article, especially for us newbs.
Great info, as always.
Wait....am I missing something? Why is the hydrometer photo indicating taking a reading at the top of the meniscus?? I always thought it was the bottom. Did I miss a memo somewhere??
@RunDadRun the photo appears to be done at the bottom of the meniscus.
Re the meniscus TOP not Base. (Read from the top* exceptions exist depending on the hydrometer)
This is only something I learned during the editing phase of this piece. Fortunately I was corrected by one of our more experienced and knowledgable members on HBT prior to publication.
https://www.homebrewtalk.com/wiki/index.php/Talk:Hydrometer
Well written article, Gavin. Thanks!
Nope...it should be the bottom of the meniscus. The image on that wiki page link even shows why: the meniscus has the upward curve on both the side of the tube and the hydrometer due to capillary action. If you're reading at the top, you're including this capillary action. True reading is at the bottom of the meniscus, same for reading any sort of calibrated vessel (e.g., graduated cylinder).
If it reads anything other than 1.000 at the bottom of the meniscus when measuring distilled water at hydrometer's calibration temperature, then your hydrometer is off.
First, Excellent article and well researched. Second, When training new lab technicians (been a few years since I've done that!!!), I've always taught the API method, which is to read the instrument based on whether the meniscus is concave or convex. Convex, top of meniscus. Concave, bottom of meniscus.
It depends on how your particular hydrometer is calibrated.
As I said. I have always read (incorrectly as I've now learned) from the bottom of the meniscus exactly in the manner y'all are describing.
After having the issue explained to me (vis a vis the meniscus and fluid inside/outside a measuring device such as a burette or hydrometer) and reading a Quote from Ray Daniels' book, my error became apparent. I made the appropriate changes to the images in the piece. It is not surprising that this point is debated here as a quick internet search will largely "prove" me wrong. On this point however I will defer to Ray Daniels and work from the top of the meniscus from now on.
[The difference between top and bottom with the narrow-range hydrometers I show is ~0.001 and ~0.0005 with a finishing hdrometer. A hydrometer with a larger range will be subject to a greater potential for measurment error]
I would hope this likely point of confusion and contention doesn't detract too much from the article. Thanks all for reading and adding your feedback. It is very much appreciated.
Great write up Gavin! So many good points made here on how to calculate.
If people are looking to increase (or - gasp - decrease) their efficiency, there are lots of good stickies in the forum, including my own.
Great article Gavin!
I'm not sure that volume adjustment need be done for actual preboil gravity readings. If the wort is consistent throughout, and you're cooling the sample, the sugar content will be appropriate no matter what, because they're all being read at the same temperature. It might affect your technical boiloff percentage, and can influence gravity points, but the difference is within the margin of error, and potentially overshot by things like insufficient mixing of preboil runnings, which can easily throw off a reading by more than 2.5%. Point is, I wouldn't bother with the correction.
Also, everything I have ever read except for Ray Daniels, including scientific texts, and the paperwork for my hydrometer, says to read at the BOTTOM of the meniscus. As hunter_la5 mentioned, it depends on your hydrometer. Some are indeed calibrated to read at the top. Most are calibrated to read at the bottom. Ray Daniels is wrong.
A few more points:
Re: Brewhouse Efficiency- Some things like loss to chiller are a known volume reduction and can be planned for all brews. Other things like hop loss are recipe dependent. You can calibrate brewhouse efficiency for the same beer over and over, or for similar "kinds" of beers from a trub/hop loss perspective (similar size/composition grain bills, similar weights of hops, etc), but the brewhouse efficiency between a Weizen and an Imperial IPA will be vastly different just because of hops alone, and this simply cannot be avoided. You have no choice but to anticipate it and plan for it.
Re: Mash efficiency- I think it's helpful to break that down into conversion efficiency and lautering efficiency. Mash pH, grain crush, water to grist ratio, mash temp/time, and such can have a big impact on conversion efficiency. Lautering efficiency is impacted by grain crush, but also by mash tun design. If you look to Kai Troester's work, you can get an idea of where your conversion efficiency falls, and then you can from there calculate your lautering efficiency. Can help you figure out exactly where your efficiency problems are.
@Qhrumphf
The volume correction is needed as I measure the gravity of the preboil wort at 60F. I also need to measure the volume at the same temperature.
If it is not done the mash efficiency number is inflated. I was seeing this discrepancy between predicted and actual mash efficiency till i started making these required adjustments. Once I started making the correction the mysteriously missing sugars from reboil to post-boil vanished and my data was making more sense.
The hydrometer meniscus is something I was just informed about. After doing some more reading on the topic on the AHA website's forum it seems that it is entirely dependent on the hydrometer. I was not aware there was any debate as I had always read from the meniscus' base.
I did not breakdown the mash efficiency further as those numbers are not something most folks will measure. I was trying to keep the piece as approachable as possible. The factors affecting mash efficiency have their effects on either conversion or lautering and are listed together. Knowing the mash efficiency number is low is enough to prompt more thorough investigations by the brewer.
My lautering is rather simpler than what Kai undoubtedly covers. (I BIAB) My lautering is solely impacted by how much wort the grain absorbs. (Typically 0.045 gallons/pound)
I agree that some factors affecting BH efficiency are fixed and some are variable. Hops absorption as a variable can be easily calculated and factored in in the planning stage. What I have done is to allow for 1 quart of dead-space and trub loss in the kettle-chiller combo. Depending on the hop bill this more or less of this volume will be absorbed by the hops. Either way the amount remains largely unchanged at 0.25 gallons. If no hops were used it would be all wort/break material. With more hops its less wort. The planned BH efficiency remains unchanged. Probably if i did a massive DIPA or something I would need to rework the planned losses at a new fixed amount. Hope I'm making some semblance of sense.
Thanks for reading and providing such in-depth feedback. Much appreciated
Nice job Gavin!
Maybe you should use some sort of software that does adjust for thermal expansion at mash temperatures...
@pricelessbrewing
I reckon I know of a great online tool that does just that and more.
Very good point.
You're not going to plug it so I will.
http://pricelessbrewing.github.io/BiabCalc/
@Gavin C I just read the 'talk:Hydrometer" link you posted, and I still have to disagree. That person is making the basis of their decision that it should be read at the top because their hydrometer appears to be miscalibrated. I just checked 4 hydrometers I own with distilled water, and 3 of them read 1.000 at the BOTTOM of the meniscus. The other is off by 2 points (and it was sold to me with that knowledge), and sure enough it reads 1.002 at the BOTTOM of the meniscus as well.
Occam's razor - which is more likely, that one guy (Ray Daniels) in one book was wrong, or every scientist in the world is wrong?
P.S. If you're calibrating your hydrometer with tap water, that will account for why you are getting 1.000 at the top of the meniscus instead of the bottom. There are trace salts in tap water that will cause your reading to be off by ~1 point. So your tap water is actually reading 1.001 at the bottom, which is indeed correct and thus you should be reading from the bottom for all samples.
@MagicMatt
Yea, I'm beginning to think I was too hasty in making the changes to the images prior to the article being published. When I did my calibrations (distilled at 60F) I read from the bottom at 1.000 for all 3 of my hydrometers that have a 1.000.
I assumed I had made a trifecta of errors. It was too late in the day so I just made the changes anyway. I will need to recheck my devices. Like I said, I have always read from the menisus' base. The image is at odds with my past practices. thanks for reading .
Great article, thank you for the work you put in!
As far as the meniscus, in college I was taught to read at the bottom. However, I believe that being CONSISTENT is the key. Those that prefer to read from the top, keep at it and don't change! Those that prefer to read from the bottom, keep at it and don't change! In the end it will be beer.
I believe that the scientific community has "rules" or "best practices" for taking measurements for consistency, not because it is the ONLY or RIGHT way. It just creates repeatable consistency!
RDWHAHB
Where has this post been all my brewing life? Thank you for putting this up.
Great write up, Gavin. We never stop learning, especially with so many knowledgeable people on HBT dedicated to advancing our craft.
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# Probability that 2 of t strings of length n are equal
Given t bit-strings of length n that are generated randomly. What is the probability that at least 2 of these strings are equal. I've seen someone who wrote that the probability is $\le \frac{t^2}{2^n}$
The reasoning is that you have $t$ options to choose the first string. The $t$ options to choose the second string, and given a string chosen at the first stage, the probability that the other chosen string is equal to the first one is $\frac{1}{2^n}$. The you sum all these options of choosing 2 equal string ($t^2$) and you get $\frac{t^2}{2^n}$ This just seems wrong because for each such choice of 2 strings there are many other options to the other strings, but I can't find a proof to show that it is wrong.
Does this reasoning make sense?
We can actually be more precise and say that the probability is at most $\frac{\binom t2}{2^n}$.
For $1 \le i < j \le n$, let $A_{ij}$ be the event that the $i^{\text{th}}$ string is equal to the $j^{\text{th}}$ string. Then $\Pr[A_{ij}] = \frac{1}{2^n}$, because the $j^{\text{th}}$ string is equally likely to be any of $2^n$ possibilities, one of which is equal to the $i^{\text{th}}$ string.
We have two equal strings if any of the $\binom t2$ events $$A_{12}, A_{13}, A_{14}, \dots, A_{t-1,t}$$ occur. If these events were all disjoint - no two could happen at the same time - the probability that any of them occur would be exactly $\binom t2 \cdot \frac1{2^n}$. They're not all disjoint; if the $1^{\text{st}}$ string is equal to the $2^{\text{nd}}$, and the $3^{\text{rd}}$ string is equal to the $4^{\text{th}}$, then $A_{12}$ and $A_{34}$ both occur. So this is not the actual probability we want.
However, it's still an upper bound. For any two events $A$ and $B$, we have \begin{align} \Pr[A \text{ or } B] &= \Pr[A] + \Pr[B] - \Pr[A \text{ and }B] \\ &\le \Pr[A] + \Pr[B] \end{align} and by induction, we can show that for any $m$ events $A_1, A_2, \dots, A_m$, we have $$\Pr[A_1 \text{ or } A_2 \text{ or } \dots \text{ or } A_m] \le \Pr[A_1] + \Pr[A_2] + \dots + \Pr[A_m].$$ This is known as the union bound. In our case, it says that \begin{align} \Pr[A_{12} \text{ or } A_{13} \text{ or } \dots \text{ or } A_{t-1,t}] &\le \Pr[A_{12}] + \Pr[A_{13}] + \dots + \Pr[A_{t-1,t}] \\ &= \binom t2 \frac1{2^n}. \end{align}
We can find the exact probability by first finding the probability that all $t$ strings are different. This is $$\frac{(2^n-1)(2^n-2) \cdots (2^n-t+1)}{(2^n)^{t-1}}.$$ (If all strings must be different, there are $2^n-1$ possibilities for the second string, $2^n-2$ for the third, and so on.) So the exact answer is $1 - \frac{(2^n-1)(2^n-2) \cdots (2^n-t+1)}{(2^n)^{t-1}}$.
But the union bound is often useful when the exact answer cannot be found: for example, if the dependence between the events is very hard to describe. Even here, the bound $\binom t2 \frac1{2^n}$ may be more useful than an exact answer: it makes it much easier to see that if $t$ is much smaller than $2^{n/2}$, then $\binom t2$ is much smaller than $2^n$, and so the probability that any two strings are equal is very small.
• I only have concern about the order - since the denominator, $\binom{2^n -1 +t }{t}$ does not consider the order of selected strings, and I'm not sure how to account for it. Though the problem seems to imply that order does not matter.
– Alex
May 16 '17 at 19:52
• @Alex - I've edited my answer to have the correct exact answer. Indeed, $\binom{2^n-1+t}{t}$ cannot possibly be the correct denominator, since the total number of possibilities for the $t$ strings is $2^{nt}$, so the correct denominator must be a divisor of $2^{nt}$: a power of $2$. May 16 '17 at 21:40
• The question says 'generated randomly', so it's not quite clear if the order matters; my answer holds if the order doesn't matter. Could you explain how you got $2^{n(t-1)}$?
– Alex
May 18 '17 at 5:31
• For each of the $t-1$ remaining strings after the first, we get a factor of $2^t$ in the denominator. I'm assuming "generated randomly" to mean "each string is uniformly and independently chosen from all $2^n$ strings" and don't see what other distribution we could reasonably take. May 18 '17 at 5:46
• Could you have a look at my edit pls
– Alex
May 19 '17 at 15:47
This reminds of the Birthday Problem!
In a room with $n$ students, what is the probability that at least two students share the same birthday? The probability that two students share a birthday is $\frac{1}{365}$, but you have to compare all pairs of n students.
To make an analogy, the probability that two strings are equal is $\frac{1}{2^n}$, but you have to compare all pairs of t strings.
I think this probability is $1-\frac{\binom{2^n}{t}}{\binom{2^n -1 +t}{2^n -1}}$. Essentially this means 1- probability to have all strings different. The numerator is all different strings. The denominator is all cases of selecting t out of $2^n$ with repetitions.
EDIT:
OK @MishaLavrov's solution $\frac{\binom{2^n}{t} t!}{2^{nt}}$ is better. What my solution gives is $\frac{\text{number of ways to get$t$unique strings length$n$from a set of$2^n$strings, order does not matter}}{\text{number of ways to get$t$strings length$n$from a set of$2^n$strings, order does not matter}}$.
For example, if $t=3$, one possible denominator would be $aba, aab, baa$ counted as one. In the numerator we can have $abc,bac,...,cba$ all counted as one outcome.
My confusion is that, even if strings are 'generated randomly', do we account for repeated sets or not? | HuggingFaceTB/finemath | |
## Unipotent Hecke algebras: the structure, representation theory, and combinatorics
Last updated: 26 March 2015
## Preliminaries
### Hecke algebras
This section gives some of the main representation theoretic results used in this thesis, and defines Hecke algebras. Three roughly equivalent structures are commonly used to describe the representation theory of an algebra $A\text{:}$ (1) Modules. Modules are vector spaces on which $A$ acts by linear transformations. (2) Representations. Every choice of basis in an $n\text{-dimensional}$ vector space on which $A$ acts gives a representation, or an algebra homomorphism from $A$ to the algebra of $n×n$ matrices. (3) Characters. The trace of a matrix is the sum of its diagonal entries. A character is the composition of the algebra homomorphism of (2) with the trace map. The following discussion will focus on modules and characters.
#### Modules and characters
Let $A$ be a finite dimensional algebra over the complex numbers $ℂ\text{.}$ An $A\text{-module}$ $V$ is a finite dimensional vector space over $ℂ$ with a map $A×V ⟶ V (a,v) ⟼ av$ such that $(c1)v = cv, c∈ℂ, (ab)v = a(bv), a,b∈A,v∈V, (a+b) (c1v1+c2v2) = c1av1+ c2av2+ c1bv1+ c2bv2, c1,c2∈ℂ.$ An $A\text{-module}$ $V$ is irreducible if it contains no nontrivial, proper subspace $V\prime$ such that $av\prime \in V\prime$ for all $a\in A$ and $v\prime \in V\prime \text{.}$
An $A\text{-module}$ homomorphism $\theta :V\to V\prime$ is a $ℂ\text{-linear}$ transformation such that $aθ(v)=θ(av), for a∈A,v∈V,$ and an $A\text{-module}$ isomorphism is a bijective $A\text{-module}$ homomorphism. Write $HomA(V,V′) = {A-module homomorphisms V→V′} EndA(V) = HomA(V,V).$
Let $\stackrel{ˆ}{A}$ be an indexing set for the irreducible modules of $A$ (up to isomorphism), and fix a set ${Aγ}γ∈Aˆ$ of isomorphism class representatives. An algebra $A$ is semisimple if every $A\text{-module}$ $V$ decomposes $V≅⨁γ∈Aˆ mγ(V)Aγ, wheremγ(V)∈ ℤ≥0,mγ (V)Aγ= Aγ⊕⋯⊕Aγ⏟mγ(V) terms.$ In this thesis, all algebras (though not necessarily all Lie algebras) are semisimple.
(Schur's Lemma). Suppose $\gamma ,\mu \in \stackrel{ˆ}{A}$ with corresponding irreducible modules ${A}^{\gamma }$ and ${A}^{\mu }\text{.}$ Then $dim(HomA(Aγ,Aμ))= δγμ= { 1, if γ=μ, 0, otherwise.$
Suppose $A\subseteq B$ is a subalgebra of $B$ and let $V$ be an $A\text{-module}\text{.}$ Then $B{\otimes }_{A}V$ is the vector space over $ℂ$ presented by generators $\left\{b\otimes v | b\in B,v\in V\right\}$ with relations $ba⊗v = b⊗av, a∈A,b∈B,v∈V, (b1+b2)⊗ (v1+v2) = b1⊗v1+ b1⊗v2+ b2⊗v1+ b2⊗v2, b1,b2∈B, v1,v2∈V.$ Note that the map $B×(B⊗AV) ⟶ B⊗AV (b′,b⊗v) ⟼ (b′b)⊗v$ makes $B{\otimes }_{A}V$ a $B\text{-module.}$ Define induction and restriction (respectively) as the maps $IndAB: {A-modules} ⟶ {B-modules} V ⟼ B⊗AV ResAB: {B-modules} ⟶ {A-modules} V ⟼ V$
Suppose $V$ is an $A\text{-module.}$ Every choice of basis $\left\{{v}_{1},{v}_{2},\dots ,{v}_{n}\right\}$ of $V$ gives rise to an algebra homomorphism $ρ:A→Mn(ℂ), where Mn(ℂ)= {n×n matrices with entries in ℂ}.$ The character ${\chi }_{V}:A\to ℂ$ associated to $V$ is $χV(a)= tr(ρ(a)).$ The character ${\chi }_{V}$ is independent of the choice of basis, and if $V\cong V\prime$ are isomorphic $A\text{-modules,}$ then ${\chi }_{V}={\chi }_{V\prime }\text{.}$ A character ${\chi }_{V}$ is irreducible if $V$ is irreducible. The degree of a character ${\chi }_{V}$ is ${\chi }_{V}\left(1\right)=\text{dim}\left(V\right),$ and if $\chi \left(1\right)=1,$ then a character is linear; note that linear characters are both irreducible characters and algebra homomorphisms.
Define $R[A]=ℂ-span {χγ | γ∈Aˆ} ,where χγ= χAγ,$ with a $ℂ\text{-bilinear}$ inner product given by $⟨χγ,χμ⟩ =δγμ. (2.1)$ Note that the semisimplicity of $A$ implies that any character is contained in the subspace $ℤ≥0-span {χγ | γ∈Aˆ}.$ If ${\chi }_{V}$ is a character of a subalgebra $A$ of $B,$ then let $IndAB(χV)= χB⊗AV$ and if ${\chi }_{V\prime }$ is a character of $B,$ then ${\text{Res}}_{A}^{B}\left({\chi }_{V\prime }\right)$ is the restriction of the map ${\chi }_{V\prime }$ to $A\text{.}$
(Frobenius Reciprocity). Let $A$ be a subalgebra of $B\text{.}$ Suppose $\chi$ is a character of $A$ and $\chi \prime$ is a character of $B\text{.}$ Then $⟨IndAB(χ),χ′⟩= ⟨χ,ResAB(χ′)⟩.$
#### Weight space decompositions
Suppose $A$ is a commutative algebra. Then $dim(Aγ)=1, for all γ∈Aˆ.$ The corresponding character ${\chi }^{\gamma }:A\to ℂ$ is an algebra homomorphism, and we may identify the label $\gamma$ with the character ${\chi }^{\gamma },$ so that $av=χγ(a)v= γ(a)v,for a∈A,v∈Aγ.$
If $A$ is a commutative subalgebra of $B$ and $V$ is an $B\text{-module,}$ then as a $A\text{-module}$ $V=⨁γ∈Aˆ Vγ,where Vγ ={v∈V | av=γ(a)v,a∈A}.$ The subspace ${V}_{\gamma }$ is the $\gamma \text{-weight}$ space of $V,$ and if ${V}_{\gamma }\ne 0$ then $V$ has a weight $\gamma \text{.}$
For large subalgebras $A,$ such a decomposition can help construct the module $V\text{.}$ For example, 1. If $0\ne {v}_{\gamma }\in {V}_{\gamma },$ then ${\left\{{v}_{\gamma }\right\}}_{{V}_{\gamma }\ne 0}$ is a linearly independent set of vectors. In particular, if $\text{dim}\left({V}_{\gamma }\right)=1$ for all ${V}_{\gamma }\ne 0,$ then $\left\{{v}_{\gamma }\right\}$ is a basis for $V\text{.}$ 2. If $V$ is irreducible, then $B{V}_{\gamma }=V\text{.}$ Chapter 6 uses this idea to construct irreducible modules of the Yokonuma algebra.
#### Characters in a group algebra
Let $G$ be a finite group. The group algebra $ℂG$ is the algebra $ℂG=ℂ-span{g∈G}$ with basis $G$ and multiplication determined by the group multiplication in $G\text{.}$ A $G\text{-module}$ $V$ is a vector space over $ℂ$ with a map $G×V ⟶ V (g,v) ⟼ gv$ such that $(gg′)v=g(g′v), g(cv+c′v′)=cgv+c′gv′, for c,c′∈ℂ,g,g′∈G,v,v′∈V.$ Note that there is a natural bijection ${ℂG-modules} ⟶ {G-modules} V ⟼ V ,$ so I will use $G\text{-modules}$ and $ℂG\text{-modules}$ interchangeably. Let $\stackrel{ˆ}{G}$ index the irreducible $ℂG\text{-modules.}$
The inner product (2.1) on $R\left[G\right]$ has an explicit expression $⟨·,·⟩: R[G]×R[G] ⟶ ℂ (χ,χ′) ⟼ 1∣G∣∑g∈Gχ(g)χ′(g-1),$ and if $U$ is a subgroup of $G,$ then induction is given by $IndUG(χ): G ⟶ ℂ g ⟼ 1∣U∣∑x∈Gxgx-1∈Uχ(xgx-1).$
#### Hecke algebras
An idempotent $e\in ℂG$ is a nonzero element that satisfies ${e}^{2}=e\text{.}$ For $\gamma \in \stackrel{ˆ}{G},$ let $eγ= χλ(1)∣G∣ ∑g∈Gχγ (g-1)g∈ ℂG.$ These are the minimal central idempotents of $ℂG,$ and they satisfy $1=∑γ∈Gˆ eγ,eγ2= eγ,eγ eμ=δγμ eγ,ℂGeγ ≅dim(Gγ)Gγ.$ In particular, as a $G\text{-module,}$ $ℂG≅⨁γ∈Gˆ ℂGeγ.$
Let $U$ be a subgroup of $G$ with an irreducible $U\text{-module}$ $M\text{.}$ The Hecke algebra $ℋ\left(G,U,M\right)$ is the centralizer algebra $ℋ(G,U,M)= EndℂG (IndUG(M)).$ If ${e}_{M}\in ℂU$ is an idempotent such that $M\cong ℂU{e}_{M},$ then $IndUG(M)= ℂG⊗ℂU ℂUeM=ℂG ⊗ℂUeM ≅ℂGeM$ and the map $θM: eMℂGeM ⟶∼ ℋ(G,U,M) eMgeM ⟼ ϕg where ϕg: ℂGeM ⟶ ℂGeM keM ⟼ keMgeM, (2.2)$ is an algebra anti-isomorphism (i.e. ${\theta }_{M}\left(ab\right)={\theta }_{M}\left(b\right){\theta }_{M}\left(a\right)\text{).}$
The following theorem connects the representation theory of $G$ to the representation theory of $ℋ\left(G,U,M\right)\text{.}$ Theorem 2.3 and the Corollary are in [CRe1981, Theorem 11.25].
(Double Centralizer Theorem). Let $V$ be a $G\text{-module}$ and let $ℋ={\text{End}}_{ℂG}\left(V\right)\text{.}$ Then, as a $G\text{-module,}$ $V≅⨁γ∈Gˆ dim(ℋγ)Gγ$ if and only if, as an $ℋ\text{-module,}$ $V≅⨁γ∈Gˆ dim(Gγ)ℋγ$ where $\left\{{ℋ}^{\gamma } | \gamma \in \stackrel{ˆ}{G},{ℋ}^{\gamma }\ne 0\right\}$ is the set of irreducible $ℋ\text{-modules.}$
Suppose $U$ is a subgroup of $G\text{.}$ Let $M\cong ℂU{e}_{M}$ be an irreducible $U\text{-module.}$ Write $ℋ=ℋ\left(G,U,M\right)\text{.}$ Then the map ${G-submodules of ℂGeM} ⟶ {ℋ-modules} V ⟼ eMV$ is a bijection that sends ${G}^{\gamma }↦{ℋ}^{\gamma }$ for every irreducible ${G}^{\gamma }$ that is isomorphic to a submodule of ${\text{Ind}}_{U}^{G}\left(M\right)$ $\text{(}{G}^{\gamma }↪{\text{Ind}}_{U}^{G}\left(M\right)\text{).}$
### Chevalley Groups
There are two common approaches used to define finite Chevalley groups. One strategy considers the subgroup of elements in an algebraic group fixed under a Frobenius map (see [Car1985]). The approach here, however, begins with a pair $\left(𝔤,V\right)$ of a Lie algebra $𝔤$ and a $𝔤\text{-module}$ $V,$ and constructs the Chevalley group from a variant of the exponential map. This perspective gives an explicit construction of the elements, which will prove useful in the computations that follow (see also [Ste1967]).
#### Lie algebra set-up
A finite dimensional Lie algebra $𝔤$ is a finite dimensional vector space over $ℂ$ with a $ℂ\text{-bilinear}$ map $[·,·]: 𝔤×𝔤 ⟶ 𝔤 (X,Y) ⟼ [X,Y]$ such that (a) $\left[X,X\right]=0,$ for all $X\in 𝔤,$ (b) $\left[X,\left[Y,Z\right]\right]+\left[Y,\left[Z,X\right]\right]+\left[Z,\left[X,Y\right]\right]=0,$ for $X,Y,Z\in 𝔤\text{.}$ Note that if $A$ is any algebra, then the bracket $\left[a,b\right]=ab-ba$ for $a,b\in A$ gives a Lie algebra structure to $A\text{.}$
A $𝔤\text{-module}$ is a vector space $V$ with a map $𝔤×V ⟶ V (X,v) ⟼ Xv$ such that $(aX+bY)v = aXv+bYv, for all a,b∈ℂ, X,Y∈𝔤,v∈V, X(av1+bv2) = aXv1+ bXv2, for all a,b∈ℂ, X∈𝔤,v1,v2∈V, [X,Y]v = X(Yv)- Y(Xv), for all X,Y∈𝔤,v∈V.$ A Lie algebra $𝔤$ acts diagonally on $V$ if there exists a basis $\left\{{v}_{1},\dots ,{v}_{r}\right\}$ of $V$ such that $Xvi∈ℂvi, for all X∈𝔤 and i=1,2,…,r.$
The center of $𝔤$ is $Z(𝔤)= {X∈𝔤 | [X,Y]=0 for all Y∈𝔤}.$ Let $\gamma \in \stackrel{ˆ}{𝔤}$ index the irreducible $𝔤\text{-modules}$ ${𝔤}^{\gamma }\text{.}$ A Lie algebra $𝔤$ is reductive if for every finite-dimensional module $V$ on which $Z\left(𝔤\right)$ acts diagonally, $V=⨁γ∈𝔤ˆ mγ(V)𝔤γ, where mγ(V)∈ ℤ≥0.$ If $𝔤$ is reductive, then $𝔤=Z(𝔤)⊕𝔤s where 𝔤s=[𝔤,𝔤] is semisimple.$
Let $𝔤$ be a reductive Lie algebra. A Cartan subalgebra $𝔥$ of $𝔤$ is a subalgebra satisfying (a) ${𝔥}^{k}=\left[{𝔥}^{k-1},𝔥\right]=0$ for some $k\ge 1$ $\text{(}𝔥$ is nilpotent), (b) $𝔥=\left\{X\in 𝔤 | \left[X,H\right]\in 𝔥,H\in 𝔥\right\}$ $\text{(}𝔥$ is its own normalizer). If ${𝔥}_{s}$ is a Cartan subalgebra of ${𝔤}_{s},$ then $𝔥=Z\left(𝔤\right)\oplus {𝔥}_{s}$ is a Cartan subalgebra of $𝔤\text{.}$ Let $𝔥*=Homℂ (𝔥,ℂ)and 𝔥s*= Homℂ(𝔥s,ℂ).$
As an ${𝔥}_{s}\text{-module,}$ ${𝔤}_{s}$ decomposes $𝔤s≅𝔥s⊕ ⨁0≠α∈𝔥s* (𝔤s)α,where (𝔤s)α= ⟨X∈𝔤s | [H,X]=α(H)X,H∈𝔥s⟩.$ The set of roots of ${𝔤}_{s}$ is $R=\left\{\alpha \in {𝔥}^{*} | \alpha \ne 0,{\left({𝔤}_{s}\right)}_{\alpha }\ne 0\right\}\text{.}$ A set of simple roots is a subset $\left\{{\alpha }_{1},{\alpha }_{2},\dots ,{\alpha }_{\ell }\right\}\subseteq R$ of the roots such that (a) $ℚ{\otimes }_{ℂ}{𝔥}_{s}^{*}=ℚ\text{-span}\left\{{\alpha }_{1},{\alpha }_{2},\dots ,{\alpha }_{\ell }\right\}$ with $\left\{{\alpha }_{1},\dots ,{\alpha }_{\ell }\right\}$ linearly independent, (b) Every root $\beta \in R$ can be written as $\beta ={c}_{1}{\alpha }_{1}+{c}_{2}{\alpha }_{2}+\cdots +{c}_{\ell }{\alpha }_{\ell }$ with either $\left\{{c}_{1},\dots ,{c}_{\ell }\right\}\subseteq {ℤ}_{\ge 0}$ or $\left\{{c}_{1},\dots ,{c}_{\ell }\right\}\subseteq {ℤ}_{\le 0}\text{.}$ Every choice of simple roots splits the set of roots $R$ into positive roots $R+= {β∈R | β=c1α1+c2α2+⋯+cℓαℓ,ci∈ℤ≥0}$ and negative roots $R-= {β∈R | β=c1α1+c2α2+⋯+cℓαℓ,ci∈ℤ≤0} .$ In fact, ${R}^{-}=-{R}^{+}\text{.}$
Example. Consider the Lie algebra ${𝔤𝔩}_{2}={𝔤𝔩}_{2}\left(ℂ\right)=\left\{2×2 \text{matrices with entries in} ℂ\right\},$ and bracket $\left[·,·\right]$ given by $\left[X,Y\right]=XY-YX\text{.}$ Write $𝔤𝔩2=Z(𝔤𝔩2) ⊕𝔰𝔩2,$ where $Z\left({𝔤𝔩}_{2}\right)=\left\{\left(\begin{array}{cc}c& 0\\ 0& c\end{array}\right) | c\in ℂ\right\}$ and ${𝔰𝔩}_{2}=\left\{X\in {𝔤𝔩}_{2} | \text{tr}\left(X\right)=0\right\}\text{.}$ Also, $𝔰𝔩2= {(a00-a) | a∈ℂ}⊕ {(00c0) | c∈ℂ}⊕ {(0c00) | c∈ℂ}.$ Let $\alpha \in {𝔥}^{*}={\left\{\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right) | a,b\in ℂ\right\}}^{*}$ be the simple root given by $\alpha \left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)=a-b,$ so that ${R}^{+}=\left\{\alpha \right\}$ and ${R}^{-}=\left\{-\alpha \right\}\text{.}$
#### From a Chevalley basis of ${𝔤}_{s}$ to a Chevalley group
For every pair of roots $\alpha ,-\alpha ,$ there exists a Lie algebra isomorphism ${\varphi }_{a}:{𝔰𝔩}_{2}\to ⟨{𝔤}_{\alpha },{𝔤}_{-\alpha }⟩\text{.}$ Under this isomorphism, let $Xα=ϕa (0100) ∈(𝔤s)α, Hα=ϕα (100-1) ∈𝔥s, X-α=ϕα (0010) ∈(𝔤s)-α.$ Note that $\left[{X}_{\alpha },{X}_{-\alpha }\right]={H}_{\alpha }\text{.}$ The following classical result can be found in [Hum1972, Theorem 25.2].
(Chevalley Basis). There is a choice of the ${\varphi }_{\alpha }$ such that the set $\left\{{X}_{\alpha },{H}_{{\alpha }_{i}} | \alpha \in R,1\le i\le \ell \right\}$ is a basis of ${𝔤}_{s}$ satisfying (a) ${H}_{\alpha }$ is a $ℤ\text{-linear}$ combination of ${H}_{{\alpha }_{1}},{H}_{{\alpha }_{2}},\dots ,{H}_{{\alpha }_{\ell }}\text{.}$ (b) Let $\alpha ,\beta \in R$ such that $\beta \ne ±\alpha ,$ and suppose $l,r\in {ℤ}_{\ge 0}$ are maximal such ${β-lα,…,β-α,β,β+α,…,β+rα} ⊆R.$ Then $[Xα,Xβ]= { ±(l+1) Xα+β, if r≥1, 0 otherwise.$
A basis as in Theorem 2.5 is a Chevalley basis of ${𝔤}_{s}$ (For an analysis of the choices involved see [Sam1969] and [Tit1966]).
Let $V$ be a finite dimensional $𝔤\text{-module}$ such that $V$ has a $ℂ\text{-basis}$ $\left\{{v}_{1},{v}_{2},\dots ,{v}_{r}\right\}$ that satisfies (a) There exists a $ℂ\text{-basis}$ $\left\{{H}_{1},\dots ,{H}_{n}\right\}$ of $𝔥$ such that (1) ${H}_{{\alpha }_{i}}\in {ℤ}_{\ge 0}\text{-span}\left\{{H}_{1},\dots ,{H}_{n}\right\},$ (2) ${H}_{i}{v}_{j}\in ℤ{v}_{j}$ for all $i=1,2,\dots ,n$ and $j=1,2,\dots ,r\text{.}$ (3) ${\text{dim}}_{ℤ}\left(ℤ\text{-span}\left\{{H}_{1},{H}_{2},\dots ,{H}_{n}\right\}\right)\le {\text{dim}}_{ℂ}\left(𝔥\right)\text{.}$ (b) $\frac{{X}_{\alpha }^{n}}{n!}{v}_{i}\in ℤ\text{-span}\left\{{v}_{1},{v}_{2},\dots ,{v}_{r}\right\}$ for $\alpha \in R,$ $n\in {ℤ}_{\ge 0}$ and $i=1,2,\dots ,r\text{.}$ (c) ${\text{dim}}_{ℤ}\left(ℤ\text{-span}\left\{{v}_{1},{v}_{2},\dots ,{v}_{r}\right\}\right)\le {\text{dim}}_{ℂ}\left(V\right)\text{.}$ (Condition (a) guarantees that $Z\left(𝔤\right)$ acts diagonally. If $Z\left(𝔤\right)=0,$ then the existence of such a basis is guaranteed by a theorem of Kostant [Hum1972, Theorem 27.1]).
Let $V$ be a finite dimensional $𝔤\text{-module}$ that satisfies (a)-(c) above. For $H\in 𝔥,$ let $(Hn)= (H-1)(H-2)⋯(H-(n+1))n! ∈End(V).$ As a transformation of the basis ${v}_{1},{v}_{2},\dots ,{v}_{r},$ the element ${H}_{i}=\text{diag}\left({h}_{1},{h}_{2},\dots ,{h}_{r}\right)\in \text{End}\left(V\right)$ with ${h}_{j}\in ℤ\text{.}$ Let $y\in {ℂ}^{*}\text{.}$ Temporarily abandon precision (i.e. allow infinite sums) to obtain $∑n≥0 (y-1)n (Hin) = diag ( ∑n≥0(y-1)n (h1n), ∑n≥0(y-1)n (h2n),…, ∑n≥0(y-1)n (hrn), ) = diag(yh1,yh2,…,yhr) ∈End(V)(by the binomial theorem).$ This computation motivates some of the definitions below.
Let $𝔥ℤ=ℤ-span {H1,H2,…,Hn}. (2.3)$ The finite field ${𝔽}_{q}$ with $q$ elements has a multiplicative group ${𝔽}_{q}^{*}$ and an additive group ${𝔽}_{q}^{+}\text{.}$ Let $Vq=𝔽q-span {v1,v2,…,vr}. (2.4)$
The finite reductive Chevalley group ${G}_{V}\subseteq GL\left({V}_{q}\right)$ is $GV= ⟨ xα(a),hH(b) | α∈R,H∈𝔥ℤ,α ∈𝔽q,b∈𝔽q* ⟩ ,$ where $xα(a) = ∑n≥0 anXαnn!, and (2.5) hH(b) = diag(bλ1(H),bλ2(H),…,bλr(H)), where Hvi=λi(H)vi. (2.6)$
Remarks. 1. If we “allow” infinite sums, then $hHi(b)= ∑n≥0 (b-1)n (Hin).$ 2. If $𝔤={𝔤}_{s},$ then ${G}_{V}=⟨{x}_{\alpha }\left(t\right)⟩\text{.}$
Example. Suppose (as before) $𝔤={𝔤𝔩}_{2}$ and let $V=ℂ-span {(10),(01)}$ be the natural $𝔤\text{-module}$ given by matrix multiplication. Then $𝔥$ has a basis $𝔥={(a00b) | a,b∈ℂ}= ℂ-span{(1000),(0001)}.$ By direct computation, $xα(t)= (1t01) andh(a00b) (t)=(ta00tb) for a,b∈ℤ,$ and ${G}_{V}={GL}_{2}\left({𝔽}_{q}\right)$ (the general linear group).
### Some combinatorics of the symmetric group
This section describes some of the pertinent combinatorial objects and techniques.
#### A pictorial version of the symmetric group ${S}_{n}$
Let ${s}_{i}\in {S}_{n}$ be the simple reflection that switches $i$ and $i+1,$ and fixes everything else. The group ${S}_{n}$ can be presented by generators ${s}_{1},{s}_{2},\dots ,{s}_{n-1}$ and relations $si2=1, sisi+1si= si+1sisi+1, sisj=sjsi, for ∣i-j∣>1.$ Using two rows of $n$ vertices and strands between the top and bottom vertices, we may pictorially describe permutations in the following way. View $sias \cdots \cdots i .$ Multiplication in ${S}_{n}$ corresponds to concatenation of diagrams, so for example, ${s}_{1} {s}_{2} = = =s2s1.$ Therefore, ${S}_{n}$ is generated by ${s}_{1},{s}_{2},\dots ,{s}_{n-1}$ with relations $= , = , = .$
#### Compositions, partitions, and tableaux
A composition $\mu =\left({\mu }_{1},{\mu }_{2},\dots ,{\mu }_{r}\right)$ is a sequence of positive integers. The size of $\mu$ is $\mid \mu \mid ={\mu }_{1}+{\mu }_{2}+\cdots +{\mu }_{r},$ the length of $\mu$ is $\ell \left(\mu \right)=r$ and $μ≤= {μ≤1,μ≤2,…,μ≤r} ,where μ≤i= μ1+μ2+⋯+μi. (2.7)$ If $\mid \mu \mid =n,$ then $\mu$ is a composition of $n$ and we write $\mu \models n\text{.}$ View $\mu$ as a collection of boxes aligned to the left. If a box $x$ is in the $i\text{th}$ row and $j\text{th}$ column of $\mu ,$ then the content of $x$ is $c\left(x\right)=i-j\text{.}$ For example, if $μ=(2,5,3,4)= ,$ then $\mid \mu \mid =14,$ $\ell \left(\mu \right)=4,$ ${\mu }_{\le }=\left(2,7,10,14\right),$ and the contents of the boxes are $0 1 -1 0 1 2 3 -2 -1 0 -3 -2 -1 0 .$ Alternatively, ${\mu }_{\le }$ coincides with the numbers in the boxes at the end of the rows in the diagram $1 2 3 4 5 6 7 8 9 10 11 12 13 14 .$
A partition $\nu =\left({\nu }_{1},{\nu }_{2},\dots ,{\nu }_{r}\right)$ is a composition where ${\nu }_{1}\ge {\nu }_{2}\ge \cdots \ge {\nu }_{r}>0\text{.}$ If $\mid \nu \mid =n,$ then $\nu$ is a partition of $n$ and we write $\nu ⊢n\text{.}$ Let $𝒫={partitions} and𝒫n= {ν⊢n}. (2.8)$ Suppose $\nu \in 𝒫\text{.}$ The conjugate partition $\nu \prime =\left({\nu }_{1}^{\prime },{\nu }_{2}^{\prime },\dots ,{\nu }_{\ell }^{\prime }\right)$ is given by $νi′=Card {j | νj≥i}.$ In terms of diagrams, $\nu \prime$ is the collection of boxes obtained by flipping $\nu$ across its main diagonal. For example, $ifν= ,thenν′= .$
Suppose $\nu ,\mu$ are partitions. If ${\nu }_{i}\ge {\mu }_{i}$ for all $1\le i\le \ell \left(\mu \right),$ then the skew partition $\nu /\mu$ is the collection of boxes obtained by removing the boxes in $\mu$ from the upper left-hand corner of the diagram $\lambda \text{.}$ For example, if $ν= andμ= ,thenν/μ= .$ A horizontal strip $\nu /\mu$ is a skew shape such that no column contains more than one box. Note that if $\mu =\varnothing ,$ then $\nu /\mu$ is a partition.
A column strict tableau $Q$ of shape $\nu /\mu$ is a filling of the boxes of $\nu /\mu$ by positive integers such that (a) the entries strictly increase along columns, (b) the entries weakly increase along rows. The weight of $Q$ is the composition $\text{wt}\left(Q\right)=\left(\text{wt}{\left(Q\right)}_{1},\text{wt}{\left(Q\right)}_{2},\dots \right)$ given by $wt(Q)i=number of i in Q.$ For example, $Q= 1 1 3 1 2 3 3 5 haswt(Q)= (3,1,3,0,1).$
#### Symmetric functions
The symmetric group ${S}_{n}$ acts on the set of variables $\left\{{x}_{1},{x}_{2},\dots ,{x}_{n}\right\}$ by permuting the indices. The ring of symmetric polynomials in the variables $\left\{{x}_{1},{x}_{2},\dots ,{x}_{n}\right\}$ is $Λn(x)= {f∈ℤ[x1,x2,…,xn] | w(f)=f,w∈Sn}.$ For $r\in {ℤ}_{\ge 0},$ the $r\text{th}$ elementary symmetric polynomial is $er(x:n)= ∑1≤i1n,$ and the $r\text{th}$ power sum symmetric polynomial is $pr(x:n)= x1r+x2r+ ⋯+xnr.$ For any partition $\nu \in 𝒫,$ let $ev(x:n) = eν1(x:n) eν2(x:n)⋯ eνℓ(x:n) pv(x:n) = pν1(x:n) pν2(x:n)⋯ pνℓ(x:n).$ The Schur polynomial corresponding to $\nu$ is $sν(x:n)= det(eνi′-i+j(x:n)), (2.9)$ for which Pieri’s rule gives $sν(x:n) s(n)(x:n)= ∑horizontal strip γ/ν∣γ/ν∣=n sγ(x:n) [Mac1995, I.5.16]. (2.10)$
For each $t\in ℂ,$ the Hall-Littlewood symmetric function is $Pν(x:n;t) = ∑wν∈Snνw ( x1ν1 x2ν2⋯ xnνn ∏νi>νj xi-txjxi-xj ) = 1vν(t) ∑w∈Snw ( x1ν1 x2ν2⋯ xnνn ∏i where ${v}_{\nu }\left(t\right)\in ℂ$ is a constant as in [Mac1995, III.2]; and they satisfy $Pν(x:n;0)= sν(x:n), P(1n) (x:n;t)= en(x:n).$ For $t\in ℂ,$ $Λn(x)=ℤ-span {eν(x:n)}= ℤ-span{sν(x:n)} =ℤ-span{Pν(x:n;t)}, (2.11)$ and if we extend coefficients to $ℚ,$ then $ℚ⊗Λn(x)= ℚ-span{pν(x:n)}.$
Note that for $n>m$ there is a map $Λn(x) ⟶ Λm(x) f(x1,x2,…,xn) ⟼ f(x1,…,xm,0,…,0)$ which sends ${s}_{\nu }\left(x:n\right)↦{s}_{\nu }\left(x:m\right),$ ${e}_{\nu }\left(x:n\right)↦{e}_{\nu }\left(x:m\right),$ etc. Let $\left\{{x}_{1},{x}_{2},\dots \right\}$ be an infinite set of variables. The Schur function ${s}_{\nu }\left(x\right)$ is the infinite sequence $sν(x)= ( sν(x:1), sν(x:2), sν(x:3),… ) ,$ and one can define elementary symmetric functions ${e}_{r}\left(x\right),$ power sum symmetric functions ${p}_{r}\left(x\right),$ and Hall-Littlewood symmetric functions ${P}_{\nu }\left(x;t\right),$ analogously. The ring of symmetric functions in the variables $\left\{{x}_{1},{x}_{2},\dots \right\}$ is $Λ(x)=ℤ-span {sν(x) | ν∈𝒫},$ and let $Λℂ(x)=ℂ -span{sν(x) | ν∈𝒫}.$
#### RSK correspondence
The classical RSK correspondence provides a combinatorial proof of the identity $∏i,j>0 11-xiyj =∑ν⊢nn≥0 sν(x)sν(y) [Knu1970]$ by constructing for each $\ell \ge 0$ a bijection between the matrices $b\in {M}_{\ell }\left({ℤ}_{\ge 0}\right)$ and the set of pairs $\left(P\left(b\right),Q\left(b\right)\right)$ of column strict tableaux with the same shape. The bijection is as follows.
If $P$ is a column strict tableau and $j\in {ℤ}_{>0},$ let $P←j$ be the column strict tableau given by the following algorithm (a) Insert $j$ into the the first column of $P$ by displacing the smallest number $\ge j\text{.}$ If all numbers are $ then place $j$ at the bottom of the first column. (b) Iterate this insertion by inserting the displaced entry into the next column. (c) Stop when the insertion does not displace an entry.
A two-line array $\left(\begin{array}{cccc}{i}_{1}& {i}_{2}& \cdots & {i}_{n}\\ {j}_{1}& {j}_{2}& \cdots & {j}_{n}\end{array}\right)$ is a two-rowed array with ${i}_{1}\le {i}_{2}\le \cdots \le {i}_{n}$ and ${j}_{k}\ge {j}_{k+1}$ if ${i}_{k}={i}_{k+1}\text{.}$ If $b\in {M}_{\ell }\left({ℤ}_{\ge 0}\right),$ then let $\stackrel{\to }{b}$ be the two-line array with ${b}_{ij}$ pairs $\left(\genfrac{}{}{0}{}{i}{j}\right)\text{.}$
For $b\in {M}_{\ell }\left({ℤ}_{\ge 0}\right),$ suppose $b→= (i1i2⋯inj1j2⋯jn).$ Then the pair $\left(P\left(b\right),Q\left(b\right)\right)$ is the final pair in the sequence $(∅,∅)= (P0,Q0), (P1,Q1), (P2,Q2),…, (Pn,Qn),= (P(b),Q(b)),$ where $\left({P}_{k},{Q}_{k}\right)$ is a pair of column strict tableaux with the same shape given by $Pk=Pk-1←jk and Qk is defined by sh(Qk)=sh(Pk) with ik in the new box sh(Qk)/sh(Qk-1).$ For example, $b=(110002010) corresponds tob→= (1122321332)$ and provides the sequence $(∅,∅), ( 2 , 1 ) , ( 1 2 , 1 1 ) , ( 1 2 3 , 1 1 2 ) , ( 1 2 3 3 , 1 1 2 2 ) , ( 1 2 3 2 3 , 1 1 3 2 2 )$ so that $(P(b),Q(b))= ( 1 2 3 2 3 , 1 1 3 2 2 ) .$
## Notes and References
This is an excerpt of the PhD thesis Unipotent Hecke algebras: the structure, representation theory, and combinatorics by F. Nathaniel Edgar Thiem.
page history | HuggingFaceTB/finemath | |
# Question #8da2c
Nov 26, 2017
A Central Force is a special kind of force that originates from a single point in space. A particle under the influence of this force will have its force vector always directed towards that origin point and the magnitude of this force will only depend on the distance of the object from the origin.
It is convenient to work with polar coordinates with such forces, placing the source of the force at the coordinate origin. In such a coordinate system the central force has a magnitude that depends only on the radial coordinate $r$ and is always directed radially inward ($- \hat{r} \setminus \quad$; attractive) or radially outward ($+ \hat{r} \setminus \quad$; repulsive).
$\vec{F} \left(r\right) = \setminus \pm f \left(r\right) \hat{r}$ - positive sign for repulsion and negative sign for attraction.
Classic Examples of Central Force are:
[a] Gravitation: $\setminus q \quad {\vec{F}}_{g} \left(r\right) = - G \frac{M m}{r} ^ 2 \hat{r}$
[b] Electrostatic Force: $\setminus q \quad {\vec{F}}_{E} \left(r\right) = k \frac{{q}_{1} {q}_{2}}{r} ^ 2 \hat{r}$
Properties of Central Forces:
[1] Central Forces are always conservative forces and so the force vector can be written as the negative of the gradient of a scalar potential function.
$\vec{F} \left(r\right) = - \setminus \nabla U \left(r\right)$
[2] The motion of a particle under the influence of a central force is confined to a plane. This is a consequence of Angular Momentum Conservation. | HuggingFaceTB/finemath | |
+0
# 0.4^((6-5x)/(2+5x))<(25/4)
0
795
5
0.4^((6-5x)/(2+5x))<(25/4)
How could I solve this :) ? thanx
Guest May 17, 2014
#4
+94105
+8
hi Blaster0
***How did you get -2 from log6,25/log0,4***
To be perfectly honest I just put the whole thing straight into a calculator and it spat out -2
However, since I am on a roll here I will explain why it is so!
$$\frac{log6.25}{log0.4}=log_{0.4}6.25$$ I did this by applying the change of base law the opposite way around to usual.
Let
$$x=log_{0.4}6.25$$
Rearranging this we get
$$6.25=0.4^x\\ 2.5^2=0.4^x\\ \left(\frac{5}{2}\right)^2=\left(\frac{2}{5}\right)^x\\ \left(\frac{2}{5}\right)^{-2}=\left(\frac{2}{5}\right)^x\\ x\:=\:-2$$
Hey Blaster0,
When you are happy with your answer can we have a thank you and some thumbs up please.
Speaking for myself I put a lot of effort into these answers!
(Don't mind me I am just trying to train people to be polite. I figure that the more people who do it then the more normal it becomes and hopefully the 'idea' will catch on. )
Melody And no chris you can not have your glasses back - I like them!
Melody May 18, 2014
#1
+94105
+8
this should be the last one for the night - It is 1:30 am
$$0.4^{((6-5x)/(2+5x))}<6.25\\\\ log0.4^{((6-5x)/(2+5x))} \frac{6-5x}{2+5x}log0.4 \frac{6-5x}{2+5x}>\frac{log6.25}{log0.4}\\\\ \frac{6-5x}{2+5x}>\frac{log6.25}{log0.4}\\\\ \frac{6-5x}{2+5x}>-2\\\\ (2+5x)^2\frac{6-5x}{2+5x}>-2(2+5x)^2\\\\ (2+5x)(6-5x)>-2(25x^2+20x+4)\\\\ 12-10x+30x-25x^2>-50x^2-40x-8\\\\$$
$$25x^2+60x+20>0\\\\ 5x^2+12x+4>0\\\\ 5x^2+10x+2x+4>0\\\\ 5x(x+2)+2(x+2)>0\\\\ (5x+2)(x+2)>0\\\\ \mbox{concave up parabola roots are -2 and -2/5}\\\\ x<-2 \mbox{ OR }x>\frac{-2}{5}$$
Okay I decided to check this by plotting the original question. And to my delight it appears to be correct!
Melody May 17, 2014
#2
+38
+5
How did you get -2 from log6,25/log0,4
thx
blaster0 May 17, 2014
#3
+92673
+8
$${log}_{10}\left({\mathtt{6.25}}\right) = {\mathtt{0.795\: \!880\: \!017\: \!344\: \!075\: \!2}}$$
$${log}_{10}\left({\mathtt{0.4}}\right) = -{\mathtt{0.397\: \!940\: \!008\: \!672\: \!037\: \!6}}$$
If you divide the first by the second, the result is -2.
CPhill May 17, 2014
#4
+94105
+8
hi Blaster0
***How did you get -2 from log6,25/log0,4***
To be perfectly honest I just put the whole thing straight into a calculator and it spat out -2
However, since I am on a roll here I will explain why it is so!
$$\frac{log6.25}{log0.4}=log_{0.4}6.25$$ I did this by applying the change of base law the opposite way around to usual.
Let
$$x=log_{0.4}6.25$$
Rearranging this we get
$$6.25=0.4^x\\ 2.5^2=0.4^x\\ \left(\frac{5}{2}\right)^2=\left(\frac{2}{5}\right)^x\\ \left(\frac{2}{5}\right)^{-2}=\left(\frac{2}{5}\right)^x\\ x\:=\:-2$$
Hey Blaster0,
When you are happy with your answer can we have a thank you and some thumbs up please.
Speaking for myself I put a lot of effort into these answers!
(Don't mind me I am just trying to train people to be polite. I figure that the more people who do it then the more normal it becomes and hopefully the 'idea' will catch on. )
Melody And no chris you can not have your glasses back - I like them!
Melody May 18, 2014
#5
+38
+5
I got it
Melody and CPhill thank you very much and sorry for that thumbs
blaster0 May 18, 2014 | HuggingFaceTB/finemath | |
Strand: MEASUREMENT AND DATA (2.MD)
Measure and estimate lengths in standard units (Standards 2.MD.1-4) and relate addition and subtraction to length (Standards 2.MD.5-6). They work with time and money (Standards 2.MD.7-8). They represent and interpret data (Standards 2.MD.9-10).
Standard 2.MD.10
Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and comparison problems using information presented in a bar graph.
• Bar Graph Sorter
The objectives of this lesson are to introduce bargraphs and ask students to sort them by a particular attribute.
• Favorite Ice Cream Flavor
The purpose of this task is for students to represent and interpret categorical data. So this task could be used with advanced first graders or second graders just beginning to work with bar graphs.
• Grade 2 Math Module 7: Problem Solving with Length, Money, and Data (Engage NY)
Module 7 presents an opportunity for students to practice addition and subtraction strategies within 100 and problem-solving skills as they learn to work with various types of units within the contexts of length, money, and data. Students represent categorical and measurement data using picture graphs, bar graphs, and line plots. They revisit measuring and estimating length from Module 2, though now using both metric and customary units.
