question stringlengths 200 50k | answer stringclasses 1
value | source stringclasses 2
values |
|---|---|---|
# Equation of Ellipse - Problems
This is a tutorial with detailed solutions to problems related to the ellipse equation. An HTML5 Applet to Explore Equations of Ellipses is also included in this website.
## Review
An ellipse with center at the origin (0,0), is the graph of with a > b > 0
The length of the major axis is 2a, and the length of the minor axis is 2b. The two foci (foci is the plural of focus) are at (~+mn~ c , 0) or at (0 , ~+mn~ c), where c2 = a2 - b2.
### Problem 1
Given the following equation
9x
2 + 4y2 = 36
a) Find the x and y intercepts of the graph of the equation.
b) Find the coordinates of the foci.
c) Find the length of the major and minor axes.
d) Sketch the graph of the equation.
Solution to Example 1
a)
We first write the given equation in standard form by dividing both sides of the equation by 36 and simplify
9 x
2 / 36 + 4 y2 / 36 = 1
x
2 / 4 + y2 / 9 = 1
x
2 / 22 + y2 / 32 = 1
We now identify the equation obtained with one of the standard equation in the review above and we can say that the given equation is that of an ellipse with a = 3 and b = 2
NOTE: a > b
Set y = 0 in the equation obtained and find the x intercepts.
x2 / 22 = 1
Solve for x.
x2 = 22
x = ~+mn~ 2
Set x = 0 in the equation obtained and find the y intercepts.
y2 / 32 = 1
Solve for y.
y2 = 32
y = ~+mn~ 3
b) We need to find c first.
c2 = a2 - b2
a and b were found in part a)
c2 = 32 - 22
c2 = 5
Solve for c to obtain
c = ~+mn~ (5)1/2
The foci are F1 (0 , √5) and F2 (0 , - √5 )
c) The major axis length is given by 2 a = 6
The minor axis length is given by 2 b = 4
d) Locate the x and y intercepts, find extra points if needed and sketch. Matched Problem: Given the following equation
4x
2 + 9y2 = 36
a) Find the x and y intercepts of the graph of the equation.
b) Find the coordinates of the foci.
c) Find the length of the major and minor axes.
d) Sketch the graph of the equation.
## More Links and References on Ellipses
College Algebra Problems With Answers - sample 8: Equation of Ellipse
Points of Intersection of an Ellipse and a line
HTML5 Applet to Explore Equations of Ellipses
Find the Points of Intersection of a Circle and an Ellipse
Ellipse Area and Perimeter Calculator
Find the Points of Intersection of Two Ellipses | HuggingFaceTB/finemath | |
x Turn on thread page Beta
You are Here: Home
# P1: How Was It For You? watch
1. you can apply to have your marked paper sent to you at a price. btw how much does it cost to request your paper back?
2. (Original post by chats)
wow 80% and you only got a low B. how do you know you got 60/75?
i got the original sent back to me, cuz i didn't think id done that badly....
3. (Original post by CoolDude13)
i got the original sent back to me, cuz i didn't think id done that badly....
lol you think that was bad! i would love that.
4. SO SO SO EASY>>>>>>>>>>>this is great
5. Thought I did OK, Only things I got Wrong were parts a & d of geometric series question. Forgot to revise how to prove the formula and know I got d wrong cos I put a value of r which was > 1. (I put 11/10 which means something went wrong with it but was close ot 10/11 which was correct)
1
b: y = 1.5x + 0.5
2
a: 17 - 12 root 2
b: 0.5 + 0.25 root 2
3
a: Draw Graphs
b: Meets Axis at (-2,0), (2,0) & (0,-4)
C: (xx or XX = X squared)
Use quadratic equation to find points of crossing (y= (xx) -4) and (y=-3x)
So: xx + 3x -4 = 0
Meet at: (1,-3) and (-4,12)
4
a: X< 65 (NOT X>65 or answer to c) does not make sense)
b: xx + 20x - 4800 > 0
c: 60 < X < 65
Second part of quadratic to do with -80 can be ignored as X>0 from first
line of question
5
a Tan x = 8/3
b Tan x = 8/3
x = 69.4 and x = 249.4
c Change Sin sqaured into 1 - cos squared
Part where cos x = -3 ignored as no answers
cos x = 1/3
x = 70.5 and x = 289.5
6
do dy/dx etc
prove by taking second derivative is > 0 (I think cos it was a minimum) value, otherwise do opposite
7
Screwed up by not revising how to prove formulae but otherwise:
b: 87.8 to 3sf
c: 100
d: answer is 10/11 but I got that wrong
8
coordinates of Q are (3,12)
y coordinate of P is also 12
Line PQ is parallel to x axis as y coordinates are the same (gradient of PQ = 0)
Area = 4/3
Hopefully only dropped 7 marks for those two geometric questions
6. does anyone hav a copy of the p1 paper yesterday that they can scan in?
7. AQA pure 1 was so easy, i think i got 100%.
Btw the grade boundry for an A is never more than 80%
8. (Original post by ram)
AQA pure 1 was so easy, i think i got 100%.
Btw the grade boundry for an A is never more than 80%
for edexcel it is. see http://www.edexcel.org.uk/VirtualCon...port_Jun03.pdf
last yr for p1, you needed 66/75 to get an A, which is more than 80%
9. well i am pretty sure i got 100%, my teacher even went over the answers after and as i remember i think i got them all right. So i guess 100% means A.
10. How modest.
BTW the grade boundary is different each year for raw percentages.
11. (Original post by mik1a)
How modest.
BTW the grade boundary is different each year for raw percentages.
so do you think this years was easier than last years?
i think last years was easier so this year 80 ums will be 64 marks
12. has anyone got a good prediction of the marks needed for an A?
Does anyone have a scan of the paper?
13. (Original post by chats)
so do you think this years was easier than last years?
i think last years was easier so this year 80 ums will be 64 marks
Was it not 66 marks last year?
14. (Original post by chats)
so do you think this years was easier than last years?
i think last years was easier so this year 80 ums will be 64 marks
I can't remember last year's. But I think on average this year's was, on average, the same as past papers.
15. (Original post by MBR50)
Was it not 66 marks last year?
yeh it was 66 last year but this year i think 64.
p3 was 51 last year! it must be damnnnn hard
16. (Original post by chats)
p3 was 51 last year! it must be damnnnn hard
it wasn't too bad looking at it, but then again, i wasn't in the exam hall doing it under timed conditions - that can be much more daunting.
this year's was quite tough, but still doable. but judging from the reaction on here, i think the grade boundaries for this year's p3 will be close to 51
17. The Grade boundries for P1 for June 2003 Edexcel were:
A = 80UMS = 66
B = 70UMS = 57
C = 60UMS = 48
D = 50UMS = 40
E = 40UMS = 32
I dont think they will be too different this year. I dont think the boundry for an A will be any higher than 67 / 68
18. (Original post by JHJH)
The Grade boundries for P1 for June 2003 Edexcel were:
A = 80UMS = 66
B = 70UMS = 57
C = 60UMS = 48
D = 50UMS = 40
E = 40UMS = 32
I dont think they will be too different this year. I dont think the boundry for an A will be any higher than 67 / 68
its out of 75 yea
19. how much UMS approximately needed for an A altogether is it 80% of 300 i.e 240
is it that also subject to change in accordance with the difficulty fo the papers?
20. (Original post by TheWolf)
its out of 75 yea
yo
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: June 12, 2004
Today on TSR
### Loughborough better than Cambridge
Loughborough at number one
### Can I date a girl with no boobs?
Poll
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | HuggingFaceTB/finemath | |
Question
A farmer with 1200 meters of fencing wishes to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. What are the dimensions of the field that produce the largest area
1. The perimeter of a given figure is a measure of the addition of each individual length of the sides of the figure. Thus the dimensions that would produce the largest area of the field are; length = 300 m, and width = 200 meters.
The perimeter of a given figure is a measure of the addition of each length of the sides of the figure. It always has the unit as that of the given sides of the figure.
So in the given question, the perimeter of the fence required = 1200 meters.
Thus, let the length of the enclosed rectangle be represented by l and its width by w. Thus,
Perimeter = 2l + 3w
1200 = 2l + 3w
Thus, let l be equal to 300, we have;
1200 = 2(300) + 3w
1200 – 600 = 3w
w = 200
Thus the dimensions of the field that would produce the largest area are; length = 300 m and width = 200 m.
For more clarifications on the perimeter of a rectangle, visit: https://brainly.com/question/10452031
#SPJ1 | HuggingFaceTB/finemath | |
# How to substitute an equation into another equation to simplify it?
I solved a system of equations. A solution of one of the coefficients can be plugged into another solution of a separate coefficient. How do I do this?
System of Equations
eqn1 = B0*Q0 == A0*P0 + C*(3*sigma0**2 + 1)
eqn2 = B2*Q2 == A2*P2 + 2*C
eqn3 = B0*kf*Q0d == A0*kp*P0d + 6*kp*C*sigma0
eqn4 = B2*kf*Q2d == A2*kp*P2d
sol = solve( [eqn1, eqn2, eqn3, eqn4], [B0, A0, B2, A2] )
I separated the solutions to the coefficients from the list.
sola = sol
B0 = sola.rhs(); B0 = B0.substitute(P0d = 0)
A0 = sola.rhs(); A0 = A0.substitute(P0d = 0, P0 =1)
print(B0)
print(A0)
Output
6*C*kp*sigma0/(Q0d*kf)
-(3*C*Q0d*kf*sigma0^2 - 6*C*Q0*kp*sigma0 + C*Q0d*kf)/(Q0d*kf)
Now how can I substitute B0 into A0 to simplify the coefficient A0?
Thanks!
enter code here
edit retag close merge delete
Sort by » oldest newest most voted
Using the solution_dict option of solve makes it easier to use the solutions afterwards.
Here is an example based on the code in the question.
First we need to declare variables.
sage: var('B0 Q0 A0 P0 sigma0 B2 Q2 A2 P2 kf Q0d kp P0d C Q2d P2d')
Define equations:
sage: eqn1 = B0*Q0 == A0*P0 + C*(3*sigma0**2 + 1)
sage: eqn2 = B2*Q2 == A2*P2 + 2*C
sage: eqn3 = B0*kf*Q0d == A0*kp*P0d + 6*kp*C*sigma0
sage: eqn4 = B2*kf*Q2d == A2*kp*P2d
Solve and check out solutions:
sage: sol = solve([eqn1, eqn2, eqn3, eqn4], [B0, A0, B2, A2], solution_dict=True)
sage: len(sol)
1
sage: sol
[{B0: -(3*C*P0d*kp*sigma0^2 - 6*C*P0*kp*sigma0 + C*P0d*kp)/(P0*Q0d*kf - P0d*Q0*kp),
A0: -(3*C*Q0d*kf*sigma0^2 - 6*C*Q0*kp*sigma0 + C*Q0d*kf)/(P0*Q0d*kf - P0d*Q0*kp),
B2: -2*C*P2d*kp/(P2*Q2d*kf - P2d*Q2*kp),
A2: -2*C*Q2d*kf/(P2*Q2d*kf - P2d*Q2*kp)}]
Take the first solution (a dictionary):
sage: sola = sol
sage: sola
{B0: -(3*C*P0d*kp*sigma0^2 - 6*C*P0*kp*sigma0 + C*P0d*kp)/(P0*Q0d*kf - P0d*Q0*kp),
A0: -(3*C*Q0d*kf*sigma0^2 - 6*C*Q0*kp*sigma0 + C*Q0d*kf)/(P0*Q0d*kf - P0d*Q0*kp),
B2: -2*C*P2d*kp/(P2*Q2d*kf - P2d*Q2*kp),
A2: -2*C*Q2d*kf/(P2*Q2d*kf - P2d*Q2*kp)}
Values of the solutions:
sage: sola[A0]
-(3*C*Q0d*kf*sigma0^2 - 6*C*Q0*kp*sigma0 + C*Q0d*kf)/(P0*Q0d*kf - P0d*Q0*kp)
sage: sola[B0]
-(3*C*P0d*kp*sigma0^2 - 6*C*P0*kp*sigma0 + C*P0d*kp)/(P0*Q0d*kf - P0d*Q0*kp)
sage: sola[A2]
-2*C*Q2d*kf/(P2*Q2d*kf - P2d*Q2*kp)
sage: sola[B2]
-2*C*P2d*kp/(P2*Q2d*kf - P2d*Q2*kp)
Simple substitutions:
sage: B = sola[B0].subs({P0d: 0})
sage: B
6*C*kp*sigma0/(Q0d*kf)
sage: A = sola[A0].subs({P0d: 0, P0: 1})
sage: A
-(3*C*Q0d*kf*sigma0^2 - 6*C*Q0*kp*sigma0 + C*Q0d*kf)/(Q0d*kf)
Noticing that one of the terms in A looks like B, we are tempted to substitute.
Unfortunately:
sage: A.subs({B: B0})
-(3*C*Q0d*kf*sigma0^2 - 6*C*Q0*kp*sigma0 + C*Q0d*kf)/(Q0d*kf)
Expanding and matching an exact term works:
sage: A.expand()
-3*C*sigma0^2 + 6*C*Q0*kp*sigma0/(Q0d*kf) - C
sage: A.expand().subs({B*Q0: B0*Q0})
-3*C*sigma0^2 + B0*Q0 - C
more
I appreciate you taking the time to help me out, but the last line of code doesn't actually substitute B0 into A0. It would look something like this:
-(3*C*Q0d*kf*sigma0^2 + C*Q0d*kf)/(Q0d*kf) + Q0*B0
Or simplified further
B0*Q0 - C*(3*sigma0^2 + 1)
Or is there anyway to simplify (the dictionary) as a whole when it comes out of the solve function?
1
You can simply get A0 in terms of B0 from the first equation:
sage: A0.subs(solve(eqn1,A0)).subs(P0=1)
-3*C*sigma0^2 + B0*Q0 - C
1
Edited my answer to name A and B the results of simple substitutions, and to add the "Noticing... Unfortunately ... Expanding and matching ..." part.
What @Juanjo proposes is probably more satisfactory!
Thank you, this was giving me a headache! You guys are good and timely! @Juanjo, @slelievre | HuggingFaceTB/finemath | |
# Capacity Lesson for Kids: Definition & Facts
Instructor: Emily Hume
Emily is a Reading Specialist and Literacy coach in a public elementary school with a Master's Degree in Elementary Education.
You may not even know it, but you use capacity every day! It's all around you - in the glass of milk you pour, the bin you put your toys in, even in your bowl of cereal! In this lesson, you will learn the definition of and important facts about capacity.
## What Is Capacity?
You decide to pour a nice glass of chocolate milk! Grabbing a glass and the milk, you begin to pour - when something on TV catches your eye. You keep pouring and pouring until...splash! Milk all over the counter! What happened?
You just learned a lesson about capacity, or the amount a container can hold. There is a limit to how much milk will fit in your glass, and if you try to go past that limit, the milk will start to overflow!
## Two Ways to Measure Capacity
There are two ways to measure capacity: One is called customary, which is the measurement used for the most part in the United States, and the other is called metric and is used throughout the world.
• Customary measurement can be a gallon, quart, pint, or cup. You can even divide that cup into smaller capacities like half cup, quarter cup, tablespoon, or even teaspoon!
• Gallons, quarts, and pints are typically used to measure the capacity for liquids, while cups, tablespoons, and teaspoons can be used to measure liquids like milk or solids like flour.
• Gallons are the biggest measurement tool; 1/8 of a teaspoon is usually the smallest.
• There are only three countries in the whole world that don't typically use the metric system to measure capacity! The United States is one of those countries, although you can find metric measurements in some places.
• Liters and milliliters are the two most common metric units of capacity. When you buy a bottle of soda at the grocery store, those are usually 2 liters. A milliliter would be just a tiny drop of that soda!
To unlock this lesson you must be a Study.com Member.
### Register to view this lesson
Are you a student or a teacher?
#### See for yourself why 30 million people use Study.com
##### Become a Study.com member and start learning now.
Back
What teachers are saying about Study.com
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | HuggingFaceTB/finemath | |
# Thread: How to solve p when p/q - 3
1. ## How to solve p when p/q - 3
Hi,
I have another problem that requires a fresh pair of eyes.
given that p/q - 3 =p, find p
my working so far
p/q - 3 = p
p/q - p = 3
(p - pq)/q = 3
p - pq = 3q
p = 3q + pq
I stopped right at p = 3q + pq. Checked the answer and it turned out to be
3q/ 1 + {q}^{2 }
How do I simplify p = 3q + pq to the answer above? can anyone show me step by step how it is done?
many thanks.
2. You are correct up to p - pq = 3q. Now take out the common factor in the LHS so that you can isolate p.
3. Great hint! manage to finish this problem and got the right answer.. | HuggingFaceTB/finemath | |
Standard Deviation
What it is:
Standard deviation is a measure of how much an investment's returns can vary from its average return. It is a measure of volatility and in turn, risk. The formula for standard deviation is:
Standard Deviation = [1/n * (ri - rave)2]½
where:
ri = actual rate of return
rave = average rate of return
n = number of time periods
For math-oriented readers, standard deviation is the square root of the variance.
How it works (Example):
Let's assume that you invest in Company XYZ stock, which has returned an average 10% per year for the last 10 years. How risky is this stock compared to, say, Company ABC stock? To answer this, let's first take a closer look at the year-by-year returns that compose that average:
At first look, we can see that the average return for both stocks over the last 10 years was indeed 10%. But let's look in a different way at how close XYZ's returns in any given year were to the average 10%:
As you can see, only during year 9 did XYZ return the average 10%. In the other years, the return was higher or lower -- sometimes much higher (as in year 7) or much lower (as in year 2). Now look at the annual returns on Company ABC stock, which also had a 10% average return for the last 10 years:
As you can see, Company ABC also averaged 10% return over 10 years but did so with far less variance than Company XYZ. Its returns are more tightly clustered around that 10% average. Thus, we can say that Company XYZ is more volatile than Company ABC stock. Standard deviation seeks to measure this volatility by calculating how "far away" the returns tend to be from the average over time.
For instance, let's calculate the standard deviation for Company XYZ stock. Using the formula above, we first subtract each year's actual return from the average return, then square those differences (that is, multiply each difference by itself):
Next, we add up column D (the total is 3,850). We divide that number by the number of time periods minus one (10-1=9; this is called the "nonbiased" approach and it is important to remember that some calculate standard deviation using all time periods -- 10 in this case rather than 9). Then we take the square root of the result. It looks like this:
Standard deviation = √(3,850/9) = √427.78 = 0.2068
Using the same process, we can calculate that the standard deviation for the less volatile Company ABC stock is a much lower 0.0129.
Why it Matters:
Standard deviation is a measure of risk that an investment will not meet the expected return in a given period. The smaller an investment's standard deviation, the less volatile (and hence risky) it is. The larger the standard deviation, the more dispersed those returns are and thus the riskier the investment is.
Many technical indicators, such as Bollinger Bands, incorporate the notion of standard deviation as a way to determine whether to buy or sell a stock, but it is important to remember that standard deviation is only one of many measures of risk and should not be the last word in deciding whether a stock is "too risky" or "not risky enough." | HuggingFaceTB/finemath | |
NextShark
# This Seemingly Simple Children’s Brain Teaser is Dividing the Entire Internet
February 17, 2016
SHARE
A seemingly simple brainteaser has netizens bending their brains trying to figure out the tricky answer.
The “fruit algebra” riddle posted by Facebook user Lisa Woelke has since been shared nearly 160,000 times and commented on by almost 2.3 million people.
While the answer may seem obvious at first, there’s more to the puzzle you probably missed.
For the answer to the puzzle, scroll down past the GIFS.
In the first line, the three apples add up to 30. Thirty divided by three equals 10, and so each apple is worth 10.
In the second line, an apple and two banana bunches are 18. That means each banana is worth four.
The third line shows that a banana bunch minus a coconut is 2, so a coconut equals 2.
The fourth line asks what a coconut (2) plus an apple (10) and a banana bunch (4) equal. The answer seems obvious: 14.
But not so fast! The banana bunch in the third line only includes three bananas — the others have four individual bananas each — so the banana bunch in the fourth line is only worth three.
In total, the coconut plus the apple and banana bunch (with only three bananas) equals 15.
###### HAPPENING NOW
Editorial Staff
Follow NextShark on Facebook and Twitter to keep up-to-date on our posts! Send us tips, press releases, and story ideas to info[at]nextshark[dot]com. | HuggingFaceTB/finemath | |
## Control System Questions and Answers Part-1
1. Which of the following is not the feature of modern control system?
a) Quick response
b) Accuracy
c) Correct power level
d) No oscillation
Explanation: For a good control system the speed of response and stability must be high and for the slow and sluggish response is not used and undesirable.
2. The output of the feedback control system must be a function of:
a) Reference input
b) Reference output
c) Output and feedback signal
d) Input and feedback signal
Explanation: Feedback control system has the property of reducing the error and that is by differencing the output with the desired output and as the equation of the output of the system is C=GR/1+GH.
3. The principle of homogeneity and superposition are applied to:
a) Linear time invariant systems
b) Nonlinear time invariant systems
c) Linear time variant systems
d) Nonlinear time invariant systems
Explanation: Superposition theorem states that for two signals additivity and homogeneity property must be satisfied and that is applicable for the LTI systems.
4. In continuous data systems:
a) Data may be continuous function of time at all points in the system
b) Data is necessarily a continuous function of time at all points in the system
c) Data is continuous at the inputs and output parts of the system but not necessarily during intermediate processing of the data
d) Only the reference signal is continuous function of time
Explanation: Continuous signals are the signals having values for the continuous time and if impulse response decays to zero as time approaches infinity, the system is stable.
5. A linear system at rest is subject to an input signal r(t)=1-e-t. The response of the system for t>0 is given by c(t)=1-e-2t. The transfer function of the system is:
a) (s+2)/(s+1)
b) (s+1)/(s+2)
c) 2(s+1)/(s+2)
d) (s+1)/2(s+2)
Explanation: c(t)=1-e-2t
R(s)=1/s-1/s+1
C(s)=1/s-1/s+2
Tf=2(s+1)/(s+2).
6. In regenerating the feedback, the transfer function is given by
a) C(s)/R(s)=G(s)H(s)/1+G(s)H(s)
b) C(s)/R(s)=G(s)H(s)/1-G(s)H(s)
c) C(s)/R(s)=G(s)/1+G(s)H(s)
d) C(s)/R(s)=G(s)/1-G(s)H(s)
Explanation: Regenerating feedback is positive feedback and it increases the infinitely and hence the speed of response of the system reduces.
7. A control system whose step response is -0.5(1+e-2t) is cascaded to another control block whose impulse response is e-t. What is the transfer function of the cascaded combination?
a) 1/(s+2)(s+1)
b) 1/(s+1)s
c) 1/(s+3)
d) 0.5/(s+1)(s+2)
Explanation: Laplace transform is the transformation that transforms the time domain into frequency domain and of both the cascaded systems are 1/(s+1)(s+2).
8. A transfer function has two zeroes at infinity. Then the relation between the numerator(N) and the denominator degree(M) of the transfer function is:
a) N=M+2
b) N=M-2
c) N=M+1
d) N=M-1
Explanation: Zeroes at infinity implies two poles at origin hence the type of the system is two and degree of denominator is M=N+2.
9. When deriving the transfer function of a linear element
b) Initial conditions are taken into account but the element is assumed to be not loaded
c) Initial conditions are assumed to be zero but loading is taken into account
d) Initial conditions are assumed to be zero and the element is assumed to be not loaded | HuggingFaceTB/finemath | |
# Re: about that 5% - Microsoft Windows
This is a discussion on Re: about that 5% - Microsoft Windows ; Obviously, there's a better metric than the one that came up 5%. I imagine it was used because they did not wish to explain a more appropriate one to a public unwilling to learn it. If 5% or more crash ...
1. ## Re: about that 5%
Obviously, there's a better metric than the one that came up 5%. I imagine
it was used because they did not wish to explain a more appropriate one to a
public unwilling to learn it. If 5% or more crash daily, how many crash once
every other day? Shouldn't that somehow be included? Perhaps they would
count as half a percentage each. However, we want a metric that starts at
zero for no crashes ever, and maxes out at 100% for the worst possible case
considered. Let us begin.
the series 1/2 + 1/4 + 1/8 + 1/16 + 1/32... = 1. This is fairly well known.
Let N be the number of computers out there running MS-Windows of the version
which is to be considered.
Let M be the number of computers which crash daily. These are each assigned
a weight of 1.
Then, for the rest of the computers, each is assigned a weight according to
the formula:
Let p_c = # of crashes per # of days observed for computer c (which must be
less than one, otherwise we throw them in with the M because on average they
are crashing more than once per day).
Then, the given c is assigned a weight, W_c, of 1 / ( 1 + log {base 2} p_c).
This will be less than one, and scaled so that for every halving of crash
time, the denominator will increase by one. For a computer that crashed once
a day, W_c would be one.
The final metric given to the public would then be:
pain_MS = (M + Sum {over all c not included in M count} W_c) / N.
You will find that 0% <= pain_MS <= 100%.
If you wish to represent computers that crash several times per day, I'm
sure we could design a metric for that as well. The one I have included here
will be higher in value than 5%, with just cause: it is likely truer to what
is going on in the system. We wish to rate the pain caused by the crashes
rather than some abstract statement based on the abstract concept of unit,
day. We should be even more pessimistic and include somehow ones that are
crashing multiple times daily, but perhaps someone could give a suggestion
on how to rate these, because it is not clearly evident how to do this. I am
currently thinking of a proof system, as in alcohol that goes up to 200,
rather than a 100% max. This would make it possible, probably, while
retaining a 100 central zone for "bad," and higher than that for "really
// u l i e n
--
As you brace for what's to come
We embrace the millenium
We evolve into info overlords
2. ## Re: about that 5%
talk.bizarre removed from newsgroups. I may be weird, but
not *that* weird. Yet. :-)
wrote
on Thu, 7 Aug 2003 09:40:29 +0000 (UTC)
<785da1860f80eac915d5483da0916dec.48257@mygate.mail gate.org>:
> "// u l i e n" wrote:
>
>> If you wish to represent computers that crash several times per day,
>> I'm sure we could design a metric for that as well.
>
> Better get on it then, the 5% figure is for computers whose MS-Windows
> Operating Systems, WinXP or newer, crash _three or more_ times a day!
>
> Obviously, any crashes at all, absent power failure ones, are a total
> condemnation of M\$-Windows quality and usability.
>
> I wish I knew more about Poisson distributions; that one figure would
> probably suffice to estimate the "at least one crash per day", which
> by guess would be around 37% (rough cube root of 5%) to yield a 5%
> crash three or more times a day result.
>
> xanthian.
>
A Poisson distribution is a standard mathematical/statistical tool, that
much I know. Google should help (I'd have to :-) ).
Another problem may be that 'on average; 5% crash per day' is not all
that great a statistic; what, precisely, is it telling us?
[1] That an "average computer" has a probability of 5% of crashing
during a given day? (More precisely, the number of crashes divided
by (the number of computers times operating hours) is 1 in 480 (the
number of hours in 20 days)?)
[2] That, if 100 computers are running Windows at 12 midnight
and left to run doing "average" things, that at the succeeding
12 midnight on average 5 will reboot, hang, or crash at least
once, and possibly more than once?
[3] Something else?
There's also the issue of what the machines are doing.
Are they heavily loaded (simulations, raytrace rendering,
serious gameplaying), lightly loaded (Solitaire, MS Word
as the writer suffers mental block, web browsing), or
sitting there playing the modern equivalent of "find the
path through the maze and leak" on their screensavers? :-)
(That bug, at least, was fixed some time ago. But I wonder
what new bug has shown up in its place.)
I can't say Linux is the ultimate in reliability but certainly
Windows isn't either. :-)
As for all crashes being a condemnation if they're not related
to power failure, I'll simply point out some obvious
other possibilities:
[1] Insufficient power. This is arguably why many consumer
computers crash in the first place; if the power supply
can't keep its voltage up under the current load things are
going to break. Most likely this is the issue for broken
hardware upgrades (e.g., placing a new drive in a machine),
for example.
[2] Hardware duds. This is an obvious one but happens occasionally.
[3] Badly-written drivers. This one's a 50-50, in
some ways; who is ultimately responsible? It may
be Microsoft for publishing ambiguous specifications.
It may be the hardware for being not quite what the driver
writer expects, or the hardware designer, for designing
the hardware. It may be the driver writer. It may be
someone else's driver entirely, or two cards interfering
as they're fighting over a resource (probably INT or DMA).
And of course:
[4] Badly-written system DLLs, and a kernel that (somehow)
allows the system to become unstable if an application
crashes. (How can we tell? We can't read the kernel
source code or the system DLL source either.) In a good
OS a crashing application should not render the entire
system unstable, even if a system DLL is the culprit.
Windows 2k and XP aspire to that and for the most part
succeed, as far as I know. However, Linux, Unix and
mainframe systems have had that from the beginning.
ObJava: Not sure what the issues are regarding Java crashes.
There are a number of things that could happen: thread deaths
are probably the most obvious. Since Java depends on a JVM
the reliability of Java is tied to that JVM, and the OS it
sits on. Java should not crash but I've seen a few too
many emergency dumps (hs-error.log, IIRC) to be entirely
satisified. (On Linux and Solaris, yet.)
-- | HuggingFaceTB/finemath | |
# 3-sphere
Last updated Stereographic projection of the hypersphere's parallels (red), meridians (blue) and hypermeridians (green). Because this projection is conformal, the curves intersect each other orthogonally (in the yellow points) as in 4D. All curves are circles: the curves that intersect ⟨0,0,0,1⟩ have infinite radius (= straight line). In this picture, the whole 3D space maps the surface of the hypersphere, whereas in the previous picture the 3D space contained the shadow of the bulk hypersphere. Direct projection of 3-sphere into 3D space and covered with surface grid, showing structure as stack of 3D spheres (2-spheres)
In mathematics, a 3-sphere, or glome, is a higher-dimensional analogue of a sphere. It may be embedded in 4-dimensional Euclidean space as the set of points equidistant from a fixed central point. Analogous to how the boundary of a ball in three dimensions is an ordinary sphere (or 2-sphere, a two-dimensional surface), the boundary of a ball in four dimensions is a 3-sphere (an object with three dimensions). A 3-sphere is an example of a 3-manifold and an n-sphere.
## Definition
In coordinates, a 3-sphere with center (C0, C1, C2, C3) and radius r is the set of all points (x0, x1, x2, x3) in real, 4-dimensional space (R4) such that
$\sum _{i=0}^{3}(x_{i}-C_{i})^{2}=(x_{0}-C_{0})^{2}+(x_{1}-C_{1})^{2}+(x_{2}-C_{2})^{2}+(x_{3}-C_{3})^{2}=r^{2}.$ The 3-sphere centered at the origin with radius 1 is called the unit 3-sphere and is usually denoted S3:
$S^{3}=\left\{(x_{0},x_{1},x_{2},x_{3})\in \mathbb {R} ^{4}:x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1\right\}.$ It is often convenient to regard R4 as the space with 2 complex dimensions (C2) or the quaternions (H). The unit 3-sphere is then given by
$S^{3}=\left\{(z_{1},z_{2})\in \mathbb {C} ^{2}:|z_{1}|^{2}+|z_{2}|^{2}=1\right\}$ or
$S^{3}=\left\{q\in \mathbb {H$ :\|q\|=1\right\}.} This description as the quaternions of norm one identifies the 3-sphere with the versors in the quaternion division ring. Just as the unit circle is important for planar polar coordinates, so the 3-sphere is important in the polar view of 4-space involved in quaternion multiplication. See polar decomposition of a quaternion for details of this development of the three-sphere. This view of the 3-sphere is the basis for the study of elliptic space as developed by Georges Lemaître.
## Properties
### Elementary properties
The 3-dimensional surface volume of a 3-sphere of radius r is
$SV=2\pi ^{2}r^{3}\,$ while the 4-dimensional hypervolume (the volume of the 4-dimensional region bounded by the 3-sphere) is
$H={\frac {1}{2}}\pi ^{2}r^{4}.$ Every non-empty intersection of a 3-sphere with a three-dimensional hyperplane is a 2-sphere (unless the hyperplane is tangent to the 3-sphere, in which case the intersection is a single point). As a 3-sphere moves through a given three-dimensional hyperplane, the intersection starts out as a point, then becomes a growing 2-sphere that reaches its maximal size when the hyperplane cuts right through the "equator" of the 3-sphere. Then the 2-sphere shrinks again down to a single point as the 3-sphere leaves the hyperplane.
### Topological properties
A 3-sphere is a compact, connected, 3-dimensional manifold without boundary. It is also simply connected. What this means, in the broad sense, is that any loop, or circular path, on the 3-sphere can be continuously shrunk to a point without leaving the 3-sphere. The Poincaré conjecture, proved in 2003 by Grigori Perelman, provides that the 3-sphere is the only three-dimensional manifold (up to homeomorphism) with these properties.
The 3-sphere is homeomorphic to the one-point compactification of R3. In general, any topological space that is homeomorphic to the 3-sphere is called a topological 3-sphere.
The homology groups of the 3-sphere are as follows: H0(S3,Z) and H3(S3,Z) are both infinite cyclic, while Hi(S3,Z) = {0} for all other indices i. Any topological space with these homology groups is known as a homology 3-sphere. Initially Poincaré conjectured that all homology 3-spheres are homeomorphic to S3, but then he himself constructed a non-homeomorphic one, now known as the Poincaré homology sphere. Infinitely many homology spheres are now known to exist. For example, a Dehn filling with slope 1/n on any knot in the 3-sphere gives a homology sphere; typically these are not homeomorphic to the 3-sphere.
As to the homotopy groups, we have π1(S3) = π2(S3) = {0} and π3(S3) is infinite cyclic. The higher-homotopy groups (k ≥ 4) are all finite abelian but otherwise follow no discernible pattern. For more discussion see homotopy groups of spheres.
k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 πk(S3) 0 0 0 Z Z2 Z2 Z12 Z2 Z2 Z3 Z15 Z2 Z2⊕Z2 Z12⊕Z2 Z84⊕Z2⊕Z2 Z2⊕Z2 Z6
### Geometric properties
The 3-sphere is naturally a smooth manifold, in fact, a closed embedded submanifold of R4. The Euclidean metric on R4 induces a metric on the 3-sphere giving it the structure of a Riemannian manifold. As with all spheres, the 3-sphere has constant positive sectional curvature equal to 1/r2 where r is the radius.
Much of the interesting geometry of the 3-sphere stems from the fact that the 3-sphere has a natural Lie group structure given by quaternion multiplication (see the section below on group structure). The only other spheres with such a structure are the 0-sphere and the 1-sphere (see circle group).
Unlike the 2-sphere, the 3-sphere admits nonvanishing vector fields (sections of its tangent bundle). One can even find three linearly independent and nonvanishing vector fields. These may be taken to be any left-invariant vector fields forming a basis for the Lie algebra of the 3-sphere. This implies that the 3-sphere is parallelizable. It follows that the tangent bundle of the 3-sphere is trivial. For a general discussion of the number of linear independent vector fields on a n-sphere, see the article vector fields on spheres.
There is an interesting action of the circle group T on S3 giving the 3-sphere the structure of a principal circle bundle known as the Hopf bundle. If one thinks of S3 as a subset of C2, the action is given by
$(z_{1},z_{2})\cdot \lambda =(z_{1}\lambda ,z_{2}\lambda )\quad \forall \lambda \in \mathbb {T}$ .
The orbit space of this action is homeomorphic to the two-sphere S2. Since S3 is not homeomorphic to S2 × S1, the Hopf bundle is nontrivial.
## Topological construction
There are several well-known constructions of the three-sphere. Here we describe gluing a pair of three-balls and then the one-point compactification.
### Gluing
A 3-sphere can be constructed topologically by "gluing" together the boundaries of a pair of 3-balls. The boundary of a 3-ball is a 2-sphere, and these two 2-spheres are to be identified. That is, imagine a pair of 3-balls of the same size, then superpose them so that their 2-spherical boundaries match, and let matching pairs of points on the pair of 2-spheres be identically equivalent to each other. In analogy with the case of the 2-sphere (see below), the gluing surface is called an equatorial sphere.
Note that the interiors of the 3-balls are not glued to each other. One way to think of the fourth dimension is as a continuous real-valued function of the 3-dimensional coordinates of the 3-ball, perhaps considered to be "temperature". We take the "temperature" to be zero along the gluing 2-sphere and let one of the 3-balls be "hot" and let the other 3-ball be "cold". The "hot" 3-ball could be thought of as the "upper hemisphere" and the "cold" 3-ball could be thought of as the "lower hemisphere". The temperature is highest/lowest at the centers of the two 3-balls.
This construction is analogous to a construction of a 2-sphere, performed by gluing the boundaries of a pair of disks. A disk is a 2-ball, and the boundary of a disk is a circle (a 1-sphere). Let a pair of disks be of the same diameter. Superpose them and glue corresponding points on their boundaries. Again one may think of the third dimension as temperature. Likewise, we may inflate the 2-sphere, moving the pair of disks to become the northern and southern hemispheres.
### One-point compactification
After removing a single point from the 2-sphere, what remains is homeomorphic to the Euclidean plane. In the same way, removing a single point from the 3-sphere yields three-dimensional space. An extremely useful way to see this is via stereographic projection. We first describe the lower-dimensional version.
Rest the south pole of a unit 2-sphere on the xy-plane in three-space. We map a point P of the sphere (minus the north pole N) to the plane by sending P to the intersection of the line NP with the plane. Stereographic projection of a 3-sphere (again removing the north pole) maps to three-space in the same manner. (Notice that, since stereographic projection is conformal, round spheres are sent to round spheres or to planes.)
A somewhat different way to think of the one-point compactification is via the exponential map. Returning to our picture of the unit two-sphere sitting on the Euclidean plane: Consider a geodesic in the plane, based at the origin, and map this to a geodesic in the two-sphere of the same length, based at the south pole. Under this map all points of the circle of radius π are sent to the north pole. Since the open unit disk is homeomorphic to the Euclidean plane, this is again a one-point compactification.
The exponential map for 3-sphere is similarly constructed; it may also be discussed using the fact that the 3-sphere is the Lie group of unit quaternions.
## Coordinate systems on the 3-sphere
The four Euclidean coordinates for S3 are redundant since they are subject to the condition that x02 + x12 + x22 + x32 = 1. As a 3-dimensional manifold one should be able to parameterize S3 by three coordinates, just as one can parameterize the 2-sphere using two coordinates (such as latitude and longitude). Due to the nontrivial topology of S3 it is impossible to find a single set of coordinates that cover the entire space. Just as on the 2-sphere, one must use at least two coordinate charts. Some different choices of coordinates are given below.
### Hyperspherical coordinates
It is convenient to have some sort of hyperspherical coordinates on S3 in analogy to the usual spherical coordinates on S2. One such choice — by no means unique — is to use (ψ, θ, φ), where
{\begin{aligned}x_{0}&=r\cos \psi \\x_{1}&=r\sin \psi \cos \theta \\x_{2}&=r\sin \psi \sin \theta \cos \varphi \\x_{3}&=r\sin \psi \sin \theta \sin \varphi \end{aligned}} where ψ and θ run over the range 0 to π, and φ runs over 0 to 2π. Note that, for any fixed value of ψ, θ and φ parameterize a 2-sphere of radius r sin ψ, except for the degenerate cases, when ψ equals 0 or π, in which case they describe a point.
The round metric on the 3-sphere in these coordinates is given by[ citation needed ]
$ds^{2}=r^{2}\left[d\psi ^{2}+\sin ^{2}\psi \left(d\theta ^{2}+\sin ^{2}\theta \,d\varphi ^{2}\right)\right]$ and the volume form by
$dV=r^{3}\left(\sin ^{2}\psi \,\sin \theta \right)\,d\psi \wedge d\theta \wedge d\varphi .$ These coordinates have an elegant description in terms of quaternions. Any unit quaternion q can be written as a versor:
$q=e^{\tau \psi }=\cos \psi +\tau \sin \psi$ where τ is a unit imaginary quaternion; that is, a quaternion that satisfies τ2 = −1. This is the quaternionic analogue of Euler's formula. Now the unit imaginary quaternions all lie on the unit 2-sphere in Im H so any such τ can be written:
$\tau =(\cos \theta )i+(\sin \theta \cos \varphi )j+(\sin \theta \sin \varphi )k$ With τ in this form, the unit quaternion q is given by
$q=e^{\tau \psi }=x_{0}+x_{1}i+x_{2}j+x_{3}k$ where x0,1,2,3 are as above.
When q is used to describe spatial rotations (cf. quaternions and spatial rotations), it describes a rotation about τ through an angle of 2ψ.
### Hopf coordinates The Hopf fibration can be visualized using a stereographic projection of S to R and then compressing R to a ball. This image shows points on S and their corresponding fibers with the same color.
For unit radius another choice of hyperspherical coordinates, (η, ξ1, ξ2), makes use of the embedding of S3 in C2. In complex coordinates (z1, z2) ∈ C2 we write
{\begin{aligned}z_{1}&=e^{i\,\xi _{1}}\sin \eta \\z_{2}&=e^{i\,\xi _{2}}\cos \eta .\end{aligned}} This could also be expressed in R4 as
{\begin{aligned}x_{0}&=\cos \xi _{1}\sin \eta \\x_{1}&=\sin \xi _{1}\sin \eta \\x_{2}&=\cos \xi _{2}\cos \eta \\x_{3}&=\sin \xi _{2}\cos \eta .\end{aligned}} Here η runs over the range 0 to π/2, and ξ1 and ξ2 can take any values between 0 and 2π. These coordinates are useful in the description of the 3-sphere as the Hopf bundle
$S^{1}\to S^{3}\to S^{2}.\,$ A diagram depicting the poloidal (ξ1) direction, represented by the red arrow, and the toroidal (ξ2) direction, represented by the blue arrow, although the terms poloidal and toroidal are arbitrary in this flat torus case.
For any fixed value of η between 0 and π/2, the coordinates (ξ1, ξ2) parameterize a 2-dimensional torus. Rings of constant ξ1 and ξ2 above form simple orthogonal grids on the tori. See image to right. In the degenerate cases, when η equals 0 or π/2, these coordinates describe a circle.
The round metric on the 3-sphere in these coordinates is given by
$ds^{2}=d\eta ^{2}+\sin ^{2}\eta \,d\xi _{1}^{2}+\cos ^{2}\eta \,d\xi _{2}^{2}$ and the volume form by
$dV=\sin \eta \cos \eta \,d\eta \wedge d\xi _{1}\wedge d\xi _{2}.$ To get the interlocking circles of the Hopf fibration, make a simple substitution in the equations above
{\begin{aligned}z_{1}&=e^{i\,(\xi _{1}+\xi _{2})}\sin \eta \\z_{2}&=e^{i\,(\xi _{2}-\xi _{1})}\cos \eta .\end{aligned}} In this case η, and ξ1 specify which circle, and ξ2 specifies the position along each circle. One round trip (0 to 2π) of ξ1 or ξ2 equates to a round trip of the torus in the 2 respective directions.
### Stereographic coordinates
Another convenient set of coordinates can be obtained via stereographic projection of S3 from a pole onto the corresponding equatorial R3 hyperplane. For example, if we project from the point (−1, 0, 0, 0) we can write a point p in S3 as
$p=\left({\frac {1-\|u\|^{2}}{1+\|u\|^{2}}},{\frac {2\mathbf {u} }{1+\|u\|^{2}}}\right)={\frac {1+\mathbf {u} }{1-\mathbf {u} }}$ where u = (u1, u2, u3) is a vector in R3 and ||u||2 = u12 + u22 + u32. In the second equality above, we have identified p with a unit quaternion and u = u1i + u2j + u3k with a pure quaternion. (Note that the numerator and denominator commute here even though quaternionic multiplication is generally noncommutative). The inverse of this map takes p = (x0, x1, x2, x3) in S3 to
$\mathbf {u} ={\frac {1}{1+x_{0}}}\left(x_{1},x_{2},x_{3}\right).$ We could just as well have projected from the point (1, 0, 0, 0), in which case the point p is given by
$p=\left({\frac {-1+\|v\|^{2}}{1+\|v\|^{2}}},{\frac {2\mathbf {v} }{1+\|v\|^{2}}}\right)={\frac {-1+\mathbf {v} }{1+\mathbf {v} }}$ where v = (v1, v2, v3) is another vector in R3. The inverse of this map takes p to
$\mathbf {v} ={\frac {1}{1-x_{0}}}\left(x_{1},x_{2},x_{3}\right).$ Note that the u coordinates are defined everywhere but (−1, 0, 0, 0) and the v coordinates everywhere but (1, 0, 0, 0). This defines an atlas on S3 consisting of two coordinate charts or "patches", which together cover all of S3. Note that the transition function between these two charts on their overlap is given by
$\mathbf {v} ={\frac {1}{\|u\|^{2}}}\mathbf {u}$ and vice versa.
## Group structure
When considered as the set of unit quaternions, S3 inherits an important structure, namely that of quaternionic multiplication. Because the set of unit quaternions is closed under multiplication, S3 takes on the structure of a group. Moreover, since quaternionic multiplication is smooth, S3 can be regarded as a real Lie group. It is a nonabelian, compact Lie group of dimension 3. When thought of as a Lie group S3 is often denoted or U(1, H).
It turns out that the only spheres that admit a Lie group structure are , thought of as the set of unit complex numbers, and S3, the set of unit quaternions (The degenerate case S0 which consists of the real numbers 1 and -1 is also a Lie group, albeit a 0-dimensional one). One might think that S7, the set of unit octonions, would form a Lie group, but this fails since octonion multiplication is nonassociative. The octonionic structure does give S7 one important property: parallelizability . It turns out that the only spheres that are parallelizable are S1, S3, and S7.
By using a matrix representation of the quaternions, H, one obtains a matrix representation of S3. One convenient choice is given by the Pauli matrices:
$x_{1}+x_{2}i+x_{3}j+x_{4}k\mapsto {\begin{pmatrix}\;\;\,x_{1}+ix_{2}&x_{3}+ix_{4}\\-x_{3}+ix_{4}&x_{1}-ix_{2}\end{pmatrix}}.$ This map gives an injective algebra homomorphism from H to the set of 2 × 2 complex matrices. It has the property that the absolute value of a quaternion q is equal to the square root of the determinant of the matrix image of q.
The set of unit quaternions is then given by matrices of the above form with unit determinant. This matrix subgroup is precisely the special unitary group SU(2). Thus, S3 as a Lie group is isomorphic to SU(2).
Using our Hopf coordinates (η, ξ1, ξ2) we can then write any element of SU(2) in the form
${\begin{pmatrix}e^{i\,\xi _{1}}\sin \eta &e^{i\,\xi _{2}}\cos \eta \\-e^{-i\,\xi _{2}}\cos \eta &e^{-i\,\xi _{1}}\sin \eta \end{pmatrix}}.$ Another way to state this result is if we express the matrix representation of an element of SU(2) as a linear combination of the Pauli matrices. It is seen that an arbitrary element U ∈ SU(2) can be written as
$U=\alpha _{0}I+\sum _{i=1}^{3}\alpha _{i}J_{i}.$ The condition that the determinant of U is +1 implies that the coefficients α1 are constrained to lie on a 3-sphere.
## In literature
In Edwin Abbott Abbott's Flatland , published in 1884, and in Sphereland , a 1965 sequel to Flatland by Dionys Burger, the 3-sphere is referred to as an oversphere, and a 4-sphere is referred to as a hypersphere.
Writing in the American Journal of Physics, Mark A. Peterson describes three different ways of visualizing 3-spheres and points out language in The Divine Comedy that suggests Dante viewed the Universe in the same way.
## Related Research Articles In mathematics and physics, Laplace's equation is a second-order partial differential equation named after Pierre-Simon Laplace who first studied its properties. This is often written as In geometry, the stereographic projection is a particular mapping (function) that projects a sphere onto a plane. The projection is defined on the entire sphere, except at one point: the projection point. Where it is defined, the mapping is smooth and bijective. It is conformal, meaning that it preserves angles at which curves meet. It is neither isometric nor area-preserving: that is, it preserves neither distances nor the areas of figures.
In mechanics and geometry, the 3D rotation group, often denoted SO(3), is the group of all rotations about the origin of three-dimensional Euclidean space under the operation of composition. By definition, a rotation about the origin is a transformation that preserves the origin, Euclidean distance, and orientation. Every non-trivial rotation is determined by its axis of rotation and its angle of rotation. Composing two rotations results in another rotation; every rotation has a unique inverse rotation; and the identity map satisfies the definition of a rotation. Owing to the above properties, the set of all rotations is a group under composition. Rotations are not commutative, making it a nonabelian group. Moreover, the rotation group has a natural structure as a manifold for which the group operations are smoothly differentiable; so it is in fact a Lie group. It is compact and has dimension 3. In physics and mathematics, the Lorentz group is the group of all Lorentz transformations of Minkowski spacetime, the classical and quantum setting for all (non-gravitational) physical phenomena. The Lorentz group is named for the Dutch physicist Hendrik Lorentz. In the mathematical field of differential topology, the Hopf fibration describes a 3-sphere in terms of circles and an ordinary sphere. Discovered by Heinz Hopf in 1931, it is an influential early example of a fiber bundle. Technically, Hopf found a many-to-one continuous function from the 3-sphere onto the 2-sphere such that each distinct point of the 2-sphere is mapped from a distinct great circle of the 3-sphere. Thus the 3-sphere is composed of fibers, where each fiber is a circle — one for each point of the 2-sphere. In quantum mechanics and computing, the Bloch sphere is a geometrical representation of the pure state space of a two-level quantum mechanical system (qubit), named after the physicist Felix Bloch.
Spatial rotations in three dimensions can be parametrized using both Euler angles and unit quaternions. This article explains how to convert between the two representations. Actually this simple use of "quaternions" was first presented by Euler some seventy years earlier than Hamilton to solve the problem of magic squares. For this reason the dynamics community commonly refers to quaternions in this application as "Euler parameters".
In mathematics, the group of rotations about a fixed point in four-dimensional Euclidean space is denoted SO(4). The name comes from the fact that it is the special orthogonal group of order 4. Conical coordinates are a three-dimensional orthogonal coordinate system consisting of concentric spheres and by two families of perpendicular cones, aligned along the z- and x-axes, respectively.
In special functions, a topic in mathematics, spin-weighted spherical harmonics are generalizations of the standard spherical harmonics and—like the usual spherical harmonics—are functions on the sphere. Unlike ordinary spherical harmonics, the spin-weighted harmonics are U(1) gauge fields rather than scalar fields: mathematically, they take values in a complex line bundle. The spin-weighted harmonics are organized by degree l, just like ordinary spherical harmonics, but have an additional spin weights that reflects the additional U(1) symmetry. A special basis of harmonics can be derived from the Laplace spherical harmonics Ylm, and are typically denoted by sYlm, where l and m are the usual parameters familiar from the standard Laplace spherical harmonics. In this special basis, the spin-weighted spherical harmonics appear as actual functions, because the choice of a polar axis fixes the U(1) gauge ambiguity. The spin-weighted spherical harmonics can be obtained from the standard spherical harmonics by application of spin raising and lowering operators. In particular, the spin-weighted spherical harmonics of spin weight s = 0 are simply the standard spherical harmonics:
In geometry, various formalisms exist to express a rotation in three dimensions as a mathematical transformation. In physics, this concept is applied to classical mechanics where rotational kinematics is the science of quantitative description of a purely rotational motion. The orientation of an object at a given instant is described with the same tools, as it is defined as an imaginary rotation from a reference placement in space, rather than an actually observed rotation from a previous placement in space.
In mathematics, the Prolate spheroidal wave functions (PSWF) are a set of orthogonal bandlimited functions. They are eigenfunctions of a timelimiting operation followed by a lowpassing operation. Let denote the time truncation operator, such that if and only if is timelimited within . Similarly, let denote an ideal low-pass filtering operator, such that if and only if is bandlimited within . The operator turns out to be linear, bounded and self-adjoint. For we denote with the n-th eigenfunction, defined as
In mathematics, the spectral theory of ordinary differential equations is the part of spectral theory concerned with the determination of the spectrum and eigenfunction expansion associated with a linear ordinary differential equation. In his dissertation Hermann Weyl generalized the classical Sturm–Liouville theory on a finite closed interval to second order differential operators with singularities at the endpoints of the interval, possibly semi-infinite or infinite. Unlike the classical case, the spectrum may no longer consist of just a countable set of eigenvalues, but may also contain a continuous part. In this case the eigenfunction expansion involves an integral over the continuous part with respect to a spectral measure, given by the Titchmarsh–Kodaira formula. The theory was put in its final simplified form for singular differential equations of even degree by Kodaira and others, using von Neumann's spectral theorem. It has had important applications in quantum mechanics, operator theory and harmonic analysis on semisimple Lie groups.
In mathematics, a Coulomb wave function is a solution of the Coulomb wave equation, named after Charles-Augustin de Coulomb. They are used to describe the behavior of charged particles in a Coulomb potential and can be written in terms of confluent hypergeometric functions or Whittaker functions of imaginary argument. In general relativity, a point mass deflects a light ray with impact parameter by an angle approximately equal to
In fluid dynamics, the Oseen equations describe the flow of a viscous and incompressible fluid at small Reynolds numbers, as formulated by Carl Wilhelm Oseen in 1910. Oseen flow is an improved description of these flows, as compared to Stokes flow, with the (partial) inclusion of convective acceleration. In fluid dynamics, a cnoidal wave is a nonlinear and exact periodic wave solution of the Korteweg–de Vries equation. These solutions are in terms of the Jacobi elliptic function cn, which is why they are coined cnoidal waves. They are used to describe surface gravity waves of fairly long wavelength, as compared to the water depth. Symmetries in quantum mechanics describe features of spacetime and particles which are unchanged under some transformation, in the context of quantum mechanics, relativistic quantum mechanics and quantum field theory, and with applications in the mathematical formulation of the standard model and condensed matter physics. In general, symmetry in physics, invariance, and conservation laws, are fundamentally important constraints for formulating physical theories and models. In practice, they are powerful methods for solving problems and predicting what can happen. While conservation laws do not always give the answer to the problem directly, they form the correct constraints and the first steps to solving a multitude of problems.
In image analysis, the generalized structure tensor (GST) is an extension of the Cartesian structure tensor to curvilinear coordinates. It is mainly used to detect and to represent the "direction" parameters of curves, just as the Cartesian structure tensor detects and represents the direction in Cartesian coordinates. Curve families generated by pairs of locally orthogonal functions have been the best studied.
Moffatt eddies are sequences of eddies that develop in corners bounded by plane walls due to an arbitrary disturbance acting at asymptotically large distances from the corner. Although the source of motion is the arbitrary disturbance at large distances, the eddies develop quite independently and thus solution of these eddies emerges from an eigenvalue problem, a self-similar solution of the second kind.
1. Weisstein, Eric W. "Glome". MathWorld . Retrieved 2017-12-04.
2. Georges Lemaître (1948) "Quaternions et espace elliptique", Acta Pontifical Academy of Sciences 12:5778
3. Banchoff, Thomas. "The Flat Torus in the Three-Sphere".
4. Mark A. Peterson. "Dante and the 3-sphere" Archived 2013-02-23 at Archive.today , American Journal of Physics, vol 47, number 12, 1979, pp1031-1035 | HuggingFaceTB/finemath | |
# graph y 2 3x 2
## Graph of Y=2/3x+2 Get Easy Solution
The graph of Y=2/3x+2 represents a graph of a linear function. On the given graph you can find all of the important points for function Y=2/3x+2 (if they exist). You can always share this solution
## SOLUTION: graph the following linear equation y=2/3x2
Click here to see ALL problems on Linearsystems. Question : graph the following linear equation. y=2/3x2. Please help! Found 2 solutions by jim thompson5910, cartert: Answer by jim thompson5910 () ( Show Source ): You can put this solution on YOUR website! Looking at we can see that the equation is in slopeintercept form where the
## Graph y=2/3x, Mathway
y = 2 3x y = 2 3 x. Use the slopeintercept form to find the slope and yintercept. Tap for more steps... Find the values of m m and b b using the form y = m x + b y = m x + b. m = 2 3 m = 2 3. b = 0 b = 0. The slope of the line is the value of m m, and the yintercept is the value of b b. Slope: 2 3 2 3.
## Graph y=3x+2, Mathway
Graph y=3x+2. y = 3x + 2 y = 3 x + 2. Use the slopeintercept form to find the slope and yintercept. Tap for more steps... The slopeintercept form is y = m x + b y = m x + b, where m m is the slope and b b is the yintercept. y = m x + b y = m x + b. Find the values of m m and b b using the form y = m x + b y
## Graph y=3x^2, Mathway
Graph y=3x^2. Find the properties of the given parabola. Tap for more steps... Rewrite the equation in vertex form. Tap for more steps... Complete the square for . The focus of a parabola can be found by adding to the ycoordinate if the parabola opens up or down. Substitute the known values of , ,
## How do you graph y=3x^22, Socratic
First, graph the parent function y = x2. Then vertically stretch this graph by a factor of 3 and translate it down 2 units. graph {3x^2 2 [10, 10, 5, 5]} You could also make a table of x and y values, graph the points, and connect them.
## Graph y=2/3x1, Mathway
Algebra. Graph y=2/3x1. y = 2 3 x 1 y = 2 3 x 1. Rewrite in slopeintercept form. Tap for more steps... The slopeintercept form is y = m x + b y = m x + b, where m m is the slope and b b is the yintercept. y = m x + b y = m x + b. Reorder terms. y = 2 3 x 1 y = 2 3 x 1.
## Graph y=2/3x+6, Mathway
Graph y=2/3x+6. Rewrite in slopeintercept form. Tap for more steps... The slopeintercept form is , where is the slope and is the yintercept. Reorder terms. Use the slopeintercept form to find the slope and yintercept. Tap for more steps... Find the values of and using the form .
## How do you graph y=2/3x, Socratic
See a solution process below: First, solve for two points which solve the equation and plot these points: First Point: For x = 0 y = 2/3 xx 0 y = 0 or (0, 0) Second Point: For x = 3 y = 2/3 xx 3 y = 6/3 y = 2 or (3, 2) We can next plot the two points on the coordinate plane: graph{(x^2+y^20.035)((x3)^2+(y+2)^20.035)=0 [10, 10, 5, 5]} Now, we can draw a straight line through the two | HuggingFaceTB/finemath | |
# Pre Kindergarten Math Teacher Toolkit Math Worksheets Algebraic Reasoning
``` Pre-Kindergarten
Math Teacher Toolkit
Math Worksheets
Algebraic Reasoning
Pattern Matcher- match the pattern to the one given (1.1.1)
Zoe’s Pet Shelter- sort animals by color, shape, size (1.1.1)
Flower- choose the correct flower based on certain characteristics given (1.1.2)
Numerical and Proportional Reasoning
Tickets Please- Count the number of children to place in each ride (2.1.1)
Count the Ants- count and find the corresponding number (2.1.1)
Animal Lost and Found- determine whether the animal is the correct one based on its
characteristics (2.2.7)
Recognizing numbers- with verbal directions (2.1.1)
More Less- determining which item holds more or less (2.1.2)
Guess the Number- find a number given a range and clues
One and One More- worksheets to use (2.2.6)
Counting and Ordering- various activities (2.2.5)
Matching Halves- match the half of an image to the other half (2.1.4)
Exploring Whole and Half- also has thirds, fourths (2.1.4)
Geometry and Measurement
Recognizing Shapes- with verbal directions (3.1.1)
Coloring Shapes- listening to directions to color shapes (3.1.3)
Constructing Shapes- Geoboards (3.1.3)
Hats- following directions using below, above, right left (3.2.4)
Which One- determine which object is longer, shorter, bigger, smaller, above, below (verbal
directions) (3.3.8)
The Shapes Book- will read out loud and ask questions about finding shapes in life (3.1.2)
Oddball- which object is different (3.1.2)
Working with Data
Grapher- create your own graph (4.1.1)
Starter- click and drag people into categories (4.1.1-4.3.3)
Kindergarten
EManipulatives
Math Lingo Game
Math Enrichment Problems
Math Teacher Toolkit
Math Worksheets
Algebraic Reasoning
Color Me Hungry (1.1.1)
Check Out Cookie- Help Cookie Monster find the missing pattern (1.1.3)
Patterns- what comes next (1.1.4)
Help Bobbie Bear make different outfits using different colored shirts and pants (1.1.1)
Curious George Banana 411- dial the phone based on three numbers verbally expressed (1.1.4)
Flower- choose the correct flower based on certain characteristics given (1.1.4)
Numerical and Proportional Reasoning
Math Bars (1-10 to make your own)- great for showing addition and subtraction (2.2.8)
Recognizing Numbers with the Word (2.2.9)
Number Sequence (2.1.3)
Number Fun (1-10 and then reverse) (2.1.3)
Counting down from 10- determine the next number (2.1.3)
Ten in the Bed- the story will read itself (2.2.7)
Recognizing numbers- with verbal directions (2.2.9)
Tickets Please- Count the number of children to place in each ride (2.1.1)
Ordinal Numbers- place the balloons in the correct carts (2.1.4)
Ordinal Numbers- various activities (2.1.4)
Hang the Shirts- hang the shirts on the line in the correct order (2.1.4)
Cats in Line- determine where the orange cat is in line (2.1.4)
Animal Lost and Found- determine whether the animal is the correct one based on its
characteristics (2.1.1)
Count the Ants- count and find the corresponding number (2.2.9)
Dartboard- make doubles with the dartboard (2.2.10)
Big Count- count by ones, two’s, five’s, and ten’s
Ten Frame- Practice counting and adding to twenty
Addition- with verbal directions (2.2.10)
Addition/Subtraction with Frog- number line (2.2.11)
Mucky- determining more or less (2.1.2)
Carnival Countdown- sort by greater than or equal to (2.1.2)
Guess the Number- find a number given a range and clues (2.1.12)
Matching Halves- match the half of an image to the other half (2.1.6)
Exploring Whole and Half- also had thirds, fourths (2.1.6)
Geometry and Measurement
Oscars Trash Challenge- sorting (3.1.2)
Constructing Shapes- Geoboards (3.1.3)
Time (interactive digital and analog clock)
Time to the Hour- with verbal directions
Zoe’s Pet Shelter- sort animals by color, shape, size (3.1.2)
Which One- determine which object is longer, shorter, bigger, smaller, above, below (verbal
directions) (3.2.4)
Leeky Compares- comparing three things using biggest, smallest, tallest, shortest (3.3.9)
Take the Trash Challenge printable (over, under, around) (3.2.4)
Shapes Activity – locating shapes by color with Curious George (3.1.2)
Math Playground Pattern Blocks (3.1.3)
Build a Sandcastle- figure out the shapes used to make the sandcastles (3.2.5)
Hats- following directions using below, above, right left (3.2.4)
The Shapes Book- will read out loud and ask questions about finding shapes in life (3.1.2)
Measuring Length- quick tutorial on measuring objects (3.3.8)
Simple Scales- weigh various toys (can hide various objects on the scale) (3.3.8)
Calendar- locate days on a calendar (3.3.7)
Days and Dates- read information from a calendar (3.3.6)
Working with Data
Grapher- create your own graph (4.1.1)
Sorting into Graphs- sort the various objects into graph categories (4.1.1-4.3.6)
Counting Objects- sort by object type to make the graph (4.1.1-4.3.6)
EManipulatives
Math Lingo Game
Math Teacher Toolkit
Math Worksheets
Algebraic Reasoning
Color Me Hungry (1.1.1)
Color Patterns (1.1.1)
Patterns- what comes next (1.1.3)
Attribute Blocks (1.1.2)
Adding Animals on a See-Saw to Balance (1.3.6)
Pan Balancing Bar (1.3.6)
Sequencing with Shapes and Colors (1.1.1)
Numerical and Proportional Reasoning
Place Value Party (2.1.1)
Math Bars (1-10 to make your own)- great for showing addition and subtraction (2.2.8)
Base Block Addition (1.2.5, 2.1.1)
Base Blocks Subtraction (1.2.5, 2.1.1) (2.2.14)
Chip Abacus- showing carrying and digits using chips (1.2.5)
Make a Really Big Number- click the number and place it in the value boxes (nice way of showing
place value) (2.1.1)
Number Jumbler More or Less- determine whether the number is more or less based on the
sentence (2.1.2)
Cats in Line- determine where the orange cat is in line (2.1.4)
100 Hunt- Finding numbers up to 100 (2.2.12)
Mend the number square- fill in up to 100 (2.2.9)
Game Bone- give the dog a bone on an empty 100 board, figuring out where the amount is
(2.2.12)
Number Square- Type the next number (verbally counts) (2.1.4)
Number Chart- can fill with colors to use for counting by 2, 5, 10 and so on (2.2.9)
Count Hoot’s Subtraction Game (2.2.14)
Hidden Picture Math Game (addition and subtraction to uncover a hidden picture) (2.2.14)
Estimate- determine what the number is on a hidden number line (2.1.3)
Guess the Number- find a number given a range and clues (2.1.3)
Counting Money- pennies, nickels, dimes, quarters
Counting by 10’s Dimes
Money Gallery (for Mimio) (for SMARTboard)
Geometry and Measurement
Oscars Trash Challenge- sorting (3.1.3)
Constructing Shapes- Geoboards (3.1.3)
Isometric Geoboard (3.1.3)
Time (interactive digital and analog clock) (3.3.7)
Time to the Hour- with verbal directions (3.3.7)
Time to Half Hour- match the times to the clocks (3.3.7)
Clock- move the hands by one, five, ten, fifteen, or thirty minutes and also by the hour (3.3.7)
Ladybug Maze- program the ladybug to move through the maze (3.2.5)
Bug Coordinates- help locate where the ladybug should go (3.2.5)
Time to Move- catch fish and decide what tank they belong in based on their size (3.2.9)
Tessellation Town (3.1.4)
Tessellations with regular and semi-regular tesselations
Polygon Playground (3.1.4)
Math Playground Pattern Blocks (3.1.2)
Find the Most- determine which object is holding the most liquid (3.3.8)
Simple Scales- weigh various toys (can hide various objects on the scale) (3.3.8)
Days and Dates and Days of Fun- read information from a calendar (3.3.6)
Working with Data
Grapher- create your own graph (4.1.2)
Counting Objects- sort by object type to make the graph (4.1.2 and 4.2.3)
```
DOCUMENT INFO
Shared By:
Categories:
Stats:
views: 273 posted: 11/19/2010 language: English pages: 7
Description: Pre Kindergarten Worksheets document sample | HuggingFaceTB/finemath | |
# Counterexamples around connected spaces
A connected space is a topological space that cannot be represented as the union of two or more disjoint nonempty open subsets. We look here at unions and intersections of connected spaces.
### Union of connected spaces
The union of two connected spaces $$A$$ and $$B$$ might not be connected “as shown” by two disconnected open disks on the plane.
However if the intersection $$A \cap B$$ is not empty then $$A \cup B$$ is connected.
### Intersection of connected spaces
The intersection of two connected spaces $$A$$ and $$B$$ might also not be connected. An example is provided in the plane $$\mathbb R^2$$ by taking for $$A$$ the circle centered at the origin with radius equal to $$1$$ and for $$B$$ the segment $$\{(x,0) \ : \ x \in [-1,1]\}$$. The intersection $$A \cap B = \{(-1,0),(1,0)\}$$ is the union of two points which is not connected.
# Playing with interior and closure
Let’s play with the closure and the interior of sets.
To start the play, we consider a topological space $$E$$ and denote for any subspace $$A \subset E$$: $$\overline{A}$$ the closure of $$A$$ and $$\overset{\circ}{A}$$ the interior of $$A$$.
### Warm up with the closure operator
For $$A,B$$ subsets of $$E$$, the following results hold: $$\overline{\overline{A}}=\overline{A}$$, $$A \subset B \Rightarrow \overline{A} \subset \overline{B}$$, $$\overline{A \cup B} = \overline{A} \cup \overline{B}$$ and $$\overline{A \cap B} \subset \overline{A} \cap \overline{B}$$.
Let’s prove it.
$$\overline{A}$$ being closed, it is equal to its closure and $$\overline{\overline{A}}=\overline{A}$$.
Suppose that $$A \subset B$$. As $$B \subset \overline{B}$$, we have $$A \subset \overline{B}$$. Also, $$\overline{B}$$ is closed so it contains $$\overline{A}$$, which proves $$\overline{A} \subset \overline{B}$$.
Let’s consider $$A,B \in E$$ two subsets. As $$A \subset A \cup B$$, we have $$\overline{A} \subset \overline{A \cup B}$$ and similarly $$\overline{B} \subset \overline{A \cup B}$$. Hence $$\overline{A} \cup \overline{B} \subset \overline{A \cup B}$$. Conversely, $$A \cup B \subset \overline{A} \cup \overline{B}$$ and $$\overline{A} \cup \overline{B}$$ is closed. So $$\overline{A \cup B} \subset \overline{A} \cup \overline{B}$$ and finally $$\overline{A \cup B} = \overline{A} \cup \overline{B}$$.
Regarding the inclusion $$\overline{A \cap B} \subset \overline{A} \cap \overline{B}$$, we notice that $$A \cap B \subset \overline{A} \cap \overline{B}$$ and that $$\overline{A} \cap \overline{B}$$ is closed to get the conclusion.
However, the implication $$\overline{A} \subset \overline{B} \Rightarrow A \subset B$$ doesn’t hold. For a counterexample, consider the space $$E=\mathbb R$$ equipped with the topology induced by the absolute value distance and take $$A=[0,1)$$, $$B=(0,1]$$. We have $$\overline{A}=\overline{B}=[0,1]$$.
The equality $$\overline{A} \cap \overline{B} = \overline{A \cap B}$$ doesn’t hold as well. For the proof, just consider $$A=[0,1)$$ and $$B=(1,2]$$. Continue reading Playing with interior and closure
# Counterexamples to Banach fixed-point theorem
Let $$(X,d)$$ be a metric space. Then a map $$T : X \to X$$ is called a contraction map if it exists $$0 \le k < 1$$ such that $d(T(x),T(y)) \le k d(x,y)$ for all $$x,y \in X$$. According to Banach fixed-point theorem, if $$(X,d)$$ is a complete metric space and $$T$$ a contraction map, then $$T$$ admits a fixed-point $$x^* \in X$$, i.e. $$T(x^*)=x^*$$.
We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled.
First, let’s consider $\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & x+1 \end{array}$ For all $$x,y \in \mathbb R$$ we have $$\vert f(x)-f(y) \vert = \vert x- y \vert$$. $$f$$ is not a contraction, but an isometry. Obviously, $$f$$ has no fixed-point.
We now prove that a map satisfying $d(g(x),g(y)) < d(x,y)$ might also not have a fixed-point. A counterexample is the following map $\begin{array}{l|rcl} g : & [0,+\infty) & \longrightarrow & [0,+\infty) \\ & x & \longmapsto & \sqrt{1+x^2} \end{array}$ Since $g^\prime(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for all } \xi \in [0, +\infty),$ by the mean value theorem $\vert g(x) - g(y)| = \vert g^\prime(\xi)\vert |x-y| < |x-y| \text{ for all } x, y \in [0, +\infty).$ However $$g$$ has no fixed-point. Finally, let's have a look to a space $$(X,d)$$ which is not complete. We take $$a,b \in \mathbb R$$ with $$0 < a < 1$$ and for $$(X,d)$$ the space $$X = \mathbb R \setminus \{\frac{b}{1-a}\}$$ equipped with absolute value distance. $$X$$ is not complete. Consider the map $\begin{array}{l|rcl} h : & X & \longrightarrow & X \\ & x & \longmapsto & ax + b \end{array}$ $$h$$ is well defined as for $$x \neq \frac{b}{1-a}$$, $$h(x) \neq \frac{b}{1-a}$$. $$h$$ is a contraction map as for $$x,y \in \mathbb R$$ $\vert h(x)-h(y) \vert = a \vert x - y \vert$ However, $$h$$ doesn't have a fixed-point in $$X$$ as $$\frac{b}{1-a}$$ is the only real for which $$h(x)=x$$.
# A counterexample to Krein-Milman theorem
In the theory of functional analysis, the Krein-Milman theorem states that for a separated locally convex topological vector space $$X$$, a compact convex subset $$K$$ is the closed convex hull of its extreme points.
For the reminder, an extreme point of a convex set $$S$$ is a point in $$S$$ which does not lie in any open line segment joining two points of S. A point $$p \in S$$ is an extreme point of $$S$$ if and only if $$S \setminus \{p\}$$ is still convex.
In particular, according to the Krein-Milman theorem, a non-empty compact convex set has a non-empty set of extreme points. Let see what happens if we weaken some hypothesis of Krein-Milman theorem. Continue reading A counterexample to Krein-Milman theorem
# Two algebraically complemented subspaces that are not topologically complemented
We give here an example of a two complemented subspaces $$A$$ and $$B$$ that are not topologically complemented.
For this, we consider a vector space of infinite dimension equipped with an inner product. We also suppose that $$E$$ is separable. Hence, $$E$$ has an orthonormal basis $$(e_n)_{n \in \mathbb N}$$.
Let $$a_n=e_{2n}$$ and $$b_n=e_{2n}+\frac{1}{2n+1} e_{2n+1}$$. We denote $$A$$ and $$B$$ the closures of the linear subspaces generated by the vectors $$(a_n)$$ and $$(b_n)$$ respectively. We consider $$F=A+B$$ and prove that $$A$$ and $$B$$ are complemented subspaces in $$F$$, but not topologically complemented. Continue reading Two algebraically complemented subspaces that are not topologically complemented
# A homeomorphism of the unit ball having no fixed point
Let’s recall Brouwer fixed-point theorem.
Theorem (Brouwer): Every continuous function from a convex compact subset $$K$$ of a Euclidean space to $$K$$ itself has a fixed point.
We here describe an example of a homeomorphism of the unit ball of a Hilbert space having no fixed point. Let $$E$$ be a separable Hilbert space with $$(e_n)_{n \in \mathbb{Z}}$$ as a Hilbert basis. $$B$$ and $$S$$ are respectively $$E$$ closed unit ball and unit sphere.
There is a unique linear map $$u : E \to E$$ for which $$u(e_n)=e_{n+1}$$ for all $$n \in \mathbb{Z}$$. For $$x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$$ we have $$u(x)= \sum_{n \in \mathbb{Z}} \xi_n e_{n+1}$$. $$u$$ is isometric as $\Vert u(x) \Vert^2 = \sum_{n \in \mathbb{Z}} \vert \xi_n \vert^2 = \Vert x \Vert^2$ hence one-to-one. $$u$$ is also onto as for $$x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$$, $$\sum_{n \in \mathbb{Z}} \xi_n e_{n-1} \in E$$ is an inverse image of $$x$$. Finally $$u$$ is an homeomorphism. Continue reading A homeomorphism of the unit ball having no fixed point
# Counterexample around Arzela-Ascoli theorem
Let’s recall Arzelà–Ascoli theorem:
Suppose that $$F$$ is a Banach space and $$E$$ a compact metric space. A subset $$\mathcal{H}$$ of the Banach space $$\mathcal{C}_F(E)$$ is relatively compact in the topology induced by the uniform norm if and only if it is equicontinuous and and for all $$x \in E$$, the set $$\mathcal{H}(x)=\{f(x) \ | \ f \in \mathcal{H}\}$$ is relatively compact.
We look here at what happens if we drop the requirement on space $$E$$ to be compact and provide a counterexample where the conclusion of Arzelà–Ascoli theorem doesn’t hold anymore.
We take for $$E$$ the real interval $$[0,+\infty)$$ and for all $$n \in \mathbb{N} \setminus \{0\}$$ the real function
$f_n(t)= \sin \sqrt{t+4 n^2 \pi^2}$ We prove that $$(f_n)$$ is equicontinuous, converges pointwise to $$0$$ but is not relatively compact.
According to the mean value theorem, for all $$x,y \in \mathbb{R}$$
$\vert \sin x – \sin y \vert \le \vert x – y \vert$ Hence for $$n \ge 1$$ and $$x,y \in [0,+\infty)$$
\begin{align*}
\vert f_n(x)-f_n(y) \vert &\le \vert \sqrt{x+4 n^2 \pi^2} -\sqrt{y+4 n^2 \pi^2} \vert \\
&= \frac{\vert x – y \vert}{\sqrt{x+4 n^2 \pi^2} +\sqrt{y+4 n^2 \pi^2}} \\
&\le \frac{\vert x – y \vert}{4 \pi}
\end{align*} using multiplication by the conjugate.
Which enables to prove that $$(f_n)$$ is equicontinuous.
We also have for $$n \ge 1$$ and $$x \in [0,+\infty)$$
\begin{align*}
\vert f_n(x) \vert &= \vert f_n(x) – f_n(0) \vert \le \vert \sqrt{x+4 n^2 \pi^2} -\sqrt{4 n^2 \pi^2} \vert \\
&= \frac{\vert x \vert}{\sqrt{x+4 n^2 \pi^2} +\sqrt{4 n^2 \pi^2}} \\
&\le \frac{\vert x \vert}{4 n \pi}
\end{align*}
Hence $$(f_n)$$ converges pointwise to $$0$$ and for $$t \in [0,+\infty), \mathcal{H}(t)=\{f_n(t) \ | \ n \in \mathbb{N} \setminus \{0\}\}$$ is relatively compact
Finally we prove that $$\mathcal{H}=\{f_n \ | \ n \in \mathbb{N} \setminus \{0\}\}$$ is not relatively compact. While $$(f_n)$$ converges pointwise to $$0$$, $$(f_n)$$ does not converge uniformly to $$f=0$$. Actually for $$n \ge 1$$ and $$t_n=\frac{\pi^2}{4} + 2n \pi^2$$ we have
$f_n(t_n)= \sin \sqrt{\frac{\pi^2}{4} + 2n \pi^2 +4 n^2 \pi^2}=\sin \sqrt{\left(\frac{\pi}{2} + 2 n \pi\right)^2}=1$ Consequently for all $$n \ge 1$$ $$\Vert f_n – f \Vert_\infty \ge 1$$. If $$\mathcal{H}$$ was relatively compact, $$(f_n)$$ would have a convergent subsequence with $$f=0$$ for limit. And that cannot be as for all $$n \ge 1$$ $$\Vert f_n – f \Vert_\infty \ge 1$$.
# A topological vector space with no non trivial continuous linear form
We consider here the $$L^p$$- spaces of real functions defined on $$[0,1]$$ for which the $$p$$-th power of the absolute value is Lebesgue integrable. We focus on the case $$0 < p < 1$$. We'll prove that those $$L^p$$-spaces are topological vector spaces on which there exists no continuous non-trivial linear forms (i.e. not vanishing identically). Continue reading A topological vector space with no non trivial continuous linear form
# Distance between a point and a hyperplane not reached
Let’s investigate the following question: “Is the distance between a point and a hyperplane always reached?”
In order to provide answers to the question, we consider a normed vector space $$(E, \Vert \cdot \Vert)$$ and a hyperplane $$H$$ of $$E$$. $$H$$ is the kernel of a non-zero linear form. Namely, $$H=\{x \in E \text{ | } u(x)=0\}$$.
## The case of finite dimensional vector spaces
When $$E$$ is of finite dimension, the distance $$d(a,H)=\inf\{\Vert h-a \Vert \text{ | } h \in H\}$$ between any point $$a \in E$$ and a hyperplane $$H$$ is reached at a point $$b \in H$$. The proof is rather simple. Consider a point $$c \in H$$. The set $$S = \{h \in H \text{ | } \Vert a- h \Vert \le \Vert a-c \Vert \}$$ is bounded as for $$h \in S$$ we have $$\Vert h \Vert \le \Vert a-c \Vert + \Vert a \Vert$$. $$S$$ is equal to $$D \cap H$$ where $$D$$ is the inverse image of the closed real segment $$[0,\Vert a-c \Vert]$$ by the continuous map $$f: x \mapsto \Vert a- x \Vert$$. Therefore $$D$$ is closed. $$H$$ is also closed as any linear subspace of a finite dimensional vector space. $$S$$ being the intersection of two closed subsets of $$E$$ is also closed. Hence $$S$$ is compact and the restriction of $$f$$ to $$S$$ reaches its infimum at some point $$b \in S \subset H$$ where $$d(a,H)=\Vert a-b \Vert$$. Continue reading Distance between a point and a hyperplane not reached | HuggingFaceTB/finemath | |
# FIN4750 HW1 MT1 Review
```FIN4750
Options
Department of Economics and Finance
Baruch College
FIN 4750
Midterm (Practice) Exam I
09/29/2023 (due on Friday 6 PM)
Name
___________________________________
ID Number
___________________________________
Signature
___________________________________
1. The exam is closed book and closed notes. You can bring in one page, double-sided, 8×11
formula sheet.
2. You can (and probably have to) use a calculator.
3. You have a total of 90 minutes for the exam.
4. The whole exam has a total of 60 points. It will count 30% for your final course grade.
5. Do not separate the exam book. Turn in the entire exam at the end.
7. Good luck.
Page 1
Q1. You believe stock price by year end will have the following multinomial distribution (15 points):
Price Probability
60
10%
80
20%
100
40%
120
20%
140
10%
Q1a. What should be the stock price TODAY? (3 points)
Q1b. what is the prob that a 110 strike CALL will expire ITM? (3 points)
Q1c. what is the conditional average price of underlying stock when 110 strike CALL expires ITM? (3
points)
Q1d. what is the conditional average payment from the 110 strike CALL option when the CALL expires
ITM? (3 points)
Q1e. based on Q1b-Q1d, how much should the 110 CALL be priced at? (3 points)
Page 2
Q2a. What is the probability of option expiring ITM for a 160 CALL? (3 points)
Q2b. What is the average underlying price when CALL expires ITM? (3 points)
Q2c. How much should the 160 CALL be priced at? (3 points)
Q2d. Out of the price in Q2c, how much of that is intrinsic value and how much is time value? (3
points) Hint: intrinsic value is the value of option if option expires NOW. Time value is the remaining.