• Grade 2 Unit 1: Extending Base Ten Understanding (Georgia Standards)
In this unit, students will understand the value placed on the digits within a three-digit number, recognize that a hundred is created from ten groups of ten, use skip counting strategies to skip count by 5s, 10s, and 100s within 1,000 and represent numbers to 1,000 by using numbers, number names, and expanded form, compare two-digit number using >, =, <.
• Grade 2 Unit 2: Becoming Fluent with Addition and Subtraction (Georgia Standards)
In this unit students will cultivate an understanding of how addition and subtraction affect quantities and are related to each other, will reinforce the multiple meanings for addition (combine, join, and count on) and subtraction (take away, remove, count back, and compare), further develop their understanding of the relationships between addition and subtraction, recognize how the digits 0-9 are used in our place value system to create numbers and manipulate amounts and continue to develop their understanding solving problems with money.
• Grade 2 Unit 3: Understanding Measurement, Length, and Time (Georgia Standards)
In this unit students will know the following customary units for measuring length: inch, foot, yard, recognize the need for standard units of measure, understand the relationship of hours and days and nderstand the importance, usefulness of reasonable estimations and e able to represent the length of several objects by making a line plot.
• Grade 2 Unit 4: Applying Base Ten Understanding (Georgia Standards)
In this unit students will continue to develop their understanding of and facility with addition and subtraction, add up to 4 two-digit numbers, use a variety of models (base ten blocks- ones, tens, and hundreds only; diagrams; number lines; place value strategies; etc.) to add and subtract within one thousand and continue to develop their understanding of, and facility with, money.
• Grade 2 Unit 5: Understanding Plane and Solid Figures (Georgia Standards)
In this unit students will cultivate spatial awareness by further developing understandings of basic geometric figures, identifying plane figures and solid figures based on geometric properties, understand what an array is and how it can be used as a model for repeated addition and organize and record data using tallies, simple tables and charts, picture graphs, and bar graphs.
• Grade 2 Unit 6: Developing Multiplication (Georgia Standards)
In this unit students will understand and model multiplication as repeated addition and as rectangular arrays, determine if a number is odd or even (within twenty) and create and interpret picture graphs and bar graphs.
• Graphing with Colors
This Teaching Channel video and lesson plan has students find colored squares around the room and then create a bar graph. (4 minutes)
• Growing Bean Plants
This task adds some rigor to this common activity, by collecting actual growth data, providing practice for students in measuring and recording length measurements.
• Hand Span Measures
This could be used as a class activity, or students could gather and plot data on separate line plots from different age groups.
• Histogram
This activity will help students understand histograms by allowing them to make their own.
• Longest Walk
Given a map, students in this task locate two points and draw a line between them and then calculate the length of the line.
• Measurement and Data (2.MD) - Second Grade Core Guide
The Utah State Board of Education (USBE) and educators around the state of Utah developed these guides for Second Grade Mathematics - Measurement and Data (2.MD)
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialists - Trish French or Molly Basham and see the Mathematics - Elementary website. For general questions about Utah's Core Standards contact the Director - Jennifer Throndsen.
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200. | HuggingFaceTB/finemath | |
# Surface Area and Volume
## Presentation on theme: "Surface Area and Volume"— Presentation transcript:
Surface Area and Volume
Surface Area of Prisms Surface Area = The total area of the surface of a three-dimensional object (Or think of it as the amount of paper you’ll need to wrap the shape.) Prism = A solid object that has two identical ends and all flat sides. We will start with 2 prisms – a rectangular prism and a triangular prism.
Triangular Prism Rectangular Prism
Surface Area (SA) of a Rectangular Prism
Like dice, there are six sides (or 3 pairs of sides)
Prism net - unfolded
Add the area of all 6 sides to find the Surface Area.
6 - height 5 - width 10 - length
SA = 2lw + 2lh + 2wh 6 - height 5 - width 10 - length
SA = 2 (10 x 5) + 2 (10 x 6) + 2 (5 x 6) = 2 (50) + 2(60) + 2(30) = = 280 units squared
Practice = 1208 ft squared 12 ft 10 ft 22 ft SA = 2lw + 2lh + 2wh
= 2(22 x 10) + 2(22 x 12) + 2(10 x 12) = 2(220) + 2(264) + 2(120) = = 1208 ft squared
Surface Area of a Triangular Prism
2 bases (triangular) 3 sides (rectangular)
Unfolded net of a triangular prism
2(area of triangle) + Area of rectangles
Area Triangles = ½ (b x h) = ½ (12 x 15) = ½ (180) = 90 Area Rect. 1 = b x h = 12 x 25 = 300 Area Rect = 25 x 20 = 500 15ft SA = SA = 1480 ft squared
Practice Triangles = ½ (b x h) = ½ (8 x 7) = ½ (56) = 28 cm
Rectangle 1 = 10 x 8 = 80 cm Rectangle 2 = 9 x 10 = 90 cm Add them all up SA = SA = 316 cm squared 9 cm 7 cm 8 cm 10 cm
Surface Area of Pyramids
Pyramids A pyramid has 2 shapes: One (1) square & Four (4) triangles
Since you know how to find the areas of those shapes and add them.
Or…
Where S is the Slant Height and p is the perimeter of the base and
you can use a formula… SA = ½ Sp + B Where S is the Slant Height and p is the perimeter of the base and B is the area of the Base
SA = ½ Sp + B = ½ (8 x 26) + (7 x 6) *area of the base* SA = ½ Sp + B
5 Perimeter = (2 x 7) + (2 x 6) = 26 Slant height S = 8 ; SA = ½ Sp + B = ½ (8 x 26) + (7 x 6) *area of the base* = ½ (208) + (42) = = 146 units 2
SA = ½ Sp + B = ½ (18 x 24) + (6 x 6) Practice = ½ (432) + (36)
10 SA = ½ Sp + B = ½ (18 x 24) + (6 x 6) = ½ (432) + (36) = = 252 units2 Slant height = 18 Perimeter = 6x4 = 24 What is the extra information in the diagram?
#3 Finding the Surface Area of a Cone
The radius of the base of a cone is 5 m. Its slant height is 13 m. Find the surface area.
Formula for Prisms VOLUME OF A PRISM The volume V of a prism is the area of its base B times its height h. V = Bh Note – the capital letter stands for the AREA of the BASE not the linear measurement.
Try It V = Bh = (8 x 4) x 3 = (32) x 3 = 96 ft3 Find area of the base
Multiply it by the height = 96 ft3 3 ft - height 4 ft - width 8 ft - length
Practice V = Bh = (22 x 10) x 12 = (220) x 12 = 2640 cm3 12 cm 10 cm
Notice that r2 is the formula for area of a circle.
Cylinders VOLUME OF A CYLINDER The volume V of a cylinder is the area of its base, r2, times its height h. V = r2h Notice that r2 is the formula for area of a circle.
Try It V = r2h The radius of the cylinder is 5 m, and the height is 4.2 m V = 3.14 · 52 · 4.2 Substitute the values you know. V = 329.7
Practice 13 cm - radius 7 cm - height V = r2h Start with the formula V = 3.14 x 132 x 7 substitute what you know = 3.14 x 169 x 7 Solve using order of Ops. = cm3
Lesson Quiz Find the volume of each solid to the nearest tenth. Use 3.14 for . 1. 2. 4,069.4 m3 861.8 cm3 3. triangular prism: base area = 24 ft2, height = 13 ft 312 ft3
Remember that Volume of a Prism is B x h where b is the area of the base.
You can see that Volume of a pyramid will be less than that of a prism. How much less? Any guesses?
If you said 2/3 less, you win!
Volume of a Pyramid: V = (1/3) Area of the Base x height V = (1/3) Bh Volume of a Pyramid = 1/3 x Volume of a Prism + = +
Find the volume of the square pyramid with base edge length 9 cm and height 14 cm.
The base is a square with a side length of 9 cm, and the height is 14 cm. V = 1/3 Bh = 1/3 (9 x 9)(14) = 1/3 (81)(14) = 1/3 (1134) = 378 cm3 14 cm
Practice V = 1/3 Bh = 1/3 (5 x 5) (10) = 1/3 (25)(10) = 1/3 250 = units3
The Rectangle This has 2 steps. To find the area we need base and height. Height is given (6) but the base is not as easy. Notice that the base is the same as the distance around the circle (or the Circumference).
Find Circumference Formula is C = x d = 3.14 x 6 (radius doubled)
= 18.84 Now use that as your base. A = b x h = x 6 (the height given) = units squared
Add them together Now add the area of the circles and the area of the rectangle together. = units squared The total Surface Area!
Quiz Find the volume of each figure. a rectangular pyramid with length 25 cm, width 17 cm, and height 21 cm 2975 cm3 2. a triangular pyramid with base edge length 12 in. a base altitude of 9 in. and height 10 in. 360 in3
Surface Area of a Cylinder
Review Surface area is like the amount of paper you’ll need to wrap the shape. You have to “take apart” the shape and figure the area of the parts. Then add them together for the Surface Area (SA)
Parts of a cylinder A cylinder has 2 main parts. A rectangle and
A circle – well, 2 circles really. Put together they make a cylinder.
The Soup Can Think of the Cylinder as a soup can.
You have the top and bottom lid (circles) and you have the label (a rectangle – wrapped around the can). The lids and the label are related. The circumference of the lid is the same as the length of the label.
Area of the Circles Formula for Area of Circle A= r2 = 3.14 x 32
= 28.26 But there are 2 of them so 28.26 x 2 = units squared
Area of Rectangle Area of Circles
Formula SA = ( d x h) + 2 ( r2) Label Lids (2) Area of Rectangle Area of Circles
Practice Be sure you know the difference between a radius and a diameter!
SA = ( d x h) + 2 ( r2) = (3.14 x 22 x 14) + 2 (3.14 x 112) = (367.12) + 2 (3.14 x 121) = (367.12) + 2 (379.94) = (367.12) + (759.88) = 1127 cm2
More Practice! SA = ( d x h) + 2 ( r2)
= (3.14 x 11 x 7) + 2 ( 3.14 x 5.52) = (241.78) + 2 (3.14 x 30.25) = (241.78) + 2 (3.14 x 94.99) = (241.78) + 2 (298.27) = (241.78) + (596.54) = cm2 11 cm 7 cm
Similar presentations | HuggingFaceTB/finemath | |
## Algebra 1: Common Core (15th Edition)
The possible dimensions of the rectangle are $n+4$ and $n-7$.
To factor $n^{2}-3n-28$, identify the pair of factors of $-28$ that has a sum of $-3$. $\left[\begin{array}{lll} \text{Factors of -28 } & \text{Sum of factors} & \\ 1\text{ and }-28 & -27 & \\ -1\text{ and }28 & 27 & \\ 2\text{ and }-14 & -12 & \\ -2\text{ and }14 & 12 & \\ 4\text{ and }-7 & -3 & \text{...is what we need}\\ -4\text{ and }7 & 3 & \end{array}\right]$ $n^{2}-3n-28=(n-4)(n+7)$ The possible dimensions of the rectangle are $n+4$ and $n-7$. | HuggingFaceTB/finemath | |
1. proof with graphs
The graph G contains at least two vertices. One vertex has degree one, and every other vertex has degree greater than 1. Show that G contains a cycle.
I was able to see that it was the case by drawing, but I need help getting started with the proof.
I tried to make a connection with the theorem that says any walk starting with an odd degree must end in another odd degree vertex, but failed.
2. Re: proof with graphs
If G does not have a cycle, then it is a tree, and in a tree, |V| = |E| + 1 where V is the set of vertices and E is the set of edges. Used together with the handshaking lemma and with the assumption about the degrees, this leads to a contradiction.
3. Re: proof with graphs
So $\displaystyle |V|=|E|+1 \\ \Rightarrow 2|E|=|E|+1 \\ \Rightarrow |E|=1$
and this is a contradiction because we assumed G has more than two vertices?
4. Re: proof with graphs
No, it does not follow from anywhere that 2|E| = |V| (this fact, I assume, is used in going from the first to the second line). Rather, the handshaking lemma says that $\displaystyle \sum_{v\in V} \mathop{\text{deg}}(v)=2|E|$. Since |E| = |V| - 1, we have $\displaystyle \sum_{v\in V} \mathop{\text{deg}}(v) =2(|V|-1)$. Now find a lower bound for the left-hand side and compare it with the right-hand side.
5. Re: proof with graphs
From the assumption about the degrees, $\displaystyle \sum_{v\in V} \mathop{\text{deg}}(v) \ge 1+2(|V|-1)$.
But then $\displaystyle \sum_{v\in V} \mathop{\text{deg}}(v) =2(|V|-1) \ge 1+2(|V|-1)$ which is impossible?
6. Re: proof with graphs
That's correct. | HuggingFaceTB/finemath | |
# Search by Topic
#### Resources tagged with Interactivities similar to Up, Down, Flying Around:
Filter by: Content type:
Stage:
Challenge level:
### There are 152 results
Broad Topics > Information and Communications Technology > Interactivities
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Got it Article
##### Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
### Fifteen
##### Stage: 2 and 3 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Drips
##### Stage: 2 and 3 Challenge Level:
An animation that helps you understand the game of Nim.
### Multiplication Tables - Matching Cards
##### Stage: 1, 2 and 3 Challenge Level:
Interactive game. Set your own level of challenge, practise your table skills and beat your previous best score.
### Semi-regular Tessellations
##### Stage: 3 Challenge Level:
Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations?
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Factors and Multiples Game
##### Stage: 2, 3 and 4 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent?
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Subtended Angles
##### Stage: 3 Challenge Level:
What is the relationship between the angle at the centre and the angles at the circumference, for angles which stand on the same arc? Can you prove it?
### Shuffles Tutorials
##### Stage: 3 Challenge Level:
Learn how to use the Shuffles interactivity by running through these tutorial demonstrations.
### Lost
##### Stage: 3 Challenge Level:
Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Have You Got It?
##### Stage: 3 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Triangles in Circles
##### Stage: 3 Challenge Level:
Can you find triangles on a 9-point circle? Can you work out their angles?
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Balancing 2
##### Stage: 3 Challenge Level:
Meg and Mo still need to hang their marbles so that they balance, but this time the constraints are different. Use the interactivity to experiment and find out what they need to do.
### Diamond Mine
##### Stage: 3 Challenge Level:
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### An Unhappy End
##### Stage: 3 Challenge Level:
Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line...
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### Flip Flop - Matching Cards
##### Stage: 1, 2 and 3 Challenge Level:
A game for 1 person to play on screen. Practise your number bonds whilst improving your memory
### Balancing 1
##### Stage: 3 Challenge Level:
Meg and Mo need to hang their marbles so that they balance. Use the interactivity to experiment and find out what they need to do.
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Balancing 3
##### Stage: 3 Challenge Level:
Mo has left, but Meg is still experimenting. Use the interactivity to help you find out how she can alter her pouch of marbles and still keep the two pouches balanced.
### Archery
##### Stage: 3 Challenge Level:
Imagine picking up a bow and some arrows and attempting to hit the target a few times. Can you work out the settings for the sight that give you the best chance of gaining a high score?
### Top Coach
##### Stage: 3 Challenge Level:
Carry out some time trials and gather some data to help you decide on the best training regime for your rowing crew.
### Right Angles
##### Stage: 3 Challenge Level:
Can you make a right-angled triangle on this peg-board by joining up three points round the edge?
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### See the Light
##### Stage: 2 and 3 Challenge Level:
Work out how to light up the single light. What's the rule?
### Rollin' Rollin' Rollin'
##### Stage: 3 Challenge Level:
Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P?
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Bow Tie
##### Stage: 3 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Circuit Maker
##### Stage: 3, 4 and 5 Challenge Level:
Investigate how logic gates work in circuits.
### Khun Phaen Escapes to Freedom
##### Stage: 3 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Jam
##### Stage: 4 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Cogs
##### Stage: 3 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### Volume of a Pyramid and a Cone
##### Stage: 3
These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.
### Nine Colours
##### Stage: 3 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Interactive Spinners
##### Stage: 3 Challenge Level:
This interactivity invites you to make conjectures and explore probabilities of outcomes related to two independent events.
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas. | HuggingFaceTB/finemath | |
# Correspondence between ODE and difference equation
In Wikipedia about difference equations, there is some description about correspondence between ODE and difference equation:
If you consider the Taylor series of the solution to a linear differential equation:$$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$$ you see that the coefficients of the series are given by the nth derivative of $f(x)$ evaluated at the point $a$. The differential equation provides a linear difference equation relating these coefficients.
The rule of thumb (for equations in which the polynomial multiplying the first term is non-zero at zero) is that:$$y^{[k]} \to f[n+k]$$and more generally $$x^m*y^{[k]} \to n(n-1)(n-m+1)f[n+k-m]$$
Example: The recurrence relationship for the Taylor series coefficients of the equation: $$(x^2 + 3x -4)y^{[3]} -(3x+1)y^{[2]} + 2y = 0\,$$ is given by$$n(n-1)f[n+1] + 3nf[n+2] -4f[n+3] -3nf[n+1] -f[n+2]+ 2f[n] = 0\,$$ or $$-4f[n+3] +2nf[n+2] + n(n-4)f[n+1] +2f[n] = 0.\,$$
My questions are:
1. I was wondering what the rationale behind this conversion from an ODE to a difference equation is? Although having tried to read it several times, I was not able to understand it.
2. In reverse direction, can a difference equation be converted to an ODE using this correspondence? How to?
3. Is the conversion of an ODE into a difference equation in numerical methods for solving an ODE related to the correspondence between the two mentioned above?
4. This equivalence can be used to quickly solve for the recurrence relationship for the coefficients in the power series solution of a linear differential equation.
Problems generally solved using the power series solution method taught in normal differential equation classes can be solved in a much easier way.
I was wondering how exactly the correspondence can make solving an ODE or a difference equation easier?
Thanks and regards!
All that's going on here is that when you differentiate a Taylor series expansion
$$f(x) = f^{(0)}(a) + \frac{f^{(1)}(a)}{1!} (x - a)^1 + \frac{f^{(2)}(a)}{2!} (x - a)^2 + ...$$
you get the Taylor series expansion
$$f'(x) = f^{(1)}(a) + \frac{f^{(2)}(a)}{1!} (x - a)^1 + \frac{f^{(3)}(a)}{2!} (x - a)^2 + ....$$
In other words, what you've done is precisely shifted the Taylor coefficients one to the left. That's all the Wikipedia article is saying.
I disagree that this makes ODEs any easier to solve, though. ODEs were already this easy to solve, you just weren't taught the right language for seeing this.
• In fact, when the Taylor expansion exists and is convergent (so the function is locally real analytic), this also gives easy solutions to partial differential equations. This basic machinery in the PDE case is called the Cauchy-Kovalevsky theorem. – Willie Wong Mar 7 '11 at 13:11
• This is the Principle of Conservation of Difficulty at work, against which we are in general defenseless! :) – Mariano Suárez-Álvarez Mar 7 '11 at 18:50 | HuggingFaceTB/finemath | |
Texas Go Math Grade 4 Lesson 7.1 Answer Key Multiply Tens, Hundreds, and Thousands
Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 7.1 Answer Key Multiply Tens, Hundreds, and Thousands.
Texas Go Math Grade 4 Lesson 7.1 Answer Key Multiply Tens, Hundreds, and Thousands
Essential Question
How does understanding place value help you multiply tens, hundreds, and thousands?
Unlock the Problem
Each car on a train has 200 seats. How many seats are on a train with 8 cars?
Find 8 × 200.
One Way Draw a quick picture. The number of seats is on a train with 8 cars is 1600 seats.
Explanation:
Given that each car on a train has 200 seats. So the number of seats are on a train with 8 cars is 200 × 8 which is 1600 seats. Here, we have represented with a quick picture.
Another Way Use place value.
8 × 200 = 8 × ____________ hundreds
= _____ hundreds
= ______ Think: 16 hundreds is 1 thousand, 6 hundreds.
So, there are ____________ seats on a train with 8 cars.
8 × 200 = 8 × 2 hundreds
= 16 hundred
= 1600 Think: 16 hundred is 1 thousand, 6 hundred.
So, there are 1600 seats on a train with 8 cars.
Explanation:
Given that each car on a train has 200 seats. So the number of seats are on a train with 8 cars is 200 × 8 which is 1600 seats. Here, we have represented with a place value
8 × 200 = 8 × 2 hundreds
= 16 hundred
= 1600 Think: 16 hundred is 1 thousand, 6 hundred.
So, there are 1600 seats on a train with 8 cars.
Math Talk
Mathematical Processes
Explain how finding 8 × 2 can help you find 8 × 200.
Here 8 × 2 = 16, to find 8 × 200 we will add two zeros to the value.
Use patterns.
Basic fact:
3 × 7 = 21 ← basic fact
3 × 70 = 210 .
3 × 700 = _____________
3 × 7,000 = _____________
3 × 7 = 21 ← basic fact
3 × 70 = 210 .
3 × 700 = 2,100.
3 × 7,000 = 21,000.
Explanation:
Here, the number of zeros in a factor increases, the number of zeros in the product increases.
3 × 7 = 21 ← basic fact
3 × 70 = 210 .
3 × 700 = 2,100
3 × 7,000 = 21,000.
Basic fact with a zero:
8 × 5 = 40
8 × 50 = 400
8 × 500 = _____________
8 × 5,000= _____________
8 × 5 = 40
8 × 50 = 400
8 × 500 = 4,000
8 × 5,000= 40,000.
Explanation:
Here, the number of zeros in a factor increases, the number of zeros in the product increases.
8 × 5 = 40
8 × 50 = 400
8 × 500 = 4,000
8 × 5,000= 40,000.
Share and show
Question 1.
Use the drawing to find 2 × 500. 2 × 500 = _____________
2 × 500 = 1,000.
Explanation:
Here, 2 × 500 is 1,000.
Math talk
Mathematical Process
Explain how to use place value to find 2 × 500.
Here, 2 × 500 is 2 times 5 hundred, which is equal to 10 hundred and 10 hundred equals 1,000.
Complete the pattern.
Question 2.
3 × 8 = 24
3 × 80 = _____________
3 × 800 = _____________
3 × 8,000 = _____________
3 × 8 = 24
3 × 80 = 240
3 × 800 = 2,400
3 × 8,000 = 24,000.
Explanation:
Here, the number of zeros in a factor increases, the number of zeros in the product increases.
3 × 8 = 24
3 × 80 = 240
3 × 800 = 2,400
3 × 8,000 = 24,000.
Question 3.
6 × 2 = 12
6 × 20 = _____________
6 × 200 = _____________
6 × 2000 = _____________
6 × 2 = 12
6 × 20 = 120
6 × 200 = 1,200
6 × 2000 = 12,000.
Explanation:
Here, the number of zeros in a factor increases, the number of zeros in the product increases.
6 × 2 = 12
6 × 20 = 120
6 × 200 = 1,200
6 × 2000 = 12,000.
Question 4.
4 × 5 = _____________
4 × 50 = _____________
4 × 500 = _____________
4 × 5,000 = _____________
4 × 5 = 20
4 × 50 = 200
4 × 500 = 2,000
4 × 5,000 = 20,000.
Explanation:
Here, the number of zeros in a factor increases, the number of zeros in the product increases.
4 × 5 = 20
4 × 50 = 200
4 × 500 = 2,000
4 × 5,000 = 20,000.
Find the product.
Question 5.
6 × 500 = 6 × _____________ hundreds
= _____________ hundreds
= _____________
6 × 500 = 3,000.
Explanation:
The product of 6 × 500 is
6 × 500 = 6 × 5 hundreds
= 30 hundreds
= 3,000.
Question 6.
9 × 5,000 = 9 × _____________ thousands
= _____________ thousands
= _____________
9 × 5,000 = 45,000.
Explanation:
The product of 9 × 5,000 is
9 × 5,000 = 9 × 5 thousands
= 45 thousands
= 45,000.
H.O.T. Algebra Find the missing factor.
Question 7.
_____________ × 9,000 = 63,000
7 × 9,000 = 63,000.
Explanation:
Here we need to find out the missing factor, so we will perform division which is 63,000÷9,000 = 7.
Question 8.
7 × _____________ = 56,000
7 × 8,000 = 56,000
Explanation:
Here we need to find out the missing factor, so we will perform division which is 56,000÷7 = 8,000.
Question 9.
8 × _____________ = 3,200
8 × 400 = 3,200
Explanation:
Here we need to find out the missing factor, so we will perform division which is 3,200÷8 = 400.
Unlock the Problem Question 10.
H.O.T. Multi-Step Apply Joe’s Fun and Sun rents beach chairs. The store rented 300 beach chairs each month in April and in May. The store rented 600 beach chairs each month from June through September. How many beach chairs did the store rent during the 6 months?
(A) 1,200
(B) 2,400
(C) 3,000
(D) 5,400
A.
Explanation:
Given that the store rented 300 beach chairs each month in April and in May which is 300+300 = 600 chairs. The store rented 600 beach chairs each month from June through September. So the total number of beach chairs did the store rent during the 6 months is 600+600 which is 1200 chairs.
a. What do you need to know?
Here, we need to know about the total number of beach chairs did the store rent during the 6 months.
b. How will you find the number of beach chairs?
We will find the number of beach chairs by adding the chairs for every month.
c. Show the steps you use to solve the problem.
600+600 = 1200 chairs.
Explanation:
Given that the store rented 300 beach chairs each month in April and in May which is 300+300 = 600 chairs. The store rented 600 beach chairs each month from June through September. So the total number of beach chairs did the store rent during the 6 months is 600+600 which is 1200 chairs.
d. Complete the sentences.
For April and May, a total of ___________ beach chairs were rented.
For June through September, a total of ___________ beach chairs were rented.
Joe’s Fun and Sun rented ___________ beach chairs during the 6 months.
For April and May, a total of 600 beach chairs were rented.
For June through September, a total of 600 beach chairs were rented.
Joe’s Fun and Sun rented 1200 beach chairs during the 6 months.
e. Fill in the bubble for the correct answer choice above.
Question 11.
H.O.T. Use Math Language How does the number of zeros in the product of 8 and 5,000 compare to the number of zeros in the factors? Explain.
Question 12.
True Blue plans to decorate backpacks. They will make a profit of $6 for each backpack they decorate. How much profit will they make all together if they decorate and sell 300 backpacks? (A)$300
(B) $1,800 (C)$180
(D) $600 Answer: B. Explanation: Given that they will make a profit of$6 for each backpack they decorate. So the profit that will they make all together if they decorate and sell 300 backpacks is 300 × 6 which is \$1,800.
Question 13.
A whistle company packs 4,000 whistles in each box that it sends to stores. How many whistles are In 8 boxes?
(A) 240
(B) 24,000
(C) 32,000
(D) 3,200
C.
Explanation:
Given that a whistle company packs 4,000 whistles in each box that it sends to stores. So the number of whistles are In 8 boxes is 4,000 × 8 which is 32,000.
Question 14
MultiStep Joshua spends 20 hours walking dogs. He also spends 2 hours on each garden he weeds. He weeds 50 gardens. How much time does Joshua spend walking dogs and weeding gardens?
(A) 90 hours
(B) 100 hours
(C) 70 hours
(D) 120 hours
D.
Explanation:
Given that Joshua spends 20 hours walking dogs and he also spends 2 hours on each garden he weeds. He weeds 50 gardens which is 50×2 = 100, so the time does Joshua spends walking dogs and weeding gardens is 20+100 = 120 hours.
TEXAS Test Prep
Question 15.
Carmen has three books of 20 stamps and five books of 10 stamps. How many stamps does she have?
(A) 110
(B) 60
(C) 50
(D) 100
A.
Explanation:
Given that Carmen has three books of 20 stamps which is 3×20 which is 60 and five books of 10 stamps 5×10 which is 50 . So the number of stamps does she have is 60+50 is 110.
Texas Go Math Grade 4 Lesson 7.1 Homework and Practice Answer Key
Multiply Tens, Hundreds, and Thousands
Complete the pattern.
Question 1.
6 × 9 = ___________
6 × 90 = ___________
6 × 900 = ___________
6 × 9,000 = ___________
6 × 9 = 54
6 × 90 = 540
6 × 900 = 5,400
6 × 9,000 = 54,000.
Explanation:
Here, the number of zeros in a factor increases, the number of zeros in the product increases.
6 × 9 = 54
6 × 90 = 540
6 × 900 = 5,400
6 × 9,000 = 54,000.
Question 2.
5 × 8 = ___________
5 × 80 = ___________
5 × 800 = ___________
5 × 8,000 = ___________
5 × 8 = 40
5 × 80 = 400
5 × 800 = 4,000
5 × 8,000 = 40,000.
Explanation:
Here, the number of zeros in a factor increases, the number of zeros in the product increases.
5 × 8 = 40
5 × 80 = 400
5 × 800 = 4,000
5 × 8,000 = 40,000.
Find the product.
Question 3.
3 × 40 = 3 × ___________ tens
= ___________ tens
= ___________
3 × 40 = 3 × 4 tens
= 12 tens
= 120.
Explanation:
The product of 3 × 40 is
3 × 40 = 3 × 4 tens
= 12 tens
=120.
Question 4.
4 × 700 = 4 × ___________ hundreds
= ___________ hundreds
= ___________
4 × 700 = 4 × 7 hundreds
= 28 hundreds
= 2,800.
Explanation:
The product of 4 × 700 is
4 × 700 = 4 × 7 hundreds
= 28 hundreds
= 2,800.
Find the missing factor.
Question 5.
9 × ___________ = 4,500
9 × 500 = 4,500.
Explanation:
Here we need to find out the missing factor, so we will perform division which is 4,500÷9 = 500.
Question 6.
6 × ___________ = 12,000
6 × 2,000 = 12,000.
Explanation:
Here we need to find out the missing factor, so we will perform division which is 12,000÷6 = 2,000.
Problem Solving
Mr. Hathaway bought 4 boxes of small paper clips and 3 boxes of large paper clips for his office. There are 300 small paper clips in a box. There are 200 large paper clips in a box. How many paper clips did Mr. Hathaway buy?
The number of paper clips did Mr. Hathaway bought is 1,800 paper clips.
Explanation:
Given that Mr. Hathaway bought 4 boxes of small paper clips and 3 boxes of large paper clips for his office and there are 300 small paper clips in a box which is 300×4 = 1200. There are 200 large paper clips in a box which is 200×3 = 600. So the number of paper clips did Mr. Hathaway bought is 1200+600 = 1800 paper clips.
Question 7.
How will you find the total number of paper clips?
By adding the small paper clips and large paper clips we have found the total number of paper clips.
Question 8.
Show the steps you use to solve the problem.
1200+600 = 1800 paper clips.
Explanation:
Given that Mr. Hathaway bought 4 boxes of small paper clips and 3 boxes of large paper clips for his office and there are 300 small paper clips in a box which is 300×4 = 1200. There are 200 large paper clips in a box which is 200×3 = 600. So the number of paper clips did Mr. Hathaway bought is 1200+600 = 1800 paper clips.
Lesson Check
Question 9.
What is the product of 8 × 80?
(A) 640
(B) 64
(C) 6,400
(D) 64,000
A.
Explanation:
The product of 8 × 80 is 640.
Question 10.
Which multiplication problem has a product of 24,000?
(A) 3 × 800
(B) 6 × 4,000
(C) 3 × 80
(D) 6 × 400
A.
Explanation:
The multiplication problem has a product of 24,000 is 3 × 800.
Question 11.
What is the missing factor?
7 × ___________ = 6,300
(A) 9
(B) 9,000
(C) 90
(D) 900
D.
Explanation:
Here we need to find out the missing factor, so we will perform division which is 6,300 ÷ 7 = 900.
Question 12.
There are 50 pennies in a roll. How many pennies are in 6 rolls?
(A) 30
(B) 3,000
(C) 300
(D) 30,000
C.
Explanation:
Given that there are 50 pennies in a roll, so the number of pennies are in 6 rolls is 50×6 = 300.
Question 13.
Multi-Step Mariah makes bead necklaces. Beads are packaged in bags of 50 and bags of 200. Mariah bought 4 bags of 50 beads and 3 bags of 200 beads. How many beads did Mariah buy?
(A) 11,000
(B) 8,000
(C) 1,100
(D) 800
D.
Explanation:
Given that Mariah makes bead necklaces and beads are packaged in bags of 50 and bags of 200. Mariah bought 4 bags of 50 beads which is 50×4 = 200 beads and 3 bags of 200 beads which is 3×200 = 600 beads. So the number of beads did Mariah bought is 200+600 = 800 beads.
Question 14.
Multi-Step Baseball cards are sold in packages of 20 cards and 50 cards. On Monday, Terence bought 6 packages of 50 cards. On Saturday, he bought 4 packages of 20 cards. How many baseball cards did Terence buy altogether?
(A) 380
(B) 1,100
(C) 110
(D) 3,800 | HuggingFaceTB/finemath | |
The difference between two numbers is 10. If the smaller number is increased by six times the larger number, the result is 4. Find the two numbers?
1
by Edidiong
2014-01-18T21:48:38+01:00
Let the unknown nos be A & B
A-B=10
(6A+A)-B=4
7A-B=4
-7A-B=4 ..... (2)
-6A=6
A=-1
Putting A=-1 into (2)
-7-B=4
B=-11 | HuggingFaceTB/finemath | |
# Program to find largest sum of any path of a binary tree in Python
Suppose we have a binary tree, we have to find the largest sum of any path that goes from the root node to the leaf node.
So, if the input is like
then the output will be 29 as from root, if we follow the path 5-<9-<7-<8 it will be 29 after addition.
To solve this, we will follow these steps−
• Define a function walk(). This will take node, s
• if node is null, then
• max_sum := maximum of max_sum and s
• return
• s := s + data of node
• walk(left of node, s)
• walk(right of node, s)
• From the main method do the following−
• max_sum := 0
• walk(root, 0)
• return max_sum
Let us see the following implementation to get better understanding−
## Example
Live Demo
from collections import defaultdict
class TreeNode:
def __init__(self, data, left = None, right = None):
self.data = data
self.left = left
self.right = right
class Solution:
def walk(self, node, s):
if not node:
self.max_sum = max(self.max_sum, s)
return
s += node.data
self.walk(node.left, s)
self.walk(node.right, s)
def solve(self, root):
self.max_sum = 0
self.walk(root, 0)
return self.max_sum
ob = Solution()
root = TreeNode(5)
root.left = TreeNode(1)
root.right = TreeNode(9)
root.right.left = TreeNode(7)
root.right.right = TreeNode(10)
root.right.left.left = TreeNode(6)
root.right.left.right = TreeNode(8)
print(ob.solve(root))
## Input
root = TreeNode(5)
root.left = TreeNode(1)
root.right = TreeNode(9)
root.right.left = TreeNode(7)
root.right.right = TreeNode(10)
root.right.left.left = TreeNode(6)
root.right.left.right = TreeNode(8)
## Output
29
Updated on: 05-Oct-2020
99 Views | HuggingFaceTB/finemath | |
# Viscosity Unit
## Viscosity Unit:
Viscosity – As already explained, a fluid is a material which will continue to deform with the application of shear stress. However, different fluids deform at different rates when the same shear stress is applied.
Viscosity is the property of a fluid which determines the amount of resistance to a shearing stress. A real fluid has no viscosity but it is nonexistent.
Viscosity can also be defined as the property of a fluid due to which it offers resistance to the movement of one layer of fluid over another adjacent layer.
Viscosity increases with increase in temperature in case of gases whereas it decreases in case of liquid.
Consider, a plate is placed at a distance of ‘Y’ from the fixed surface. The space in between is filled with a fluid. The plate moves with a velocity U by a force F as shown in fig 1.2. A fluid layer at a distance of y from surface moves with a velocity of `u’ and a layer at a distance of dy from y moves with a velocity of u+du.
According to Newton’s law of viscosity the shear force, F acting between two layers of fluid is proportional to difference in their velocities du and area A of the plate and inversely proportional to the distance dy between them.
Where,
μ (Mu) is the constant of proportionality or co-efficient of dynamic viscosity or viscosity. Its unit can be derived as
### Dynamic Viscosity (μ):
As explained earlier, the dynamic viscosity (μ) is defined as the shear stress required causing unit rate of shear deformation.
### Kinematic Viscosity (v):
The kinematic viscosity ( v) viscosity is defined as the ratio of dynamic viscosity to mass density.
The name kinematic viscosity has been given to the ratio (µ/p) because kinematics is defined as the study of motion without regard to the cause of motion and it concerned with length and time only. Likewise kinematic viscosity also involves the magnitudes of length and rime only.
#### Units:
In case of liquids, kinematic viscosity decreases with increase in temperature. In case of gases, it increases with increase in temperature.
Relative or Specific viscosity is the ratio of dynamic viscosity of any fluid to the dynamic viscosity of water at 20°C.
Since, water has a viscosity of 1, it is taken as standard substance for relative viscosity.
### Newton’s law of viscosity:
It states that the shear stress τ on fluid element, layer is directly proportional to the rate of shear strain. The constant of proportionality is called coefficient of viscosity.
The fluids which follow this law is known as Newtonian, fluids otherwise, it is known as Non-Newtonian fluids.
### Effects of temperature on viscosity:
Variation of viscosity with respect o temperature for some common fluids are given in fig 1.2(b).
In case of liquids, when the temperature increases the distance between molecules increases and the cohesive force decreases. So, viscosity of liquids decreases when the temperature increases.
In case of gases, the contribution to viscosity is more due to momentum transfer. When temperature increases, more molecules cross over with higher momentum differences. Hence, viscosity of gases increases with temperature. | HuggingFaceTB/finemath | |
Start typing, then use the up and down arrows to select an option from the list.
1. 1. Intro to General Chemistry 2. Temperature
Problem
# At a certain point, the Celsius and Fahrenheit scales “cross,” giving the same numerical value on both. At what tempera-ture does this crossover occur?
Relevant Solution 3m
Play a video:
Hi everyone. This problem reads the Celsius and Fahrenheit scales cross at a point where their numerical values are the same. What is the temperature at which the crossover happens? Okay, so this is what we want to figure out the temperature at which the crossover happens. So let's go ahead and write our equation for the two scales where they relate to each other and that equation is degrees Fahrenheit is equal to degrees Celsius times 9/5 plus 32. So the question is asking us for the point where the numerical values are the same. So this is where the crossover happens. So what that means is at the crossover degrees Fahrenheit is going to equal degrees Celsius and we want to know what is that value when they both equal each other. Okay, and so what we can do is we can set our equation, we can set it equal to each other, we can choose one of them to set equal to each other. So let's go ahead and choose degrees Fahrenheit. So wherever we see degrees Celsius, we're going to set that equal two F. And what that looks like when we write it out is degrees Fahrenheit is going to equal degrees Fahrenheit times 9/ plus 32. Okay, so we just replaced our degrees Celsius with degrees Fahrenheit, we set it equal to each other. So now what we want to do is we want to solve for degrees Fahrenheit. So let's go ahead and start off by bringing degrees Fahrenheit to one side. Alright, so what that's gonna look like when we bring degrees Fahrenheit to one side is we're going to have negative 32 is equal to negative degrees Fahrenheit times 9/5 minus degrees Fahrenheit. So see here we have our degrees Fahrenheit on one side of our equation. Now that we have it on one side of our equation, we're going to go ahead and take our minus degrees Fahrenheit and that's going to become minus one. Ok, so negative 32 equals negative degrees Fahrenheit times 9/5 minus one. Okay, so we can simplify this to be negative, 32 degrees equals negative degrees Fahrenheit, 9/5 minus one is 4/5. Okay, and now what we want to do is isolate our degrees Fahrenheit. Okay, so let's bring it over to the left side degrees. Fahrenheit is going to equal negative 32. Now we're going to flip the fraction because we're bringing it to the other side. So negative 32 times 5/4. So now we can go ahead and simplify this out to solve for degrees Fahrenheit and the value we get is negative 40. So our answer for this problem is that negative 40 F Which is also negative 40°C is where the crossover is going to happen. This is the temperature at which they're both going to equal or where the this is going to be the point at where the numerical values are the same. Okay, so that's it for this problem. I hope this was helpful. | HuggingFaceTB/finemath | |
#### What is 121 percent of 298?
How much is 121 percent of 298? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 121% of 298 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 121% of 298 = 360.58
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating one hundred and twenty-one of two hundred and ninety-eight How to calculate 121% of 298? Simply divide the percent by 100 and multiply by the number. For example, 121 /100 x 298 = 360.58 or 1.21 x 298 = 360.58
#### How much is 121 percent of the following numbers?
121 percent of 298.01 = 36059.21 121 percent of 298.02 = 36060.42 121 percent of 298.03 = 36061.63 121 percent of 298.04 = 36062.84 121 percent of 298.05 = 36064.05 121 percent of 298.06 = 36065.26 121 percent of 298.07 = 36066.47 121 percent of 298.08 = 36067.68 121 percent of 298.09 = 36068.89 121 percent of 298.1 = 36070.1 121 percent of 298.11 = 36071.31 121 percent of 298.12 = 36072.52 121 percent of 298.13 = 36073.73 121 percent of 298.14 = 36074.94 121 percent of 298.15 = 36076.15 121 percent of 298.16 = 36077.36 121 percent of 298.17 = 36078.57 121 percent of 298.18 = 36079.78 121 percent of 298.19 = 36080.99 121 percent of 298.2 = 36082.2 121 percent of 298.21 = 36083.41 121 percent of 298.22 = 36084.62 121 percent of 298.23 = 36085.83 121 percent of 298.24 = 36087.04 121 percent of 298.25 = 36088.25
121 percent of 298.26 = 36089.46 121 percent of 298.27 = 36090.67 121 percent of 298.28 = 36091.88 121 percent of 298.29 = 36093.09 121 percent of 298.3 = 36094.3 121 percent of 298.31 = 36095.51 121 percent of 298.32 = 36096.72 121 percent of 298.33 = 36097.93 121 percent of 298.34 = 36099.14 121 percent of 298.35 = 36100.35 121 percent of 298.36 = 36101.56 121 percent of 298.37 = 36102.77 121 percent of 298.38 = 36103.98 121 percent of 298.39 = 36105.19 121 percent of 298.4 = 36106.4 121 percent of 298.41 = 36107.61 121 percent of 298.42 = 36108.82 121 percent of 298.43 = 36110.03 121 percent of 298.44 = 36111.24 121 percent of 298.45 = 36112.45 121 percent of 298.46 = 36113.66 121 percent of 298.47 = 36114.87 121 percent of 298.48 = 36116.08 121 percent of 298.49 = 36117.29 121 percent of 298.5 = 36118.5
121 percent of 298.51 = 36119.71 121 percent of 298.52 = 36120.92 121 percent of 298.53 = 36122.13 121 percent of 298.54 = 36123.34 121 percent of 298.55 = 36124.55 121 percent of 298.56 = 36125.76 121 percent of 298.57 = 36126.97 121 percent of 298.58 = 36128.18 121 percent of 298.59 = 36129.39 121 percent of 298.6 = 36130.6 121 percent of 298.61 = 36131.81 121 percent of 298.62 = 36133.02 121 percent of 298.63 = 36134.23 121 percent of 298.64 = 36135.44 121 percent of 298.65 = 36136.65 121 percent of 298.66 = 36137.86 121 percent of 298.67 = 36139.07 121 percent of 298.68 = 36140.28 121 percent of 298.69 = 36141.49 121 percent of 298.7 = 36142.7 121 percent of 298.71 = 36143.91 121 percent of 298.72 = 36145.12 121 percent of 298.73 = 36146.33 121 percent of 298.74 = 36147.54 121 percent of 298.75 = 36148.75
121 percent of 298.76 = 36149.96 121 percent of 298.77 = 36151.17 121 percent of 298.78 = 36152.38 121 percent of 298.79 = 36153.59 121 percent of 298.8 = 36154.8 121 percent of 298.81 = 36156.01 121 percent of 298.82 = 36157.22 121 percent of 298.83 = 36158.43 121 percent of 298.84 = 36159.64 121 percent of 298.85 = 36160.85 121 percent of 298.86 = 36162.06 121 percent of 298.87 = 36163.27 121 percent of 298.88 = 36164.48 121 percent of 298.89 = 36165.69 121 percent of 298.9 = 36166.9 121 percent of 298.91 = 36168.11 121 percent of 298.92 = 36169.32 121 percent of 298.93 = 36170.53 121 percent of 298.94 = 36171.74 121 percent of 298.95 = 36172.95 121 percent of 298.96 = 36174.16 121 percent of 298.97 = 36175.37 121 percent of 298.98 = 36176.58 121 percent of 298.99 = 36177.79 121 percent of 299 = 36179 | HuggingFaceTB/finemath | |
# Circulation and lift (Kutta-Zhukovsky theorem)
In Eqn (3.52) it was shown that the lift / per unit span and the circulation Г of a spinning circular cylinder are simply related by
l = pVT
where p is the fluid density and V is the speed of the flow approaching the cylinder. In fact, as demonstrated independently by Kutta[12] and Zhukovsky*, the Russian physicist, at the beginning of the twentieth century, this result applies equally well to a cylinder of any shape and, in particular, applies to aerofoils. This powerful and useful result is accordingly usually known as the Kutta-Zhukovsky Theorem. Its validity is demonstrated below.
The lift on any aerofoil moving relative to a bulk of fluid can be derived by direct analysis. Consider the aerofoil in Fig. 4.7 generating a circulation of Г when in a stream of velocity V, density p, and static pressure po. The lift produced by the aerofoil must be sustained by any boundary (imaginary or real) surrounding the aerofoil.
For a circuit of radius r, that is very large compared to the aerofoil, the lift of the aerofoil upwards must be equal to the sum of the pressure force on the whole periphery of the circuit and the reaction to the rate of change of downward momentum of the air through the periphery. At this distance the effects of the aerofoil thickness distribution may be ignored, and the aerofoil represented only by the circulation it generates.
Fig. 4.7
The vertical static pressure force or buoyancy 4 on the circular boundary is the sum of the vertical pressure components acting on elements of the periphery. At the element subtending 66 at the centre of the aerofoil the static pressure is p and the local velocity is the resultant of V and the velocity v induced by the circulation. By Bernoulli’s equation
Po+jpV2 =p + jp[V2 + v2 + 2Fvsin 6]
giving
p = po – pVvsin#
if v2 may be neglected compared with V2, which is permissible since r is large.
The vertical component of pressure force on this element is
—pr sin 6 66
and, on substituting for p and integrating, the contribution to lift due to the force acting on the boundary is
л2тг
k = ~ (po – pFvsin6)rsin 6d6
Jo (4.7)
+pVvnr
with po and r constant.
The mass How through the elemental area of the boundary is given by pVr cos ObO. This mass flow has a vertical velocity increase of v cos 9, and therefore the rate of change of downward momentum through the element is —pVvr cos2 9b9 therefore by integrating round the boundary, the inertial contribution to the lift, /j, is
/і = + / /э Kit cos2 9d9
J о
= pVvrTT
Thus the total lift is:
I = 2pVvrir
From Eqn (4.5):
2nr
giving, finally, for the lift per unit span, /:
I = pVT
This expression can be obtained without consideration of the behaviour of air in a boundary circuit, by integrating pressures on the surface of the aerofoil directly.
It can be shown that this lift force is theoretically independent of the shape of the aerofoil section, the main effect of which is to produce a pitching moment in potential flow, plus a drag in the practical case of motion in a real viscous fluid. | HuggingFaceTB/finemath | |
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A recent survey showed that 50 percent of people polled
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A recent survey showed that 50 percent of people polled believe that elected officials should resign if indicted for a crime, whereas 35 percent believe that elected officials should resign only if they are convicted of a crime. Therefore, more people believe that elected officials should resign if indicted than believe that they should resign if convicted.
The reasoning above is flawed because it
(A) draws a conclusion about the population in general based only on a sample of that population
(B) confuses a sufficient condition with a required condition
(C) is based on an ambiguity of one of its terms
(D) draws a conclusion about a specific belief based on responses to
queries about two different specific beliefs
(E) contains premises that cannot all be true
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03 May 2006, 14:52
Clear A!
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03 May 2006, 15:55
I was also inclined to A... though OA is B...Don't know why?
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03 May 2006, 17:10
My guess was B...
For a conviction of a crime it is necessary for a person to be indicted for the crime. So of the 50% who believe an indictment is reason enough to resign 35% believe the person needs to be convicted in order to resign.
so 50-35=15% only believe an indictment is the only pre requisite for resignation.
Here the premise is in order for a person to be convicted an indictment is a required condition which is 35% of the people sampled and only 15% believe only an indictment is necessary which is the sufficient condition.
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03 May 2006, 17:51
Darrenbatty,thanks for the wonderful expanation!!
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03 May 2006, 19:29
darrenbatty wrote:
My guess was B...
For a conviction of a crime it is necessary for a person to be indicted for the crime. So of the 50% who believe an indictment is reason enough to resign 35% believe the person needs to be convicted in order to resign.
so 50-35=15% only believe an indictment is the only pre requisite for resignation.
Here the premise is in order for a person to be convicted an indictment is a required condition which is 35% of the people sampled and only 15% believe only an indictment is necessary which is the sufficient condition.
Wow. Awesome explaination man...
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03 May 2006, 21:13
thanks for explaination.
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Re: A recent survey showed that 50 percent of people polled [#permalink]
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29 Jun 2013, 18:45
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B is the correct answer, but not for the reasons outlined above.
This is a question from the June 1997 LSAT. This is the last question within section II of the exam. This was, for me, a difficult question, but after diagramming the argument out, it was fairly easy to identify the flaw within the reasoning.
The stimulus states:
"A recent survey showed that 50 percent of people polled believe that elected officials should resign if indicted for a crime, whereas 35 percent believe that elected officials should resign only if they are convicted of a crime. Therefore, more people believe that elected officials should resign if indicted than believe that they should resign if convicted."
Words like "if" and "only if" help to set off sufficient and necessary conditions within conditional reasoning. In this case, the first sentence states "A recent survey showed that 50 percent of people polled believe that elected officials should resign if indicted for a crime" (the "if" sets off a sufficient condition). Rearranging this statement, we can see that "if indicted for a crime (I), then that elected official should resign (R)"
When diagrammed, the statements above should look similar to this:
I--->R
The second sentence states that "35 percent believe that elected officials should resign (R) only if they are convicted of a crime (C)."
The phrase "only if" sets off a necessary condition. The diagram of this statement should look as follows:
R--->C
The conclusion of the argument says "Therefore, more people believe that elected officials should resign if indicted than believe that they should resign if convicted."
This (flawed) conditional statement would be diagrammed as follows:
Therefore, more people believe that I--->R than believe C--->R; or more simply I--->R, C--->R
The sufficient and necessary (required) conditions have been confused. A sound conclusion based on these statements would read:
"Therefore, more people believe that elected officials should resign if indicted than believe that they should resign only if convicted."
The correct diagram of these statements would look like this:
I--->R, R--->C
In short, the conclusion confuses a sufficient condition with a necessary (required) condition, as reflected in answer choice B.
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Re: A recent survey showed that 50 percent of people polled [#permalink] 29 Jun 2013, 18:45
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## College Algebra 7th Edition
$\frac{u^{2}+3u+1}{u+1}$
We form a common denominator: $\displaystyle u+1+\frac{u}{u+1}=\frac{(u+1)(u+1)}{u+1}+\frac{u}{u+1}=\frac{u^{2}+2u+1+u}{u+1}=\frac{u^{2}+3u+1}{u+1}$ | HuggingFaceTB/finemath | |
Zeke Zalcstein worked, before he retired a few years ago, on the boundary between mathematics and computational complexity. He started his career in mathematics, and then Zeke moved into computational complexity theory. His PhD advisor was John Rhodes who is an expert on many things, including the structure of semigroups; Rhodes has recently worked on the P=NP question himself; more on that in another post. The picture is John Rhodes since Zeke is camera shy.
I have known Zeke since I was a graduate student at Carnegie-Mellon and he was faculty there. Later, Zeke worked at a number of universities, including Stony-Brook, and then spend the latter part of his career as a program director at CISE/NSF.
Zeke loves to laugh, he is funny, he knows tons of mathematics–especially algebra, but he is a character in the best sense of the word. For example, he is very particular about what he eats and drinks, even his water cannot be just any water. When Zeke was at Stony-Brook he discovered that there was a water source in a certain open field that had the best water in the world. Not just Stony-Brook, the world. The water source was a faucet coming out of the ground, in an otherwise empty field. I asked Zeke where the water came from, the faucet was simply connected to a pipe that came out of the ground, no label, no sign, nothing. Did it come from a spring, from a well, or from the city water main? Zeke said he just knew it was great water.
Unfortunately, one day Zeke fell and broke his arm so it was impossible for him to drive for several weeks. While Zeke was incapacitated, a certain graduate student was kind enough to make periodic trips out to the faucet, fill up plastic jugs with the special water, and deliver them to Zeke.
I was once telling this story to John Hennessy, who is the President of Stanford University. He started to laugh, and confirmed the story: John had been the graduate student who was kind enough to help Zeke.
Enough, on to the main topic of today’s post. Zeke and I worked together on a number of problems over the years, and today I will talk about one that has a curious history, a neat proof, and a interesting application that never really happen.
The Problem and Our Results
The problem is: what is the space complexity of the word problem for the free group on two letters? This is not how we first heard the problem, but it is equivalent to what we were asked. The free group $F_2$ on two letters “a” and “b” is the group where the only relationships are:
$\displaystyle aa^{-1} = a^{-1}a = bb^{-1} = b^{-1}b = 1.$
It is called the free group since these are the minimal relations that can hold in any group. As usual the word problem is to determine given any word over ${a,a^{-1},b,b^{-1}}$ whether or not it equals ${1}$. For the rest, when we say word problem, we mean “the word problem for the free group on two letters.”
Zeke and I proved:
Theorem: The word problem is in ${\mathsf{DLOG}}$.
Actually we can proved more:
Theorem: A probabilistic log-space machine that has a one-way read only input tape can solve the word problem with error at most ${\epsilon}$, for any ${\epsilon>0}$.
There is a simple linear time algorithm for the word problem. The algorithm uses a pushdown store, which is initially empty. The algorithm processes each input symbol ${x}$ as follows: If the top of the pushdown is ${y}$ and ${xy=1}$, then pop off the top of the pushdown; if not, then push ${x}$ onto the pushdown. Then, go to the next input symbol. When there is no more input symbols accept only if the pushdown is empty. The algorithm clearly runs in linear time, and it is not hard to show that it is correct.
This algorithm uses linear space: a string that starts with many ${a}$‘s will, for example, require the pushdown to hold many symbols. Thus, the goal is to find a different algorithm that avoids using so much space.