Page 3
Q3. Implied MAD with Uniform Distribution
Underlying price currently at 200, and follows a uniform distribution with mean of 200. You
observed 80 strike PUT priced at \$5.00. What is the implied MAD? (10 points)
A: put price equation
Which one to choose? Or need to keep one
A: need to double check to make sure strike price X stays inside the uniform form distribution
boundaries. We throw away the smaller solution and keep the larger one.
Page 4
Q4. Current underlying price at 100, and you expect price at expiration follows uniform distribution
with mean absolute deviation of 20.
You short 10 PUTs with strike at 90.
Q4a. What is the TOTAL delta of your 10 short PUT position? (2 points)
Q4b. How many shares do you need in order to offset PUT delta from Q4a? Do you long or short the
underlying shares (2 points)?
Q4c. What is the total gamma value of your hedged PUT positions from Q4a and Q4b? (2 points)
Q4d. For the hedged put position (short 10 puts, hedged with shares), what is the PnL from
(starting) delta, and from gamma when underlying moves from 100 to 80, respectively? (4 points)
Page 5
Q4e. Calculate option price when stock moves \$80. How much of the total PnL of Q4d is from option
position, and how much is from the stock position? (2 points)
Q4f. If underlying moves down from 100 to 80, what is the new delta at stock price of 80 for your
overall position based on gamma? If you need to re-hedge to flatten delta with underlying shares,
what trade do you need to do in the shares to flatten the delta? (2 points)
Q4g. What is the PnL from the delta and gamma when underlying moves from 100 to 120,
respectively? (2 points)
Q4h. For Q4g, how much of the total PnL is from option position, and how much is stocks, when
underlying moves from 100 to 120? (2 points)
Q4i. If underlying moves up from 100 to 120, If you need to re-hedge to flatten delta with
underlying shares, what trade do you need to do in the shares? (2 points)
Page 6
``` | HuggingFaceTB/finemath | |
# Calcium chloride when dissolved in water
Question:
Calcium chloride when dissolved in water dissociates into its ions according to the following equation.
$\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)$
Calculate the number of ions obtained from $\mathrm{CaCl}_{2}$ when $222 \mathrm{~g}$ of it is dissolved in water.
Solution:
Molar mass of $\mathrm{CaCl}_{2}=40+(2 \times 35 \cdot 5)=111 \mathrm{~g} \mathrm{~mol}^{-1}$
$111 \mathrm{~g}$ of $\mathrm{CaCl}_{2}$ represent $=1 \mathrm{~mol}$
$222 \mathrm{~g}$ of $\mathrm{CaCl}_{2}$ represent $=\frac{(222 \mathrm{~g})}{(111 \mathrm{~g})} \times 1 \mathrm{~mol}=2 \mathrm{~mol}$
No. of molcules in 2 moles of $\mathrm{CaCl}_{2}=2 \times \mathrm{N}_{\mathrm{A}}=2 \times 6.022 \times 10^{23}$.
1 molecule of $\mathrm{CaCl}_{2}$ in solution form ions $=3$
$2 \times 6.022 \times 10^{23}$ molecules of $\mathrm{CaCl}_{2}$ in solution form ions $=3 \times 2 \times 6.022 \times 10^{23}$
$=36.132 \times 10^{23}$
$=3.6132 \times 10^{24}$ | HuggingFaceTB/finemath | |
# A Thinking Man
Professor Percent was a maths lecturer with an interest for new ways to express mathematical expressions. The traditional symbols (+, -, *, /, etc) were not enough anymore, to convey his superior numeric operations, so he had to invent new symbols, and only a superior brain would be able to understand the need for his new symbols.
The first symbol he invented was §; between two numbers, it meant that, if the first number was greater than the second, then the second should be subtracted from the first one; otherwise the two numbers should be added. Therefore 5 § 2 = 3, while 2 § 5 = 7.
The poor people that had to put up with this were, of course, his students. In the last test they were faced with:
5 ¿ 2 = 27
6 ¿ 3 = 27
8 ¿ 4 = 36
and also with:
5 ¤ 2 = 15
6 ¤ 4 = 12
3 ¤ 8 = 40
What are the meanings of the symbols ¿ and ¤?
Notes:
There are at least 3 different solutions for ¿.
### A Thinking Man Puzzle Solution
The symbol ¿ means the difference between the number made up of the all digits of the operation, and the mirror of this last number. i.e,
5 ¿ 2 = 52 - 25 = 27
6 ¿ 3 = 63 - 36 = 27
8 ¿ 4 = 84 - 48 = 36
An alternative solution for this symbol (submitted by Alfa Chan... many thanks!) is simply the difference between the two numbers multiplied by 9. i.e,
5 ¿ 2 = (5 - 2) × 9 = 27
6 ¿ 3 = (6 - 3) × 9 = 27
8 ¿ 4 = (8 - 4) × 9 = 36
Another alternative solution submitted by Mickey Kawick... thanks!! We have x ¿ y; If x is odd, then the result is 5x + y, otherwise it's 5x - y. i.e,
5 ¿ 2 = 5 × 5 + 2 = 27 (5 is odd, so we add the 2)
6 ¿ 3 = 6 × 5 - 3 = 27 (6 is even, so we subtract the 3)
8 ¿ 4 = 8 × 5 - 4 = 36 (8 is even, so we subtract the 4)
The symbol ¤ means the difference between the two numbers multiplied by the larger of the two numbers. i.e,
5 ¤ 2 = (5 - 2) * 5 = 3 * 5 = 15
6 ¤ 4 = (6 - 4) * 6 = 2 * 6 = 12
3 ¤ 8 = (8 - 3) * 8 = 5 * 8 = 40
An alternative solution for this symbol, as submitted by Jeff Schall (many thanks!), is the difference between the square of the bigger of the two numbers and their product. i.e,
5 ¤ 2 = (5 ^ 2) - (5 * 2) = 25 - 10 = 15
6 ¤ 4 = (6 ^ 2) - (6 * 4) = 36 - 24 = 12
3 ¤ 8 = (8 ^ 2) - (3 * 8) = 64 - 24 = 40 | HuggingFaceTB/finemath | |
##### 70 calories of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is
We know that, Q = nCpdT… (i),
Where
Q = heat required to raise the temperature
n = number of moles
Cp = specific heat capacity at constant pressure
dT = rise in temperature.
In this first case,
Amount of heat required(Q) = 70 cal
Number of moles(n) = 2 mol
Rise in temperature(dT) = (35-30)0C = 50C
Hence, from (i), we get Cp = 7 calmol-1 0C-1
Now, we know, Cp - Cv = R,
Where
Cp = specific heat capacity at constant pressure
Cv = specific heat capacity at constant volume
R = universal gas constant = 1.98 calmol-1 0C-1
Therefore, we get Cv = Cp - R = (7-1.98) cal mol-1 0C-1
= 5.02 cal mol-1 0C-1
Now, in the new case, change in temperature dT = 50C as before. Number of moles(n) = 2, and Cv = 5.02 calmol-1 0C-1
Hence, amount of heat required to raise the temperature of the same gas through the same range
dQ = nCvdT = (2 x 5.02 x 5) cal = 50.2 cal which is approximately equal to 50 cal.
Hence, the correct option is (b).
1 | HuggingFaceTB/finemath | |
# INTRO
In this blog, I will be describing all things to do with Table Calculations. The pretence behind this blog is to help with a teaching session I will give soon about Table Calculations.
So, what are Table Calculations? Directly from Tableau help, “Table Calculations are a special type of calculated field that computes on the local data in Tableau. They are calculated based on what is currently in the visualisation and do not consider any measures or dimensions that are filtered out of the visualisation”
This means that they are a way of manipulating data on Tableau, for example, if you wanted a running sum of sales for a year then this could be done with Tables Calculations. Below is Picture of where Table Calculations can be found in the calculated field menu. I have grouped the Table Calculations into different groups, as they all perform different tasks. For example, in Blue, we see the position-based Table Calculations. This work based on where the value is located. In red we see the Running expressions. In Brown the Scripts. In purple, the other Table Calculations. Lastly in Black, the Window Table Calculations. Below I give examples of when to use these.
# EXPRESSIONS
### position-based Expressions
• INDEX – Returns the Position of the Current row in a partition (group). Linked to FIRST, as it is derived from the first index position subtract row position. So, the 10th Customer, in this case, would have a value of 10.
• FIRST – Returns the positional difference from the first row in a partition (group). So 10th Customer would have a difference of -9. This Calculation is useful with Visualisations that involve Calendars or Planning.
• LAST – Returns the positional value from the Last value. Using Superstore Sample data found on Tableau, I get the first customer as having a “LAST” value of 792.
• LOOKUP – This Calculation can be used in conjunction with Table Calculations like FIRST, for example, ZN(SUM([Profit])) – LOOKUP(ZN(SUM([Profit])), FIRST ()) gives the difference in profit from the first value. This Calculation does not have to be memorised as it can be made using a quick-Table Calculation, these are made by dragging your profit variable to the marks section right clicking selecting quick Table Calculations and selecting whatever you would like. Once you have done that you can create a Calculation and drag the quick Tabled “Green Pill” into the Calculation box. Rename it to whatever you would like.
• PREVIOUS_VALUE – This Calculation can be used to get the running sum. We can use a Calculation like SUM([Profit]) + PREVIOUS_VALUE (0). This Method is a bit more intuitive than Lookup.
• RANK – The Rank Calculation uses positions like before but based on values, for example having the largest profit, would result in a rank of 1, e.g. 1st. RANK(SUM([Profit]),’ASC’) would give rank in ascending The default behaviour of RANK is very important as any two or more same values would get the same rank, so the very possible outcome could be 1st, 2nd, 2nd, 4th – Skipping 3rd place completely.
• RANK_DENSE – This rank Calculation instead allows third place after, so 1st, 2nd,2nd,3rd
• RANK_MODIFIED – This rank Calculation instead feels that if two values are identical, that the rank after is best to use, so 1st, 3rd, 3rd, 4th – skipping 2nd
• RANK_PERCENTILE – This rank behaves similarly to RANK_DENSE as the positions are divided by the end rank, for example, 1st, 2nd,2nd,3rd would give a result of 0, 0.67, 0.67, and 1. This is because 0/3 here is 0 , 2/3 is 0.67 and 3/3 = 1. Why would this ever be useful? Well, the rank that yields 0.5 e.g. 50% would be the median value.
• RANK_UNIQUE – This Rank gives a value of 1st,2nd, 3rd, 4th even if the values are identical.
### Running Expressions
• RUNNING_AVG – Gives the running average of expression, so that a clustered line chart with many peaks becomes smoothed out. I personally prefer the quick Table Calculation of a moving average as I can control more aspects of how many periods the value is averaged by. With enough previous values to average the expression, the running average and the moving average become the same.
• RUNNING_COUNT – This Calculation is essentially a running sum of the count of values, useful for showing how the count changes over time.
• RUNNING_MAX – This Calculation returns the max value over time, so as a line chart the max value is updating on whatever level of granularity. As time moves on the value will remain as a straight line until a new maximum value appears.
• RUNNING_MIN – Does the same thing as RUNNING_MAX except in reverse, so the minimum value is updated over time.
• RUNNING_SUM – Returns the running sum from the first row to the current row
### Script Expressions
• SCRIPT_BOOL – The Script Table Calculations use R and Python programming language, without downloading an external package, these will not work. BOOL implies a true-false statement, for example, SCRIPT_BOOL (“. argl>0”, SUM([Profit])) would return true if profit was above 0. For Python it would be SCRIPT_BOOL (“return [x > 0 for x in_argl]”, SUM([Profit]))
• SCRIPT_INT – Is like other SCRIPT expressions but returns an integer
• SCRIPT_REAL – Is like other SCRIPT expressions but returns a numeric value
• SCRIPT_STR – Is like other SCRIPT expressions but returns a string
### Other Expressions
• SIZE – This returns the number of rows in a partition, useful in correlation matrices. Quite like “Number of Records”
• TOTAL – Works in a similar way to Level of Detail Calculations, for example, if you wanted total average profit over three categories then TOTAL(AVG([Profit])) would give you the average across all rows, whereas AVG([Profit]) would yield average profit per category.
### Window Expressions
• WINDOW_ – The below expression with WINDOW_ appended in front of them (e.g. WINDOW_AVG) return the value of the expression in the whole window. The expressions used below can be found in the AGGREGATE section of calculated fields. These can be useful when your Level of Detail Calculations won’t allow you to use the Calculations.
• AVG
• CORR
• COUNT
• COVAR
• COVARP
• MAX
• MEDIAN
• MIN
• PERCENTILE
• STDEV
• STDEVP
• SUM
• VAR
• VARP
In my opinion, there will be specific cases when to use some of these Table Calculations. However, some of my personal favourites are:
• WINDOW_AVG – with WINDOW_AVG(SUM([Profit]), -2, 0)
• LOOKUP – with (ZN(SUM([Profit])) – LOOKUP(ZN(SUM([Profit])), -1)) / ABS(LOOKUP(ZN(SUM([Profit])), -1))
• RANK_DENSE – with RANK_DENSE(SUM([Profit]))
# QUICK TABLE CALCULATIONS
Very briefly there are a few different calculations that are “pre-built” into Tableau. By right-clicking a measure mark pill, a few options will appear:
• Running Total
• Difference
• Percent Difference
• Percent of Total
• Rank
• Percentile
• Moving Average
• YTD Total
• Compound Growth Rate
• Year over Year Growth
• YTD Growth
I hope I was able to help with any brooding questions or clarifications that people may have had or wanted before reading this blog. | HuggingFaceTB/finemath | |
### - Art Gallery -
A laser beam profiler captures, displays, and records the spatial intensity profile of a laser beam at a particular plane transverse to the beam propagation path. Since there are many types of lasers — ultraviolet, visible, infrared, continuous wave, pulsed, high-power, low-power — there is an assortment of instrumentation for measuring laser beam profiles. No single laser beam profiler can handle every power level, pulse duration, repetition rate, wavelength, and beam size.
Overview
Laser beam profiling instruments measure the following quantities:
Beam width: There are over five definitions of beam width.
Beam quality: Quantified by the beam quality parameter, M2.
Beam divergence: This is a measure of the spreading of the beam with distance.
Beam profile: A beam profile is the 2D intensity plot of a beam at a given location along the beam path. A Gaussian or flat-top profile is often desired. The beam profile indicates nuisance high-order spatial modes in a laser cavity as well as hot spots in the beam.
Beam astigmatism: The beam is astigmatic when the vertical and horizontal parts of the beam focus in different locations along the beam path.
Beam wander or jitter: The amount that the centroid or peak value of the beam profile moves with time.
Instruments and techniques were developed to obtain the beam characteristics listed above. These include:
Camera techniques: These include the direct illumination of a camera sensor. The maximum spot size that will fit onto a CCD sensor is on the order of 10 mm. Alternatively, illuminating a flat diffuse surface with the laser and imaging the light onto a CCD with a lens allows profiling of larger-diameter beams. Viewing lasers off diffuse surfaces is excellent for large beam widths but requires a diffuse surface that has uniform reflectivity (<1% variation) over the illuminated surface.
Knife-edge technique: A spinning blade or slit cuts the laser beam before detection by a power meter. The power meter measures the intensity as a function of time. By taking the integrated intensity profiles in a number of cuts, the original beam profile can be reconstructed using algorithms developed for tomography. This usually does not work for pulsed lasers, and does not provide a true 2D beam profile, but it does have excellent resolution, in some cases <1 μm.
Phase-front technique: The beam is passed through a 2D array of tiny lenses in a Shack–Hartmann wavefront sensor. Each lens will redirect its portion of the beam, and from the position of the deflected beamlet, the phase of the original beam can be reconstructed.
Historical techniques: These include the use of photographic plates and burn plates. For example, high-power carbon dioxide lasers were profiled by observing slow burns into acrylate blocks.
As of 2002, commercial knife-edge measurement systems cost $5,000–$12,000 USD and CCD beam profilers cost $4,000–9,000 USD.[1] The cost of CCD beam profilers has come down in recent years, primarily driven by lower silicon CCD sensor costs, and as of 2008 they can be found for less than$1000 USD.
Applications
The applications of laser beam profiling include:
Laser cutting: A laser with an elliptical beam profile has a wider cut along one direction than along the other. The width of the beam influences the edges of the cut. A narrower beam width yields high fluence and ionizes, rather than melts, the machined part. Ionized edges are cleaner and have less knurling than melted edges.
Nonlinear optics: Frequency conversion efficiency in nonlinear optical materials is proportional to the square (sometimes cubed or more) of the input light intensity. Therefore, to get efficient frequency conversion, the input beam waist must be as small as possible. A beam profiler can help minimize the beam waist in the nonlinear crystal.
Alignment: Beam profilers align beams with orders of magnitude better angular accuracy than irises.
Laser monitoring: It is often necessary to monitor the laser output to see whether the beam profile changes after long hours of operation. Maintaining a particular beam shape is critical for adaptive optics, nonlinear optics, and laser-to-fiber delivery. In addition, laser status can be measured by imaging the emitters of a pump diode laser bar and counting the number of emitters that have failed or by placing several beam profilers at various points along a laser amplifier chain.
Laser and laser amplifier development: Thermal relaxation in pulse-pumped amplifiers causes temporal and spatial variations in the gain crystal, effectively distorting the beam profile of the amplified light. A beam profiler placed at the output of the amplifier yields a wealth of information about transient thermal effects in the crystal. By adjusting the pump current to the amplifier and tuning the input power level, the output beam profile can be optimized in real-time.
Far-field measurement: It is important to know the beam profile of a laser for laser radar or free-space optical communications at long distances, the so-called "far-field". The width of the beam in its far-field determines the amount of energy collected by a communications receiver and the amount of energy incident on the ladar's target. Measuring the far-field beam profile directly is often impossible in a laboratory because of the long path length required. A lens, on the other hand, transforms the beam so that the far-field occurs near its focus. A beam profiler placed near the focus of the lens measures the far-field beam profile in significantly less benchtop space.
Education: Beam profilers can be used for student laboratories to verify diffraction theories and test the Fraunhofer or Fresnel diffraction integral approximations. Other student laboratory ideas include using a beam profiler to measure Poisson's spot of an opaque disk and to map out the Airy disk diffraction pattern of a clear disk.
Measurements
Beam width
Main article: Beam diameter
The beam width is the single most important characteristic of a laser beam profile. At least five definitions of beam width are in common use: D4σ, 10/90 or 20/80 knife-edge, 1/e2, FWHM, and D86. The D4σ beam width is the ISO standard definition and the measurement of the M² beam quality parameter requires the measurement of the D4σ widths.[2][3][4] The other definitions provide complementary information to the D4σ and are used in different circumstances. The choice of definition can have a large effect on the beam width number obtained, and it is important to use the correct method for any given application.[5] The D4σ and knife-edge widths are sensitive to background noise on the detector, while the 1/e2 and FWHM widths are not. The fraction of total beam power encompassed by the beam width depends on which definition is used.
Beam quality
Beam quality parameter, M2
Main article: M squared
The M2 parameter is a measure of beam quality; a low M2 value indicates good beam quality and ability to be focused to a tight spot. The value M is equal to the ratio of the beam's angle of divergence to that of a Gaussian beam with the same D4σ waist width. Since the Gaussian beam diverges more slowly than any other beam shape, the M2 parameter is always greater than or equal to one. Other definitions of beam quality have been used in the past, but the one using second moment widths is most commonly accepted.[6]
Beam quality is important in many applications. In fiber-optic communications beams with an M2 close to 1 are required for coupling to single-mode optical fiber. Laser machine shops care a lot about the M2 parameter of their lasers because the beams will focus to an area that is M4 times larger than that of a Gaussian beam with the same wavelength and D4σ waist width before focusing; in other words, the fluence scales as 1/M4. The rule of thumb is that M2 increases as the laser power increases. It is difficult to obtain excellent beam quality and high average power (100 W to kWs) due to thermal lensing in the laser gain medium.
The M2 parameter is determined experimentally as follows:[2]
Measure the D4σ widths at 5 axial positions near the beam waist (the location where the beam is narrowest).
Measure the D4σ widths at 5 axial positions at least one Rayleigh length away from the waist.
Fit the 10 measured data points to $$\sigma ^{2}(z)=\sigma _{0}^{2}+M^{4}\left({\frac {\lambda }{\pi \sigma _{0}}}\right)^{2}(z-z_{0})^{2}$$ ,[7] where$$\sigma ^{2}(z)$$ is the second moment of the distribution in the x or y direction (see section on D4σ beam width), and $$z_{0}$$ is the location of the beam waist with second moment width of 2 σ 0 {\displaystyle 2\sigma _{0}} 2\sigma _{0}. Fitting the 10 data points yields M2, $$z_{0}$$, and $$\sigma _{0}$$. Siegman showed that all beam profiles — Gaussian, flat top, TEMXY, or any shape — must follow the equation above provided that the beam radius uses the D4σ definition of the beam width. Using the 10/90 knife-edge, the D86, or the FWHM widths does not work.
Complete E-field beam profiling
Beam profilers measure the intensity, |E-field|2, of the laser beam profile but do not yield any information about the phase of the E-field. To completely characterize the E-field at a given plane, both the phase and amplitude profiles must be known. The real and imaginary parts of the electric field can be characterized using two CCD beam profilers that sample the beam at two separate propagation planes, with the application of a phase recovery algorithm to the captured data. The benefit of completely characterizing the E-field in one plane is that the E-field profile can be computed for any other plane with diffraction theory.
Power-in-the-bucket or Strehl definition of beam quality
The M2 parameter is not the whole story in specifying beam quality. A low M2 only implies that the second moment of the beam profile expands slowly. Nevertheless, two beams with the same M2 may not have the same fraction of delivered power in a given area. Power-in-the-bucket and Strehl ratio are two attempts to define beam quality as a function of how much power is delivered to a given area. Unfortunately, there is no standard bucket size (D86 width, Gaussian beam width, Airy disk nulls, etc.) or bucket shape (circular, rectangular, etc.) and there is no standard beam to compare for the Strehl ratio. Therefore, these definitions must always be specified before a number is given and it presents much difficulty when trying to compare lasers. There is also no simple conversion between M2, power-in-the-bucket, and Strehl ratio. The Strehl ratio, for example, has been defined as the ratio of the peak focal intensities in the aberrated and ideal point spread functions. In other cases, it has been defined as the ratio between the peak intensity of an image divided by the peak intensity of a diffraction-limited image with the same total flux.[8][9] Since there are many ways power-in-the-bucket and Strehl ratio have been defined in the literature, the recommendation is to stick with the ISO-standard M2 definition for the beam quality parameter and be aware that a Strehl ratio of 0.8, for example, does not mean anything unless the Strehl ratio is accompanied by a definition.
Beam divergence
Main article: Beam divergence
The beam divergence of a laser beam is a measure for how fast the beam expands far from the beam waist. It is usually defined as the derivative of the beam radius with respect to the axial position in the far field, i.e., in a distance from the beam waist which is much larger than the Rayleigh length. This definition yields a divergence half-angle. (Sometimes, full angles are used in the literature; these are twice as large.) For a diffraction-limited Gaussian beam, the beam divergence is λ/(πw0), where λ is the wavelength (in the medium) and w0 the beam radius (radius with 1/e2 intensity) at the beam waist. A large beam divergence for a given beam radius corresponds to poor beam quality. A low beam divergence can be important for applications such as pointing or free-space optical communications. Beams with very small divergence, i.e., with approximately constant beam radius over significant propagation distances, are called collimated beams. For the measurement of beam divergence, one usually measures the beam radius at different positions, using e.g. a beam profiler. It is also possible to derive the beam divergence from the complex amplitude profile of the beam in a single plane: spatial Fourier transforms deliver the distribution of transverse spatial frequencies, which are directly related to propagation angles. See US Laser Corps application note[10] for a tutorial on how to measure the laser beam divergence with a lens and CCD camera.
Beam astigmatism
Astigmatism in a laser beam occurs when the horizontal and vertical cross sections of the beam focus at different locations along the beam path. Astigmatism can be corrected with a pair of cylindrical lenses. The metric for astigmatism is the power of cylindrical lens needed to bring the focuses of the horizontal and vertical cross sections together. Astigmatism is caused by:
Thermal lensing in Nd:YAG slab amplifiers. A slab that is sandwiched between two metal heat sinks will have a temperature gradient between the heat sinks. The thermal gradient causes an index of refraction gradient that is very similar to a cylindrical lens. The cylindrical lensing caused by the amplifier will make the beam astigmatic.
Unmatched cylindrical lenses or error in placement of these optics.
Propagation through a nonlinear uniaxial crystal (common in nonlinear optic crystals). The x- and y-polarized E-fields experience different refractive indices.
Not propagating through the center of a spherical lens or mirror.
Astigmatism can easily be characterized by a CCD beam profiler by observing where the x and y beam waists occur as the profiler is translated along the beam path.
Beam wander or jitter
Every laser beam wanders and jitters — albeit a small amount. The typical kinematic tip-tilt mount drifts by around 100 μrad per day in a laboratory environment (vibration isolation via optical table, constant temperature and pressure, and no sunlight that causes parts to heat). A laser beam incident upon this mirror will be translated by 100 m at a range of 1000 km. This could make the difference between hitting or not hitting a communications satellite from Earth. Hence, there is a lot of interest in characterizing the beam wander (slow time scale) or jitter (fast time scale) of a laser beam. The beam wander and jitter can be measured by tracking the centroid or peak of the beam on a CCD beam profiler. The CCD frame rate is typically 30 frames per second and therefore can capture beam jitter that is slower than 30 Hz — it can't see fast vibrations due to one's voice, 60 Hz fan motor hum, or other sources of fast vibrations. Fortunately, this is usually not a great concern for most laboratory laser systems and the frame rates of CCDs are fast enough to capture the beam wander over the bandwidth that contains the greatest noise power. A typical beam wander measurement involves tracking the centroid of the beam over several minutes. The rms deviation of the centroid data gives a clear picture of the laser beam pointing stability. The integration time of the beam jitter measurement should always accompany the computed rms value. Even though the pixel resolution of a camera may be several micrometres, sub-pixel centroid resolution (possibly tens of nanometer resolution) is attained when the signal to noise ratio is good and the beam fills most of the CCD active area.[11]
Beam wander is caused by:
Slow thermalization of the laser. Laser manufacturers usually have a warm-up specification to allow the laser to drift to an equilibrium after startup.
Tip-tilt and optical mount drift caused by thermal gradients, pressure, and loosening of springs.
Non-rigidly mounted optics
Vibration due to fans, people walking/sneezing/breathing, water pumps, and movement of vehicles outside the laboratory.
Misrepresentation of beam profiler measurements for laser systems
It is to most laser manufacturers' advantage to present specifications in a way that shows their product in the best light, even if this involves misleading the customer. Laser performance specifications can be clarified by asking questions such as:
Is the specification typical or worst-case performance?
What beam width definition was used?
Is the M2 parameter for both vertical and horizontal cross sections, or just for the better cross section?
Was M2 measured using the ISO-standard technique or some other way — e.g. power in the bucket.
Over how long was the data taken to come up with the specified rms beam jitter. (RMS beam jitter gets worse as the measurement interval increases.) What was the laser environment (optical table, etc.)?
What is the warm-up time needed to achieve the specified M2, beam width, divergence, astigmatism, and jitter?
Techniques
Beam profilers generally fall into two classes: the first uses a simple photodetector behind an aperture which is scanned over the beam. The second class uses a camera to image the beam.[12]
Scanning-aperture techniques
The most common scanning aperture techniques are the knife-edge technique and the scanning-slit profiler. The former chops the beam with a knife and measures the transmitted power as the blade cuts through the beam. The measured intensity versus knife position yields a curve that is the integrated beam intensity in one direction. By measuring the intensity curve in several directions, the original beam profile can be reconstructed using algorithms developed for x-ray tomography. The measuring instrument is based on high precision multiple knife edges each deployed on a rotating drum and having a different angle with respect to beam orientation. Scanned beam is then reconstructed using tomographic algorithms and provides 2D or 3D high resolution energy distribution plots. Because of the special scanning technique the system automatically zooms in onto the current beam size enabling high resolution measurements of sub micron beams as well as relative large beams of 10 or more millimeters. To obtain measurement of various wavelength different detectors are used to allow laser beam measurements from deep UV to far IR. Unlike other camera based systems this technology also provides accurate power measurement in real time Scanning-slit profilers use a narrow slit instead of a single knife edge. In this case, the intensity is integrated over the slit width. The resulting measurement is equivalent to the original cross section convolved with the profile of the slit.
These techniques can measure very small spot sizes down to 1 μm, and can be used to directly measure high power beams. They do not offer continuous readout, although repetition rates as high as twenty hertz can be achieved. Also, the profiles give integrated intensities in the x and y directions and not the actual 2D spatial profile (integrating intensities can be hard to interpret for complicated beam profiles). They do not generally work for pulsed laser sources, because of the extra complexity of synchronizing the motion of the aperture and the laser pulses.[13]
CCD camera technique
The CCD camera technique is simple: attenuate and shine a laser onto a CCD and measure the beam profile directly. It is for this reason that the camera technique is the most popular method for laser beam profiling. The most popular cameras used are silicon CCDs that have sensor diameters that range up to 25 mm (1 inch) and pixel sizes down to a few micrometres. These cameras are also sensitive to a broad range of wavelengths, from deep UV, 200 nm, to near infrared, 1100 nm; this range of wavelengths encompass a broad range of laser gain media. The advantages of the CCD camera technique are:
It captures the 2D beam profile in real-time
High dynamic range. Even a webcam's CCD chip has a dynamic range of around 2⁸.[14]
Software typically displays critical beam metrics, such as D4σ width, in real-time
Sensitive CCD detectors can capture the beam profiles of weak lasers
Resolution down to about 4 μm, depending on the pixel size. In a special case a resolution of ±1.1 μm was demonstrated.[14]
CCD cameras with trigger inputs can be used to capture beam profiles of low-duty-cycle pulsed lasers
CCD's have broad wavelength sensitivities from 200 to 1100 nm
The disadvantages of the CCD camera technique are:
Attenuation is required for high-power lasers
CCD sensor size is limited to about 1 inch.
CCDs are prone to blooming when used near the edge of their sensitivity (e.g. close to 1100 nm)[15] [16]
Baseline subtraction for D4σ width measurements
The D4σ width is sensitive to the beam energy or noise in the tail of the pulse because the pixels that are far from the beam centroid contribute to the D4σ width as the distance squared. To reduce the error in the D4σ width estimate, the baseline pixel values are subtracted from the measured signal. The baseline values for the pixels are measured by recording the values of the CCD pixels with no incident light. The finite value is due to dark current, readout noise, and other noise sources. For shot-noise-limited noise sources, baseline subtraction improves the D4σ width estimate as N {\displaystyle {\sqrt {N}}} {\sqrt {N}}, where N {\displaystyle N} N is the number of pixels in the wings. Without baseline subtraction, the D4σ width is overestimated.
Averaging to get better measurements
Averaging consecutive CCD images yields a cleaner profile and removes both CCD imager noise and laser beam intensity fluctuations. The signal-to-noise-ratio (SNR) of a pixel for a beam profile is defined as the mean value of the pixel divided by its root-mean-square (rms) value. The SNR improves as square root of the number of captured frames for shot noise processes – dark current noise, readout noise, and Poissonian detection noise. So, for example, increasing the number of averages by a factor of 100 smooths out the beam profile by a factor of 10.
Attenuation techniques
Since CCD sensors are highly sensitive, attenuation is almost always needed for proper beam profiling. For example, 40 dB (ND 4 or 10−4) of attenuation is typical for a milliwatt HeNe laser. Proper attenuation has the following properties:
It does not result in multiple reflections leaving a ghost image on the CCD sensor
It does not result in interference fringes due to reflections between parallel surfaces or diffraction by defects
It does not distort the wavefront and will be an optical element with sufficient optical flatness (less than one tenth of a wavelength) and homogeneity
It can handle the required optical power
For laser beam profiling with CCD sensors, typically two types of attenuators are used: neutral density filters, and wedges or thick optical flats.
Neutral density filters
Main article: Neutral density filter
Neutral density (ND) filters come in two types: absorptive and reflective.
Absorptive filters are usually made of tinted glass. They are useful for lower-power applications that involve up to about 100 mW average power. Above those power levels, thermal lensing may occur, causing beam size change or deformation, because of the low thermal conductivity of the substrate (usually a glass). Higher power may result in melting or cracking. Absorptive filter attenuation values are usually valid for the visible spectrum (500–800 nm) and are not valid outside of that spectral region. Some filters can be ordered and calibrated for near-infrared wavelengths, up to the long wavelength absorption edge of the substrate (around 2.2 μm for glasses). Typically, one can expect about 5-10% variation of the attenuation across a 2-inch (51 mm) ND filter, unless specified otherwise to the manufacturer. The attenuation values of ND filters are specified logarithmically. A ND 3 filter transmits 10−3 of the incident beam power. Placing the largest attenuator last before the CCD sensor will result in the best rejection of ghost images due to multiple reflections.
Reflective filters are made with a thin metallic coating and hence operate over a larger bandwidth. An ND 3 metallic filter will be good over 200–2000 nm. The attenuation will rapidly increase outside this spectral region because of absorption in the glass substrate. These filters reflect rather than absorb the incident power, and hence can handle higher input average powers. However, they are less well suited to the high peak powers of pulsed lasers. These filters work fine to about 5 W average power (over about 1 cm2 illumination area) before heating causes them to crack. Since these filters reflect light, one must be careful when stacking multiple ND filters, since multiple reflections among the filters will cause a ghost image to interfere with the original beam profile. One way to mitigate this problem is by tilting the ND filter stack. Assuming that the absorption of the metallic ND filter is negligible, the order of the ND filter stack doesn't matter, as it does for the absorptive filters.
Diffractive beam sampler
Diffractive beam samplers are used to monitor high power lasers where optical losses and wavefront distortions of the transmitted beam need to be kept to a minimum. In most applications, most of the incident light must continue forward, "unaffected," in the "zero order diffracted order" while a small amount of the beam is diffracted into a higher diffractive order, providing a "sample" of the beam. By directing the sampled light in the higher order(s) onto a detector, it is possible to monitor, in real time, not only the power levels of a laser beam, but also its profile, and other laser characteristics.
Optical wedges
Optical wedges and reflections from uncoated optical glass surfaces are used to attenuate high power laser beams. About 4% is reflected from the air/glass interface and several wedges can be used to greatly attenuate the beam to levels that can be attenuated with ND filters. The angle of the wedge is typically selected so that the second reflection from the surface does not hit the active area of the CCD, and that no interference fringes are visible. The farther the CCD is from the wedge, the smaller the angle required. Wedges have the disadvantage of both translating and bending the beam direction — paths will no longer lie on convenient rectangular coordinates. Rather than using a wedge, an optical-quality thick glass plate tilted to the beam can also work — actually, this is the same as a wedge with a 0° angle. The thick glass will translate the beam but it will not change the angle of the output beam. The glass must be thick enough so that the beam does not overlap with itself to produce interference fringes, and if possible that the secondary reflection does not illuminate the active area of the CCD. The Fresnel reflection of a beam from a glass plate is different for the s- and p-polarizations (s is parallel to the surface of the glass, and p is perpendicular to s) and changes as a function of angle of incidence – keep this in mind if you expect that the two polarizations have different beam profiles. To prevent distortion of the beam profile, the glass should be of optical quality — surface flatness of λ/10 (λ=633 nm) and scratch-dig of 40-20 or better. A half-wave plate followed by a polarizing beam splitter form a variable attenuator and this combination is often used in optical systems. The variable attenuator made in this fashion is not recommended for attenuation for beam profiling applications because: (1) the beam profile in the two orthogonal polarizations may be different, (2) the polarization beam cube may have a low optical damage threshold value, and (3) the beam can be distorded in cube polarizers at very high attenuation. Inexpensive cube polarizers are formed by cementing two right angle prisms together. The glue does not stand up well to high powers — the intensity should be kept under 500 mW/mm2. Single-element polarizers are recommended for high powers.
Optimal beam size on the CCD detector
There are two competing requirements that determine the optimal beam size on the CCD detector. One requirement is that the entire energy — or as much of it as possible — of the laser beam is incident on the CCD sensor. This would imply that we should focus all the energy in the center of the active region in as small a spot as possible using only a few of the central pixels to ensure that the tails of the beam are captured by the outer pixels. This is one extreme. The second requirement is that we need to adequately sample the beam profile shape. As a rule of thumb, we want at least 10 pixels across the area that encompasses most, say 80%, of the energy in the beam. Therefore, there is no hard and fast rule to select the optimal beam size. As long as the CCD sensor captures over 90% of the beam energy and has at least 10 pixels across it, the beam width measurements will have some accuracy.
Pixel size and number of pixels
The larger the CCD sensor, the larger the size of beam that can be profiled. Sometimes this comes at the cost of larger pixel sizes. Small pixels sizes are desired for observing focused beams. A CCD with many megapixels is not always better than a smaller array since readout times on the computer can be uncomfortably long. Reading out the array in real-time is essential for any tweaking or optimization of the laser profile.
Far-field beam profiler
A far-field beam profiler is nothing more than profiling the beam at the focus of a lens. This plane is sometimes called the Fourier plane and is the profile that one would see if the beam propagated very far away. The beam at the Fourier plane is the Fourier transform of the input field. Care must be taken in setting up a far-field measurement. The focused spot size must be large enough to span across several pixels. The spot size is approximately fλ/D, where f is the focal length of the lens, λ is the wavelength of the light, and D is the diameter of the collimated beam incident upon the lens. For example, a helium-neon laser (633 nm) with 1 mm beam diameter would focus to a 317 μm spot with a 500 mm lens. A laser beam profiler with a 5.6 μm pixel size would adequately sample the spot at 56 locations.
Special applications
The prohibitive costs of CCD laser beam profilers in the past have given way to low-cost beam profilers. Low-cost beam profilers have opened up a number of new applications: replacing irises for super-accurate alignment and simultaneous multiple port monitoring of laser systems.
Iris replacement with microradian alignment accuracy
In the past, alignment of laser beams was done with irises. Two irises uniquely defined a beam path; the farther apart the irises and the smaller the iris holes, the better the path was defined. The smallest aperture that an iris can define is about 0.8 mm. In comparison, the centroid of a laser beam can be determined to sub-micrometre accuracy with a laser beam profiler. The laser beam profiler's effective aperture size is three orders of magnitude smaller than that of an iris. Consequently, the ability to define an optical path is 1000 times better when using beam profilers over irises. Applications that need microradian alignment accuracies include earth-to-space communications, earth-to-space ladar, master oscillator to power oscillator alignment, and multi-pass amplifiers.
Simultaneous multiple port monitoring of laser system
Experimental laser systems benefit from the use of multiple laser beam profilers to characterize the pump beam, the output beam, and the beam shape at intermediate locations in the laser system, for example, after a Kerr-lens modelocker. Changes in the pump laser beam profile indicate the health of the pump laser, which laser modes are excited in the gain crystal, and also determine whether the laser is warmed up by locating the centroid of the beam relative to the breadboard. The output beam profile is often a strong function of pump power due to thermo-optical effects in the gain medium.
References
R. Bolton, "Give your laser beam a checkup," Photonics Spectra, June 2002. Table 1.
ISO 11146-1:2005(E), "Lasers and laser-related equipment — Test methods for laser beam widths, divergence angles and beam propagation ratios — Part 1: Stigmatic and simple astigmatic beams."
ISO 11146-2:2005(E), "Lasers and laser-related equipment — Test methods for laser beam widths, divergence angles and beam propagation ratios — Part 2: General astigmatic beams."
ISO 11146-1:2005(E), "Lasers and laser-related equipment — Test methods for laser beam widths, divergence angles and beam propagation ratios — Part 3: Intrinsic and geometrical laser beam classification, propagation and details of test methods."
Ankron. "Standard definition of beam width" Technical Note, 13 Sep 2008,
A. E. Siegman, "How to (Maybe) Measure Laser Beam Quality," Tutorial presentation at the Optical Society of America Annual Meeting Long Beach, California, October 1997.
A. E. Siegman, "How to (Maybe) Measure Laser Beam Quality," Tutorial presentation at the Optical Society of America Annual Meeting Long Beach, California, October 1997, p.9.