My on-going agenda, in these posts, is to explain the role people play in doing research. In the next two sections I will explain the “curious history” of the question: who I “think” first asked the question, and why they may have asked the question. Then, I will explain the how we solved the problem and proved a theorem. You can skip the next two sections and get right to the proof method. No history. No background. But, I hope that you find the history and motivation interesting. Your choice.
My memory for technical conversations is usually pretty good, but this question on the space complexity of the word problem has a murky history. My version is that at a STOC conference, years ago, Juris Hartmanis mentioned a related problem to a number of us over a coffee break. I instantly liked the problem, and during the next few months Zeke and I worked on the problem, until we found a solution.
At the next conference, FOCS, we told Juris our solution to “his” problem. Juris said that he liked the problem, liked our solution, but he had never asked us the problem. Who knows. At least we did not call the paper: “On a Conjecture of Hartmanis.” I still believe, sometimes, that maybe Juris told it to us, but I must be confused. Anyway I think you will like the method we used to solve it.
The problem that someone asked us, was not what is the space complexity of the word problem for the free group on two letters. Instead we were asked a more “language” type question that is the same as the word problem. We were asked: what is the space complexity of a certain language ${D2}$?
Before defining ${D2}$ it may help to explain where ${D2}$ comes from. Define the following language ${D}$ over the alphabet that contains two types of parentheses: “${[}$” and “${]}$” and “${(}$” and “${)}$”. A string ${w}$ is in ${D}$ provided the application of the following rules eventually lead to the empty string:
$\displaystyle u[]v \rightarrow uv \text{ and } u()v \rightarrow uv.$
Thus,
$\displaystyle [ \quad [ \quad ( \quad ) \quad ] \quad]$
is in the language, but
$\displaystyle [ \quad ( \quad [ \quad ) \quad ] \quad ]$
is not. Note, ${D}$ is closed under concatenation.
The language ${D}$ consists of expressions with two types of parentheses that are well nested: sometimes it is called the Dyck language. Each ${[}$ must have a matching ${]}$ and also each ${(}$ must have a matching ${)}$. Moreover, the two types of parentheses must not get in each others way, thus
$\displaystyle \big[ \quad \big( \quad \big] \quad \big)$
is not in ${D}$.
The Dyck language ${D}$ is a context-free language, and is central to the theory of context-free language theory. Thus, another way to define ${D}$ is by the following context free grammar: (The symbol ${\lambda}$ is the empty string.)
$\displaystyle S \rightarrow \lambda \mid S \rightarrow SS \mid S \rightarrow [S] \mid S \rightarrow (S).$
The following is an easy theorem:
Theorem: The language ${D}$ is in ${\mathsf{DLOG}}$.
I will now explain how to prove this theorem; to do this we need the notion of matching parentheses. In a string ${w}$ say that ${w_{i}=[}$ matches ${w_{j}=]}$ provided ${i and as we count left-to-right ${j}$ is the first time that starting with ${i}$ that the number of ${[}$ and ${]}$ are equal. There is a similar definition for the parentheses ${(}$ and ${)}$. We call ${[}$ and ${(}$ left parentheses and ${]}$ and ${)}$ right parentheses.
Call a string ${w}$ good provided it satisfies the following:
1. for every left parentheses of either type there is a matching right one of the same type;
2. for every right parentheses of either type there is a matching left one of the same type;
3. if ${w_{i}}$ matches ${w_{j}}$ and ${w_{k}}$ matches ${w_{l}}$, then $(i,j)$ and $(k,l)$ are disjoint intervals or one wholly contains the other.
The last means that there is no overlapping matches. It should be clear that checking whether or not a string is good can be done easily in log-space.
Lemma: A string ${w}$ is good if and only if ${w}$ is in ${D}$.
It is easy to see that all strings in ${D}$ are good. So suppose that there is a good string that is not in ${D}$. Select the shortest possible string ${w}$. The first symbol of ${w}$ must be a left parentheses by (2). So assume that ${w_{1} = [}$. There must be a right parentheses in ${w}$ by (1) so assume that ${w_{i}}$ is the first one. Clearly, ${i>1}$. Suppose that ${w_{i} = )}$. Then, the previous ${w_{i-1}}$ must equal ${[}$; otherwise, ${w}$ would not be the shortest possible counterexample. Thus,
$\displaystyle w=[ \quad x \quad [_{1} \quad )_{2} \quad y$
where ${x}$ are all left parentheses. By (2) the ${)_{2}}$ matches some earlier ${(}$, but the ${[_{1}}$ matches some later ${]}$ and this contradicts (3).
Next suppose that ${w_{i} = ]}$. Then, the previous ${w_{i-1}}$ must equal ${(}$; otherwise, ${w}$ would not be the shortest possible counterexample. Thus,
$\displaystyle w=[ \quad x \quad (_{1} \quad ]_{2} \quad y$
where again ${x}$ are all left parentheses. By (2) the ${]_{2}}$ matches some earlier ${[}$, but the ${(_{1}}$ matches some later ${)}$ and this contradicts (3).
Now let us finally define the language ${D2.}$ It is a variation of the Dyck language that allows a different type of pairing of the symbols. This difference makes proving that it lies in ${\mathsf{DLOG}}$ harder. This is the question, someone asked us–I guess not Juris. Sometimes ${D2}$ is called a “two-sided” Dyck language.
A string ${w}$ is in ${D2}$ provided the application of the following rules eventually lead to the empty string:
$\displaystyle u[]v \rightarrow uv \text{ and } u()v \rightarrow uv$
and
$\displaystyle u][v \rightarrow uv \text{ and } u)(v \rightarrow uv.$
The point of ${D2}$ is that the parentheses have a type but no “direction.”
The trouble with ${D2}$ is there does not seem to be a counterpart to the notion of matching; a string of the form:
$\displaystyle \dots [ w ] \dots$
could have the ${[}$ match with a ${]}$ that comes earlier. This is why I do not know a direct counting method to recognize ${D2}$. I guess this is why someone asked us the question. I wish I could remember who.
Our Solution
Zeke and I knew two insights, one trivial and the other based on a classic result from group theory. The first is that ${D2}$ is really the same as accepting the words from a free group on two letters.
$\displaystyle [ \rightarrow a \text{ and } ] \rightarrow a^{-1}$
and
$\displaystyle ( \rightarrow b \text{ and } ) \rightarrow b^{-1}.$
The second insight is that the free group on two letters has a faithful representation over ${2 \times 2}$ integer valued matrices; this result was proved by Ivan Sanov in 1947.
Theorem: There are two integer matrices $A,B$ so that the mapping ${a \rightarrow A}$ and ${b \rightarrow B}$ is a faithful representation of the free group on ${a,b}$.
Actually we can construct the matrices explicitly. Consider the following two matrices: $A$ is
1 2 0 1
and $B$ is:
1 0 2 1
Both matrices are invertible and further their inverses are also integer matrices The map that sends
$\displaystyle a \rightarrow A \text{ and } a^{-1} \rightarrow A^{-1} \text{ and } b \rightarrow B \text{ and } b^{-1} \rightarrow B^{-1}$
defines a mapping $\rho$ from the free group on two letters, $F_2$ to $\mathsf{SL}(2,\mathsf{Z})$: the latter is the $2 \times 2$ matrices over the integers with determinant $1.$ This mapping $\rho : F_2 \to \mathsf{SL}(2,\mathsf{Z})$ is a group isomorphism.
That means that we can replace the word problem by: does a sequence of matrices over ${A,B,A^{-1},B^{-1}}$ equal the identity matrix ${I}$. This transformation of the word problem into a problem about matrices is the key to our results. For example,
$\displaystyle aba^{-1}b \rightarrow ABA^{-1}B.$
Here is how Zeke and I use this transformation to prove the our theorem. We show how to check whether or not
$\displaystyle M = M_{1} \times M_{2} \times \dots \times M_{n}$
is equal to ${I}$ where each ${M_{i}}$ is from ${A,B,A^{-1},B^{-1}}$. The obvious way to do this is to compute matrix ${M}$ and see if it is equal to ${I}$. The difficulty with this approach is that the matrix ${M}$ may have entries that are too large and cannot be stored in log-space.
We solve this with the Chinese Remainder Theorem. Suppose that ${p}$ is a prime with at most ${O(\log n)}$ bits. Let ${M_{p} = M \bmod p}$. A log-space machine can pass over the input and compute the product ${M_{p}}$: this means that we do all the arithmetic modulo ${p}$, but we still are multiplying ${2 \times 2}$ matrices. Then, the machine checks whether or not ${M_{p} = I}$. If it does not, then clearly ${M \neq I}$, and we can reject the input. The machine does this for all primes ${p}$ of the given size. If all ${M_{p} \equiv I \bmod p}$, then the machine accepts.
We claim that this algorithm is correct. The insight is that if ${M}$ is not equal to ${I}$, then ${M-I}$ has some non-zero entry, which cannot be too large. Then, by the Chinese Remainder Theorem we have a contradiction, and so the algorithm is correct. This uses the well known fact that
$\displaystyle \prod_{p \le t} p \ge c^{t}$
where the product is over primes and ${c>1}$.
Finally, the probabilistic result follows in essentially the same way. The only difference is that now the machine randomly selects one prime ${p}$ to check. We then argue that the non-zero entry of ${M-I}$ is unlikely to be divisible by a randomly selected prime. This relies the simple fact that the non-zero entry can only have a polynomial number of prime factors. Thus, as long as we randomly select the prime from a large enough set, it is unlikely that the non-zero entry will be $0$ modulo the prime.
Karp-Rabin
What does the Karp-Rabin pattern matching algorithm have to do with our result on the word problem? Indeed. Dick Karp and Michael Rabin are both famous for so many things, but one of my favorites of their results is their randomized pattern matching algorithm.
Rabin has told how they first thought of their algorithm. They needed a way to hash strings that had a certain property, and our mapping from the free group to matrices modulo a prime worked perfectly. So at one time they were using the same technology that we used to solve the word problem in log-space.
Unfortunately for Zeke and I, they quickly got rid of the matrix ideas and replaced them by a much simpler method. But Michael has repeatedly told me that the matrix ideas played a role in his initial thinking.
Open Problems
Our proof method could handle a larger class of word problems, than just the word problem for free groups. Suppose that ${G}$ is any linear group over the rationals. Then, the word problem for this group can also be solved in ${\mathsf{DLOG}}$. One can even prove more, but see our paper for the details. Thus, many infinite groups have word problems that can be done in log-space.
The power of matrix representation theory is something that you may be able to use to solve other problems. I believe that we have not made as much use of the power of representation methods in other parts of computer science as perhaps we could. In any event the mapping $\rho$ may itself be useful to you.
Now if I could only recall who initially asked me the problem ${\dots}$
April 16, 2009 9:49 am
Lovely post!
April 16, 2009 3:53 pm
Great post! I love the Karp-Rabin algorithm, it’s beautifully simple, one of the many examples Rabin has given of the power of randomness (your post that mentioned his O(n) closest pair in 2d space algorithm was also interesting). If you pay attention to parameters, it’s also a useful building block for many neat string matching algotithms (finding all palindromes in a long string for example).
Interesting to know the relation to your work on the word problem. Somehow the more complicated solution came first.
April 16, 2009 4:06 pm
One question about balance lemma: suppose we have w = “(((((“, it seems it satisfies mentioned condition, but indeed it doesn’t belong to D.
April 16, 2009 7:20 pm
I am sorry, I made a silly error in the post. The definition and the Balance Lemma are slightly off. First, the definition needs to allow [][][] to be in the language–I have fixed this. Second, the Balance Lemma needs to be fixed too: the string w itself must be [-balanced and (-balanced.
I will update post in am. Thanks for reading so closely. Again sorry for the errors…rjlipton
Did you ever think that a post is unlucky? I seem to be having a great deal of trouble getting this one right. I again an sorry for any trouble or confusion.
• September 24, 2016 5:18 pm
Is this post unlucky? At least the following statement is also wrong: “This mapping $$\rho : F_2 \to \mathsf{SL}(2,\mathsf{Z})$$ is a group isomorphism.” It is not an isomorphism, but only a monomorphism. The matrices in the image are congruent to the identity modulo 2…
4. April 16, 2009 4:52 pm
for D and D2 and in particular the Balance Lemma, is it correct that you’re also assuming the languages are closed under concatenation? e.g. is ([]()) in D?
April 18, 2009 5:10 am
“A probabilistic log-space machine that has a one-way read only input tape”
Thought provoking! So this is the first streaming algorithm 🙂
April 18, 2009 7:41 am
Ah, in your fix, you seem to have cut out the part where you defined D2…
April 19, 2009 3:27 pm
It could be added that the one-sided Dyck languages are in TC0
(Lynch’s algorithm seen as a circuit) while the two-sided ones
are NC1-hard as shown in Robinson’s thesis (1993, San Diego).
April 20, 2009 7:58 am
Is the group generated by the two 2×2 matrices
[3 5]
[0 5]
and
[2 0]
[0 3]
free?
April 20, 2009 8:07 am
I do not know if these matrices are free? I see the issue. Is there an application in mind. As you may know the prove that the ones I mention are free relies on showing that the sequence of matrices is coded into the one of the entries of the matrix product.
April 20, 2009 10:37 am
It’s just an open problem that I don’t know how to solve. Seems amazing that such a simple problem could cause difficulties. | open-web-math/open-web-math | |
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# So previously we talked about expected values, and their properties, and a little bit of how to calculate them.
Now let's talk about the expected value operator itself, and the expected value operator is in fact a linear operator, and that'll greatly simplify calculating expected values for some relatively complicated things. So, loomis in the'A' and'B' are not random numbers. So, when you think about'A' and'B', you think'A' as some thing like five, it's this number you can plug in. And'x' and'Y' are two random variables. Then expective value of'AX plus B' works out to be'A' expective value of'X' plus'B', exactly what you hope it work out to be. And expective value of'X' plus'Y' works out to be expective value of'X' and expective value of'Y'. And the reason it works out in these cases is because the expected is a linear operator, it is not always the case that expected value of g of x is equal to g of expected value of x, where g is some general function that's not linear. This can happen with specific random variables and specific values of g, but in general it's not the case. The most, sort of, famous example where it's not the case is that expected value of x squared is not equal to expected value of x, whole thing quantity squared. Now, let's talk about what the difference of these two entities is. What do we mean by the difference between these two things? Here x is a random variable. X-squared is the random variable you obtain by squaring x. So, for example, if x is a die roll, it can take values one, two, three, four, five, six. X-squared then can take the values one, four, nine, and it takes those values with probability one-sixth each. So the expected value of X squared represents the expected value of the squared random variable. On the other hand, expected value of X quantity squared, represents what you obtain if you first calculate the expected value of X, and then square the result. And these two things are not equal. This is a, well-known example where expected value of g of x is not equal to g of expected value of x. And we'll see in a couple slides why it is
a well-known example of that property. But in general, I would like you to remember that if g is not a linear function then you can't just commute expected values outside of g to the inside of g. And that, that rule would generally hold. If it's, if it's linear, if g is a linear function you can always do it. If g is not a linear, just in general things that you cannot. The expected value rule. Hold, no matter what. Constitutes X and Y. X could be discrete, continuous, mixed discrete and continuous. Y could be discrete, continuous, and mixed and the rules still holds. So, let me go through an example, supposed you flip a coin X, and as we normally do, X is zero is it's tails and one if it's head and you simulate a uniform random number Y. The random number is between zero and one. What's the expected value of their sum. Well the sum of a coin flip and a uniform random variable is weird distribution. It's not obvious, especially if all you've had is the handful of lectures from this class, how you would calculate that distribution, and then from that distribution then calculate the expected value. However, we do know how to calculate the expected value of a uniform random variable and the expected value of a coin flip, and so the expected value of their sum is the sum of their expected values. We know the expected value of the coin flip is.5. We know that the expected value of the uniform random variable is.5, so that expected value is one. So you can see how these. Expected value operator rules make calculating things associated with expected values a lot easier. Another example is, suppose you role a die twice. What is the expected value of the average of two die roles. So you often roll two dies when you're playing a board game, for example. Okay, let's let x1 be the result of the first die and x2 be the result of the second die. Now the variable that we're interested in, let's call it y equal to x1 plus x2, divided by two. Now, one way you could calculate the
expected value of y is to figure out what the distribution was of the average of two die rolls. So, let me give you a sense of this really quickly. The reason we think the distribution of a single die roll is one sixth at each number, is if you roll a die a lot of times you get about, one sixth of the, of the die rolls are one, one sixth are two, one sixth are three, one sixth are four, and so on. And, and, and then kind of geometrically we are modelling the process as if they're all equally likely, and so that's why we're going to model the population of die rolls as having probability one sixth on each number. Now this implies a distribution on the average of two die rolls, right? That the smallest number it could take is one, right? One plus one divided by two, this, the average of if you were to get two 1s. And the largest it could take is, is six plus six divided by two or six if you were to roll two 6s. But it takes different values in between, and it, and it's not equally likely. For all the, all the numbers in between. A one has probability 136, but some of the middle values have higher probabilities. So this, any rate, our variable y itself has a distribution. And you could get a pretty good sense of it. Maybe you could do this by taking two dice, rolling them, taking the average, rolling them again, taking the average, doing that over and over and over again, and prob, plotting, you know, a bar plot of the frequency of the, the averages that you get. And that would give you a good sense of what the population distribution is. Or you could work it out on pen and paper as to what, what the distribution actually is. And then once you get that worked out, then you could use your expected value formula to calculate the expected value of y directly by doing summation overall the possible values of y times p of y and calculate its expected value. Another way to do it is to directly use the expected value of linear operator rule. So in this case, expected value of x1 plus x2 divided by two is one half expected value of x1 plus the expected value of x2
because the one half is the non random variable that we could just pull out and the expected value goes across the two sums here to get expected value of x1 plus expected value of x2. That then yields 3.5 plus 3.5 divided by two, which is 3.5. Now you might be wanting, wondering. After hearing this it's, "oh, that's interesting." You'd expect a value of the average of two die rolls is the exact same as the expected value of an individual die roll. And that is exactly the case, but you're probably thinking, "Maybe does this extend beyond that. Is the expected value of the average of N die rolls equal to the 3.5 as well." And the, the answer is yes, that is, that's exactly true. In, in fact, as a nice segue way into our next slide. Where we actually derive. The property that we were hinting at in the previous slide. Namely that the expected value of the average of a collection of random variables from the same distribution, is the same as the, the expected value of the individual random variables. So lets let XI, for I equal one to N, be a collection of random variables, each. Each from a distribution would mean mu. I just wanna also point out that we tend to use Greek letters to represent, population quantities, in this case the population mean of the distribution is mu. So lets calculate the expected value of the sample average of the XI. Well, we want the expected value of the sample average which is one over N, summation I equals one to the N to the XI's. The one over N pulls out because its not random. The expected value commutes across the sum. And the expected value of each of those x I's, is itself mu. So we get the summation I equals one to n of mu. We get mu added up n times then which is n mu divided by n on the outside, so we get mu. So what this says is, it doesn't matter what the distribution of the individual x's is The distribution of the mean of the Xes, has the same mean as the, the, individual means. So let me just summarize one more time.
The expected value of the sample mean, is the population mean that it's trying to estimate. The population mean of the distribution of the sample mean of N observations is exactly the population mean that it's trying to estimate. And so when this happens. When the expected value of an estimator is what it's trying to estimate, that's a good thing. We say that the estimator is itself unbiased. So sample means are unbiased estimators of population means. And again, there were some assumptions for this to be true, right? All the axis have to be from a distribution that has mean . U being the value you want estimate and then the. The sample mean is, is an unbiased estimator of the population mean and we finally getting to the point where we can talk about how we're going to connect our probability modeling to the data that we observed. We're not quite there yet but we're getting closer and closer to this and I want you to remember that we're throwing around the term mean a lot. And I want, and so if you get confused, I want you to qualify the mean that we're talking about, whether it's a population quantity. By component of the probability distribution, or a sample quantity, an empirical quantity that you connect from the data. And remember our goal in probability modeling is to connect our sample observations to the population using our probability model. | HuggingFaceTB/finemath | |
# Prove that two triangles are congruent if a median and the angles between the median and the including Sides of the one are respectively congruent to the median and the angles between it and the including sides of the other.
Prove that two triangles are congruent if a median and the angles between the median and theincluding Sides of the one are respectively congruent to the median and the angles between it andthe including sides of the other.
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A triangle having two sides of equal length is called an isosceles triangle. The two angles adjacent to the third side of an isosceles triangle are equal. If a triangle has two equal angles, it is an isosceles triangle, i.e. it has two equal sides too.
In an isosceles triangle, the median, bisectrix, and height of the vertex between the equal sides all coincide. This median/bisectrix/height divides the isosceles triangle into two congruent right triangles. If a median coincides with a height, or if a bisectrix coincides with a height, or if a median coincides with a bisectrix), then the triangle is isosceles - the adjacent sides are equal to each other..
If BD is the median and the height of the triangle ABC, then triangles ABD and CBD are equal because they are right triangles whose both legs are equal, respectively. Therefore, AB = BC. If BD is a bisectrix and is also a height, then triangles ABD and CBD are equal because they are right triangles with one common leg and equal acute angles. If BD is a median and a bisectrix of triangle ABC, then triangles ABD and CBD are equal because all three angles of ABD and CBD are equal, respectively, and AD = CD.
On to the problem, in order to compare congruency between to different triangles we need to remember that two triangles are congruent (equal) if they have identical size and shape so that they can be exactly superimposed. Two congruent triangles ABC and CDA with a common side and equal angles located at two different ends of the common side always form a parallelogram. | HuggingFaceTB/finemath | |
## Essential University Physics: Volume 1 (3rd Edition)
$\omega_{min}=\sqrt{\omega_{max}^2-1.24\times \frac{1}{R}}$
When the disk is at the top of its path, the center of mass is shifted up by $\frac{1}{30}R$. The mass of the disk is $\frac{15}{16}M$, since $M-(\frac{1}{4})^2M =\frac{15}{16}M$. Thus, using conservation of energy, it follows: $I\omega_{max}^2=I\omega_{min}^2+\frac{15}{16}M(\frac{1}{30}R)g$ Thus, using the result from problem 65 and simplifying, we find: $\omega_{min}=\sqrt{\omega_{max}^2-1.24\times \frac{1}{R}}$ | HuggingFaceTB/finemath | |
# Quant Quiz For SSC CGL Exam 2017
Q1. The price of a commodity has increased by 60%. By what percent must a consumer reduce the consumption of the commodity so as not to increase the expenditure?
(a) 37%
(b) 37.5%
(c) 40.5%
(d) 60%
Q2. Two pipes can fill a tank with water in 15 and 12 hours respectively and a third pipe can empty it in 4 hours. If the pipes be opened in order at 8, 9 and 11 a.m. respectively, the tank will be emptied at
(a) 11:40 a.m.
(b) 12:40 p.m.
(c) 1:40 p.m.
(d) 2:40 p.m.
Q4. A book vendor sold a book at a loss of 20%. Had he sold it for Rs. 108 more, he would have earned a profit of 30%. Find the cost price of the book?
(a) Rs. 216
(b) Rs. 648
(c) Rs. 240
(d) Rs. 432
Q5. An article is sold at a loss of 10%. Had it been sold for Rs. 9 more there would have been a gain of 12(1/2)% on it. The cost price of the article is:
(a) Rs. 40
(b) Rs. 45
(c) Rs. 50
(d) Rs. 35
Q6. A student finds the average of ten 2 digit numbers. While copying numbers, by mistake, he writes one number with its digits interchanged. As a result, his answer is 1.8 less than the correct answer. What is the difference of the digits of the number, in which he made mistake, is:
(a) 2
(b) 3
(c) 4
(d) 6
Q7. If a man walks at the rate of 5 km/hr, he misses a train by 7 minutes. However, if he walks at the rate of 6 km/hr, he reaches the station 5 minutes before the arrival of the train. The distance covered by him to reach the station is
(a) 6 km
(b) 7 km
(c) 6.25 km
(d) 4 km
Q8. Ramesh bought 10 cycle for Rs. 500 each. He sold five of them for Rs. 750 each and the remaining for Rs. 550 each. Then the total gain or loss percentage is
(a) Gain of 81/3%
(b) Loss of 30%
(c) Gain of 30%
(d) Loss of 7(1/7)%
Q9. A, B and C share the profit in the ratio of 1/4 ∶1/6 ∶7/12. If C retires, then A and B share the profit of C in the ratio of 4 : 5 respectively. The new profit sharing ratio of A and B will be:
(a) 55 : 53
(b) 53 : 55
(c) 5 : 3
(d) 3 : 5
Q10. A runs 1(2/3) times faster than B. A gives 60 metres start to B and both finish the race at same time. Then find the length of race-course.
(a) 156 m
(b) 150 m
(c) 180 m
(d) 160 m
Q11. The ratio of the present ages of a son and his father is 1 : 5 and that of his mother and father is 4 : 5. After 2 years the ratio of the age of the son to that of his mother becomes 3 : 10. What is the present age of the father?
(a) 30 years
(b) 28 years
(c) 37 years
(d) 35 years
Q12. The ratio of the present ages of P and Q is 8 : 5. After 6 years their ages will be in the ratio of 3 : 2. Find the ratio of the sum and difference of the present ages of P and Q.
(a) 13 : 3
(b) 39 : 19
(c) 13 : 2
(d) 13 : 5
Q13. Two types of alloy possess gold and silver in the ratio of 7 : 22 and 21 : 37. In what ratio should these alloys be mixed so as to have a new alloy in which gold and silver would exist in the ratio 25 : 62?
(a) 13 : 8
(b) 8 : 13
(c) 13 : 12
(d) 6 : 9
Q14. For a triangle base is 6√3 cm and two base angles are 30° and 60°. Then height of the triangle is
(a) 3√3 cm
(b) 4.5 cm
(c) 4√3 cm
(d) 2√3 cm
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# Problem Solving Using Equations So the equation will look like this: (c 3) (2c 3) = 33 Now that we have an equation, we can solve for the unknown variable, c.
In other words, let c represent Charlie's age today.
Since Brian is twice as old as Charlie today, then let 2c represent Brian's age today.
Let's try θ = 30°: sin(−30°) = −0.5 and −sin(30°) = −0.5 So it is true for θ = 30° Let's try θ = 90°: sin(−90°) = −1 and −sin(90°) = −1 So it is also true for θ = 90° Is it true for all values of θ?
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In this problem, we don't know Jack's or Diane's age.
Since Jack's age is expressed in relation to Diane's age (in this problem, Jack is two years older than Diane), so then our variable will be based on Diane's age. If Jack is two years older than Diane, then Jack's age must be: d 2.
We must substitute into Brian's equation to determine his current age. Therefore, Brian must be 18 years old (2c = 2 x 9).
To check our answer, we must substitute into the original equation to verify that the left side of the equation equals the right side.
You can solve these types of problems in four steps.
First, we can express what we don't know as a variable, an alphabetical representation of what we're trying to solve for. | HuggingFaceTB/finemath | |
# Fundamental Theorem of Calculus (FTC)
Author: John J Weber III, PhD Corresponding Textbook Sections:
• Section 4.9 – Antiderivatives
• Section 5.3 – The Fundamental Theorem of Calculus
• Section 5.4 – Indefinite Integrals and the Net Change Theorem
### Expected Educational Results
• Objective 01–01: I can state the Fundamental Theorem of Calculus.
• Objective 01–02: I can explain the meaning of the Fundamental Theorem of CalculusPart I.
• Objective 01–03: I can explain the meaning of the Fundamental Theorem of CalculusPart II.
• Objective 01–04: I can determine the properties of an area function using FTC-I and Calculus I.
• Objective 01–05: I can evaluate indefinite integrals using the Fundamental Theorem of CalculusPart I.
• Objective 01–06: I can evaluate definite integrals using the Fundamental Theorem of CalculusPart II.
• Objective 01–07: I can use the Net Change Theorem to identify the meaning of ${\int }_{a}^{b}{f}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(x\right)\phantom{\rule{0.167em}{0ex}}\text{d}x$$\int_a^b{f^{\,\prime}(x)\,\text{d}x}$.
• Objective 01–08: Given velocity of an object in motion along a line, I can find the object’s displacement and the total distance traveled.
• Objective 01–09: Given acceleration of an object in motion along a line, I can find the object’s velocity, displacement, and the total distance traveled.
### Bloom’s Taxonomy
A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.
## Using FTC-I
#### Investigation 12
Use what you learned from Activity 11 to answer the following:
Let $g\left(x\right)={\int }_{a}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$$g(x) = \int_a^x{f(t)\,dt}$, where the graph of $f\left(t\right)$$f(t)$ is shown below:
1. Identify the graph of ${g}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(x\right)$$g^{\,\prime}(x)$. Explain.
2. For what value(s) of $x$$x$, does $g\left(x\right)$$g(x)$ have relative (local) minimum(s)? Explain.
3. For what value(s) of $x$$x$, does $g\left(x\right)$$g(x)$ have relative (local) maximum(s)? Explain.
4. For what value(s) of $x$$x$, does $g\left(x\right)$$g(x)$ have absolute (global) minimum(s)? Explain.
5. For what value(s) of $x$$x$, does $g\left(x\right)$$g(x)$ have absolute (global) maximum(s)? Explain.
6. On what interval(s) of $x$$x$, is $g\left(x\right)$$g(x)$ increasing? Explain.
7. On what interval(s) of $x$$x$, is $g\left(x\right)$$g(x)$ decreasing? Explain.
8. On what interval(s) of $x$$x$, is $g\left(x\right)$$g(x)$ concave up? Explain.
9. On what interval(s) of $x$$x$, is $g\left(x\right)$$g(x)$ concave down? Explain.
10. For what value(s) of $x$$x$, does $g\left(x\right)$$g(x)$ have inflection point(s)? Explain.
### Homework
At this time, you should be able to complete the following assignments:
• Section 5.3: # 3. | HuggingFaceTB/finemath | |
1. 4^(2log(base8)7), equivalent expression, integers
Simplify 4^(2log(base8)7) into the form a^(b/c), all integers.
Could someone point me in the right direction?
2. Originally Posted by jsmith90210
Simplify 4^(2log(base8)7) into the form a^(b/c), all integers.
Could someone point me in the right direction?
First, convert 2Log8(7) > Log8(49)
Then, I believe you can use the following rule:
$\displaystyle a^{\text{Logb}(c)}=c^{\text{Logb}(a)}$
Therefore,
$\displaystyle 4^{\text{Log8}(49)}=49^{\text{Log8}(4)}$
Log8(4) = 3/2
So, you have
49^(3/2)
---------------------------------
Josh
---------------------------------
3. Originally Posted by Invertible
First, convert 2Log8(7) > Log8(49)
Then, I believe you can use the following rule:
$\displaystyle a^{\text{Logb}(c)}=c^{\text{Logb}(a)}$
Therefore,
$\displaystyle 4^{\text{Log8}(49)}=49^{\text{Log8}(4)}$
Log8(4) = 3/2 ...... Unfortunately no.
So, you have
49^(3/2)
---------------------------------
Josh
---------------------------------
......
4. Originally Posted by earboth
......
My mistake, I believe that:
Log8(4) = 2/3
$\displaystyle 49^{(2/3)}$
5. what is the name of that rule?
6. Hello, jsmith90210!
A different approach . . .
Simplify $\displaystyle 4^{2\log_87}$ into the form $\displaystyle a^{\frac{b}{c}}$, all integers.
We have: .$\displaystyle 4^{\log_8(7^2)} \:=\:4^{\log_8(49)}$ .[1]
Let $\displaystyle \log_8(49) = p \quad\Rightarrow\quad 8^p \,=\,49\quad\Rightarrow\quad \left(2^3\right)^p \,=\,49 \quad\Rightarrow\quad 2^{3p} \,=\,49$
. . Then: .$\displaystyle 3p \,=\,\log_2(49) \quad\Rightarrow\quad p \:=\:\tfrac{1}{3}\log_2(49)$
. . Hence: .$\displaystyle \log_8(49) \:=\:\tfrac{1}{3}\log_2(49)$
Substitute into [1]: .$\displaystyle 4^{\frac{1}{3}\log_2(49)} \;=\;\left(2^2\right)^{\frac{1}{3}\log_2(49)}$
. . . . . $\displaystyle =\;2^{\frac{2}{3}\log_2(49)} \;=\; 2^{\log_2(49^{\frac{2}{3}})} \;=\;\boxed{49^{\frac{2}{3}}}$
This answer is correct ... it satisfies the requirements of the problem.
But I bet that "they" simplified it further: .$\displaystyle 49^{\frac{2}{3}} \:=\:\left(7^2\right)^{\frac{2}{3}} \:=\:7^{\frac{4}{3}}$ | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## Subtract 1/8 from 9/35
9/35 - 1/8 is 37/280.
#### Steps for subtracting fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 35 and 8 is 280
2. For the 1st fraction, since 35 × 8 = 280,
9/35 = 9 × 8/35 × 8 = 72/280
3. Likewise, for the 2nd fraction, since 8 × 35 = 280,
1/8 = 1 × 35/8 × 35 = 35/280
4. Subtract the two fractions:
72/280 - 35/280 = 72 - 35/280 = 37/280
#### Subtract Fractions Calculator
-
Use fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS | HuggingFaceTB/finemath | |
# Thread: Limits
1. ## Limits
So unfortunately, I missed one class and that was when we were learning about limits. The only thing I understand is that the limit does not exist if going towards an x value, the y value's are positive and negative.
So I understand that this:
would equal 1, right? (I hope, or I really need to get help with this stuff)
But I don't understand what I'm supposed to do with this:
Would it equal 2? it says x doesn't equal two... but does. I'm confused.
Thanks to all who reply
2. Originally Posted by forsheezy
So unfortunately, I missed one class and that was when we were learning about limits. The only thing I understand is that the limit does not exist if going towards an x value, the y value's are positive and negative.
So I understand that this:
would equal 1, right? (I hope, or I really need to get help with this stuff)
But I don't understand what I'm supposed to do with this:
Would it equal 2? it says x doesn't equal two... but does. I'm confused.
Thanks to all who reply
For the first one direct substitution of the limit yields a determinant form. Determinant form just means it has no problems like divide by zero etc.
So $\lim_{x\to{1}}x+2=(1)+2=3$
For the second one a limit is defined as the value as $x\to{c}$ what $f(x)\to$
So the exact value at x=2 is not equivalent to limit since f(x) is not continous at x=2
But $\lim_{x\to{c}}f(x)\text{ }\exists\text{ }\text{iff}\text{ }\lim_{x\to{c^{-}}}f(x)=\lim_{x\to{c^{+}}}f(x)$
And as it does $\lim_{x\to{2^{-}}}f(x)=4-(2)=2=\lim_{x\to{2^{+}}}f(x)$
$\therefore\lim_{x\to{2}}f(x)=2$
3. Thank You
I somewhat understand it now! | open-web-math/open-web-math | |
# How To Find The Value Of X In Angles In Transversal References
How To Find The Value Of X In Angles In Transversal References. If ∠cmq = 60°, find all other angles. X° angle pairs created by parallel lines cut by a transversal.
Alternate exterior angles are the pair of angles that are formed on the outer side of two lines but on the opposite side of the transversal. Thus, the corresponding angles are. Because the lines are parallel, the measures of these angles are equal.
### ∠ U A N D ∠ Z.
Algebra can be used to find unknown values in angles formed by a transversal and parallel lines. 📈the image shows parallel lines cut by a. They appear on opposite sides of the.
### Here We Have Been Provided With A Figure Shown Below And We Are Asked To Find The Values Of X.
In the following diagram, two parallel lines are cut by a transversal. (ii) ∠apd (iii) ∠bpd (iv) ∠bpc solution: X = 20 to find the measure of each angle, substitute the value of “x” back in the algebraic expressions.
### Discover Creative Projects Anyone Can Make.
Well, the first thing that we might realize is that, look, corresponding angles are equivalent. Now, since angle x was split into angles p and q so the value of x will be equal to the sum of the value of these two angles obtained. Find the value of the variable(s) in each figure.
### Alternating Angles Are Pairs Of Angles In Which Both Angles Are Either Interior Or Exterior.
Finding the measure of an angle given parallel lines. Steps involved in finding the value of x calculator is as follows: ∠ q a n d ∠ v.
### ∠1 And The 75° Angle Are Vertical Angles.
How to find the value of x in angles in transversal. Angle x and y must be equal since k and l are parallel with a single line transecting them.angles 1 and 2 are corresponding angles, m∠1 = 45°, and m∠2 = (x + 25)°.angles 3 and 4 are alternate inter If ∠cmq = 60°, find all other angles. | HuggingFaceTB/finemath | |
Resurrectionofgavinstonemovie.com
Live truth instead of professing it
# What is the maximum volume of a cylinder inscribed in a sphere?
## What is the maximum volume of a cylinder inscribed in a sphere?
A sphere of fixed radius (R) is given. Let r and h be the radius and the height of the cylinder respectively. Hence, the volume of the cylinder is the maximum when the height of the cylinder is 2 R / √ 3.
## What is the maximum volume of a cylinder?
Therefore the volume is a maximum when 2r−2h+h=0, so h=2r and hr=2. Show activity on this post. Show activity on this post. Let the ratio of height to radius be ρ, then h=ρr.
How many sphere can fill a cylinder?
All Answers (5) One can from there estimate how many such spheres will fit. Your cylinder has volume V_c=pi*D^3. Your spheres are each V_s= 4/3*pi*d^3.
How many spheres can fit in a sphere?
Sphere packing in a sphere
Number of inner spheres Maximum radius of inner spheres Packing density
Approximate
1 1.0000 1
2 0.5000 0.25
3 0.4641… 0.29988…
### What is the volume of the sphere to the cylinder?
Today, we are blessed with the understanding Calculus, and formulas for the volumes of basic shapes are taught to all in High School. What pleased Archimedes so much about the solution is that the ratio of the volume of the sphere to the cylinder is the same as the ratio of the surface area of the sphere to the surface area of the cylinder!
### How to find the volume of a cylinder using the Pythagorean theorem?
By the Pythagorean theorem, the radius of the cylinder is given by r2 = R2 − (h 2)2. The volume of the cylinder is hence V = πr2h = π (hR2 − h3 4). Differentiating with respect to h and equating to 0 to find extrema gives dV dh = π (R2 − 3h2 4) = 0 ∴ h0 = 2R √3.
What is the volume of a hollow cylinder?
Tadaaam! The volume of a hollow cylinder is equal to 742.2 cm 3. Remember that the result is the volume of the paper and the cardboard. If you want to calculate how much plasticine you can put inside the cardboard roll, use the standard formula for the volume of a cylinder – the calculator will calculate it in the blink of an eye!
How do you find the volume of a slanted cylinder?
Find the radius, side length and slant angle of the cylinder. Square the radius. Multiply the result by pi. Take the sin of the angle. Multiply the sin by the side length. Multiply the result from step 3 and 5 together. The result is the slanted volume. | HuggingFaceTB/finemath | |
# 39 Polynomial Functions and Graphs Objective Students will
• Slides: 12 3/9 Polynomial Functions and Graphs Objective: Students will be able to determine the characteristics of a polynomial function. Essential Question: If your smile was represented by a polynomial function, what are the characteristics of that function? Higher Degree Polynomial Functions and Graphs Polynomial Function A polynomial function of degree n in the variable x is a function defined by where each ai is real, an 0, and n is a whole number. ¡ ¡ ¡ an is called the leading coefficient n is the degree of the polynomial a 0 is called the constant term Polynomial Functions Polynomial Function in General Form Degree Name of Function 1 2 3 4 Linear Quadratic Cubic Quartic Degree: largest exponent Number of solutions: degree Turning Points: degree – 1 Leading coefficient and degree determine the end behaviors. End Behavior Let a be the leading coefficient of a polynomial function P. n odd n even 1. If a > 0, the graph of P falls on the left and rises on the right. 1. If a > 0, the graph of P opens up. 2. If a < 0, the graph of P rises on the left and falls on the right. 2. If a < 0, the graph of P opens down. Polynomial Functions f(x) = 3 Constant. Function Degree: Max. # of Turning Points: Max(s): Min(s): End behaviors: Polynomial Functions f(x) = x + 2 Linear. Function Degree: Number of Solutions: X-intercept(s): Y-intercepts: Max. # of Turning Points: Max(s): Min(s): End behaviors: Polynomial Functions f(x) = x 2 + 3 x + 2 Quadratic. Function Degree: Number of Solutions: X-intercept(s): Y-intercepts: Max. # of Turning Points: Max(s): Min(s): End behaviors: Polynomial Functions f(x) = x 3 + 4 x 2 + 2 Cubic Function Degree: Number of Solutions: X-intercept(s): Y-intercepts: Max. # of Turning Points: Max(s): Min(s): End behaviors: Polynomial Functions f(x) = -x 3 + 2 x 2 – x – 2 Cubic Function Degree: Number of Solutions: X-intercept(s): Y-intercepts: Max. # of Turning Points: Max(s): Min(s): End behaviors: Polynomial Functions f(x) = x 4 - 5 x 2 + 4 Quartic Function Degree: Number of Solutions: X-intercept(s): Y-intercepts: Max. # of Turning Points: Max(s): Min(s): End behaviors: Polynomial Functions f(x) = -2 x 4 + 8 x 3 – 8 x 2 + 2 Quartic Function Degree: Number of Solutions: X-intercept(s): Y-intercepts: Max. # of Turning Points: Max(s): Min(s): End behaviors: Polynomial Functions f(x) = Quintic Function Degree: Number of Solutions: X-intercept(s): Y-intercepts: Max. # of Turning Points: Max(s): Min(s): End behaviors: | HuggingFaceTB/finemath | |
# 19) The Ka of benzoic acid is 6.30 × 10-5. The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassiu?
0
19) The Ka of benzoic acid is 6.30 × 10-5. The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassium benzoate and 50.0 mL of 1.00 M benzoic acid is __.
A) 1.705
B) 0.851
C) 3.406
D) 4.201
E) 2.383
• pKa = -log(6.30e-5)
pKa = 4.20
0.050L * 1.00M = 0.050mols benzoic acid
0.050L * 1.00M = 0.050mols salt
0.050mols / 0.1L = 0.50M
0.050mols / 0.1 = 0.50M
pH = 4.20 + log(0.50/0.50)
pH = 4.20
Choice D
Source(s): Live4
• use the H-H equation for first area. you may extremely sparkling up for the ratio Then sparkling up for the A-, it extremely is likewise the quantity of salt that replaced into created and is likewise the quantity of base that replaced into further. Convert the molarity to moles
• moles benzoic acid = 0.05 L x 1 mol/L = 0.05 moles
moles benzoate = 0.05 L x 1 mol/L = 0.05 moles
total volume = 50 ml + 50 ml = 100 ml = 0.1 L
[benzoate] = 0.05 mol/0.1 L = 0.5 M
[benzoic acid] = 0.05 mol/0.1 L = 0.05 M
pH = pKa + log [salt]/[acid]
pKa = -log Ka = 4.2
pH = 4.2 + log 0.5/0.5
pH = 4.2
Also Check This Solve this inequality: 3q + 11 + 8q > 99.? | HuggingFaceTB/finemath | |
# Solve $(2^x+3^x+5^x)^3=160^x$
The question is to find all real solutions of the equation:
$$(2^x+3^x+5^x)^3=160^x$$
Using Wolfram, I can check that $x=3$ is the only solution, but I'm having trouble trying to find it by hand.
First idea was try to use AM-GM inequality. Some algebric manipulation also didn't work.
Any idea?
• Isn't it enough to divide both sides by $160^x$ and exploit convexity? – Jack D'Aurizio Jun 13 '17 at 19:49
## 2 Answers
It's $f(x)=0$, where $$f(x)=\left(\frac{2}{\sqrt[3]{160}}\right)^x+\left(\frac{3}{\sqrt[3]{160}}\right)^x+\left(\frac{5}{\sqrt[3]{160}}\right)^x-1.$$ But $f$ is a decreasing function and the rest for you.
• Is there any way to find the answer without testing? – Arnaldo Jun 13 '17 at 19:58
• @Arnaldo Since $f$ is a decreasing function, our equation has one root maximum. But $3$ is a root, which gives the answer: $\{3\}$. It's not testing, it's just solution of your problem. – Michael Rozenberg Jun 13 '17 at 20:01
• What I meant is once you know that you have just one solution, you then started to test which value of $x$ could give you the solution. Did you do that process? – Arnaldo Jun 13 '17 at 20:10
• @Arnaldo The graph of a decreasing function can cross the $x$- axis in one point maximum. Draw it! We see that $(3,0)$ is a common, Thus, it's an unique point. – Michael Rozenberg Jun 13 '17 at 20:13
• Thanks Michael, we are talking the same thing but just using different words. – Arnaldo Jun 13 '17 at 20:14
The equation can be written as
$$\left(\frac{2}{160^{1/3}}\right)^x + \left(\frac{3}{160^{1/3}}\right)^x + \left(\frac{5}{160^{1/3}}\right)^x = 1$$
The left side is a decreasing function of $x$, since $$\frac{2}{160^{1/3}} < \frac{3}{160^{1/3}} < \frac{5}{160^{1/3}} < 1$$
• Is there any way to find the answer without testing? – Arnaldo Jun 13 '17 at 19:59
• @Arnaldo: This answer proves there can only be one solution. So once we have found one, then we are sure to be done. – mathreadler Jun 13 '17 at 20:50
• @mathreadler: You are right. My question was how to find such solution. After talk a little with Michael, I'm convinced that if we don't use a numerical method we must have luck to have a easy solution and find it by testing some values of $x$ or use some graph interpretation. – Arnaldo Jun 13 '17 at 21:01
• If you want a closed-form solution, there will be very little hope unless $x$ is an integer. Since $160^x$ must be the cube of an integer, that integer must be divisible by $3$. So the first thing to try is $x=3$. – Robert Israel Jun 13 '17 at 21:20
• Thank you Robert. You got my point. – Arnaldo Jun 13 '17 at 21:30 | HuggingFaceTB/finemath | |
# Using Lagrange's Mean Value theorem find a point on the curve y=(x-3)^2 where tangent is parallel to chord joining (3,0) and (4,1).
1
by komalarora22
2015-05-29T22:51:55+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Lagrange''s theorem states that for a function f (x) = y, defined over [a, b] which is continuous in the above range and the derivative f '(x) is defined over (a , b) :
there exists a point X, at which : the derivative is such that:
f '(X) = [ f (b) - f(a) ] / (b - a) = slope of the function at x = X
given f (x) = (x - 3)²
f '(x) = 2 (x - 3) ---- (1)
the interval is x = [ 3, 4 ]
b = 4 and a = 3
f(a) = 0 and f(b) = 1
f '(X) = [ 1 - 0 ] / [ 4 - 3] = 1
So using equation (1), 2 (x - 3) = 1
=> x = 7/2
The point on the curve y = (x - 3)² needed is (7/2, 1/4)
click on thanks button above please
tnxx kvnmurty | HuggingFaceTB/finemath | |
# What are the critical points, if any, of f(x,y) = 2x^3 - 6x^2 + y^3 + 3y^2 - 48x - 45y?
Mar 26, 2017
$\left.\begin{matrix}f \left(- 2 - 5\right) & = 231 & \implies \left(- 2 - 5 231\right) \\ f \left(- 2 3\right) & = - 25 & \implies \left(- 2 3 25\right) \\ f \left(4 - 5\right) & = 15 & \implies \left(4 - 5 15\right) \\ f \left(4 3\right) & = - 241 & \implies \left(4 3 - 241\right)\end{matrix}\right.$
#### Explanation:
The theory to identify the extrema of $z = f \left(x , y\right)$ is:
1. Solve simultaneously the critical equations
$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0 \setminus \setminus \setminus$ (ie ${f}_{x} = {f}_{y} = 0$)
2. Evaluate ${f}_{x x} , {f}_{y y}$ and ${f}_{x y} \left(= {f}_{y x}\right)$ at each of these critical points. Hence evaluate $\Delta = {f}_{x x} {f}_{y y} - {f}_{x y}^{2}$ at each of these points
3. Determine the nature of the extrema;
$\left.\begin{matrix}\Delta > 0 & \text{There is minimum if " f_(x x)>0 \\ \null & "and a maximum if " f_(x x)<0 \\ Delta<0 & "there is a saddle point" \\ Delta=0 & "Further analysis is necessary}\end{matrix}\right.$
So we have:
$f \left(x , y\right) = 2 {x}^{3} - 6 {x}^{2} + {y}^{3} + 3 {y}^{2} - 48 x - 45 y$
Let us find the first partial derivatives:
$\frac{\partial f}{\partial x} = 6 {x}^{2} - 12 x - 48$
$\frac{\partial f}{\partial y} = 3 {y}^{2} + 6 y - 45$
So our critical simultaneous equations are:
$\frac{\partial f}{\partial x} = 0 \implies 6 {x}^{2} - 12 x - 48 = 0$
$\frac{\partial f}{\partial y} = 0 \implies 3 {y}^{2} + 6 y - 45 = 0$
$\therefore {x}^{2} - 2 x - 8 = 0$
$\therefore \left(x - 4\right) \left(x + 2\right) = 0$
$\therefore x = - 2 , 4$
$\therefore {y}^{2} + 2 y - 15 = 0$
$\therefore \left(y + 5\right) \left(y - 3\right) = 0$
$\therefore y = - 5 , 3$
Any permutation of these solutions will simultaneously make both partial derivatives vanish, so the critical values are:
$\left(x , y\right) = \left(- 2 , - 5\right) , \left(- 2 , 3\right) , \left(4 , - 5\right) , \left(4 , 3\right)$
So we can now calculate the coordinates of the critical points:
$\left.\begin{matrix}f \left(- 2 - 5\right) & = 231 & \implies \left(- 2 - 5 231\right) \\ f \left(- 2 3\right) & = - 25 & \implies \left(- 2 3 25\right) \\ f \left(4 - 5\right) & = 15 & \implies \left(4 - 5 15\right) \\ f \left(4 3\right) & = - 241 & \implies \left(4 3 - 241\right)\end{matrix}\right.$
We can visualise these critical points on a 3D-plot:
As we were not asked to determine the nature of the critical points I will omit that analysis. | open-web-math/open-web-math | |
# Solve the IVP: 1. (e^x + 2y)dx + (2x - siny)dy = 0, y(0) = 2 pi 2. {dy}/{dx}={e^x}/{1 + e^x},...
## Question:
Solve the IVP:
{eq}1. (e^x + 2y)dx + (2x - siny)dy = 0, y(0) = 2 \pi\\ 2. \frac{dy}{dx}=\frac{e^x}{1 + e^x}, y(0)=4 {/eq}
## Separable Differential Equation:
An equation involving a dependent variable {eq}y {/eq}, an independent variable {eq}x {/eq} and derivative of the dependent variable with respect to an independent variable is known as a differential equation.
A separable differential equation is of the type {eq}\dfrac{dy}{dx} = f \left( x \right) g \left( y \right) {/eq}.
Then, this differential equation can be solved by separating {eq}x \text{ and } y {/eq} variables and then integrating with respect to their derivatives.
Become a Study.com member to unlock this answer!
1]
Given the differential equation is
{eq}\left( e^{x}+2y \right)dx+ \left( 2x- \sin y \right)dy = 0 {/eq}
This... | HuggingFaceTB/finemath | |
Home » How Many Times Does 4 Go Into 7? New
# How Many Times Does 4 Go Into 7? New
Let’s discuss the question: how many times does 4 go into 7. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## How many times can 4 go into?
Simple, 4 goes into 4, 8, 12, and 16.
7 into 28
7 into 28
## How many times does 4 become 2?
We know that 2 goes into 4 twice (4 ÷ 2 = 2) and we know that 1 goes into 4 four times (4 ÷ 1 = 4), but 5 does not go into 4 because 5 is larger than 4.
## How many times can 4 go into 15?
Consider the math: 4 goes into 15 three times with remainder 3.
## What can go into 8?
What are the factors of 8? The factors of 8 are 1, 2, 4 and 8.
## What is 4 divided by half?
In other words – four divided by one half = eight.
## How many fours is 60?
Answer. therefore 15 times 4 goes in 60.
## How many times can 5 go 40?
If we go back to our multiplying, we might be able to remember that 5 goes into 40 exactly 8 times.
## How many times can 4 go 18?
Instead of saying 18 divided by 4 equals 4.5, you could just use the division symbol, which is a slash, as we did above. Also note that all answers in our division calculations are rounded to three decimals if necessary.
## How do you write 7 divided by 4?
7 divided by 4 is 1 ¾ as a fraction or 1.75 as a decimal.
## What is the 7th multiple of 7?
Multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, … and so on. How do we get to know a number is multiple of another number?
List of Multiples of 7.
Multiplication Multiples of 7
7 x 8 56
7 x 9 63
7 x 10 70
7 x 11 77
## Can 4 be divided by 2?
Using a calculator, if you typed in 4 divided by 2, you’d get 2.
### Learning Unit 2-2 Video
Learning Unit 2-2 Video
Learning Unit 2-2 Video
## What is the answer for 7 divided by 2?
Using a calculator, if you typed in 7 divided by 2, you’d get 3.5. You could also express 7/2 as a mixed fraction: 3 1/2. If you look at the mixed fraction 3 1/2, you’ll see that the numerator is the same as the remainder (1), the denominator is our original divisor (2), and the whole number is our final answer (3).
See also How To Make Money Online Fast From Home make money online at home
## How many 4s is 84?
Answer: If you go by the number’s face value, there one 4. If you go by the number’s place value, there are 84 / 4 fours i.e. 21 fours.
## Can 6 be divided by 4?
Answer: 6 divided by 4 is 1.5.
## What can go into 10?
The factors of 10 are 1, 2, 5, and 10. You can also look at this the other way around: if you can multiply two whole numbers to create a third number, those two numbers are factors of the third. 2 x 5 = 10, so 2 and 5 are factors of 10. 1 x 10 = 10, so 1 and 10 are also factors of 10.
## What is the factor of 7?
The factors of 7 are 1 and 7. As the number 7 is a prime number, the factors of 7 are one and the number itself.
## What can go into 81?
Factors of 81 are 1, 3, 9, 27, and 81. 1 is a universal factor. It is a factor of all numbers.
## What is 7 divided by half?
Half of 7 would be 3 1/2, or in decimal form, 3.5. Finding ‘half’ of a number is the same as dividing by 2.
## What is the answer when 4 1 2?
The correct answer for this is 9/2. To figure out the answer you first multiply the whole number times the denominator.
## What is ¼ divided by 4?
In other words – one quarter divided by four = one sixteenth.
## How many times can 6 Enter 50?
How many times does 6 go into 50 and what is the remainder? Answer. It goes in 8 times with remainder 2 (because 50 = 6 × 8 + 2).
Long division
Long division
## How many times does 6 go 60?
How many times does 6 go 60? There are 10 times 6 in 60.