M. Born and E. Wolf, Principles of Optics: Electromagnetic Theory of Propagation, Interference and Diffraction of Light, 6th edition, Cambridge University Press, 1997.
Strehl meter W.M. Keck Observatory.
Measuring laser beam divergence US Laser Corps application note
Ankron. "Technical Note 5: How to measure beam jitter with nanometer accuracy using a CCD sensor with 5.6 μm pixel size".
Aharon. "Laser beam profiling and measurement"
Aharon. "High Power Beam Analysis"
G. Langer et al., "A webcam in Bayer-mode as a light beam profiler for the near infra-red," Optics and Lasers in Engineering, 51 (2013) 571–575.
Aharon. "Wide spectral band beam analysis"
Aharon. "Metrology system for inter-alignment of lasers, telescopes, and mechanical datum"
Laser beam profiling measurement
vte
Lasers
List of laser articles List of laser types List of laser applications Laser acronyms
Laser physics
Laser optics
Beam expander Beam homogenizer B Integral Chirped pulse amplification Gain-switching Gaussian beam Injection seeder Laser beam profiler M squared Mode-locking Multiple-prism grating laser oscillator Multiphoton intrapulse interference phase scan Optical amplifier Optical cavity Optical isolator Output coupler Q-switching Regenerative amplification
Laser spectroscopy
Cavity ring-down spectroscopy Confocal laser scanning microscopy Laser-based angle-resolved photoemission spectroscopy Laser diffraction analysis Laser-induced breakdown spectroscopy Laser-induced fluorescence Noise-immune cavity-enhanced optical heterodyne molecular spectroscopy Raman spectroscopy Second-harmonic imaging microscopy Terahertz time-domain spectroscopy Tunable diode laser absorption spectroscopy Two-photon excitation microscopy Ultrafast laser spectroscopy
Laser ionization
Above-threshold ionization Atmospheric-pressure laser ionization Matrix-assisted laser desorption/ionization Resonance-enhanced multiphoton ionization Soft laser desorption Surface-assisted laser desorption/ionization Surface-enhanced laser desorption/ionization
Laser fabrication
Laser beam welding Laser bonding Laser converting Laser cutting Laser cutting bridge Laser drilling Laser engraving Laser-hybrid welding Laser peening Multiphoton lithography Pulsed laser deposition Selective laser melting Selective laser sintering
Laser medicine
Computed tomography laser mammography Laser capture microdissection Laser hair removal Laser lithotripsy Laser coagulation Laser surgery Laser thermal keratoplasty LASIK Low-level laser therapy Optical coherence tomography Photorefractive keratectomy Photorejuvenation
Laser fusion
Civil applications
3D laser scanner CD DVD Blu-ray Laser lighting display Laser pointer Laser printer Laser tag
Military applications
Advanced Tactical Laser Boeing Laser Avenger Dazzler (weapon) Electrolaser Laser designator Laser guidance Laser-guided bomb Laser guns Laser rangefinder Laser warning receiver Laser weapon LLM01 Multiple Integrated Laser Engagement System Tactical High Energy Laser Tactical light ZEUS-HLONS (HMMWV Laser Ordnance Neutralization System)
Physics Encyclopedia
World
Index | open-web-math/open-web-math | |
## Tuesday, 28 January 2014
### 28 Jan 2014: Homework
(A) Written Homework
Do the following on foolscap papers; to be submitted on 4 February 2014
• Remember to copy questions
• Show all working clearly to demonstrate how you could systematically solve the problem.
Workbook (p13)
Question 19 (a), (b) and (c)
• Show how you would compare the numbers (of the different forms)
• Illustrate these numbers on a number line
• Write your final answer: Organise the given numbers in descending order
Workbook (p14) Addition & Subtraction of Integers
Question 20 (a), (c), (e), (g), (i)
Workbook (p14) Addition, Subtraction, Multiplication and Division of Integers
Question 22 (b), (e), (g), (i)
Complete the Quizzes assigned in AceLearning
### (Arithmetic Operations) What's Wrong: Diagnostic Test 2 Q4c
Here's the presentation of working amongst those who submitted.
Can you identify the error(s) in each of the following?
• Are they purely careless mistakes or conceptual errors?
• Do you know the correct way of working out the solution?
### (Arithmetic Operations) What's Wrong: Diagnostic Test 2 Q4d
Here's the presentation of working amongst those who submitted.
Can you identify the error(s) in each of the following?
• Are they purely careless mistakes or conceptual errors?
• Do you know the correct way of working out the solution?
## Monday, 27 January 2014
### Numbers... How they look like on Number Lines
Take note how the number lines are drawn and how are the numbers, in each case are represented on the number line.
Do the following questions on foolscap as HOMEWORK
### "What's Wrong?" with these Number Line
Question:
Represent the following numbers on a number line: 5, -1/11, π, 0.9
Below are 4 number lines that have "gone wrong".
Do you know what's wrong with these number lines?
Line (1)
Line (2)
Line (3)
Line (4)
Think through before clicking at "Comments" to find out "What's Wrong".
## Sunday, 26 January 2014
Q4
Q7
### Math Study Notes pg 10-11
Mathematics Notebook page 10 and 11(Questions 5 and 8)
Group 1
Members:Shanice,Everi,Min Quan,Sophia,Jin Yu,Joshua
5.It is given that 648 = 23 × 34.
(a) Express 84 as a product of its prime factors.
84=2x2x3x7
=2^2x3x7
The product of 84’s prime factors is 2^2x3x7
(b)Hence find smallest value of x such that 648x is a multiple of 84.
648 = 2^3 x 3^4
84 = 2^2 x 3 x 7
LCM = 2^3 x 3^4 x 7
= 4536
4536/648=7
The smallest value of x is 7
8.A school plans to donate 1400 packs of instant noodles, 350 packets of rice and \$700 in cash to the elderly in the neighbourhood. All items to be distributed are packed in gift bags such that there are equal amount of cash in each gift bag.
• (a) How many gift bags are needed?
• Using the ladder method (to find HCF)
• HCF of 1400, 700, 350 is 350
• 350 gift bags are needed
• (b) Hence, write down the number of packets of rice and the amount of cash in each gift bag
Instant noodles 1400/350 =4
Money \$700/350 =\$2
Rice 350/350 =1
There will be 1 packets of rice and \$2 in each bag.
## Saturday, 25 January 2014
### Group Activity: Which Tribe do I belong to?
Here's the complete solution:
Group 1: 0 Points - Non-submission
Table: 6 points - Check 5th & 8th columns
Venn Diagram: 2 points
Group 2: 8 Points
Table: 4 points - Check 2nd, 3rd, 4th & 7th columns
Venn Diagram: 0 point - missing
Group 2: 4 Points
Table: 7 points - Check 4th column
Venn Diagram: 2 points
Group 2: 9 Points
### HCF & LCM: Concluding Exercise - Discussion
Dear S1-01
As assigned at the end of yesterday's lesson (24 Jan), the groups will post the worked solution of the questions in this blog.
Study Notes (p10-11) Q3 to Q10
You should pick the most complete solution.
Subject title of the post: Study Notes (p10)
To be submitted by 26 January 2014 (Sunday).
### More than 6 AM Quiz: Are you ready for Brilliant Mathematics?
Note: If you are signing up manually, it will prompt to check that you are 13 years old. Alternatively, you may sign in with your Facebook account (if you already have one).
### 6 AM Quiz: Bridge Crossing
The 6 AM Quiz is a platform to engage those who would like to seek deeper exploration and understanding of selected topics. It is therefore not compulsory.
While you tried to solve the puzzle, what mathematical knowledge and skills did you apply to solve the problem?
Done by: Grp 2
## Thursday, 23 January 2014
### 23 Jan 2014: Homework on HCF and LCM
Dear S1-01
Attempt the following for discussion on Friday (24 Jan):
Study Notes:
• Tier A (p10) Q3 and Q4
• Tier B (p10) Q5, Q6 and Q7
• Tier C (p11) Q8, Q9 and Q10
You may do the questions in the notebook if you have not printed the study notes.
If you have already printed the study notes, you may do your work directly on the notes.
## Wednesday, 22 January 2014
### We have a Real World... Application in HCF and LCM
As a group, you will propose 2 scenarios (each) to illustrate the application of HCF and LCM in real world application.
• Each group is assigned 2 slides (e.g. Group 1: Slides 4 & 5. Group 2: Slides 6 & 7. ...)
• Insert your suggested answer as a Comment to the slide where the scenario is.
Lowest Common Multiple
Highest Common Factor
### HCF and LCM: Is this TipSheet Useful?
Some teachers had a discussion about when to use HCF and LCM. Here's one suggestion by the teacher named "Funny".
• Do you agree with what she suggested?
• Are those pointers useful?
Share under "Comments" if you have other insights on this :)
Note:
In overseas context, Highest Common Factor (HCF) is also known as Greatest Common Factor (GCF)
### Venus Transit - what has it got to do with Maths?
How do astronomers & scientists predict when Venus Transit takes place?
Let's read what NASA says :)
### Finding Cube Root... What's wrong with this working?
Question: Find the cube root of 3824
• Highlight area(s) that you think is incomplete
• Identify an 'critical' mistake - because of the way the working is presented
• Point out what's 'wrong' with the final statement
## Tuesday, 21 January 2014
### 21 Jan 2014 Homework
Dear S1-01
Here's the homework for today - to be ready for submission when we meet tomorrow:
• Homework Set (2): Handout - from Study Notes (p16, 17) Q7, Q8, Q9, Q10, Q11
## Monday, 20 January 2014
### Recap: Prime Numbers
Try this interactive application at http://anshula.com/sieve/ to find all prime numbers that are between 1 and 100.
Note that this does not work in Chrome Browser.
Here's another one :) Do you know how this one works?
http://www.hbmeyer.de/eratclass.htm
### Prime Factorisation - Doing it...
Prime factorisation is the process of expressing a composite number as the product of prime factors.
There are 2 methods to do this:
(a) Repeated Division
(b) Factor Tree
Click HERE to view the illustrations
(
### Pondering over Primes 01
A number's prime factorisation is
23 X 32 X 52
Is the number even or odd? Explain your reasoning.
Name four other factors of this number, other than 2, 3 and 5.
### Pondering over Primes 02
Given that 74 088 000 = 2^m X 3^3 X 7^3 X 5^n, find the values of m and n.
Key in the values of m and n under Comments
### Review: Homework Chap 1 Q4
Workbook (p2) Q4: Find the prime factorisation of each of the following numbers, leaving your answer in index notation
Examine the following working:
(1) Has the student answered to the question? If not, what's wrong in the way the answer is presented?
(2) Is the overall working clearly presented to demonstrate the student knows what he/she needs to find? If not, how could the working/ presentation be improved?
* Afternote was added to the end of each case, after discussion with the class on 20 January.
Case 1:
Afternote:
The student had done the Prime Factorisation correctly.
The question asked for prime factorisation and answers be presented in Index Notation.
However, the student did not answer to the question. Instead, he/ she listed all the factors of the number, which is not required.
Case 2:
Afternote:
The student had done the Prime Factorisation correctly.
However, he/ she did not understand what is meant by 'index notation' (see the arrow he/ she drew, pointing at "1").
In addition, though he/ she had written "2^3 x 7 x 11 x 13", his/ her final answer was 8008.
Case 3:
Afternote:
The student had done the Prime Factorisation correctly.
3^2 (5 x 7) is a mixed presentation of "multiplication". He/ She should have written it as 3^2 x 5 x 7.
In addition, from the presentation, he/ she had 315 as the final answer. So, it is an indication of not understanding what "index notation" means.
Case 4:
Afternote:
The student had done the Prime Factorisation correctly.
(2^3 x 7) x (11 x 13) is a mixed presentation of "multiplication" and the "operation"/ working seemed incomplete. He/ She should have written it as 2^3 x 7 x 11 x 13
Case 5:
Afternote:
The student had done the Prime Factorisation correctly; however, he/ she had not carried it out in a systematic manner.
As a good practice, we should always test the division with the smallest possible prime number (divisor) before moving on to the next one so that we need not to do further "random checks" in subsequent divisions - for the ease of checking and accounting for all possible prime factors involved.
Similarly, when writing the product of prime factors in index notation, It would also be a good practice to start with the smallest prime factor (in ascending order) - for the ease of checking.
Discuss it as a group and comment for all the cases
Remember to sign off with your Group number.
## Friday, 17 January 2014
### On Your Own - For Practice, Revision & Acceleration
For your information, the Khan Academy comes with a vast collection of video clips on almost all topics (& sub-topics) in our curriculum. It also comes with quizzes that auto-mark and therefore enables you to check your understanding and mastery of the skills. This will complement what we do in class.
Note that the quizzes are largely Multiple Choice Questions or require you to enter numerical values only. You must also keep in mind the importance of writing the working/ steps in a logical manner. Hence, practices on papers should continue.
This is a useful resource that you can use for practice, revision... and for those of you who are would like to accelerate your learning, you may pace yourself accordingly - e.g. pick a topic that would be taught this year and start to learn on your own.
Note of caution: Always check your textbook on the presentation of the mathematical notation.
For example, at secondary level, 4 x 7 should not be written as 4.7 (by inserting a dot between 4 and 7).
Set-up Guide for Parent:
1. Parent to set up an account
2. Student to invite Parent as Coach
3. Parent to login to monitor child's progress
4. Parent can also assign tasks for child to attempt
## Monday, 13 January 2014
### Email to Parents
Dear S1-01
On Sunday, I sent an email to your parents to provide them an overview of the lessons and learning materials. You were included in the "bcc", too.
Some of you have not submitted the emails to me. As a result, your parents might not have received it:
• [16] Neol Lim
• [17] Siddhartha Jaruhar
• [23] Yap Clement
• [19] Xavien Teo - Mother's email address incorrectly keyed in
Please update the Contact info at http://sst2014-s101maths.blogspot.sg/p/info-gathering.html
On the other hand, you are also kept informed in the "bcc" of the email.
Please forward & share the email to your parents if they mentioned that the email has not arrived.
## (Done by Group 1)
Math Workbook, Page 2.
Reference can be found in the math textbook on page 11.
Question 5 a), c), e) and g)
Find each of the following using prime factorisation.
a) square root of 2 025
using prime factorisation, we get
3 x 3 x 3 x 3 x 5 x 5 =2025
(3x3x5) x (3x3x5) =2025
45 x 45 =2025
Ans: 45
c) square root of 3 969
3969=3x3x3x3x7x7
=(3x3x7)x(3x3x7)
=63 x 63
e) cube root of 5 832
using prime factorisation, we get
2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3 =5832
(2x3x3) x (2x3x3) x (2x3x3) =5832
18 x 18 x 18 =5832
Ans: 18
g) cube root of 17 576
using prime factorisation, we get
2 x 2 x 2 x 13 x 13 x 13 =17576
(2x13) x (2x13) x (2x13) =17576
26 x 26 x 26 =17576
Ans: 26
### For Info: Grouping
Please note that the Homework assigned on last Friday should be completed with the Group Members as at during our Maths lesson.
Leaders of the respective groups - Please discuss and update the seating plan that was shared with you in our 1st lesson. If necessary, re-draw the 'boxes'.
Thanks.
### Homework assigned in Term 1 Week 1 (Friday)
Dear S1-01
The following were assigned to the class on Friday:
(1) 2 Handouts about Space Maths was issued to all
(a) Planetary Conjuctions
(b) Planetary Alignments
These two activities give you a "preview" of what we will be doing in Term 1 Week 3.
Read the questions carefully - Think through the strategies. It is something that you are familiar with and there are several ways to solve the problem.
Some of you have already attempted and shared how you solved the problem.
Now, think if there are more ways to solve the same problem?
(2) Homework in Maths Workbook (p2) You have been assigned to solve Question 5 (a)(c)(e)(g) as a group. Afterwhich, post your answers up in the blog.
In this exercise, you will apply what we discussed on Friday (Prime Factorisation) and apply this to find the square root and cube root of a number in a systematic manner.
[Hint: Check the resources you have]
This will be the first topic we discuss when we meet again on Monday (20 Jan).
Post title: Square Root & Cube Root (by Group...)
## Saturday, 11 January 2014
### My answers for Planetary Conjunctions and Alignments
Done with the bonus worksheet but tho will there be points for right answers? For planetary conjunctions, my workings are messy and I have redrawn them. :) | HuggingFaceTB/finemath | |
## Flash Anzan: an amazing new number game
The Guardian‘s science section reports on a new numbers game, “Flash Anzan.” It’s based on the Japanese abacus, or soroban, which a million Japanese kids learn to use every year. The game requires mental representations of an abacus; the game, according to author Alex Bellos, goes like this:
. . . 15 numbers are flashed consecutively on a giant screen. Each number is between 100 and 999. The challenge is to add them up.
Simple, right? Except the numbers are flashed so fast you can barely read them.
Takeo Sasano, a school clerk in his 30s, broke his own world record: he got the correct answer when the numbers were flashed in 1.70 seconds. In the clip below, taken shortly before, the 15 numbers flash in 1.85 seconds. The speed is so fast I doubt you can even read one of the numbers.
Apparently they flash different sets of numbers at different rates, and the winner is the one who gets the right answer first at the fastest flashing rate.
Amazing. Here’s another clip,
. . . showing Sasano break the world record at 1.80 seconds. Note that the format of the competition is a bit like an arithmetical version of a spelling bee. The remaining contestants are sitting in chairs. The numbers are flashed. The contestants write down their answers and exchange papers for marking. The result is displayed on the screen, and those who got the correct answer stand up.
How does it work? Bellos explains more, but go over to his piece to see a bunch of other interesting stuff and one other amazing video in which the mental addition is done simultaneously with a word game.
Anzan was invented a few years ago by an abacus teacher, Yoji Miyamoto, who wanted to design a maths game that was only solvable by calculation with an imaginary abacus, a skill known as anzan.
When the contestant sees the first number he or she instantly visualizes the number on the imaginary abacus. When they see the second number they instantly add it to the number already visualized, and so on. At the end of the game the contestants cannot remember any of the numbers, or the intermediate sums. They only retain the final answer on the imaginary abacus.
Performing arithmetic using an imaginary abacus is the fastest way to perform mental calculations.
h/t: Michael
1. morkindie
Posted November 2, 2012 at 7:38 am | Permalink
I think my brain just broke.
2. Jim
Posted November 2, 2012 at 7:44 am | Permalink
That is fucking awesome.
• elsburymk14
Posted November 2, 2012 at 9:41 am | Permalink
My thoughts exactly
3. Posted November 2, 2012 at 9:52 am | Permalink
The result of academic freedom. If math were biology, that would be the work of the Devil.
4. Posted November 2, 2012 at 11:13 am | Permalink
Here is software for learning this method: http://figur8.net/baby/2011/04/18/math-secret-teach-your-child-to-do-math-faster-than-a-calculator/
5. Neil Schipper
Posted November 2, 2012 at 11:15 am | Permalink
Startling what trained brains can do. A few years ago I reviewed Bellos’ fun book about math in the big world for my local paper.
Another interesting things was that in Japan, basic arithmetic facts are taught to children with songs (probably not unique in itself), and in order to expand the possibilities for good retention-building rhymes, some of the single-digit numbers have two different Japanese words. (Any Jap. speakers out there? When did this “synonymity” start?)
Also of interest, but maybe not so surprising in hindsight, was that a driving force for the pioneers of probability theory was success at the European gambling houses.
Then there was the story of the math genius who determined conditions under which it was favorable to purchase elebenty gazillion lottery tickets. He had investors, formed a company, and had some big wins.
Fun book.
• Flaffer
Posted November 2, 2012 at 11:26 am | Permalink
On the different names for numbers, they are used to count different objects. The difference is in Chinese and is millenia old, I think. See here for more info on this very confusing aspect of Japanese language (it sometimes can confuse Japanese!): http://en.wikipedia.org/wiki/Japanese_counter_word
• eric
Posted November 2, 2012 at 11:38 am | Permalink
Another interesting things was that in Japan, basic arithmetic facts are taught to children with songs
You mean like this?
Its not de riguer in the English-speaking world, but we do have them.
6. Posted November 2, 2012 at 12:07 pm | Permalink
Put’s me in mind of the Speed Arithmetic method invented by Jacow Trachtenberg while he was in concentration camp. You have to work at it, but I remember multiplication by 11 being fantastically fast and easy.
7. eric
Posted November 2, 2012 at 12:25 pm | Permalink
I can’t even visualize 796 as an abacus in 1.8 seconds, let alone add stuff to it.
8. Old Rasputin
Posted November 2, 2012 at 12:38 pm | Permalink
It’s still easier than “Number Wang”.
• Posted November 2, 2012 at 1:24 pm | Permalink
I am SO relieved I wasn’t the only one thinking of Numberwang. (and Wordwang)
• swences2003
Posted November 2, 2012 at 3:35 pm | Permalink
Yes, it’s time for wangon’on.
Let’s rotate the board!
• Posted November 3, 2012 at 6:39 am | Permalink
Just don’t ask me to think of The Event. Remain indoors!
b&
9. Tom Phoenix
Posted November 2, 2012 at 1:03 pm | Permalink
“Performing arithmetic using an imaginary abacus is the fastest way to perform mental calculations.”
Citation needed.
• Tyle
Posted November 2, 2012 at 1:50 pm | Permalink
Yes, it depends on the calculation. For sums it might well be true. However, nonlinear calculations, such as taking a cube root, are slow on an imaginary abacus and much faster with something like a mental Taylor series.
• dth
Posted November 2, 2012 at 3:38 pm | Permalink
I just started an hour ago and can already do it (slowly)! And you can too!
I had no prior knowledge of the soroban and had to look soroban addition up on the internet. I have not looked up specific instructions on how to do the visualization.
The first thing I did, was to visualize only the pebbles that are out of their default position. I imagine the lower pebbles as connected chunks. That reduces visual clutter a lot.
It’s kind of amazing to be able to comfortably look and think about a second number, while comfortably holding the first one as a mental picture at the same time. This works presumably because different parts of the brain are used.
That means that I now have an accumulator register in my head, which is a basic necessity for any CPU or calculator.
The mechanics of the addition are easy to imagine.
So it is simple to start and I am sure that over time all of the transformations (arabic digit -> soroboan digit, arabic digit -> tens complement in soroban and additon) can be memorized by rote. First for one digit, then chunks of two, etc.
This should relly be standard teaching in all schools. I plan to develop this skill for practical use.
• dth
Posted November 2, 2012 at 4:41 pm | Permalink
There are some reasons for why this might be close to optimal for the decimal system.
10 is only divisible by 2 and 5. The soroban breaks a decimal digit into a combined binary and a quinary digit, while maintaining compatibility with the decimal system at the same time.
There are lots of mnemonic device techniques to memorize numbers by shifting them out of your verbal system. But you can’t add an elephant to the eiffel tower.
In visual design there is the rule of seven. That is based on the notion that you can only perceive 7±2 items in a group before it gets confusing.
So you need a transformation that
.) frees up your verbal/symbolic system
.) maintains compatibility and interoperability with it
.) can be used for calculations
.) has manageable individual digits
.) but not too many digits to handle | HuggingFaceTB/finemath | |
# How To Draw Mohr’s Circle?
## How do you use Mohr’s circle?
Suggested clip 95 seconds
Mohr’s Circle (1/2 – explanation and how to draw) – Mechanics of
Start of suggested clip
End of suggested clip
## How do you draw a Mohr’s strain circle?
Suggested clip 117 seconds
08.3-2 Plane strain Mohr’s circle – EXAMPLE – YouTube
Start of suggested clip
End of suggested clip
## How do you calculate Mohr’s circle?
Suggested clip 68 seconds
mohr’s circle example 1 (2/2 – principal and max shear stresses
Start of suggested clip
End of suggested clip
## What does Mohr’s circle represent?
Mohr’s circle is a graphical representation of a general state of stress at a point. It is a graphical method used for evaluation of principal stresses, maximum shear stress; normal and tangential stresses on any given plane. The normal stresses are plotted along the abscissa.
## Why Mohr circle is a circle?
be represented in graphical form by a plot known as Mohr’s Circle. because it enables you to visualize the relationships between the normal and shear stresses acting on various inclined planes at a point in a stressed body.
## What are the signs of shear stress?
A shear stress acting on a positive face is positive if it acts in the positive direction of an axis, and negative if it acts in the negative direction of an axis. A shear stress acting on a negative face is positive if it acts in the negative direction of an axis and negative if it acts in the positive direction.
## How do you draw Mohr’s Circle PDF step by step?
Suggested clip 107 seconds
We recommend reading: How To Draw A Pony?
How to draw Mohr’s Circle – GATE 2020 examination preparation
Start of suggested clip
End of suggested clip
## What are the principal stresses on a Mohr’s circle?
Mohr’s circle also tells you the principal angles (orientations) of the principal stresses without your having to plug an angle into stress transformation equations. Starting with a stress or strain element in the XY plane, construct a grid with a normal stress on the horizontal axis and a shear stress on the vertical.
## How do you draw stress elements?
Suggested clip 109 seconds
Stress Element – Brain Waves.avi – YouTube
Start of suggested clip
End of suggested clip
## What is maximum shear stress?
The maximum shear stress is the maximum concentrated shear force in a small area. The neutral axis of a cross section is the axis at which the value of the normal stress and strain are equal to zero.
## Why is shear stress maximum at 45?
So simply put, it depends on how you orientate your sensor, at 45 degrees, shear is maximum, tensile stress is minimum. This is how we can find out if the material will fail under due to tensile or shear. Some of the compressive stress forces them to slide against each another, resulting in a shear stress on the glue.
## How do you calculate shear stress?
Suggested clip 112 seconds
Shear Stress in Beams Example – YouTube | HuggingFaceTB/finemath | |
# Compatible pair of actions
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
## Definition
### Definition with left action convention
Suppose $G$ and $H$ are groups. Suppose $\alpha:G \to \operatorname{Aut}(H)$ is a homomorphism of groups, defining a group action of $G$ on $H$. Suppose $\beta:H \to \operatorname{Aut}(G)$ is a homomorphism of groups, defining a group action of $H$ on $G$. For $g \in G$, denote by $c_g: G \to G$ the conjugation map by $g$. See group acts as automorphisms by conjugation. Then, we say that the actions $\alpha,\beta$ form a compatible pair if both these conditions hold:
• $\beta(\alpha(g_1)(h))(g_2) = c_{g_1}(\beta(h)(c_{g_1^{-1}}(g_2)))) \ \forall \ g_1,g_2 \in G, h \in H$
• $\alpha(\beta(h_1)(g))(h_2) = c_{h_1}(\alpha(g)(c_{h_1^{-1}}(h_2))) \ \forall \ h_1,h_2 \in H, g \in G$
The above expressions are easier to write down if we use $\cdot$ to denote all the actions. In that case, the conditions read:
• $(g_1 \cdot h) \cdot g_2 = g_1 \cdot (h \cdot (g_1^{-1} \cdot g_2)) \ \forall \ g_1,g_2 \in G, h \in H$
• $(h_1 \cdot g) \cdot h_2 = h_1 \cdot (g \cdot (h_1^{-1} \cdot h_2)) \ \forall \ h_1,h_2 \in H, g \in G$
Here is an equivalent formulation of these two conditions that is more convenient:
• $c_{g_1}(\beta(h)g_2) = \beta(\alpha(g_1)h)(c_{g_1}(g_2)) \ \forall \ g_1,g_2 \in G, h \in H$ (the $g_2$ here is the $c_{g_1^{-1}}g_2$ of the preceding formulation).
• $c_{h_1}(\alpha(g)h_2) = \alpha(\beta(h_1)g)(c_{h_1}(h_2)) \ \forall \ g \in G, h_1, h_2 \in H$ (the $h_2$ here is the $c_{h_1^{-1}}h_2$ of the preceding formulation)
In the $\cdot$ notation, these become:
• $g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H$ (the $g_2$ here is the $g_1^{-1} \cdot g_2$ of the preceding formulation).
• $h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H$ (the $h_2$ here is the $h_1^{-1} \cdot h_2$ of the preceding formulation)
### Definition with right action convention
We can give a corresponding definition using the right action convention, but the literature uses the left action convention, so this definition is intended purely as an illustrative exercise. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
## Symmetry in the definition
Given groups $G$ and $H$ with actions $\alpha:G \to \operatorname{Aut}(H)$ and $\beta:H \to \operatorname{Aut}(G)$, $\alpha$ is compatible with $\beta$ if and only if $\beta$ is compatible with $\alpha$. In other words, the definition of compatibility is symmetric under interchanging the roles of the two groups.
## Particular cases
• Trivial pair of actions is compatible: If both the actions are trivial, i.e., both the homomorphisms $\alpha,\beta$ are trivial maps, then they form a compatible pair.
• Compatible with trivial action iff image centralizes inner automorphisms: Suppose the homomorphism $\alpha$ is trivial. In that case, the homomorphism $\beta$ is compatible with $\alpha$ if and only if the image of $\beta$ in $\operatorname{Aut}(G)$ is in the centralizer $C_{\operatorname{Aut}(G)}(\operatorname{Inn}(G))$.
• Conjugation actions between subgroups that normalize each other are compatible: If $G,H$ are both subgroups of some group $Q$ that normalize each other (i.e., each is contained in the normalizer of the other), and $\alpha,\beta$ are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in $Q$ and checking the conditions simply amounts to checking two words to be equal. | HuggingFaceTB/finemath | |
# Question: How many grams of potassium chloride can be dissolved in 200g of water at 80 c?
## How many grams of KCl can be dissolved in 100g of water at 60 C?
Substance Solubility in 100g water
KCl at 75°C 50g
NH4Cl at 65°C 60g
HCl at 10°C 78g
NaNO3at 70°C 135g
## How many grams of potassium chloride can be dissolved in 100g of water?
The problem provides you with the solubility of potassium chloride, KCl, in water at 20∘C, which is said to be equal to 34 g / 100 g H2O. This means that at 20∘C, a saturated solution of potassium chloride will contain 34 g of dissolved salt for every 100 g of water.
## When 20 grams of potassium chlorate KClO3 is dissolved in 100 grams of water at 80 ºC the solution can be correctly described as?
Do I Know about Solubility Graphs?
A B
When 20 grams of potassium chlorate, KClO3, is dissolved in 100 grams of water at 80 ºC, the solution can be correctly described as:, unsaturated
At approximately what temperature does the solubility of sodium chloride, NaCl, match the solubility of potassium dichromate, K2Cr2O7?, 60 ºC
## How many grams of potassium nitrate can dissolve in 100g of water at 50 C?
(d) Initially there is 100 g of potassium nitrate in solution, at 50 °C the solubility is 84 g/100g (original data or graph), so 16 g will have crystallised out.
## Which compound is least soluble at 60 C?
Seeing the above records we can easily say that the solubility of KClO3 is the lowest at 60 degree centigrade.
## How many grams of NaNO3 will dissolve in 100 grams of water at 30 degrees C?
Explanation: This site lists the solubility of NaNO3 in water at 30C as 94.9⋅g per 100⋅mL.
We recommend reading: FAQ: How soon after exposure can you test for chlamydia?
## What is the mass of potassium chloride that will dissolve in 250g of water at 70 degrees Celsius?
2 Answers By Expert Tutors. Solubility of KCl in 70C water is about 47 g/100g water. Thus in 250 grams of water, you should be able to dissolve about 117.5 grams KCl. the units gm H2O and Kg H2O cancel, and you are left with gm KCl.
## What is the solubility of potassium chloride at 10 C?
At 10 ∘C, the solubility of potassium chloride is measured to be 31.2⋅g per 100⋅g of water.
## Which substance is most soluble at 20 C?
Do I Know about Solubility Graphs?
A B
At what temperature can 100 mL of water first dissolve 50 grams of potassium chloride?, about 80 ºC
Which compound is least soluble at 50 ºC?, NaCl (sodium chloride)
Which compound is most soluble at 20 ºC?, NaCl (sodium chloride)
## How do you know which salt is more soluble?
Solubility Rules
1. Salts containing Group I elements (Li+, Na+, K+, Cs+, Rb+) are soluble.
2. Salts containing nitrate ion (NO3) are generally soluble.
3. Salts containing Cl , Br , or I are generally soluble.
4. Most silver salts are insoluble.
5. Most sulfate salts are soluble.
6. Most hydroxide salts are only slightly soluble.
## Is sodium or potassium nitrate more soluble?
sodium chloride and lead(II) nitrate are more soluble than potassium nitrate at low temperatures. the solubility of lead(II) nitrate and potassium nitrate is the same at about 50°C (approximately 80 g per 100 g water, where the solubility curves of these substances cross)
We recommend reading: Quick Answer: How can i start my own website?
## How many grams of KNO3 will dissolve in 100g of water at 70 C?
All 80 g KNO3 will dissolve in 100 g H₂O to form a saturated solution at so°C. If the solution is heated to 70°c an additional sog KNO3 must be dissolved to form a saturated solution.
## What is the solubility of KNO3 at 50 C?
The solubility of KNO3 at 50 degrees C is 80 g/100 g water.
## Which is most soluble sugar salt or fertilizer?
sugar is so soluble in water that there can be much more sugar than water in the mixture. fertilizers are a kind of salt, so I am not sure you mean salt to be sodium chloride? this link will describe solubility of a large number of ionic compounds.. | HuggingFaceTB/finemath | |
Is $\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi ixj \left/\phantom\vert\!{N} \right.}|j\rangle |j\rangle$ a valid entangled quantum state?
$$\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi i x j \left/ {\phantom\vert\!\!} N \right.}\,\left|j\right\rangle_h\left|j\right\rangle_t$$
For a valid state we should have sum of probabilities = 1. However, when I compute the sum of the squares of the amplitudes of this state, I get zero. Am I making a mistake?
• You have to take the sum of squares of the moduli of the coefficients. In your example every coefficient has modulus $1/\sqrt{N}$. Squaring and summing gives 1. – smapers Sep 3 at 11:02
For $$\left|\psi\right> = \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi i\frac{x_1}{N}j}\left|j\right>_h\left|j\right>_t$$ to be normalized. It must satisfy $$\left<\psi|\psi\right> = 1$$. Let's check, $$\left<\psi|\psi\right> = \frac{1}{N}\sum_{j=0}^{N-1}e^{-2\pi i\frac{x_1}{N}j}\left_h\left|j\right>_t$$. Assuming the $$\left|j\right>_i$$ states are orthonormal, then $$\left<\psi|\psi\right> = \frac{1}{N}N = 1$$
Another approach is $$\exp(ix) = \cos x + i\sin x$$, so $$\left|\cos x + i\sin x\right| = \sqrt{\left(\cos x + i\sin x\right)\left(\cos x - i\sin x\right)} = \sqrt{\cos^2x + \sin^2x} = 1$$. so each term in $$|ψ⟩$$ has coefficient amplitude $$1/√N$$, so sum of square of all the terms $$N\times(1/√N)^2 =1$$. | HuggingFaceTB/finemath | |
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
# Integration using completing the square and the derivative of arctan(x)
AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.3 (EK)
## Video transcript
- [Instructor] All right, let's see if we can find the indefinite integral of one over five x squared minus 30x plus 65 dx. Pause this video and see if you can figure it out. All right, so this is going to be an interesting one. And it'll be a little bit hairy, but we're gonna work through it together. So, immediately you might try multiple integration techniques and be hitting some walls. And what we're going to do here is actually try to complete the square in this denominator right over here. And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan. And if that's a big hint to you, once again pause the video and try to move forward. All right, now let's do this together. So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one. And so I could just factor a five out of the denominator. If I did that, then this integral will become 1/5 times the integral of one over, so I've factored a five out of the denominator so it is x squared minus six x plus 13 dx. And then as I mentioned I'm gonna complete the square down here. So let me rewrite it, so this is equal to 1/5 times the integral of one over, and so x squared minus six x is clearly not a perfect square the way it's written. Let me write this plus 13 out here. Now what could I add and then I'm gonna to have to subtract if I don't want to change the value of the denominator. In order to make, In order to make this part right over here a perfect square. Well we've done this before, you take half of your coefficient here, which is negative three and you square that. So you want to add a nine here. But if you add a nine then you have to subtract a nine as well. And so this part is going to be x minus three squared and then this part right over here is going to be equal to a positive four and we of course don't want to forget our dx out here. And so let me write it in this form. So this is going to be equal to 1/5 times the integral of one over, get myself some space, x minus three squared plus four, which we could also write as plus two squared. Actually let me do it that way, plus two squared dx. Now many of y'all might already say hey this looks a lot like arc tangent, but I'm gonna to try to simplify it even more so it becomes very clear that it looks like arc tangent is going to be involved. I'm actually gonna do some u substitution in order to do it. So the first thing I'm going to do is let's factor a four out of the denominator here. So if we do that, then this is going to become 1/5 times 1/4 which is going to be 1/20 times the integral of one over x minus three squared over two squared. And this is going to be a plus one, and of course we have our dx. And then we could write this as, and I'm trying to just do every step here. A lot of these you might have been able to do in your head. One over, and I'll just write this as, x minus three over two squared plus one. Plus one and then dx. And now the u substitution is pretty clear. I'm just going to make the substitution that u is equal to x minus three over two. Or we could even say that's u is equal to one half x minus three halves, that's just x minus three over two. And du is going to be equal to one half dx. And so what I can do here is, actually let me start to re-engineer this integral a little bit so that we see a one half here. So if I make this a one half and then I multiply the outside by two. So I divide by two multiply by two. It's one way to think about it. This becomes 1/10 and so doing my u substitution I get 1/10, that's the 1/10 there times the integral of. Well I have one half dx right over here which is the same thing as du. So I can put the du either in the numerator or I could put it out here. And then I have one over, this is u squared, u squared plus one. Now you might immediately recognize, what's the derivative of arctan of u? Well that would be one over u squared plus one. So this is going to be equal to 1/10 times the arc tangent of u, and of course we can't forget our big constant C because we're taking an indefinite integral. And now we just want to do the reverse substitution. We know that u is equal to this business right over here. So we deserve a little bit of a drum roll. This is going to be equal to 1/10 times the arc tangent of u. Well u is just x minus three over two, which could also be written like this. So arctan of x minus three over two and then plus C. And we are done.
AP® is a registered trademark of the College Board, which has not reviewed this resource. | HuggingFaceTB/finemath | |
```Question 213298
Find the maximum possible area of a rectangle with a perimeter of 100 feet.
I've drawn a picture, labeled both the length and width on the rectangle, and written down the formulas for perimeter and area.
------------------------------
Perimeter = 2(L + W)
100 = 2(L+W)
L+W = 50
---
Let width be W
Then length = 50-W
--------
Area = length * width
A = (50-W)W
A = 50W-W^2
--
This is a quadratic with a = -1, b = 50:
Maximum area occurs when W = -b/2a = -50/-2 = 25
Since W = 25 and L+W=50, L = 25
----------------
Maximum area is 25*25 = 625 sq. units
=========================================
Cheers,
Stan H.``` | HuggingFaceTB/finemath | |
## Key Concepts
• If $a,b$, and $c$ are numbers where $c\ne 0$ , then ${\Large\frac{a}{c}}+{\Large\frac{b}{c}}={\Large\frac{a+c}{c}}$ .
• To add fractions, add the numerators and place the sum over the common denominator.
• Fraction Subtraction
• If $a,b$, and $c$ are numbers where $c\ne 0$ , then ${\Large\frac{a}{c}}-{\Large\frac{b}{c}}={\Large\frac{a-b}{c}}$ .
• To subtract fractions, subtract the numerators and place the difference over the common denominator.
• Find the prime factorization of a composite number using the tree method.
1. Find any factor pair of the given number, and use these numbers to create two branches.
2. If a factor is prime, that branch is complete. Circle the prime.
3. If a factor is not prime, write it as the product of a factor pair and continue the process.
4. Write the composite number as the product of all the circled primes.
• Find the prime factorization of a composite number using the ladder method.
1. Divide the number by the smallest prime.
2. Continue dividing by that prime until it no longer divides evenly.
3. Divide by the next prime until it no longer divides evenly.
4. Continue until the quotient is a prime.
5. Write the composite number as the product of all the primes on the sides and top of the ladder.
• Find the LCM using the prime factors method.
1. Find the prime factorization of each number.
2. Write each number as a product of primes, matching primes vertically when possible.
3. Bring down the primes in each column.
4. Multiply the factors to get the LCM.
• Find the LCM using the prime factors method.
1. Find the prime factorization of each number.
2. Write each number as a product of primes, matching primes vertically when possible.
3. Bring down the primes in each column.
4. Multiply the factors to get the LCM.
• Find the least common denominator (LCD) of two fractions.