## How many times 4 is written in numbers from 1 to 60?
The digit four appears once in each of the following: 4, 14, 24, 34, 40, 41, 42, 43, 45, 46, 47, 48, 49, 54, 64, 74, 84 and 94 for a total of 18 times.
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## Thomas' Calculus 13th Edition
$\dfrac{1}{3} \ln |y^3-2|= x^3+c$
Rewrite the given equation and then integrate. We have $\int 3x^2 dx =\int \dfrac{y^2}{(y^3-2)}dy$ This can also be written as: $\int 3x^2 dx =(\dfrac{1}{3}) \int \dfrac{3y^2}{(y^3-2)}dy$ Now, $x^3+c =\dfrac{1}{3} \int \dfrac{3y^2}{(y^3-2)}dy$ Suppose $a=y^3 \implies da =3y^2 dy$ $x^3+c =\dfrac{1}{3} \int \dfrac{3y^2}{(y^3-2)}dy \implies \dfrac{1}{3} \int \dfrac{da}{a-2}=x^3+c \implies \dfrac{1}{3} \ln |a-2|= x^3+c$ Hence, $\dfrac{1}{3} \ln |y^3-2|= x^3+c$ | HuggingFaceTB/finemath | |
Eliciting a Dirichlet Distribution
We illustrate the process of eliciting a Dirichlet distribution using the methodology and case study in Zapata-Vazquez, R., O’Hagan, A. and Bastos, L. S. (2014). Eliciting expert judgements about a set of proportions. Journal of Applied Statistics 41, 1919-1933. Quoting from their Section 3.4,
This example concerns the efficacy of a new antibiotic in patients who are hospitalised in the Pediatric Intensive Care Unit (PICU) and who are severely infected by pneumococci (which is associated with pneumonia, meningitis, and septicaemia, among other conditions). The possible results after the infection are: to survive in good condition, to have a sequel, or to die. An expert is asked to provide judgements about the proportions of patients who will have each of these possible results. Denoting these proportions by $$\pi_1$$, $$\pi_2$$, $$\pi_3$$, these form a set of proportions that must sum to 1.
The Dirichlet distribution is parameterised by $f(\pi_1,\pi_2,\pi_3)\propto \prod_{i=1}^3\pi_i^{d_i-1},$ with $$n=\sum_{i=1}^3 d_i$$.
The elicited judgements for the three marginal proportions were
Lower quartile 0.50 0.22 0.11
Median 0.55 0.30 0.15
Upper quartile 0.60 0.35 0.20
For each marginal proportion $$\pi_i$$, the expert has provided a lower quartile, a median and an upper quartile, so we define a single vector of probabilities, specifying which quantiles have been elicited.
p1 <- c(0.25, 0.5, 0.75)
We then define one vector for each marginal proportion, giving the values of the elicited quantiles.
v.good <- c(0.5, 0.55, 0.6)
v.seql <- c(0.22, 0.3, 0.35)
v.dead <- c(0.11, 0.15, 0.2)
Next we fit probability distributions to each set of elicited quantiles.
library(SHELF)
fit.good <- fitdist(vals = v.good, probs = p1, lower = 0, upper = 1)
fit.seql <- fitdist(vals = v.seql, probs = p1, lower = 0, upper = 1)
fit.dead <- fitdist(vals = v.dead, probs = p1, lower = 0, upper = 1)
The objects and all include parameters of fitted beta distributions, for example,
fit.good\$Beta
## shape1 shape2
## 1 24.88927 20.4087
We can now fit the Dirichlet distribution to the elicited marginals.
d.fit <- fitDirichlet(fit.good, fit.seql, fit.dead,
n.fitted = "opt")
##
## Directly elicted beta marginal distributions:
##
## shape1 24.9000 6.2000 4.6000
## shape2 20.4000 14.7000 24.4000
## mean 0.5490 0.2960 0.1590
## sd 0.0731 0.0975 0.0667
## sum 45.3000 20.9000 29.0000
##
## Sum of elicited marginal means: 1.004
##
## Beta marginal distributions from Dirichlet fit:
##
## shape1 16.6000 8.9600 4.8000
## shape2 13.8000 21.4000 25.6000
## mean 0.5470 0.2950 0.1580
## sd 0.0889 0.0814 0.0651
## sum 30.4000 30.4000 30.4000
The above plot shows both the marginal distributions that were elicited directly, and the the marginal distributions resulting from the Dirichlet fit. Parameters and summaries from these two sets of distributions are shown as output. We see that the marginal distribution for the ‘Dead’ proportion hasn’t changed appreciably, but that the Dirichlet fit has resulted in a little more uncertainty for the ‘Good outcome’ proportion, and a little less uncertainty for the ‘Sequel’ proportion.
The Dirichlet parameters are stored in , but can be read off from the row: we have $$d_1=16.6, d_2=8.96, d_3=4.8$$ (the values have been rounded for display purposes).
We can report feedback from the marginal distributions of the fitted Dirichlet:
feedbackDirichlet(d.fit, quantiles = c(0.1, 0.5, 0.9))
## [1,] 0.1 0.43 0.19 0.08
## [2,] 0.5 0.55 0.29 0.15
## [3,] 0.9 0.66 0.40 0.25
so, for example, after fitting the Dirichlet distribution, the fitted median and 90th percentile for the proportion of ‘good outcomes’ are 0.55 and 0.66 respectively.
The parameter $$n$$ was chosen by minimising the sum of squared differences between the marginal standard deviations in the elicited marginal beta distributions and the marginals from the fitted Dirichlet. An alternative, more conservative choice is to set $$n$$ as the minimum of the sum of the beta parameters in each elicited marginal. From the output above, we can see that this will correspond to the ‘Sequel’ proportion.
d.fit <- fitDirichlet(fit.good, fit.seql, fit.dead,
n.fitted = "min")
##
## Directly elicted beta marginal distributions:
##
## shape1 24.9000 6.2000 4.6000
## shape2 20.4000 14.7000 24.4000
## mean 0.5490 0.2960 0.1590
## sd 0.0731 0.0975 0.0667
## sum 45.3000 20.9000 29.0000
##
## Sum of elicited marginal means: 1.004
##
## Beta marginal distributions from Dirichlet fit:
##
## sum 20.900 20.9000 20.9000 | HuggingFaceTB/finemath | |
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2224-Sec13_3-HWT
# 2224-Sec13_3-HWT - Mat h 22 24 Multiva ri a ble C alc ul us...
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Math 2224 Multivariable Calculus – Sec. 13.3: Area by Double Integration I. Review from 1206 A. Area Between the Curves Using Vertical Rectangles; Integrating wrt x 1. If f and g are continuous on [ a,b ] and g(x) < f(x) for all x in [ a,b ], then the area of the region bounded by the graphs of f and g and the vertical lines x=a and x=b is A = f x ( ) ! g x ( ) [ ] a b " dx ; where f(x) is the upper curve & g(x) is the lower curve. 2. Steps to Find the Area Between Two Curves (wrt x ) a. Graph the curves and draw a representative rectangle. This reveals which curve is f (upper curve) and which is g (lower curve). It also helps find the limits of integration if you do not already know them. b. Find the limits of integration; you may need to find the pts of intersection. c. Write a formula for f(x)–g(x). Simplify it if you can. d. Integrate [ f(x)–g(x) ] from a to b . The number you get is the area. B. Area of a Region Between Two Curves - Integrating with respect to y .
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Associated Topics || Dr. Math Home || Search Dr. Math
### Graphing the Absolute Value/Square Root of a Function
```Date: 07/26/2002 at 09:32:33
From: Daniel Brostin
Subject: Abs value/square root of functions
Can you illustrate and discuss how taking the absolute value or the
square root of a function affects the graph of the function?
Thanks for the help!
```
```
Date: 07/26/2002 at 13:04:48
From: Doctor Peterson
Subject: Re: Abs value/square root of functions
Hi, Daniel.
Think about what the absolute value does to a number. If the number is
positive, its absolute value is the number itself: the absolute value
function does not change it. If the number is negative, its absolute
value is the negative of the number (which is positive): the absolute
value function reflects the number in the origin, sending it to the
point the same distance away but on the opposite side.
So what happens if you take the absolute value of a function? Wherever
the function is positive (or zero), the graph stays right where it
was. Wherever the function is negative, the graph is reflected up over
the x-axis, so that its y coordinate is positive rather than negative:
| ___ | ___
| __/ \ + __/ \ /
| / \ |\ / \ /
+-+---------+-- ----> +-+---------+--
|/ \ |
+ \ |
| |
Now how about the square root? In this case, if f(x) is negative, its
square root does not exist (since we are presumably talking about real
functions). You can just erase that part of the graph! If f(x) is
positive, the square root will squash it or stretch it vertically,
depending on its value. Look at the graph of y=sqrt(x), compared with
the graph of y=x. You will see that if x<1, sqrt(x) is larger, while
when x>1, sqrt(x) is smaller. So when f(x)<1, sqrt(f(x)) will be
above f(x), and when f(x)>1, sqrt(f(x)) will be below f(x):
| ___ | ___
1+ __/ \ 1+ __/ \
| / \ | / \
+-+---------+-- ----> +-+---------+--
|/ \ |
+ \ |
| |
I can't get enough detail in my graph to show the stretching. But you
can put a dot on the graph where f(x)=1, so the graph of the square
root will go through that point; then you can warp the given graph
upward where it is below y=1 and downward where it is above y=1. The
peaks and valleys of the new graph will lie at the same x coordinates
as in the original graph, but will be at different heights.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Equations, Graphs, Translations
High School Functions
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# Théorème du point fixe de Kleene
Kleene fixed-point theorem This article is about Kleene's fixed-point theorem in lattice theory. For the fixed-point theorem in computability theory, see Kleene's recursion theorem. Computation of the least fixpoint of f(X) = 1 / 10 x2+atan(X)+1 using Kleene's theorem in the real interval [0,7] with the usual order In the mathematical areas of order and lattice theory, the Kleene fixed-point theorem, named after American mathematician Stephen Cole Kleene, states the following: Kleene Fixed-Point Theorem. Supposer {style d'affichage (L,sqsubseteq )} is a directed-complete partial order (dcpo) with a least element, et laissez {style d'affichage f:Lto L} be a Scott-continuous (and therefore monotone) fonction. Alors {style d'affichage f} has a least fixed point, which is the supremum of the ascending Kleene chain of {displaystyle f.} The ascending Kleene chain of f is the chain {displaystyle bot sqsubseteq f(bot )sqsubseteq f(F(bot ))sqsubseteq cdots sqsubseteq f^{n}(bot )sqsubseteq cdots } obtained by iterating f on the least element ⊥ of L. Expressed in a formula, le théorème dit que {style d'affichage {textrm {lfp}}(F)=sup left(la gauche{f ^{n}(bot )mid nin mathbb {N} droit}droit)} où {style d'affichage {textrm {lfp}}} denotes the least fixed point.
Although Tarski's fixed point theorem does not consider how fixed points can be computed by iterating f from some seed (aussi, it pertains to monotone functions on complete lattices), this result is often attributed to Alfred Tarski who proves it for additive functions [1] En outre, Kleene Fixed-Point Theorem can be extended to monotone functions using transfinite iterations.[2] Preuve[3] We first have to show that the ascending Kleene chain of {style d'affichage f} exists in {displaystyle L} . To show that, we prove the following: Lemme. Si {displaystyle L} is a dcpo with a least element, et {style d'affichage f:Lto L} is Scott-continuous, alors {style d'affichage f^{n}(bot )sqsubseteq f^{n+1}(bot ),nin mathbb {N} _{0}} Preuve. We use induction: Assume n = 0. Alors {style d'affichage f^{0}(bot )=bot sqsubseteq f^{1}(bot ),} puisque {displaystyle bot } is the least element. Assume n > 0. Then we have to show that {style d'affichage f^{n}(bot )sqsubseteq f^{n+1}(bot )} . By rearranging we get {style d'affichage f(f ^{n-1}(bot ))sqsubseteq f(f ^{n}(bot ))} . By inductive assumption, we know that {style d'affichage f^{n-1}(bot )sqsubseteq f^{n}(bot )} détient, and because f is monotone (property of Scott-continuous functions), the result holds as well.
As a corollary of the Lemma we have the following directed ω-chain: {style d'affichage mathbb {M} ={bot ,F(bot ),F(F(bot )),ldots }.} From the definition of a dcpo it follows that {style d'affichage mathbb {M} } has a supremum, appeler {displaystyle m.} What remains now is to show that {style d'affichage m} is the least fixed-point.
Première, nous montrons que {style d'affichage m} is a fixed point, c'est à dire. ce {style d'affichage f(m)=m} . Car {style d'affichage f} is Scott-continuous, {style d'affichage f(sup(mathbb {M} ))= sup(F(mathbb {M} ))} , C'est {style d'affichage f(m)= sup(F(mathbb {M} ))} . Aussi, puisque {style d'affichage mathbb {M} =f(mathbb {M} )Coupe {bot }} and because {displaystyle bot } has no influence in determining the supremum we have: {displaystyle sup(F(mathbb {M} ))= sup(mathbb {M} )} . Il s'ensuit que {style d'affichage f(m)=m} , fabrication {style d'affichage m} a fixed-point of {style d'affichage f} .
The proof that {style d'affichage m} is in fact the least fixed point can be done by showing that any element in {style d'affichage mathbb {M} } is smaller than any fixed-point of {style d'affichage f} (because by property of supremum, if all elements of a set {displaystyle Dsubseteq L} are smaller than an element of {displaystyle L} then also {displaystyle sup(ré)} is smaller than that same element of {displaystyle L} ). This is done by induction: Présumer {style d'affichage k} is some fixed-point of {style d'affichage f} . We now prove by induction over {style d'affichage i} ce {displaystyle forall iin mathbb {N} :f ^{je}(bot )sqsubseteq k} . The base of the induction {style d'affichage (je=0)} obviously holds: {style d'affichage f^{0}(bot )=bot sqsubseteq k,} puisque {displaystyle bot } is the least element of {displaystyle L} . As the induction hypothesis, we may assume that {style d'affichage f^{je}(bot )sqsubseteq k} . We now do the induction step: From the induction hypothesis and the monotonicity of {style d'affichage f} (encore, implied by the Scott-continuity of {style d'affichage f} ), we may conclude the following: {style d'affichage f^{je}(bot )sqsubseteq k~implies ~f^{je+1}(bot )sqsubseteq f(k).} À présent, by the assumption that {style d'affichage k} is a fixed-point of {style d'affichage f,} we know that {style d'affichage f(k)=k,} and from that we get {style d'affichage f^{je+1}(bot )sqsubseteq k.} See also Other fixed-point theorems References ^ Alfred Tarski (1955). "A lattice-theoretical fixpoint theorem and its applications". Pacific Journal of Mathematics. 5:2: 285–309., page 305. ^ Patrick Cousot and Radhia Cousot (1979). "Constructive versions of Tarski's fixed point theorems". Pacific Journal of Mathematics. 82:1: 43–57. ^ Stoltenberg-Hansen, V; Lindstrom, JE.; Griffor, E. R. (1994). Mathematical Theory of Domains by V. Stoltenberg-Hansen. la presse de l'Universite de Cambridge. pp. 24. est ce que je:10.1017/cbo9781139166386. ISBN 0521383447. Catégories: Order theoryFixed-point theorems
Si vous voulez connaître d'autres articles similaires à Théorème du point fixe de Kleene vous pouvez visiter la catégorie Théorèmes du point fixe.
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Matrix Theory...showing that matrix is Unitary
cylers89
New member
Let x be an element in space C be a given unit vector (x*x=1) and write x=[x,yT]T, where x1 is element in space C and y is element in Cn-1. Choose theta (element in space R) such that ei(theta)x1 greater than or equal to 0 and define z=ei(theta)x =[z1, BT]T, where z1 is element in R is non negative and B is element in Cn-1. Show that the matrix V is unitary.
V= [z1 B* ]
[B -I+((1)/(1+z1)) BB*]
Can someone help me get off on the right foot?
I know that to show unitary, I can prove that VTV=I.
So I can do the VTV.
What I don't understand is how to impliment what z= into my VTV. Would it be better to substitute in z1 in the beginning, or simplify VTV first, then plug in? Does (x*x=1) imply that it works for (B*B) as well?
Opalg
MHB Oldtimer
Staff member
Let $x$ be an element in space C be a given unit vector ($x^*x=1$) and write $x=[x_1,y^{\text{T}}]^{\text{T}}$, where $x_1$ is element in space $C$ and $y$ is element in $C^{n-1}$. Choose $\theta$ (element in space $\mathbb R$) such that $e^{i\theta}x_1$ greater than or equal to 0 and define $z=e^{i\theta}x =[z_1, B^{\text{T}}]^{\text{T}}$, where $z_1$ is element in $\mathbb R$ is non negative and $B$ is element in $C^{n-1}$. Show that the matrix $V$ is unitary.
$$V = \begin{bmatrix}z_1&B^* \\ B & -I + (1/(1+z_1))BB^* \end{bmatrix}$$
Can someone help me get off on the right foot?
I know that to show unitary, I can prove that $V^*V=I$.
So I can do the $V^*V$.
What I don't understand is how to impliment what z= into my $V^*V$. Would it be better to substitute in $z_1$ in the beginning, or simplify $V^*V$ first, then plug in? Does $x*x=1$ imply that it works for $B^*B$ as well?
The condition $xx^*x = 1$ tells you that $|x_1|^2 + y^*y=1$. Since $z$ is obtained from $x$ by multiplying by a scalar of modulus 1, it follows that the analogous result holds for $z$, namely $z_1^2 + B^*B = 1$. Notice that in this case we do not need to put mod signs around $z_1$, because $z_1$ is real and nonnegative. Now compute $V^*V$, and then substitute $1-z_1^2$ for $B^*B.$ You should find that $V^*V$ is the identity matrix.
Last edited:
cylers89
New member
Okay, that makes sense. I do remember covering that in class. So I also know that BB*=(1-z1)(1+z1).
So for the matrix, after I multiply V*V...row1column1 easily is substituted to 1, since z12+B*B = 1.
Now I am getting tripped up on the canceling of row2column2 to 1.
The first one is: BB*+(-I+(1/1+z1)BB*)2
Substituting and squaring the binomial gets: (1-z12)+(I2-I+Iz1-I+Iz1+1-z1-z1+z12)
Combining like terms gives me: 2+I2-2I+2Iz1-2z1
This is where I am getting fuzzy...I cannot seem to figure out how to cancel this down to 1. It seems to overlap just enough that I am going in loops.
Now once I finish that part up, V*V will in fact be shown to be equal to the Identity matrix.
What does the directions mean by "choose theta in R such that ei(theta)x1 is greater than or equal to 0. and define z=ei(theta)x "...I'm not understanding what it is asking I guess.
Opalg
MHB Oldtimer
Staff member
Okay, that makes sense. I do remember covering that in class. So I also know that BB*=(1-z1)(1+z1).
So for the matrix, after I multiply V*V...row1column1 easily is substituted to 1, since z12+B*B = 1.
Now I am getting tripped up on the canceling of row2column2 to 1.
The first one is: BB*+(-I+(1/1+z1)BB*)2
(I may have confused you in my previous comment by writing $BB^*$ where it should have been $B^*B$. I have edited it to correct that.)
Okay, so the (2,2)-element of your matrix is $$BB^* + \Bigl(-I + \tfrac1{1+z_1}BB^*\Bigr)^2 = BB^* + I - \tfrac2{1+z_1}BB^* + \tfrac1{(1+z_1)^2}B(B^*B)B^*.$$
Replace the $B^*B$ that I have bracketed in that last term by $1-z_1^2$, and you will see that the whole thing reduces to $I$, as required. In a similar way, the off-diagonal terms in the matrix are both 0.
What does the directions mean by "choose theta in R such that ei(theta)x1 is greater than or equal to 0. and define z=ei(theta)x "...I'm not understanding what it is asking I guess.
In this problem, $x_1$ is a complex number, and any complex number can be multiplied by a complex number of modulus 1 (that is, a number of the form $e^{i\theta}$) so as to become real and nonnegative. The point here is that the computation to show that $V$ is unitary relies on the fact the element in the top left corner of the matrix is real. So you need a construction that replaces $x_1$ by $z_1$ in order to make $V$ unitary.
• cylers89 | HuggingFaceTB/finemath | |
1.8k views
A subnet has been assigned a subnet mask of $255.255.255.192$. What is the maximum number of hosts that can belong to this subnet?
1. $14$
2. $30$
3. $62$
4. $126$
edited | 1.8k views
(C) is answer since you have $6$ zeroes so you can make $64-2$ hosts
by Active (2.5k points)
edited by
0
Here class is not given , how one could find the hostid
nid + hostID = All 1's
Without nid how we could be able to find the hostID .
Plz explain .
0
there is slight mistake. no. of 1's is equal to netid+subnetid
0
@ashwina here no need of class,
255.255.255.192 = 11111111.11111111.11111111.11000000
we have to find number of zero i.e 6 here,
maximum no. of hosts = 2^ (no. of bits in hid) - 2
= 2^6 - 2
= 64 - 2
= 62
0
its a classless ip address(if not mention explicitly in the question)
to calculate the network address(NID) if one of the host IP address belonging to that subnet is given,the one have to perform the bitwise '&' operation between ip address and subnet mask
for eg: host ip address ==>192.168.1.0.134
&
------------------------
192.168.0.128
out of the 64 hosts ip address in the subnet ,one is given to network address(router inteface) and one is directed broadcast address in that network.
so the available host address are 64 - 2 = 62
maximum no. of hosts = 2(no. of bits in hid) - 2
= 26 - 2
= 64 - 2
= 62
by Junior (757 points) by Junior (505 points)
0
why any host cannot take base Id and broadcast Id??
+2
@Vishnathan your answer is the only one which explains why $2$ is being subtracted from $2^6$.
If we only write the last byte into binary, then we get,
255.255.255.11000000
∴ Number of 0's in subnet mask = Number of bits in host ID = 6
∴ Maximum number of hosts = 2^ (Number of bits in host Id) - 2 [since the 1st and last IP address is used for Network ID and Directed Broadcast Adddress]
= 2^6 - 2
= 64 - 2
= 62
Option (c) is the right answer
by (259 points) | HuggingFaceTB/finemath | |
# CAT 2022 Set-3 | Data Interpretation and Logical Reasoning | Question: 13
381 views
Comprehension:
All the first-year students in the computer science (CS) department in a university take both the courses (i) $\mathrm{Al}$ and (ii) $\mathrm{ML}$. Students from other departments (non-CS students) can also take one of these two courses, but not both. Students who fail in a course get an $\mathrm{F}$ grade; others pass and are awarded $\mathrm{A}$ or $\mathrm{B}$ or $\mathrm{C}$ grades depending on their performance. The following are some additional facts about the number of students who took these two courses this year and the grades they obtained.
1. The numbers of non-CS students who took $\mathrm{Al}$ and $\mathrm{ML}$ were in the ratio $2: 5$.
2. The number of non-CS students who took either $\mathrm{Al}$ or $\mathrm{ML}$ was equal to the number of CS students.
3. The numbers of non-CS students who failed in the two courses were the same and their total is equal to the number of CS students who got a $\mathrm{C}$ grade in $\mathrm{ML}$.
4. In both the courses, $50 \%$ of the students who passed got a $\mathrm{B}$ grade. But, while the numbers of students who got $\mathrm{A}$ and $\mathrm{C}$ grades were the same for $\mathrm{Al}$, they were in the ratio $3 : 2$ for $\mathrm{ML}$.
5. No CS student failed in $\mathrm{Al}$, while no non-CS student got an $\mathrm{A}$ grade in $\mathrm{Al}$.
6. The numbers of CS students who got $\mathrm{A, B}$ and $\mathrm{C}$ grades respectively in $\mathrm{Al}$ were in the ratio $3: 5: 2$, while in $\mathrm{ML}$ the ratio was $4: 5: 2$.
7. The ratio of the total number of non-CS students failing in one of the two courses to the number of CS students failing in one of the two courses was $3: 1$.
8. $30$ students failed in $\mathrm{ML}$.
How many non-CS students got $\mathrm{A}$ grade in $\mathrm{ML}$?
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Comprehension:In the following, a year corresponds to $1^{\text {st }}$ of January of that year.A study to determine the mortality rate for a disease began in $1980$. The... | HuggingFaceTB/finemath | |
Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #10201 2013-04-16 07:20:09
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Okay, see you later.
## #10202 2013-04-16 12:01:42
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Also I think I know the solution to the cake problem, it is a lot easier than I thought. I'll post the solution tomorrow.
## #10203 2013-04-16 16:21:49
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Hi;
Okay thanks, I have been unable to find any similar problem.
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10204 2013-04-17 02:59:51
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Nothing from adriana yet, 4 days ago now.
Here's the problem again just as a reminder:
---
The cakes in our canteen each contain exactly four currants, each currant being randomly placed in the cake. I take a proportion X of a cake where X is a random variable with density function
f(x) = Ax
for 0 ≤ x ≤ 1, where A is a constant.
(i) What is the expected number of currants in my portion?
(ii) If I find all four currants in my portion, what is the probability that I took more than
half the cake?
---
(i)
So f(x) = 2x.
The 4 currents are randomly distributed, so the expected number of currents in my portion is 4*(2/3) = 8/3.
(ii) If I take a proportion x of cake, the probability that a currant fell into it is clearly also x. The currant is either in the slice, or not in the slice. So we have a binomial distribution for the number of currants C ~ B(n, p), with p = x and n = 4. So P(C = 4) = x[sup]4[/sup]. The probability distribution for a proportion X AND having 4 currants, is 2x*x[sup]4[/sup] = 2x[sup]5[/sup].
Then we just use P(A n B) = P(A|B)P(B), for which I am getting 63/64.
## #10205 2013-04-17 03:09:14
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Uh oh! I like the reasoning for the 2x^5. I think that solves that. But why not use the integral idea that toasted fellow did?
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10206 2013-04-17 03:15:31
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
That's what I did -- it's a little strange because here we are kind of using individual values in a continuous distribution. You obtain P(4 currants in portion AND more than half a portion) by integrating 2x[sup]5[/sup] between 1/2 and 1. P(4 currants in portion) is obtained just by integrating 2x[sup]5[/sup] between 0 and 1, since the size of the portion doesn't matter for that probability. Putting that together yields the same answer toasted-lion got.
## #10207 2013-04-17 03:19:26
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Sure wish we had a few more of these type to try out your idea on..
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10208 2013-04-17 03:20:50
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
I might come across some more in the 916 more STEP questions I have to do. It is going to be a long summer...
## #10209 2013-04-17 03:22:40
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
I do not know how to phrase the problem to search for more. In the meantime would you mind writing up your full solution for it.
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10210 2013-04-17 03:24:31
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
My solution is basically the same as toasted-lion's on The Student Room, the only difference is I explained where the x^4 bit came from...
## #10211 2013-04-17 03:29:51
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Okay, then I will use that one with your explanation. I think it is done.
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10212 2013-04-17 03:43:32
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Okay, I have ticked it off my spreadsheet now.
No comment from adriana about the question. I think I will leave her to figure it out. (That being said, she is scared stiff of STEP III.)
## #10213 2013-04-17 03:47:07
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Good, good, you will be there to maybe comfort her. Tell her it is not possible to explain them with an email or chat. It must be done with paper and pencil and ...
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
zetafunc.
Guest
And...?
## #10215 2013-04-17 03:57:56
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Hmmm, can only be done in person is the point.
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10216 2013-04-17 04:06:55
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
It sounds like a nigh impossible task...
## #10217 2013-04-17 04:10:46
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Well if you have explained that too her what other choice does she have but to plan something more intimate?
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10218 2013-04-17 04:12:24
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
She could say that she will just not do the question, as there are 1166 others and she is not trying to do all the questions at any rate...
## #10219 2013-04-17 04:13:59
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Then you are alleviated from having to explain it to her.
The point is to arrange a meeting not to discuss math by email, is it not?
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10220 2013-04-17 11:15:11
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
I don't know, at the moment I'm not sure what I want to do. Part of me just wants to permanently ignore her, completely forget about her. I don't think she is worth the pursuit. There is now 0 benefit from knowing her.
## #10221 2013-04-17 11:29:37
anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,657
### Re: Linear Interpolation FP1 Formula
Hi zetafunc.
You are studying at the Cambridge university, right?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
Offline
## #10222 2013-04-17 12:51:01
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Hi;
I would say -5 is closer to it.
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
## #10223 2013-04-17 20:53:26
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Nope, I was rejected by Cambridge -- for a reason my school found very controversial. I am going to UCL instead, and then maybe go there for my Masters degree instead.
Knowing her does mean she won't be put out of my mind; but I agree, nothing good really comes from talking to her anymore. I wonder if she's told her BF about me at all. Some of the stuff she used to say to me, you'd think that I was her BF! Oh well, I failed this time, but I will get another opportunity at some point, probably.
## #10224 2013-04-18 01:54:20
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,513
### Re: Linear Interpolation FP1 Formula
Probably? I have known mere bumpkins, guys that were barely human that had dozens of opportunities.
In mathematics, you don't understand things. You just get used to them.
I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.
Online
zetafunc.
Guest | HuggingFaceTB/finemath | |
Math 166 - Week in Review #2
Section 2.4 - Matrices
• The size of a matrix is always number of rows × number of columns.
• cijre presents the entry of the matrix C in row i and column j.
• To add and subtract matrices , they must be the same size.
• When adding or subtracting matrices, add or subtract corresponding entries.
• A scalar product is computed by multiplying each entry of a matrix by a scalar (a number).
• Transpose - The rows of the matrix A become the columns of AT .
Section 2.5 - Multiplication of Matrices
• The matrix product AB can be computed only if the number of columns of A equals the number of rows of B.
• If C = AB, then cij is computed by multiplying the ith row of A by the jth column of B.
• Identity Matrix - Denoted by In, the identity matrix is the n×n matrix with 1’s down the main diagonal (from upper
left corner to lower right corner) and 0’s for all other entries.
• If A is m×n, then and
Section 2.6 - The Inverse of a Square Matrix
• Only square matrices can have inverses, but not all square matrices have inverses.
• A square matrix that does not have an inverse is called a singular matrix.
• The inverse of A, denoted A-1, is the square n×n matrix such that
Systems of equations can be represented as a matrix equation of the form AX = B where A is the coefficient matrix ,
X is a column vector containing the variables, and B is a column vector containing constants.
• If A has an inverse, the solution to the matrix equation is X = A-1B.
• If A does not have an inverse (i.e., if A is singular), this does NOT imply the system has no solution. It simply
means that you must use another method to solve the system.
Section 2.7 - Leontief Input-Output Model
• The input-output model is used to study the relationship between industrial production and con sumer demand .
• If A is the input-output matrix and X represents the total output of all industries, then AX represents the internal
consumption for that economy.
• Subtracting the internal consumption from the total output gives the net output of goods and services for consumer
demand, D:
X −AX = D
• To find the amount of goods and services that must be produced to satisfy consumer demand, solve for X in the
above
equation to obtain
X = (I−A)-1D
1. Let and .Compute each of
the following:
(a) B+3D
(b) 2C+B
(e) DB
2. Using matrix algebra , solve for the matrix D:
3. Solve for x and y:
4. The times (in minutes) required for assembling, testing, and packaging large and small capacity food processors
are shown in the following table:
Assembling Testing Packaging Large Small 45 30 15 10 10 5
(a) Define a matrix T that summarizes the above data.
(b) Let M =[100 200] represent the number of large and small food processors ordered , respectively. Find
MT and explain the meaning of its entries.
(c) If assembling costs \$3 per minute, testing costs \$1 per minute, and packaging costs \$2 per minute, find a
matrix C that, when multiplied with T, gives the total cost for making each size of food processor.
5. If , find A-1.
6. If find B-1.
7. Solve the following system of equations using matrix inverses.
8. A small village has two major industries: steel and electronics. For each unit of steel produced, 0.02 units of steel
and 0.15 units of electronices are used by the village. For each unit of electronics produced, 0.1 units of steel and
0.01 units of electronics are used by the village. The remaining steel and electronics products are then available to
export to a local city. If that city demands 500 units of steel and 800 units of electronics, how many units of steel
and electronics pruducts should be produced by the village to meet its own needs and those of the city?
9. Consider three sectors of the US economy: crude petroleum (crude), petroleum-refining and related industries
(refining), and chemical production (chemical). The following table gives the number of units of crude, refining,
and chemical products consumed in the production of one unit of crude product, one unit of refining product, and
one unit of chemical product.
To Crude Refining Chemical From Crude 0.31 0.42 0.050 Refining 0.0086 0.11 0.13 Chemical 0.010 0.47 0.38
(a) What is the input-output matrix for this problem?
(b) Explain the meaning of the entries in row 1 of this matrix.
(c) How many units of refining products are consumed in the production of 7,500 units of crude product?
(d) How many units of chemical products are required to produce 500 units of each sector in this economy?
(e) If a neighboring city demands 5,500 units of crude products, 6,750 units of refining products, and 1,250 units
of chemical products, how much should this economy produce to satisfy internal consumption and meet the
city’s demand?
(f) Referring to (e), how many units of each product are consumed internally in meeting the other city’s demands?
Prev Next
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Term Rewriting System R:
[x, y]
f(x, x) -> a
f(g(x), y) -> f(x, y)
Innermost Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
F(g(x), y) -> F(x, y)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Forward Instantiation Transformation`
Dependency Pair:
F(g(x), y) -> F(x, y)
Rules:
f(x, x) -> a
f(g(x), y) -> f(x, y)
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
F(g(x), y) -> F(x, y)
one new Dependency Pair is created:
F(g(g(x'')), y'') -> F(g(x''), y'')
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳FwdInst`
` →DP Problem 2`
` ↳Forward Instantiation Transformation`
Dependency Pair:
F(g(g(x'')), y'') -> F(g(x''), y'')
Rules:
f(x, x) -> a
f(g(x), y) -> f(x, y)
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
F(g(g(x'')), y'') -> F(g(x''), y'')
one new Dependency Pair is created:
F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳FwdInst`
` →DP Problem 2`
` ↳FwdInst`
` ...`
` →DP Problem 3`
` ↳Polynomial Ordering`
Dependency Pair:
F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')
Rules:
f(x, x) -> a
f(g(x), y) -> f(x, y)
Strategy:
innermost
The following dependency pair can be strictly oriented:
F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')
There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(g(x1)) = 1 + x1 POL(F(x1, x2)) = 1 + x1
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳FwdInst`
` →DP Problem 2`
` ↳FwdInst`
` ...`
` →DP Problem 4`
` ↳Dependency Graph`
Dependency Pair:
Rules:
f(x, x) -> a
f(g(x), y) -> f(x, y)
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes | HuggingFaceTB/finemath | |
# Maths CBSE Class 12 Problem Exercise for DETERMINANTS
Best and Important Solved Practice Questions and unsolved Questions with answers for Maths CBSE Class 12 Problem Exercise for DETERMINANTS.
## DETERMINANTS: CBSE Class 12 Problem Exercise
### You may also like
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d=document.createElement("script");d.addEventListener("load",e),d.addEventListener("error",e),t.getAttributeNames().forEach(e=>{"type"!=e&&d.setAttribute("data-src"==e?"src":e,t.getAttribute(e))});let a=!(d.type="text/javascript");!d.src&&t.textContent&&(d.src=litespeed_inline2src(t.textContent),a=!0),t.after(d),t.remove(),a&&e()}function litespeed_inline2src(t){try{var d=urlCreator.createObjectURL(new Blob([t.replace(/^(?:<!--)?(.*?)(?:-->)?\$/gm,"\$1")],{type:"text/javascript"}))}catch(e){d="data:text/javascript;base64,"+btoa(t.replace(/^(?:<!--)?(.*?)(?:-->)?\$/gm,"\$1"))}return d} ``` | HuggingFaceTB/finemath | |
# P2197R0Formatting for std::complex
## Working Draft, 2019-12-22
This version:
http://fmt.dev/papers/p2197r0.html
Authors:
Audience:
LEWG
Project:
ISO/IEC JTC1/SC22/WG21 14882: Programming Language — C++
## Abstract
This paper discusses extending coverage of the formatting functionality of [P0645] to std::complex.
## 1. Introduction
[P0645] has proposed a text formatting facility that provides a safe and extensible alternative to the printf family of functions. This paper explores how to format complex numbers represented by std::complex.
## 2. Examples
Default format:
std::string s = std::format("{}", 1.0 + 2i); // s == "(1+2i)"
iostreams-compatible format (optional):
std::string s = std::format("{:p}", 1.0 + 2i); // s == "(1,2)"
Format specifiers:
std::string s = std::format("{:.2f}", 1.0 + 2i); // s == "1.00+2.00i"
## 3. Motivation
This is a proposal defining formatting of complex numbers represented by the library type std::complex. The default notation (3+4i) is proposed, as it is common in mathematics, the physical sciences, and many other popular mathematical software environments. This form is also more consistent with the standard library literals for std::complex [N3660]. In addition to defining the new format and discussing design choices, this proposal attempts to address questions around introducing a format which differs from the existing iostreams format, and why the aforementioned advantages outweigh the disadvantages of introducing a potentially incompatible format. An option to produce iostreams-compatible output is also provided.
The formatting of std::complex should be simple, consistent with existing conventions of <format>, and conveniently support the most common use cases of std::complex. As the first nested format specified for <format>, it can also serve as an example for how format nesting can be done.
Mathematics generally follows the convention that complex numbers consist of a real part and an orthogonal imaginary part which is identified by multiplication of the imaginary unit vector $$i$$. Extending the set of unit vectors in this way furthermore implies straightforward extensions to other useful algebras such as quaternions $$i, j, k$$, dual numbers $$\epsilon$$, etc.
For the types std::complex<{float,double,long double}>, C++14 introduced string literals to the standard library in the namespace std::complex_literals. These string literals acknowledge the common use cases of these types and provide a convenient way to write complex numbers in code, for example the number $$1 + 1i$$ can be written in code as as 1.0f + 1if, 1.0 + 1i, or 1.0l + 1il, depending on the desired underlying type.
Sometimes it is possible to omit one part in a symbolic representation yet retain bijectivity in the machine representation to symbolic mapping. For example, the complex number $$0 + 0i$$ can be unambiguously written as either 0 or 0i. The convention of mathematics is the former, although the latter has the advantage of implying the underlying field.
As specified in [N4849], the existing iostreams formatting of a complex number x is essentially
s << '(' << x.real() << "," << x.imag() << ')';
where s is a stream object.
This embedded comma can cause silent unexpected generation of ambiguous output, which can happen e.g. when the locale’s decimal separator is set to comma. This ambiguity does not exist in the imaginary unit notation, even when an unusual locale is used.
## 4. Design Considerations
With an eye to providing a replacement for all the functionality of iostreams, the following considerations are made.
### 4.1. Numeric form
The question of how to represent the numeric type T of std::complex<T> is simply delegated to the formatter<T> for that type. Special alignment, fill, and sign rules may apply when T is float, double or long double, but other custom value types are accomodated. This is done by optionally forwarding a designated portion of the formatter<std::complex<T>> format spec to formatter<T>.
Although the standard does not specify behavior of std::complex<T> for types other than float, double, long double, it is not uncommon to use a type for T which provides functionality such as extended precision or automatic differentiation. The formatting specification should therefore be recursive, so that arbitrary numerical types for T are properly formatted.
### 4.2. Imaginary unit
As previously mentioned, mathematics notation typically uses $$i$$ as the complex unit vector, however it is very common in electrical engineering to use $$j$$ instead. Mathematica uses the Unicode character ⅈ for the imaginary unit. Another common written form of complex numbers puts the imaginary unit in front of the imaginary part rather than after it. Julia uses the dual-character symbol im, and it it easy to imagine wanting to explicitly specify the usually-omitted implied real unit-vector, result in a format like 3re + 4im. Supporting these use cases would be nice, but not with significant implementation difficulty.
### 4.3. Omission of a part
Because the complex number is always a pair of real part and imaginary part, it is not necessary to print both parts if one of the parts is identical to a known quantity: typically (nonnegative) zero; in this case omission implies the value uniquely. Either the real or the imaginary part can be omitted when this condition is satisfied, although clearly not both.
Should a part be dropped?
The benefits of part dropping include: shorter conversions in the special but common cases of purely real or imaginary numbers, adherence to common notation. There is also a tie-in with the design consideration discussed below of whether surrounding parenthesis are necessary: a single numeric value does not need to be surrounded by parenthesis in order to recognize it as the value for an entire complex number.
What are the conditions under which a part can be dropped?
A simple comparison with zero is usually insufficient to decide whether a part can be omitted. While C++ does not specify the underlying floating-point format, for correct round-trip conversions, the omitted part must be binary equivalent to T(0). The function std::signbit<T> is used to distinguish between -0 and 0, so the type T must have both a defined std::formatter<T> and std::signbit<T> to distinguish the two cases.
This nuance is demonstrated by the result of sqrt(-1. + 0i) vs sqrt(-1. - 0i).
Which part should be dropped?
Either part of an imaginary number could be dropped if it is binary equal to T(0), but in the special case of $$0 + 0i$$ dropping both parts would lead to the absurdity of an empty string. This is an open question, but it is the opinion of the author that the real part should be dropped, so that the remaining symbolic representation retains the imaginary unit vector, indicating use of the complex field $$\mathbb{C}$$.
## 5. Parentheses
Should parentheses be mandatory?
Are parentheses always neccesary to unambiguously specify a complex number?
Do mandatory parentheses significantly improve ease or speed of complex number parsing?
If parentheses are not mandatory, when should they be omitted?
## 6. Backwards Compatibility
To maintain backward compatibility we propose an easy-to-use format specifier that exactly reproduces the legacy iostreams output format.
The ios specifiers that affect complex number output are precision and width, these can not be easily guessed, but can be specified manually in the nested format specifier. Otherwise the compatibilty format the output will produce roughly the same output (modulo locale and default format for formatter<T>) that iostreams produces.
## 7. Parsing
This paper does not address parsing (scan’ing) for the type std::complex<T> but does aim to produce formatted output that can unambiguously round trip formatted and parsed.
## 8. Survey of other languages
The following programming languages/environments similarly use the imaginary-unit notation as their default: Python, Julia, R, MATLAB, Mathematica, Go. If you know the type of the data, these languages offer round-trip conversion from complex -> text -> complex, but because some of them drop the complex part in their textual output when the complex part is zero (or even negative zero!) some arguably pertinent information can be lost during formatting.
Language Basic Format Result of sqrt(-1) Result of sqrt(-1) - sqrt(-1) C++ iostreams (3,4) (0,1) (0,0) NumPy (3+4j) 1j 0j Julia 3.0 + 4.0im 0.0 + 1.0im 0.0 + 0.0im Octave 3 + 4i 0 + 1i 0 Mathematica* 1+ⅈ ⅈ 0 R (3+4i) (0+1i) (0+0i) C++14 literals 3.0 + 4i 1i 0i Go (3+4i) (0+1i) (0+0i)
* - checked via wolframalpha
Haskell provides a :+ b notation - this choice does not need much commentary, this much is offered: it is quite unique.
C# does not provide this functionality, but the doc page for complex includes an example code for creating an appropriate formatter.
## 9. Wish List
Feature wish list:
• nested specification of real and imaginary parts via formatter<T>
• easy substitution of "old style" iostreams format with simply {:p}
• defineable symbol for imaginary unit (j, im)
• option to prefix the imaginary part with the imaginary unit
• control over which (real/imag) part omission (0 or 0j)
• default to minimalist unique parseable format: 1, 1i, 0j, (1+1i)
• toggle to turn off surrounding parens: 1+1i
• toggle to turn off outputting - on negative zero - addressed by P2021
• center alignment ^ aligns output around the connecting +/-
• option for polar formatted output, ie (1.41421*exp(i*3.14159))
## 10. Proposed Wording
Modify [complex.syn] as follows:
template<class T, class charT, class traits>
basic_ostream<charT, traits>& operator<<(basic_ostream<charT, traits>&, const complex<T>&);
// 26.4.?, formatting
template<class charT> struct formatter<complex<float>, charT>;
template<class charT> struct formatter<complex<double>, charT>;
template<class charT> struct formatter<complex<long double>, charT>;
Add a new section 26.4.? Formatting [complex.format]:
Each formatter<complex<T>, charT> (format.formatter) specialization in this section meets the Formatter requirements (formatter.requirements). The parse member functions of these formatters interpret the format specification as std-format-spec (format.string.std) except that the 0 option is invalid.
template<class charT> struct formatter<complex<T>, charT> {
typename basic_format_parse_context<charT>::iterator
parse(basic_format_parse_context<charT>& ctx);
template<class FormatContext>
typename FormatContext::iterator
format(const complex<T>& c, FormatContext& ctx);
};
template<class FormatContext>
typename FormatContext::iterator
format(const complex<T>& c, FormatContext& ctx);
Let real = format(ctx.locale(), "{:<format-specs>}", c.real()) and imag = format(ctx.locale(), "{:<format-specs>}", c.imag()), where <format-specs> is std-format-spec with fill-and-align and width removed.
Effects: Equivalent to:
format_to(ctx.out(), "{:<fill-align-width>}",
format(c.real() != 0 ? "({0}+{1}i)" : "{1}i", real, imag))
where <fill-align-width> is the fill-and-align and width part of std-format-spec. If alignment is not specified > is used.
## 11. Questions
Q1: Do we want any of this?
Q2: The strategy of this paper is to include a laundry list of possibilities, which parts do we want?
## References
### Informative References
[N3660]
Peter Sommerlad. User-defined Literals for std::complex, part 2 of UDL for Standard Library Types (version 4). 19 April 2013. URL: https://wg21.link/n3660
[N4849]
Richard Smith. Working Draft, Standard for Programming Language C++. URL: https://wg21.link/n4849
[P0645]
Victor Zverovich. Text Formatting. URL: https://wg21.link/p0645 | open-web-math/open-web-math | |
# Is 1743 Divisible By Anything?
Okay, so when we ask if 1743 is divisible by anything, we are looking to see if there are any WHOLE numbers that can be divided into 1743 that will result in a whole number as the answer. In this short guide, we'll walk you through how to figure out whether 1743 is divisible by anything. Let's go!
Fun fact! All whole numbers will have at least two numbers that they are divisible by. Those would be the actual number in question (in this case 1743), and the number 1.
So, the answer is yes. The number 1743 is divisible by 8 number(s).
Let's list out all of the divisors of 1743:
• 1
• 3
• 7
• 21
• 83
• 249
• 581
• 1743
When we list them out like this it's easy to see that the numbers which 1743 is divisible by are 1, 3, 7, 21, 83, 249, 581, and 1743.
You might be interested to know that all of the divisor numbers listed above are also known as the Factors of 1743.
Not only that, but the numbers can also be called the divisors of 1743. Basically, all of those numbers can go evenly into 1743 with no remainder.
As you can see, this is a pretty simple one to calculate. All you need to do is list out all of the factors for the number 1743. If there are any factors, then you know that 1743 is divisible by something.
Give this a go for yourself and try to calculate a couple of these without using our calculator. Grab a pencil and a piece of paper and pick a couple of numbers to try it with.
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "Is 1743 Divisible By Anything?". VisualFractions.com. Accessed on February 9, 2023. http://visualfractions.com/calculator/divisible-by-anything/is-1743-divisible-by-anything/.
• "Is 1743 Divisible By Anything?". VisualFractions.com, http://visualfractions.com/calculator/divisible-by-anything/is-1743-divisible-by-anything/. Accessed 9 February, 2023. | HuggingFaceTB/finemath | |
## Points A, B, and C are collinear. Find the value of x given AB = 3x, BC = 5x + 10. and AC = 74.
Question
Points A, B, and C are collinear. Find the value of x given AB = 3x, BC = 5x + 10. and AC = 74.
in progress 0
3 years 2021-09-02T04:06:15+00:00 1 Answers 2 views 0
x = 8
Step-by-step explanation:
since it’s a collinerar
or, AB+BC=AC
or, 3x+5x+10 = 74
or, 8x = 74-10
or, 8x = 64
or, x = 64/8
x = 8 | HuggingFaceTB/finemath | |
## Sunday, 19 October 2008
### Even More on Pythagorean Triples and Barning Trees A while back I wrote a couple of blogs on Primitive Pythagorean triples and Matrices and Barning Trees of Primitive Pythagorean triples. Shortly afterward I got a nice note from H. Lee Price with a link to a paper he wrote on "The Pythagorean Tree: A New Species". A couple of things I learned from it were too good to keep to myself, so here are some of the things I thought were amazing.. or read the paper for yourself to find your own favorites.
One of the clever things that I learned was a way of creating an informative 2x2 matrix for any triple that holds some interesting information. I will use the 5,12,13 triangle as an example, but any of them produce the same sorts of information. Take one leg, and write it as a ratio to the sum of the hypotenuse and other leg, then simplify. Using 5 over 12+13 gives 5/25 or 1/5. Using the other leg we get 12/18 = 2/3. One of the first amazing things is that there will always be one, and only one even number in the two fractions created. Put the numerator and denominator of the fraction with the even number in the left column of a 2x2 matrix, and the other fraction makes the right column. For the 5, 12, 13 Pythagorean triple we get: "SO WHAT?", you ask. Well, amazingly, you can use these to quickly find the radii of the in-circle, and all three ex-circles of the triangle. If you multiply across the two rows the two products formed will give you the radius of the in-circle and also the radius of the ex-circle on the hypotenuse of the triangle. Then if you multiply the two diagonals, you get the other two ex-circles radii. In the case of the 5,12, 13 triangle, the in-circle has a radius of 2, and the three ex-circles have radii of 3, 10, and 15 units respectively. I had never observed until it was pointed out to me, that the radius of the ex-circle on the hypotenuse is always the sum of the radii of the in-circle and the other two ex-circles...2 + 3 + 10 = 15....
Finally, these in-circle and ex-circle relations are tied back to the Barning tree. Remember the diagram that shows that each of the Pythagorean triples produces three offspring in the tree. Mr. Price points out that in each case, the new Triangle has an in-circle that is one of the three ex-circles of the parent triangle. The 3,4,5 triangle has ex-circle radii of 2, 3 and 6 units. The off spring triangles are the 5, 12, 13 triangle with in-center radius of 2, the 8, 15, 17 triangle with in-center radius of 3, and the 20,21,29 triangle with in-center radius of 6.
Somehow, all that geometry packed into a simple matrix seems incredible. By the way, the original two fractions we used to form the matrix, they are the tangents of 1/2 the two acute angles of the triangle. The focus of Mr. Price's paper was to point out that from these 2x2 matrices, you can easily build a different tree to produce all the Primitive Pythagorean triples as well. I will return to that when time permits, and I feel I understand it well enough to do it justice.
#### 1 comment: Anonymous said...
help me. | HuggingFaceTB/finemath | |
Home » Divided by 6 » One Million Divided by 6
# One Million Divided by 6
Welcome to one million divided by 6, our post which explains the division of a million by six to you. 🙂
The numeral one million (1000000) is called the numerator or dividend, and the number 6 is called the denominator or divisor.
The quotient of one million and 6, the ratio of one million and 6, as well as the fraction of one million and 6 all mean (almost) the same:
a million divided by six, commonly written as 1000000/6.
Read on to find the result in various notations, along with its properties.
## Calculator
Show Steps
166666.666666666
=
166666 Remainder 4
The Long Division Steps are explained here. Read them now!
## What is One Million Divided by 6??
We provide you with the result of the division of one million by 6 straightaway:
one million divided by 6 = 166666.6
The result of one million divided by 6 is a non-terminating, repeating decimal.
The repeating pattern above, 6, is called repetend, and denoted overlined with a vinculum.
The notation in parentheses is also common: 166666.(6): However, in daily use it’s likely you come across the reptend indicated as ellipsis: 166666.6… .
• one million divided by 6 in decimal = 166666.6
• one million divided by 6 in fraction = 1000000/6
• one million divided by 6 in percentage = 16666666.6666667%
Note that you may use our state-of-the-art calculator above to obtain the quotient of any two numerals, integers or decimals, including one million and six, of course.
Repetends, if any, are denoted in ().
The conversion is done automatically once the nominator, e.g. one million, and the denominator, e.g. six, have been inserted in decimal notation. To start over overwrite the values of our calculator.
Give it a try now with a similar division by 6.
## What is the Quotient and Remainder of 1000000 Divided by 6?
Here we provide you with the result of the division with remainder, also known as Euclidean division, including the terms in a nutshell:
The quotient and remainder of 1000000 divided by 6 = 166666 R 4
The quotient (integer division) of 1000000/6 equals 166666; the remainder (“left over”) is 4.
1000000 is the dividend, and 6 is the divisor.
In the next section of this post you can find the additional information in the context of one million over six, followed by the summary of our information.
Observe that you may also locate many calculations such as one million ÷ 6 using the search form in the sidebar.
The result page lists all entries which are relevant to your query.
Give the search box a go now, inserting, for instance, a million divided by six, or what’s one million over 6 in decimal, just to name a few potential search terms.
Further information, such as how to solve the division of one million by six, can be found in our article Divided by, along with links to further readings.
## Conclusion
To sum up, one million over 6 = 166666.(6). The indefinitely repeating sequence of this decimal is 6.
As division with remainder the result of 1000000 ÷ 6 = 166666 R 4.
You may want to check out What is a Long Division?
For questions and comments about the division of one million by 6 fill in the comment form at the bottom, or get in touch by email using a meaningful subject line.
If our content has been helpful to you, then you might also be interested in the Remainder of One Million Divided by 8.
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Thanks for visiting our article explaining the division of one million by 6.
For a list of our similar sites check out the sidebar of our home page. – Article written by Mark | HuggingFaceTB/finemath | |
# Volume
Photo by: LuckyPhoto
Volume is the amount of space occupied by an object or a material. Volume is said to be a derived unit, since the volume of an object can be known from other measurements. In order to find the volume of a rectangular box, for example, one only needs to know the length, width, and depth of the box. Then the volume can be calculated from the formula, V = l · w · d.
The volume of most physical objects is a function of two other factors: temperature and pressure. In general, the volume of an object increases with an increase in temperature and decreases with an increase in pressure. Some exceptions exist to this general rule. For example, when water is heated from a temperature of 32°F (0°C) to 39°F (4°C), it decreases in volume. Above 39°F, however, further heating of water results in an increase in volume that is more characteristic of matter.
## Units of volume
The term unit volume refers to the volume of "one something": one quart, one milliliter, or one cubic inch, for example. Every measuring system that exists defines a unit volume for that system. Then, when one speaks about the volume of an object in that system, what he or she means is how many times that unit volume is contained within the object. If the volume of a glass of water is said to be 35.6 cubic inches, for example, what is meant is that 35.6 cubic inch unit volumes could be placed into that glass.