1. Factor each denominator into its primes.
2. List the primes, matching primes in columns when possible.
3. Bring down the columns.
4. Multiply the factors. The product is the LCM of the denominators.
5. The LCM of the denominators is the LCD of the fractions.
• Equivalent Fractions Property
• If $a,b$ , and $c$ are whole numbers where $b\ne 0$ , $c\ne 0$ then$\Large\frac{a}{b}=\Large\frac{a\cdot c}{b\cdot c}$
and $\Large\frac{a\cdot c}{b\cdot c}=\Large\frac{a}{b}$
• Convert two fractions to equivalent fractions with their LCD as the common denominator.
1. Find the LCD.
2. For each fraction, determine the number needed to multiply the denominator to get the LCD.
3. Use the Equivalent Fractions Property to multiply the numerator and denominator by the number from Step 2.
4. Simplify the numerator and denominator.
• Add or subtract fractions with different denominators.
1. Find the LCD.
2. Convert each fraction to an equivalent form with the LCD as the denominator.
3. Add or subtract the fractions.
4. Write the result in simplified form.
## Glossary
composite number
A composite number is a counting number that is not prime
divisibility
If a number $m$ is a multiple of $n$ , then we say that $m$ is divisible by $n$
least common denominator (LCD)
The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators
multiple of a number
A number is a multiple of $n$ if it is the product of a counting number and $n$
ratio
A ratio compares two numbers or two quantities that are measured with the same unit. The ratio of $a$ to $b$ is written $a$ to $b$ , $\Large\frac{a}{b}$ , or $a:b$
prime number
A prime number is a counting number greater than 1 whose only factors are 1 and itself | HuggingFaceTB/finemath | |
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 25 May 2017, 22:07
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# In tests for pironoma, a serious disease, a false positive
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
### Hide Tags
Senior Manager
Joined: 27 May 2009
Posts: 268
Followers: 3
Kudos [?]: 421 [0], given: 18
In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
22 Aug 2009, 07:35
4
This post was
BOOKMARKED
00:00
Difficulty:
75% (hard)
Question Stats:
53% (02:26) correct 47% (01:54) wrong based on 391 sessions
### HideShow timer Statistics
87. In tests for pironoma, a serious disease, a false positive result indicates that people have pironoma when, in fact, they do not; a false negative result indicates that people do not have pironoma when, in fact, they do. To detect pironoma most accurately, physicians should use the laboratory test that has the lowest proportion of false positive results.
Which of the following, if true, gives the most support to the recommendation above?
(A) The accepted treatment for pironoma does not have damaging side effects.
(B) The laboratory test that has the lowest proportion of false positive results causes the same minor side
effects as do the other laboratory tests used to detect pironoma.
(C) In treating pironoma patients, it is essential to begin treatment as early as possible, since even a week of
delay can result in loss of life.
(D) The proportion of inconclusive test results is equal for all laboratory tests used to detect pironoma.
(E) All laboratory tests to detect pironoma have the same proportion of false negative results.
[Reveal] Spoiler:
I am not satisfied with the explanation.. OA is E.. I did it first time perhaps and now i am not able to corelate howis it corelating with the argument..pls explain
[Reveal] Spoiler: OA
If you have any questions
you can ask an expert
New!
Senior Manager
Joined: 29 Jul 2009
Posts: 312
Followers: 5
Kudos [?]: 338 [14] , given: 9
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
22 Aug 2009, 08:23
14
KUDOS
2
This post was
BOOKMARKED
Premise 1: a false positive result indicates that people have pironoma when, in fact, they do not;
Premise 2: a false negative result indicates that people do not have pironoma when, in fact, they do.
Conclusion: To detect pironoma most accurately, physicians should use the laboratory test that has the lowest proportion of
false positive results.
The conclusion that the author draws is unexpected because we should look for laboratories that have the lowest proportion of false negative results. However, the questions is asking for an answer choice that strengthens the argument.
Option E says that all the laboratories have the same proportion of false negative results. If all have the same proportion, then the one with the lowest proportion of false positive results would be the best.
Consider the following example
Height is the most important feature for a basketball player, whereas speed is the second one. However, Richard is the best player in his team because he is the fastest.
What would make this conclusion be properly drawn?
All the players in Richard's team have the same height.
Last edited by mikeCoolBoy on 23 Aug 2009, 01:14, edited 1 time in total.
Manager
Status: Berkeley Haas 2013
Joined: 23 Jul 2009
Posts: 191
Followers: 1
Kudos [?]: 39 [0], given: 16
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
22 Aug 2009, 08:59
mikeCoolBoy wrote:
Premise 1: a false positive result indicates that people have pironoma when, in fact, they do not;
Premise 2: a false negative result indicates that people do not have pironoma when, in fact, they do.
Conclusion: To detect pironoma most accurately, physicians should use the laboratory test that has the lowest proportion of
false positive results.
The conclusion that the author draws is unexpected because we should look for laboratories that have the lowest proportion of false negative results. However, the questions is asking for an answer choice that strengthens the argument.
Option E says that all the laboratories have the same proportion of false negative. If all have the same proportion, then the one with the lowest proportion of false positive results would be the best.
Consider the following example
Height is the most important feature for a basketball player, whereas speed is the second one. However, Richard is the best player in his team because he is the fastest.
What would make this conclusion be properly drawn?
All the players in Richard's team have the same height.
Perfect. +1 for ya
Director
Joined: 01 Apr 2008
Posts: 882
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 30
Kudos [?]: 704 [0], given: 18
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
22 Aug 2009, 10:48
2
This post was
BOOKMARKED
In other words,
X and Y are two scenarios/situations that affect a conclusion.
When can we say that the conclusion depends ONLY on how X varies?
=>Only when Y is constant or when it varies proportionately we can say that the conclusion entirely depends on how X varies.
Senior Manager
Joined: 27 May 2009
Posts: 268
Followers: 3
Kudos [?]: 421 [0], given: 18
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
23 Aug 2009, 00:18
mikeCoolBoy wrote:
Option E says that all the laboratories have the same proportion of false negative. If all have the same proportion, then the one with the lowest proportion of false positive results would be the best.
All the players in Richard's team have the same height.
Economist wrote:
In other words,
X and Y are two scenarios/situations that affect a conclusion.
When can we say that the conclusion depends ONLY on how X varies?
=>Only when Y is constant or when it varies proportionately we can say that the conclusion entirely depends on how X varies.
But if the lowest proportion of false positive results are to be selected how can we say strengthen it by saying all have the same proportion. IF it is the case we can not select th lowest coz all are the same. ..
Pls let me know what i am missing on?
Senior Manager
Joined: 29 Jul 2009
Posts: 312
Followers: 5
Kudos [?]: 338 [1] , given: 9
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
23 Aug 2009, 01:13
1
KUDOS
rohansherry wrote:
mikeCoolBoy wrote:
Option E says that all the laboratories have the same proportion of false negative. If all have the same proportion, then the one with the lowest proportion of false positive results would be the best.
All the players in Richard's team have the same height.
Economist wrote:
In other words,
X and Y are two scenarios/situations that affect a conclusion.
When can we say that the conclusion depends ONLY on how X varies?
=>Only when Y is constant or when it varies proportionately we can say that the conclusion entirely depends on how X varies.
But if the lowest proportion of false positive results are to be selected how can we say strengthen it by saying all have the same proportion. IF it is the case we can not select th lowest coz all are the same. ..
Pls let me know what i am missing on?
watch out
all laboratories have the same proportion of FALSE NEGATIVE RESULTS NOT FALSE POSITIVE RESULTS
Senior Manager
Joined: 27 May 2009
Posts: 268
Followers: 3
Kudos [?]: 421 [0], given: 18
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
23 Aug 2009, 01:16
mikeCoolBoy wrote:
rohansherry wrote:
mikeCoolBoy wrote:
Option E says that all the laboratories have the same proportion of false negative. If all have the same proportion, then the one with the lowest proportion of false positive results would be the best.
All the players in Richard's team have the same height.
Economist wrote:
In other words,
X and Y are two scenarios/situations that affect a conclusion.
When can we say that the conclusion depends ONLY on how X varies?
=>Only when Y is constant or when it varies proportionately we can say that the conclusion entirely depends on how X varies.
But if the lowest proportion of false positive results are to be selected how can we say strengthen it by saying all have the same proportion. IF it is the case we can not select th lowest coz all are the same. ..
Pls let me know what i am missing on?
watch out
all laboratories have the same proportion of FALSE NEGATIVE RESULTS NOT FALSE POSITIVE RESULTS
Thanks mike Kudos to you
Senior Manager
Joined: 29 Jul 2009
Posts: 312
Followers: 5
Kudos [?]: 338 [0], given: 9
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
23 Aug 2009, 01:39
you're welcome.
Manager
Joined: 14 Dec 2008
Posts: 166
Followers: 1
Kudos [?]: 25 [0], given: 39
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
06 Oct 2009, 00:30
well explained mikeCoolBoy, even I was very confused with the options.
Manager
Joined: 30 Sep 2009
Posts: 59
Followers: 1
Kudos [?]: 26 [0], given: 6
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
06 Oct 2009, 12:19
I got it this way....
Since none of the labs have a chance of saying the negative results perfectly, the only one who would atleast test the positive results closest should be chosen, sorta elimination for the labs given all other criterion being equal
Intern
Joined: 03 Oct 2009
Posts: 38
Schools: NY Stern; Columbia; Cornell; Waterloo; U of T
Followers: 0
Kudos [?]: 7 [0], given: 8
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
06 Oct 2009, 15:18
Uupppsss...
Was going over my antena. Thanks to everyone of the contributors!
Intern
Joined: 18 Sep 2009
Posts: 9
Followers: 1
Kudos [?]: 2 [0], given: 1
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
22 Oct 2009, 23:02
Please tell me what is the correct answer.
Senior Manager
Joined: 21 Dec 2009
Posts: 265
Location: India
Followers: 10
Kudos [?]: 252 [1] , given: 25
In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
20 Jan 2010, 08:25
1
KUDOS
9
This post was
BOOKMARKED
In tests for pironoma, a serious disease, a false positive result indicates that people have pironoma when, in fact, they do not; a false negative result indicates that people do not have pironoma when, in fact, they do. To detect pironoma most accurately, physicians should use the laboratory test that has the lowest proportion of false positive results.
Which of the following, if true, gives the most support to the recommendation above?
(A) The accepted treatment for pironoma does not have damaging side effects.
(B) The laboratory test that has the lowest proportion of false positive results causes the same minor side effects as do the other laboratory tests used to detect pironoma.
(C) In treating pironoma patients, it is essential to begin treatment as early as possible, since even a week of delay can result in loss of life.
(D) The proportion of inconclusive test results is equal for all laboratory tests used to detect pironoma.
(E) All laboratory tests to detect pironoma have the same proportion of false negative results.
try this but give your logic to support your answer
_________________
Cheers,
SD
Last edited by mikemcgarry on 09 Jan 2017, 11:21, edited 2 times in total.
Manager
Joined: 20 Mar 2007
Posts: 59
Followers: 1
Kudos [?]: 16 [1] , given: 0
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
20 Jan 2010, 09:06
1
KUDOS
1
This post was
BOOKMARKED
SudiptoGmat wrote:
In tests for pironoma, a serious disease, a false positive result indicates that people have pironoma when, in fact, they do not; a false negative result indicates that people do not have pironoma when, in fact, they do. To detect pironoma most accurately, physicians should use the laboratory test that has the lowest proportion of false positive results.
Which of the following, if true, gives the most support to the recommendation above?
(A) The accepted treatment for pironoma does not have damaging side effects.
(B) The laboratory test that has the lowest proportion of false positive results causes the same minor side effects as do the other laboratory tests used to detect pironoma.
(C) In treating pironoma patients, it is essential to begin treatment as early as possible, since even a week of delay can result in loss of life.
(D) The proportion of inconclusive test results is equal for all laboratory tests used to detect pironoma.
(E) All laboratory tests to detect pironoma have the same proportion of false negative results.
try this but give your logic to support your answer
Choosing Option E here.
Reasoning : am assuming each lab tests consists of false postive results,accurate results and false negative results. Providing treatment to a person not afflicted by the disease ( a false positive) may have serious side effects ,while in the other 2 cases- false negative and accurate,the treatment works in the person's favour.
So ,if all lab tests have the same proportion of false -ve results,it make sense to use the laboratory test that has the lowest proportion of false positive results for minimising the risk of any adverse side effects ,while maximising treatment benefit to the rest of the population.
Senior Manager
Joined: 21 Dec 2009
Posts: 265
Location: India
Followers: 10
Kudos [?]: 252 [1] , given: 25
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
20 Jan 2010, 11:31
1
KUDOS
OA is
[Reveal] Spoiler:
E
.
I wanted to know what others think about the question. Thanks for explanation. I am in sync with you.
_________________
Cheers,
SD
Senior Manager
Joined: 23 May 2010
Posts: 430
Followers: 5
Kudos [?]: 117 [0], given: 112
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
15 Aug 2010, 19:22
marking E by POE
Senior Manager
Joined: 14 Jun 2010
Posts: 317
Followers: 2
Kudos [?]: 24 [0], given: 7
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
15 Aug 2010, 23:25
+1 for E. E says False negative results are the same in all the labs and hence it is the false positive results that would determine which lab to use!!
Manager
Joined: 11 May 2010
Posts: 218
Followers: 4
Kudos [?]: 128 [0], given: 11
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
16 Aug 2010, 00:27
E it is
Manager
Joined: 29 Dec 2009
Posts: 118
Location: india
Followers: 2
Kudos [?]: 21 [0], given: 10
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
17 Aug 2010, 23:52
E
Manager
Joined: 28 Feb 2010
Posts: 170
WE 1: 3 (Mining Operations)
Followers: 7
Kudos [?]: 34 [0], given: 33
Re: In tests for pironoma, a serious disease, a false positive [#permalink]
### Show Tags
18 Aug 2010, 00:14
Ya, I went for C via pre-phrasing , and coud'nt concentrate on E.
E should be the correct one
_________________
Regards,
Invincible...
"The way to succeed is to double your error rate."
"Most people who succeed in the face of seemingly impossible conditions are people who simply don't know how to quit."
Re: In tests for pironoma, a serious disease, a false positive [#permalink] 18 Aug 2010, 00:14
Go to page 1 2 Next [ 34 posts ]
Similar topics Replies Last post
Similar
Topics:
23 In tests for pironoma, a serious disease, a false positive 6 24 Mar 2016, 05:42
22 In tests for Feline Leukemia Virus (FeLV), a serious disease 11 16 Oct 2015, 23:57
5 In tests for pironoma, a serious disease, a false positive 9 09 Mar 2015, 05:11
2 In tests for pironoma, a serious disease, a false positive 5 22 Nov 2014, 06:36
In tests for pironoma, a serious disease, a false positive 0 01 Sep 2016, 14:12
Display posts from previous: Sort by
# In tests for pironoma, a serious disease, a false positive
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | HuggingFaceTB/finemath | |
# Limit of (tan(x))^(tan(2x)) as x approaches pi/4
## Homework Statement
$\frac{Lim}{x-> \pi/4}$ tan(x)$^{tan(2x)}$
$\frac{Lim}{x-> \pi/4}tan(2x)$ does not exist.
However, Wolfram Alpha and my TI-89 say that $\frac{Lim}{x-> \pi/4}$ tan(x)$^{tan(2x)}$does exist, and that it's $\frac{1}{e}$
I submitted this answer (it's web based homework and calculators are allowed) and found it was correct. I still don't understand how though.
## Homework Equations
tan(2x) = $\frac{2tan(x)}{1-tan^{2}(x)}$
## The Attempt at a Solution
I attempted to split it in to
$\frac{sin(x)^{tan(2x)}}{cos(x)^{tan(2x)}}$, and then use L'Hospital's rule, but I can't seem to get the tan(2x) to go away. Both appear to be indeterminate, but neither one is 0 at the same time, so it doesn't appear that I should be using L'Hopital's rule in this case. However, I can't see any other way to proceed.
Related Calculus and Beyond Homework Help News on Phys.org
Thanks.
So people can't do these tasks without calculators anymore? :-)
Hey, at the very least my professor doesn't allow them on tests.
jbunniii
Homework Helper
Gold Member
Have you tried using
$$\tan(2x) = \frac{2 \tan(x)}{1 - \tan^2(x)}$$
Have you tried using
$$\tan(2x) = \frac{2 \tan(x)}{1 - \tan^2(x)}$$
There is the good response here, and I particularly agree with this because this generally makes the simplification a bit simple.
Don't forget that you have the function as the exponent of the another function! Here is the hint:
Let y = lim x→π/4 (tan(x))^(tan(2x)). Then, perform logarithms, and we have...
ln(y) = lim x→π/4 tan(2x) * ln(tan(x))
Mod note: Removed intermediate steps students should work out on their own.
Also don't forget to set both sides by e. You should get the results. Let me know if this helps.
Key: ♪ Practice, practice! You will get better with limits like this! ♫
Last edited by a moderator:
SammyS
Staff Emeritus
Homework Helper
Gold Member
## Homework Statement
$\frac{Lim}{x-> \pi/4}$ tan(x)$^{tan(2x)}$
$\frac{Lim}{x-> \pi/4}tan(2x)$ does not exist.
However, Wolfram Alpha and my TI-89 say that $\frac{Lim}{x-> \pi/4}$ tan(x)$^{tan(2x)}$does exist, and that it's $\frac{1}{e}$
I submitted this answer (it's web based homework and calculators are allowed) and found it was correct. I still don't understand how though.
## Homework Equations
tan(2x) = $\frac{2tan(x)}{1-tan^{2}(x)}$
## The Attempt at a Solution
I attempted to split it in to
$\frac{sin(x)^{tan(2x)}}{cos(x)^{tan(2x)}}$, and then use L'Hospital's rule, but I can't seem to get the tan(2x) to go away. Both appear to be indeterminate, but neither one is 0 at the same time, so it doesn't appear that I should be using L'Hopital's rule in this case. However, I can't see any other way to proceed.
Yes, it's true that$\displaystyle \lim_{x\to \pi/4} \tan(2x)$ does not exist.
But $\displaystyle \lim_{x\to \pi/4} \tan(x)=1\,,$ and $\displaystyle \lim_{x\to (\pi/4)^+} \tan(2x)=+\infty\ .$
Find the limit of the log of that expression.
If $\displaystyle \lim_{x\to \pi/4} \ln\left(\left(\tan(x)\right)^{\tan(2x)}\right)=L\,,$
then $\displaystyle \lim_{x\to \pi/4} \left(\tan(x)\right)^{\tan(2x)}=e^L\ .$
Last edited: | HuggingFaceTB/finemath | |
Search All of the Math Forum:
Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic: It is a very bad idea and nothing less than stupid to define 1/3
= 0.333...
Replies: 4 Last Post: Oct 4, 2017 4:05 PM
Messages: [ Previous | Next ]
netzweltler Posts: 473 From: Germany Registered: 8/6/10
Re: It is a very bad idea and nothing less than stupid to define 1/3
= 0.333...
Posted: Oct 4, 2017 4:05 PM
Am Mittwoch, 4. Oktober 2017 13:43:34 UTC+2 schrieb FromTheRafters:
> netzweltler was thinking very hard :
> > Am Mittwoch, 4. Oktober 2017 11:44:35 UTC+2 schrieb Zelos Malum:
> >>> In fact it means exactly infinitely many commands.
> >>> But of course if you define a series to be equal to its limit, then that's
> >>> like defining an apple equal to an orange. That is your problem, not ours.
> >>
> >> It doesn't because it is not operations upon operations, it is just a
> >> representation of one element in real numbers.
> >
> > It does. And 0.875 is representing 3 operations, e.g.
> > 0.8 + 0.07 + 0.005 or
> > 0.5 + 0.25 + 0.125.
> >
> > It can be an element AND represent some number of operations.
>
> I agree. The thing is that a finite number of steps (or commands) can,
> at best, give a good enough approximation of some numbers.
The 3-step operations above give the number exactly. You can get this number after any finite number of operations. Whereas you can't get the number 0.874999... after a finite number of operations.
> 'ad infinitum' to the 'end' of these, and assuming it can be completed
> in that (NaN) number of steps, you can arrive at the exact answer. This
> is a case where it is not about the trip, but about the (eventual)
> destination being defined exactly.
>
> As you already know, an arrow traveling from zero toward a target at
> two which can be described as going halfway there then halfway the
> remaining distance, then halfway again, may seem like an unending
> process with only better and better approximations being attainable.
> But, if I can get that same step by step process by stating that the
> process is actually pulling the arrow half the distance (from zero to
> two for example) then I don't need the unending process anymore as I
> already have the destination and the fact that the process is the same
> confirms that two is the answer I want.
>
> Insisting on the process while standing two units in front of the
> archer will not save you from getting the point. :)
Date Subject Author
10/4/17 zelos.malum@gmail.com
10/4/17 netzweltler
10/4/17 FromTheRafters
10/4/17 netzweltler | HuggingFaceTB/finemath | |
\$ 8 for adults & \$ 5 for seniors. 325 people paid, the total receipts were \$ 2569. How many paid were adults? How many paid were seniors? // //
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
Let us assume that the number os adults = x
And assume that the number os seniors = y
We are given that total people paid is 325.
==> x + y = 325
==> x= 325 - y ...........(1)
We are given that :
cost for an adults = 8
= 8*x
Cost for a senior = 5
==> total cost for seniors = 5*y
But total cost ( adults and seniors) = 2569
==> 8x + 5y = 2569 .........(2)
From (1), we will substitute in (2):
==> 8(325-y) + 5y = 2569
==> 2600 - 8y + 5y = 2569
==> 2600 - 3y = 2569
==> 3y = 31
==> y= 31/3 = 10.33 which is impossible because number of seniors should be an integer.
We will assume that y= 10 and total cost where 2570
==> total number of seniors = 10
total number of adults = 315.
william1941 | College Teacher | (Level 3) Valedictorian
Posted on
This problem can be solved in the following way using the information provided.
The total number of people who paid: 325
Let the number of adults be A
The number of seniors = 325 - A
The amount paid by one adult was \$8, So the total collection from the adults is 8A
The amount paid by one senior was \$5, so the total collection from the seniors is (325 - A )*5
As the total collection is 2569 we have
8A + (325 - A )*5 = 2569
=> 8A + 1625 - 5A = 2569
=> 3A = 2569 - 1625
=> 3A = 944
=> A = 944 / 3
As 944 / 3 is not an integer but A which is the number of adults has to be an integer, there is an error in the information provided. It is not possible to find the number of seniors and number of adults.
neela | High School Teacher | (Level 3) Valedictorian
Posted on
Let a be the number of adults s be the number of seniors.
The amount paid by a number of adults @ \$8 per person = 8a
The amount paid by the s sinior @ 5 per person = 5s.
The total amouny paid by (a+s = 325) persons = (8a+5s) = 2569
So the required equation is setup.
a+s = 325..........(1)
(28a+5s = 2569.........(2)
(2)- 5*(1) gives: (8a+5s) -5(a+s) = 2569-5*325
3a = 944
a =944/3 = 314.4...
Similarly ,
Eq(2) -8*Eq(1) gives: (8a+5s)-8*(a+s) = -3s = -31
So s = -31/-3 = 10.33..
The fractional solution is emberassing.
So we suggest the total bill collection to be \$2569 +\$1 = \$2570
Then a = 315 and b = 10 and a+b = 315+10 = 325.
And the recieved amount = 315*8+10*5 = \$2570. | HuggingFaceTB/finemath | |
Question
The difference between the sum of first 100 natural numbers and the sum of even numbers from 1 to 100.
A
2200 B
2400 C
2500 D
2700 Solution
The correct option is B 2500 Sum of first 100 natural numbers = 1002(2+99)=5050 Sum of even numbers from 1 to 100, there are 50 even numbers between 1 and 100. So, the sum of the 50 even numbers between 1 and 100 =502(2×2+49×2)=2550. =502(2×2+49×2)=2550. So, the difference is 5050 - 2550 = 2500 Mathematics
Suggest Corrections 0 Similar questions
View More People also searched for
View More | HuggingFaceTB/finemath | |
location: Publications → journals
Search results
Search: MSC category 11F ( Discontinuous groups and automorphic forms [See also 11R39, 11S37, 14Gxx, 14Kxx, 22E50, 22E55, 30F35, 32Nxx] {For relations with quadratic forms, see 11E45} )
Expand all Collapse all Results 1 - 25 of 91
1. CJM Online first
Bijakowski, Stephane
Partial Hasse invariants, partial degrees, and the canonical subgroup If the Hasse invariant of a $p$-divisible group is small enough, then one can construct a canonical subgroup inside its $p$-torsion. We prove that, assuming the existence of a subgroup of adequate height in the $p$-torsion with high degree, the expected properties of the canonical subgroup can be easily proved, especially the relation between its degree and the Hasse invariant. When one considers a $p$-divisible group with an action of the ring of integers of a (possibly ramified) finite extension of $\mathbb{Q}_p$, then much more can be said. We define partial Hasse invariants (they are natural in the unramified case, and generalize a construction of Reduzzi and Xiao in the general case), as well as partial degrees. After studying these functions, we compute the partial degrees of the canonical subgroup. Keywords:canonical subgroup, Hasse invariant, $p$-divisible groupCategories:11F85, 11F46, 11S15
2. CJM Online first
Varma, Sandeep
On Residues of Intertwining Operators in Cases with Prehomogeneous Nilradical Let $\operatorname{P} = \operatorname{M} \operatorname{N}$ be a Levi decomposition of a maximal parabolic subgroup of a connected reductive group $\operatorname{G}$ over a $p$-adic field $F$. Assume that there exists $w_0 \in \operatorname{G}(F)$ that normalizes $\operatorname{M}$ and conjugates $\operatorname{P}$ to an opposite parabolic subgroup. When $\operatorname{N}$ has a Zariski dense $\operatorname{Int} \operatorname{M}$-orbit, F. Shahidi and X. Yu describe a certain distribution $D$ on $\operatorname{M}(F)$ such that, for irreducible unitary supercuspidal representations $\pi$ of $\operatorname{M}(F)$ with $\pi \cong \pi \circ \operatorname{Int} w_0$, $\operatorname{Ind}_{\operatorname{P}(F)}^{\operatorname{G}(F)} \pi$ is irreducible if and only if $D(f) \neq 0$ for some pseudocoefficient $f$ of $\pi$. Since this irreducibility is conjecturally related to $\pi$ arising via transfer from certain twisted endoscopic groups of $\operatorname{M}$, it is of interest to realize $D$ as endoscopic transfer from a simpler distribution on a twisted endoscopic group $\operatorname{H}$ of $\operatorname{M}$. This has been done in many situations where $\operatorname{N}$ is abelian. Here, we handle the standard examples' in cases where $\operatorname{N}$ is nonabelian but admits a Zariski dense $\operatorname{Int} \operatorname{M}$-orbit. Keywords:induced representation, intertwining operator, endoscopyCategories:22E50, 11F70
3. CJM 2016 (vol 69 pp. 186)
Pan, Shu-Yen
$L$-Functoriality for Local Theta Correspondence of Supercuspidal Representations with Unipotent Reduction The preservation principle of local theta correspondences of reductive dual pairs over a $p$-adic field predicts the existence of a sequence of irreducible supercuspidal representations of classical groups. Adams/Harris-Kudla-Sweet have a conjecture about the Langlands parameters for the sequence of supercuspidal representations. In this paper we prove modified versions of their conjectures for the case of supercuspidal representations with unipotent reduction. Keywords:local theta correspondence, supercuspidal representation, preservation principle, Langlands functorialityCategories:22E50, 11F27, 20C33
4. CJM Online first
Lei, Antonio; Loeffler, David; Zerbes, Sarah Livia
On the asymptotic growth of Bloch-Kato--Shafarevich-Tate groups of modular forms over cyclotomic extensions We study the asymptotic behaviour of the Bloch--Kato--Shafarevich--Tate group of a modular form $f$ over the cyclotomic $\mathbb{Z}_p$-extension of $\mathbb{Q}$ under the assumption that $f$ is non-ordinary at $p$. In particular, we give upper bounds of these groups in terms of Iwasawa invariants of Selmer groups defined using $p$-adic Hodge Theory. These bounds have the same form as the formulae of Kobayashi, Kurihara and Sprung for supersingular elliptic curves. Keywords:cyclotomic extension, Shafarevich-Tate group, Bloch-Kato Selmer group, modular form, non-ordinary prime, p-adic Hodge theoryCategories:11R18, 11F11, 11R23, 11F85
5. CJM Online first
Xu, Bin
On Moeglin's parametrization of Arthur packets for p-adic quasisplit $Sp(N)$ and $SO(N)$ We give a survey on Moeglin's construction of representations in the Arthur packets for $p$-adic quasisplit symplectic and orthogonal groups. The emphasis is on comparing Moeglin's parametrization of elements in the Arthur packets with that of Arthur. Keywords:symplectic and orthogonal group, Arthur packet, endoscopyCategories:22E50, 11F70
6. CJM 2016 (vol 68 pp. 1382)
Zydor, Michał
La Variante infinitésimale de la formule des traces de Jacquet-Rallis pour les groupes unitaires We establish an infinitesimal version of the Jacquet-Rallis trace formula for unitary groups. Our formula is obtained by integrating a truncated kernel à la Arthur. It has a geometric side which is a sum of distributions $J_{\mathfrak{o}}$ indexed by classes of elements of the Lie algebra of $U(n+1)$ stable by $U(n)$-conjugation as well as the "spectral side" consisting of the Fourier transforms of the aforementioned distributions. We prove that the distributions $J_{\mathfrak{o}}$ are invariant and depend only on the choice of the Haar measure on $U(n)(\mathbb{A})$. For regular semi-simple classes $\mathfrak{o}$, $J_{\mathfrak{o}}$ is a relative orbital integral of Jacquet-Rallis. For classes $\mathfrak{o}$ called relatively regular semi-simple, we express $J_{\mathfrak{o}}$ in terms of relative orbital integrals regularised by means of zêta functions. Keywords:formule des traces relativeCategories:11F70, 11F72
7. CJM 2016 (vol 68 pp. 961)
Greenberg, Matthew; Seveso, Marco
$p$-adic Families of Cohomological Modular Forms for Indefinite Quaternion Algebras and the Jacquet-Langlands Correspondence We use the method of Ash and Stevens to prove the existence of small slope $p$-adic families of cohomological modular forms for an indefinite quaternion algebra $B$. We prove that the Jacquet-Langlands correspondence relating modular forms on $\textbf{GL}_2/\mathbb{Q}$ and cohomomological modular forms for $B$ is compatible with the formation of $p$-adic families. This result is an analogue of a theorem of Chenevier concerning definite quaternion algebras. Keywords:modular forms, p-adic families, Jacquet-Langlands correspondence, Shimura curves, eigencurvesCategories:11F11, 11F67, 11F85
8. CJM 2016 (vol 68 pp. 1227)
Brasca, Riccardo
Eigenvarieties for Cuspforms over PEL Type Shimura Varieties with Dense Ordinary locus Let $p \gt 2$ be a prime and let $X$ be a compactified PEL Shimura variety of type (A) or (C) such that $p$ is an unramified prime for the PEL datum and such that the ordinary locus is dense in the reduction of $X$. Using the geometric approach of Andreatta, Iovita, Pilloni, and Stevens we define the notion of families of overconvergent locally analytic $p$-adic modular forms of Iwahoric level for $X$. We show that the system of eigenvalues of any finite slope cuspidal eigenform of Iwahoric level can be deformed to a family of systems of eigenvalues living over an open subset of the weight space. To prove these results, we actually construct eigenvarieties of the expected dimension that parameterize finite slope systems of eigenvalues appearing in the space of families of cuspidal forms. Keywords:$p$-adic modular forms, eigenvarieties, PEL-type Shimura varietiesCategories:11F55, 11F33
9. CJM 2016 (vol 68 pp. 908)
Sugiyama, Shingo; Tsuzuki, Masao
Existence of Hilbert Cusp Forms with Non-vanishing $L$-values We develop a derivative version of the relative trace formula on $\operatorname{PGL}(2)$ studied in our previous work, and derive an asymptotic formula of an average of central values (derivatives) of automorphic $L$-functions for Hilbert cusp forms. As an application, we prove the existence of Hilbert cusp forms with non-vanishing central values (derivatives) such that the absolute degrees of their Hecke fields are arbitrarily large. Keywords:automorphic representations, relative trace formulas, central $L$-values, derivatives of $L$-functionsCategories:11F67, 11F72
10. CJM 2016 (vol 68 pp. 571)
Gras, Georges
Les $\theta$-régulateurs locaux d'un nombre algébrique : Conjectures $p$-adiques Let $K/\mathbb{Q}$ be Galois and let $\eta\in K^\times$ be such that $\operatorname{Reg}_\infty (\eta) \ne 0$. We define the local $\theta$-regulators $\Delta_p^\theta(\eta) \in \mathbb{F}_p$ for the $\mathbb{Q}_p\,$-irreducible characters $\theta$ of $G=\operatorname{Gal}(K/\mathbb{Q})$. A linear representation ${\mathcal L}^\theta\simeq \delta \, V_\theta$ is associated with $\Delta_p^\theta (\eta)$ whose nullity is equivalent to $\delta \geq 1$. Each $\Delta_p^\theta (\eta)$ yields $\operatorname{Reg}_p^\theta (\eta)$ modulo $p$ in the factorization $\prod_{\theta}(\operatorname{Reg}_p^\theta (\eta))^{\varphi(1)}$ of $\operatorname{Reg}_p^G (\eta) := \frac{ \operatorname{Reg}_p(\eta)}{p^{[K : \mathbb{Q}\,]} }$ (normalized $p$-adic regulator). From $\operatorname{Prob}\big (\Delta_p^\theta(\eta) = 0 \ \& \ {\mathcal L}^\theta \simeq \delta \, V_\theta\big ) \leq p^{- f \delta^2}$ ($f \geq 1$ is a residue degree) and the Borel-Cantelli heuristic, we conjecture that, for $p$ large enough, $\operatorname{Reg}_p^G (\eta)$ is a $p$-adic unit or that $p^{\varphi(1)} \parallel \operatorname{Reg}_p^G (\eta)$ (a single $\theta$ with $f=\delta=1$); this obstruction may be lifted assuming the existence of a binomial probability law confirmed through numerical studies (groups $C_3$, $C_5$, $D_6$). This conjecture would imply that, for all $p$ large enough, Fermat quotients, normalized $p$-adic regulators are $p$-adic units and that number fields are $p$-rational. We recall some deep cohomological results that may strengthen such conjectures. Keywords:$p$-adic regulators, Leopoldt-Jaulent conjecture, Frobenius group determinants, characters, Fermat quotient, Abelian $p$-ramification, probabilistic number theoryCategories:11F85, 11R04, 20C15, 11C20, 11R37, 11R27, 11Y40
11. CJM 2016 (vol 68 pp. 422)
Kohen, Daniel; Pacetti, Ariel
Heegner Points on Cartan Non-split Curves Let $E/\mathbb{Q}$ be an elliptic curve of conductor $N$, and let $K$ be an imaginary quadratic field such that the root number of $E/K$ is $-1$. Let $\mathscr{O}$ be an order in $K$ and assume that there exists an odd prime $p$, such that $p^2 \mid\mid N$, and $p$ is inert in $\mathscr{O}$. Although there are no Heegner points on $X_0(N)$ attached to $\mathscr{O}$, in this article we construct such points on Cartan non-split curves. In order to do that we give a method to compute Fourier expansions for forms on Cartan non-split curves, and prove that the constructed points form a Heegner system as in the classical case. Keywords:Cartan curves, Heegner pointsCategories:11G05, 11F30
12. CJM 2015 (vol 68 pp. 179)
Takeda, Shuichiro
Metaplectic Tensor Products for Automorphic Representation of $\widetilde{GL}(r)$ Let $M=\operatorname{GL}_{r_1}\times\cdots\times\operatorname{GL}_{r_k}\subseteq\operatorname{GL}_r$ be a Levi subgroup of $\operatorname{GL}_r$, where $r=r_1+\cdots+r_k$, and $\widetilde{M}$ its metaplectic preimage in the $n$-fold metaplectic cover $\widetilde{\operatorname{GL}}_r$ of $\operatorname{GL}_r$. For automorphic representations $\pi_1,\dots,\pi_k$ of $\widetilde{\operatorname{GL}}_{r_1}(\mathbb{A}),\dots,\widetilde{\operatorname{GL}}_{r_k}(\mathbb{A})$, we construct (under a certain technical assumption, which is always satisfied when $n=2$) an automorphic representation $\pi$ of $\widetilde{M}(\mathbb{A})$ which can be considered as the `tensor product'' of the representations $\pi_1,\dots,\pi_k$. This is the global analogue of the metaplectic tensor product defined by P. Mezo in the sense that locally at each place $v$, $\pi_v$ is equivalent to the local metaplectic tensor product of $\pi_{1,v},\dots,\pi_{k,v}$ defined by Mezo. Then we show that if all of $\pi_i$ are cuspidal (resp. square-integrable modulo center), then the metaplectic tensor product is cuspidal (resp. square-integrable modulo center). We also show that (both locally and globally) the metaplectic tensor product behaves in the expected way under the action of a Weyl group element, and show the compatibility with parabolic inductions. Keywords:automorphic forms, representations of covering groupsCategory:11F70
13. CJM 2014 (vol 67 pp. 893)
Mok, Chung Pang; Tan, Fucheng
Overconvergent Families of Siegel-Hilbert Modular Forms We construct one-parameter families of overconvergent Siegel-Hilbert modular forms. This result has applications to construction of Galois representations for automorphic forms of non-cohomological weights. Keywords:p-adic automorphic form, rigid analytic geometryCategories:11F46, 14G22
14. CJM 2014 (vol 66 pp. 993)
Beuzart-Plessis, Raphaël
Expression d'un facteur epsilon de paire par une formule intégrale Let $E/F$ be a quadratic extension of $p$-adic fields and let $d$, $m$ be nonnegative integers of distinct parities. Fix admissible irreducible tempered representations $\pi$ and $\sigma$ of $GL_d(E)$ and $GL_m(E)$ respectively. We assume that $\pi$ and $\sigma$ are conjugate-dual. That is to say $\pi\simeq \pi^{\vee,c}$ and $\sigma\simeq \sigma^{\vee,c}$ where $c$ is the non trivial $F$-automorphism of $E$. This implies, we can extend $\pi$ to an unitary representation $\tilde{\pi}$ of a nonconnected group $GL_d(E)\rtimes \{1,\theta\}$. Define $\tilde{\sigma}$ the same way. We state and prove an integral formula for $\epsilon(1/2,\pi\times \sigma,\psi_E)$ involving the characters of $\tilde{\pi}$ and $\tilde{\sigma}$. This formula is related to the local Gan-Gross-Prasad conjecture for unitary groups. Keywords:epsilon factor, twisted groupsCategories:22E50, 11F85
15. CJM 2014 (vol 67 pp. 424)
Samart, Detchat
Mahler Measures as Linear Combinations of $L$-values of Multiple Modular Forms We study the Mahler measures of certain families of Laurent polynomials in two and three variables. Each of the known Mahler measure formulas for these families involves $L$-values of at most one newform and/or at most one quadratic character. In this paper, we show, either rigorously or numerically, that the Mahler measures of some polynomials are related to $L$-values of multiple newforms and quadratic characters simultaneously. The results suggest that the number of modular $L$-values appearing in the formulas significantly depends on the shape of the algebraic value of the parameter chosen for each polynomial. As a consequence, we also obtain new formulas relating special values of hypergeometric series evaluated at algebraic numbers to special values of $L$-functions. Keywords:Mahler measures, Eisenstein-Kronecker series, $L$-functions, hypergeometric seriesCategories:11F67, 33C20
16. CJM 2014 (vol 66 pp. 1078)
Lanphier, Dominic; Skogman, Howard
Values of Twisted Tensor $L$-functions of Automorphic Forms Over Imaginary Quadratic Fields Let $K$ be a complex quadratic extension of $\mathbb{Q}$ and let $\mathbb{A}_K$ denote the adeles of $K$. We find special values at all of the critical points of twisted tensor $L$-functions attached to cohomological cuspforms on $GL_2(\mathbb{A}_K)$, and establish Galois equivariance of the values. To investigate the values, we determine the archimedean factors of a class of integral representations of these $L$-functions, thus proving a conjecture due to Ghate. We also investigate analytic properties of these $L$-functions, such as their functional equations. Keywords:twisted tensor $L$-function, cuspform, hypergeometric seriesCategories:11F67, 11F37
17. CJM 2013 (vol 67 pp. 214)
Szpruch, Dani
Symmetric Genuine Spherical Whittaker Functions on $\overline{GSp_{2n}(F)}$ Let $F$ be a p-adic field of odd residual characteristic. Let $\overline{GSp_{2n}(F)}$ and $\overline{Sp_{2n}(F)}$ be the metaplectic double covers of the general symplectic group and the symplectic group attached to the $2n$ dimensional symplectic space over $F$. Let $\sigma$ be a genuine, possibly reducible, unramified principal series representation of $\overline{GSp_{2n}(F)}$. In these notes we give an explicit formulas for a spanning set for the space of Spherical Whittaker functions attached to $\sigma$. For odd $n$, and generically for even $n$, this spanning set is a basis. The significant property of this set is that each of its elements is unchanged under the action of the Weyl group of $\overline{Sp_{2n}(F)}$. If $n$ is odd then each element in the set has an equivariant property that generalizes a uniqueness result of Gelbart, Howe and Piatetski-Shapiro. Using this symmetric set, we construct a family of reducible genuine unramified principal series representations which have more then one generic constituent. This family contains all the reducible genuine unramified principal series representations induced from a unitary data and exists only for $n$ even. Keywords:metaplectic group, Casselman Shalika FormulaCategory:11F85
18. CJM 2013 (vol 66 pp. 566)
Choiy, Kwangho
Transfer of Plancherel Measures for Unitary Supercuspidal Representations between $p$-adic Inner Forms Let $F$ be a $p$-adic field of characteristic $0$, and let $M$ be an $F$-Levi subgroup of a connected reductive $F$-split group such that $\Pi_{i=1}^{r} SL_{n_i} \subseteq M \subseteq \Pi_{i=1}^{r} GL_{n_i}$ for positive integers $r$ and $n_i$. We prove that the Plancherel measure for any unitary supercuspidal representation of $M(F)$ is identically transferred under the local Jacquet-Langlands type correspondence between $M$ and its $F$-inner forms, assuming a working hypothesis that Plancherel measures are invariant on a certain set. This work extends the result of MuiÄ and Savin (2000) for Siegel Levi subgroups of the groups $SO_{4n}$ and $Sp_{4n}$ under the local Jacquet-Langlands correspondence. It can be applied to a simply connected simple $F$-group of type $E_6$ or $E_7$, and a connected reductive $F$-group of type $A_{n}$, $B_{n}$, $C_n$ or $D_n$. Keywords:Plancherel measure, inner form, local to global global argument, cuspidal automorphic representation, Jacquet-Langlands correspondenceCategories:22E50, 11F70, 22E55, 22E35
19. CJM 2012 (vol 66 pp. 170)
Guitart, Xavier; Quer, Jordi
Modular Abelian Varieties Over Number Fields The main result of this paper is a characterization of the abelian varieties $B/K$ defined over Galois number fields with the property that the $L$-function $L(B/K;s)$ is a product of $L$-functions of non-CM newforms over $\mathbb Q$ for congruence subgroups of the form $\Gamma_1(N)$. The characterization involves the structure of $\operatorname{End}(B)$, isogenies between the Galois conjugates of $B$, and a Galois cohomology class attached to $B/K$. We call the varieties having this property strongly modular. The last section is devoted to the study of a family of abelian surfaces with quaternionic multiplication. As an illustration of the ways in which the general results of the paper can be applied we prove the strong modularity of some particular abelian surfaces belonging to that family, and we show how to find nontrivial examples of strongly modular varieties by twisting. Keywords:Modular abelian varieties, $GL_2$-type varieties, modular formsCategories:11G10, 11G18, 11F11
20. CJM 2012 (vol 65 pp. 544)
Deitmar, Anton; Horozov, Ivan
Iterated Integrals and Higher Order Invariants We show that higher order invariants of smooth functions can be written as linear combinations of full invariants times iterated integrals. The non-uniqueness of such a presentation is captured in the kernel of the ensuing map from the tensor product. This kernel is computed explicitly. As a consequence, it turns out that higher order invariants are a free module of the algebra of full invariants. Keywords:higher order forms, iterated integralsCategories:14F35, 11F12, 55D35, 58A10
21. CJM 2012 (vol 64 pp. 497)
Li, Wen-Wei
22. CJM 2011 (vol 65 pp. 22)
Blomer, Valentin; Brumley, Farrell
Non-vanishing of $L$-functions, the Ramanujan Conjecture, and Families of Hecke Characters We prove a non-vanishing result for families of $\operatorname{GL}_n\times\operatorname{GL}_n$ Rankin-Selberg $L$-functions in the critical strip, as one factor runs over twists by Hecke characters. As an application, we simplify the proof, due to Luo, Rudnick, and Sarnak, of the best known bounds towards the Generalized Ramanujan Conjecture at the infinite places for cusp forms on $\operatorname{GL}_n$. A key ingredient is the regularization of the units in residue classes by the use of an Arakelov ray class group. Keywords:non-vanishing, automorphic forms, Hecke characters, Ramanujan conjectureCategories:11F70, 11M41
23. CJM 2011 (vol 64 pp. 1248)
Gärtner, Jérôme
Darmon's Points and Quaternionic Shimura Varieties In this paper, we generalize a conjecture due to Darmon and Logan in an adelic setting. We study the relation between our construction and Kudla's works on cycles on orthogonal Shimura varieties. This relation allows us to conjecture a Gross-Kohnen-Zagier theorem for Darmon's points. Keywords:elliptic curves, Stark-Heegner points, quaternionic Shimura varietiesCategories:11G05, 14G35, 11F67, 11G40
24. CJM 2011 (vol 64 pp. 588)
Nekovář, Jan
Level Raising and Anticyclotomic Selmer Groups for Hilbert Modular Forms of Weight Two In this article we refine the method of Bertolini and Darmon and prove several finiteness results for anticyclotomic Selmer groups of Hilbert modular forms of parallel weight two. Keywords:Hilbert modular forms, Selmer groups, Shimura curvesCategories:11G40, 11F41, 11G18
25. CJM 2011 (vol 64 pp. 1122)
$p$-adic $L$-functions and the Rationality of Darmon Cycles Darmon cycles are a higher weight analogue of Stark--Heegner points. They yield local cohomology classes in the Deligne representation associated with a cuspidal form on $\Gamma _{0}( N)$ of even weight $k_{0}\geq 2$. They are conjectured to be the restriction of global cohomology classes in the Bloch--Kato Selmer group defined over narrow ring class fields attached to a real quadratic field. We show that suitable linear combinations of them obtained by genus characters satisfy these conjectures. We also prove $p$-adic Gross--Zagier type formulas, relating the derivatives of $p$-adic $L$-functions of the weight variable attached to imaginary (resp. real) quadratic fields to Heegner cycles (resp. Darmon cycles). Finally we express the second derivative of the Mazur--Kitagawa $p$-adic $L$-function of the weight variable in terms of a global cycle defined over a quadratic extension of $\mathbb{Q}$. Categories:11F67, 14G05 | open-web-math/open-web-math | |
1
AP EAPCET 2022 - 4th July Evening Shift
+1
-0
If $$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x}) d x=e^{\sqrt{x}}[A x+B \sqrt{x}+C]+K$$ then $$A+B+C=$$
A
$$-$$2
B
2
C
4
D
$$-$$4
2
AP EAPCET 2022 - 4th July Evening Shift
+1
-0
If $$\int \frac{1+\sqrt{\tan x}}{\sin 2 x} d x=A \log \tan x+B \tan x+C$$, then $$4 A-2 B=$$
A
$$-$$1
B
2
C
1
D
$$-$$2
3
AP EAPCET 2022 - 4th July Evening Shift
+1
-0
$$\int \frac{1+\tan x \tan (x+a)}{\tan x \tan (x+a)} d x=$$
A
$$\tan a(\log (\sec (x+a))+\log \sec x+C$$
B
$$\cot a(\log |\sin x|-\log |\sin (x+a)|)+C$$
C
$$\tan a\left(\log \left(\frac{\cos x}{\sin (x+a)}\right)\right)+C$$
D
$$\cot a\left(\log \frac{\sin (x+a)}{\cos (x+a)}\right)+C$$
4
AP EAPCET 2022 - 4th July Morning Shift
+1
-0
Assertion (A) If $$I_n=\int \cot ^n x d x$$, then $$I_6+I_4=\frac{-\cot ^5 x}{5}$$
Reason (R) $$\int \cot ^n x d x=\frac{-\cot ^{n-1} x}{n} -\int \cot ^{n-2} x d x$$
A
A is false, R is false
B
A is true, R is true
C
A is true, R is false
D
A is false, R is true
EXAM MAP
Medical
NEET | HuggingFaceTB/finemath | |
Finding the Schwarzchild radius of a star of solar mass 30
I am currently trying to determine the Schwarzchild radius of a star with solar mass 30. I am calculating it both with respect to solar mass, and w.r.t kilograms, however I am getting conflicting answers. (of a factor of 10)
$$1 \text{ solar mass} \sim 1.9891 \cdot 10^{31}\,\text{kg}$$ so I calulated
$$30\,\text{SM}\sim 5.97 \cdot 10^{32}\,\text{kg}$$
Using the formula for the Sch Radius:
$$R_s =\frac{2GM}{c^2}$$
I determined that you can calculate this using both the solar mass, and the kg mass to confirm.