The units in which volume is measured depend on a variety of factors, such as the system of measurement being used and the type of material being measured. For example, volume in the British system of measurement may be measured in barrels, bushels, drams, gills, pecks, teaspoons, or other units. Each of these units may have more than one meaning, depending on the material being measured. For example, the precise size of a barrel ranges anywhere from 31 to 42 gallons, depending on federal and state statutes. The more standard units used in the British system, however, are the cubic inch or cubic foot and the gallon.
Variability in the basic units also exists. For example, the quart differs in size depending on whether it is being used to measure a liquid or dry volume and whether it is a measurement made in the British or customary U.S. system. As an example, 1 customary liquid quart is equivalent to 57.75 cubic inches, while 1 customary dry quart is equivalent to 67.201 cubic inches. In contrast, 1 British quart is equivalent to 69.354 cubic inches.
The basic unit of volume in the metric system is the liter (abbreviated as L), although the cubic centimeter (cc or cm 3 ) and milliliter (mL) are also widely used as units for measuring volume. The fundamental relationship between units in the two systems is given by the fact that 1U.S. liquid quart is equivalent to 0.946 liter or, conversely, 1 liter is equivalent to 1.057 customary liquid quarts.
## Words to Know
British system: A system of measurement long used in many parts of the world but now used commonly only in the United States among the major nations of the world.
Displacement method: A method for determining the volume of an irregularly shaped solid object by placing it in a measured amount of water or other liquid and noting the increase in volume of the liquid.
Metric system: A system of measurement used by all scientists and in common practice by almost every nation of the world.
Unit volume: The basic size of an object against which all other volumes are measured in a system.
## The volume of solids
The volumes of solids are relatively less affected by pressure and temperature changes than are the volumes of most liquids and all gases. For example, heating a liter of iron from 0°C to 100°C causes an increase in volume of less than 1 percent. Heating a liter of water through the same temperature range causes an increase in volume of less than 5 percent. But heating a liter of air from 0°C to 100°C causes an increase in volume of nearly 140 percent.
The volume of a solid object can be determined in one of two general ways, depending on whether or not a mathematical formula can be written for the object. For example, the volume of a cube can be determined if one knows the length of one side. In such a case, V = s 3 , or the volume of the cube is equal to the cube of the length of any one side (all sides being equal in length). The volume of a cylinder, on the other hand, is equal to the product of the area of the base multiplied by the height of the cylinder.
Many solid objects have irregular shapes for which no mathematical formula exists. One way to find the volume of such objects is to subdivide them into recognizable shapes for which formulas do exist (such as many small cubes) and then approximate the total volume by summing the volumes of individual sub-divisions. This method of approximation can become exact by using calculus.
Another way is to calculate the volume by water displacement, or the displacement of some other liquid. Suppose, for example, that one wishes to calculate the volume of an irregularly shaped piece of rock. One way to determine that volume is first to add water to some volume-measuring instrument, such as a graduated cylinder. The exact volume of water added to the cylinder is recorded. Then, the object whose volume is to be determined is also added to the cylinder. The water in the cylinder will rise by an amount equivalent to the volume of the object. Thus, the final volume read on the cylinder less the original volume is equal to the volume of the submerged object.
This method is applicable, of course, only if the object is insoluble in water. If the object is soluble in water, then another liquid, such as alcohol or cyclohexane, can be substituted for the water.
## The volume of liquids and gases
Measuring the volume of a liquid is relatively straightforward. Since liquids take the shape of the container in which they are placed, a liquid whose volume is to be found can simply be poured into a graduated container, that is, a container on which some scale has been etched. Graduated cylinders of various sizes ranging from 10 milliliters to 1 liter are commonly available in science laboratories for measuring the volumes of liquids. Other devices, such as pipettes and burettes (small measuring tubes), are available for measuring exact volumes, especially small volumes.
The volume of a liquid is only moderately affected by pressure, but it is often quite sensitive to changes in temperature. For this reason, volume measurements made at temperatures other than ambient (the surrounding) temperature are generally so indicated when they are reported, as V = 35.89 milliliters (35°C).
The volume of gases is very much influenced by temperature and pressure. Thus, any attempt to measure or report the volume of the gas must always include an indication of the pressure and temperature under which that volume was measured. Indeed, since gases expand to fill any container into which they are placed, the term volume has meaning for a gas only when temperature and pressure are indicated.
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Ryden McLeod
This site is so good for information using it next time for info. (Science) | HuggingFaceTB/finemath | |
# How do you prove sin4x = 4sinxcos^3x - 4sin^3xcosx?
May 28, 2015
I'll start from the double angle identities:
$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta$
$\sin 2 \theta = 2 \sin \theta \cos \theta$
Then:
$\sin 4 x = 2 \sin 2 x \cos 2 x$
$= 2 \left(2 \sin x \cos x\right) \left({\cos}^{2} x - {\sin}^{2} x\right)$
$= 2 \left(2 \sin x \cos x {\cos}^{2} x - 2 \sin x \cos x {\sin}^{2} x\right)$
$= 4 \sin x {\cos}^{3} x - 4 {\sin}^{3} x \cos x$ | HuggingFaceTB/finemath | |
# Largest Rectangle in Histogram
Given n non-negative integers representing the histogram’s bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
Above is a histogram where width of each bar is 1, given height = `[2,1,5,6,2,3]`.
The largest rectangle is shown in the shaded area, which has area = `10` unit.
Example:
```Input: [2,1,5,6,2,3]
Output: 10```
## Solution
```public class Solution {
public int LargestRectangleArea(int[] heights) {
Stack<int> stack = new Stack<int>();
int maxArea = 0;
int i = 0;
for(i=0;i<heights.Length;i++)
{
int currentElement = heights[i];
if(stack.Count == 0)
stack.Push(i);
else
{
int topElementInStack = heights[stack.Peek()];
if(topElementInStack <= currentElement)
{
stack.Push(i);
}
else
{
//Pop stack until topElementInStack <= currentElement
while(stack.Count > 0 && heights[stack.Peek()] > currentElement)
{
int poppedElement = heights[stack.Pop()];
//Find Area
if(stack.Count == 0)
{
int area = poppedElement * i;
maxArea = Math.Max(area,maxArea);
}
else
{
int area = poppedElement * (i - stack.Peek() - 1);
maxArea = Math.Max(area,maxArea);
}
}
stack.Push(i);
}
}
}
while(stack.Count > 0)
{
int poppedElement = heights[stack.Pop()];
//Find Area
if(stack.Count == 0)
{
int area = poppedElement * i;
maxArea = Math.Max(area,maxArea);
break;
}
else
{
int area = poppedElement * (i - stack.Peek() - 1);
maxArea = Math.Max(area,maxArea);
}
}
return maxArea;
}
}
```
Time Complexity: O(n)
Space Complexity: O(n) | HuggingFaceTB/finemath | |
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```
In a class some of the students are planned for a picnic,each student collect the money twice their number .If the total amount is Rs.7200/- then find the each person share?
In a class some of the students are planned for a picnic,each student collect the money twice their number .If the total amount is Rs.7200/- then find the each person share?
```
2 years ago
Rrr
13 Points
``` Given that each student collect money twice their number .to find each person share total amount is equal to twice their number7200=2nn=7200/2n=3600
```
2 years ago
Arun
25768 Points
``` Let there are n students hence 2n + 2n +2n +2n ….....n times = 7200 2 n^2 = 7200 n^2 = 3600 n = 60 rupees hence per student share is 120 rupees
```
2 years ago
Think You Can Provide A Better Answer ?
## Other Related Questions on Algebra
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Add and Subtract Fractions with the Same Denominator | Year 3 Maths
## What you need to know
Things to remember:
• If we add or subtract two fractions with the same denominator, we can just add or subtract the top numbers.
• If we add or subtract two fractions with the same denominator, the denominator in the answer will be the same as the denominator in the question.
Let’s recap a little, below we have a fraction of a grid shaded in and what this means in terms of numbers.
$=\frac{3}{6}$ Here, the numerator is 3, and the denominator is 6.
If two fractions have the same denominator, we can actually just add them straight away
$$\frac{3}{6} + \frac{2}{6}= \frac{5}{6}$$ Because our two fractions have been split into the same amount, so is the answer, so the denominators stay the same! We have added 3 sixths and 2 sixths, to give us a total of 5 sixths. We can do the same with subtraction.
$$\frac{3}{6} - \frac{2}{6}= \frac{1}{6}$$ ## Example Questions
$$\frac{3}{8} + \frac{1}{8}=\frac{4}{8}$$ So, the answer is $\frac{4}{8}$
$$\frac{7}{9} - \frac{5}{9}=\frac{2}{9}$$ So, the answer is $\frac{2}{9}$
## KS2 SATs Flash Cards
(43 Reviews) £8.99
• All of the major KS2 Maths SATs topics covered
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# Simple Lie group
Last updated
In mathematics, a simple Lie group is a connected non-abelian Lie group G which does not have nontrivial connected normal subgroups. The list of simple Lie groups can be used to read off the list of simple Lie algebras and Riemannian symmetric spaces.
## Contents
Together with the commutative Lie group of the real numbers, ${\displaystyle \mathbb {R} }$, and that of the unit-magnitude complex numbers, U(1) (the unit circle), simple Lie groups give the atomic "blocks" that make up all (finite-dimensional) connected Lie groups via the operation of group extension. Many commonly encountered Lie groups are either simple or 'close' to being simple: for example, the so-called "special linear group" SL(n) of n by n matrices with determinant equal to 1 is simple for all n > 1.
The simple Lie groups were first classified by Wilhelm Killing and later perfected by Élie Cartan. This classification is often referred to as Killing-Cartan classification.
## Definition
Unfortunately, there is no universally accepted definition of a simple Lie group. In particular, it is not always defined as a Lie group that is simple as an abstract group. Authors differ on whether a simple Lie group has to be connected, or on whether it is allowed to have a non-trivial center, or on whether ${\displaystyle \mathbb {R} }$ is a simple Lie group.
The most common definition is that a Lie group is simple if it is connected, non-abelian, and every closed connected normal subgroup is either the identity or the whole group. In particular, simple groups are allowed to have a non-trivial center, but ${\displaystyle \mathbb {R} }$ is not simple.
In this article the connected simple Lie groups with trivial center are listed. Once these are known, the ones with non-trivial center are easy to list as follows. Any simple Lie group with trivial center has a universal cover, whose center is the fundamental group of the simple Lie group. The corresponding simple Lie groups with non-trivial center can be obtained as quotients of this universal cover by a subgroup of the center.
### Alternatives
An equivalent definition of a simple Lie group follows from the Lie correspondence: A connected Lie group is simple if its Lie algebra is simple. An important technical point is that a simple Lie group may contain discrete normal subgroups, hence being a simple Lie group is different from being simple as an abstract group.
Simple Lie groups include many classical Lie groups, which provide a group-theoretic underpinning for spherical geometry, projective geometry and related geometries in the sense of Felix Klein's Erlangen program. It emerged in the course of classification of simple Lie groups that there exist also several exceptional possibilities not corresponding to any familiar geometry. These exceptional groups account for many special examples and configurations in other branches of mathematics, as well as contemporary theoretical physics.
As a counterexample, the general linear group is neither simple, nor semisimple. This is because multiples of the identity form a nontrivial normal subgroup, thus evading the definition. Equivalently, the corresponding Lie algebra has a degenerate Killing form, because multiples of the identity map to the zero element of the algebra. Thus, the corresponding Lie algebra is also neither simple nor semisimple. Another counter-example are the special orthogonal groups in even dimension. These have the matrix ${\displaystyle -I}$ in the center, and this element is path-connected to the identity element, and so these groups evade the definition. Both of these are reductive groups.
### Simple Lie algebras
The Lie algebra of a simple Lie group is a simple Lie algebra. This is a one-to-one correspondence between connected simple Lie groups with trivial center and simple Lie algebras of dimension greater than 1. (Authors differ on whether the one-dimensional Lie algebra should be counted as simple.)
Over the complex numbers the semisimple Lie algebras are classified by their Dynkin diagrams, of types "ABCDEFG". If L is a real simple Lie algebra, its complexification is a simple complex Lie algebra, unless L is already the complexification of a Lie algebra, in which case the complexification of L is a product of two copies of L. This reduces the problem of classifying the real simple Lie algebras to that of finding all the real forms of each complex simple Lie algebra (i.e., real Lie algebras whose complexification is the given complex Lie algebra). There are always at least 2 such forms: a split form and a compact form, and there are usually a few others. The different real forms correspond to the classes of automorphisms of order at most 2 of the complex Lie algebra.
### Symmetric spaces
Symmetric spaces are classified as follows.
First, the universal cover of a symmetric space is still symmetric, so we can reduce to the case of simply connected symmetric spaces. (For example, the universal cover of a real projective plane is a sphere.)
Second, the product of symmetric spaces is symmetric, so we may as well just classify the irreducible simply connected ones (where irreducible means they cannot be written as a product of smaller symmetric spaces).
The irreducible simply connected symmetric spaces are the real line, and exactly two symmetric spaces corresponding to each non-compact simple Lie group G, one compact and one non-compact. The non-compact one is a cover of the quotient of G by a maximal compact subgroup H, and the compact one is a cover of the quotient of the compact form of G by the same subgroup H. This duality between compact and non-compact symmetric spaces is a generalization of the well known duality between spherical and hyperbolic geometry.
### Hermitian symmetric spaces
A symmetric space with a compatible complex structure is called Hermitian. The compact simply connected irreducible Hermitian symmetric spaces fall into 4 infinite families with 2 exceptional ones left over, and each has a non-compact dual. In addition the complex plane is also a Hermitian symmetric space; this gives the complete list of irreducible Hermitian symmetric spaces.
The four families are the types A III, B I and D I for p = 2, D III, and C I, and the two exceptional ones are types E III and E VII of complex dimensions 16 and 27.
### Notation
${\displaystyle \mathbb {R,C,H,O} }$ stand for the real numbers, complex numbers, quaternions, and octonions.
In the symbols such as E626 for the exceptional groups, the exponent 26 is the signature of an invariant symmetric bilinear form that is negative definite on the maximal compact subgroup. It is equal to the dimension of the group minus twice the dimension of a maximal compact subgroup.
The fundamental group listed in the table below is the fundamental group of the simple group with trivial center. Other simple groups with the same Lie algebra correspond to subgroups of this fundamental group (modulo the action of the outer automorphism group).
## Full classification
Simple Lie groups are fully classified. The classification is usually stated in several steps, namely:
One can show that the fundamental group of any Lie group is a discrete commutative group. Given a (nontrivial) subgroup ${\displaystyle K\subset \pi _{1}(G)}$ of the fundamental group of some Lie group ${\displaystyle G}$, one can use the theory of covering spaces to construct a new group ${\displaystyle {\tilde {G}}^{K}}$ with ${\displaystyle K}$ in its center. Now any (real or complex) Lie group can be obtained by applying this construction to centerless Lie groups. Note that real Lie groups obtained this way might not be real forms of any complex group. A very important example of such a real group is the metaplectic group, which appears in infinite-dimensional representation theory and physics. When one takes for ${\displaystyle K\subset \pi _{1}(G)}$ the full fundamental group, the resulting Lie group ${\displaystyle {\tilde {G}}^{K=\pi _{1}(G)}}$ is the universal cover of the centerless Lie group ${\displaystyle G}$, and is simply connected. In particular, every (real or complex) Lie algebra also corresponds to a unique connected and simply connected Lie group ${\displaystyle {\tilde {G}}}$ with that Lie algebra, called the "simply connected Lie group" associated to ${\displaystyle {\mathfrak {g}}.}$
### Compact Lie groups
Every simple complex Lie algebra has a unique real form whose corresponding centerless Lie group is compact. It turns out that the simply connected Lie group in these cases is also compact. Compact Lie groups have a particularly tractable representation theory because of the Peter–Weyl theorem. Just like simple complex Lie algebras, centerless compact Lie groups are classified by Dynkin diagrams (first classified by Wilhelm Killing and Élie Cartan).
For the infinite (A, B, C, D) series of Dynkin diagrams, a connected compact Lie group associated to each Dynkin diagram can be explicitly described as a matrix group, with the corresponding centerless compact Lie group described as the quotient by a subgroup of scalar matrices. For those of type A and C we can find explicit matrix representations of the corresponding simply connected Lie group as matrix groups.
## Overview of the classification
Ar has as its associated simply connected compact group the special unitary group, SU(r + 1) and as its associated centerless compact group the projective unitary group PU(r + 1).
Br has as its associated centerless compact groups the odd special orthogonal groups, SO(2r + 1). This group is not simply connected however: its universal (double) cover is the Spin group.
Cr has as its associated simply connected group the group of unitary symplectic matrices, Sp(r) and as its associated centerless group the Lie group PSp(r) = Sp(r)/{I, −I} of projective unitary symplectic matrices. The symplectic groups have a double-cover by the metaplectic group.
Dr has as its associated compact group the even special orthogonal groups, SO(2r) and as its associated centerless compact group the projective special orthogonal group PSO(2r) = SO(2r)/{I, −I}. As with the B series, SO(2r) is not simply connected; its universal cover is again the spin group, but the latter again has a center (cf. its article).
The diagram D2 is two isolated nodes, the same as A1 A1, and this coincidence corresponds to the covering map homomorphism from SU(2) × SU(2) to SO(4) given by quaternion multiplication; see quaternions and spatial rotation. Thus SO(4) is not a simple group. Also, the diagram D3 is the same as A3, corresponding to a covering map homomorphism from SU(4) to SO(6).
In addition to the four families Ai, Bi, Ci, and Di above, there are five so-called exceptional Dynkin diagrams G2, F4, E6, E7, and E8; these exceptional Dynkin diagrams also have associated simply connected and centerless compact groups. However, the groups associated to the exceptional families are more difficult to describe than those associated to the infinite families, largely because their descriptions make use of exceptional objects. For example, the group associated to G2 is the automorphism group of the octonions, and the group associated to F4 is the automorphism group of a certain Albert algebra.
## List
### Abelian
DimensionOuter automorphism groupDimension of symmetric spaceSymmetric spaceRemarks
${\displaystyle \mathbb {R} }$ (Abelian)1${\displaystyle \mathbb {R} ^{*}}$1${\displaystyle \mathbb {R} }$
#### Notes
^† The group ${\displaystyle \mathbb {R} }$ is not 'simple' as an abstract group, and according to most (but not all) definitions this is not a simple Lie group. Further, most authors do not count its Lie algebra as a simple Lie algebra. It is listed here so that the list of "irreducible simply connected symmetric spaces" is complete. Note that ${\displaystyle \mathbb {R} }$ is the only such non-compact symmetric space without a compact dual (although it has a compact quotient S1).
### Compact
DimensionReal rankFundamental
group
Outer automorphism
group
Other namesRemarks
An (n ≥ 1) compactn(n + 2)0Cyclic, order n + 11 if n = 1, 2 if n > 1. projective special unitary group
PSU(n + 1)
A1 is the same as B1 and C1
Bn (n ≥ 2) compactn(2n + 1)021 special orthogonal group
SO2n+1(R)
B1 is the same as A1 and C1.
B2 is the same as C2.
Cn (n ≥ 3) compactn(2n + 1)021projective compact symplectic group
PSp(n), PSp(2n), PUSp(n), PUSp(2n)
Hermitian. Complex structures of Hn. Copies of complex projective space in quaternionic projective space.
Dn (n ≥ 4) compactn(2n 1)0Order 4 (cyclic when n is odd).2 if n > 4, S3 if n = 4projective special orthogonal group
PSO2n(R)
D3 is the same as A3, D2 is the same as A12, and D1 is abelian.
E678 compact78032
E7133 compact133021
E8248 compact248011
F452 compact52011
G214 compact14011This is the automorphism group of the Cayley algebra.
### Split
DimensionReal rankMaximal compact
subgroup
Fundamental
group
Outer automorphism
group
Other namesDimension of
symmetric space
Compact
symmetric space
Non-Compact
symmetric space
Remarks
An I (n ≥ 1) splitn(n + 2)nDn/2 or B(n1)/2Infinite cyclic if n = 1
2 if n ≥ 2
1 if n = 1
2 if n ≥ 2.
projective special linear group
PSLn+1(R)
n(n + 3)/2Real structures on Cn+1 or set of RPn in CPn. Hermitian if n = 1, in which case it is the 2-sphere.Euclidean structures on Rn+1. Hermitian if n = 1, when it is the upper half plane or unit complex disc.
Bn I (n ≥ 2) splitn(2n + 1)nSO(n)SO(n+1)Non-cyclic, order 41identity component of special orthogonal group
SO(n,n+1)
n(n + 1)B1 is the same as A1.
Cn I (n ≥ 3) splitn(2n + 1)nAn1S1Infinite cyclic1projective symplectic group
PSp2n(R), PSp(2n,R), PSp(2n), PSp(n,R), PSp(n)
n(n + 1)Hermitian. Complex structures of Hn. Copies of complex projective space in quaternionic projective space.Hermitian. Complex structures on R2n compatible with a symplectic form. Set of complex hyperbolic spaces in quaternionic hyperbolic space. Siegel upper half space.C2 is the same as B2, and C1 is the same as B1 and A1.
Dn I (n ≥ 4) splitn(2n - 1)nSO(n)SO(n)Order 4 if n odd, 8 if n even2 if n > 4, S3 if n = 4identity component of projective special orthogonal group
PSO(n,n)
n2D3 is the same as A3, D2 is the same as A12, and D1 is abelian.
E66 I split786C4Order 2Order 2E I42
E77 V split1337A7Cyclic, order 4Order 270
E88 VIII split2488D821E VIII128@ E8
F44 I split524C3 × A1Order 21F I28Quaternionic projective planes in Cayley projective plane.Hyperbolic quaternionic projective planes in hyperbolic Cayley projective plane.
G22 I split142A1 × A1Order 21G I8Quaternionic subalgebras of the Cayley algebra. Quaternion-Kähler.Non-division quaternionic subalgebras of the non-division Cayley algebra. Quaternion-Kähler.
### Complex
Real dimensionReal rankMaximal compact
subgroup
Fundamental
group
Outer automorphism
group
Other namesDimension of
symmetric space
Compact
symmetric space
Non-Compact
symmetric space
An (n ≥ 1) complex2n(n + 2)nAnCyclic, order n + 12 if n = 1, 4 (noncyclic) if n ≥ 2.projective complex special linear group
PSLn+1(C)
n(n + 2)Compact group AnHermitian forms on Cn+1
with fixed volume.
Bn (n ≥ 2) complex2n(2n + 1)nBn2Order 2 (complex conjugation)complex special orthogonal group
SO2n+1(C)
n(2n + 1)Compact group Bn
Cn (n ≥ 3) complex2n(2n + 1)nCn2Order 2 (complex conjugation)projective complex symplectic group
PSp2n(C)
n(2n + 1)Compact group Cn
Dn (n ≥ 4) complex2n(2n 1)nDnOrder 4 (cyclic when n is odd)Noncyclic of order 4 for n > 4, or the product of a group of order 2 and the symmetric group S3 when n = 4.projective complex special orthogonal group
PSO2n(C)
n(2n 1)Compact group Dn
E6 complex1566E63Order 4 (non-cyclic)78Compact group E6
E7 complex2667E72Order 2 (complex conjugation)133Compact group E7
E8 complex4968E81Order 2 (complex conjugation)248Compact group E8
F4 complex1044F41252Compact group F4
G2 complex282G21Order 2 (complex conjugation)14Compact group G2
### Others
DimensionReal rankMaximal compact
subgroup
Fundamental
group
Outer automorphism
group
Other namesDimension of
symmetric space
Compact
symmetric space
Non-Compact
symmetric space
Remarks
A2n1 II
(n ≥ 2)
(2n 1)(2n + 1)n 1CnOrder 2SLn(H), SU(2n)Quaternionic structures on C2n compatible with the Hermitian structureCopies of quaternionic hyperbolic space (of dimension n 1) in complex hyperbolic space (of dimension 2n 1).
An III
(n ≥ 1)
p + q = n + 1
(1 ≤ pq)
n(n + 2)pAp1Aq1S1SU(p,q), A III2pq Hermitian.
Grassmannian of p subspaces of Cp+q.
If p or q is 2; quaternion-Kähler
Hermitian.
Grassmannian of maximal positive definite
subspaces of Cp,q.
If p or q is 2, quaternion-Kähler
If p=q=1, split
If |pq| ≤ 1, quasi-split
Bn I
(n > 1)
p+q = 2n+1
n(2n + 1)min(p,q)SO(p)SO(q) SO(p,q) pqGrassmannian of Rps in Rp+q.
If p or q is 1, Projective space
If p or q is 2; Hermitian
If p or q is 4, quaternion-Kähler
Grassmannian of positive definite Rps in Rp,q.
If p or q is 1, Hyperbolic space
If p or q is 2, Hermitian
If p or q is 4, quaternion-Kähler
If |pq| ≤ 1, split.
Cn II
(n > 2)
n = p+q
(1 ≤ pq)
n(2n + 1)min(p,q)CpCqOrder 21 if pq, 2 if p = q.Sp2p,2q(R)4pqGrassmannian of Hps in Hp+q.
If p or q is 1, quaternionic projective space
in which case it is quaternion-Kähler.
Hps in Hp,q.
If p or q is 1, quaternionic hyperbolic space
in which case it is quaternion-Kähler.
Dn I
(n ≥ 4)
p+q = 2n
n(2n 1)min(p,q)SO(p)SO(q)If p and q ≥ 3, order 8.SO(p,q)pqGrassmannian of Rps in Rp+q.
If p or q is 1, Projective space
If p or q is 2 ; Hermitian
If p or q is 4, quaternion-Kähler
Grassmannian of positive definite Rps in Rp,q.
If p or q is 1, Hyperbolic Space
If p or q is 2, Hermitian
If p or q is 4, quaternion-Kähler
If p = q, split
If |pq| ≤ 2, quasi-split
Dn III
(n ≥ 4)
n(2n 1)n/2⌋An1R1Infinite cyclicOrder 2SO*(2n)n(n 1)Hermitian.
Complex structures on R2n compatible with the Euclidean structure.
Hermitian.
E62 II
(quasi-split)
784A5A1Cyclic, order 6Order 2E II40Quaternion-Kähler.Quaternion-Kähler.Quasi-split but not split.
E614 III782D5S1Infinite cyclicTrivialE III32Hermitian.
Rosenfeld elliptic projective plane over the complexified Cayley numbers.
Hermitian.
Rosenfeld hyperbolic projective plane over the complexified Cayley numbers.
E626 IV782F4TrivialOrder 2E IV26Set of Cayley projective planes in the projective plane over the complexified Cayley numbers.Set of Cayley hyperbolic planes in the hyperbolic plane over the complexified Cayley numbers.
E75 VI1334D6A1Non-cyclic, order 4TrivialE VI64Quaternion-Kähler.Quaternion-Kähler.
E725 VII1333E6S1Infinite cyclicOrder 2E VII54Hermitian.Hermitian.
E824 IX2484E7 × A1Order 21E IX112Quaternion-Kähler.Quaternion-Kähler.
F420 II521B4 (Spin9(R))Order 21F II16Cayley projective plane. Quaternion-Kähler.Hyperbolic Cayley projective plane. Quaternion-Kähler.
## Simple Lie groups of small dimension
The following table lists some Lie groups with simple Lie algebras of small dimension. The groups on a given line all have the same Lie algebra. In the dimension 1 case, the groups are abelian and not simple.
DimGroupsSymmetric spaceCompact dualRankDim
1${\displaystyle \mathbb {R} }$, S1 = U(1) = SO2(${\displaystyle \mathbb {R} }$) = Spin(2)AbelianReal line01
3S3 = Sp(1) = SU(2)=Spin(3), SO3(${\displaystyle \mathbb {R} }$) = PSU(2)Compact
3SL2(${\displaystyle \mathbb {R} }$) = Sp2(${\displaystyle \mathbb {R} }$), SO2,1(${\displaystyle \mathbb {R} }$)Split, Hermitian, hyperbolicHyperbolic plane ${\displaystyle \mathbb {H} ^{2}}$Sphere S212
6SL2(${\displaystyle \mathbb {C} }$) = Sp2(${\displaystyle \mathbb {C} }$), SO3,1(${\displaystyle \mathbb {R} }$), SO3(${\displaystyle \mathbb {C} }$)ComplexHyperbolic space ${\displaystyle \mathbb {H} ^{3}}$Sphere S313
8SL3(${\displaystyle \mathbb {R} }$)SplitEuclidean structures on ${\displaystyle \mathbb {R} ^{3}}$Real structures on ${\displaystyle \mathbb {C} ^{3}}$25
8SU(3)Compact
8SU(1,2)Hermitian, quasi-split, quaternionicComplex hyperbolic planeComplex projective plane14
10Sp(2) = Spin(5), SO5(${\displaystyle \mathbb {R} }$)Compact
10SO4,1(${\displaystyle \mathbb {R} }$), Sp2,2(${\displaystyle \mathbb {R} }$)Hyperbolic, quaternionicHyperbolic space ${\displaystyle \mathbb {H} ^{4}}$Sphere S414
10SO3,2(${\displaystyle \mathbb {R} }$), Sp4(${\displaystyle \mathbb {R} }$)Split, HermitianSiegel upper half spaceComplex structures on ${\displaystyle \mathbb {H} ^{2}}$26
14G2Compact
14G2Split, quaternionicNon-division quaternionic subalgebras of non-division octonionsQuaternionic subalgebras of octonions28
15SU(4) = Spin(6), SO6(${\displaystyle \mathbb {R} }$)Compact
15SL4(${\displaystyle \mathbb {R} }$), SO3,3(${\displaystyle \mathbb {R} }$)Split${\displaystyle \mathbb {R} }$3 in ${\displaystyle \mathbb {R} }$3,3Grassmannian G(3,3)39
15SU(3,1)HermitianComplex hyperbolic spaceComplex projective space16
15SU(2,2), SO4,2(${\displaystyle \mathbb {R} }$)Hermitian, quasi-split, quaternionic${\displaystyle \mathbb {R} }$2 in ${\displaystyle \mathbb {R} }$2,4Grassmannian G(2,4)28
15SL2(${\displaystyle \mathbb {H} }$), SO5,1(${\displaystyle \mathbb {R} }$)HyperbolicHyperbolic space ${\displaystyle \mathbb {H} ^{5}}$Sphere S515
16SL3(${\displaystyle \mathbb {C} }$)ComplexSU(3)28
20SO5(${\displaystyle \mathbb {C} }$), Sp4(${\displaystyle \mathbb {C} }$)ComplexSpin5(${\displaystyle \mathbb {R} }$)210
21SO7(${\displaystyle \mathbb {R} }$)Compact
21SO6,1(${\displaystyle \mathbb {R} }$)HyperbolicHyperbolic space ${\displaystyle \mathbb {H} ^{6}}$Sphere S6
21SO5,2(${\displaystyle \mathbb {R} }$)Hermitian
21SO4,3(${\displaystyle \mathbb {R} }$)Split, quaternionic
21Sp(3)Compact
21Sp6(${\displaystyle \mathbb {R} }$)Split, hermitian
21Sp4,2(${\displaystyle \mathbb {R} }$)Quaternionic
24SU(5)Compact
24SL5(${\displaystyle \mathbb {R} }$)Split
24SU4,1Hermitian
24SU3,2Hermitian, quaternionic
28SO8(${\displaystyle \mathbb {R} }$)Compact
28SO7,1(${\displaystyle \mathbb {R} }$)HyperbolicHyperbolic space ${\displaystyle \mathbb {H} ^{7}}$Sphere S7
28SO6,2(${\displaystyle \mathbb {R} }$)Hermitian
28SO5,3(${\displaystyle \mathbb {R} }$)Quasi-split
28SO4,4(${\displaystyle \mathbb {R} }$)Split, quaternionic
28SO8(${\displaystyle \mathbb {R} }$)Hermitian
28G2(${\displaystyle \mathbb {C} }$)Complex
30SL4(${\displaystyle \mathbb {C} }$)Complex
## Simply laced groups
A simply laced group is a Lie group whose Dynkin diagram only contain simple links, and therefore all the nonzero roots of the corresponding Lie algebra have the same length. The A, D and E series groups are all simply laced, but no group of type B, C, F, or G is simply laced.
## Related Research Articles
In mathematics, a Lie group is a group that is also a differentiable manifold. A manifold is a space that locally resembles Euclidean space, whereas groups define the abstract concept of a binary operation along with the additional properties it must have to be a group, for instance multiplication and the taking of inverses (division), or equivalently, the concept of addition and the taking of inverses (subtraction). Combining these two ideas, one obtains a continuous group where multiplying points and their inverses are continuous. If the multiplication and taking of inverses are smooth (differentiable) as well, one obtains a Lie group.
In mathematics, G2 is the name of three simple Lie groups (a complex form, a compact real form and a split real form), their Lie algebras as well as some algebraic groups. They are the smallest of the five exceptional simple Lie groups. G2 has rank 2 and dimension 14. It has two fundamental representations, with dimension 7 and 14.
In mathematics, F4 is the name of a Lie group and also its Lie algebra f4. It is one of the five exceptional simple Lie groups. F4 has rank 4 and dimension 52. The compact form is simply connected and its outer automorphism group is the trivial group. Its fundamental representation is 26-dimensional.
In mathematics, E6 is the name of some closely related Lie groups, linear algebraic groups or their Lie algebras , all of which have dimension 78; the same notation E6 is used for the corresponding root lattice, which has rank 6. The designation E6 comes from the Cartan–Killing classification of the complex simple Lie algebras (see Élie Cartan § Work). This classifies Lie algebras into four infinite series labeled An, Bn, Cn, Dn, and five exceptional cases labeled E6, E7, E8, F4, and G2. The E6 algebra is thus one of the five exceptional cases.
In mathematics, a generalized flag variety is a homogeneous space whose points are flags in a finite-dimensional vector space V over a field F. When F is the real or complex numbers, a generalized flag variety is a smooth or complex manifold, called a real or complexflag manifold. Flag varieties are naturally projective varieties.
In mathematics, E8 is any of several closely related exceptional simple Lie groups, linear algebraic groups or Lie algebras of dimension 248; the same notation is used for the corresponding root lattice, which has rank 8. The designation E8 comes from the Cartan–Killing classification of the complex simple Lie algebras, which fall into four infinite series labeled An, Bn, Cn, Dn, and five exceptional cases labeled G2, F4, E6, E7, and E8. The E8 algebra is the largest and most complicated of these exceptional cases.
In mathematics, the ADE classification is a situation where certain kinds of objects are in correspondence with simply laced Dynkin diagrams. The question of giving a common origin to these classifications, rather than a posteriori verification of a parallelism, was posed in. The complete list of simply laced Dynkin diagrams comprises
In mathematics, E7 is the name of several closely related Lie groups, linear algebraic groups or their Lie algebras e7, all of which have dimension 133; the same notation E7 is used for the corresponding root lattice, which has rank 7. The designation E7 comes from the Cartan–Killing classification of the complex simple Lie algebras, which fall into four infinite series labeled An, Bn, Cn, Dn, and five exceptional cases labeled E6, E7, E8, F4, and G2. The E7 algebra is thus one of the five exceptional cases.
In mathematics, a compact (topological) group is a topological group whose topology realizes it as a compact topological space. Compact groups are a natural generalization of finite groups with the discrete topology and have properties that carry over in significant fashion. Compact groups have a well-understood theory, in relation to group actions and representation theory.
In mathematics, a reductive group is a type of linear algebraic group over a field. One definition is that a connected linear algebraic group G over a perfect field is reductive if it has a representation with finite kernel which is a direct sum of irreducible representations. Reductive groups include some of the most important groups in mathematics, such as the general linear group GL(n) of invertible matrices, the special orthogonal group SO(n), and the symplectic group Sp(2n). Simple algebraic groups and semisimple algebraic groups are reductive.
In mathematics, a symmetric space is a Riemannian manifold whose group of symmetries contains an inversion symmetry about every point. This can be studied with the tools of Riemannian geometry, leading to consequences in the theory of holonomy; or algebraically through Lie theory, which allowed Cartan to give a complete classification. Symmetric spaces commonly occur in differential geometry, representation theory and harmonic analysis.
In mathematics, a Hermitian symmetric space is a Hermitian manifold which at every point has an inversion symmetry preserving the Hermitian structure. First studied by Élie Cartan, they form a natural generalization of the notion of Riemannian symmetric space from real manifolds to complex manifolds.
In mathematics, the main results concerning irreducible unitary representations of the Lie group SL(2,R) are due to Gelfand and Naimark (1946), V. Bargmann (1947), and Harish-Chandra (1952).
In mathematics, a prehomogeneous vector space (PVS) is a finite-dimensional vector space V together with a subgroup G of the general linear group GL(V) such that G has an open dense orbit in V. Prehomogeneous vector spaces were introduced by Mikio Sato in 1970 and have many applications in geometry, number theory and analysis, as well as representation theory. The irreducible PVS were classified by Sato and Tatsuo Kimura in 1977, up to a transformation known as "castling". They are subdivided into two types, according to whether the semisimple part of G acts prehomogeneously or not. If it doesn't then there is a homogeneous polynomial on V which is invariant under the semisimple part of G.
In mathematics, the notion of a real form relates objects defined over the field of real and complex numbers. A real Lie algebra g0 is called a real form of a complex Lie algebra g if g is the complexification of g0:
In mathematics, Borel–de Siebenthal theory describes the closed connected subgroups of a compact Lie group that have maximal rank, i.e. contain a maximal torus. It is named after the Swiss mathematicians Armand Borel and Jean de Siebenthal who developed the theory in 1949. Each such subgroup is the identity component of the centralizer of its center. They can be described recursively in terms of the associated root system of the group. The subgroups for which the corresponding homogeneous space has an invariant complex structure correspond to parabolic subgroups in the complexification of the compact Lie group, a reductive algebraic group.
In mathematics, the complexification or universal complexification of a real Lie group is given by a continuous homomorphism of the group into a complex Lie group with the universal property that every continuous homomorphism of the original group into another complex Lie group extends compatibly to a complex analytic homomorphism between the complex Lie groups. The complexification, which always exists, is unique up to unique isomorphism. Its Lie algebra is a quotient of the complexification of the Lie algebra of the original group. They are isomorphic if the original group has a quotient by a discrete normal subgroup which is linear.
This is a glossary of representation theory in mathematics.
In algebra, a simple Lie algebra is a Lie algebra that is non-abelian and contains no nonzero proper ideals. The classification of real simple Lie algebras is one of major achievements of Wilhelm Killing and Élie Cartan.
In mathematics, the representation theory of semisimple Lie algebras is one of the crowning achievements of the theory of Lie groups and Lie algebras. The theory was worked out mainly by E. Cartan and H. Weyl and because of that, the theory is also known as the Cartan–Weyl theory. The theory gives the structural description and classification of a finite-dimensional representation of a semisimple Lie algebra ; in particular, it gives a way to parametrize irreducible finite-dimensional representations of a semisimple Lie algebra, the result known as the theorem of the highest weight.
## References
• Jacobson, Nathan (1971). Exceptional Lie Algebras. CRC Press. ISBN 0-8247-1326-5.
• Fulton, William; Harris, Joe (2004). Representation Theory: A First Course. Springer. doi:10.1007/978-1-4612-0979-9. ISBN 978-1-4612-0979-9.
• Besse, Einstein manifolds ISBN 0-387-15279-2
• Helgason, Differential geometry, Lie groups, and symmetric spaces. ISBN 0-8218-2848-7
• Fuchs and Schweigert, Symmetries, Lie algebras, and representations: a graduate course for physicists. Cambridge University Press, 2003. ISBN 0-521-54119-0 | open-web-math/open-web-math | |
# Math websites that give answers
## The Best Math websites that give answers
Math can be a challenging subject for many students. But there is help available in the form of Math websites that give answers. Algebra math solvers are one of the most useful tools that a student can have in their arsenal. They can help you solve problems that you might otherwise be stuck on, and can save you a lot of time and frustration. There are many different algebra math solvers out there, so it is important to find one that suits your needs. Some are better for solving simple equations, while others are better for more complex problems. There are even some that can help you understand the concepts behind
There are a few key steps to solving any word problem. First, read the problem and identify what information is given and what is being asked. Next, create a visual representation of the problem, whether that be a drawing or a simple equation. Lastly, solve the problem and check your work to make sure the answer makes sense. With practice, solving word problems can become second nature!
There are a few steps that you can follow to solve linear inequalities. First, you need to simplify the inequality by removing any extraneous operations. Next, you need to solve for the variable by isolating it on one side of the inequality. Once you have done that, you can then graph the inequality on a number line to find the solution set. Finally, you need to check your work by plugging in a few points from the solution set to make sure that they satisfy the inequality
There are a few different methods that can be used to solve quadratics by factoring. One method is to factor the quadratic equation into two linear equations. Another method is to use the quadratic formula. To factor a quadratic equation, one must first determine the greatest common factor of the terms. Then, the terms are divided by the greatest common factor and the equation is rewritten. The next step is to find two factors of the leading coefficient that add
There is no one definitive answer to this question, as there are a number of different apps that can be used to solve math problems. Some of the more popular options include DragonBox, Mathway, and PhotoMath, among others. Each of these apps has its own unique features and functionality, so it is important to choose one that will best meet your individual needs. | HuggingFaceTB/finemath | |
# Thread: marginal variance of discrete joint distribution
1. ## marginal variance of discrete joint distribution
$\\\text{Given }Pr(X=i,\text{ }Y=j)=\frac{i+j-1}{8}\text{ for }i=1,\text{ }2\text{ and }j=1,\text{ }2.\\\\\text{Find }Var[Y].$
I know how to find the solution through finding $E[Y]$ and $E[Y^2]$, but my book gives an alternative solution that I haven't seen before.
Alternative solution:
$\\\text{Note:}\\\\\text{Alternatively, since }Y\text{ can take only two values, 1 and 2, with probabilities }\frac{3}{8}\text{ and }\frac{5}{8}\text{ respectively, we can use the Bernoulli shortcut to find }Var[Y].\\\\Var[Y]=(1-2)^2\left(\frac{3}{8}\right) \left(\frac{5}{8}\right)=0.2344$
I know that the variance of the Bernoulli distribution is $p(1-p)$, but if that's what the book is referring to, what's with the $(1-2)^2$ part?
2. ## Re: marginal variance of discrete joint distribution
I think you are confusing "Bernoulli distribution" with "Bernoulli shortcut". If your book says it is using the "Bernoulli shortcut", it must have an explanation of that. Have you tried looking it up in the index?
3. ## Re: marginal variance of discrete joint distribution
Thanks, there is no index
4. ## Re: marginal variance of discrete joint distribution
Nevermind, I figured it out. Since Y can only take on two values, you can find the variance the same way it's found in the Bernoulli distribution.
5. ## Re: marginal variance of discrete joint distribution
$\\\text{Let } q=1-p\\\\f(x;p;q)=\begin{cases}p, & \text{for }x=a\\q, & \text{for }x=b \end{cases}$
$\\E[X]=ap+bq\\\\E[X^2]=a^2p+b^2q\\\\Var[X]=a^2p+b^2q-(ap)^2-2abpq-(bq)^2$
$=a^2p(1-p)+b^2q(1-q)-2abpq$
$=a^2pq+b^2pq-2abpq$
$=pq(a^2+b^2)-2abpq$
$=pq(a^2-2abpq+b^2)$
$=pq(a-b)^2$ | HuggingFaceTB/finemath | |
# Question about a Symmetric relation
I have the following relation R ⊆ N × N and is defined as R = {(x, y); 3 | (x − y)}
A symmetric relation is defined as follows ∀x, y ∈ A : (xRy ⇒ yRx)
What i did was, i defined R = {(x, y); 3 | (x − y)} as x - y = 3k and y - x = -3k to get 3 | (y - x) in order to prove the implication xRy ⇒ yRx
I know this relation is symmetric i just dont understand the reason behind it, so any help is greatly appreciated.
Your proof is right. That is the the reason behind it. If $$x,y \in \mathbb{N}$$ and $$xRy$$, then there exists $$k \in \mathbb{Z}$$ such that $$x-y = 3k$$ so $$y-x = -3k$$. Hence $$yRx$$.
• @newplayer $y-x=-(x-y)=-(3k)=-3k$ – Grešnik Jun 18 at 9:29
• Multiply the equation $3k = x-y$ by $-1$. Then you get $y-x$ on the right hand side and $3 \cdot (-k)$ on the left hand side. The thing is that you need to change $k$. For example $3$ is a divisor of $9-6$ since $3 \cdot 1 = 9-6 = 3$. But $3$ is also a divisor of $6-9$! Here you can see that since $3 \cdot (-1) = 6-9 = -3$. – ThorWittich Jun 18 at 9:31
What you did was basically correct. Let $$(x,y) \in R$$, i.e. we have that $$3$$ is a divisor of $$x - y$$, such that we can write $$3k = x-y$$ for a suitable $$k$$. Since $$x - y = -(y - x)$$ and $$3 \mid x-y$$ we also know that $$3$$ is a divisor of $$y-x$$, since $$3 \cdot (-k) = y - x$$. | open-web-math/open-web-math | |
# 154 (number)
154 (one hundred fifty-four) is an even three-digits composite number following 153 and preceding 155. In scientific notation, it is written as 1.54 × 102. The sum of its digits is 10. It has a total of 3 prime factors and 8 positive divisors. There are 60 positive integers (up to 154) that are relatively prime to 154.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 3
• Sum of Digits 10
• Digital Root 1
## Name
Short name 154 one hundred fifty-four
## Notation
Scientific notation 1.54 × 102 154 × 100
## Prime Factorization of 154
Prime Factorization 2 × 7 × 11
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 154 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 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(n) 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 154 is 2 × 7 × 11. Since it has a total of 3 prime factors, 154 is a composite number.
## Divisors of 154
1, 2, 7, 11, 14, 22, 77, 154
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 288 Sum of all the positive divisors of n s(n) 134 Sum of the proper positive divisors of n A(n) 36 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 12.4097 Returns the nth root of the product of n divisors H(n) 4.27778 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 154 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 154) is 288, the average is 36.
## Other Arithmetic Functions (n = 154)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 60 Total number of positive integers not greater than n that are coprime to n λ(n) 30 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 36 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 60 positive integers (less than 154) that are coprime with 154. And there are approximately 36 prime numbers less than or equal to 154.
## Divisibility of 154
m n mod m 2 3 4 5 6 7 8 9 0 1 2 4 4 0 2 1
The number 154 is divisible by 2 and 7.
## Classification of 154
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Nonagonal
• Square Free
### Other numbers
• CentralPolygonal
• Sphenic
## Base conversion (154)
Base System Value
2 Binary 10011010
3 Ternary 12201
4 Quaternary 2122
5 Quinary 1104
6 Senary 414
8 Octal 232
10 Decimal 154
12 Duodecimal 10a
20 Vigesimal 7e
36 Base36 4a
## Basic calculations (n = 154)
### Multiplication
n×y
n×2 308 462 616 770
### Division
n÷y
n÷2 77 51.333 38.5 30.8
### Exponentiation
ny
n2 23716 3652264 562448656 86617093024
### Nth Root
y√n
2√n 12.4097 5.36011 3.52274 2.73845
## 154 as geometric shapes
### Circle
Diameter 308 967.611 74506
### Sphere
Volume 1.52986e+07 298024 967.611
### Square
Length = n
Perimeter 616 23716 217.789
### Cube
Length = n
Surface area 142296 3.65226e+06 266.736
### Equilateral Triangle
Length = n
Perimeter 462 10269.3 133.368
### Triangular Pyramid
Length = n
Surface area 41077.3 430423 125.74
## Cryptographic Hash Functions
md5 1d7f7abc18fcb43975065399b0d1e48e 06349be70bd2d5dd98d36b9b8dba0a057500fdac 1d0ebea552eb43d0b1e1561f6de8ae92e3de7f1abec52399244d1caed7dbdfa6 dc76224ce103959668fe797ec4184bc5605c323faa77b22ee62df9d1d139dfe6b9aa61cb42f7ee9707ae1d10143a180a8a48b012ef6d4d63f5187bd9e2b9ada9 64c4161aa8bc032ea8d1070dd61986841896ee5a | HuggingFaceTB/finemath | |
## What is the probability of getting 1 head in 5 tosses?
With 5 coins to flip you just times 16 by 2 and then minus 1, so it would result with a 31 in 32 chance of getting at least one heads.
What is the probability of getting a head when a coin is tossed 5 times?
Probability of getting utmost 4 heads is 1−Probability of getting 5 heads =1−321=3231.
What is the probability of 1 head?
In counting the number of heads in 4 coin flips, the probability that we get exactly one head is the probability that we get anyone of the following 4 outcomes: HTTT, THTT, TTHT, or TTTH. Each has probability 1/16, so the probability to get exactly one head in 4 flips is 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4.
### What is the expected number of heads in 5 tosses?
2.5
1. What is the expected number of heads that come up when a fair coin is flipped five times? 5. That comes out to be 2.5.
What is the probability of getting for 5 heads for a coin flip 5 times?
So the odds of flipping a coin 5 times and getting 5 heads are 1/2 ^5 (half to the power of 5). Which gives us 1/32 or just over a 3% chance.
What is the probability of getting 0 heads in 5 tosses?
1/32
Answer: The probability of getting zero heads in five tosses is 1/32.
## What is the probability of getting at most 5 heads?
For the other part, at most 5 heads we add the results for 0, 1, 2, 3, 4 and 5 giving 225 so the probability of getting at most 5 heads is 225/256.
What is the probability of getting 1 or 2 heads?
For example, the probability of two heads is 1/2 · 1/2 = 1/4, and the probability of two tails is the same. Since there are two leaves corresponding to one head and one tail, each of probability 1/4, the probability of this event is 1/4 + 1/4 = 1/2.
Is it possible to get a probability of 1 for head in a coin toss?
For example, the probability of an outcome of heads on the toss of a fair coin is ½ or 0.5. The probability of an event can also be expressed as a percentage (e.g., an outcome of heads on the toss of a fair coin is 50% likely) or as odds (e.g., the odds of heads on the toss of a fair coin is 1:1).
### What is the probability of winning the toss in five consecutive hockey matches?
But if you keep calling heads five consecutive times, like what Kohli did in this series, then the odds of winning the toss on the fifth time are a meagre 3%.
What is the expected number of heads flips in 5 flips of a fair coin given that the number of heads flips is greater than 2?
The formula is easy to extract: you need 2N−1 flips to get either N heads or N tails, or 2N+1−2 to get N heads only. If N=5 we get the answer: 62.
What is the probability of 5 heads in a row?
For 20 trials we obtain that the probability of throwing at least five successive Heads is equal to 0.2499. | HuggingFaceTB/finemath | |
## Elementary Statistics (12th Edition)
a) $P(X=0)={5\choose 0} \cdot (0.2)^0\cdot (0.8)^5=0.328.$ b)$P(X=1)={5\choose 1} \cdot (0.2)^1\cdot (0.8)^4=0.41.$ c)$P(X=0 \ or X=1)=P(X=0)+P(X=1)={5\choose 0} \cdot (0.2)^0\cdot (0.8)^5+{5\choose 1} \cdot (0.2)^1\cdot (0.8)^4=0.328+0.41=0.738$ d) No, it is not an unusually low probability, because 0.738 is really high, more than 0.05. | HuggingFaceTB/finemath | |
## You need to have an account to continue
Things to remember for Paper 2 HL
Junior Cert Mathematics — 16/11/16 111
I will post things that you should remember for Paper 2 - HL. Feel free to post your own stuff also.
SryanBruen — 12/06/16
Remember Soh Cah Toa Sin = O / H (Opposite / Hypotenuse) Cos = A / H (Adjacent / Hypotenuse) Tan = O / A (Opposite / Adjacent) You MUST know Soh Cah Toa as it is not in the tables book.
SryanBruen — 12/06/16
Relative frequency = Number of successful trials / number of trials
SryanBruen — 12/06/16
Pythagoras' Theorem = a(squared) + b(squared) = c(squared) This is in the tables book but you must know that a and b are the opposite and adjacent sides of the triangle whilst c is the hypotenuse.
SryanBruen — 12/06/16
To show that lines are perpendicular to each other, the answer should be -1.
mathswhiz101 — 12/06/16
are you going for straight A's Syran?
SryanBruen — 12/06/16
If the line is rising, the slope is positive. If the line is falling, the slope is negative.
Whackojacko — 12/06/16
Silly Old Harry (son) Caught A Herring (cah) Trawling Off Africa (toa)
Elisha447 — 12/06/16
I'm not able to understand when to use sin cos and tan. Like if you put it as a fraction or how to know when to use second function. Any advice?
SryanBruen — 12/06/16
Instead of using the slope formula, on a diagram, you can find the slope by doing the rise / run. So make a right angled triangle on the diagram between the two points, then count the number of squares rising and running. It don't matter if the squares are not full.
SryanBruen — 12/06/16
No mathswhiz101, after all it's only the JC.
Danielle1358 — 12/06/16
The probability of something NOT happening : 1 - the probability of it happening
SryanBruen — 12/06/16
When given equations of lines, always remember that the number in front of the x is the slope. So for example, y = 8 - 2x. The slope is -2. y = x + 5. The slope is 1.
Danielle1358 — 12/06/16
to find the intersection of two lines - use simultaneous equations
SryanBruen — 12/06/16
Oh yeah I forgot about that ^. Thank you.
mathswhiz101 — 12/06/16
Also syran, i see where your prediction come from for theorom 19 but i dont think maths will be that predictable. As paper 1 was easy i imagine they might ask an awkward variation of the theorom rather than just it straight.
SryanBruen — 12/06/16
The question would be with that ^, how could they make it anymore awkward? hahaha (the original is awkward already LOL).
Aoife_9776 — 12/06/16
Which one is theorem 19?
SryanBruen — 12/06/16
To find missing sides or angles in a right-angled triangle, we use Sin, Cos or Tan. Follow these steps whilst doing so: 1. Label the sides of the triangle with all the information you're given. 2. Select which formula to use. 3. Substitute values into the formula. 4. Solve the resulting equation to find the missing value. So for example of a past question, ABC is a right-angled triangle |ACB (triangle)| = 50 degrees and |AC| = 10cm Calculate the length of [AB], correct to two decimal places. Label the sides first! We have the hypotenuse and we want the opposite. Which formula has H and O in it? Soh Cah or Toa? Soh has. So you use Sin's formula. Sin 50 degrees = |AB| / 10 Bring the ten over and multiply (because it was dividing before you brought it over) with Sin of 50 degrees. Fill it into your calculator and you should get: 7.66044 = |AB| Correct to two decimal places, 7.66 = |AB|.
SryanBruen — 12/06/16
The angle at the centre of a circle standing on a given arc is twice the angle at any point of the circle standing on the same arc Aoife
Danielle1358 — 12/06/16
what constructions will come up do you think?
Name — 12/06/16
sound these notes are great
Danielle1358 — 12/06/16
Sin 25° = Cos x 90-25= 65 Sin 25° = Cos 65°
SryanBruen — 12/06/16
Theorem 19 answer for anybody here Given: Circle, centre 0, containing points A, B and C To prove: |BOC (triangle)| = 2|BAC (triangle)| Construction: Join A to 0 and continue to D Label angles 1, 2, 3, 4 and 5 Proof: Consider |AOB (triangle)|: |1| = |2| + |3| But |2| = |3| Similarly, |4| = 2|5| |1| + |4| = 2|2| + 2|5| Therefore |BOC (triangle)| = 2|BAC (triangle)|
SryanBruen — 12/06/16
When it asks you for the probability of something, DO NOT answer in words unless they specifically say to. If they ask probability, it has to be a fraction. Like what's the probability of getting a King out of a 52 card pack: The probability is 4/52 which simplifies to 2/26 which simplifies to 1/13. Unless they specifically say to, it is not advised to simplify probabilities. You should keep it as the original probability, in this case, 4/52 but there is NO HARM in simplifying it.
Study2000 — 12/06/16
Actually sometimes u lose marks if u do not simplify ur answer, so it is better to simplify.
SryanBruen — 12/06/16
My teacher even told us to not simplify unless they say to.
Study2000 — 12/06/16
In some of the marking schemes, for probability questions if u leave ur answer as it is, u are deducted 1 mark. Even in my mock i was deducted one mark for it.
SryanBruen — 12/06/16
Yeah I'm just saying what I have been told. I do agree with you.
needhelp101 — 12/06/16
I'm lost with some of the volume questions. The way that sometimes they ask you about surface area's or something of shapes but the formula's aren't in the log tables. And how do you learn off the formula's? Thanks.
needhelp101 — 12/06/16
Whoops. I meant how to you learn off the proofs of theorems.
aisling31 — 12/06/16
thanks for posting so much advice throughout the year :)
SryanBruen — 12/06/16
Volume is always something cubed whether it's cm or m etc. So if the output is something cubed, then there are 3 inputs that you multiply to make it. This is the clue I do < Volume (not Cylinder, Sphere & Cone which you're given in the tables book) = Length x Base x Height (which you could also write as lbh or l (cubed) Surface area (not Cylinder, Sphere & Cone which you're given in the tables book) = All the sides added with Base and Height (Otherwise: 6l (squared) (squared because it's area) You MUST KNOW these formulae as they are not in the tables book.
SryanBruen — 12/06/16
You MUST REMEMBER the following conversions: 1 litre = 1,000 cm (cubed) To convert litres to cubic centimetres (cm cubed), multiply by 1,000. 1 litre = 1,000 ml > 1cm (cubed) = 1 ml
SryanBruen — 12/06/16
When the question has multiple units of measurement, make sure to convert each one all to the same unit as it will make the question 10x easier for you to understand.
nms2001 — 12/06/16
Soh cah to
nms2001 — 12/06/16
Ignore my last comment soh cah toa is in the log tables
belle1234 — 12/06/16
will you ever get question asking to explain or define collaries and theorems, etc
naoise123 — 12/06/16
Danielle1358 — 12/06/16
constructions predictions??
SryanBruen — 12/06/16
I double it Belle but EXPLAIN THE TERM COROLLARY came up on my mock.
Aoife_9776 — 12/06/16
What is the definition of a corollary?
eveen — 12/06/16
corollary - a statement that follows from a previous thereom
SryanBruen — 12/06/16
A corollary follows after a theorem and is a statement which must be true because of that theorem. More or less what eveen said ^.
Aoife_9776 — 12/06/16
Have you an example of one?
Study2000 — 12/06/16
Know the Converse of a theorem, an axiom and especially corollary 3.
Study2000 — 12/06/16
Corollary 3: Each angle in a semi-circle is a right angle.
Aoife_9776 — 12/06/16
Have you got definitions and examples for those please?
Aoife_9776 — 12/06/16
Just for theorem and axiom
Study2000 — 12/06/16
Axiom 1: There is exactly one line through any two given points
Study2000 — 12/06/16
An axiom is a statement accepted without any proof
Study2000 — 12/06/16
A theorem is a statement deduced from the axioms by logical argument e.g Pythagoras Theorem.
Aoife_9776 — 12/06/16
Thanks!
Study2000 — 12/06/16
No problem:)
rocky123 — 12/06/16
To find an angle if you know the sin,cos or tan of it you use the inverse of it Eg Sin X =3/4 X=Sin^-1 (3/4)
rocky123 — 12/06/16
In an equation of a line in the format y=mx+c , m is the slope and c the y intercept
rocky123 — 12/06/16
Know the different geometry statements that apply to junior cert geometry E.g 3 sums of a triangle =180 degrees In a cyclic quadrilateral opposite angles sum to 180 degrees
rocky123 — 12/06/16
Know how to prove congruency E.g. ASA SSS SAS RHS
SryanBruen — 12/06/16
Well here's past constructions Danielle: Bisector of a line segment - 2015, 2013 Constructing a right-angled triangle - 2014, 2012 Drawing line segments, 2014 Maybe constructing a right-angle triangle... but also maybe one of the other 12 constructions..
Danielle1358 — 12/06/16
thanks
ST@1 — 12/06/16
(Sin) Oh Hell (Cos) Another Hour (Tan) Of Algebra Sin= O/H Cos= A/H Tan= O/A
Elisha447 — 12/06/16
For 2015 paper 2 question 5 how do you know which equation matches with each?? :)
123HOLOPER456 — 12/06/16
how many constructions are there in total?
123HOLOPER456 — 12/06/16
@SryanBruen could u post the theorem 14 asnwer as u did above for theorem 19 thanks!
Aoife_9776 — 12/06/16
To remember sin, tan and cos there's also Silly old Harry (sin= opp /hyp ) caught a herring (cos= adj/hyp ) trawling off America (tan=opp/adj)
Kristine_5238 — 12/06/16
you're litch my saviour not even messing
SryanBruen — 12/06/16
Theorem 14 Given: A right-angled triangle To prove: a(squared) + b(squared) = c(squared) Construction: Draw a square PQRS with sides of length a + b. Draw four congruent right-angled triangles in the square with sides of length a and b and hypotenuse c. Label angles 1, 2, 3 and 4 Proof: Each of the four inscribed triangles is congruent to the original triangle. Each side of the inner quadrilateral has length c. |1| + |2| = 90 degrees (Remaining angles in the triangle) |1| = |3| (Corresponding angles in congruent triangles) |2| + |3| = 90 degrees |4| = 90 degrees (Straight angle) The inscribed quadrilateral is a square. Area of square PQRS = 4 (a + b) (squared) = 4(half x ab) + c(squared) a(squared) + 2ab + 2b(squared) = 2ab + c(squared) a(squared) + b(squared) = c(squared)
Study2000 — 12/06/16
Ur lucky u have a short and easy one.
carolinaaplasencia — 12/06/16
SryanBruen — 12/06/16
Study2000 — 12/06/16
Carolina these ^^ are the ones in my book
OisinR — 12/06/16
What theorem do you think will come up? 19 came up last year!
SryanBruen — 12/06/16
19 did not come up last year Oisin
123HOLOPER456 — 12/06/16
thanks
123HOLOPER456 — 12/06/16
thanks
123HOLOPER456 — 12/06/16
thanks
123HOLOPER456 — 12/06/16
@carolinaaplasencia you only need to know theorems 4,6,9,14 and 19 for the exam
carolinaaplasencia — 12/06/16
yeah but ppl wanted the link to them online 123HOLOPER456 and Study2000 yeah Ikr active maths?
rocky123 — 12/06/16
The only acceptable form of the theorem of pythagaros is where you divide a right angled triangle in 2 then use similar triangles to prove the theorem The one that sryan bruen has posted above isn't acceptable as far as i know as we did that one but after the mocks we were told we had to learn the version project maths wants The correct version is in this link http://www.projectmathsbooks.com/wp-content/uploads/2014/02/Theorems-1-21.pdf
SryanBruen — 13/06/16
Neither theorems appeared hahaha. But the one that did was so easy to proof and explain. I loved paper 2. It was so easy.
Elisha447 — 13/06/16
How did you get the diagonal in the two cubes in square root form??
SryanBruen — 13/06/16
Oh I meant to also say that only 2 questions caught me out, that ^ and the spinner 8 euro back thing.
SryanBruen — 13/06/16
Point of intersection came up and I wouldn't have a clue how to do it if it weren't for the person who posted how to do it on this thread. Thank you very much - sorry I cannot see your post right now. Plus, Rocky how? Less Stress More Success and my teacher told me to learn off the theorem that way? And yeah my Less Stress More Success book says PROJECT MATHS on it, not just Maths. And no constructions came up after all that Danielle haha.
Elisha447 — 13/06/16
For making equations in the co ord geometry question I used to slope formula to get slopes but the slope of a and b didn't work out which affected other questions will I loose many marks?!
SryanBruen — 13/06/16
No I got the same Elisha (negative fractions). It won't!
Elisha447 — 13/06/16
Okay great thanks :)
jenny_1000 — 13/06/16
@sryanbruen constructions did come up when you had to draw the net of the prism you had to CLEARLY show construction lines. But its all g cause loads ofpeople didn't get it !
jenny_1000 — 13/06/16
@sryanbruen constructions did come up when you had to draw the net of the prism you had to CLEARLY show construction lines. But its all g cause loads ofpeople didn't get it !
SryanBruen — 13/06/16
I don't really consider that a construction (the 15 ones that you study I do) but I did get it! I'm not normally great with nets but I got it.
Aoife_9776 — 13/06/16
It was a combination of the rectangle and triangle constructions. I realised that 5 mins before the end (and then I was frantically rubbing out lol). I just drew it with a ruler the first time but the second time I used the constructions methods
SryanBruen — 13/06/16
I just drew it with a ruler....
jenny_1000 — 13/06/16
you will probs still get marks
Theresa_3825 — 13/06/16
Same, except for the triangles.
Theresa_3825 — 13/06/16
I used the boxes on the page as a guide and I didn't measure with a ruler. Will I be okay though? ��
SryanBruen — 13/06/16
I measured with a ruler....
Theresa_3825 — 13/06/16
Or is 1cm equal to the length of one of the boxes?
Theresa_3825 — 13/06/16
Nevermind, I think it was.
Elisha447 — 13/06/16
Ya it was I did the same! I measured the box afterwards!
Aoife_9776 — 13/06/16
I'm sure they'll give you some marks
barcelona — 13/06/16
all the squares were 1cm by 1cm so it would be the same with a ruler!!!!!
Aoife_9776 — 13/06/16
It did say show all construction lines clearly though
Danielle1358 — 13/06/16
oh no i just drew a net?? no constructions or anything?
Danielle1358 — 13/06/16
@SryanBruen you're very welcome :)
Aoife_9776 — 13/06/16
I drew the net using constructions
Danielle1358 — 13/06/16
how?
SryanBruen — 13/06/16
Higher Level Maths paper 2 is up now! Just thought I'd let you know.
Aoife_9776 — 13/06/16
I used the triangle construction for each end and the rectangle one for the sides
Danielle1358 — 13/06/16
i dont think i ever learnt those constructions! will i still get marks for just drawing with a ruler?
SryanBruen — 13/06/16
Well I did Danielle... I don't see why not. I have exactly constructed as they told me to with the exact cm they instructed..
Danielle1358 — 13/06/16
ok good thanks
Danielle1358 — 13/06/16 | HuggingFaceTB/finemath | |
2-dimensional Lebesgue-Stieltjes measure (Bivariate Density to Distribution Function)
Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ along with a random variable $X:\Omega\rightarrow\mathbb{R}$ such that $(\mathbb{R},\mathcal{B},\mu)$ is an induced probability space where $\mathcal{B}$ is to denote the Borel $\sigma$-algebra and $\mu$ is a pushforward measure such that $\mu=\mathbb{P}\circ X^{-1}$. Hence, $\mathbb{P}(X\in B)=\mu(B)$ for all $B\in\mathcal{B}$.
It then follows that $\mu$ - as a Lebesgue-Stieltjes measure $\mu(a,b]=F(b)-F(a)$ - is linked to a probability distribution function $F$ such that $F$ is monotonically increasing, right-continuous, and bounded to take values in $\,[0,1]$ with $F(-\infty)=0$ and $F(\infty)=1$.
I do not really feel comfortable with this result, but I was told that
$$\text{d}\mu=\mu(x,x+\text{d}x]=F(x+\text{d}x)-F(x)=\text{d}F.$$
Therefore, one can define the Lebesgue-Stieltjes integral of an arbitrary function $g$ as
$$\int_Bg\,\text{d}F=\int_{B}g\,\text{d}\mu\qquad\forall\quad B\in\mathcal{B}$$
where the righthandside is just the conventional Lebesgue integral w.r.t $\mu$.
Assuming $\mu\ll\lambda$ where $\lambda$ is the 1-dim Lebesgue measure, $F$ is differentiable such that $\text{d}F=F^\prime\,\text{d}x$. It thus follows that $$\int_Bg\,F^\prime\,\text{d}x=\int_{B}g\,\text{d}\mu\qquad\forall\quad B\in\mathcal{B}.$$ Moreover, since $\mu\ll\lambda$ the Radon-Nikodym theorem dictates that there exists a density function $f$ such that $$\int_{B}g\,\text{d}\mu=\int_{B}g\,f\,\text{d}\lambda$$
And finally, since $\lambda$ is a Lebesgue-Stieltjes measure with $F$ such that $x\mapsto x$ such that $\text{d}\lambda=\text{d}x$
$$\int_{B}g\,F^\prime\,\text{d}x=\int_{B}g\,f\,\text{d}\lambda=\int_Bg\,f\,\text{d}x$$ Since this statement holds true for an arbitrary function $g$ as well as for every set $B\in\mathcal{B}$ it would seem to follow that $F^\prime=f$ such that the pdf of a continuous random variable $X$ is simply the derivative of the corresponding cdf.
I would like to be able to transfer this line of reasoning to a bivariate case such that where a vector-valued measurable function $\boldsymbol{X}:\Omega\rightarrow\mathbb{R}^2$ gives a random variable that allows for the construction of an induced probability space $(\mathbb{R}^2,\mathcal{B}_2,\mu)$. At this point $\mathcal{B}_2=\mathcal{B}\otimes\mathcal{B}$ is the product $\sigma$-algebra of $\mathcal{B}$ and $\mu$ is a pushforward measure $\mu=\mathbb{P}\circ \boldsymbol{X}^{-1}$.
Starting from
$\mu(a,b]\times(c,d]=F(b,d)-F(a,d)-F(b,c)+F(a,c)$
(which is just my starting point since $\mathbb{P}(a<X\leq b,c<Y\leq d)=F(b,d)-F(a,d)-F(b,c)+F(a,c)$
and not because I would have read anywhere that the 2-dim Lebegue-Stieltjes measure is defined this way)
I would like to see why $$\text{d}\mu=\text{d}F$$
such that I can arrive at a statement to the effect that
$$\int_{B}g\,\dfrac{\partial^2 F}{\partial x\partial y}\,\text{d}x\,\text{dy}=\int_Bg\,f\,\text{d}x\,\text{d}y.$$
Thank you so much for taking the time to read my question. I would appreciate any pointers you might have for me - especially on why $\text{d}\mu=\text{d}F$ should be true in the bivariate case.
Best regards,
Jon
It seems to me that you are going in circles.
Let $\mu$ denote some probability measure on $(\mathbb R^2,\mathcal B_2)$.
Then $\mu$ induces a function $F:\mathbb R^2\to\mathbb R$ that is prescribed by: $$(b,d)\mapsto P((-\infty,b]\times(-\infty,d])$$
This function on its turn induces a (unique) measure $\mu_F$ that is characterized by: $$\mu_F((a,b]\times(c,d])=F(b,d)-F(a,d)-F(b,c)+F(a,c)$$ Observe that we could also write this as:$$\mu_F((a,b]\times(c,d])=\mu((a,b]\times(c,d])$$
So actually now we have two measures $\mu$ and $\mu_F$ that coincide on the collection: $$\mathcal A=\{(a,b]\times(c,d]\mid -\infty<a<b<\infty, -\infty<c<d<\infty\}$$
This collection is closed under intersection so $\mu$ and $\mu_F$ will also coincide on the sigma-algebra generated by $\mathcal A$ which is $\mathcal B_2$.
In a way this closes the circle: $\mu=\mu_F$.
The expression $\text{d}\mu=\text{d}F$ is just a notation for $\mu=\mu_F$.
• Thank you so very much for your answer. So $\text{d}\mu=\text{d}F$ is essentially notational justification for writing $$\int_{B}\,g\,\text{d}\mu=\int_B\,g\,\text{d}\mu_F=\int_B\,g\,\text{d}F?$$ I am afraid, however, since the lhs is evaluated as a Lebesgue-integral and the rhs is a Stieltjes-integral and thus an integral w.r.t. a function $F:\mathbb{R}^2\rightarrow[0,1]$ rather than a measure $\mu:\mathcal{B}\rightarrow[0,1]$, I have to admit, that I still have a hard time seeing why the two need to be equal just as, by definition, $F(b,d)=\mu(-\infty,b]\times(-\infty,d]$. – J.Beck Jul 24 '18 at 14:30
• According to my view every integral is based on a measure. We can write $\int gdF$ but actually that is only a notation for $\int gd\mu_F$ where $\mu_F$ is a unique measure generated by $F$. The notation $d\mu=dF$ tells us that we are dealing with the same measure. Also have a look here. – drhab Jul 24 '18 at 14:55
• If I may be a bit tenacious. Presuming $\mu_F$ is absolutely continuous w.r.t. to $\lambda$, then $F$ is a real-valued continuous function and the Lebesgue-Stieltjes integral specializes to a Rieman-Stieltjes integral. Hence, we can write $\text{d}F=F^\prime\,\text{d}x$ and end up with $$\int\,g\,\text{d}\mu=\int\,g\,\text{d}\mu_F=\int\,g\,\text{d}F=\int\,g\,F^\prime\,\text{d}x$$ which is a simple Rieman integral. I fail to see, how it can just be a question of notation to go from $\int g\,\text{d}\mu$ to $\int g\,F^\prime\,\text{d}x$ – J.Beck Jul 24 '18 at 15:05
• For the last equality $\mu_F$ must be absolutely continuous wrt Lebesguemeasure $\lambda$. Then a density $f$ exists with $\int gdF=\int gfd\lambda$ or in other notation $=\int gfdx$. For this the derivative of $F$ does not have to exist. If it does then $g=F'$ can be chosen as density (there are more than one candidates). – drhab Jul 24 '18 at 15:13
• What I basically want to say is that I feel that there must be a proof to show that this is true iff $F(b,d)=\mu(-\infty,b]\times(-\infty,d].$ – J.Beck Jul 24 '18 at 15:13
Thank to the helpful input of drhab, I believe I have pinpointed my problem:
For the sake of argument I confine my attention to the one-dim case. Let g be a continuous function and $\mu\ll\lambda$ (and therefore also $\mu_F\ll\lambda$). It follows that $F$ is continuous and differentiable.
By the Radon-Nikodym theorem there exists (more than one) density function $f$ such that $\text{d}\mu=f\,\text{d}\lambda$ and $\text{d}\mu_F=f\,\text{d}\lambda$. Hence one can write $$\int_Bg\,\text{d}\mu=\int_Bg\,f\,\text{d}\lambda.$$
Moreover, since $F$ is a differentiable function, it is also true that - in a usual calculus sense - $\text{d}F=F^\prime\,\text{d}x$. Just substituting that in yields $$\int_Bg\,\text{d}F=\int_Bg\,F^\prime\,\text{d}x.$$ On their own, I am okay with both of these statements. I am struggling with the fact, that the notational convention above says that they are equal such that since $$\int_Bg\,\text{d}\mu=\int_Bg\,\text{d}F\quad\text{ it follows that }\quad\int_Bg\,f\,\text{d}\lambda=\int_Bg\,F^\prime\,\text{d}x.$$
I suppose it would already help, if I could see a rigorous proof that
$\,\,\,$a) $\quad\text{d}\lambda=\text{d}x$
(I suppose this says that Rieman and Lebesgue integral coincide provided the Rieman integral exists)
$\,\,\,$b) $\quad F^\prime$ is, in fact, a Radon-Nikodym derivative of $\mu_F$ with respect to $\lambda$.
I am afraid, I could show b) only by relying on the notational convention that $\text{d}\mu_F=\text{d}F$ and $\text{d}\lambda=\text{d}x$ (where $\text{d}x$ is essentially $\text{d}F$ of the distribution function prescribed by $x\mapsto x$, i.e. the distribution function corresponding to the Lebesgue measure). Then one could that argue that $$\dfrac{\text{d}\mu_F}{\text{d}\lambda}=f\quad\text{and}\quad\dfrac{\text{d}F}{\text{d}x}=F^\prime$$ But since $\text{d}\mu_F=\text{d}F$ and $\text{d}\lambda=\text{d}x$ $$\dfrac{\text{d}\mu_F}{\text{d}\lambda}=\dfrac{\text{d}F}{\text{d}x}\quad\text{such that}\quad f=F^\prime$$ However, this does not answer the question why the derivative of a distribution function is a Radon-Nikodym derivative of the measure induced by said distribution function with respect to Lebesgue measure.
Once again, thank you so much.
Best regards,
Jon
PS. Having read the wikipedia article on "Absolute Continuity", the third paragraph in the section on the "Relation between the two notions of absolute continuity" asserts that
If the absolute continuity holds then the Radon–Nikodym derivative of μ is equal almost everywhere to the derivative of F .
This is precisely what I need to see proved in order to resolve my struggle. Wikipedia cites Royden 1988, Problem 12.17(b) on page 303. | open-web-math/open-web-math | |
+0
0
90
1
It is reported that 77% of workers aged 16 and over drive to work alone. Choose 8 workers at random. If X represents the random variable of driving to work alone, the following table depicts the probabilities for this case.
Find the following probabilities as percentages to 1 d.p.
a) Probability that none drives to work alone
b)Probability that exactly 5 drive to work alone
c)Probability that at most 4 drive to work alone.
Apr 15, 2020
#1
+21004
+1
Probability of driving to work alone = 0.77
Probability of not driving to work alone = 0.23
Total number of workers = 8
Probability that none drive to work alone: 8C0·(0.77)0·(0.23)8 =
Probability that exactly 5 drive to work alone: 8C5·(0.77)5·(0.23)3 =
Probability that at most 4 drive to work alone: [this would be either 4 or 3 or 2 or 1 or none of them]
= 8C4·(0.77)4·(0.23)4 + 8C3·(0.77)3·(0.23)5 + 8C2·(0.77)2·(0.23)6 + 8C1·(0.77)1·(0.23)7 + 8C0·(0.77)0·(0.23)8 =
Apr 15, 2020 | HuggingFaceTB/finemath | |
We think you are located in United States. Is this correct?
# 4.3 Atomic mass and diameter
## 4.3 Atomic mass and diameter (ESAAU)
It is difficult sometimes to imagine the size of an atom, or its mass, because we cannot see an atom and also because we are not used to working with such small measurements.
### How heavy is an atom? (ESAAV)
It is possible to determine the mass of a single atom in kilograms. But to do this, you would need special instruments and the values you would get would be very clumsy and difficult to work with. The mass of a carbon atom, for example, is about $$\text{1,99} \times \text{10}^{-\text{26}}$$ $$\text{kg}$$, while the mass of an atom of hydrogen is about $$\text{1,67} \times \text{10}^{-\text{27}}$$ $$\text{kg}$$. Looking at these very small numbers makes it difficult to compare how much bigger the mass of one atom is when compared to another.
To make the situation simpler, scientists use a different unit of mass when they are describing the mass of an atom. This unit is called the atomic mass unit ($$\text{amu}$$). We can abbreviate (shorten) this unit to just $$\text{u}$$. Scientists use the carbon standard to determine $$\text{amu}$$. The carbon standard gives carbon an atomic mass of $$\text{12,0}$$ $$\text{u}$$. Compared to carbon the mass of a hydrogen atom will be $$\text{1}$$ $$\text{u}$$. Atomic mass units are therefore not giving us the actual mass of an atom, but rather its mass relative to the mass of one (carefully chosen) atom in the periodic table. In other words it is only a number in comparison to another number. The atomic masses of some elements are shown in Table 4.1.
Element Atomic mass ($$\text{u}$$) Carbon ($$\text{C}$$) $$\text{12,0}$$ Nitrogen ($$\text{N}$$) $$\text{14,0}$$ Bromine ($$\text{Br}$$) $$\text{79,9}$$ Magnesium ($$\text{Mg}$$) $$\text{24,3}$$ Potassium ($$\text{K}$$) $$\text{39,1}$$ Calcium ($$\text{Ca}$$) $$\text{40,1}$$ Oxygen ($$\text{O}$$) $$\text{16,0}$$
Table 4.1: The atomic mass number of some of the elements.
The actual value of 1 atomic mass unit is $$\text{1,66} \times \text{10}^{-\text{24}}$$ $$\text{g}$$ or $$\text{1,66} \times \text{10}^{-\text{27}}$$ $$\text{kg}$$. This is a very tiny mass! If we write it out it looks like this: $$\text{0,00000000000000000000000000166}$$ $$\text{kg}$$. An atom is therefore very very small.
### Rutherford's alpha-particle scattering experiment (ESAAW)
Radioactive elements emit different types of particles. Some of these are positively charged alpha (α) particles. Rutherford wanted to find out where the positive charge in an atom is. He carried out a series of experiments where he bombarded sheets of gold foil with alpha particles (since these would be repelled by the positive nucleus). A simplified diagram of his experiment is shown in Figure 4.6. Figure 4.6: Rutherford's gold foil experiment. Figure (a) shows the path of the α particles after they hit the gold sheet. Figure (b) shows the arrangement of atoms in the gold sheets and the path of the α particles in relation to this.
Rutherford set up his experiment so that a beam of alpha particles was directed at the gold sheets. Behind the gold sheets was a screen made of zinc sulfide. This screen allowed Rutherford to see where the alpha particles were landing. Rutherford knew that the electrons in the gold atoms would not really affect the path of the alpha particles, because the mass of an electron is so much smaller than that of a proton. He reasoned that the positively charged protons would be the ones to repel the positively charged alpha particles and alter their path.
If Thomson's model of the atom was correct then Rutherford would have observed mostly path A in Figure 4.6. (A represents alpha particles that pass straight through the positive nucleus.) He would not have observed any deflection of the alpha particles. What he found instead was that while most of the alpha particles passed through the foil undisturbed and could be detected on the screen directly behind the foil (path A), some of the particles ended up being slightly deflected onto other parts of the screen (path B)! The fact that some particles were deflected suggested that the positive charge was concentrated in one part of the atom only.
Through this experiment he concluded that the nucleus of the atom is positively charged and situated in the middle of the atom.
### How small is the nucleus? (ESAAX)
The nucleus of an atom is very small. We can compare an atom with a soccer stadium. If the whole atom is as big as a soccer stadium, the nucleus is only the size of a pea in the middle of the stadium! This should help you see that the atom is mainly made up of empty space. If you removed all the empty space from the atoms in your body, you would become the size of a grain of salt!
### Relative atomic mass (ESAAY)
Relative atomic mass
The relative atomic mass of an element is the average mass of all the naturally occurring isotopes of that element. The units for relative atomic mass are atomic mass units.
The relative atomic mass of an element is the number you will find on the periodic table.
temp text | HuggingFaceTB/finemath | |
Custom math worksheets at your fingertips
# Details for problem "Long division of two whole numbers"
Quickname: 4913
Suitable for K-12 grades: Grade 2 Grade 3 Grade 4
Elementary School, Primary School.
## Summary
Division of two whole numbers, with remainder, written long division method illustrated.
## Description
Two positive whole numbers are presented for a division problem with our without a remainder.
The number of problems can be configured, as can the number of places of the divisor and dividend. It can be specified whether remainders are allowed, forbidden, or will always occur.
In the problem description, it may be requested that the division is executed and written down using the long division written method. The first problem may be output as an example with the solution to give some guidance.
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## Customization options for this problem
Parameter
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10
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1, 2, 3, 4, 5, 6
Remainder
Never, May, Always
Written method
no, yes, yes, with sample problem
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As division using place values, splitting the dividend
Divide by splitting the dividend.
Special case: Halving a number
Divide numbers by two, whole numbers or with decimal places.
With blanks to be filled in
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# 86 inches to cm
0
732
Are you looking to convert 86 inches to centimeters quickly and accurately? You’ve come to the right place! In this concise guide, we’ll provide you with a simple conversion formula that makes the process a breeze.
Whether you’re working on a home improvement project, comparing clothing sizes, or just curious about the metric system, understanding the relationship between inches and centimeters is essential.
By the end of this post, you’ll have all the information you need to convert any length from inches to centimeters with confidence. So, let’s dive into the world of unit conversions and make your life just a little bit easier!
## Conversion Process from 86 Inches to Centimeters
To convert 86 inches to centimeters, you simply need to use the following conversion formula:
1 inch = 2.54 centimeters
So, to convert a length in 86 inches to centimeters, you can use this formula:
Length (in centimeters) = Length (in inches) × 2.54
For example, if you want to convert 86 inches to centimeters, the calculation would be:
Length (in centimeters) = 86 inches × 2.54 = 218.44 centimeters
This means that 86 inches is equal to 218.44 centimeters. Using this formula, you can easily convert any length from inches to centimeters. | HuggingFaceTB/finemath | |
# More pixel to hex approaches
from Red Blob Games
31 Jan 2015
There are several different ways to convert a pixel location into a hex coordinate. On the main page I cover what I consider to be the simplest approach, reusing the `hex_round()` function that we already used in line drawing. But there are many other approaches that I’ve found or that readers have sent to me. Some are faster. Some are more flexible. Some handle edge cases better. I’m collecting them on this page.
## 1 Tested algorithms#
These are the algorithms I’ve tested myself, with code on this page, and the output of running that code. All of these tests are for pointy-topped hexes, and some of the algorithms use a helper function from the main page:
```function pixelToPointyHexFraction(x, y) {
let q = sqrt(3) / 3 * x + -1/3 * y;
let r = 2/3 * y;
return {q, r};
}
```
The first algorithm is the `hex_round` function from the main page. It uses three `round()` calls, three `abs()` calls, and two if/else branches.
If you look closely at the lines below or above the white hexagon, the boundary between the blue and red hexagons is not perfectly straight. There’s a (literal) edge case there that `hex_round()` treats inconsistently due to numerical precision / float errors. A reader says that changing `2/3 * y` to `(1 - 1/3) * y` in the `pixelToPointyHexFraction` code will help.
### 1.1. Charles Chambers#
Back in 2013, Charles Chambers sent me pixel to hex conversion code that has five `floor()` calls. First, divide `x` and `y` by `size * sqrt(3)`; then find `q,r` with:
```temp = floor(x + sqrt(3) * y + 1)
q = floor((floor(2*x+1) + temp) / 3);
r = floor((temp + floor(-x + sqrt(3) * y + 1))/3);
```
How does it work? Charles uses the `floor()` function to divide space into rows of rhombuses. Two rhombuses and two half-rhombuses make one hexagon. If you look at subexpressions in the “inner” calls to `floor()`, `2 * x`, `x + sqrt(3) * y`, and `-x + sqrt(3) * y`, you’ll notice that they’re dot products of `(x, y)` with the three basis vectors aligned with the hex grid. Each “outer” call to `floor()` combines two of the dot products.
If you’re using this for your own project, note that the `q, r` here are for a different axial form than the one I use in my guide, but I’ve modified my sample code here to match my guide.
### 1.2. Kenneth Shaw#
Kenneth Shaw posted a simplification of Charles Chambers’s algorithm:
### 1.3. James McNeill#
James McNeill’s site playtechs has a great hex grid guide including pixel to hex. He uses two separate affine transformations, one that makes `q` easy to calculate and one that makes `r` easy to calculate. There are six `floor()` calls.
#### 1.3.1. Eric Gradman#
Eric Gradman’s numpy code is an implementation of the playtechs algorithm:
#### 1.3.2. Saevax#
Saevax’s method is based on the playtechs hex guide, inlining and simplifying the code:
### 1.4. Justin Pombrio#
Justin Pombrio has a visual explanation of his algorithm, which elegantly has five calls (three `ceil`, two `round`):
### 1.5. BorisTheBrave#
BorisTheBrave’s algorithm is a variant of Justin Pombrio’s (two `ceil`, one `floor`, two `round`):
### 1.6. Mark Steere#
Mark Steere’s method looks elegant, with five `floor()` calls:
### 1.7. Jacob Rus#
Jacob Rus’s algorithm has three `round()`, two `abs()`, and one `if`. His page has diagrams, interactive demos, and code showing how it works.
### 1.8. vaxquis#
vaxquis sent me some code for pixel to hex, with each hexagon formed out of one rhombus and two triangles. There are two `floor()` for the rhombus and two `if` to handle the triangles.
## 2 Guess and test#
If you can somehow guess a location within a hex of the true answer, you can then correct the guess. Scan its neighbors and itself, and pick the one closest with Euclidean distance. This works because a regular hex grid also happens to satisfy the Voronoi property: the hexagon contains all points that are closer to its center than to any other hexagon center.
Unlike the geometric algorithms listed above, guess and test doesn’t require you to use cube/axial coordinates. It works the same with offset or doubled coordinates.
## 3 Pixel map#
The geometric algorithms above work on pure polygons, but they don’t always match up with a pixel grid. There are three issues, two of which are fixed easily:
The first issue is that you may need to adjust the x, y to point to the center of the pixel instead of the top left of the pixel. Do this by calling `pixelToHex(x + 0.5, y + 0.5)`.
The second issue is that if you want to align to integer pixel boundaries you will need to adjust the width or height of the hexagon, one of which has a sqrt(3) in it. In this diagram you can see that of the hexagons is slightly different in shape and alignment because the width has a sqrt(3) (non-integer value) in it:
To solve this we can stretch or shrink the hexagons slightly to make them line up with the pixel grid, so that they all are the same shape. The math on the main page doesn’t support this, but the implementation page does. In this case I set the width to be 10 ÷ (sqrt(3) ⨉ 6) times what it normally would be:
The third issue is that these hexagons may not look nice (symmetric, etc.), and may not match your desired pixel boundaries, especially if you have pixel art. The flaws are more easily seen when the pixels are large relative to the hexagons:
For more control, work directly with pixels. Compute or decide on the answer per pixel and store it in an array or bitmap. Hand tune the boundaries to make them look nice. You can get precise control so that it can fit your pixel grid.
A global pixel map would be to store the answer in an array or bitmap the size of the screen or game map. An array would contain pairs (`q:int, r:int`), or a bitmap could contain `q` in the red channel and `r` in the green channel. However, this array may be large.
For many hexagon sizes, especially if adjusting for pixel art, the pixel boundaries will repeat in an offset pattern:
If you have a repeating pattern, you can use a local pixel map. Each pixel inside the pixel map will be an offset relative to the hexagon at the top left of the pixel map. You’ll need to:
1. Calculate which rectangle the pixel is in.
2. Calculate which hexagon is at the top left of that rectangle.
3. Calculate which pixel within the rectangle.
4. Look up in the pixel map the hexagon offset relative to the top left hexagon.
5. Add that pixel’s offset to the top left hexagon.
For example you might determine rectangle’s top left is at hexagon `q:5, r: 1`. Then you look at the pixel within the rectangle, and see that its offset is `q:-1, r: +1`. Then you add the two, and get `q:4, r: 2` as the final hexagon in the lookup.
It’s more steps but the main advantage is that a local pixel map will be much smaller than a global pixel map, and it will also work on an infinite map. You may be able to use a rectangle even smaller than the one I show in the diagram. The disadvantage of the local pixel map is that it only works when all the hexagons follow a repeating pattern.
Another advantage of a pixel-perfect approach, whether global or local, is that you can handle the borders between hexagons. David Johnson has a multipage guide for hex grids that builds a global pixel map. He takes into account the width of the borders, explains the theory, and has interactive demos.
A system I used for an old game encodes the pixel boundary in an array. Many of the algorithms listed below use the repeating rectangle approach for lookups.
## 4 More algorithms#
By looking at the geometry of hexagons you can narrow down a pixel location to either one, two, or three hexagons, typically by dividing hexagons into smaller shapes. You can then use the edges between hexes to determine which subshape the pixel coordinate is on. There are a variety of shapes you can use. Here are some articles to read with details:
Do you have any other algorithms for this list?
The visual tests on this page are generated by this source code.
Email me , or tweet @redblobgames, or comment: | HuggingFaceTB/finemath | |
[ Home ] [ Up ] [ Current Grades ] [ MidTerm1-Practice ] [ MidTerm2-Practice ]
ENGR 315 MIDTERM #1 – PRACTICE TEST QUESTIONS
April 28, 1998
Instructions for the Test: This test will be closed book. However, you will be allowed to use up to 6 pages of your own notes. You will have 50 minutes. For each problem, you should show how you got your answer. The method of the solution will count more than the numerical values. We encourage you to draw diagrams and to state, explicitly and precisely, the definition of your random variables. Good luck!
Practice Test Notes: Below are 9 practice test questions. The nine questions should provide you with an understanding of the types of concepts that may be on the test. The test will consist of four questions.
Suppose that a consumer testing service rates lawn mowers using a simple binary rating scale: 0 or 1, where 1 is the preferred rating. There are three categories being rated: Ease of operation - 0 means difficult to operate, 1 means easy to operate; Cost – 0 means very expensive and 1 means inexpensive; and Ability to be repaired – 0 means difficult to repair and 1 means easy to repair
1. In how many ways can a lawn mower be rated by this testing service?
2. The total score for a lawn mower, Y, is the sum of the ratings over three categories. If a group of lawn mowers have equal probability to be rated as a 0 or a 1 in each category, and each category is considered independent, find the probability distribution for Y (i.e., find P(Y=0), P(Y=1), P(Y=2) and P(Y=3)). Show your work.
3. Find the expected score, E[X], and the standard deviation, s , of the score for the lawn mowers described in part b.
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 120 passengers. The probability that a passenger does not show up is 0.10, and the passengers behave independently.
1. What is the probability that every passenger who shows up can take the flight?
2. What is the probability that the flight departs with empty seats?
3. What is the expected number of passengers who will show up?
4. What is the standard deviation in the number of passengers who will show up?
For each of the following situations, identify the appropriate probability distribution and parameter values and set up the problem. You do not need to find a final numerical solution. Hint – start by precisely defining the random variable.
1. Printed circuit cards are placed in a functional test after being populated with semiconductor chips. A lot contains 150 cards, and 30 are selected without replacement for functional testing. If 12 cards are defective in the lot, find the probability that at least one defective card appears in the sample.
2. The phone lines to an airline system are occupied 50% of the time. Assume that 15 calls are made to the airline independently. Find the probability that exactly 8 calls are occupied.
3. You are playing basketball with a friend and taking turns shooting free throws. The first one to make 10 free throws buys lunch. If your probability of making a successful free throw is 0.6, what is the probability you make your tenth free throw on your 13th turn?
4. A particularly long traffic light on your morning commute is green 20% of the time that you approach it. Assume that each morning represents a random trial. Find the probability that the first time the light is green is on the fifth morning you approach it.
The edge roughness of slit paper products increases as knife blades wear. Only 1% of products slit with new blades have rough edges, 4% of products slit with blades of average sharpness have rough edges, and 6% of products slit with worn blades exhibit roughness. Currently, 30% of the blades in manufacturing are new, 60% are of average sharpness, and 10% are worn.
1. Draw a diagram to represent this scenario.
2. What is the proportion of products that exhibit edge roughness?
3. If a paper slit has a rough edge, what is the probability that it was caused by a blade of average sharpness?
A pop-quiz is given unannounced in the first 10 minutes of a statistics class. The quiz has 10 multiple choice questions with four answers each. Assume that Brian comes to class completely unprepared, not having had time to study and thus will simply guess at the correct answer to every question.
1. What is the probability that he will get a perfect score, that is, all correct?
2. What is the probability that he will get exactly two questions correct?
3. What is the probability that he will get at least two questions correct?
A particularly long traffic light on your morning commute is green 20% of the time that you approach it. Assume that each morning represents an independent trial. Hint: Define the random variable that is appropriate in order to determine the appropriate probability density function.
1. What is the probability that the light would be green upon approach for an entire week (5 days in a row)?
2. What is the probability you would need to go five days before getting a day with a green light?
3. What is the probability that for any given week (5 days), you would get 2 green lights and three red lights.
Three members of a private country club have been nominated for the office of president. The probability that Mr. Adams will be elected is 0.3. The probability that Mr. Brown will be elected is 0.5. The probability that Ms. Cooper will be elected is 0.2. Should Mr. Adams be elected, the probability of an increase in membership fees is 0.8. Should Mr. Brown or Ms. Cooper be elected, the corresponding probabilities for an increase in fees are 0.1 and 0.4.
1. Draw a diagram that represents captures this situation.
2. What is the probability that there will be an increase in membership fees?
3. If it is known that membership fees have increased, what is the probability that Ms. Cooper was elected?
The highway department has published a model for the number of cracks in Interstate 5. The number of cracks in a highway section that are significant enough to require repair is assumed to follow a Poisson distribution, with a mean of one crack per mile.
1. What is the probability that there are 3 cracks in one mile of highway?
2. What is the expected number of cracks in 5 miles, and the standard deviation in the number of cracks in 5 miles.
3. What is the probability that there is at least one crack that requires repair in 5 miles of highway?
4. You are on a long car trip and decide to pass time by recording some data about the number of cracks in the highway per mile. How would you transform the data in order to compare your findings with those of the highway department’s model?
Your company has recently purchase a new sorting machine that can concurrently sort five streams of letters in its five sorting compartments. However, because your company did not want to spend the extra money, the sorting machine you purchased has an undesirable behavior. If any of the five sorting compartments gets jammed, the entire machine shuts down. If the probability that a sorting component gets jammed during a day is 0.01 and the sorting compartments jam independently, what is the probability that the sorting machine will shut down during any given day? | HuggingFaceTB/finemath | |
# How does a geometric series converge, or have a sum?
1. Apr 21, 2015
### RyanTAsher
1. The problem statement, all variables and given/known data
How does a geometric series have a sum, or converge?
2. Relevant equations
Sum of Geometric Series = $\frac {a} {1-r}$
If r ≥ ±1, the series diverges. If -1 < r < 1, the series converges.
3. The attempt at a solution
How exactly does a infinite geometric series have a sum, or converge (tend to) a specific limit?
I understand that it is due to partial sums that we are able to derive the formula for the sum of a geometric series, yet at the same time I don't understand how a sequence that will be always multiplied by itself to infinity can ever STOP and have a final sum, or how it converges (tends toward) a specific limit.
For example $\sum\limits_{k=1}^{∞} (\frac {2} {3})^k$. This sum works out to 3, and does converge as r > -1 and r < 1. Why??? The partial sums proof makes no sense to me, and how on earth do things ever cancel to end up summing to 3?
Please explain to me if I have some conceptual misunderstanding of converges or sums of series, or if I'm just overlooking something...
2. Apr 21, 2015
### SammyS
Staff Emeritus
What proof are you familiar with?
What is it in that proof that you don't understand, or what is it that you can't 'buy'?
3. Apr 21, 2015
### kiritee Gak
Converging to something means it doesn't matter how much to add to that series(by next terms) (since they get smaller and smaller in our case) the sum does not exceed a particular value(the converging point).what is it u specifically want?? cheers.
4. Apr 21, 2015
### RyanTAsher
I don't exactly understand how at the end of the proof you can just ignore all the terms in between the kth terms and first few non-kth terms, and just divide those out. I understand the factoring out the a of the general equation ar^k, but how can you ignore the quotient of all the in between terms.
5. Apr 21, 2015
### RyanTAsher
That actually helps a lot to think of it that way... In other words since r ≤ ±1, the terms become so insignificant that their effect on the sum is negligible?
6. Apr 21, 2015
### SammyS
Staff Emeritus
For a series in general, having successive terms get smaller does not guarantee convergence of the series.
7. Apr 21, 2015
### kiritee Gak
i think i specifically mentioned "in our case" which is 2/3 series Ryan talking.
8. Apr 21, 2015
### Staff: Mentor
Please don't use "text speak" ("u" for "you") here at PF. From the forum rules (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/)
9. Apr 21, 2015
### Staff: Mentor
Their effect is not insignificant, but they don't cause the partial sums to go past a specific value. Here's an example of a geometric series with r = 0.9.
$$\sum_{n = 1}^{\infty} \frac 9 {10^n}$$
The partial sums go like this:
S1 = .9
S2 = .9 + .09 = .99
S3 = .9 + .09 + .009 = .999
and so forth.
Each partial sum merely adds one more 9 onto the end of the decimal fraction. All of the partial sums are bounded above by 1, but successive partial sums get closer and closer to 1 as more terms are added, allowing us to conclude that the sum of the series is 1.
10. Apr 23, 2015
### Fredrik
Staff Emeritus
The sum is 3 because for each n, the nth partial sum is
$$\frac{1-\big(\frac 2 3\big)^{n+1}}{1-\frac 2 3} = 3-3\left(\frac 2 3\right)^{n+1}$$ and the second term on the right goes to zero as n goes to infinity. If that doesn't make any sense to you, there's probably something missing in your understanding of limits of sequences. Perhaps you can be more specific about what's bothering you.
Note in particular that every partial sum is less than 3.
11. Apr 23, 2015
### RyanTAsher
Thank you that makes a lot more sense when you use the partial sums equation, rather than just the sum of an infinite geometric series equation. It makes sense now!
12. Apr 24, 2015
### epenguin
Your problem seemed to be you could not argue with the algebra yet could not imagine how it could be. Attached (excuse crudity of my first attempt with whiteboard but I hope it gives the idea) a visualisation convincingly showing
1/2 + 1/4 + 1/8 + ... → 1
View attachment 82466
#### Attached Files:
• ###### Screenshot 2015-04-24 09.55.50.png
File size:
44 KB
Views:
101
Last edited: Apr 24, 2015
13. Apr 24, 2015
### RyanTAsher
Thank you for the help, the visual representation helps a lot more too, actually being able to see something, essentially converge.
14. Apr 25, 2015 | HuggingFaceTB/finemath | |
# Physics - Maranda High School Form 1 End Term 1 2018
232/COMPOSITE
PHYSICS
FORM ONE 2018
TIME: 2HRS
MARANDA HIGH SCHOOL
END TERM ONE EXAM 2018
INSTRUCTIONS TO CANDIDATES
• This paper consist of Two sections; A and B
• Answer ALL the questions in section A and B in the spaces provided.
• All tables and Electronic calculators may be used.
Take g = 10N/kg
SECTION A; Answer all questions in this section
1.
1. What is surface tension? (1 mk)
2. Figure 3 below shows a funnel dipped into a liquid soap solution.
Explain what happens to the soap bubble when the funnel is removed (2mk)
2. 50 drops of a liquid were released from a burette which was originally reading 22cm3 to give a new reading of 56cm3. Calculate the volume of each drop. (2mks)
3. A sphere of diameter 4.5cm is molded into a thin uniform wire of diameter 0.15mm; calculate the length of the wire in meters. (Take π = 22/7) (3mks)
4. The diagram below shows a wire loop with two threads tied across it. The loop is dipped into a soap solution such that the soap film covers it as shown.
Region B is punctured such that the soap film in that section is broken. On the space alongside the diagram sketch the resulting shape of the wine loop. Give a reason for the shape (2mks)
5. A uniform mixture consists of 30cm3 of water and 40cm3 of ethanol. If the densities of water and ethanol are 1g/cm3 and 0.85g/cm3 respectively. Determine the density of the mixture. (3mks)
6. Two identical spring balances A and B each weighing 0.8N are arranged as shown below.
What are the readings of A and B? (2mks)
7.
1. Differentiate between scalar and vector quantities (1 mk)
2. State two examples of scalar quantities (2 mks)
8. Figure 2 shows water placed in a measuring cylinder calibrated in cm3
An object of mass 50.1g and density 16.7 g/cm3 is lowered gently in the water. Indicate on the diagram the new level (2mks)
9. An object is attached to a spring balance and its weight determined in air. It is then gently lowered into a liquid in a beaker. State what will happen to the reading (1 mk)
10. A butcher has a beam balance and masses 0.5 kg and 2 kg. How would he measure 1.5 kg of meat on the balance at once? (2 mks)
11. A thin wire was wound 30 times closely over a boiling tube. The total length of the windings was found to be 9.3mm. Calculate the radius of the wire (2mks)
SECTION B: Answer all the questions in this section
1.
1. Define force and give its SI unit (2 mks)
2. Differentiate between cohesive force and adhesive force (2mks)
3. Explain how increase in temperature affects surface tension (2mks)
4. Explain why a drop of oil spreads on a water surface (2mks)
5. An astronaut weighs 900N on earth. On the moon, he weighs 150N. Calculate the moon's gravitational strength. (Take g=10Nkg-1) (3mks)
2.
1. State two types of non-contact forces (2mks)
2. A body weighs 100N in air and 80N when submerged in water. Calculate the uptrust acting on the body in water (2mks)
3. The diagram below shows sets of tubes dipped inside a basin of water. Explain why the level of water is higher in tube A than in tube B. (2mks)
4. State three ways of reducing frictional force between two moving bodies (3mks)
5. Define the term 'Resultant Vector' as used in force. (1mk)
6. The figure below shows a body of mass m acted upon by different forces. On the space provided sketch the final resultant force on the body (2mks)
3. In an experiment to determine the density of sand using a density bottle, the following measurements were recorded;
Mass of empty density bottle = 43.2g
Mass of density bottle full of water = 66.4g
Mass of density with some sand = 67.5g
Mass of density bottle with the sand filled up with water 82.38
Use the above data to determine the;
1. Volume of water that completely filled the bottle; (3mks)
2. The volume of the density bottle; (1 mk)
3. Volume of water that filled the space above the sand (2mks)
4. Mass of the sand (2mks)
5. Volume of the sand (2mks)
6. Density of the sand (2mks)
7. State two precautions that should be taken while using a density bottle (2mks)
4.
1. Explain how physics is related to Biology as a subject. (2mks)
2. Differentiate between metre rule and metre ruler (1mk)
3. State the accuracy level of a metre ruler. (1 mk)
4. A form one student from Maranda High School was burnt with an acid in the laboratory while taking out an experiment. Give one fist aid measures he should take. (1 mk)
5. Highlight two major systems that must be installed in a laboratory for efficient functioning (2mks)
6. Estimate the area of the irregular surface shown below by counting the small squares. (3mks)
5.
1. Define pressure and state its SI unit (2mks)
2. A block of copper of density 8.9gcm-3 measures 5cm by 3.5cm by 2.1 cm. Given that the force of gravity is 10Nkg-1, Calculate,
1. The maximum pressure exerted by the block on the floor (4mks)
2. The minimum pressure that it can exert. (2mk)
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# Thread: Another induction problem help
1. ## Another induction problem help
I'm having really hard time about the induction. I have no idea how to start. Please help this problem!
2. ## Re: Another induction problem help
Base case is P(1) and as the problem says no proof is necessary since a single square
is already a square. The problem asks to assume that P(2). Now let $n\ge 1$
be arbitrary. Suppose P(n). To prove P(n+1) , assume that you have n+1 squares given. Use the inductive hypothesis on the first n squares. What do you get ?
3. ## Re: Another induction problem help
What is the inductive hypothesis here?
I'm new to this material so I'm so confused..
Does I.H means P(1)??
4. ## Re: Another induction problem help
Inductive hypothesis is P(n). It means that given n squares, we can dissect
them in finite ways and then rearrange them in a single square.
When you have n+1 squares, you can take first n squares and use inductive hypothesis to cut first n squares in finite ways and them rearrange them in a single
square. Now you have this rearranged square and the (n+1) th square. So you have
two squares. But the problem asks you to assume P(2). So given these two squares, you can cut them in finite number of ways and then rearrange them in a single square. So given n+1 squares, you have been able to arrange them in a single square after cutting them in finite number of ways. So we prove P(n+1). So given
P(n), we proved P(n+1). But since n was taken as arbitrary to begin with, using the principle of induction, this proves the result for all n (1,2,3.....).
If you are just beginning to learn mathematical induction then this problem is bit difficult. What text is this problem from ? | HuggingFaceTB/finemath | |
# Rotate an array to the right by a given number of steps
Recently, I encountered this coding exercise on codility and the idea is
A zero-indexed array A consisting of N integers is given. Rotation of the array means that each element is shifted right by one index, and the last element of the array is also moved to the first place.
For example, the rotation of array A = [3, 8, 9, 7, 6] is [6, 3, 8, 9, 7]. The goal is to rotate array A K times; that is, each element of A will be shifted to the right by K indexes.
Write a function:
class Solution { public int[] CyclicRotation(int[] A, int K); }
that, given a zero-indexed array A consisting of N integers and an integer K, returns the array A rotated K times.
For example, given array A = [3, 8, 9, 7, 6] and K = 3
the function should return [9, 7, 6, 3, 8].
Assume that:
N and K are integers within the range [0..100]; each element of array A is an integer within the range [−1,000..1,000].
My approach was if the array has zero or 1 element, the array given is returned else thne for loop is executed. I also added the new array to extract a subset of the array which was meant to be shifted and the first element of Array A is swapped with the last element. Finally, a new list is created with aim to achieve the insert the first element A after being swapped.
public static int[] CyclicRotation(int[] A, int K)
{
//Rotate an array to the right by a given number of steps.
// eg k= 1 A = [3, 8, 9, 7, 6] the result is [6, 3, 8, 9, 7]
// eg k= 3 A = [3, 8, 9, 7, 6] the result is [9, 7, 6, 3, 8]
if(A.Length== 0 || A.Length ==1)
{
return A;
}
int lastElement;
int[] newArray = new int[A.Length];
List<int> listOfNumbers = new List<int>();
for (int i = 1; i < K+1; i++)
{
lastElement = A[A.Length - 1];
newArray = A.Take(A.Length - 1).ToArray();
listOfNumbers = newArray.ToList<int>();
listOfNumbers.Insert(0, lastElement);
A = listOfNumbers.ToArray();
newArray = A;
}
return newArray;
}
I believe there is room for improvements. Any suggestions would be appreciated.