Using given proportionality constants for $2G/c^2$:
$$= 2.95\,\text{km/solar mass}\\ = 1.48 \cdot 10^{-27}\,\text{m/kg}$$
Using the formula above, I have obtained:
$$\text{using solar mass: }R_s=88.5\,\text{km}\\ \text{using kg: } R_s=883\,\text{km}$$
If someone could work this out and help me clarify I would be very grateful!
• You will need to show us the details of the two calculations for us to comment usefully. – John Rennie Jan 9 '14 at 13:46
• WolframAlpha confirms the 88 km number – user23660 Jan 9 '14 at 14:17
Your method is correct, but you've got lost in the numbers. This is a good opportunity to use some neat web tools.
Answer: 30 solar masses = 5.9673 × $10^{31}$ kg | HuggingFaceTB/finemath | |
Algebra and Trigonometry
# 12.5Conic Sections in Polar Coordinates
Algebra and Trigonometry12.5 Conic Sections in Polar Coordinates
### Learning Objectives
In this section, you will:
• Identify a conic in polar form.
• Graph the polar equations of conics.
• Define conics in terms of a focus and a directrix.
Figure 1 Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr)
Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits.
In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.
### Identifying a Conic in Polar Form
Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola $x=2+ y 2 x=2+ y 2$ shown in Figure 2.
Figure 2
In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus $P(r,θ) P(r,θ)$ at the pole, and a line, the directrix, which is perpendicular to the polar axis.
If $F F$ is a fixed point, the focus, and $D D$ is a fixed line, the directrix, then we can let $e e$ be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points $P P$ such that $e= PF PD e= PF PD$ is a conic. In other words, we can define a conic as the set of all points $P P$ with the property that the ratio of the distance from $P P$ to $F F$ to the distance from $P P$ to $D D$ is equal to the constant $e. e.$
For a conic with eccentricity $e, e,$
• if $0≤e<1, 0≤e<1,$ the conic is an ellipse
• if $e=1, e=1,$ the conic is a parabola
• if $e>1, e>1,$ the conic is an hyperbola
With this definition, we may now define a conic in terms of the directrix, $x=±p, x=±p,$ the eccentricity $e, e,$ and the angle $θ. θ.$ Thus, each conic may be written as a polar equation, an equation written in terms of $r r$ and $θ. θ.$
### The Polar Equation for a Conic
For a conic with a focus at the origin, if the directrix is $x=±p, x=±p,$ where $p p$ is a positive real number, and the eccentricity is a positive real number $e, e,$ the conic has a polar equation
$r= ep 1±ecosθ r= ep 1±ecosθ$
For a conic with a focus at the origin, if the directrix is $y=±p, y=±p,$ where $p p$ is a positive real number, and the eccentricity is a positive real number $e, e,$ the conic has a polar equation
$r= ep 1±esinθ r= ep 1±esinθ$
### How To
Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.
1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
2. Identify the eccentricity $e e$ as the coefficient of the trigonometric function in the denominator.
3. Compare $e e$ with 1 to determine the shape of the conic.
4. Determine the directrix as $x=p x=p$ if cosine is in the denominator and $y=p y=p$ if sine is in the denominator. Set $ep ep$ equal to the numerator in standard form to solve for $x x$ or $y. y.$
### Example 1
#### Identifying a Conic Given the Polar Form
For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.
1. $r= 6 3+2sinθ r= 6 3+2sinθ$
2. $r= 12 4+5cosθ r= 12 4+5cosθ$
3. $r= 7 2−2sinθ r= 7 2−2sinθ$
Try It #1
Identify the conic with focus at the origin, the directrix, and the eccentricity for $r= 2 3−cosθ . r= 2 3−cosθ .$
### Graphing the Polar Equations of Conics
When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine $e e$ and, therefore, the shape of the curve. The next step is to substitute values for $θ θ$ and solve for $r r$ to plot a few key points. Setting $θ θ$ equal to $0, π 2 ,π, 0, π 2 ,π,$ and $3π 2 3π 2$ provides the vertices so we can create a rough sketch of the graph.
### Example 2
#### Graphing a Parabola in Polar Form
Graph $r= 5 3+3cosθ . r= 5 3+3cosθ .$
#### Analysis
We can check our result with a graphing utility. See Figure 4.
Figure 4
### Example 3
#### Graphing a Hyperbola in Polar Form
Graph $r= 8 2−3sinθ . r= 8 2−3sinθ .$
### Example 4
#### Graphing an Ellipse in Polar Form
Graph $r= 10 5−4cosθ . r= 10 5−4cosθ .$
#### Analysis
We can check our result using a graphing utility. See Figure 7.
Figure 7 $r= 10 5−4cosθ r= 10 5−4cosθ$ graphed on a viewing window of $[ –3,12,1 ] [ –3,12,1 ]$ by and
Try It #2
Graph $r= 2 4−cosθ . r= 2 4−cosθ .$
### Defining Conics in Terms of a Focus and a Directrix
So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.
### How To
Given the focus, eccentricity, and directrix of a conic, determine the polar equation.
1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of $y, y,$ we use the general polar form in terms of sine. If the directrix is given in terms of $x, x,$ we use the general polar form in terms of cosine.
2. Determine the sign in the denominator. If $p<0, p<0,$ use subtraction. If $p>0, p>0,$ use addition.
3. Write the coefficient of the trigonometric function as the given eccentricity.
4. Write the absolute value of $p p$ in the numerator, and simplify the equation.
### Example 5
#### Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix
Find the polar form of the conic given a focus at the origin, $e=3 e=3$ and directrix $y=−2. y=−2.$
### Example 6
#### Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix
Find the polar form of a conic given a focus at the origin, $e= 3 5 , e= 3 5 ,$ and directrix $x=4. x=4.$
Try It #3
Find the polar form of the conic given a focus at the origin, $e=1, e=1,$ and directrix $x=−1. x=−1.$
### Example 7
#### Converting a Conic in Polar Form to Rectangular Form
Convert the conic $r= 1 5−5sinθ r= 1 5−5sinθ$ to rectangular form.
Try It #4
Convert the conic $r= 2 1+2cosθ r= 2 1+2cosθ$ to rectangular form.
### Media
Access these online resources for additional instruction and practice with conics in polar coordinates.
### 12.5 Section Exercises
#### Verbal
1.
Explain how eccentricity determines which conic section is given.
2.
If a conic section is written as a polar equation, what must be true of the denominator?
3.
If a conic section is written as a polar equation, and the denominator involves $sinθ, sinθ,$ what conclusion can be drawn about the directrix?
4.
If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph?
5.
What do we know about the focus/foci of a conic section if it is written as a polar equation?
#### Algebraic
For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.
6.
$r= 6 1−2cosθ r= 6 1−2cosθ$
7.
$r= 3 4−4sinθ r= 3 4−4sinθ$
8.
$r= 8 4−3cosθ r= 8 4−3cosθ$
9.
$r= 5 1+2sinθ r= 5 1+2sinθ$
10.
$r= 16 4+3cosθ r= 16 4+3cosθ$
11.
$r= 3 10+10cosθ r= 3 10+10cosθ$
12.
$r= 2 1−cosθ r= 2 1−cosθ$
13.
$r= 4 7+2cosθ r= 4 7+2cosθ$
14.
$r(1−cosθ)=3 r(1−cosθ)=3$
15.
$r(3+5sinθ)=11 r(3+5sinθ)=11$
16.
$r(4−5sinθ)=1 r(4−5sinθ)=1$
17.
$r(7+8cosθ)=7 r(7+8cosθ)=7$
For the following exercises, convert the polar equation of a conic section to a rectangular equation.
18.
$r= 4 1+3sinθ r= 4 1+3sinθ$
19.
$r= 2 5−3sinθ r= 2 5−3sinθ$
20.
$r= 8 3−2cosθ r= 8 3−2cosθ$
21.
$r= 3 2+5cosθ r= 3 2+5cosθ$
22.
$r= 4 2+2sinθ r= 4 2+2sinθ$
23.
$r= 3 8−8cosθ r= 3 8−8cosθ$
24.
$r= 2 6+7cosθ r= 2 6+7cosθ$
25.
$r= 5 5−11sinθ r= 5 5−11sinθ$
26.
$r(5+2cosθ)=6 r(5+2cosθ)=6$
27.
$r(2−cosθ)=1 r(2−cosθ)=1$
28.
$r(2.5−2.5sinθ)=5 r(2.5−2.5sinθ)=5$
29.
$r= 6secθ −2+3secθ r= 6secθ −2+3secθ$
30.
$r= 6cscθ 3+2cscθ r= 6cscθ 3+2cscθ$
For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci.
31.
$r= 5 2+cosθ r= 5 2+cosθ$
32.
$r= 2 3+3sinθ r= 2 3+3sinθ$
33.
$r= 10 5−4sinθ r= 10 5−4sinθ$
34.
$r= 3 1+2cosθ r= 3 1+2cosθ$
35.
$r= 8 4−5cosθ r= 8 4−5cosθ$
36.
$r= 3 4−4cosθ r= 3 4−4cosθ$
37.
$r= 2 1−sinθ r= 2 1−sinθ$
38.
$r= 6 3+2sinθ r= 6 3+2sinθ$
39.
$r(1+cosθ)=5 r(1+cosθ)=5$
40.
$r(3−4sinθ)=9 r(3−4sinθ)=9$
41.
$r(3−2sinθ)=6 r(3−2sinθ)=6$
42.
$r(6−4cosθ)=5 r(6−4cosθ)=5$
For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.
43.
Directrix: $x=4;e= 1 5 x=4;e= 1 5$
44.
Directrix: $x=−4;e=5 x=−4;e=5$
45.
Directrix: $y=2;e=2 y=2;e=2$
46.
Directrix: $y=−2;e= 1 2 y=−2;e= 1 2$
47.
Directrix: $x=1;e=1 x=1;e=1$
48.
Directrix: $x=−1;e=1 x=−1;e=1$
49.
Directrix: $x=− 1 4 ;e= 7 2 x=− 1 4 ;e= 7 2$
50.
Directrix: $y= 2 5 ;e= 7 2 y= 2 5 ;e= 7 2$
51.
Directrix: $y=4;e= 3 2 y=4;e= 3 2$
52.
Directrix: $x=−2;e= 8 3 x=−2;e= 8 3$
53.
Directrix: $x=−5;e= 3 4 x=−5;e= 3 4$
54.
Directrix: $y=2;e=2.5 y=2;e=2.5$
55.
Directrix: $x=−3;e= 1 3 x=−3;e= 1 3$
#### Extensions
Recall from Rotation of Axes that equations of conics with an $xy xy$ term have rotated graphs. For the following exercises, express each equation in polar form with $r r$ as a function of $θ. θ.$
56.
$xy=2 xy=2$
57.
$x 2 +xy+ y 2 =4 x 2 +xy+ y 2 =4$
58.
$2 x 2 +4xy+2 y 2 =9 2 x 2 +4xy+2 y 2 =9$
59.
$16 x 2 +24xy+9 y 2 =4 16 x 2 +24xy+9 y 2 =4$
60.
$2xy+y=1 2xy+y=1$
Order a print copy
As an Amazon Associate we earn from qualifying purchases. | HuggingFaceTB/finemath | |
### Math 1620: Statistics
#### Professor Richard Kenyon Tel. 863-6406 rkenyon -at- math dot brown dot edu office: Kassar 305 Office hours: M 1-3pm, Thurs 3-4pm
Text: Statistics, 4th ed. Freedman, Pisani, Purves
There will be homeworks collected weekly and one midterm.
The final grade will be weighted as follows: Homework 30%, midterm 25%, final 45%.
# final: May 6, 9am, Barus-Holley 141. (Note room change!)
You may bring one page of notes and a calculator.
midterm: March 10 in class
Late homework will not be accepted.
Your lowest homework grade will be dropped.
Homework 1 due Tues Feb 3 in class:
Page 24, Review exercises, 1-12.
Page 50ff: 6,7,10,11,12
Homework 2 due Tues Feb 10 in class:
Page 94: 3
Page 106: 7,13
Page 137: 9a, 12
Page 154: 4,7,9,10
Homework 3 due Thurs Feb 19 in class:
Page 176: 4,5,8
Page 198: 3,4,7,11
Page 214: 2,5,7,9,12
Extra problem: n fair dice are rolled. Compute the correlation coefficient between the sum of the first k and the sum of the last k, for all k from 1 to n.
Homework 4 due Thurs Feb 26 in class:
Page 154: 4 (assume there are 500 men and 500 women; compute the correlation explicitly).
Page 215: 10
Page 286: 9
Page 329: 15
X1. (X,Y) is a joint Gaussian RV with density proportional to exp(-2x^2-xy-4y^2). Compute the covariance matrix. What is the correlation between X and Y?
X2. Class grades on three tests X,Y,Z each average 60 with SD=15. The correlations between the grades are r(X,Y)=.6, r(X,Z)=.7, r(Y,Z)=.8; given that a student scores 75 on tests X and Y, what is his expected score on Z?
X3. In the previous problem, suppose we don't know r(Y,Z). Can Y and Z be negatively correlated?
X4. 100 independent draws from an exponential RV are given. Is the correlation between the largest and second largest going to be positive, zero, or negative?
Homework 5 due Thurs March 12 in class:
Page 352: 5,6,7,12
Page 372: 5,6,8
Page 391: 2,3,7,9,14
Page 406:3,6
Homework 6 due Thurs March 19 in class:
Page 426: 2,3,4
Page 432: 20
Page 436: 24,28
Homework 7 due Tues April 7 in class:
Homework 7
Homework 8 due Tues April 14 in class:
Homework 8
Homework 9 due Tues April 21 in class:
Homework 9 | HuggingFaceTB/finemath | |
Convert Ø dia- part to sign | diameter part to angle signs
# angle units conversion
## Amount: 1 diameter part (Ø dia- part) of angle Equals: 0.032 angle signs (sign) in angle
Converting diameter part to angle signs value in the angle units scale.
TOGGLE : from angle signs into diameter parts in the other way around.
## angle from diameter part to sign angle conversion results
### Enter a new diameter part number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other angle measuring units - complete list.
How many angle signs are in 1 diameter part? The answer is: 1 Ø dia- part equals 0.032 sign
## 0.032 sign is converted to 1 of what?
The angle signs unit number 0.032 sign converts to 1 Ø dia- part, one diameter part. It is the EQUAL angle value of 1 diameter part but in the angle signs angle unit alternative.
Ø dia- part/sign angle conversion result From Symbol Equals Result Symbol 1 Ø dia- part = 0.032 sign
## Conversion chart - diameter parts to angle signs
1 diameter part to angle signs = 0.032 sign
2 diameter parts to angle signs = 0.064 sign
3 diameter parts to angle signs = 0.095 sign
4 diameter parts to angle signs = 0.13 sign
5 diameter parts to angle signs = 0.16 sign
6 diameter parts to angle signs = 0.19 sign
7 diameter parts to angle signs = 0.22 sign
8 diameter parts to angle signs = 0.25 sign
9 diameter parts to angle signs = 0.29 sign
10 diameter parts to angle signs = 0.32 sign
11 diameter parts to angle signs = 0.35 sign
12 diameter parts to angle signs = 0.38 sign
13 diameter parts to angle signs = 0.41 sign
14 diameter parts to angle signs = 0.45 sign
15 diameter parts to angle signs = 0.48 sign
Convert angle of diameter part (Ø dia- part) and angle signs (sign) units in reverse from angle signs into diameter parts.
## Angles
This calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units.
# Converter type: angle units
First unit: diameter part (Ø dia- part) is used for measuring angle.
Second: sign angle (sign) is unit of angle.
QUESTION:
15 Ø dia- part = ? sign
15 Ø dia- part = 0.48 sign
Abbreviation, or prefix, for diameter part is:
Ø dia- part
Abbreviation for sign angle is:
sign
## Other applications for this angle calculator ...
With the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool:
1. in practicing diameter parts and angle signs ( Ø dia- part vs. sign ) measures exchange.
2. for conversion factors between unit pairs.
3. work with angle's values and properties.
To link to this angle diameter part to angle signs online converter simply cut and paste the following.
The link to this tool will appear as: angle from diameter part (Ø dia- part) to angle signs (sign) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | HuggingFaceTB/finemath | |
OpenStudy (loser66):
find the solution of the given initial value problem $t^3y'+4t^2y =e^{-t}$y(-1) =0, t<0
OpenStudy (loser66):
$y = \frac{1}{t^4}\int t^3e^{-t}dt + C\frac{1}{t^4}$
OpenStudy (anonymous):
i'm assuming you solved that using integrating factors, right?
OpenStudy (loser66):
I got that, and use tabular method to take that integral, but I am wrong,
OpenStudy (loser66):
yes
OpenStudy (loser66):
would you please check my y?
OpenStudy (anonymous):
give me a sec
OpenStudy (anonymous):
i just had my final exam today for this class so it's pretty fresh on my mind haha
OpenStudy (loser66):
I saw you solve it before. today is my first day for this course. ha!! 2 chapters/ day . we are gonna fail......
OpenStudy (anonymous):
so your integral is a little wrong... it should be$\int\limits_{}^{}te^{-t}$ not $\int\limits_{}^{}t^{3}e^{-t}$
OpenStudy (loser66):
$e^{\int P(t)dt}= t^4$
OpenStudy (anonymous):
correct
OpenStudy (loser66):
and $e^{-\int P(t)dt}= \frac{1}{t^4}$
OpenStudy (loser66):
correct?
OpenStudy (anonymous):
yes, but what is the need for that?
OpenStudy (loser66):
because of the formula need them to plug in
OpenStudy (loser66):
$y = e^{-\int(P(t)dt}\int(g(t)e^{\int P(t)dt})dt + Ce^{-\int P(t)dt}$
OpenStudy (anonymous):
$e^{\int\limits\limits_{}^{}P(t)dt}y = \int\limits\limits_{}^{}e^{\int\limits\limits_{}^{}P(t)dt} *e^{-t}*t^{-3}dt$
OpenStudy (loser66):
I know this method, too. I watched videos on youtube. Ok, go ahead friend.
OpenStudy (anonymous):
well, you have $t^{4}y = \int\limits{}^{}e^{-t}tdt$
OpenStudy (anonymous):
integrate by parts, then divide your t^4 to solve for y
OpenStudy (loser66):
solve by your way: $t^3y' +4t^2y = e^{-t}$divide both sides by t^3 $y' +\frac{4}{t}y = \frac{e^{-t}}{t}$ integrating fractor is t^4 , therefore $t^4y' +4t^3y=t^3e^{-t}$ $\int \frac{d}{dt}(t^4+y)=\int t^3e^{-t}dt$ $t^4 +y = \int t^3e^{-t}dt$ $y = \int t^3e^{-t}dt -t^3 +C$ apply tabular for int part, he hehe... it's equal but anyway, that's what I have, not yours. either way is the same. but the answer is not in book. hehehe
OpenStudy (anonymous):
sorry i'm not familiar with your 'tabular' method
OpenStudy (anonymous):
so now that you solved for y, you know how to find C now, right? you should be all set!
OpenStudy (anonymous):
Not trying to step on robz8 toes, or anything. Loser66 sent me a private message and asked my input on this problem so let me take a look: $t^3y'+4t^2y=e^{-t}$ To go to standard linear form divide both sides by t^3: $y'+\frac{ 4 }{ t }y=\frac{ e^{-t} }{ t^3 }$ This is a linear equation so we need to find the integrating factor. this is $e^{\int\limits{\frac{ 4 }{ t }}dt}=e^{4\ln(t)}=t^4$ Multiplying through gives the equation: $t^4y'+4t^3y=te^{-t}$ which is $\frac{ d }{ dt }[t^4y]=te^{-t}$ The left side is solved using the FTC. The right side requires integration by parts which gives: $t^4y=-te^{-t}-e^{-t}+C$ So $y= \frac{ e^{-t} }{ t^3 }-\frac{e^{-t}+C}{t^4}$ Substitute the initial conditions y(-1)=0 and solve you'll see that C=0. Thus the solution is: $y= \frac{ e^{-t} }{ t^3 }-\frac{e^{-t}}{t^4}$ You can check on your own that it satisfies both the initial conditions and the differential equation.
OpenStudy (anonymous):
loser66 in your post above where you use an integrating factor, there is a plus sign where there should be a times sign.
OpenStudy (anonymous):
anytime, sorry it taken a while i haven't logged in for a few days. rob was on the right track to helping you
OpenStudy (anonymous):
that formula is just a generalized way to solve the problem and it makes more errors than its worth using sometimes. but you have to use the formula when the right hand side becomes a function you can't integrate | HuggingFaceTB/finemath | |
234
Q:
# Tea worth of Rs. 135/kg & Rs. 126/kg are mixed with a third variety in the ratio 1: 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be____?
A) Rs. 169.50 B) Rs.1700 C) Rs. 175.50 D) Rs. 180
Explanation:
Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50
So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.
Cost of 1 kg tea of 1st kind Cost of 1 kg tea of 2nd kind
x-153/22.50 = 1 => x - 153 = 22.50 => x=175.50.
Hence, price of the third variety = Rs.175.50 per kg.
Q:
Shiva purchased 280 kg of Rice at the rate of 15.60/kg and mixed it with 120 kg of rice purchased at the rate of 14.40/kg. He wants to earn a profit of Rs. 10.45 per kg by selling it. What should be the selling price of the mix per kg?
A) Rs. 22.18 B) Rs. 25.69 C) Rs. 26.94 D) Rs. 27.54
Explanation:
Rate of rice of quantity 280 kg = Rs. 15.60/kg
Rate of rice of quantity 120 kg = Rs. 14.40/kg
He want to earn a profit of Rs. 10.45/kg
Rate of Mix to sell to get profit of 10.45 =
3 378
Q:
Two bottles contains mixture of milk and water. First bottle contains 64% milk and second bottle contains 26% water. In what ratio these two mixtures are mixed so that new mixture contains 68% milk?
A) 3 : 2 B) 2 : 1 C) 1 : 2 D) 2 : 3
Explanation:
% of milk in first bottle = 64%
% of milk in second bottle = 100 - 26 = 74%
Now, ATQ
64% 74%
68%
6 4
Hence, by using allegation method,
Required ratio = 3 : 2
7 1268
Q:
In what ratio must wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg so that the mixture be worth Rs. 3/kg?
A) 1:2 B) 3:2 C) 2:1 D) 2:3
Explanation:
Given rate of wheat at cheap = Rs. 2.90/kg
Rate of wheat at cost = Rs. 3.20/kg
Mixture rate = Rs. 3/kg
Ratio of mixture =
2.90 3.20
3
(3.20 - 3 = 0.20) (3 - 2.90 = 0.10)
0.20 : 0.10 = 2:1
Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.
7 887
Q:
An alloy contains gold and silver in the ratio 5 : 8 and another alloy contains gold and silver in the ratio 5 : 3. If equal amount of both the alloys are melted together, then the ratio of gold and silver in the resulting alloy is ?
A) 113/108 B) 105/103 C) 108/115 D) 103/113
Explanation:
As given equal amounts of alloys are melted, let it be 1 kg.
Required ratio of gold and silver =
Hence, ratio of gold and silver in the resulting alloy = 105/103.
9 1546
Q:
A tin a mixture of two liquids A and B in the proportion 4 : 1. If 45 litres of the mixture is replaced by 45 litres of liquid B, then the ratio of the two liquids becomes 2 : 5. How much of the liquid B was there in the tin? What quantity does the tin hold?
A) 58 l B) 65 l C) 50 l D) 62 l
Explanation:
Let the tin contain 5x litres of liquids
=> 5(4x - 36) = 2(x + 36)
=> 20x - 180 = 2x + 72
=> x = 14 litres
Hence, the initial quantity of mixture = 70l
Quantity of liquid B
= 50 litres.
12 2432
Q:
An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%?
A) 45 gm B) 50 gm C) 55 gm D) 60 gm
Explanation:
Initial quantity of copper = = 40 g
And that of Bronze = 50 - 40 = 10 g
Let 'p' gm of copper is added to the mixture
=> = 40 + p
=> 45 + 0.9p = 40 + p
=> p = 50 g
Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.
8 2020
Q:
A man pays Rs. 6.40 per litre of milk. He adds water and sells the mixture at Rs. 8 per litre, thereby making 37.5% profit. The proportion of water to milk received by the customers is
A) 1 : 10 B) 10 : 1 C) 9 : 11 D) 11 : 9
Explanation:
Customer ratio of Milk and Water is given by
Milk :: Water
6.4 0
$6411$
=> Milk : Water = 110 : 11 = 10 : 1
Therefore, the proportionate of Water to Milk for Customer is 1 : 10
14 2161
Q:
In a 40 litre mixture of alcohol & water, the ratio of alcohol and water is 5 : 3. If 20% of this mixture is taken out and the same amount of water is added then what will be the ratio of alcohol and water in final mixture?
A) 1:1 B) 2:1 C) 3:1 D) 1:2
Explanation:
Quantity of alohol in the mixture = 40 x 5/8 = 25 lit
Quantity of water = 40 - 25 = 15 lit
According to question,
Required ratio = | HuggingFaceTB/finemath | |
# How do you solve x-24=60?
Dec 20, 2016
$x = 84$
#### Explanation:
Add $24$ to each side of the equation to isolate $x$ and keep the equation balanced:
$x - 24 + \textcolor{red}{24} = 60 + \textcolor{red}{24}$
$x - 0 = 84$
$x = 84$ | HuggingFaceTB/finemath | |
## Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
35 Lesson(s)
### Making M&M Treats Part I
4th Grade Math » Unit: Multi-Digit Division
4th Grade Math » Unit: Multi-Digit Division
Environment: Urban
Big Idea:
Standards:
Favorites (40)
Resources (63)
Reflections (4)
### Place Value Review
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (95)
Resources (9)
Reflections (1)
### Ten Times
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
#### My neighbor is just 10 times as much as me.
Standards:
Favorites (40)
Resources (14)
Reflections (1)
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (24)
Resources (9)
Reflections (1)
### Taking it Back to the Old School
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (27)
Resources (10)
Reflections (1)
### Powers of Ten (Day 1)
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (23)
Resources (9)
Reflections (1)
### 1/10 Of...
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (21)
Resources (11)
Reflections (1)
### Powers of Ten (Day 2)
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (10)
Resources (9)
Reflections (1)
### Decimal Operations
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (11)
Resources (9)
Reflections (1)
### Double Digit Division
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (22)
Resources (9)
Reflections (1)
### Powers of Ten Applications
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (13)
Resources (11)
Reflections (1)
### Division with Area Models
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
#### One for you. One for you. And, one for you.
Standards:
Favorites (10)
Resources (8)
Reflections (1)
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (12)
Resources (8)
Reflections (1)
### Remainder Riddles
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (14)
Resources (11)
Reflections (1)
### Operations with Decimals & Whole Numbers Assessment
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
5th Grade Math » Unit: Operations with Decimals and Whole Numbers
Grand Rapids, MI
Environment: Urban
Big Idea:
Standards:
Favorites (5)
Resources (7)
Common Core Math | HuggingFaceTB/finemath | |
### Bend
What is the longest stick that can be carried horizontally along a narrow corridor and around a right-angled bend?
### Exponential Trend
Find all the turning points of y=x^{1/x} for x>0 and decide whether each is a maximum or minimum. Give a sketch of the graph.
### Slide
This function involves absolute values. To find the slope on the slide use different equations to define the function in different parts of its domain.
# Generally Geometric
##### Stage: 5 Challenge Level:
Andy from Clitheroe Royal Grammar School sent us his work on this problem. He's given us two methods; can you see why he prefers the second one?
We begin by summing the series
$x+2x^2+3x^3+4x^4+\cdots$
$x$ + $x^2$ + $x^3$ + $x^4$ + $\cdots$ + $x^2$ + $x^3$ + $x^4$ + $\cdots$ + $x^3$ + $x^4$ + $\cdots$ + $\cdots$
In other words, we are writing it as a sum of geometric series!
Now, let us factorise the above sum as follows:
$(x + x^2 + x^3 + x^4+\ldots)(1 + x + x^2 + x^3 + x^4+\ldots)$
Wow, a product of geometric series!
We can then take a factor of $x$ out the first bracket to leave us with
$x(1 + x + x^2 + x^3+\ldots)^2$
Using the geometric sum given in the question, this comes to $$x\times \left(\frac{1}{1-x}\right)^2 = \frac{x}{(1-x)^2}$$ __
A similar method could be used for the series $x + 4x^2 + 9x^3 + 16x^4 +\ldots$, factorising it as $(x + 3x^2 + 5x^3 + 7x^4+\ldots)(1 + x + x^2 + x^3 +\ldots)$, then writing the left hand bracket as $(x + x^2 + x^3+\ldots + 2x^2 + 4x^3 + 6x^4+\ldots)$, from which point we can use our previous sum to obtain an answer. Unfortunately this doesn't generalise easily into higher powers, the amount of working needed growing much larger at each stage.
A more elegant solution is differentiation. If we differentiate our first series, we get $1 + 4x + 9x^2 + 16x^3+\ldots + n^2x^{n-1}+\ldots$. Multiplying through by $x$ gives us $x + 4x^2 + 9x^3+\ldots + n^2 x^n+\ldots$, which is the $n^2 x^n$ series we need.
If $x + 2x^2 + 3x^3 + 4x^4+\ldots = x/(1-x)^2$ then $x(d[x + 2x^2 + 3x^3+\ldots]/dx) = x(d[x/(1-x)^2]/dx)$.
But the left-hand side is equal to $x + 4x^2 + 9x^3 + 16x^4 +\ldots$, the sequence we want to sum.
We can resolve the right-hand using the quotient rule, and it comes to $x(1+x)/(1-x)^3$.
__
To take it into higher powers, notice that
$d[x + 4x^2 + 9x^3+\ldots]/dx = 1 + 8x + 27x^2+\ldots$.
Therefore $x d[x + 4x^2 + 9x^3+\ldots]/dx = x + 8x^2 + 27x^3+\ldots$, our next sequence. We can differentiate the previous infinite sum and multiply by $x$ at each stage to get the sum for the next power, and by applying the same process to the closed-form expression, we can obtain a closed-form expression for the next power. | HuggingFaceTB/finemath | |
# Quadratics by taking square roots calculator
Quadratics by taking square roots calculator keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website
## Keyword Suggestions
Quadratics by taking square roots calculator
Quadratics by taking square roots with steps
Quadratics by taking square roots intro
Solving quadratics by completing the square
How to solve quadratics by factoring
( Please select at least 2 keywords )
#### Most Searched Keywords
## Websites Listing
We found at least 10 Websites Listing below when search with quadratics by taking square roots calculator on Search Engine
### Quadratic Equation Completing the Square Calculator
Symbolab.com DA: 16 PA: 50 MOZ Rank: 66
• Solve quadratic equations using completing the square step-by-step
Mathpapa.com DA: 16 PA: 27 MOZ Rank: 44
• Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.) Take the Square Root
• About quadratic equations Quadratic equations have an x^2 term, and can be rewritten to have the form: a x 2 + b x + c = 0
### Square Root Calculator
Quadraticformulacalculator.net DA: 30 PA: 35 MOZ Rank: 67
• Square root calculator is a free online tool that shows the square root of the assign numbers
• It is the fast and easier approach for finding square root of any number
• The radical (√) symbol is normally used to indicate the square root. The radical symbol is also known root symbol or surd
• The square root of a function is represented by f (x) = √x
Equationcalc.com DA: 20 PA: 37 MOZ Rank: 60
• This quadratic equation root calculator lets you find the roots or zeroes of a quadratic equation
• A quadratic is a second degree polynomial of the form: ax2 + bx + c = 0 where a ≠ 0
• To solve an equation using the online calculator, simply enter the math problem in the text area provided
• Hit the calculate button to get the roots.
### Square Roots Calculator
Symbolab.com DA: 16 PA: 31 MOZ Rank: 51
• Free Square Roots calculator - Find square roots of any number step-by-step
• Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge
### Solving quadratics by taking square roots
Khanacademy.org DA: 19 PA: 50 MOZ Rank: 74
• Not all quadratic equations are solved by immediately taking the square root. Sometimes we have to isolate the squared term before taking its root
• For example, to solve the equation we should first isolate
• We do this exactly as we would isolate the term in a linear equation. Now solve a few similar equations on your own
Calculatorsoup.com DA: 22 PA: 50 MOZ Rank: 78
• This online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax 2 + bx + c = 0 for x, where a ≠ 0, using the quadratic formula
• The calculator solution will show work using the quadratic formula to solve the entered equation for real and complex roots.
Mathpapa.com DA: 16 PA: 19 MOZ Rank: 42
Learn step-by-step how to use the quadratic formula! Quadratic Formula Calculator Watch on Example (Click to try) 2 x 2 − 5 x − 3 = 0 About the quadratic formula Solve an equation of the form a x 2 + b x + c = 0 by using the quadratic formula: x = − b ± √ b 2 − 4 a c 2 a Step-By-Step Video Lesson
### Equations with square roots Calculator & Solver
Snapxam.com DA: 15 PA: 50 MOZ Rank: 73
We need to isolate the dependent variable t t, we can do that by subtracting -2 −2 from both sides of the equation \sqrt {3}\sqrt {t}=2 3 t = 2 4 Eliminate the \sqrt {3} 3 from the left side, multiplying both sides of the equation by the inverse of \sqrt {3} 3 \sqrt {t}=\frac {2\sqrt {3}} {3} …
Wolframalpha.com DA: 20 PA: 17 MOZ Rank: 46
• Added Feb 25, 2012 by Saurav in Mathematics
• Enter the real coeffecients and get the roots of the the equation along with the graph and solution represented on the respective plane
• Send feedback | Visit Wolfram|Alpha.
### Quadratic Equation Calculator With Steps
Equationcalc.com DA: 20 PA: 39 MOZ Rank: 69
• A quadratic equation of the form ax2 + bx + c = 0 for x, where a ≠ 0 can be solved using the completing the square formula with this calculator. It is an online calculator capable of completing the square of a quadratic equation with all the steps
• Naturally some quadratics can …
### Quadratic equation solver that shows work
Mathportal.org DA: 18 PA: 50 MOZ Rank: 79
• Method 2 : Completing the square
• This method can be used to solve all types of quadratic equations, although it can be complicated for some types of equations
• The method involves seven steps
• Example 04: Solve equation $2x^2 + 8x - 10= 0$ by completing the square
• Step 1: Divide the equation by the number in front of the square term.
### Solve by taking square roots calculator
Softmath.com DA: 12 PA: 50 MOZ Rank: 74
Algebra Parabolas Worksheets | how to solve systems of linear equations by graphing | formula for fraction from least to greatest | two step algebra calculator | real analysis i: hints for problems of chapter 3 | solve a wronskian with ti-89 | quadratic formula calculator | roots on ti 83 | square in hieroglyphics | math investigatory project.com
Calculator.net DA: 18 PA: 34 MOZ Rank: 65
• In algebra, a quadratic equation is any polynomial equation of the second degree with the following form: ax 2 + bx + c = 0
• Where x is an unknown, a is referred to as the quadratic coefficient, b the linear coefficient, and c the constant
• The numerals a, b, and c are coefficients of the equation, and they represent known numbers
• For example, a cannot be 0, or the equation would be linear
### Solving Quadratic Equations by Square Root Method
Chilimath.com DA: 17 PA: 50 MOZ Rank: 81
• The solutions to this quadratic formula are x = 3 x = 3 and x = - \,3 x = −3
• Example 4: Solve the quadratic equation below using the Square Root Method
• The two parentheses should not bother you at all
• The fact remains that all variables come in the squared form, which is what we want.
### Completing the Square Calculator
Calculatorsoup.com DA: 22 PA: 50 MOZ Rank: 87
• This calculator is a quadratic equation solver that will solve a second-order polynomial equation in the form ax 2 + bx + c = 0 for x, where a ≠ 0, using the completing the square method
• The calculator solution will show work to solve a quadratic equation by completing the square to solve the entered equation for real and complex roots.
### Quadratics by taking square roots: with steps
Khanacademy.org DA: 19 PA: 50 MOZ Rank: 85
• Solving quadratics by taking square roots: with steps
• Our mission is to provide a free, world-class education to anyone, anywhere
• Khan Academy is a 501(c)(3) nonprofit organization
• Donate or volunteer today! Site Navigation
### Solve Quadratic Equations By Taking Square Roots Calculator
Tessshebaylo.com DA: 20 PA: 50 MOZ Rank: 87
Solving quadratic equations by square root method chilimath how to program the formula into a calculator rider legacy hs student media extracting roots solve equation tessshlo completing examples calculate using 22 best free factoring quadratics for windows wolfram alpha Solving Quadratic Equations By Square Root Method Chilimath How To Program The Quadratic Formula Into A …
### Quadratic Equation Solver with Steps
Mathcracker.com DA: 15 PA: 29 MOZ Rank: 62
• There are several steps you have to follow in order to successfully solve a quadratic equation: Step 1: Identify the coefficients
• Examine the given equation of the form
• Ax^2+bx+c ax2 +bx+c, and determine the coefficients
Wolframalpha.com DA: 20 PA: 33 MOZ Rank: 72
• Wolfram|Alpha can apply the quadratic formula to solve equations coercible into the form ax2 +bx+c= 0 a x 2 + b x + c = 0
• In doing so, Wolfram|Alpha finds both the real and complex roots of these equations
• It can also utilize other methods helpful to solving quadratic equations, such as completing the square, factoring and graphing.
### Roots of Quadratic Equation Calculator
Cuemath.com DA: 15 PA: 50 MOZ Rank: 85
• The word "Quadratic" is derived from the word "Quad" which means square
• In other words, a quadratic equation is an “equation of degree 2 Solved Example on Roots of Quadratic Equation Calculator
• Example 1: Find the roots of given quadratic equation x 2 + 5x + 6 =0
• Solution: Given: a = 1, b = 5, c = 6
### Completing The Square Method To Solve The Quadratic Equation
Quadraticformulacalculator.net DA: 30 PA: 50 MOZ Rank: 29
Now taking square root on both sides of the equation, $$\sqrt{(x-1)^2} = \sqrt{3}$$ $$x-1 = \pm \sqrt{3}$$ function calculator or finding factoring quadratics calculator then you are at the right place because our factoring quadratics calculator or our Quadratic function calculator will solve all your Quadratic related problems.
### Solving Quadratic Equations By Taking The Square Root Calculator
Tessshebaylo.com DA: 20 PA: 50 MOZ Rank: 92
Solving quadratic equations by square root method chilimath calculate using how to program the formula into a calculator rider legacy hs student media extracting roots deriving wolfram alpha quadratics taking article khan academy mathematics gcse revision equation solve fx 991ex casio tutorials george garside Solving Quadratic Equations By Square Root Method Chilimath Solving …
### Solving Quadratics By Taking The Square Root Calculator …
Essaybots.com DA: 13 PA: 50 MOZ Rank: 86
• While a solving quadratics by taking the square root calculator dissertation is a substantially bigger piece of work than an essay or a investigation paper usually is, proofreading of this style of an assignment traditionally involves the solving quadratics by taking the square root calculator identical aspects.
### Solving quadratic equations by taking square roots calculator
Visusventures.org DA: 21 PA: 50 MOZ Rank: 95
• Solving quadratic equations by taking square roots calculator
• By | May 25, 2022 | north carolina mental health statistics 2021 | ivanka trump young model | May 25, 2022 | north carolina mental health statistics 2021 | ivanka trump young model
### Quadratic equation square root property +online calculator
Rational-equations.com DA: 26 PA: 50 MOZ Rank: 25
• Posted: Saturday 30th of Dec 13:04
• Right! May God save us students from the wrath of quadratic equation square root property +online calculator
• I used to face same problems that you do when I was there
• I always used to be confused in Remedial Algebra, Intermediate algebra and College Algebra
• I was worst in quadratic equation square root
Cuemath.com DA: 15 PA: 37 MOZ Rank: 78
• The roots of quadratic equation are the values of the variable that satisfy the equation
• They are also known as the "solutions" or "zeros" of the quadratic equation
• For example, the roots of the quadratic equation x 2 - 7x + 10 = 0 are x = 2 and x = 5 because they satisfy the equation
• I.e., when x = 2, 2 2 - 7 (2) + 10 = 4 - 14 + 10 = 0
### H 3 1 Solving Quadratic Equations By Taking Square Roots
Proxy-18.interexchange.org DA: 26 PA: 50 MOZ Rank: 23
• H 3.1 Solving Quadratic Equations by Ta king Square Roots.notebook 7 October 15, 2015 H 3.1 Solving Quadrati c Equations by Taking Square Roots.note book 8 October 15, 2015 Solve = 16 by taking squ are roots
• A real number x is a square root Of a nonnegative real number a provided x2 a
• H 3.1 Solving Quadratic Equations by Taking Square Roots
### Solve quadratic equations by taking square roots calculator
Elev8-va.com DA: 12 PA: 50 MOZ Rank: 90
• Solve quadratic equations by taking square roots calculator
• January 13, 2022; penguin linux commands; cnn boris sanchez eyebrows
### Quadratic by square root method
Softmath.com DA: 12 PA: 50 MOZ Rank: 91
• Taks test-taking strategies prentice hall prentice hall solving absolute value functions video , Pre-Algebra with Pizzazz Pg CC-19, second order quadratic difference, square root calculator online
• Dimensional analysis practice problems for middle school, sample papers of viii class, do 17 and 13 have common factors, FREE online rational
### Square Root Equation Calculator
Byjus.com DA: 9 PA: 33 MOZ Rank: 72
• The procedure to use the square root equation calculator is as follows: Step 1: Enter the equation in the input field
• Step 2: Now click the button “Solve” to get the variable value
• Step 3: Finally, the solution for the given square root equation will be displayed in the output field.
### Solving Quadratics: By the Square Root
Coolmath.com DA: 16 PA: 50 MOZ Rank: 97
• Solving Quadratics: By the Square Root - Cool Math has free online cool math lessons, cool math games and fun math activities
• Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too.
### Essaybots Solving Quadratics By Taking The Square Root …
Essaybots.com DA: 13 PA: 50 MOZ Rank: 95
It truly solving quadratics by taking the square root calculator is needed to point out the simple fact that many people today have the identical passwords for their e-mails, accounts at solving quadratics by taking the square root calculator the financial institution and in the social networks.
### Solve quadratic equations by taking square roots calculator
Latviadental.lv DA: 15 PA: 50 MOZ Rank: 98
• Solve quadratic equations by taking square roots calculator
• От Опубликовано в: portable virtualbox without admin rights В my cafe equipment lemonade
### Solving Quadratics By Taking Square Roots: Intro
Youtube.com DA: 15 PA: 6 MOZ Rank: 55
• This video introduces the idea of solving quadratic equations by taking a square root of each side
• The general strategy is to solve for the squared term and
### Solving Quadratic Equations by Taking Square Roots
Purplemath.com DA: 18 PA: 31 MOZ Rank: 84
• Then the solution is x = ±2, just like it was when I solved by factoring the difference of squares.
• Why did I need the "±" (that is, the "plus-minus") sign on the 2 when I took the square root of the 4?Because I'm trying to find all values of the variable which make the original statement true, and it could have been either a positive 2 or a negative 2 that was squared to get that 4 in the
### Solve quadratic equations by taking square roots calculator
Web.think201.com DA: 16 PA: 22 MOZ Rank: 74
• Solve quadratic equations by taking square roots calculatorsolve quadratic equations by taking square roots calculator
• Solve quadratic equations by taking square roots calculator in
• How far is south padre from houston; Thursday Jan 20, read
### Solve quadratic equations by taking square roots calculator
Distrito.uy DA: 11 PA: 42 MOZ Rank: 90
• Recall that a quadratic equation is an equation that can be written in the form ax bx c2 + + = 0, with a≠0
• Sum of two numbers is 15, if sum of their reciprocals is 3
• Solving Quadratic Equations with Square Roots Date_____ Period____ Solve each equation by …
### 4.3 Solving Quadratic Equations Using Square Roots
Static.bigideasmath.com DA: 23 PA: 49 MOZ Rank: 12
• Section 4.3 Solving Quadratic Equations Using Square Roots 211 Solving a Quadratic Equation Using Square Roots Solve (x − 1)2 = 25 using square roots.SOLUTION (x − 1)2 = 25 Write the equation.x − 1 = ±5 Take the square root of each side
• X = 1 ± 5 Add 1 to each side
• So, the solutions are x = 1 + 5 = 6 and x = 1 − 5 = −4
• Check Use a graphing calculator to check
### Solving quadratic equations by taking square roots calculator
Timarucourier.co.nz DA: 23 PA: 21 MOZ Rank: 83
• Solving quadratic equations by taking square roots calculator
• Which ivy league should i go to quiz; solving quadratic equations by taking square roots calculator
• New homes for sale in chattanooga, tn
• Andalusia city schools job openings.
### Solve quadratic equations by taking square roots calculator
Drinkexpert.com.br DA: 18 PA: 50 MOZ Rank: 10
• I can solve by taking the square root
• H 3 1 Solving Quadratic Equations By Taking Squar | HuggingFaceTB/finemath | |
# Algorithm
An algorithm is a method, or a set of step-by-step instructions for performing some task or calculation, especially for computer programs.
For example, suppose we want an algorithm to find the average of a list of numbers. We must add them up and then divide by n, where n is the count of how many numbers there are. The algorithm for this might look like:
```start with a list of numbers (eg 2, 5, 99, -4, 20)
set the total to zero
start at the top of the list
repeat the following until end of list
get the next number from list
add it to the running total
At the end of the list
the average is the total divided
by the length of the list
```
When an algorithm is written out in this English-like way it is called 'pseudo-code'. It looks a little like computer code (programming), but is really meant to describe the algorithm to another person.
## As a program
The algorithm is a step-by-step method for solving a problem. It can then be translated into a program that a computer could perform. So for example, the above algorithm might be put into a program form in the Javascript language as follows:
```var list = [2, 5, 99, -4, 20];
var total = 0;
var average;
for(var i=0; i<list.length; i++);
{ total = total + list[i];
}
average = total/list.length;
```
So, as you can see, the algorithm is the method for getting the job done, which is then written in a computer language of your choice so it can performed by the computer. | HuggingFaceTB/finemath | |
# Square Root
## Learn The Definition And How To Find The Square Roots Of Numbers With Formulas And Examples
As kids progress through the grades, they start learning more and more complex math concepts. Square root and squares are one such topic that kids start learning in grades 4 and 5. But, what are they? the square root of a number is another number, which you get when the original number is multiplied by itself. Before introducing this concept to the little ones, ensure they know the multiplication table for kids. The square root can be expressed as a x where a is the natural number and the x is the square root of the number.
## What is Square Root?
In math, square roots are the factor of a number, which gives the original number when it’s multiplied by itself. It is represented by the symbol √. Consider this formula, if p is the square root of q, it implies that p x p = q.
Here is an example to help you understand better, 4 is the square root of 16. To get 16, 4 is multiplied by itself. In this context, the exponent is 2, it is known as a square.
According to mathematical concepts, square roots and squares are reverse operations. The square of any numeral is the value of power 2 of the number. Another fact about squares and square roots is that they are always a positive number. Here is a list of square roots to help your child learn the concept easily given below:
## List Of Square Roots
### Definition:
The square root of a digit is the value of power ½ of that number. In simple words, the square root is the number that is multiplied by itself to obtain the actual number. It is expressed as ‘ ‘.
Also, refer to Multiplication Worksheets, available at Osmo.
## How to Find The Square Root?
Once your child has understood the definition, it’s time to teach how to find the square root of a number. Some kids might find it difficult to grasp the concept easily. Therefore, you can start with simpler numbers that are perfect squares, which is a number that can be expressed as the product of two equal integers. Finding the square roots of perfect square is quite easy and kids will learn it faster.
Here are four different methods to determine the square roots of numbers:
• Repeated Subtraction Method
• Estimation Method
• Prime Factorization Method
• Long Division Method
### Table Of Square Roots
Here is a table to help your kids learn easily. The table has a list of the square roots of numbers from 1 to 10. This list includes both perfect squares and numbers that are non-perfect squares.
## Square Root Formula
One thing that makes it easy to determine square roots of numbers is a formula. Here is a simple formula you can use: √x= x1/2
According to this formula, 41/2 = √4 = √(2×2) = 2.
### Determining Square Roots Of Negative Numbers
Now that you know how to find the square roots of positive integers, how do you find the same for a negative integer? A square is always 0 or positive, so negative numbers don’t have real square roots.
But, here is an easy formula you can use: √(-y)= i√y. Here, i is the square root of -1.
For instance, let’s determine the √ -25.
√(-25)= √25 × √(-1)
Since negative numbers don’t have real square roots, let’s consider √(-1)= i
√(-25)= √25 × √(-1) = 5i
Therefore, the √-25 is 5i.
Getting kids to learn math can be difficult. Involve kids in fun activities like math games for kids and worksheets to help them learn in a fun and engaging way. Check Osmo’s kids learning section to find more fun ways to help kids learn.
## Frequently Asked Questions on Square Root
### What Is A Square Root?
A square root is a number that is obtained when a number is multiplied by itself. For example, 10 is the square root of 100; that is, 10 x 10 = 100. The square root is denoted by a symbol, √.
### How is the square root of a number calculated?
The square root of a number is calculated using this formula. The formula for the square root is used to determine the square root of a digit. The exponent formula to find out the square root is, n√y y n = y1/n. Here, n= 2, is called a square root. We can use any of these simple ways for determining the square root, like, long division, prime factorization, and many more.
30% OFF* | HuggingFaceTB/finemath | |
# Why can't i apply the simple average velocity formula?
by rainstom07
Tags: apply, average, formula, simple, velocity
P: 16 I'm doing a homework problem (i already know the answer) and i came across an error in my logic/application of the formula Vavg = (v + v0)/2. Hopefully you can help me understand why it's incorrect to use the formula. x = 12t2-2t3 describes a particle position. the derivative of x = 24t-6t2. The homework question asked me find the average velocity between t = 0 and t = 3. Using the formula Vavg = Δx/Δt yields 18 m/s... the correct answer. --- When i use the simpler formula: Vavg = (vf + vi) / 2 = (x'(3.0)+x'(0.0))/2 = 18/2. I get 9 m/s which is incorrect. -- Adding the velocity at t=3 with the velocity at t=0 and then dividing by 2 should've produced 18 m/s... My logic is clearly wrong, but how? x' describes the velocity of the particle at t seconds? right? thanks.
P: 5,462 Hello rainstom, welcome to Physics Forums Your average forumla (t0+t3)/2 is only correct if x is a linear function of t ie a straight line. The function x = 12t2-2t3 is decidedly non linear The way to derive an average for a non linear function (works for linear as well but is trivial) is to integrate the function and divide by the interval or number of points or samples. So average = Area/Interval Does this help and can you now obtain the correct answer?
P: 16
Your average forumla (t0+t3)/2 is only correct if x is a linear function of t ie a straight line.
Thank you! i knew there was some sort of condition attached to the simpler version of average velocity.
Does this help and can you now obtain the correct answer?
^^ yup
P: 1,284
## Why can't i apply the simple average velocity formula?
Quote by Studiot Hello rainstom, welcome to Physics Forums Your average forumla (t0+t3)/2 is only correct if x is a linear function of t ie a straight line. The function x = 12t2-2t3 is decidedly non linear
I suppose you meant v_avg = (v_0 + v_3)/2 is only correct if v is a linear function of t.?
X must then be a quadratic function of t, and x = 12t2-2t3 isn't a quadratic.
P: 5,462 Hello willem2 does this attachment help? The average velocity is the number which if you multiplied it by the time would give you the total distance travelled. Attached Thumbnails
Related Discussions Introductory Physics Homework 11 Introductory Physics Homework 2 Introductory Physics Homework 2 Introductory Physics Homework 1 Introductory Physics Homework 1 | HuggingFaceTB/finemath | |
Subject: Maths, asked on 23/5/18
## Please help me.... Q. There are wo classrooms A and B. If 15 students are sent from A to B, then the number of students in each classroom will be same. If 5 students are sent from B to A, then the number or students in A will be the double the number students of in B. Represent this information algebrically.
Subject: Maths, asked on 22/5/18
## 7x-15y=2 x-2y=3 I got the value of x=41 y=19 ARE THESE VALUES CORRECT ?
Subject: Maths, asked on 21/5/18
Subject: Maths, asked on 21/5/18
## Solve the equation 10x+3y=75 and 6x-5y=11
Subject: Maths, asked on 21/5/18
## Solve the equation 10x+3y=75 6x-5y=11
Subject: Maths, asked on 21/5/18
## I didn't understand elimination method
Subject: Maths, asked on 21/5/18
## Solve:2x-y=4 Y-z=6 X-z=10 Give explanation to solve these kind of equations
Subject: Maths, asked on 21/5/18
Subject: Maths, asked on 20/5/18
## Please solve the 3rd and the 4th one. 2. Find the value of a and b for which each of the following systems of linear equations has a infinte number of solutions :
Subject: Maths, asked on 19/5/18
## Pls tell question no 12,13,24,25,26,27,28 Q.12 If are the zeroes Of the polynomial then find value Of b. . Q. 14. If (x+p) is a factor of the polynomial 2x2+2px+5x+10 find p
Subject: Maths, asked on 18/5/18
## Answer 3 questions
Subject: Maths, asked on 18/5/18
## solve this by substitution method 3x - y=3 9x - 3y=9
Subject: Maths, asked on 18/5/18
## solve 3x+5y=12 and 5x+4y=14 by elimination method
Subject: Maths, asked on 17/5/18
## 2x/a+y/b=2 x/a-y/b=4 solve by sustitution method
Subject: Maths, asked on 17/5/18
## Solve this: 14. For what values of a and b, will be the following pair of linear equations have infinitely many solutions? 2y + x = 1; (a - b) y = a + b - 2
What are you looking for? | HuggingFaceTB/finemath | |
9445 (number)
9,445 (nine thousand four hundred forty-five) is an odd four-digits composite number following 9444 and preceding 9446. In scientific notation, it is written as 9.445 × 103. The sum of its digits is 22. It has a total of 2 prime factors and 4 positive divisors. There are 7,552 positive integers (up to 9445) that are relatively prime to 9445.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 4
• Sum of Digits 22
• Digital Root 4
Name
Short name 9 thousand 445 nine thousand four hundred forty-five
Notation
Scientific notation 9.445 × 103 9.445 × 103
Prime Factorization of 9445
Prime Factorization 5 × 1889
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 9445 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 9,445 is 5 × 1889. Since it has a total of 2 prime factors, 9,445 is a composite number.
Divisors of 9445
1, 5, 1889, 9445
4 divisors
Even divisors 0 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 11340 Sum of all the positive divisors of n s(n) 1895 Sum of the proper positive divisors of n A(n) 2835 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 97.1854 Returns the nth root of the product of n divisors H(n) 3.33157 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 9,445 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 9,445) is 11,340, the average is 2,835.
Other Arithmetic Functions (n = 9445)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7552 Total number of positive integers not greater than n that are coprime to n λ(n) 1888 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1170 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 7,552 positive integers (less than 9,445) that are coprime with 9,445. And there are approximately 1,170 prime numbers less than or equal to 9,445.
Divisibility of 9445
m n mod m 2 3 4 5 6 7 8 9 1 1 1 0 1 2 5 4
The number 9,445 is divisible by 5.
Classification of 9445
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
Other numbers
• LucasCarmichael
Base conversion (9445)
Base System Value
2 Binary 10010011100101
3 Ternary 110221211
4 Quaternary 2103211
5 Quinary 300240
6 Senary 111421
8 Octal 22345
10 Decimal 9445
12 Duodecimal 5571
20 Vigesimal 13c5
Basic calculations (n = 9445)
Multiplication
n×y
n×2 18890 28335 37780 47225
Division
n÷y
n÷2 4722.5 3148.33 2361.25 1889
Exponentiation
ny
n2 89208025 842569796125 7958071724400625 75163987436963903125
Nth Root
y√n
2√n 97.1854 21.1382 9.85827 6.23793
9445 as geometric shapes
Circle
Diameter 18890 59344.7 2.80255e+08
Sphere
Volume 3.52935e+12 1.12102e+09 59344.7
Square
Length = n
Perimeter 37780 8.9208e+07 13357.2
Cube
Length = n
Surface area 5.35248e+08 8.4257e+11 16359.2
Equilateral Triangle
Length = n
Perimeter 28335 3.86282e+07 8179.61
Triangular Pyramid
Length = n
Surface area 1.54513e+08 9.92978e+10 7711.81
Cryptographic Hash Functions
md5 f74412c3c1c8899f3c130bb30ed0e363 baa4408def659ddf803d68ad5b94d6ca29597483 0c957eac7ed9468d27b0101bc81d68a95399d62dfe90ceb3b067b0083b8e9214 702df6f9395152142abc3f1a9d766c49dcd0dac2062b02fecd4df95a81f4e11d7b193518e4b6b83699cb1f62669fa6063fdc9e4eee0ccb73e2e3924de92726e2 f267736b9cef8e45f559627afd40749d0421e559 | HuggingFaceTB/finemath | |
# 893: 65 Years
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
65 Years Title text: The universe is probably littered with the one-planet graves of cultures which made the sensible economic decision that there's no good reason to go into space--each discovered, studied, and remembered by the ones who made the irrational decision.
## Explanation
Randall is showing the number of still living humans who have walked on another world for the 65-year period that begins in 1969 (when a human first walked on the moon). Up to 2011 (when the comic was drawn), he has drawn a single line for the actual figures.
For the subsequent years, he has drawn three lines using actuarial tables or life tables (such tables show, for each age, the probability that a certain person will die within the next year).
The line marked "5th Percentile" indicates that there is a 95% probability that the number alive in a given year will be above that line and a 5% probability that the number alive will be below that line. For example, this line indicates a 5% chance that all Apollo moon walkers will be dead by 2023, and a 95% chance that at least one will still be alive by that year.
The line marked "95th Percentile" indicates that there is a 5% probability that the number alive in a given year will be above that line and a 95% probability that the number alive will be below that line. For example, this line indicates a 95% chance that all Apollo moon walkers will be dead by 2035, and a 5% chance that at least one will still be alive by that year.
The middle line is not identified, but is probably the "50th Percentile" (see these tables). If so, it indicates that there is a 50% probability that the number alive in a given year will be above that line and a 50% probability that the number alive will be below that line. For example, this line indicates a 50% chance that all Apollo moon walkers will be dead by 2028 (see previous link), and a 50% chance that at least one will still be alive by that year.
Although the term other world would include all other worlds on which humans have walked, there is currently only one other world on which humans have walked, which is the moon. The humans that have walked there are the 12 Apollo astronauts who landed on the Moon between 1969 and 1972.
In particular, Neil Armstrong and Buzz Aldrin landed in July 1969. Pete Conrad and Alan Bean landed in November. Alan Shepard and Edgar Mitchell: February 1971. David Scott and James Irwin: July 1971. John Young and Charles Duke: April 1972. Eugene Cernan and Harrison Schmitt: December 1972.
Irwin died in 1991. Shepard and Conrad died in 1998 and 1999 respectively, making the total 9 as of the date this comic was published. Since then Armstrong died in 2012, Mitchell in 2016, Cernan in 2017, Young on January 6, 2018, and Bean on May 26, 2018. The current (as of September 2022) number is 4, which lies close to the middle line (the supposed 50th Percentile). The oldest living person to have landed on the moon is Aldrin at 91. Also living are Scott at 88, Schmitt at 85, and Duke at 85.
The chart assumes that no other humans will go to walk on another world within the time-frame plotted and the title text implies that this is primarily an economically determined decision. While noting that not exploring space is a justifiable and sensible decision which may also be made by many hypothetical cultures on other worlds, the text implies a grandness to a civilization that would be given the opportunity to discover, study and memorialize the 'one-world graves' of other civilizations by choosing to explore space despite the economic difficulty. This also implies that the likely consequence of not exploring space is that a civilization which chooses to do this is doomed to go extinct eventually while those which do explore and colonize may last long enough to be safely established on multiple worlds and discover the remains of civilizations which acted on a purely economic basis and hence ensured their own collapse. High five for exoplanet archaeology.
## Transcript
[A graph titled 'Number of Living Humans Who Have Walked on Another World' - its y-axis is numbered 5, 10, 15, its x-axis increments every ten years from 1960-2040. The line of the graph has a bracket above it that says '65 Years', starting at 1969, ending in 2034.
The line starts at 1969 and increases steeply to 12 by 1972. It then plateaus until the early nineties, declines gradually to 9 between 1991-1999, and then plateaus again.
From 2011-2035, which is labeled 'Projected Actuarial Tables', the line branches into three and begins to decline more steeply to zero. The area between the first and second branch is shaded and labeled '5th percentile' and the area between the second and third branch is shaded and labeled '95th percentile.']
## Trivia
• The theme of actuarial projections was explored earlier in 493: Actuarial; Randall's morbid python script for both was given in the blag.
Name Born Died Age atfirst step Age at death Mission Lunar dates Service Alma Mater 1. Neil Armstrong 1930-08-05 2012-08-25 38y 11m 15d 82y 0m 20d Apollo 11 July 21, 1969 NASA Purdue University, University of Southern California 2. Buzz Aldrin 1930-01-20 39y 6m 0d Air Force United States Military Academy, MIT 3. Pete Conrad 1930-06-02 1999-07-08 39y 5m 17d 69y 1m 6d Apollo 12 November 19–20, 1969 Navy Princeton University 4. Alan Bean 1932-03-15 2018-05-26 37y 8m 4d 86y 2m 11d Navy University of Texas, Austin 5. Alan Shepard 1923-11-18 1998-07-21 47y 2m 18d 74y 8m 3d Apollo 14 February 5–6, 1971 Navy United States Naval Academy 6. Edgar Mitchell 1930-09-07 2016-02-04 40y 4m 19d 85y 4m 28d Navy Carnegie Mellon University, Naval Postgraduate School, MIT 7. David Scott 1932-06-06 39y 1m 25d Apollo 15 July 31 - August 2, 1971 Air Force University of Michigan (freshman year, and later, an honorary doctorate), United States Military Academy, MIT 8. James Irwin 1930-03-17 1991-08-08 41y 4m 14d 61y 4m 22d Air Force United States Naval Academy, University of Michigan 9. John W. Young 1930-09-24 2018-01-06 41y 6m 28d 87y 3m 13d Apollo 16 April 21–23, 1972 Navy Georgia Institute of Technology 10. Charles Duke 1935-10-03 36y 6m 18d Air Force United States Naval Academy, MIT 11. Eugene Cernan 1934-03-14 2017-01-16 38y 9m 7d 82y 10m 2d Apollo 17 December 11–14, 1972 Navy Purdue University, Naval Postgraduate School 12. Harrison Schmitt 1935-07-03 37y 5m 8d NASA Caltech, University of Oslo (exchange), Harvard University
# Discussion
I wonder if it would be possible to identify individual people who are behind those vertical jumps in the graph (in the not projected part)... --JakubNarebski (talk) 19:18, 14 December 2012 (UTC)
• Glad you asked! </Information Hen> Neil Armstrong and Buzz Aldrin landed in July 1969; that's two. Pete Conrad and Alan Bean joined the group that November; that's four. Alan Shepard and Edgar Mitchell in February '71; that's six. David Scott and James Irwin in July '71; that's eight. John W. Young and Charles Duke in April '72; that's ten. Eugene Cernan and Harrison Schmitt in December '72; that's twelve. Irwin died in '91, dropping it to 11. Shepard and Conrad died in '98 and '99 respectively, making it 9 as of the date this comic was published. Armstrong died in '12, so our current number is 8. The oldest living person to have landed on the moon is Aldrin, 83. There are two 82-year-olds, two 80s, one 78 and two 77s. Ekedolphin (talk) 13:28, 27 January 2013 (UTC)
Almost prophetic and very, very sad. RIP Neil Armstrong ------
Can we add the 5% and 95% columns to the table? Spongebog (talk)
i dont feel like this would add to the explanation of the comic and would require us to know a great deal about the author's calculations. rather than attempt to redo the actuarial calculations performed to make the chart and assign this to the individuals in the table we should rather explain the concepts behind the 5% and 95% and preserve the intention of actuarial information as applying to demographic groups. 5% of people in the demographic the author selected live to _ age 95% of those people live to _ age and how this affects our subject population. Mrarch (talk) 21:43, 6 December 2013 (UTC)
Why is this explanation incomplete? The second paragraph does a good job explaining what the 5th percentile and 95th percentile are referring to. String userName = new String(); (talk) 23:35, 19 June 2015 (UTC)
I prefer to think of the inhabitable planets as extensions to earth reserved for when we have learned not to kill all the inhabitants of the only inhabited planet in the universe.
I used Google News BEFORE it was clickbait (talk) 22:39, 23 January 2015 (UTC)
I see no reason this is marked as incomplete; I've tidied up the percentile explanations, but haven't really added much more. I think it's fine, and will remove the incomplete tag in a few days if nobody objects. Cosmogoblin (talk) 13:53, 24 June 2015 (UTC)
UPDATED GRAPH: I've updated the image with a red line showing actual moon walker deaths. View here: [1]. Sadly, it's right on track. 172.68.58.59 22:19, 9 August 2018 (UTC)
As of mid April, 2020, this prediction is still accurate, but I'm really scared of what it'll be by the end of 2020 or 2021. Stay healthy everyone, astronaut or not! PotatoGod (talk) 07:04, 22 April 2020 (UTC)
Interesting that 6/12 of all the people who walked on the moon were born in 1930, and all bar Alan Shepard was born 1930-1935. Reminds me of some of the ideas in Malcolm Gladwell's *Outliers* about there being especially good birth years to succeed at high levels in given fields. It seems you want to have been mid-30s to early-40s (Shephard the outlier at 47) in the late 60s/early 70s. This also makes the comic more dramatic - if there had been a wider spread of ages, then the "death curve" would be a lot more gradual. -Honeypuppy (talk) 01:15, 30 September 2020 (UTC) What? No. That's a false correlation. The moon program took place over a very short span of time, and was looking for very specific qualification. Including age.
Honorary mention: Michael Collins (1930-2021), RIP this date. 162.158.158.104 17:44, 28 April 2021 (UTC)
Also Thomas Stafford (1930-2024). 172.71.242.5 17:10, 19 March 2024 (UTC)
IMO the saddest part isn't astronauts dying - it's lack of any new people getting to walk on another planet. 162.158.90.157 14:04, 5 October 2021 (UTC)
Well, if the Artemis missions go as planned, the count might soon be increasing again for the first time in fifty-three years. Hopefully, not all of the remaining veteran astronauts will have died by then. --162.158.22.16 22:57, 4 August 2023 (UTC)
But with increasing delays (now until 2026) the window of time is closing. 172.71.242.43 17:03, 19 March 2024 (UTC) | HuggingFaceTB/finemath | |
Engineering Calculus Notes 69
# Engineering Calculus Notes 69 - 57 1.5 PLANES 6 Consider...
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: 57 1.5. PLANES 6. Consider the line ℓ in the plane defined as the locus of the linear equation in the two rectangular coordinates x and y Ax + By = C. Define the vector − → → → N = A− + B − . ı → (a) Show that ℓ is the set of points P whose position vector − p satisfies →→ −− N · p = C. → (b) Show that if − 0 is the position vector of a specific point on the p → line, then ℓ is the set of points P whose position vector − p satisfies →→→ −−− N · ( p − p 0 ) = 0. → − (c) Show that N is perpendicular to ℓ. 7. Show that if ℓ is a line given by Ax + By = C then the distance from a point Q(x, y ) to ℓ is given by the formula dist(Q, ℓ) = |Ax + By − C | √ . A2 + B 2 (1.20) → − → p (Hint: Let N be the vector given in Exercise 6, and − 0 the position vector of any point P0 on ℓ. Show that −→ − → → →→ dist(Q, ℓ) = proj− P0 Q = proj− (− − − 0 ) , and interpret this in q p N N terms of A, B , C , x and y .) 1.5 Planes Equations of Planes We noted earlier that the locus of a “linear” equation in the three rectangular coordinates x, y and z Ax + By + Cz = D (1.21) ...
View Full Document
## This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
Ask a homework question - tutors are online | HuggingFaceTB/finemath | |
## The distribution of the amount of money spent by students for textbooks in a semester is approximately normal in shape with a mean of \$240 a
Question
The distribution of the amount of money spent by students for textbooks in a semester is approximately normal in shape with a mean of \$240 and a standard deviation of \$25. According to the Standard Deviation Rule, in a semester, almost all (99.7%) of the students spent on textbooks in a semester:_________.
in progress 0
2 months 2021-10-13T15:41:41+00:00 1 Answer 0 views 0
1. The final part of the question is asking;
How much did all (99.7%) of the students spend on textbooks in a semester
almost all (99.7%) of the students spent between \$165 and \$315 on textbooks in a semester.
Step-by-step explanation:
The standard deviation rule describes to us that for distributions that have the normal shape, approximately 99.7% of the observations fall within 3 standard deviations of the mean.
In this question, we are given that; Mean = 240 and Standard deviation= 25
So, 3 standard deviation below the mean = Mean – 3(standard deviation)
= 240 – (3 × 25)
= 240 – 75 = 165
Now, 3 standard deviation above the mean = Mean + 3 standard deviation = 240 + (3 × 25)
= 240 + 75 = 315
So, almost all (99.7%) of the students spent between \$165 and \$315 on textbooks in a semester. | HuggingFaceTB/finemath | |
# 642.9 seconds in minutes
## Result
642.9 seconds equals 10.72 minutes
You can also convert 642.9 seconds to minutes and seconds.
## Conversion formula
Multiply the amount of seconds by the conversion factor to get the result in minutes:
642.9 s × 0.0166667 = 10.72 min
## How to convert 642.9 seconds to minutes?
The conversion factor from seconds to minutes is 0.0166667, which means that 1 seconds is equal to 0.0166667 minutes:
1 s = 0.0166667 min
To convert 642.9 seconds into minutes we have to multiply 642.9 by the conversion factor in order to get the amount from seconds to minutes. We can also form a proportion to calculate the result:
1 s → 0.0166667 min
642.9 s → T(min)
Solve the above proportion to obtain the time T in minutes:
T(min) = 642.9 s × 0.0166667 min
T(min) = 10.72 min
The final result is:
642.9 s → 10.72 min
We conclude that 642.9 seconds is equivalent to 10.72 minutes:
642.9 seconds = 10.72 minutes
## Result approximation:
For practical purposes we can round our final result to an approximate numerical value. In this case six hundred forty-two point nine seconds is approximately ten point seven two minutes:
642.9 seconds ≅ 10.72 minutes
## Conversion table
For quick reference purposes, below is the seconds to minutes conversion table:
seconds (s) minutes (min)
643.9 seconds 10.731688 minutes
644.9 seconds 10.748355 minutes
645.9 seconds 10.765022 minutes
646.9 seconds 10.781688 minutes
647.9 seconds 10.798355 minutes
648.9 seconds 10.815022 minutes
649.9 seconds 10.831688 minutes
650.9 seconds 10.848355 minutes
651.9 seconds 10.865022 minutes
652.9 seconds 10.881688 minutes
## Units definitions
The units involved in this conversion are seconds and minutes. This is how they are defined:
### Seconds
The second (symbol: s) (abbreviated s or sec) is the base unit of time in the International System of Units (SI). It is qualitatively defined as the second division of the hour by sixty, the first division by sixty being the minute. The SI definition of second is "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom". Seconds may be measured using a mechanical, electrical or an atomic clock. SI prefixes are combined with the word second to denote subdivisions of the second, e.g., the millisecond (one thousandth of a second), the microsecond (one millionth of a second), and the nanosecond (one billionth of a second). Though SI prefixes may also be used to form multiples of the second such as kilosecond (one thousand seconds), such units are rarely used in practice. The more common larger non-SI units of time are not formed by powers of ten; instead, the second is multiplied by 60 to form a minute, which is multiplied by 60 to form an hour, which is multiplied by 24 to form a day. The second is also the base unit of time in other systems of measurement: the centimetre–gram–second, metre–kilogram–second, metre–tonne–second, and foot–pound–second systems of units.
### Minutes
The minute is a unit of time or of angle. As a unit of time, the minute (symbol: min) is equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). As a unit of angle, the minute of arc is equal to 1⁄60 of a degree, or 60 seconds (of arc). Although not an SI unit for either time or angle, the minute is accepted for use with SI units for both. The SI symbols for minute or minutes are min for time measurement, and the prime symbol after a number, e.g. 5′, for angle measurement. The prime is also sometimes used informally to denote minutes of time. In contrast to the hour, the minute (and the second) does not have a clear historical background. What is traceable only is that it started being recorded in the Middle Ages due to the ability of construction of "precision" timepieces (mechanical and water clocks). However, no consistent records of the origin for the division as 1⁄60 part of the hour (and the second 1⁄60 of the minute) have ever been found, despite many speculations. | HuggingFaceTB/finemath | |
# Math homework help -factors, primes andcomposites, prime factorization
Students send me scans (or cell phone pictures) of their math homework with their own answers written down. I review and rectify mistakes, if any so that they have all correct answers by the time they turn them in to their school teachers. I also make efforts to explain the concepts/steps briefly.