• Yes there is a lot of room for improvement. That is not efficient. In each loop you are creating two arrays, one list, and insert(0 which are all order N operations. – paparazzo May 20 '16 at 18:41
• You subtract the lengths by one in both lastElement and newArray why not just subtract the length the second you get A? – 13aal May 20 '16 at 19:49
• How would you handle the situation of K > N? – corsiKa May 21 '16 at 1:39
• Instead of moving the actual data, why dont you just move an index at where to start reading once the array shall be extracted? A lot faster to change one integer than the whole array – Mattias Åslund May 21 '16 at 7:31
• This is a learning project for sure. As @jakub pointed the task is in use since 1986 (or earlier). When you exercise with [1,2,3] all solutions are great. Consider an array of 10B length and you will see that any memory allocation appear to be very costly. – Alex Kudryashev May 21 '16 at 23:51
## In place solution
Jon Bentley "Programming Pearls" (1986) has this task as one of problems.
The recommended solution, instead of either:
• doing K 1-element rotations, using 1 temporary variable, with O(K N) time, which is slow
• using K-element temporary array, with O(N) time, which is wasteful... but only if there is requirement of doing it in place (and not returning copy, i.e. new N-element array)
is the Aha! solution of using reversals, with the following pseudocode
reverse(A, 0, K-1);
reverse(A, K, A.length()-1);
reverse(A, 0, A.length()-1);
assuming that reverse(A, I, J) operation reverses elements of array A from I-th element to J-th element, inclusive, that is, if $$A=\{a_0,a_1,...,a_i,a_{i+1}...,a_j,a_{j+1}...,a_{N-1}\}$$ then $$\text{reverse}(A, i,j)=\{a_0,a_1,...,a_j,...,a_{i+1},a_i,a_{j+1},...,a_{N-1}\}$$
For example, given array A = [3, 8, 9, 7, 6] and K = 3, the function should return [9, 7, 6, 3, 8].
Steps, corrected for the account of different meaning for K, namely whether it is number of elements to move from left to right ("Programming Pearls"), or from right to left (an example):
1. [3, 8,| 9, 7, 6] - original array
2. [8, 3,| 9, 7, 6] - reversed first part
3. [8, 3,| 6, 7, 9] - also reversed second part
4. [9, 7, 6,| 3, 8] - the solution
The idea by @forsvarir, namely:
You're currently rotating the array by 1 over and over again. Instead, think about where each element will end up at the end of N rotations. It's end position should be something like (startPos + numberOfRotations) modulus array size.
is another possible algorithm for an in-place rotation in "Programming Pearls". It can be written in an elegant way, but coming up with correct implementation is a bit more tricky. Though it does twice less operations than the reverse based one, each operation is more complex; also it doesn't play well when N is large enough that swap must be used (nowadays probably not a problem).
## Return copy solution
If you are returning a copy, you don't need a temporary K-element array. The solution is to create a result by copying appropriate parts.
In pseudocode (and in programming languages where you can do whole (sub)array copy) it would look like this:
Anew = allocate(N)
Anew[0:K-1] = A[N-K-1:N-1]
Anew[K:N-1] = A[0:N-K-1]
Here I assume that K means number of elements from the end to move to beginning, and that A[0:K-1] means array slice (a(1:K) in Fortran).
Note: I have not checked this solution - beware off-by-one errors here.
• I guess this is the goto approach when you want to rotate it in place – WorldSEnder May 21 '16 at 0:15
• [8, 3,| 6, 7, 9] - also reversed second part you are reversing it again as in the last step it is [9,7,6 | 3,8] not [6,7,9 | 3,8] because the original is [3, 8,| 9, 7, 6] so is reversing the 2nd half necessary? – Simple Fellow Apr 22 at 6:17
Naming
Think about your variable naming. You've used listOfNumbers for your local variable, however your parameter names are A and K. Do they mean anything without looking at the rest of the code?
Break down the problem into methods
The code below looks like it represents rotating the array by 1 place. If so, put it in a method with an appropriate name, then call it. It will make the problem breakdown easier to follow.
lastElement = A[A.Length - 1];
newArray = A.Take(A.Length - 1).ToArray();
listOfNumbers = newArray.ToList<int>();
listOfNumbers.Insert(0, lastElement);
A = listOfNumbers.ToArray();
newArray = A;
Think about the algorithm
You're currently rotating the array by 1 over and over again. Instead, think about where each element will end up at the end of N rotations. It's end position should be something like (startPos + numberOfRotations) modulus array size.
I think there is easier solution without creating additional arrays which only add overhead. All you want to do can be done using your original array and a few int variables. For example:
int len = A.Length; //self explanatory
int tmp = A[len - 1]; //save last element value
for(int i = len-1;i > 0; i--) //starting from the end to begining
{
A[i] = A[i - 1]; //assign value of the previous element
}
A = tmp; //now "rotate" last to first.
This is just a fragment of code and further improvements are still possible. For example, you can rotate K times recursively.
• So you think moving one at a time K times is optimal? – paparazzo May 20 '16 at 18:35
• @Paparazzi It's definitely an improvement over the original, which also moves one at a time but creates a new array each iteration. But if your point is there are yet better ways to do things, you're free to offer your own suggestions. – brian_o May 20 '16 at 18:39
• @brian_o The original actually returns the correct answer – paparazzo May 20 '16 at 18:42
• @Paparazzi surely no. This is just first step of improvement. Next steps may include a) optimize the number of steps K % len, b) save first K elements in temp array. Also negative K is a challenge. I'm not the author of the original code and I don't want to do instead of author. – Alex Kudryashev May 20 '16 at 18:42
• @Paparazzi I probably use misleading comment. It should be ...optimize the number of "rotations" K % len and then do all rotations in a single loop. – Alex Kudryashev May 20 '16 at 18:48
My suggestion is to use the built-in Array.Copy static method to minimize allocation of additional memory:
private static void RotateArrayRight<T>(T[] array, int count)
{
if (count < 0)
throw new ArgumentOutOfRangeException("count");
if (count == 0)
return;
// If (count == array.Length) there is nothing to do.
// So we need the remainder (count % array.Length):
count %= array.Length;
// Create a temp array to store the tail of the source array
T[] tmp = new T[count];
// Copy tail of the source array to the temp array
Array.Copy(array, array.Length - count, tmp, 0, count);
// Shift elements right in the source array
Array.Copy(array, 0, array, count, array.Length - count);
// Copy saved tail to the head of the source array
Array.Copy(tmp, array, count);
}
Usage:
int[] a = { 1, 2, 3, 4, 5, 6, 7 };
RotateArrayRight(a, 2);
// Now: a = { 6, 7, 1, 2, 3, 4, 5 }
The same approach with immutable source array:
private static T[] RotateArrayRight<T>(T[] array, int count)
{
if (count < 0)
throw new ArgumentOutOfRangeException("count");
// If (count == array.Length) there is nothing to do.
// So we need the remainder (count % array.Length):
count %= array.Length;
// Create a temp array to store the result
T[] tmp = new T[array.Length];
// Copy the last {count} elements of the source array to the head of the temp array
Array.Copy(array, array.Length - count, tmp, 0, count);
// Copy the rest elements
Array.Copy(array, 0, tmp, count, array.Length - count);
return tmp;
}
Usage:
int[] a = { 1, 2, 3, 4, 5, 6, 7 };
a = RotateArrayRight(a, 2);
// Now: a = { 6, 7, 1, 2, 3, 4, 5 }
• What happens when you rotate a 4 element array 5 times? :-) – brian_o May 20 '16 at 19:41
• @brian_o, I've added a fix: count %= array.Length;. Thanks. – Dmitry May 20 '16 at 19:44
You have an interface problem... caller passes an array and receives an array as return value. Is the return value meant to be a handle to the same array or a copy? Right now the behavior is inconsistent and depends on the size of the array argument.
static void DumpArray(int[] array)
{
foreach (var element in array)
{
Console.Write(element);
Console.Write(" ");
}
Console.WriteLine();
}
static void Main(string[] args)
{
{
int[] untouchable = { 1, 2, 3 };
DumpArray(untouchable);
var rotated = CyclicRotation(untouchable, 2); // won't change "untouchable"
rotated = 9;
DumpArray(untouchable); // good, still untouched
DumpArray(rotated);
}
{
int[] untouchable = { 1 };
DumpArray(untouchable);
var rotated = CyclicRotation(untouchable, 2); // won't change "untouchable"?
rotated = 9;
DumpArray(untouchable); // oh no! mutated!
DumpArray(rotated);
}
}
public static int[] CyclicRotation(int[] A, int K)
{
...
}
To make it consistent (and I think more intuitive), you can change the guard condition to if (A.Length == 0 || A.Length == 1) return (int[])A.Clone();. Better yet, rework your algorithm a bit and you won't need a special case.
Regardless, you should document the expected behavior and make sure your implementation actually follows through.
• I am passing a copy of the array as an input- passing by value rather than reference. Hence, the newArray was returned rather than the original A – Siobhan May 21 '16 at 22:11
• @TolaniJaiye-Tikolo That may have been your intention, but what I'm saying is that in the case of a 1 element array, your original code does not return a copy of the array but rather the original array. In your original code, for 2+ element arrays, you indeed return a copy. – brian_o May 22 '16 at 4:15
You could do the following:
var sample = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
var rotatedSample = Rotate(sample, 3);
protected static IEnumerable<int> Rotate(int[] source, int rotation)
{
var temporary = source.Take(rotation);
return source.SkipWhile(index => index <= rotation).Concat(temporary);
}
The benefit of this approach is simplicity. It will take based on the rotation, causing you to hold the original position's of your contents, but it will also shift the entire contents accordingly. Then you simply append the temporary to the end of the source, causing you to shift the entire array.
The performance may not be quite as quick as some other approaches, but the simplicity of syntax I feel is a valid trade off.
As others have mentioned, your solution is more involved (and slower) than necessary.
I would do this using an Int pointer that is an offset (mod "ArraySize") into the array, rather than actually moving elements within an array.
When you want to rotate the array, k steps to the right, then simply use k mod ArraySize.
When you need to access element N (0 for example) of array A, you read A[(0 + Offset) mod ArraySize].
The code below assumes that like the example in the original post, the "exercise" requires that the function has to return the original array (A), rotated, as a "new Array". If this is not really necessary, then the code below can be simplified even further.
I have also kept the same variable names you used in you example (A, k), because I assume those are the variable names given to you in the "exercise". Otherwise, consider making these names more descriptive.
Lastly, in your code, you have made a special case for when the size of the array (A) is "0 or 1". In the code below, these are handled naturally by the code and no special case treatment is necessary.
public static int[] RotateR(int[] A, int k)
{
// Rotate an array to the right by a given number of steps.
// eg k= 0 A = [3, 8, 9, 7, 6] the result is newArray = [3, 8, 9, 7, 6]
// eg k= 1 A = [3, 8, 9, 7, 6] the result is newArray = [6, 3, 8, 9, 7]
// eg k= 3 A = [3, 8, 9, 7, 6] the result is newArray = [9, 7, 6, 3, 8]
// int iOffset;
int iArraySize=A.Length;
int[] newArray = new int[iArraySize];
for (int i = 0; i < iArraySize; i++)
{
// iOffset=(k+i) % iArraySize;
// newArray[i] = A[iOffset];
newArray[i] = A[(k+i) % iArraySize];
}
return newArray;
}
Inefficient
In a loop you create 2 array, 1 list, and 1 insert at 0.
All of those are O(n) operations
You should not have any O(n) operations in the loop
public static int[] CyclicRotation(int[] A, int K)
{
//Rotate an array to the right by a given number of steps.
// eg k= 1 A = [3, 8, 9, 7, 6] the result is [6, 3, 8, 9, 7]
// eg k= 3 A = [3, 8, 9, 7, 6] the result is [9, 7, 6, 3, 8]
if(K > arr.Lengh || k < 0) throw new Excetion();
if(A.Length <=1 || K == 0)
{
return A.Clone();
}
// get a copy of the last k
int[] last = new int[K];
for (int i = 0; i < k; i++)
{
last[i] = A[a.A.Length - k + i];
}
// shift k
int[] newArray = new int[A.Length];
for (int i = A.Length - 1; i >= k; i--)
{
newArray[i] = A[i - k];
}
// copy the last 3 back
for (int i = 0; i < k; i++)
{
newArray[i] = last[i];
}
return newArray;
}
This is an excellent opportunity to use a LinkedList<int>.
To generalize the solution a bit, we can pass a boolean to a Rotate() method that determines the direction to rotate.
using System.Collections.Generic;
using System.Linq;
public class NumberArrayRotater
{
public int[] Rotate(int[] numbers, int cycles, bool goRight)
{
if (goRight)
{
for (var i = 0; i < cycles; ++i)
{
var node = linkedList.Last;
}
}
for (var i = 0; i < cycles; ++i)
{
var node = linkedList.First;
}
}
}
Also note that we could easily accept an IEnumerable<int> instead of an array.
• I have further generalized this and some other transformations that may be done to enumerables here. – xofz Jul 3 '16 at 6:18
There are two ways to do this - by the Programming Pearls swap trick, and by recursive rotation. The swap code is simpler to write and easier to prove, but the rotation code can be faster. Here's a worked example of each algorithm.
#include <stdio.h>
#include <stdlib.h>
// Rotate a subset of a array
// 0 1 2 3 4 5 6 7 8 9
// eg rotate A B C D E F G H I J, 2, 3, 8
// \_/ \_______/
// => A B E F G H I C D J
// \_______/ \_/
void reverse(int *A, int low, int high)
{
while (low < high) {
register int temp = A[low];
A[low++] = A[high];
A[high--] = temp;
}
}
void rotate_by_reversals(int *A, int first_index, int rot_index, int last_index)
{
// f r l
// 0 1 2 3 4 5 6 7 8 9
// eg rotate A B C D E F G H I J, 2, 3, 8
// \_/ \_______/
// => A B E F G H I C D J, 2, 3, 8
// \_______/ \_/
if (first_index == rot_index || rot_index+1 == last_index) return;
reverse(A, first_index, rot_index);
// f r l
// A B C D E F G H I J
// \_/
// => A B D C E F G H I J
reverse(A, rot_index+1, last_index);
// f r l
// A B D C E F G H I J
// \_______/
// => A B D C I H G F E J
reverse(A, first_index, last_index);
// f r l
// 0 1 2 3 4 5 6 7 8 9
// A B D C I H G F E J
// \___________/
// => A B E F G H I C D J
// \___________/
}
void swap_blocks(int *A, int left, int right, int len)
{
// preserve order
while (len-- > 0) {
register int temp = A[left];
A[left++] = A[right];
A[right++] = temp;
}
}
void rotate_by_blocks(int *A, int first_index, int rot_index, int last_index)
{
int len;
while (first_index != last_index) { // while loop replaces recursion
if (rot_index-first_index+1 < last_index-rot_index) {
len = rot_index-first_index+1;
// f r l
// 0 1 2 3 4 5 6 7 8 9
// eg rotate A B C D E F G H I J, 2, 3, 8
// \_/ \_______/
// => A B E F G H I C D J
// \_______/ \_/
swap_blocks(A, first_index, rot_index+1, len);
// f r l
// 0 1 2 3 4 5 6 7 8 9
// A B C D E F G H I J
// \_/ \_/
// => A B E F C D G H I J
// \_/ \_/
// rotate_by_blocks(A, rot_index+1, rot_index+len, last_index);
first_index = rot_index+1;
rot_index += len;
// f r l
// A B E F C D G H I J
// \_/ \___/
// => A B E F G H I C D J
// \___/ \_/
} else {
len = last_index-rot_index;
// f r l
// 0 1 2 3 4 5 6 7 8 9
// eg rotate A B C D E F G H I J, 2, 6, 8
// \_______/ \_/
// => A B H I C D E F G J
// \_/ \_______/
swap_blocks(A, rot_index-len+1, last_index-len+1, len); // or rot_index, last_index, -len ???
// f r l
// 0 1 2 3 4 5 6 7 8 9
// A B C D E F G H I J
// \_/ \_/
// => A B C D E H I F G J
// rotate_by_blocks(A, first_index, rot_index-len, rot_index);
last_index = rot_index;
rot_index -= len;
// f r l
// A B C D E H I F G J
// \___/ \_/
// => A B H I C D E F G J
// \_/ \___/
}
}
}
int main(int argc, char **argv) {
int fred, i;
for (i = 0; i < 10; i++) fred[i] = i+'A';
fprintf(stdout, "rotate_by_reversals(ABCDEFGHIJ, 2, 3, 8) => ");
rotate_by_reversals(fred, 2, 3, 8);
for (i = 0; i < 10; i++) putchar(fred[i]); putchar('\n');
for (i = 0; i < 10; i++) fred[i] = i+'A';
fprintf(stdout, "rotate_by_blocks(ABCDEFGHIJ, 2, 3, 8) => ");
rotate_by_blocks(fred, 2, 3, 8);
for (i = 0; i < 10; i++) putchar(fred[i]); putchar('\n');
exit(0);
return(1);
}
• Hi. Welcome to Code Review! It is more customary to write answers in the same language as the question. Also, you don't explain why either of these would be better than the original solution or other solutions. I also find the way that you comment out a recursive solution and add an iterative solution a bit confusing. It's not clear what code is shared between the two versions and what is specific to the iterative version. – mdfst13 May 21 '16 at 0:39
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## 12 July 2010
### Tutorial 03: Flipping Bits
Now that we have access to the registers, how do we actually flip the switches (change the bits)? We'll examine the ways to do this using the following example: we'll use a G2001 with an LED connected from P1.4 to ground. If P1.4 is at 0 V, nothing happens, but if P1.4 is at Vcc, the LED will turn on. Since we've connected to P1.4, we should be able to control it by fiddling with bit 4 in the P1 registers. (Keep in mind that we're not changing the 4th bit, at least not in the way we normally think of it. The 1st bit is bit 0, so bit 4 is technically the 5th bit, but who's counting?)
Before we look at setting bit values, we should take a look at the ways we can represent the values in 8-bit registers. We could always use regular decimal notation, but unfortunately it has little advantage to counter the confusion caused by having to use 8-bit numbers in a non-multiple base (the decimal base is, of course, 10).
The clearest way to represent the value is in binary notation, or base 2. The register has 8 digits, each is a 0 or 1, corresponding to the value in each bit. To make a binary 2 clear from a decimal 10, we can preface the number with the notation 0b (first character is a zero), such as 0b10.
Another common notation used is hexadecimal (base 16). This notation is convenient because each 8-bit value is represented by exactly two digits, each a value from 0-9 or a-f. The preface for hexadecimal is 0x, so the number 2 in hex is 0x02. The number 12 would be 0x0c, since the digits a-f correspond to the numbers 10-15. 16-bit numbers are represented with 4 digits, such as 0x14da. Many values in the MSP430 are 16-bit, so hexadecimal is very convenient here. (I apologize for the change in notation from the previous tutorial; in the TI documentation they use the notation 14DAh for hex values, but here we're using 0x14da instead. The 0x notation is used in the programming, and I will stick to that notation unless I'm citing from a TI document directly that uses the h notation.)
We could also use an octal notation (base 8), with the preface 0o. I've not seen this in practice, however.
We'll start using some actual code in this post, so I want to establish a convention. Anything written in the Courier font is a code snippet or example. The coloring in the code matches the color scheme I usually see when I look at a .c file in a text editor, and for now doesn't make a difference. You may have other color schemes that you see. I'll try to use the colors that are used in CCS as best I can.
Binary notation is obviously the one that makes it clear what bits we're using and what they're set to. Unfortunately, it also takes lots of characters to write them out. When we write C programs for the MSP430, we can include a header file with information and shortcuts specific to the device we're using (such as #include <msp430g2001.h>). These header files include some shortcut names for the binary values where each digit is 0 except for a specific bit; the header file allows us to use the name BIT2 to represent 0b00000100. There is a BIT0 through BIT7 included in each header file, which makes for a nice shortcut and less typing. In the discussion below, however, I'll try to keep it to the full 8 digit binary values to make it clear what is happening in the operations described. Now, on to the fiddling!
For our LED example, we'll need to first make P1.4 an output so that we can control the voltage above the LED. We do this by changing bit 4 in P1DIR to 1, for output mode.
1. Direct Assignment
If we know exactly what the binary (or hex, or even decimal) value corresponding to the configuration we need is, we can just set it directly: P1DIR = 0b00010000; would set all of the bits in P1DIR to 0 except bit 4. This method works pretty well for initializing the processor, but later in the program can be problematic when we might not know for certain what state bits we don't want to change are in. Turning off the bit is easy, just set it to the value with bit 4 set at 0: P1DIR = 0b00000000, or just as easily for this particular case, P1DIR = 0. (Using the BITx notation, we would write P1DIR = BIT4.)
2. Addition and Subtraction
If we don't know exactly the value needed for the configuration, or if we're just trying to turn on one bit, we could just add the bit's value to P1DIR (I'm using the colon character : to represent unknown digit values): 0b:::0:::: + 0b:::1:::: = 0b:::1::::, so we could turn on bit 4 by doing P1DIR += 0b00010000. *(Confused by the +=? See the footnote below!) The problem with this method is if we accidentally turn it on twice: 0b:::1:::: + 0b00010000 is going to cause a carry over into the next bits, changing I/O directions where we don't want to! This method should be used with caution. Note that the equivalent to turn the bit off would be to subtract the value, of course. (BITx notation: P1DIR += BIT4 or P1DIR -= BIT4.)
3. Logic Operators & and |
To avoid the problems with addition and subtraction, we can use logic operations. Let's say we have a bit with the value x. (x could be 0 or 1, we'll just look at this generally.) The operations and (&) and or (|) can be used to change the value of the bit to be specifically 1 or 0. Note that x & 0 = 0 no matter what value x is. Similarly, x | 1 = 1 regardless of the value of x. Also, x & 1 = x and x | 0 = x, so the value of x is unchanged. for these operations. So what would be the result of the operation P1DIR | 0b00010000? Bits 0-3 and 5-7 would be unchanged (remember, | 0 doesn't change the value) while bit 4 is set to 1. (It may have been 1 before, but is definitely 1 after the operation!) On the other hand, P1DIR &= 0b11101111 would have the effect of setting bit 4 to zero and leaving the others unchanged. Note that 0b11101111 is the not operation (~) applied to the original 0b00010000, so we could write this as P1DIR &= ~0b00010000. (In BITx notation: P1DIR |= BIT4 turns bit 4 on, and P1DIR &= ~BIT4 turns it off.)
Reader Exercise: Sometimes you'll see example code that uses the notation a + b; and sometimes a | b; to do the same thing. Convince yourself that these are equivalent in particular circumstances; when would you not want to use one or the other?
4. Toggling with ^
Ok, I said there were three choices, and there really are only three. This fourth option is similar to the third choice, but is an effective way if we just want to toggle the value, not set it to be specifically 0 or 1. The exclusive or operator (^) has this effect: x^0 = 1 if x is 1, and 0 if x is 0. x^1 = 1 if x is 0, and 0 if x is 1. So x^0 leaves the bit unchanged, and x^1 changes the value of x to its opposite. P1DIR ^= 0b00010000 flips bit 4 (to 1 if it was 0, to 0 if it was 1) just like toggling a switch. (In BITx notation: P1DIR ^= BIT4.)
Now, in our program we need to initialize the pin to be an output:
P1DIR = BIT4;
(An equivalent way to do this is to use a logic operator rather than an assignment: P1DIR |= BIT4;)
The value in P1OUT could be either 0 or 1, so if we care what state the LED starts in we should set that:
P1OUT = 0;
(A better practice is to initialize the output state before we change the pin to an output.)
Now we have an output pin set at 0, keeping the LED off. To turn it on, we give the command:
P1OUT |= BIT4;
To turn it off, we give:
P1OUT &= ~BIT4;
To toggle its state (off if on, on if off):
P1OUT ^= BIT4;
As a final example, demonstrating some power in the other assignment methods, let's put another LED on P1.6.
First, let's configure the two pins as outputs with an initial state off:
P1OUT = 0;
P1DIR = BIT4 + BIT6;
(adding the two values together is easily verified to be 0b01010000).
We can turn each one on and off individually:
P1OUT |= BIT4; // P1.4 on
P1OUT &= ~BIT4; // P1.4 off
P1OUT |= BIT6; // P1.6 on
P1OUT &= ~BIT6; // P1.6 off
Or we can turn them on and off together:
P1OUT |= BIT4 + BIT6; // both on
P1OUT &= ~(BIT4 + BIT6); // both off
And finally, toggle them simultaneously:
P1OUT ^= BIT4 + BIT6; // toggle both
Reader Exercise: Write down the necessary code to use LEDs on pins 3 and 7 of port 1, with pin 3 initially off and pin 7 initially on. Then toggle the LEDs so that 3 is on and 7 is off.
* += is a shorthand operator in C. If I say x += 1;, the command adds one to x and reassigns the result to x. In other words, it's the same as saying x = x + 1;. While algebraically that statement makes no sense, in programming it's very common. Don't think of = as meaning the two sides are equal, read it as "x is now equal to the value currently in x plus 1".
OCY said...
1) Usually, LEDs need to have a resistor in series to limit the amount of current.
2) In order to be able to turn on/off a LED, the LED can be either connected between an output pin and ground or between an output pin and Vcc. The voltage level at the output pin has opposite effect depending on which way the LED is connected.
3) The LaunchPad has a Red LED controlled by P1.6 and a Green LED controlled by P1.0. plazma said...
Bit manipulation is a good thing to learn in the beginning. It is used a lot when programming microcontrollers. Do the math with logic operators on papers and it's learned quite fast. Shifting << or >> is also good to learn.
http://en.wikipedia.org/wiki/Bitwise_operation
http://en.wikipedia.org/wiki/Bit_manipulation
David said...
I agree, which is why I brought it up so early in the tutorial. Thanks for the wikipedia links!
Another note on the LEDs, many LEDs can handle the currents from directly connecting to Vcc, but the MSP430 might not. In general, always limit the currents for LEDs in a design. Expect something around a few hundred ohms for the voltages we use to power these chips.
David said...
One other point... the LaunchPads I received have the red LED on P1.0 and the green LED on P1.6-- opposite what was expected from the release announcement.
m.Alin said...
I'm very curious as to what are the circumstances in which one should use the a + b or the a | b notations Anonymous said...
Under the Direct Assignment section:
What does the following mean:
(Using the BITx notation, we would write P1DIR = BIT4.)
What does equating P1DIR to BIT4 achieve?
David said...
Keep in mind that in C a single = is not equating, but assigning. The statement P1DIR = BIT4; assigns the value BIT4 to the P1DIR register. BIT4 is a shorthand notation for an 8 bit value where all bits are 0 except bit 4 (counting from 0-7). So:
BIT4 is 0b00010000, which is the same as 0x10. (If you look in the header file, you'll see a line that says #define BIT4 0x10.)
The effect of the line P1DIR = BIT4; is to set the direction flags for each GIO in P1 to 0 (making them inputs) except for P1.4, which is set to 1 (making it an output).
Keep in mind this is different from P1DIR |= BIT4;. This line is different in that it leaves all bits unchanged (be they 0's or 1's) except for bit 4--this line ensures that bit is set (to 1) without changing any other bits. P1DIR = BIT4; will set bit 4 and clear all others. You'll have to choose the assignment operator used according to the needs in the program.
The whole point of the notation is that the port registers can be viewed as individual switches for each port pin-- BITx in each register controls each respective pin, eg. BITx in the P1 registers control P1.x.
Hope that helps! Feel free to ask for any other clarification. Anonymous said...
Thanks for the earlier clarification regarding P1DIR= BIT4.
In light of the dangers involved in accidentally clearing bits using the above notation, wouldn't it be better if we always use logical operators | and & to manipulate bits?
Just a thought..
David said...
I agree, but with one exception: initialization. When your program starts, it can be helpful to leave no ambiguity as to the initial state of some peripherals. In the end, they both get you there, but I do tend to only use the direct assignment method at the beginning of my code. Everywhere else I'll use logic operators.
Thanks for helping to make that point!
billabott said...
Me again. on another site I stumbled across the missing file. (REv. E) and I quote:
BigglesPiP said ...
Coming from the Microchip PIC family of uCs it's nice to have real interrupts, I miss the RAM being flash a little.
But one thing I'm finding hard to forgive is TI's diabolical datasheets, there's nothing immediately useful in either the MSP430 family datasheet or the datasheet speific to the MCU.
After days of searching I came across a rumour that Revision E of something generic to MSP430 told me what I wanted to know (what every bit in every control register does and how to operate features). But the document is a 404 on TI's site. Eventually I found it, I shall host it while TI get their act together: http://iamtheb.org/me/MSP430x2xx_Family_Users_Guide_Rev_E.pdf
TI, can we please have good datasheets, Microchip's are the business.
--BigglesPiP 07:38, 18 January 2011 (CST)
David said...
Those are interesting points; I do think I need to come in TI's defense on this one, however.
First off, there's nothing magical about Rev. E itself, it's the Family User's Guide that's key. Rev. F is just as good, possibly better. Rev. C or D will also work, however later revisions will have corrections, additions for newer devices and technologies, etc.
Second, I personally think the datasheets are very useful. As I've been learning to use the MSP430, I've found that I use both the Family Guide and the Datasheet extensively as I figure out how something works, how it's implemented on my specific device, and how to code it up.
As a disclaimer, I do not work for TI at all... I'm simply an MSP430 user. I started using this uC as opposed to one of the others on the market simply because the chief electrical engineer on my project said to use it. As time has gone on, I've found there are some great features and advantages in this uC as opposed to some of the others, and I really enjoy using it. I'm not paid by TI in any way. NKT said...
Hi, just working through this again.
Firstly, can I input a binary number rather than using the BITx system?
Also, if I use your code and make P1DIR = BIT6 + BIT0; then if I toggle P1DIR both LEDs flash together.
However, could I toggle the whole of P1DIR rather than the BITs, to do the same thing? (And, save a line or two by only setting BIT6 and toggling the entire register to get alternate LED flashing?
Or is there also an "undefined" state as well as 0 and 1?
David said...
NKT,
You certainly can use binary directly; P1OUT = 0b01000001 is equivalent to P1OUT = BIT6 + BIT0.
Perhaps it was a typo, but I think it would make more sense to toggle P1DIR rather than P1OUT. Change P1OUT only if you really want to switch to inputs. However, you should be able to toggle all of P1OUT. Any pins configured as inputs are unaffected by changes to P1OUT. So P1OUT ^= P1OUT should toggle everything at once.
There are potentially undefined states when pins are configured as inputs; TI recommends unused inputs be tied to pull-up or down resistors, which can be done internally. Another options is to make all of them outputs and ground the unused ones. Of course, that makes toggling P1OUT less useful.
Hope these thoughts help some! metaone said...
When you want to toggle many bit in the same operation it is better to use | then + because
P1OUT = BIT1 + BIT1
makes P1OUT to be equal to BIT2 and
P1OUT = BIT1 | BIT1
makes P1OUT to be equal to BIT1
Personally, I would banish the use of + except if you really want an addition (which is often not the case when doing bits operations).
Also, many thanks for these great tutos.
David said...
I agree; though I have to admit I prefer using + in one situation: initializing peripherals. I think it makes more sense in my head (visually, at least) to say P1DIR = BIT0 + BIT2 rather than P1DIR = BIT0 | BIT2, even though they have the exact same result. It just makes the code more reasonable, in my opinion.
Your comment made me go back and read the other comments, and I realized that I made a type too.. in my previous comment I intended to say it's easier to toggle P1OUT rather than P1DIR. Hopefully that was more obvious from the context. I wish there was a way to edit comments here!
David said...
Oh, and now that I think about it, my suggestion of toggling P1OUT ^= P1OUT won't work; that will always return 0. P1OUT = ~P1OUT would toggle it, however. Looks like I've learned a lot since I wrote this tutorial... it's probably time to go back and edit/improve everything. Ulric said...
Hello David,
"P1DIR = BIT4 + BIT6;
(adding the two values together is easily verified to be 0b01010000)"
How come that? I thought it would be 0x00010100.
As well as:
"You certainly can use binary directly; P1OUT = 0b01000001 is equivalent to P1OUT = BIT6 + BIT0"
Isn't it BIT2 + BIT8?
thank you very much
David said...
The bit notation can be a bit confusing if you're not familiar with programming. There are 8 bits, numbered 0-7, and counted from right to left. So BIT0 is the least significant bit. The name comes from the fact that when this bit is set, it represents a value of 2^0. Likewise, BIT3 represents 2^3, and so on.
So when we say something like BIT4, it's not the 4th bit in, but rather the 5th bit from the right: 0b00010000.
Just remember that the order always looks like this:
0b(7)(6)(5)(4)(3)(2)(1)(0)
where the values in parentheses represent the bit number. Anonymous said...
@billabot,
the documentation on TI devices is generally divided into
* datasheets, containing hard number on one or a few microcontrollers properties such as maximum power supply voltage, sink current etc, as well as what peripherals are in that specific mcu (eg X timers, Y USCI etc etc)
* the FUG/family user guide, that tells you how to use that family of mcu's, what bitfields are in what registers etc
* application notes and white papers, outlining how to solve some specific task or use some peripherals etc.
They are usually very, very easy to find. Just google for your device, eg msp430g2553, and on the TI page for that, there are links that are easy to find.
Good luck.
SB said...
I got lost somewhere. I can't make sense of the statement:
"(A better practice is to initialize the output state before we change the pin to an output.)"
If we haven't set a pin to an output first, how do we change its state?
David Olson said...
Thanks for pointing out the confusing statement; exactly the kind of thing that will be fixed in the new website I'm working on.
What you are changing is the register that holds the values of the output. Until the pin is configured as an output, it does nothing externally as it's not connected. Once the pin is configured, it takes on the value that is held in the output state register. Setting that value before making the pin an output guarantees that the pin starts in the state you want it to have. | HuggingFaceTB/finemath | |
# What is the compound interest on $68619 at 9% over 21 years? If you want to invest$68,619 over 21 years, and you expect it will earn 9.00% in annual interest, your investment will have grown to become $419,180.28. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$68,619 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 21 years, but it could be monthly, weekly, daily, or even continuously compounding.
The formula for calculating compound interest is:
$$A = P(1 + \dfrac{r}{n})^{nt}$$
• A is the amount of money after the compounding periods
• P is the principal amount
• r is the annual interest rate
• n is the number of compounding periods per year
• t is the number of years
We can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.
For this formula, we need to convert the rate, 9.00% into a decimal, which would be 0.09.
$$A = 68619(1 + \dfrac{ 0.09 }{1})^{ 21}$$
As you can see, we are ignoring the n when calculating this to the power of 21 because our example is for annual compounding, or one period per year, so 21 × 1 = 21.
## How the compound interest on $68,619 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 21 years it is compounding: Start Balance Interest End Balance 1$68,619.00 $6,175.71$74,794.71
2 $74,794.71$6,731.52 $81,526.23 3$81,526.23 $7,337.36$88,863.59
4 $88,863.59$7,997.72 $96,861.32 5$96,861.32 $8,717.52$105,578.84
6 $105,578.84$9,502.10 $115,080.93 7$115,080.93 $10,357.28$125,438.22
8 $125,438.22$11,289.44 $136,727.66 9$136,727.66 $12,305.49$149,033.14
10 $149,033.14$13,412.98 $162,446.13 11$162,446.13 $14,620.15$177,066.28
12 $177,066.28$15,935.97 $193,002.24 13$193,002.24 $17,370.20$210,372.45
14 $210,372.45$18,933.52 $229,305.97 15$229,305.97 $20,637.54$249,943.50
16 $249,943.50$22,494.92 $272,438.42 17$272,438.42 $24,519.46$296,957.88
18 $296,957.88$26,726.21 $323,684.09 19$323,684.09 $29,131.57$352,815.65
20 $352,815.65$31,753.41 $384,569.06 21$384,569.06 $34,611.22$419,180.28
We can also display this data on a chart to show you how the compounding increases with each compounding period.
As you can see if you view the compounding chart for $68,619 at 9.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure. ## How long would it take to double$68,619 at 9% interest?
Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of $68,619 assuming an interest rate of 9.00%. We can calculate this very approximately using the Rule of 72. The formula for this is very simple: $$Years = \dfrac{72}{Interest\: Rate}$$ By dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this: $$Years = \dfrac{72}{ 9 } = 8$$ Using this, we know that any amount we invest at 9.00% would double itself in approximately 8 years. So$68,619 would be worth $137,238 in ~8 years. We can also calculate the exact length of time it will take to double an amount at 9.00% using a slightly more complex formula: $$Years = \dfrac{log(2)}{log(1 + 0.09)} = 8.04\; years$$ Here, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value. As you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember. Hopefully, this article has helped you to understand the compound interest you might achieve from investing$68,619 at 9.00% over a 21 year investment period. | HuggingFaceTB/finemath | |
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## Reconstructing an ordering of a multiset from its consecutive submultisets
We have a multiset $S$ of size $t$ with $r$ distinct elements, where $t$ is much larger than $r$. We want to reconstruct an ordering $s_1, s_2, ... s_t$ of the elements of $S$ given the values of $t$ and $r$ and, for some fixed positive integer $j$, the set of multisets $\{ s_k, s_{k+1}, ... s_{k+j-1} \}$ for $1 \le k \le t-j+1$. (Note that we are not given either the order of elements in each multiset or the order of the multisets.) We also know $s_1, ... s_{\lfloor pt \rfloor}$ for some fixed $0 < p < 1$.
For what values of $t, r, j, p$ is it possible to reconstruct the entire ordering $s_1, ... s_t$? For those values, what is the average- or worst-case complexity of doing so?
Edit (Qiaochu Yuan): Here's a restatement of the problem in the language Steve Huntsman is using. The generalized de Brujin graph $B(r, j)$ (it does not seem to have a standard name) has vertices the set of all words of length $j$ from an alphabet of size $r$ and edges defined as follows: the word $w_1, ... w_j$ has an edge directed to the word $w_2 ... w_j w_{j+1}$ for all possible choices of $w_{j+1}$. There is a natural equivalence relation on words where two words are equivalent if the same letters occur in them with the same frequency. What we are trying to do is reconstruct a walk on $B(r, j)$ of length $t - j + 1$ given only the set of equivalence classes of its vertices, and also given an initial segment of the walk which is some fixed proportion of the entire walk.
-
Rob, there are some things about your question I don't understand. When you write "[0,1,...,R]", what do the square brackets mean? When you say that the ordered set S can be represented as a binary string, what do you mean? I wonder whether you're using the term "ordered set" in the usual mathematical sense, as for instance in en.wikipedia.org/wiki/Order_theory . If not, some clarification would be good. – Tom Leinster Jan 18 2010 at 21:14 Dear Tom, the square brackets have no special meaning, I was using them because jsMath was making '{}' brackets disappear (I'll fix it because it's caused concern) - for [0,1,...,R] I'm simply referring to a set with up to 'R' unique elements in it. Also, I meant 'ordered set $S$' in a loose computer-science sense (again, sloppy) - I should say 'permutation of a set $S$'. The idea is that with two unique elements in $S$, a binary string with a length equal to the cardinality of $S$ would represent a particular permutation. – Rob Grey Jan 18 2010 at 21:45 Rob, I'm still confused. How are you writing permutations as binary strings? – Qiaochu Yuan Jan 18 2010 at 22:03 Qiaochu, I was talking about representing a unique permutation of two elements as a binary string (and only talking about binary strings as an illustrative example), not trying to enumerate permutations or list permutations. My objective here is to try to recover a unique permutation in S with the scrambled subsets. Does that make more sense to you, or should I try to rewrite the question? – Rob Grey Jan 18 2010 at 22:09 Your notation is very confusing. What does R = 2 mean if there are T elements in your set? – Qiaochu Yuan Jan 18 2010 at 22:11
I have some unpublished notes that I think are relevant to this problem, and I am cherry-picking from them. (See also here for a related question.) Because the notes are unpublished I hope it will not be considered poor form to provide a rather lengthy response.
To save myself trouble while picking the relevant parts from them, I will write $q$ in place of $r$ and $k$ in place of $j$. Also, let $B = q^k$ and let $N$ be the length of a generic cyclic word (or necklace) over a $q$-ary alphabet $\mathcal{A}_q$.
As Gerhard suggests, the problem can be recast in terms of necklaces and their subwords of length $k$. We introduce the appellation dimgraph for “directed multigraph”. Let $[w]_k$ denote the generalized order $(k – 1)$ de Bruijn dimgraph of the $q$-ary word $w$: i.e., vertices correspond to the $(k – 1)$-tuples, and directed edges correspond to the $k$-tuples in $w$ in the usual way. If $[w′]_k = [w]_k$, we say that $w$ and $w′$ belong to the same order $k$ de Bruijn homology class. It is easy to see that this really is an equivalence relation, which we denote $\sim_k$ (suppressing subscripts when there is no risk of confusion). If the size of an equivalence class is greater than unity, it will be necessary to take into account the additional data associated to $p$, but I will not touch on that (or on the formula to determine the multiplicity of a nontrivial equivalence class, though if anyone wants this please leave a comment and I will try to reply with a PDF) here.
For convenience, assume first that $q = 2$ and define the frequency tuple $\alpha = \alpha(w)$ as the $B$-tuple of natural numbers indicating the frequency of occurrence in $w$ of the (lexicographically ordered) $k$-tuples over $\mathcal{A}_q$. For later convenience we take the frequency tuple to be an integral column vector. For example,
$w = 00011 \in \mathcal{A}_2^5, k = 3 \Rightarrow \alpha(w) = (\alpha_{000},\dots,\alpha_{111})^T \equiv (\alpha_0,\dots,\alpha_7)^T = (1,1,0,1,1,0,1,0)$.
Notice that we start the index at zero: this will be our default convention for all indices relating to $k$-tuples, and context should generally suffice to determine the indexing scheme for a particular object.
The Eulerian property of generalized de Bruijn graphs means that $\alpha$ is completely specified by the $B/2 + 1$ components $\alpha_0,\alpha_1,\dots,\alpha_{B/2-1},\alpha_{B-1}$: conversely, any such assignment completely specifies the corresponding generalized de Bruijn dimgraph—though notice here that we make no implicit requirement that such a dimgraph must as a matter of course be strongly connected (i.e., correspond to a word). We could make other choices for the “free” components: the essential thing is that the two “loop” components $\alpha_0$ and $\alpha_{B-1}$ must be free, and any other $B/2 – 1$ components will suffice to complete a specification. The reason for this is basically that the Eulerian property (besides its requirement of strong connectivity) provides $B/2$ independent equations in $B$ unknowns.
More explicitly, for k > 1, write the frequency tuple as a column vector:
$\alpha = (\alpha_0,\alpha_{[0]}^{(k)},\alpha_{[1]}^{(k)},\alpha_{B-1})^T; \quad \alpha_{[0]}^{(k)} = (\alpha_1,\dots,\alpha_{B/2-1})^T, \quad \alpha_{[1]}^{(k)} = (\alpha_{B/2},\dots,\alpha_{B-2})^T$.
Now there is an integral matrix—actually there are many integral matrices—satisfying
$\alpha_{[1]}^{(k)} = M_{[0 \rightarrow 1]}^{(k)} \alpha_{[1]}^{(k)}$.
There is an algorithm for computing these matrices for arbitrary $k$: it is just an exercise in index-juggling, however, so we omit the details. The point is that we can dispense with the unnecessary components of frequency tuples for the binary case, although some care must be taken. Also, it turns out to simplify matters considerably if we omit the words of all zeros and all ones from consideration: we will do so without comment when convenient. More generally, it is convenient to require that none of the components of the frequency tuple are zero (which also implies that none of them equal $N$), but we shall be explicit when making such an assumption.
We can treat some of the components of a frequency tuple in the manner just outlined as functions into the natural numbers. In turn, the matrix-tree and BEST theorems can be jointly recast as a recipe for another function—the BEST function—from this space into the natural numbers that gives the cardinality of the de Bruijn equivalence class.
...
With the binary case in hand, we seek to generalize to $k$-tuples over $\mathcal{A}_q$. The approach will be somewhat different here. Define ancestor and descendant matrices by
$M_\leftarrow := 1_{1 \times q} \otimes I_{B/q}, \quad M_\rightarrow := I_{B/q} \otimes 1_{1 \times q}$.
Now the indegree/outdegree (I/O) or weak Eulerian property reduces to
$M_\leftarrow \alpha = M_\rightarrow \alpha$.
Nothing is lost here by forming the augmented ancestor and descendant matrices
$\tilde M_\leftarrow := \delta_1 \otimes M_\leftarrow + I_B, \quad \tilde M_\rightarrow := \delta_1 \otimes M_\rightarrow + I_B$
and characterizing the I/O property instead by
$\tilde M_\leftarrow \alpha = \tilde M_\rightarrow \alpha$.
The augmented matrices are invertible, and so we can rewrite the I/O property once more as
$M_{(I/O)}\alpha \equiv \left( \tilde M_\rightarrow^{-1} \tilde M_\leftarrow -I \right)\alpha = 0$.
In other words, a proper frequency tuple must belong to the kernel of the consistency matrix on the LHS above. The dimension $d$ of this in/out (I/O) kernel $\ker_{I/O}(q,k)$ is just the number of independent variables required to determine a frequency tuple.
This number is $d = B – B/q + 1$. A sketch of a proof is as follows: of the $B$ edges (including any with multiplicity zero) in an order $(k – 1)$ generalized de Bruijn dimgraph, $q$ are loops: the corresponding frequency tuple components obviously need to be treated as free variables. For the remaining edges, we have B/q independent linear equations in $B – q$ variables. This gives a net of $B – B/q$ independent variables for a minimal integral solution to the linear equations. The +1 is accounted for by scaling.
Accordingly, if the word length is fixed to $N$ then evidently there are $B – B/q$ free components.
For small $q, k$ an exact/integral basis for the in/out kernel can be formed easily (e.g., by using the null command in MATLAB with the r option), and since these are the only cases we can realistically address, we will not bother with an algorithm.
We have two comments at this point. The first is that the reader should be convinced (or take it as an exercise to prove) that the I/O kernel contains an integer lattice whose primitive cells are relatively small (a precise formulation and attendant proof are probably rather technical, and certainly uninteresting). The second is that the lattice can be generated by a positive integral basis: the few negative entries above can clearly be made to go away with a few elementary linear operations.
That is, we consider the I/O lattice
$\Lambda_{I/O}(q,k) :- \ker_{I/O}(q,k) \cap \mathbb{Z}^B$
and the associated lattice cone
$\Lambda_{I/O}^{0+}(q,k) := \ker_{I/O}(q,k) \cap \mathbb{N}^B$.
Except for some degenerate cases corresponding to disconnected generalized de Bruijn dimgraphs (to be addressed in the sequel), points in this cone correspond to valid frequency tuples. For computational purposes we want to be able to efficiently generate the frequency tuple simplices
$\Lambda_{I/O}^{(N)}(q,k) := \Lambda_{I/O}^{0+}(q,k) \cap N \Delta_B$
where a dilation of the standard simplex is indicated on the RHS. (It will be an implicit corollary of our construction of these sets that the I/O lattice is unimodular, but we do not use this.)
We first define a bucketspace (“$s$ balls in $r$ buckets”) $X_{r,s} := \{\alpha \in \mathbb{N}^r : \sum_{j=1}^r \alpha_j = s\}$. Now it is clear that there exists a $B \times d$ integer matrix $M^{(\Lambda)}$ s.t.
$M^{(\Lambda)}X_{B,N} = \Lambda_{I/O}^{(N)}(q,k)$
(where the LHS is interpreted in the obvious way). In fact there may be several such integer matrices—each reflecting a particular choice of basis—but all we need is one. We will call it a bucketspace-I/O lattice matrix.
The problem of its construction appears subtle (but perhaps that is because we have focused on explicit examples). A useful heuristic is to invoke the LLL lattice basis reduction algorithm on an integer basis for the I/O kernel, then use the structure of the de Bruijn graph—in particular, a table of its cycles—to improve on LLL. Note that the cycle enumeration for de Bruijn graphs has only been effected for small values of $k$.
In practice there are some subtle considerations, which we sketch in the context of the two specific cases $q = 2$ and $k = 3, 4$. Since the I/O kernel is the span of the dimgraph cycle space (see the sequel), it makes sense to look for a set of cycles corresponding to a “nice” basis. A natural candidate is a reduced basis of positive lattice vectors with minimal length.
Once the simple cycles of the de Bruijn graph are enumerated, it is easy to produce such a positive basis with a greedy algorithm. Start with the cycle representatives as a proto-basis. Now pick one of the remaining simple cycles of minimal weight uniformly at random: if it is linearly independent of the existing proto-basis, add it. Keep going until a basis is formed.
For $q = 2, k = 3$ the 6 simple cycles and corresponding simple cycle vectors are
$(0) \leftrightarrow (1,0,0,0,0,0,0,0)^T$
$(001) \leftrightarrow (0,1,1,0,1,0,0,0)^T$
$(0011) \leftrightarrow (0,1,0,1,1,0,1,0)^T$
$(01) \leftrightarrow (0,0,1,0,0,1,0,0)^T$
$(011) \leftrightarrow (0,0,0,1,0,1,1,0)^T$
$(1) \leftrightarrow (0,0,0,0,0,0,0,1)^T$
and the greedy algorithm produces the unique minimal positive basis
$\{(0),(001),(01),(011),(1)\}$.
Similarly for $q=2, k=4$ we have the 19 simple cycles
$(0),(0001),(0001011),(00010111),(00011),(0001101),(000111),(00011101),(001)$, $(001011),(0010111),(0011),(001101),(00111),(0011101),(01),(011),(0111),(1)$
and the greedy algorithm produces either of the two minimal positive bases
$\{(0),(0001),(001),(001011),(0011),(01),(011),(0111),(1)\}$
$\{(0),(0001),(001),(0011),(001101),(01),(011),(0111),(1)\}$
Note that the simple vectors of length 5 are rejected by the greedy algorithm, as is the simple vector $(000111)$ of weight 6.
(Notice that Gordan’s lemma implies that the monoid [commutative semigroup with identity and cancellation] of integral points in a lattice cone is finitely generated. This in turn implies that we can use one of the minimal positive bases to produce frequency tuple simplices if we are willing to pay a large overhead factor over a more efficient [if less well-understood] method outlined below.)
By performing elementary column operations it is possible to find a workable bucketspace-I/O lattice matrix by trial and error (of course an algorithm would be preferable, but the cycle decomposition of de Bruijn graphs is known only for small values of $k$, suggesting that such a goal is too lofty). We find by direct experiment with $N = 16$ that the following matrices work:
The columns of the $k = 4$ matrix can be indicated graphically:
Finally, notice that the method outlined essentially boils down to finding the vertices of a frequency tuple simplex—but this appears to be more easily said than done.
...
Now that the I/O condition has been suitably developed, we move on to the question of (strong) connectivity. Let $B$ (with an abusive resue of notation) denote the incidence matrix of a dimgraph $G$. Without bothering to restate the common definition, we simple provide a relevant example: for $q = 2$ and $k = 4$ we have
Notice by way of passing that we can recast the I/O equations as
$B1 = 0$.
It can be shown that $\ker B$ is the cycle space of $G$.
In any event it is a standard result for dimgraphs that
$\mbox{rank}(B(G)) = |V(G)| - c(G)$
where on the RHS we indicate the number of vertices minus the number of components. In short, the rank of the incidence matrix allows us to judge whether or not the corresponding dimgraph is (strongly) connected.
Our intent here is that a generalized de Bruijn dimgraph has only a single component by definition: although we have not really produced a proper mathematical definition, this intent should suffice for the remainder. Now the requirement on a frequency tuple is that
$|V(\alpha)| - \mbox{rank}(B(\alpha)) = 1$.
The number of vertices is just the number of $(k – 1)$-tuples making an appearance; the rank can be computed by Gaussian elimination. For $q = 2$, it can quickly be seen that the rank is just the number of nonzero components in $(\alpha_1,\dots,\alpha_{B/2-1})$.
With that in hand, define the tuple canonical ensemble (TCE) by
$\hat \Lambda_{I/O}^{(N)}(q,k) := \Lambda_{I/O}^{(N)}(q,k) \cap \{\alpha: |V(\alpha)|-\mbox{rank}(B(\alpha)) = 1\}$.
The TCE gives all the frequency tuples for a given length, and no garbage points, although some care is required in its construction.
In conclusion: check to see if a given point is in the TCE. If it is, then use the matrix-tree and BEST theorems (again, ask me how and I'll provide a PDF if I see the comment) to determine its cardinality. Providing an enumeration and selecting $p$ minimal completes the solution to the problem, although I'm not sure about the value of $p$, complexity, etc.
-
added some matrix-tree/BEST details in a separate answer. – Steve Huntsman Jan 19 2010 at 5:25 Steve, thanks so much for your notes! It's going to take me a little bit to go through and understand the argument. I should mention however that I'm most interested in the case where the permutation of the multiset S is a long random sequence rather than a debruijn sequence... – Rob Grey Jan 19 2010 at 12:44 The sequences in question are arbitrary. I use the term "generalized de Bruijn" to emphasize the connection between sequences and Eulerian paths on certain associated graphs, etc. – Steve Huntsman Jan 19 2010 at 13:35 Steve, got it, thanks that makes sense after more carefully reading your post. – Rob Grey Jan 19 2010 at 15:27
(First, apologies to the administrators for not yet registering.)
If you consider S to be a debruijn sequence (a short sequence where each pair of letters from an alphabet occurs as a contiguous subword, e.g. 0120210), you will have a high amount of symmetry in the information given, and not be able to distinguish which sequence to reconstruct, even if you are told that it is a debruijn sequence. (You might be able to if you were given a long enough initial segment of the sequence.) Also, you may not be able to distinguish the string from its reverse, given only the information you list above. So I think a unique reconstruction will be impossible.
I am looking at a similar problem where I want to compress a long list of numbers by looking at all sublists of length j for small j. Here I have your information plus the ordering on all subsets of length j, but I have not found a way to reconstruct the list because for each length (j-1) prefix, I have several choices to complete the length j list. It seems I will need large j to do the reconstruction uniquely, which will defeat the intent of compressing the sequence.
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Gerhard, I'm imagining very long sequences, where the cardinality of $S$ is in the hundreds to thousands. Also, I should add a note that one is allowed some short initial stretch of the permutation in $S$. – Rob Grey Jan 18 2010 at 22:04 You have an interesting problem, I will think about it. – Rob Grey Jan 18 2010 at 22:04 To compress the list, save the adjacency matrix of the induced generalized de Bruijn graph and a number indicating which of the members of the generalized equivalence class it corresponds to. I have some notes from 2003 on this in the context of Kolmogorov complexity and its applications to phylogenetic analysis using mitochondrial DNA if you're interested. – Steve Huntsman Jan 19 2010 at 5:00
Generally speaking, I think these sequences should be non-reconstructible, at least when $j \leq r$.
For instance, consider the following two sequences:
01230123012301
01320132013201
These are indistinguishable (when j = 4) if we only look at the multisets and don't know the first few elements of the sequence. However, we can get around this problem for some fixed $p > 0$ by appending each of these sequences to some other sequence s.t. the initial fixed sequence has length $pt$. The only difference is that the first one has a consecutive subsequence of the form "x012" where the other one has one of the form "x013"; this can be remedied by appending an "x" to the end of both sequences.
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I'm just about to walk out the door, but (quickly) I believe it's true that with P(1/r^j) - i.e. where $r$ the number of unique elements in the multiset and $j$ is the size of any subset - we can create a non-uniquely reconstructable sequence. I'm interested in the limit where P(1/r^j) << 1. – Rob Grey Jan 19 2010 at 3:50 Ah, and thanks! – Rob Grey Jan 19 2010 at 3:53 The necklace ATAGTC contains the 2-lets AT, TA, AG, GT, and TC. There is another necklace that has the same initial and terminal singlets and the same multiset of doublets: namely, AGTATC. – Steve Huntsman Jan 19 2010 at 14:37
No "palindromic sandwich" is reconstructible. By this I mean an ordering of the form $aba'$ where $a'$ is $a$ reversed and $a$ is at least the length of the peek of the initial segment you get. This ordering cannot be distinguished from $ab'a'$ regardless of the value of $j$; in other words, the problem cannot be solved when $p < \frac{1}{2}$. Perhaps you want instead an algorithm that generates all possible orderings?
-
Dear Qiaochu, I don't expect every sequence to be perfectly reconstructible given arbitrary computational resources and you make a good point about the palindromic sandwiches. "Perhaps you want instead an algorithm that generates all possible orderings?" - Yes, it would be terrific to have an efficient algorithm to enumerate all 'legal' orderings so that probabilities could be assigned to them, or at least to see the growth in the number of orderings as a function of (t, r, j, p). – Rob Grey Jan 19 2010 at 15:34 And please note that I still need to take time to think about Steve Huntsman's answer. – Rob Grey Jan 19 2010 at 15:37
Dealing with the cardinality of the equivalence classes naturally fits into another answer, which I'm covering here.
We begin with the binary case. The first step in this direction is to notice that the adjacency matrix of a generalized binary de Bruijn graph has a very simple structure:
$A_{ij}(\alpha) = \alpha_{2i} \delta_{2i \mod (B/q),j} + \alpha_{2i+1} \delta_{(2i+1) \mod (B/q),j}$
The corresponding matrix/combinatorial Laplacian is prepared from the recipe
$\mathcal{L}(A) := \Delta(A1) - A$
where $\Delta$ denotes any of the various diagonal operations (context should suffice to determine which). The matrix-tree theorem gives that the number of directed spanning trees is any principal minor (they are all equal):
$t(A) = \hat{\mathcal{L}}_{i,i}(A), \quad \forall i$.
It does not appear to be worth working out an explicit formula for such a determinant for $k$ generic, even with $q = 2$. A key argument supporting this pessimism is that any finite Eulerian (hence strongly connected) dimgraph can be embedded in some generalized de Bruijn graph (just take an Euler trail to see this). In the example $q = 2$, $k = 4$ we have
$\hat{\mathcal{L}}_{B,B}(\alpha) = \alpha_1 \alpha_7 (\alpha_8 + \alpha_9)(\alpha_{12} + \alpha_{13})(\alpha_2 \alpha_5 \alpha_{11} + \alpha_3 \alpha_4 \alpha_{10} + \alpha_3 \alpha_4 \alpha_{11} + \alpha_3 \alpha_5 \alpha_{11})$
and this takes several pages to derive by hand using expansion by minors. It is not hard to see that there is no way to perform row or column permutations to cast the matrix into block diagonal form (except in degenerate cases), and in the minor expansion enough terms will probably crop up to yield (at best) a Pyrrhic victory. For special cases where a large set of terms possessing symmetries might be set to zero, searching for a formula might be fruitful: however, we will not pursue any such line here.
The BEST theorem states that the number of unlabeled Euler circuits in a de Bruijn equivalence class is given by the BEST function
$f_{BEST}(\alpha) := t(\alpha) \cdot \prod_{i=0}^{B/q-1} (\deg_i(\alpha) - 1)!$
where the vertex in- or out-degrees (they are the same) are indicated. Evaluating this function quickly becomes quite demanding as the number of evaluations and $k$ increase. The factorial terms can only add to any computational demands this function makes.
Actually, things get even worse: the unlabeled Euler circuits enumerated by the BEST function are in a many-to-one correspondence with necklaces. The correspondence is nontrivial for periodic necklaces, so dividing the BEST function by the obvious product of factorials does not quite work. The proper formula for the necklace function is (as J. Jonsson pointed out on sci.math.research)
$f_{neck}(\alpha) := \sum_{d|\gcd \alpha}\frac{\phi(d) f_{BEST}(\alpha/d)}{d\cdot(\alpha/d)!}$
where the Euler phi function is indicated and the functions on tuples are defined in the obvious (componentwise) way. In lieu of a detailed proof we merely offer a brief explanation: for each term in the summation, the $f_{BEST}(\alpha/d)$ contribution gives the number of labeled “divisor-Euler” circuits with the commensurate length; the factorial terms account for removing the labels. The $d$ term in the denominator accounts for the concatenation of these “divisor-Euler” circuits into an actual Euler circuit. Finally, the phi function accounts for inequivalent shifts amongst these various “divisor-Euler” circuits.
We can move on from here to get the q-ary BEST and necklace functions. The adjacency matrix of a generalized q-ary de Bruijn graph is not much different than in the binary case:
$A_{ij} = \sum_{\ell=0}^{q-1} \alpha_{qi+\ell} \delta_{(qi+\ell) \mod (B/q), j}$.
It is an exercise in notation to show that the Laplacian is
$\hat{\mathcal{L}}_{ij}(\alpha) = \sum_{\ell = 0}^{q-1} \alpha_{qi+\ell}\left(\delta_{ij} - \delta_{(qi+\ell) \mod (B/q),j} \right)$
and the $q$-ary form of the necklace function has the same form as the binary version.
As an example, consider the case $q = 2$: it is easy to get that
$f_{BEST}(\alpha) = \frac{\alpha_1(\alpha_0 + \alpha_1)!(\alpha_1 + \alpha_3)!}{(\alpha_0 + \alpha_1)(\alpha_1 + \alpha_3)}$
and from this that
$f_{neck}(\alpha) = \frac{\alpha_1}{(\alpha_0 + \alpha_1)(\alpha_1 + \alpha_3)} \sum_{d|\gcd(\alpha_0, \alpha_1, \alpha_3)} \phi(d) \binom{\frac{\alpha_0 + \alpha_1}{d}}{\frac{\alpha_1}{d}}\binom{\frac{\alpha_1 + \alpha_3}{d}}{\frac{\alpha_1}{d}}$.
To take the example further, let us fix the word length at 16 and compute away. We compute nontrivial values of the necklace function over the $\alpha_0$-$\alpha_1$ plane for $0 \le \alpha_0 \le 14$ and $1 \le \alpha_1 \le 7$:
4 7 4
22 56 75 56 22
42 126 210 245 210 126 42
43 120 212 280 309 280 212 120 43
22 55 90 120 140 147 140 120 90 55 22
7 12 17 20 23 24 25 24 23 20 17 12 7
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Notice also that the necklace function equals unity for $(\alpha_0, \alpha_1) = (0, 0)$, $(0, 8)$, and $(16, 0)$.
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I don't think that any p<1 is large enough to work for all t. The hardest case to reconstruct is 2 distinct elements (if r>2, there is nothing to stop you from using only 2 of them or putting the rest once each in the front). Even if you have the first pt terms there are $2^{(1-p)t}$ ways to finish. There are only j+1 multisets of size j so the information you get back is just an ordered list of j+1 integers which add to t-j. These number about $\binom{t}{j}$ which is less that $t^j$ which will be outstripped by any exponential. But maybe I don't understand, what about sequences which are all 0 with a single 1 out past the initial part you are given (but at least j away from the end)? I don't think that even having the multiset of ordered j-substrings would be much help there.
- | HuggingFaceTB/finemath | |
# NCERT Solutions for Class 8 Maths Maths - NCERT Solution Chapter 12
## Exponents and Powers Class 8
### Exercise 12.1 : Solutions of Questions on Page Number : 197
Q1 :
Evaluate
(i) 3 - 2 (ii) ( - 4) - 2 (iii) (i) (ii) (iii) Q2 :
Simplify and express the result in power notation with positive exponent.
(i) (ii) (iii) (iv) (v) (i) ( - 4)5 ÷ ( - 4)8 = ( - 4)5 - 8 (am ÷ an = am - n)
= ( - 4) - 3 (ii) (iii) (iv) (3 - 7 ÷ 3 - 10) × 3 - 5 = (3 - 7 - ( - 10)) × 3 - 5 (am ÷ an = am - n)
= 33 × 3 - 5
= 33 + ( - 5) (am × an = am + n)
= 3 - 2 (v) 2 - 3 × ( - 7) - 3 = Q3 :
Find the value of.
(i) (30 + 4 - 1) × 22 (ii) (2 - 1 × 4 - 1) ÷2 - 2
(iii) (iv) (3 - 1 + 4 - 1 + 5 - 1)0
(v) Q4 :
Evaluate (i) (ii) Q5 :
Find the value of m for which 5m ÷5-3 = 55.
Q6 :
Evaluate (i) (ii) Q7 :
Simplify. (i) (ii) ### Exercise 12.2 : Solutions of Questions on Page Number : 200
Q1 :
Express the following numbers in standard form.
(i) 0.0000000000085 (ii) 0.00000000000942
(iii) 6020000000000000 (iv) 0.00000000837
(v) 31860000000
Q2 :
Express the following numbers in usual form.
(i) 3.02 x 10-6 (ii) 4.5 x 104
(iii) 3 x 10-8 (iv) 1.0001 x 109
(v) 5.8 x 1012 (vi) 3.61492 x 106
Q3 :
Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to m.
(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm | HuggingFaceTB/finemath | |
## Graphing Stories
18 Aug
How do you and your coworkers effect change in the classroom? Two years ago, I asked our principal whether we could schedule a geometry class during first block with four teachers and about 25 students. It was our first year to implement our new CCSS Geometry standards, and we needed to try it together. I have learned over the years that it doesn’t hurt to ask – he might say no, but he might also said yes. Well, he said yes, and as you can imagine, sharing a class together has been important for us both as learners and as teachers. As one of the teachers reflected recently, “Participating in a class all year with a team of teachers is the best professional development I have ever had.” Last year, our Algebra 2 team had a shared class to implement their new standards, and this year, our Algebra 1 team shares a class. I visit as often as I can.
We are building our course as we go, using all sorts of resources. We are using the framework from EngageNY, and we are using some of the activities and tasks in their lessons. We have started with a unit on Graphing Stories.
On the first day, we used Growing Patterns from an NCTM Article Coloring Formulas for Growing Patterns.
Lesson Goals (written with Jill Gough this summer):
Level 4: I can represent the number of tiles in a figure in more than one way and show the equivalence between the expressions.
Level 3: I can represent the number of tiles in a figure using an explicit expression or a recursive process.
Level 2: I can apply patterns to predict the number of tiles in a later figure.
Level 1: I can describe the pattern and draw a figure before and after the given figures.
How do you see the pattern growing?
How many tiles are in H(5)? H(100)? H(N)?
Some students completed a table of values, and some students drew a graph.
On the second day, we used a Mathematics Assessment Project formative assessment lesson, Interpreting Time-Distance Graphs, with a focus on rate of change.
Lesson Goals:
Level 4: I can calculate average rate of change from a graph and a table.
Level 3: I can calculate average rate of change from a graph or a table.
Level 2: I can match a distance-time graph with a story and a table.
Level 1: I can annotate a graph using away from home and towards home, how fast and how slow.
On the third day, we used Graphing Stories Video 2 to begin the lesson.
Lesson Goals:
Level 4: I can create a believable story for a given graph.
Level 3: I can calculate average rates of change for an elevation graph.
Level 2: I can create an elevation graph from a video, labeling the axes with appropriate units of measure.
Level 1: I can identify time intervals for each piece of an elevation graph.
Level 1: I can calculate the slope of a line.
Whose graph is believable?
We used additional scenarios from Engage NY Algebra 1, Module 1, Lessons 1-2.
There were more videos and scenarios on Day 4 from Engage NY Algebra 1, Module 1, Lessons 3-4.
The lesson from Day 5 comes from David Wees’ webinar at the Global Math Department on Strategic Inquiry.
We have read Timothy Kanold’s blogpost on Leaving the Front of the Classroom Behind. And we are trying. And we are most grateful to our administrators for letting us try it together before we have to try it alone. | HuggingFaceTB/finemath | |
So pitch of roof is equal to = rise/run = 6/12. Therefore, mark off 12 inches on the level and place it down horizontally against the roof rafter.
Convert Roof Pitch To Degrees Solar panels roof, Pitched
What is a roof pitch?
How to find the pitch of a roof. Recalculate this value into an angle: If you want to know how to find the pitch of a roof, then, start with a calculator! Watch this short video titled area of a roof instead.
Roof pitch, or slope, is a measure of vertical rise to horizontal run expressed in inches per foot. Now it is 3:12 and can be written as 25% or 14°. Which of roof is calculated by number of inches it rises vertically for every 12 inches in extend horizontally.
Roof pitch refers to the measurement of the slope of a roof and you express this as a ratio. Pitch = tan (angle), (use this when you express the angle of the roof pitch in degrees) what is the standard pitch of a roof? Pitch = rise / run = 1.5 / 6 = 25%.
Roof pitch refers to the amount of rise a roof has compared to the horizontal measurement of the roof called the run. Even if you want to use a flat roof pitch calculator system, you will be able to find one. The pitch of your roof is 3:12.
The roof pitch is important in determining the appropriate installation method and how much roofing material will be needed. 8 feet by 15 feet answer by tutorcecilia(2152) (show source): A roof slope of 4:12 is a 1/6 pitch, a 6:12 slope is a 1/4 pitch, an 8:12 slope is a 1/3 pitch and so on.
The pitch of the roof is the slope of the roof that is due to the rafters. Put again this value into an angle: The roof pitch is a measure of roof steepness.
Finding a roof pitch angle. The rise is the height of the roof at its highest point, and the run is the horizontal span of the roof, measured from the roof ridge to the side of the building. This will allow your project to carry the depth and detail in the roof that you wanted, as well as make sure that it will maintain a sturdy and effective nature throughout.
Read How To Make Website Secure On Chrome
You can put this solution on your website! This will be a right triangle: It is basically the steepness of the roof of a house.
Roof pitch calculator results explained. The pitch is commonly defined as the ratio of rise over run in the form of x/12. The run is the distance from the outside of the wall to the inside of the ridge.
Angle = arctan (pitch) = arctan (0.25) = 14°. Calculate the pitch of a roof from the rise and rune measurements. The roof pitch is the slope of the rafter.
To see how pitch impacts the look of a garage and changes cost click the design center button on our pole barn kits page. Lastly, you can have the roof pitch in the form of x:12. Amongst the finest ways of calculating roof pitch angles is by having a definite range that the surface of the roof is going to make a horizontal plane at.
But, again we are jumping ahead of ourselves. So, a roof that rises six inches for every foot of run has a “6 in 12” slope. The rise is the distance from the top of the roof to the bottom.
Most people use it to mean the steepness of the roof. Some carpenters consider it more correct to use the term slope for the steepness, and use pitch to describe the roof's overall dimensions. If a magazine such as finehomebuilding, is going to put out an article to teach a topic, it should use the proper words and definitions.
The term pitch has two different meanings when it comes to roofs. It is expressed as an angle or as a pitch ratio. The rise is the height of the roof, and the run is the horizontal span (as pictured above).
Read How To Design A 3d Printed Plane
Pitch = rise / run = 1.5 / 6 = 25%. For example a roof that rises 6 inches vertically for every 12 inches of horizontal run. The pitch is defined as the slope of the roof, and is commonly represented as the ratio of the rise over the run in the form x/12.
X = pitch * 12 = 0.25 * 12 = 3. Roof pitch is used to describe the slope, or angle, of the roof. Finding the pitch of a roof is simple with a speed square.
The roof’s pitch is the number of inches the roof rises in 12 inches. The incline or pitch of a roof can be easily measured with a level, tape measure, and a pencil. Here is an online rise and run calculator which helps to calculate pitch, slope and angle with the help of rise and run values.
First, measure 12 inches from one end of the level and make a mark. For example, if a roof has a pitch of 4/12, then for every 12 inches the building extends horizontally, it rises 4 inches. Pitch is shown as a fraction, which divides the rise by the span.
Find the pitch of the roof shown. Finally, you can find the roof pitch in the form of x:12. Measure a horizontal distance (run) inward from the edge of the building either on top of the fascia board or below.
X = pitch * 12 = 0.25 * 12 = 3. Just make sure you measure the vertical height (rise) to the same as you did the run measurement, to the top of the fascia or the bottom. The pitch of a roof is the height of the roof (rise) it obtains after a certain distance (run).
For a roof with the ridge above one wall, the span is the same as the run. Divide the slope by two to get the pitch. Then, in the attic, place the end of the level against the bottom of a roof rafter and hold it perfectly level.
Read How To Accept Yourself Wikihow
Compute the roof pitch as the percentage of rise and run: Carpenters or civil engineers make use of diverse finders of roof pitch angles, an example being a rafter. I hear/read all too often this misuse and usually ignore it.
It can also be written down as 25% or 14°. Eg a 6 / 12 for every 12 inches of horizontal length the roof height increases by 6 inches. For a simple gable roof, the span (wall to wall) will be twice the size of the run (wall to ridge, horizontally).
Angle = arctan (pitch) = arctan (0.25) = 14°. = let the slope = the hypotenuse = c let one side = a=8 Before we get started with all that we need to find out the pitch of a roof.
Use the pythagorean theorem to solve.
My friend wanted to extend his lounge and wanted a pitched
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30' x 40' Ellington CT (8/12 Roof Pitch) Farmhouse | HuggingFaceTB/finemath | |
# Pade approximations for state space systems
I am studying the application of Pade approximations on a system with a dead time. In most cases, we consider a transfer function with a dead time and apply the Pade approximation on it.
Now here is my question: can we introduce a delay in state space? If yes, how one can apply the Pade approximation on it? If no, why is it not possible?
Yes, it is possible to consider Pade approximants for delays in linear state space systems. The simplest approach is to construct the transfer function for the delay and convert this to a state space model. As an example, the second order Pade approximation for a time delay of $T$ is $$H_{\text{delay}}(s) = \frac{s^2-6/T s+12/T^2}{s^2+6/Ts+12/T^2}$$ which corresponds to a state space system with matrices $$A_{\text{delay}} = \left[ \begin{array}{cc} 0 & -12/T \\ 1 & -6/T \end{array} \right], \quad B_{\text{delay}} = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right], \quad C_{\text{delay}} = \left[ \begin{array}{c} -12/T & 72/T^2 \end{array} \right]$$ where I have used the companion state space form.
Supposing we want to apply this delay at the input to another linear state space system with matrices $A_{\text{sys}}$, $B_{\text{sys}}$, $C_{\text{sys}}$, $D_{\text{sys}}$, we can achieve this by forming the larger system with matrices: $$A = \left[ \begin{array}{cc} A_\text{delay} & 0 \\ B_\text{sys} & A_\text{sys} \end{array} \right], \quad B = \left[ \begin{array}{c} B_{\text{delay}} \\ 0 \end{array} \right], \quad C = \left[ \begin{array}{c} D_\text{sys} & C_\text{sys} \end{array} \right], \quad D = 0$$ In practice, most of this can be carried out easily in Matlab by using the pade() and ss() functions. | HuggingFaceTB/finemath | |
How to Make a Digital Valentine's Day Card
# How to Make a Digital Valentine's Day Card
### Description
Valentine's Day is tomorrow. Here's how to make a digital card to capture the heart of your sweetie. Well, OK, that might be overstating things a bit, but Harley does show an algorithmic way using trigonometry to generate a beating heart animation in an easy to use programming environment called Processing. This was inspired by a video LeiosOS posted last week.
The original inspiration: LeiosOS: Drawing a heart
Wolfram algorithm site: Heart shaped graphs
For a written transcript, go to How to Make A Digital Valentine's Day Card
Intro/Exit: "Hot Swing" by Kevin MacLeod at http://incompetech.com
Incidental: “Carefree” by Kevin MacLeod at http://incompetech.com
livingroom_light_switch by AlienXXX at http://freesound.com
### Transcript
Today at the House of Hacks we're going to make a high-tech Valentine's Day greeting.
[Music]
Hi Makers, Builders and Do-it-yourselfers. Harley here.
I recently ran across a new-to-me programming channel who's name I won't even begin to try to pronounce but I will link to in the cards and description below.
He did a video about drawing a heart where he used a bit of C code and his own graphics library.
I want to take his idea, expand a bit on why it works and show how to use the Processing programming environment to do the same thing.
First let's look at generating shapes...
Many people get overwhelmed by the thought of trigonometry with its sine and cosine, angles and other related things, but at its core, it’s really pretty simple.
Remember the old cartesian graph from math class?
It had an x-axis and a y-axis and points could be located on this graph using just two coordinates.
On this graph, let’s draw a circle with its center at the origin with a radius of one.
Now, let’s imagine a line starting at the origin and going through the circle.
We can draw this line at any angle from the x axis.
If it is on the axis it will be 0 degrees.
Regardless of where we draw it, it’s going to intersect the circle at only one place.
That one place will have an x and y coordinate that, by definition, is the value of the cosine and sine functions at that angle.
So, for 0 degrees, the place where the circle intersects the x-axis, cosine will be 1 and sine will be 0.
As the angle increases, the x value decreases and the y value increases until we get to 90 degrees.
This lies on the y axis where cosine is 0 and sine is 1.
As the angle continues to increase, the x and y values continue to change, always between 0 and 1, always on the circle.
And that’s the fundamental theory behind trigonometry, everything else is derived from this.
Now let’s plot these points a bit differently.
On the x axis, let’s plot the angle and on the y axis, let’s plot of value of sine remembering that sine is the y value on our original circle.
At 0 degrees, sine is 0 so we start at the origin.
As the angle increases, sine increases until we get to 90 degrees and reach 1, then it starts decreasing as we move into the second quadrant of the circle.
At 180 degrees sine is back to 0.
As the angle continues to increase, the y coordinate of our circle drops into the negative values in the third quadrant until it reaches -1 at 270 degrees.
Finally in the fourth quadrant, we see y coming back up to 0.
After this, the cycle simple repeats itself as higher angles are duplicates of the previous angles.
Now, doing the same thing with cosine, at 0 degrees, cosine is 1 since it represents the x value on the circle.
As the angle increases, x decreases to 0 at 90 degrees, -1 at 180, back up to 0 at 270 and finally 1 at 360 degrees, the place we started.
These functions are great for describing a lot of things that are periodic in nature, things that fluctuate between two values.
If we multiply the results of sine or cosine we can make these graphs fluctuate by a different amount.
For example, here’s the result of multiplying the sine by 2.
And if we multiply the angle by a value, the frequency of the fluctuation will increase or decrease.
Here’s what it looks like if we multiply the angle by two.
Things can get really wild if we start adding these two curves together.
Or subtracting them.
Or multiplying them.
Or doing anything else really.
Sines and Cosines are the basis for a lot of things in our modern world.
Today we’re going to see how they can be used to send a message to our sweetheart.
Processing is a programming environment that was developed about 15 years ago and designed to teach the basics of programming within a visual context.
Since it was designed with a visual context in mind, it’s great for experimenting with graphics and drawing things.
The Processing.org website has free installers for Windows, Linux and Macs.
This gives us a a Processing environment that, once started, we can start writing programs in.
Let's look at a simple program to draw a circle.
We first define the size of our canvas, the color of our background and the color we want to draw in.
Then we’re going to define a variable to scale our circle by and an offset for the x and y coordinates.
These are needed because sine and cosine return values around the cartesian origin but the coordinate system of the Processing screen is based with 0,0 in the upper left corner of the canvas.
Then we’re going to define a loop from 0 to twice PI, going in small increments.
We used twice PI here because in computers cosine and sine typically use what are called radians instead of degrees as their inputs.
This is just fine though because there’s a one to one mapping between degrees and radians where PI is 180 degrees and two PI is 360 degrees.
Inside the loop we set x to the cosine value of the loop variable and y to the sine value of the loop variable.
Then we’re simply going to plot that point, offset and scaled by the previously defined values.
Let's save our work at this point.
When we press the arrow button in the top tool bar, our program runs and we get this nice circle.
Now that we have the basics of creating a circle, how’s this help us with our Valentine Day project?
A web search for “heart drawing formula” leads us to the Wolfram site that has a number of different formulas for drawing a heart shape.
The sixth one in particular lends itself well to the program we have for drawing a circle.
All we have to do is change the assignments to x and y to get something that looks like a heart.
One of the cool things about the Processing environment is it works well for animating things.
If we put the statements before the loop in a function called setup they will get called one time when we run our program.
Then if we put the loop inside a method called draw, it will get continually called over and over again.
But because we’re drawing the same thing over and over, it's not going to be too exciting.
So before running this, let's add a line to change the scaling each time the draw method is called and adjust a control parameter to keep the scale between a couple values.
Finally, move the background call from setup into draw so it happens every time draw is called. This will clear our canvas each time.
Now let’s run this.
Ah, a nice beating heart.
This is cool and all, but it’s hard to show our special someone.
By adding a line to the end of our draw method, we can tell Processing to save our drawing to a file.
Now let's run this for a little bit, stop it and go look at where we saved our program.
We can see a bunch of individual snapshots of each frame of our animation.
In one last step, we can go to Processing’s Tools menu and select Movie Maker.
It gives us a dialog where we can select the directory containing the shots of our animations and press the Create Movie button.
This compiles our individual images into a MOV file that we can e-mail to our intended.
[Music]
If this is your first time here at House of Hacks: Welcome, I’m glad you’re here and would love to have you subscribe.
I believe everyone has a God-given creative spark.
Sometimes this manifests through making things with a mechanical and technical bent.
Through this channel I hope to inspire, educate and encourage these types of makers in their creative endeavors.
Usually this involves various physical media like wood, metal, electronics, photography but sometimes other things, like programming.
If this sounds interesting to you, go ahead and subscribe and I’ll see you again in the next video.
Now, go make something. Perfection’s not required. Fun is! | HuggingFaceTB/finemath | |
# lim x→8 {[1+(1+x)1/2 ]1/2 -2}/x-8Pls answer. sorry for this uncomfortable font
Ashwin Muralidharan IIT Madras
290 Points
12 years ago
Hi Saideep.
Multiply Nr and Dr by the conjugate of the Nr Twice.
When you do it for the first time, the Nr will be (1+x)1/2-3.
And the next time when you do it, the Nr will become x-8.
And this will get cancelled with the x-8, in the Dr, and then you can put x=8, in the resulting expression.
The resulting expression would be 1/{[1+(1+x)1/2 ]1/2 + 2}*{(1+x)1/2 +3}.
Put x=8 in this expression and you will get the limit as, 1/24.
Hope that helps.
All the best.
Regards,
vbhootna vb
14 Points
12 years ago
You can directly differnsiate nr and dr sepratly twice and get answer
Bestin Johnson
31 Points
12 years ago
lim {{1+(1+x)^1/2}^1/2}/x-8 upon direct substituition results in 0/0 form.Therefore by appying L-Hospital's rule,the limit
x->8
reduces to lim d/dx [1/{{1+(1+x)^1/2}^1/2} / x-8]
x->8
=lim 1 / [2{{1+(1+x)^1/2}^1/2} * 1 / [2{(1+x)^1/2}
x->8
Upon putting the value of x=8,we get the required limit=1/24. | HuggingFaceTB/finemath | |
The two bases (or grids or gates) are inputs which are differentially amplified (subtracted and multiplied) by the transistor pair; they can be fed with a differential (balanced) input signal, or one input could be grounded to form a phase splitter circuit. endobj The gain is half that of the stage with differential output. {\displaystyle A_{\text{c}}} , the lower (better) is the common-mode gain The stabilizer reacts to this intervention by changing its output quantity (current, respectively voltage) that serves as a circuit output. Find (W/L) of all transistors, V G 3, V G 4, and V G 5. (µ n C when the two inputs are joined together. differential driving. V 6 0 obj When the input is zero or negative, the output is close to zero (but can be not saturated); when the input is positive, the output is most-positive, dynamic operation being the same as the amplifier use described above. 1 Differential Amplifiers â¢Single Ended and Differential Operation â¢Basic Differential Pair â¢Common-Mode Response â¢Differential Pair with MOS loads Hassan Aboushady University of Paris VI ⢠B. Razavi, âDesign of Analog CMOS Integrated Circuitsâ, a3e�����N7!�C�*a��.ӡ��ח�u o$-f��>����m��VW��zB�t��t��/w�V���0[c�Q�N1�qPU�}�B�m�vƛgh��j0����,H?5Is]�ġ)m���+���)�qC87����5Ʌ�&Ѥ���Vʠ�ڶ�Se�����"G�#��lb�l�0����]�����e��f�tUQ������?��[���\�?��X�Q��Ԅ�XЮi��K��M�vx:4|Vw�t$P۟��M�IQ���w�T�Iҧ�7Hy�1AɌ�29�Vs��Sq�� j��kH��O/V��F|��k8^�2I�2���\lX.��Jmg���� � �W�m�,��M�, ������.ǰ�(#��Z/���������c�BϬhfO�T普3/���3��Nu�����d �N)m���s?����xV�%Ӑ�겡�Q�q�)K��\U��R#/\���w��E�|ͮU] �GW]�֏��J�ґ_/8Ӿ�n��4�l�h_� �9%�� >/j��u��yI2�͓fг��iϬn�z�l�M/�� r,���Ư���UW�������e ˽�%� y(L�['��#}|%D,zh.�����ߏ��7:���m���v��!� 1. Finally, as long as the open-loop voltage gain Aol is much larger than unity, the closed-loop voltage gain is Rf / Ri, the value one would obtain through the rule-of-thumb analysis known as "virtual ground". Differential amplifiers are used as a means of suppressing common-mode noise. %�쏢 Why is the differential amplifier transfer function as in the following But letâs not get much into that. It also implies that the common-mode input bias current has cancelled out, leaving only the input offset current IÎb = 'I+b - 'Iâb still present, and with a coefficient of Ri. Differential amplifier designed using opamp. {\displaystyle V_{\text{in}}^{+}} − That is why, in more sophisticated designs, an element with high differential (dynamic) resistance approximating a constant current source/sink is substituted for the âlong tailâ (Figure 3). The output of an ideal differential amplifier is given by: Where This means, for instance, that if Thus the higher the resistance of the current source endobj This is achieved by copying the input collector current from the left to the right side where the magnitudes of the two input signals add. The differential pair can be used as an amplifier with a single-ended input if one of the inputs is grounded or fixed to a reference voltage (usually, the other collector is used as a single-ended output) This arrangement can be thought of as cascaded common-collector and common-base stages or as a buffered common-base stage. This circuit was originally implemented using a pair of vacuum tubes. [nb 4] So, due to the negative feedback, the quiescent current depends only slightly on the transistor's β. V If the source is capacitive, two resistors have to be connected between the two bases and the ground to ensure different paths for the base currents. The high-resistance emitter element does not play any role—it is shunted by the other low-resistance emitter follower. [nb 5] Some kinds of differential amplifier usually include several simpler differential amplifiers. A differential amplifier is a type of electronic amplifier that amplifies the difference between two input voltages but suppresses any voltage common to the two inputs. 16 0 obj ), where one input is used for the input signal, the other for the feedback signal (usually implemented by operational amplifiers). The long-tailed pair was developed from earlier knowledge of push-pull circuit techniques and measurement bridges. The earliest definite long-tailed pair circuit appears in a patent submitted by Alan Blumlein in 1936. out A ��ܮ^�bpLN��]�Q���5E�����~�Š���� The long-tailed pair was very successfully used in early British computing, most notably the Pilot ACE model and descendants,[nb 1] Maurice Wilkesâ EDSAC, and probably others designed by people who worked with Blumlein or his peers. Closed-loop Frequency Response (voltage feedback amplifier) Resistance Formulas Reactance Formulas are the input voltages and in is the differential gain. The current mirror copies the left collector current and passes it through the right transistor that produces the right collector current. An operational amplifier, or op-amp, is a differential amplifier with very high differential-mode gain, very high input impedance, and low output impedance. A common application is for the control of motors or servos, as well as for signal amplification applications. The two transistors Q1 and Q2 have identical characteristics. At this right output of the differential amplifier, the two signal currents (pos. It is interesting fact that the negative feedback as though has reversed the transistor behavior - the collector current has become an input quantity while the base current serves as an output one. A more realistic expression for the output of a differential amplifier thus includes a second term. Modern differential amplifiers are usually implemented with a basic two-transistor circuit called a âlong-tailedâ pair or differential pair. Grungy Algebra Yes, it's time for everyone's favorite game show, Grungy Algebra! The biasing base currents needed to evoke the quiescent collector currents usually come from the ground, pass through the input sources and enter the bases. in Single amplifiers are usually implemented by either adding the appropriate feedback resistors to a standard op-amp, or with a dedicated IC containing internal feedback resistors. c HI! We can further simplify the above equation by considering R1=R2 and R3=R4. {\displaystyle V_{\text{in}}^{-}} {\displaystyle \scriptstyle A} It is as if the input offset current is equivalent to an input offset voltage acting across an input resistance Ri, which is the source resistance of the feedback network into the input terminals. Okay, well, that's the idea, at least. With relatively small collector resistor and moderate overdrive, the emitter can still follow the input signal without saturation. and large output voltage swings. In addi-tion, there is a ⦠The biasing current will enter directly this base and indirectly (through the input source) the other one. This amplifier is basically used in industrial and instrumentation purpose because this type of amplifier are better able to reject common-mode (noise) voltage then single-input circuits such as inverting and non-inverting amplifier. The two bases (or grids or gates) are inputs which are differentially amplified (subtracted and multiplied) by the transistor pair; they can be fed with a differential (balanced) input signal, or one input could be grounded to form a phase splitter circuit. Differential Amplifier Circuit The differential amplifier can be considered as an analog circuit that consists of two inputs and one output. ���3�� 4�XGJ.�Vk��M0��NR)Fi�F����Y���ab��\�%��2龟�c�C��Hk����IL��$���U��Kb��8��M��� {\displaystyle R_{\text{e}}} e *�U@Env�'�Wu�� Overdriven. Differential amplifier is a basic building block of an op-amp. In differential mode, the emitter voltage is fixed (equal to the instant common input voltage); there is no negative feedback and the gain is maximum. As a result, the output collector voltages do not change as well. For example, a fully differential amplifier, an instrumentation amplifier, or an isolation amplifier are often built from a combination of several op-amps. The symbol shown below represents a differential amplifier. Based on the methods of providing input and taking output, differential amplifiers can have four different configurations as below. The âlong tailâ resistor circuit bias points are largely determined by Ohm's Law and less so by active component characteristics. In this video, how to use the op-amp as the differential amplifier (Difference amplifier) or as subtractor has been discussed with solved examples. The long-tailed pair has many favorable attributes if used as a switch: largely immune to tube (transistor) variations (of great importance when machines contained 1,000 tubes or more), high gain, gain stability, high input impedance, medium/low output impedance, good clipper (with a not-too-long tail), non-inverting (EDSAC contained no inverters!) (����X�: ����ȿ�+R4�{#����� y�w��˖��ٹ~+w��/[.g����r��Rr�d���Őb�)�� B(o �Vy�Ձ��/����C�����e�+�oHN)�!���(={jO�j�����J+�=�����!��誐"����� �� }��&Y���M&5�����y�B��6�縤 �6J:vo(��3�YI ��oyL�ZY�z¼d�RJ��!y������m�d}(�:�g�p�ݎ �Y�M̔�n�G�}�M��d*�j{��� 2643 In this arrangement it seems strange that a, For the closed-loop common-mode gain to be zero only requires that the ratio of resistances, "PROCEEDINGS OF THE PHYSIOLOGICAL SOCIETY", Analog Devices â AN-0990 : Terminating a Differential Amplifier in Single-Ended Input Applications, https://en.wikipedia.org/w/index.php?title=Differential_amplifier&oldid=997842163, Creative Commons Attribution-ShareAlike License, This page was last edited on 2 January 2021, at 14:08. This requirement is not so important in the case of a differential output since the two collector voltages will vary simultaneously but their difference (the output voltage) will not vary. + [:(=K4�֭�xh+�q�� A differential (long-tailed,[nb 2] emitter-coupled) pair amplifier consists of two amplifying stages with common (emitter, source or cathode) degeneration. The above formula is used to calculate the output voltage of differential amplifier. Figure 3. ��TN�)BVp�[��6���5+bx�Y)+�E�Z���ϐ���$���n��� ��z��=�D�m�9�!c1x���s�yr潲燤���,S���O���,Mo��VV*�,�l�ZG�Ĥ�7�9/�%)�LZ���t]���t�;�S��u� V������Q�aόt&h��2I�C�6�� |F)V���8[�լ+�2� is the gain of the amplifier. Ό]}�����#��d�i�>@)Ź.����*^���:�$�T��\�j� �������F���5�k�O#j7u�"o�Z�����t. One disadvantage is that the output voltage swing (typically ±10â20 V) was imposed upon a high DC voltage (200 V or so), requiring care in signal coupling, usually some form of wide-band DC coupling. Hi , I designed a Galvanic skin response meter , it works well as per the circuit attached , The difference amplifier works as per formula . So, the sources have to be galvanic (DC) to ensure paths for the biasing current and low resistive enough to not create significant voltage drops across them. If the input differential voltage changes significantly (more than about a hundred millivolts), the transistor driven by the lower input voltage turns off and its collector voltage reaches the positive supply rail. V The Thévenin equivalent for the network driving the V+ terminal has a voltage V+' and impedance R+': while for the network driving the Vâ terminal, The output of the op amp is just the open-loop gain Aol times the differential input current i times the differential input impedance 2Rd, therefore. If the differential output is not desired, then only one output can be used (taken from just one of the collectors (or anodes or drains), disregarding the other output; this configuration is referred to as single-ended output. x��Z�o� ~߿b[tu��R���Hї;��P\o6vj{�ě���KΌ�O3�N�� �E�?��·�Ru=�M���M������|���ϰ�㫻���~�]�u�M/b�:�aYv^u�����Ͷ�]��()E���{ч���u��yab4�H�~�ׁ!���9qS!�H����n�n�� va�w����t����^�\� �B�Hj��$���u��A)f�j�W"z���zg��!n/�� The circuit works the same way for all three-terminal devices with current gain. The typical op-amp 4. ��� ( �Xr!���*[�E�@��kݙ̩g����AH ��y�W� ����@����7����Zn�S�r˳K��M����hm�J"y��3w��O���TeiGH�D���h���H���� g CH 10 Differential Amplifiers 3 âHummingâ Noise in Audio Amplifier Example However, VCC contains a ripple from rectification that leaks to the output and is perceived as a âhummingâ noise by theCH 10 Differential Amplifiers 4 Supply ����n/��ʙ�#SZ�ھ���)���s�I�$�$�3F���)�{Iv4�^j�=-�Av���"� ����n�E��Hy�6Kw? is zero and the CMRR is infinite. <> �f@H���"��:Q$���u���tخ4jy�ȿK�N� in Differential summing in is called the common-mode gain of the amplifier. In this case (differential input signal), they are equal and opposite. As differential amplifiers are often used to null out noise or bias-voltages that appear at both inputs, a low common-mode gain is usually desired. Practical differential amplifier circuit with gain 5 using uA741 opamp IC. To explain the circuit operation, four particular modes are isolated below although, in practice, some of them act simultaneously and their effects are superimposed. Discrete Semiconductor Circuits: Simple Op-Amp 3. One of these more complex amplifier types that weâll be studying is called the differential amplifier . ��=gD�;K8zM��ތM�$�13���)��w8�\��4q=��rH�cЏ�6>��1=*a s�mr,N�t���F�t��~���@�J������-r8 -�z�Ǖ�[~�*�7 L�V�����c���h���>����e�j��8H��%3���� )&i)�m��&+�xp��g�@K��3��6�Uj� 1��� )�eendstream The formula for a simple differential amplifier can be expressed: Where V 0 is the output voltage V 1 and V 2 are the input voltages A d is the gain of the amplifier (i.e. Practical op-amps 6. Thus the differential collector current signal is converted to a single ended voltage signal without the intrinsic 50% losses and the gain is greatly increased. current changes) are subtracted. V Breakdown. The quiescent current has to be constant to ensure constant collector voltages at common mode. [1] It is an analog circuit with two inputs fI�7�Ldi��>���[��T�4��(�Wٯ@�Ʉ��Xh��f���+�6ΐ[����z5_|W+H�f����+�م]�����#� Bias stability and independence from variations in device parameters can be improved by negative feedback introduced via cathode/emitter resistors with relatively small resistances. But in the case of a single-ended output, it is extremely important to keep a constant current since the output collector voltage will vary. Decibel Formula (equivalent impedance) Johnson-Nyquist Noise Formula Ohm's Law (DC circuit) Figure 11. in A That is why it is used to form emitter-coupled amplifiers (avoiding Miller effect), phase splitter circuits (obtaining two inverse voltages), ECL gates and switches (avoiding transistor saturation), etc. Fig. The constant current needed can be produced by connecting an element (resistor) with very high resistance between the shared emitter node and the supply rail (negative for NPN and positive for PNP transistors) but this will require high supply voltage. '��+ͻ������ There is no negative feedback, since the emitter voltage does not change at all when the input base voltages change. Manufacturersâ specifications 5. The name "differential amplifier" must not be confused with the "differentiator", which is also shown on this page.The "instrumentation amplifier", which is also shown on this page, is a modification of the differential amplifier that also provides high input impedance. At common mode, the two parts behave as common-collector stages with high emitter loads; so, the input impedances are extremely high. The transfer function of the differential amplifier, also known as difference amplifier, can be found in articles, websites, formula tables, but where is it coming from? Biasing of Differential Amplifiers Constant Current Bias: In DC analysis of the differential amplifier, we have seen that emitter current IE depends upon value of the bdc. These equations undergo a great simplification if, which implies that the closed-loop gain for the differential signal is V+in - Vâin, but the common-mode gain is identically zero. To make the operating point stable IE current should be constant Differential Amplifier as Comparator A differential amplifier circuit is a very useful op-amp circuit, since it can be configured to either âaddâ or âsubtractâ the input voltages, by suitably adding more resistors in parallel with the input 3). The emitter-coupled amplifier is compensated for temperature drifts, VBE is cancelled, and the Miller effect and transistor saturation are avoided. In Figure 6, current generators model the input bias current at each terminal; I+b and Iâb represent the input bias current at terminals V+ and Vâ, respectively. Now it's time for a reality check. For comparison, the old-fashioned inverting single-ended op-amps from the early 1940s could realize only parallel negative feedback by connecting additional resistor networks (an op-amp inverting amplifier is the most popular example). β1 = 0 R4 R3 + + â â V + OUT V â OUT V OCM V + IN A F Figure 5. and An op-amp differential amplifier can be built with predictable and stable gain by applying negative feedback (Figure 5). where R// is the average of R+// and Râ//. c ! stream 1. As the signals propagate down the differential pair, there is a voltage pattern between each signal line and the reference plane below. {\displaystyle A_{\text{d}}} Ra = Rb = Rf = Rg = R, the amplifier will provide output that is the difference of input voltages; Vout = Vb â Va in which the output is ideally proportional to the difference between the two voltages. At differential mode, they behave as common-emitter stages with grounded emitters; so, the input impedances are low. Note that a differential amplifier is a more general form of amplifier than one with a single input; by grounding one input of a differential amplifier, a single-ended amplifier results. Single-ended to differential amplifier + + â â R1 R2 V + OUT V â OUT V OCM V + IN A F Figure 4. The common-mode input voltage can vary between the two supply rails but cannot closely reach them since some voltage drops (minimum 1 volt) have to remain across the output transistors of the two current mirrors. stream x���r+��ί�-db��/��!��S��V%'=�.j{O�+�3r���k��!���z���h4z��� \,�䇋�[���,�n��/|����?�-���-a���� �{��7��bi��Y��/~Z��� �+oU��g���b����j&Ww덀�Z���zc��'OWk9�ڏ�W=�7 are equal, the output will not be zero, as it would be in the ideal case. Figure 3 shows a block diagram The common-mode rejection ratio (CMRR), usually defined as the ratio between differential-mode gain and common-mode gain, indicates the ability of the amplifier to accurately cancel voltages that are common to both inputs. The common-mode rejection ratio is defined as: In a perfectly symmetric differential amplifier, The function of a differential amplifier is to amplify the difference between two input signals. A Crecraft, S. Gergely, in Analog Electronics: Circuits, Systems and Signal Processing, 20023.9 Offsets The ideal d.c. amplifier has a d.c. output of 0 V when the d.c. input is 0 V. If the d.c. amplifier is a differential amplifier, such as an op amp, the output is expected to be zero when the input differential voltage is zero, i.e. If the resistor at the collector is relatively large, the transistor will saturate. In contrast with classic amplifying stages that are biased from the side of the base (and so they are highly β-dependent), the differential pair is directly biased from the side of the emitters by sinking/injecting the total quiescent current. It is usually implemented by a current mirror because of its high compliance voltage (small voltage drop across the output transistor). It is possible to connect a floating source between the two bases, but it is necessary to ensure paths for the biasing base currents. and R {\displaystyle V_{\text{in}}^{+}} The output voltage of the differential amplifiershown above can be given by the below formula The above formula was obtained from the transfer function of the above circuit using superposition theorem. Dual Input Balanced Output 1. Thus a differential amplifier amplifies the difference between two input signals. Exercise 3: The differential amplifier below should achieve a differential gain of 40 with a power consumption of 2 mW. [nb 6], electronic amplifier, a circuit component, Operational amplifier as differential amplifier, Symmetrical feedback network eliminates common-mode gain and common-mode bias, Details of the long-tailed pair circuitry used in early computing can be found in. My friends advised me that it would be helpful to have on this site the most common operational amplifier configurations and transfer functions or formulas. %PDF-1.4 At high overdrive the base-emitter junction gets reversed. Derivations for voltage gain and output voltage. *��6?�"e��Ą��n�+��C�"!�߈��x���P����⾧�����g~�ilBz 9�;g�7crӚ�wɲ����_�D�xOU����� �EMCGi��w��Q� Discrete Semiconductor Circuits: Differential Amplifier 2. Inverting amplifier 9. This mode is used in differential switches and ECL gates. In case the operational amplifier's (non-ideal) input bias current or differential input impedance are a significant effect, one can select a feedback network that improves the effect of common-mode input signal and bias. An amplifier with differential output can drive a floating load or another stage with differential input. <> Primary analog circuit: principle of 9-2 differential amplifier Timeï¼2021-1-5 Back to the catalog Inside the op amp, it is usually divided into several stages, each of which performs different functions. With two inputs and two outputs, this forms a differential amplifier stage (Figure 2). [2] An early circuit which closely resembles a long-tailed pair was published by British neurologist Bryan Matthews in 1934,[3] and it seems likely that this was intended to be a true long-tailed pair but was published with a drawing error. in This is often implemented as a current mirror (Figure 3, below). The collector resistors can be replaced by a current mirror, whose output part acts as an active load (Fig. Amplifies the difference in voltage between its inputs. + There is a full (100%) negative feedback; the two input base voltages and the emitter voltage change simultaneously while the collector currents and the total current do not change. Single Input Unbalanced Output 2. In common mode (the two input voltages change in the same directions), the two voltage (emitter) followers cooperate with each other working together on the common high-resistive emitter load (the "long tail"). {\displaystyle \scriptstyle V_{\text{in}}^{-}} ^�JڑX����'լ�h���&��xP�l Computer Simulation of Op-amp circuits 7. So, here they are. where ӟ����HV*V�mŘ�1���ix����J�u�#f[&�S�S�@S�������ܗ)Ď m���R>s���g�(��.F��Bp=(*������m�zʽ�t{RP�W��;gP�6�\$�!�5L�k��s=~��T���?�ݜ��u�ݾ��� ��e��6w8���������4�c�:� �DS(;�X�*�ֱ# ��P��dE�(�8䜖d�,{�F�k�J�5�i��e�t� 4�A��z So, the common point does not change its voltage; it behaves like a virtual ground with a magnitude determined by the common-mode input voltages. {\displaystyle V_{\text{in}}^{-}} [nb 3]. D.I. V {\displaystyle \scriptstyle V_{\text{in}}^{+}} − Normal. . If the input voltage continues increasing and exceeds the base-emitter breakdown voltage, the base-emitter junction of the transistor driven by the lower input voltage breaks down. ���X��1N l�IME*:��U>��iW�l�'�mT������ A V The differential amplifier circuit can be represented as shown in the figure below. When used as a switch, the "left" base/grid is used as signal input and the "right" base/grid is grounded; output is taken from the right collector/plate. Gain by applying negative feedback introduced via cathode/emitter resistors with relatively small resistances V G 3, V G,. And as switch as switch in this case ( differential input base voltages.., there is no negative feedback and the reference plane below active load ( Fig 3, below.! Input and taking output, differential amplifiers are usually implemented with a two-transistor! V OCM V + OUT V â OUT V â OUT V â OUT V V. Avoid sacrificing gain, a differential amplifier, the quiescent current has to be constant to ensure collector. Is a full negative feedback ( Figure 5 ) serves as a circuit output for three-terminal! G 4, and V G 5 expression for the above circuit and check if the resistor at collector. Is to amplify the difference between two input terminals that are both isolated from ground by the one... 3, below ) at common mode block of an op-amp differential amplifier circuit can be considered as analog... 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Circuits that utilize series negative feedback, the quiescent current depends only slightly on input. ( W/L ) of all transistors, V G 3, V G 5 signals propagate the... ) the other one realistic expression for the two transistors and the output transistor ) points are determined... | open-web-math/open-web-math | |
## Exercises - Review Set B
1. What is the implicit domain (in set-builder notation) of $\displaystyle{f(x) = \frac{(x-7)^2\log_2(x+5)}{x-7}}$
The domain of the $\log_2 x$ is $\mathbb{R}_{>0}$, so we need $x \gt -5$. Further, we need $x \neq 7$ so that we don't divide by zero. Thus, the implicit domain of the $f$ above is $\{x \in \mathbb{R} \ | \ x \gt -5 \textrm{ and } x \neq 7\}$.
2. Does the graph below represent a function? Explain how you know.
No, it does not pass the vertical line test.
3. Given the graph of the function $f(x)$ below:
1. Determine if $f$ has an inverse, explaining how you know.
2. Find what appears to be the domain of $f$
3. Find what appears to be the image/range of $f$
1. $f$ does not have an inverse as it fails the horizontal line test.
2. The domain of $f$ appears to be $(-6,0] \cup (2,8]$.
3. The image/range of $f$ appears to be $[-4,6)$.
4. Find the following sum, expressing your answer in base $5$, doing so without converting to another base first. $$(4213)_5 + (2334)_5$$
$(12102)_5$
5. Express the following combination of polynomials (of two variables) as a single polynomial of two variables: $$(3a - 2b)(5a + ab - 4b) + (22ab + 2ab^2)$$
$\displaystyle{15a^2 + 8b^2 + 3a^2b}$
6. Factor completely: $x^5 - 9x^3 -8x^2 + 72$
Use factoring by grouping first, and then use difference of squares and difference of cubes factor what results to obtain: $$(x-3)(x+3)(x-2)(x^2+2x+4)$$
7. Decide if Eisenstein's Criterion can be used to prove the following polynomial is irreducible. If it can, find the prime $p$ in question. If not, explain why not. $$x^5 + 9x^4 - 15x^3 + 3x^2 + 21$$
Eisenstein's Criterion apples with $p=3$ (note: $p$ does not divide the coefficient on $x^5$, but divides all other coefficients (i.e., $9$, $-15$, $3$, and $21$), and $p^2$ does not divide the constant term, $21$). The polynomial given is irreducible -- it does not factor into a product of polynomials with integer coefficients.
8. Divide $f(x)$ by $g(x)$, expressing your answer in quotient-remainder form: $$f(x) = x^4 - 3x^3 + 2x^2 - 5 \quad \quad g(x) = x^2 - 2x + 1$$
$\displaystyle{x^4 - 3x^3 + 2x^2 - 5 = (x^2 - 2x + 1)(x^2 -x -1) + (-x-4)}$
9. Find both the additive and the multiplicative inverses of $4$ in a $13$-hour clock arithmetic.
$9$ is the additive inverse; $10$ is the multiplicative inverse
10. The set $S = \{A,B,C,D\}$, along with addition and multiplication as described by the tables below is a commutative ring. Explain why it is not also a field. $$\begin{array}{c|cccc} + & A & B & C & D\\\hline A & A & B & C & D\\ B & B & C & D & C\\ C & C & D & C & B\\ D & D & C & B & A \end{array} \quad \quad \begin{array}{c|cccc} \times & A & B & C & D\\\hline A & A & A & A & A\\ B & A & B & C & D\\ C & A & C & A & C\\ D & A & D & C & B \end{array}$$
Note that $A$ plays the role of zero (i.e., the additive identity), as $A+x = x+A = x$ for any $x \in S$. Further, $B$ plays the role of one (i.e., the multiplicative identity), as $Bx = xB = x$ for the same. However, this makes $C$ a non-zero element who does not have a multiplicative inverse.
11. Simplify each expression below, assuming it is defined. You may leave your answer in factored form.
1. $\displaystyle{\frac{x^2 + 6x + 9}{4x+12} \div \frac{2x^2+5x-3}{6x}}$
2. $\displaystyle{\frac{x-2}{x^3 + 6x^2 + 5x} - \frac{1}{x^2 + 3x + 2}}$
1. $\displaystyle{\frac{3x}{2(2x-1)}}$
2. $\displaystyle{\frac{-(5x+4)}{x(x+1)(x+2)(x+5)}}$
12. Find the simplified difference quotient $\cfrac{f(x+h)-f(x)}{h}$ for $f(x) = x^3$, assuming $h \neq 0$
$\displaystyle{3x^2 + 3xh + h^2}$
13. Recalling $\mathbb{Q}(\sqrt{5})$ is a field, find rational values $c$ and $d$ so that $$\frac{1}{7 + 3\sqrt{5}} = c + d\sqrt{5}$$
$c = \frac{7}{4}$ and $d = -\frac{3}{4}$
14. Express in interval notation where the graph of $f(x) = x^3 + 2x^2$ is above the graph of $g(x) = 25x+50$.
$\displaystyle{(-5,-2) \cup (5,\infty)}$ | HuggingFaceTB/finemath |
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