Note that this is a paid service but costs significantly less than live tutoring sessions. Here are a few examples from pre algebra to show what I receive and what I send back. You can use these solutions as solved examples to learn on your own. Don't miss out on the tips and tricks, if any.
#### Word problems related to factors What I received What I sent back click on pictures to enlarge
#### Prime factors and exponents What I received What I sent back click on pictures to enlarge
#### Prime factorization with exponents, primes and composites What I received What I sent back click on pictures to enlarge
#### Primes and composites What I received What I sent back click on pictures to enlarge
Tip: A quick way to figure out if 3 can divide a number without leaving a remainder is to add up all the digits in the number and check if 3 can divide the sum obtained without leaving a remainder.
For example, to check if 3 can go into 93, add 9 with 3. Since 9 + 3 = 12 and 3 can divide 12, 3 can also divide 93 without leaving a remainder.
Similarly, 3 can't divide 145 because 1+4+5 = 10 and 3 can't divide 10.
In almost the same way, if you want to quickly find out if 9 can divide a number without leaving a remainder, add up the digits in the number. If 9 can divide the sum thus obtained, then 9 can also divide the given number.
For example, to check if 9 can go into 456, do 4 + 5 + 6 = 15. Since 9 can't divide 15, it can't divide 456 either.
Note: The tricks mentioned above work only for 3 and 9. Never try to apply these rules to find out if any other number can divide a given number. For example, if you want to find out if 4 can divide 124, you don't add 1, 2 and 4. The divisibility rule for 4 is different. We can talk about that on another day. | HuggingFaceTB/finemath | |
What are the zeros of the polynominal?
If we can reduce this to a quadratic by finding an easy factor, then we can find the zeroes. If we substitute a set of values of x and work out the result for p(x), we may be able to work out a factor. Let's start with x=0, then p(0)=60. Let's try x=1. p(1)=36 which is smaller than 60, so we're on track for zero. p(2)=14, smaller again. p(3)=27-18-69+60=0. Bingo! So (x-3) is a factor. Divide the cubic by (x-3) using synthetic division and we get x^2+x-20 which, surprise, surprise, factorises into (x+5)(x-4). We have th three zeroes: x=-5, 3 and 4. Let's check out -5: -125-50+115+60=0. Success!
by Top Rated User (720k points) | HuggingFaceTB/finemath | |
# Really challenging puzzle
1. Nov 20, 2005
### theonerester
this is a puzzle from highiqsociety.org
i've been looking at this for a very long time, can anyone break this?
http://www.geocities.com/invinbotmail1/hd3
Last edited: Nov 20, 2005
2. Nov 21, 2005
### The Sphinx
very difficult :D
3. Nov 21, 2005
### neo143
The missing number is 27
Add all the numbers = 757
And this is on a quare shown
so 784 = 27* 27
4. Nov 21, 2005
### The Sphinx
You make some assumption, I think. :)
5. Nov 21, 2005
### ahrkron
Staff Emeritus
27*27 = 729, not 784.
28*28 = 784.
However, you could also say that the area of the missing piece is 84, so that the total area is 841 (=29*29).
6. Nov 21, 2005
### neo143
sorry Mr. ahrkron,
You are right......it could be 28 or 84......
7. Nov 21, 2005
### scott_alexsk
It seems like it is just to confuse you with the shapes within the square. I was able to set the square up into a grid. With the grid( I may be pulling at straws here) it seems like the answer would be 45.
-Scott
8. Nov 29, 2005
### serg
Last edited: Nov 30, 2005
9. Nov 30, 2005
### DaveC426913
or 143 (30^2-757)
or 204 (31...)
or 267 (32...)
or...
Sorry, no dice. These problems have unambiguous answers.
10. Dec 2, 2005
### Cybersteve
My guess is the numbers are related to the angles that the individual shapes contain. A different value for 90 degrees, more than 90, or less than 90.
Possibly the number is then subtracted from or added to a constant.
I
11. Dec 2, 2005
### DaveC426913
It think the shapes are arbitrary. I think the only correlation is the numbers and their neighbours. There are some tantalizing correlations when factoring the numbers, but nothing jumps out.
12. Dec 3, 2005
### Ubern0va
Ok here is the approach I took. I have too many Math finals coming up anyways so I won't be solving this myself. Anyone who knows linear algebra is welcome to try this, I have no clue if it will work. But here it goes:
After playing with the shapes and sums of numbers I realized that the numbers don't appear to describe the shapes. So I took the following outlook on the situation. Assume that the shapes represent provinces. Each province produces the same amount of rice each year, no matter its size.
This means that if the barriers changed and the provinces moved around, that, the amount of rice produced would stay the same. I then realized that the shapes can basically be anything you want them to be, because they aren't representative of the numbers (as long as there are 16 of them).
I decided to redraw the big square into 16 equal squares (since there are 16 shapes). I put the numbers randomly into any square (imagine the provinces fighting over land and moving around but still producing the same amount of rice.
Then it dawned on me that if x were to be the number behind the question mark, r were to be the amount of (lets say) rice prduced by all provinces including x, and y would be some value that is a common sum within the rows/columns of my grid, then solving the puzzle is just a matter of finding three equations in three variables (with no two equations alike).
I took [r = 757 + x], [r - x = 3y + (y - x)], and I haven't actually figured out the equation relating y and r. I'm thinking along the lines of [4r = y/4].
However, if the relation [y] between the numbers is incorrect, then the answer you get from solving the system would be correct but not the one they are looking for. Though if you can actually find a [y] variable which every row/column sum to, I'd say you're on the right track.
Graphic Illustraction [of my attempt]
Anyways, give it a try if you are good at this stuff, my brain will be overloaded from Math in a few days anyways so I'm gonna lay off of it for now. :zzz:
Last edited: Dec 3, 2005
13. Dec 4, 2005
### croxbearer
r = 804
y = 201
x = 47
here is the matrix i made:
74 66 49 57
42 33 62 41
55 42 47 56
30 60 43 47
14. Dec 4, 2005
### Ubern0va
Thanks, I was really hoping someone would pick that up. Does anyone know if it's correct?
15. Dec 4, 2005
### Cybersteve
Isn't [r - x = 3y + (y - x)] just r=4y?
So if 47 was a valid solution then so would 51 be. You'd just have to change the values of r and y.
r=808
y=202
Or am I misunderstanding you?
16. Dec 4, 2005
### Ubern0va
Yes that's the simplified formula, but I wanted to be clear on how that formula came about, hence the sort of redundant equation.
As for changing [y] and [r]: Then its just a random answer. My approach was based on the idea that all the columns add to a certain value, that value, as we now see, is 201. If you made [y] 202 then 3 of the columns would add to 201 and the last one would add to 202. Thus defeating the entire purpose of the [y] variable. | HuggingFaceTB/finemath | |
Solving system of linear equations
Solution:
k+s = 21
9s + 12k = 213
k+s = 21
9•s + 12•k = 213
k+s = 21
12k+9s = 213
k = 8
s = 13
Write each equation on a new line or separate it by a semicolon. The online calculator solves a system of linear equations (with 1,2,...,n unknowns), quadratic equation with one unknown variable, cubic equation with one unknown variable, and finally, any other equation with one variable. Even if an exact solution does not exist, it calculates a numerical approximation of roots.
Examples:
```a+b = 12
a-3b = a-b+43```
```x+y+z=100
3x-6y+2z=50
y-3z+x=(44-22)x+45```
`(x+4)(x-3)+34x+6x^2 = 256`
`(x+4)+34x+x^2-x^3 = -32`
`ln x = 1.2-x`
`sin x = cos(x-pi/3)+x`
`sin x = x^3+2x+x-1`
`sin x = 0.5`
`(cos x)^2 = tan(x-pi/3)`
`x^5+x^4 = -23+sin x`
`|log x|=2`
`|ln x+1|=x/10`
`|ln x+1|=ln x^2`
`1/(x+2)=1/x+4`
`sqrt(x^4+x^2+2)=22`
Our mission:
Provide simple, fast, and reliable mathematical service for solving any equation(s).
More info:
Unknowns (variables) write as one character a-z, i.e., a, b, x, y, z. No matter whether you want to solve an equation with a single unknown, a system of two equations of two unknowns, the system of three equations and three unknowns, or a linear system with twenty unknowns. The number of equations and the number of unknowns should be equal, and the equation should be linear (and linear independent). Then you can be expected that the equations have one solution.
It is not necessary to write equations in the basic form. The calculator quickly performs equivalent operations on the given linear system. | HuggingFaceTB/finemath | |
#### What is 7790 percent of 198?
How much is 7790 percent of 198? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 7790% of 198 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 7790% of 198 = 15424.2
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating seven thousand, seven hundred and ninety of one hundred and ninety-eight How to calculate 7790% of 198? Simply divide the percent by 100 and multiply by the number. For example, 7790 /100 x 198 = 15424.2 or 77.9 x 198 = 15424.2
#### How much is 7790 percent of the following numbers?
7790 percent of 198.01 = 1542497.9 7790 percent of 198.02 = 1542575.8 7790 percent of 198.03 = 1542653.7 7790 percent of 198.04 = 1542731.6 7790 percent of 198.05 = 1542809.5 7790 percent of 198.06 = 1542887.4 7790 percent of 198.07 = 1542965.3 7790 percent of 198.08 = 1543043.2 7790 percent of 198.09 = 1543121.1 7790 percent of 198.1 = 1543199 7790 percent of 198.11 = 1543276.9 7790 percent of 198.12 = 1543354.8 7790 percent of 198.13 = 1543432.7 7790 percent of 198.14 = 1543510.6 7790 percent of 198.15 = 1543588.5 7790 percent of 198.16 = 1543666.4 7790 percent of 198.17 = 1543744.3 7790 percent of 198.18 = 1543822.2 7790 percent of 198.19 = 1543900.1 7790 percent of 198.2 = 1543978 7790 percent of 198.21 = 1544055.9 7790 percent of 198.22 = 1544133.8 7790 percent of 198.23 = 1544211.7 7790 percent of 198.24 = 1544289.6 7790 percent of 198.25 = 1544367.5
7790 percent of 198.26 = 1544445.4 7790 percent of 198.27 = 1544523.3 7790 percent of 198.28 = 1544601.2 7790 percent of 198.29 = 1544679.1 7790 percent of 198.3 = 1544757 7790 percent of 198.31 = 1544834.9 7790 percent of 198.32 = 1544912.8 7790 percent of 198.33 = 1544990.7 7790 percent of 198.34 = 1545068.6 7790 percent of 198.35 = 1545146.5 7790 percent of 198.36 = 1545224.4 7790 percent of 198.37 = 1545302.3 7790 percent of 198.38 = 1545380.2 7790 percent of 198.39 = 1545458.1 7790 percent of 198.4 = 1545536 7790 percent of 198.41 = 1545613.9 7790 percent of 198.42 = 1545691.8 7790 percent of 198.43 = 1545769.7 7790 percent of 198.44 = 1545847.6 7790 percent of 198.45 = 1545925.5 7790 percent of 198.46 = 1546003.4 7790 percent of 198.47 = 1546081.3 7790 percent of 198.48 = 1546159.2 7790 percent of 198.49 = 1546237.1 7790 percent of 198.5 = 1546315
7790 percent of 198.51 = 1546392.9 7790 percent of 198.52 = 1546470.8 7790 percent of 198.53 = 1546548.7 7790 percent of 198.54 = 1546626.6 7790 percent of 198.55 = 1546704.5 7790 percent of 198.56 = 1546782.4 7790 percent of 198.57 = 1546860.3 7790 percent of 198.58 = 1546938.2 7790 percent of 198.59 = 1547016.1 7790 percent of 198.6 = 1547094 7790 percent of 198.61 = 1547171.9 7790 percent of 198.62 = 1547249.8 7790 percent of 198.63 = 1547327.7 7790 percent of 198.64 = 1547405.6 7790 percent of 198.65 = 1547483.5 7790 percent of 198.66 = 1547561.4 7790 percent of 198.67 = 1547639.3 7790 percent of 198.68 = 1547717.2 7790 percent of 198.69 = 1547795.1 7790 percent of 198.7 = 1547873 7790 percent of 198.71 = 1547950.9 7790 percent of 198.72 = 1548028.8 7790 percent of 198.73 = 1548106.7 7790 percent of 198.74 = 1548184.6 7790 percent of 198.75 = 1548262.5
7790 percent of 198.76 = 1548340.4 7790 percent of 198.77 = 1548418.3 7790 percent of 198.78 = 1548496.2 7790 percent of 198.79 = 1548574.1 7790 percent of 198.8 = 1548652 7790 percent of 198.81 = 1548729.9 7790 percent of 198.82 = 1548807.8 7790 percent of 198.83 = 1548885.7 7790 percent of 198.84 = 1548963.6 7790 percent of 198.85 = 1549041.5 7790 percent of 198.86 = 1549119.4 7790 percent of 198.87 = 1549197.3 7790 percent of 198.88 = 1549275.2 7790 percent of 198.89 = 1549353.1 7790 percent of 198.9 = 1549431 7790 percent of 198.91 = 1549508.9 7790 percent of 198.92 = 1549586.8 7790 percent of 198.93 = 1549664.7 7790 percent of 198.94 = 1549742.6 7790 percent of 198.95 = 1549820.5 7790 percent of 198.96 = 1549898.4 7790 percent of 198.97 = 1549976.3 7790 percent of 198.98 = 1550054.2 7790 percent of 198.99 = 1550132.1 7790 percent of 199 = 1550210 | HuggingFaceTB/finemath | |
What is the probability of getting 7 in a dice?
Contents
Is the probability of an event that 7 comes up during tossing dice?
7 is always 16 probability.
What is the probability of not having 7 in dice?
Explanation: When rolling 2 dice there are 36 possible outcomes. [to see this imagine one die is red and the other green; there are 6 possible outcomes for the red die and for each of these red outcomes there are 6 possible green outcomes]. That is 30 out of 36 outcomes will not be a 7 total.
What is the probability of getting 7 in a single throw of dice?
Answer: There is no 7 in dice at all. So, 0/6 is the Probability of getting a 7 in a single throw of a dice.
What is the probability that the sum 7 appears in a single toss of a pair of fair dice?
(d)Probability of getting a sum of 7 when a pair of fair dice tossed = 6/36 because (1,6) (2,5) (3,4) (4,3) (5,2) and (6,1) are favourable outcomes Total number of possible outcomes = 6^2 = 36 Probability of getting a sum of 11 when a pair of fair dice tossed = 2/36 because (5,6) and (6,5) are favourable outcomes …
Is the probability of rolling a 7 with two dice?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
THIS IS FUNNING: Can I make crushed tomatoes out of diced tomatoes?
How do you find the probability of two dice?
If you want to know how likely it is to get a certain total score from rolling two or more dice, it’s best to fall back on the simple rule: Probability = Number of desired outcomes ÷ Number of possible outcomes. | HuggingFaceTB/finemath | |
#### Provide solution for RD sharma maths class 12 chapter 21 Differential Equation exercise 21.1 question 13
Order=1, Degree=1, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\left ( xy^{2}+x \right )dx+\left ( y-x^{2}y \right )=0$
Solution:
The above equation can be written as
$x\left ( y^{2}+1 \right )dx=y\left ( x^{2}-1\right )dy\\ \frac{dy}{dx}\left ( x^{2}-1 \right )=x\left ( y^{2}+1 \right )\\ \frac{dy}{dx}x^{2}y-\frac{dy}{dx}y=xy^{2}+x$
Here in this question, the order of the differential equation is 1 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by y and also y is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=1, Degree=1, Non-linear | HuggingFaceTB/finemath | |
$$\require{cancel}$$
# 1.7: Absolute Magnitude
The subject of magnitude scales in astronomy is an extensive one, which is not pursued at length here. It may be useful, however, to see how magnitude is related to flux and intensity. In the standard usage of the word flux, in the sense that we have used it hitherto in this chapter, flux is related to absolute magnitude or to intensity, according to
$M_2 - M_1 = 2.5 \log (\Phi_1/\Phi_2) \label{1.7.1}$
or
$M_2 - M_1 = 2.5 \log (I_1/I_2) \label{1.7.2}$
That is, the difference in magnitudes of two stars is related to the logarithm of the ratio of their radiant fluxes or intensities.
If we elect to define the zero point of the magnitude scale by assigning the magnitude zero to a star of a specified value of its radiant flux in watts or intensity in watts per steradian, equations $$\ref{1.7.1}$$ and $$\ref{1.7.2}$$ can be written
$M = M_0 - 2.5 \log \Phi \label{1.7.3}$
or to its intensity by
$M = {M_0}' - 2.5 \log I \label{1.7.4}$
If by $$\Phi$$ and $$I$$ we are referring to flux and intensity integrated over all wavelengths, the absolute magnitudes in equations $$\ref{1.7.1}$$ to $$\ref{1.7.4}$$ are referred to as absolute bolometric magnitudes. Practical difficulties dictate that the setting of the zero points of the various magnitude scales are not quite as straightforward as arbitrarily assigning numerical values to the constants $$M_0$$ and $${M_0}'$$ and I do not pursue the subject further here, other than to point out that $$M_0$$ and $${M_0}'$$ must be related by
${M_0}' = M_0 -2.5 \log 4 \pi = M_0 - 2.748. \label{1.7.5}$ | HuggingFaceTB/finemath | |
0
# Can a quadrilateral be a rectangle if the sides are not equal?
Updated: 12/8/2022
Wiki User
11y ago
Yes, but there must be two pairs of opposite sides that are equal.
Yes, but there must be two pairs of opposite sides that are equal.
Yes, but there must be two pairs of opposite sides that are equal.
Yes, but there must be two pairs of opposite sides that are equal.
Wiki User
11y ago
Wiki User
11y ago
Yes, but there must be two pairs of opposite sides that are equal.
Earn +20 pts
Q: Can a quadrilateral be a rectangle if the sides are not equal?
Submit
Still have questions?
Related questions
### Quadrilateral that is not a rectangle?
A quadrilateral that is not a rectangle is a kite. A quadrilateral is a figure that has four straight sides. A rectangle has opposites sides that are parallel and of equal measurement. However, a kite is a quadrilateral by definition but is unlike a rectangle because it has pairs of equal adjacent sides.
A rectangle
### What is a shape that doesnt have equal sides?
Shapes that do not have equal sides is called a quadrilateral. They include an irregular quadrilateral, rectangle, or a parallelogram.
### Which quadrilateral has equal opposite sides?
Rectangle Parallelogram
### What quadrilateral has two equal sides?
rectangle, parallelogram
a rectangle
### If a quadrilateral is a rectangle then it is a equilateral?
Not necissarily. Quadrilateral simply says it has 4 sides. A rectangle has perpendicular sides. Equilateral means that the sides are of equal length (like a square, which is a special case rectangle)
rectangle
### When is a quadrilateral a rectangle but not a square?
When all the vertices of a plane quadrilateral are exactly equal to 90 degrees it is is called a rectangle and, if all its sides are of equal length, that rectangle is called a square. So when a plane quadrilateral has all its vertices exactly equal to 90 degrees, but it does not have all its sides equal in length, it must be a rectangle but not a square.
Rectangle
A rectangle.
### Can a quadrilateral be a rectangle and a square?
Yes. A quadrilateral is any plane figure bounded by four straight lines. A rectangle is a quadrilateral with each of its vertex angles equal to 90°, opposite sides are parallel and of equal length but the pairs of opposite sides are of different lengths. A square is a rectangle with all its sides equal in length. | HuggingFaceTB/finemath | |
# Determining if a system of equations has an all negative solution
Consider the equation $f(x) = cx$. It can be seen that there is some $x$ such that $f(x)<0$ iff $c\not=0$.
If we write this system of one equation as a matrix, it's just the matrix $M=(c)$. So in this case, there is a negative assignment to $f$ iff $det(M)\not=0$.
I read something which seemed to indicate that this is true more generally: If you have $n$ equations over $n$ variables, there will be an assignment such that all $f$s are $<0$ iff $det(M)\not=0$.
I'm having trouble proving this even in the 2x2 case, much less in general.
1. Is there a name for this theorem? (If it's true)
2. Any hints on how to [dis]prove it?
EDIT: The paper I'm looking at is Quantum probabilities as Bayesian probabilities. On page 2, equation (2), they say "The bookie can choose values xA , xB , and xC that lead to R < 0 in all three cases unless [$\det(M)=0$]". Maybe I have misinterpreted their argument, but it seems like they're saying this is a general property of the determinant.
-
Do you mean all the components of the vector $Mx$ negative? – matgaio May 1 '12 at 0:37
@matgaio: yes, exactly – Xodarap May 1 '12 at 0:45
Just to argue geometrically: we know that a non-singular matrix take $S^{n-1}\subset\mathbb{R}^{n}$ to an $n-1$-dimensional ellipsoid. centered at the origin (possibly with axes non-aligned to the canonical axes). Once this ellipsoid intersect the orthant with all coordinates negative, then there exist some $x$ with $Mx<0$. – matgaio May 1 '12 at 1:01
I take it you have $n$ linear forms in $n$ variables, $$f_i=a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n,\qquad1\le i\le n$$ and you want a condition to ensure that there exist $x_1,\dots,x_n$ such that $f_i(x_1,\dots,x_n)\lt0$ for all $i$. Letting $M$ be the matrix whose entries are the $a_{ij}$, it is certainly true that if $\det M\ne0$ then such $x_i$ exist, for if $\det M\ne0$ then the column space of $M$ is all of ${\bf R}^n$, including that part of ${\bf R}^n$ where all components are negative. However, even if $\det M=0$, such $x_i$ might exist, e.g., consider $f_1(x,y)=f_2(x,y)=x$. I don't know if there's any simple condition on $M$ which causes exclusion of the negative orthant.
-
This certainly seems to be in line with the edit of OP's question and the quote from the paper - it's an if statement, not an iff. – process91 May 1 '12 at 2:02
@process91: good point, it's if not iff. This makes it a lot simpler, thanks. – Xodarap May 1 '12 at 12:56 | HuggingFaceTB/finemath | |
## Conversion formula
The conversion factor from feet to yards is 0.33333333333333, which means that 1 foot is equal to 0.33333333333333 yards:
1 ft = 0.33333333333333 yd
To convert 232 feet into yards we have to multiply 232 by the conversion factor in order to get the length amount from feet to yards. We can also form a simple proportion to calculate the result:
1 ft → 0.33333333333333 yd
232 ft → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 232 ft × 0.33333333333333 yd
L(yd) = 77.333333333333 yd
The final result is:
232 ft → 77.333333333333 yd
We conclude that 232 feet is equivalent to 77.333333333333 yards:
232 feet = 77.333333333333 yards
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 0.012931034482759 × 232 feet.
Another way is saying that 232 feet is equal to 1 ÷ 0.012931034482759 yards.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred thirty-two feet is approximately seventy-seven point three three three yards:
232 ft ≅ 77.333 yd
An alternative is also that one yard is approximately zero point zero one three times two hundred thirty-two feet.
## Conversion table
### feet to yards chart
For quick reference purposes, below is the conversion table you can use to convert from feet to yards
feet (ft) yards (yd)
233 feet 77.667 yards
234 feet 78 yards
235 feet 78.333 yards
236 feet 78.667 yards
237 feet 79 yards
238 feet 79.333 yards
239 feet 79.667 yards
240 feet 80 yards
241 feet 80.333 yards
242 feet 80.667 yards | HuggingFaceTB/finemath | |
×
Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 11.4 - Problem 5
Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 11.4 - Problem 5
×
# Get answer: Cross Product of Unit Vectors In Exercises 16, find the cross product of the
ISBN: 9781285774770 141
## Solution for problem 5 Chapter 11.4
Calculus: Early Transcendental Functions | 6th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Calculus: Early Transcendental Functions | 6th Edition
4 5 1 416 Reviews
19
2
Problem 5
In Exercises 1-6, find the cross product of the unit vectors and sketch your result.
$$\mathbf{i} \times \mathbf{k}$$
Text Transcription:
i x k
Step-by-Step Solution:
Step 1 of 3
Week 4 MUL2010 Intro to Music Lit ▯1 Introduction to Music Literature ♪ WEEK 4 NOTES ♪ ▯ Main Theme - HARMONY AND KEY Harmony • organization of chords ⁃ chord: a combination of three or more notes sounded simultaneously ⁃ triad - fundamental chord group in music, consisting of three tones ⁃ root: bottom of chord ⁃ 3rd, 5th respective relations of other notes to chord ⁃ TYPES of TRIADS ⁃ tonic: first note of scale, root of tonic chord. Strongest feeling of rest and stability. ⁃ dominant: 5th note of scale, root of dominant chord. consonant but weaker than tonic, often leads to tonic. ⁃ V - I cadence is the most conclusive cadence in western music. ⁃ progression: a group of chords in succession, usually indication motion of some sort ⁃ cadences - resolutions at ends of musical phrases that mark a rest or pause before the next passage ⁃ IAC/PAC - Authentic cadences - V or vii chords that lead to the tonic. Voice
Step 2 of 3
Step 3 of 3
##### ISBN: 9781285774770
The full step-by-step solution to problem: 5 from chapter: 11.4 was answered by , our top Calculus solution expert on 11/14/17, 10:53PM. Since the solution to 5 from 11.4 chapter was answered, more than 325 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Calculus: Early Transcendental Functions, edition: 6. This full solution covers the following key subjects: . This expansive textbook survival guide covers 134 chapters, and 10738 solutions. The answer to “?In Exercises 1-6, find the cross product of the unit vectors and sketch your result.$$\mathbf{i} \times \mathbf{k}$$Text Transcription:i x k” is broken down into a number of easy to follow steps, and 20 words. Calculus: Early Transcendental Functions was written by and is associated to the ISBN: 9781285774770.
## Discover and learn what students are asking
Calculus: Early Transcendental Functions : Basic Differentiation Rules and Rates of Change
?Finding a Value In Exercises 65–70, find k such that the line is tangent to the graph of the function. Function
Statistics: Informed Decisions Using Data : The Randomized Complete Block Design
?Given the following ANOVA output, answer the questions that follow: (a) The researcher wants to test H0: m1 = m2 = m3 =
#### Related chapters
Unlock Textbook Solution | HuggingFaceTB/finemath | |
+0
# Help. Thank you.
+3
150
3
+306
Find the constant coefficient when the polynomial \(3(x - 4) + 2(x^2 - x + 7) - 5(x - 1)\) is simplified.
Feb 27, 2021
### 3+0 Answers
#1
+2202
0
The constant coefficient is the number with x^0.
So, we just need to find those.
3(x-4) has the -12. (3*-4)
2(x^2-x+7) has a 14. (2*7)
-5(x-1) has a 5. (-5*-1)
Now, we just sum everything together. :))
-12+14+5 = 7
I hope this helped.
=^._.^=
Feb 27, 2021
#2
+306
+3
Thank you!
calvinbun Feb 27, 2021
#3
+2202
0
You're welcome. :))
=^._.^=
catmg Feb 27, 2021 | HuggingFaceTB/finemath | |
# What is the probability both dice land on the same number?
Contents
## What is the probability of rolling 2 dice?
Two (6-sided) dice roll probability table
Roll a… Probability
2 1/36 (2.778%)
3 2/36 (5.556%)
4 3/36 (8.333%)
5 4/36 (11.111%)
## When 2 dice are rolled find the probability of getting?
Probabilities for the two dice
Total Number of combinations Probability
10 3 8.33%
11 2 5.56%
12 1 2.78%
Total 36 100%
## What is the probability of getting doublet when two dice are thrown simultaneously?
Probability of getting a doublet = 6/36 = 1/6.
## What is the probability of getting twin prime numbers when two dice are thrown at a time?
Hence probability of getting a prime number on both dice is 1/4.
## What are all the possible outcomes of rolling 2 dice?
Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36.
## What is the probability of rolling 2 sixes in a row?
The probability of rolling two sixes on two rolls is 16 as likely as one six in one roll.
THIS IS IMPORTANT: Does Missouri have a cap on casinos? | HuggingFaceTB/finemath | |
# The Point Of The Banach Tarski Theorem Subscribe! My latest posts can be found here:
Previous blog posts:
Additionally, some earlier writings:
This page has been Tagged As Maths.
# The Point Of The Banach-Tarski Theorem - 2015/06/06
One of my favourite Limited Audience Jokes (although technically it's actually a riddle) goes:
Added after seeing some discussion ... This is not intended to explain why the Banach-Tarski theorem is true, nor to give a proof, nor to talk about what the pieces look like, etc. The purpose of this article is to explain why the result is relevant. In short, it tells us what is and is not possible in measure theory. It tells us that things we would in general want to be true of a measure are mutually inconsistent. Now read on ...
• Q: What's an anagram of Banach-Tarski?
• A: Banach-Tarski Banach-Tarski
Now, if you already know what the Banach-Tarski theorem says, that riddle is really funny. If you don't, then you're simply not in the audience, and you'll just go: "Huh?"
Which is a perfectly reasonable reaction. Indeed, if you then have the Banach-Tarski theorem explained to you, you most likely will still go: "Huh?" It's a perplexing result, some even call it a paradox, but the fact that it's so odd actually masks the fact that it's a truly important result, with some deep implications.
So I'm going to explain here why it's important, and what some of the implications are. The hope is that even if you do know the result you will find this interesting, largely because in the shock of seeing what the result says, you've never been shown why it's interesting.
So for those of you who don't know the result, here it is in simple, non-technical terms:
• In $\Re^3$, given a solid ball $B$ of radius $R,$
it is possible to partition $B$ into finitely many
pieces such that those pieces can be reassembled
to form two solid balls $B_1$ and $B_2,$ each of
radius $R.$
Apart from using the Axiom of Choice.
Now, that's obvious nonsense, and that's why the result is so shocking. It defies common sense, and immediately makes you go looking for some kind of loophole. But there isn't one.
So in part because it's so surprising, and shocking, and nonsensical, you might think it's in a dead-end of mathematics and of no real use or interest. That's what this article is intending to address.
To do so, let's consider the idea of measurement. We can talk about the length of a line, the area of a polygon (or other shape), the volume of a lump, or whatever. The development of the concept of accurate measurement goes back millennia, and is critical in the development of commerce, engineering, and so many other things. So we're going to look at the concept of measuring something, and see what we can say about it mathematically.
To do that we'll talk about a function called a "measure". One of the problems in maths is we use ordinary words in a technical sense, so it's a bit dangerous to call this thing a "measure," but I'll try to use that word only ever in its technical sense.
And fairly obviously we'll need a different function, or measure, when we're in one dimension as compared with two dimensions, or more, so we'll talk about a measure for each dimension.
So a "measure" is a function that takes a set and returns a number that somehow represents the length, area, volume, whatever.
But regardless of dimension there are things we expect a measure to do, ways we expect it to behave. For example, we expect the measure of a zero length line to be zero. We also expect the measure of a non-zero length line to be non-zero. These seem to be obvious requirements. Also, when we're talking about length (or area or volume, etc ) we would expect the sum of the measure of the parts to be the same as the measure of the whole. In other words, if you take the measure of a set, then partition the set into two pieces and take the measure of each of those, you'd expect the sum of the measures to be the measure of the original.
Let's start writing these down. If we are working in $n$ dimensions and we have a measure, $\mu$ which is defined on subsets of $\Re^n,$ we expect that:
• $\mu(\{\})\;=\;0$ - the measure of the empty set is 0.
• If I is the unit object, then $\mu(I)=1.$
• For sets $A$ and $B$ that do not overlap, then $\mu(A{\cup}B)\;=\;\mu(A)+\mu(B).$
If we want to, we can weaken the middle condition to simply ask that $\mu(I)\;>\;0,$ because then we can just apply a scaling. This is like changing units.
The last one we can repeat over and over to get the idea of the measure being finitely additive. In fact, we'd really like to extend that to being countably additive, so that:
• $\mu(\bigcup_{i=0}^{\infty}A_i)\;=\;\sum_{i=0}^\infty\mu(A_i)$
• Provided all the sets are (pairwise) disjoint
In other words, when we have a set that's made up of countably infinitely many disjoint pieces, we can choose to take the measure of them all and add them up, or we can think of the union of them all, and take the measure of that.
But there are other things we expect to be true of a measure. We expect that if we take the measure of something, move the something around, and then take the measure again, we get the same answer. Moving something around should not change its size. In mathematics the idea of moving something around is captured by the idea of what we call an isometry. An isometry is a function that doesn't change distances, so if we have an isometry $\tau$ and apply it to a set $A$, none of the distances between points in $A$ will change, so we can think of $\tau(A)$ as being $A$ moved somewhere else (and maybe flipped over to give a mirror image).
So for any set $A,$ and any isometry $\tau,$ then we expect of a measure $\mu$ that:
• $\mu(A)\;=\;\mu(\tau(A))$
In words, we expect a measure to be isometry invariant.
So far:
• $\mu(I)\;=\;1$
• $\mu$ is countably additive
• $\mu$ is isometry invariant
Finally, we'd like $\mu$ to be defined for all sets, although it might end up being infinite if our set is unbounded (and even then not always). So if we have a bounded set (and I've not given a technical definition of what that means) then we'd like the measure of that to exist. So we finally want:
• For all bounded sets $A$, $\mu(A)$ exists.
Well, the problem is that we can't have all of these. This was shown in 1905 and is a classic example. In short, these natural and obvious requirements for a measure are mutually inconsistent. We cannot have all of:
• $\mu(I)\;=\;1$
• $\mu$ is countably additive
• $\mu$ is isometry invariant
• For all bounded sets $A$, $\mu(A)$ exists.
## Outline of the Vitali Set proof
The idea is:
• Consider the points from 0 (inclusive) to 1 (not inclusive)
• Declare two points to be equivalent if they differ by a rational
• This gives us families of points, the members of a family differing only by a rational number.
• Points from different families do not differ by a rational.
• From each equivalence class (family) choose one element
• Call that set $V$.
• Given two points in $V$ they do not differ by a rational
• Enumerate the rationals in $[0,1)$ :
• $r_0,\;r_1,\;r_2,\;\ldots$
• For each rational $r_i$ consider $V_i\;=\;V+r_i\;\;(mod\;1)$
• These are disjoint
• (Exercise: you should check that)
• For any measure $\mu$, for all $i$, $\mu(V_i)\;=\;\mu(V).$
• This requires some proof, but basically $V_i$ is just $V$ moved about.
• By the properties we have:
• $\mu([0,1))\;=\;\sum_i\mu(V_i)$
But what can $\mu(V)$ possibly be?
If $\mu(V)\;=\;0$ then when we add them all up we get zero, which would mean the measure of $[0,1)$ is zero. If $\mu(V)\;\ne\;0$ then when we add infinitely many together we definitely don't get a finite answer.
Which is wrong. Which is wrong.
So either way, using only the things we decided were obvious and natural for a measure, we've arrived at a contradiction. So we cannot have a measure that has all the properties that we quite reasonably want.
The proof is simple, and can be found all over the 'net. Look for the Vitali Set.
So what do we do?
Well, what we have to do is relax one of our requirements, and make it weaker. The obvious thing that people want to try is to reduce the power of the additivity rule. So our requirements become:
• $\mu(I)\;=\;1$
• $\mu$ is finitely additive
• $\mu$ is isometry invariant
• For all bounded sets $A$, $\mu(A)$ exists.
Can we do this?
As it happens, for $\Re^1$ we can do this. Even more, for $\Re^2$ we can do this! But for $\Re^3,$ we can't.
How do we know? Because of the Banach-Tarski theorem.
The Banach-Tarski theorem says that if $B$ is the unit ball in $\Re^3$, there exist pair-wise disjoint sets $\{A_i\}_{i=1}^n$ and isometries $\{\tau_i\}_{i=1}^n$ and $\tau$ such that:
• $\bigcup_{i=1}^nA_i\;=\;B$
• $B\cap\tau(B)=\{\}$
• $\bigcup_{i=1}^n\tau_i(A_i)\;=\;B\cup\tau(B)$
In other words, $B$ can be partitioned into sets, and those sets can be moved about and reassembled to make two balls the same as the original.
That means there can be no measure satisfying the requirements even when weakened from countable additivity to finite additivity.
It's not enough in three dimensions.
And that's what the Banach-Tarski theorem is really all about.
There's more we can say about this. Do we want or need every set to have a measure? We can define some really weird sets - should it always make sense for them to have a concept of length/area/volume?
There is one thing that could save us, and that's the Axiom of Choice. Or rather, denying the Axiom of Choice. In each case, showing that in $\Re^1$ there's an unmeasurable set, and showing that in $\Re^3$ we can have a paradoxical decomposition, we need to use the Axiom of Choice. So maybe we should choose not to believe in the Axiom of Choice.
But that's another discussion.
<<<< Prev <<<< Sieve Of Eratosthenes In Python : >>>> Next >>>> Photocopy A Mirror ...
You should follow me on twitter I've decided no longer to include comments directly via the Disqus (or any other) system. Instead, I'd be more than delighted to get emails from people who wish to make comments or engage in discussion. Comments will then be integrated into the page as and when they are appropriate.
If the number of emails/comments gets too large to handle then I might return to a semi-automated system. That's looking increasingly unlikely.
# Contents Suggest a change ( <-- What does this mean?) / Send me email | HuggingFaceTB/finemath | |
# Entropy, (ir)reversibility & Otto
I've been trying to grasp it very thoroughly (excuse me for my english, I'm belgian) as I have a course of thermodynamics now in my first year of Physics. There is one last thing bothering me about entropy:
Take the Otto-cycle (two adiabats, two ischores/isometrics). We know the cycle is even theoretically irreversible, because its efficiency (which we can calculate) is smaller than Carnot's and all reversible processes have the same efficiency. But now I want to calculate the change of entropy in the universe after one cycle to SEE its irreversibility.
I assume in some way it HAS to be possible to calculate it, because you already know the change of entropy is greater than zero (cf. last paragraph). Take an isochore (I will continue with the isochore from cold to hot [2 -> 3 in bottom image]). My teacher said, after inquiry, that you can calculate the entropy change in the universe (of this isochoric process) by assuming the process is equivalent to placing an infinite reservoir at Th in contact with the gas in a solid container, initially at Tc, until it's warmed up to Th. In this case (not really important how) it can be shown the total entropy increase of the universe is Cv(ln(Th/Tc) + Tc/Th - 1) which can be proven to be greater than zero.
BUT this process is not quasi-static and since you've drawn the cycle in a PV-diagram (indicating at each moment its T,P,V is well-defined), it has to be quasi-static (and ANY non-quasistatic process is irreversible, so it's kind of a lame and non-interesting choice in this case). So a process that would fit would be one where you first place against the gas an infinite reservoir at the temperature of Tc + delta, then Tc + 2delta, ... Th. But in this case the entropy change of the system (which is the same as the earlier case is) is Cvln(Th/Tc) and that of the reservoir/exterior/... is Cvln(Tc/Th) and thus total entropy change zero...
But the Otto cyclus is irreversible
What is the deal?
Thank you tremendously for any help, I've been breaking my head on it,
Ruben
PS: if my explenation was hard to follow, feel free to ignore it, basically my question is: how can I calculate the change in the universal entropy (i.e.: show it is irreversible) after going through one isochoric process in an Otto-cycle
Last edited:
All cycle that can be drawn on a p-V diagram ARE reversible, and so is Otto cycle.
Otto cycle is just a model;it is far from what really happens in the car engine.
All cycle that can be drawn on a p-V diagram ARE reversible, and so is Otto cycle.
Otto cycle is just a model;it is far from what really happens in the car engine.
That is NOT true. You can calculate the Otto's efficiency from the PV-diagram and see that it is less than the reversible carnot engine and it's a (proveable) theorem that all theoretically reversible processes have the same efficiency.
Andrew Mason
Homework Helper
BUT this process is not quasi-static and since you've drawn the cycle in a PV-diagram (indicating at each moment its T,P,V is well-defined), it has to be quasi-static (and ANY non-quasistatic process is irreversible, so it's kind of a lame and non-interesting choice in this case). So a process that would fit would be one where you first place against the gas an infinite reservoir at the temperature of Tc + delta, then Tc + 2delta, ... Th. But in this case the entropy change of the system (which is the same as the earlier case is) is Cvln(Th/Tc) and that of the reservoir/exterior/... is ln(Tc/Th) and thus total entropy change zero...
But the Otto cyclus is irreversible
What is the deal?
The entropy change of the gas is 0 in one complete cycle. This is because entropy is a state function and the gas returns to the initial state after one complete cycle. However, the change in entropy of the surroundings is greater than 0.
To calculate the change in entropy of the surroundings during the isochoric processes use:
$$\Delta S_h = \int_{P_i}^{P_f} dQ/T_h = \int_{P_i}^{P_f} nC_vdT/T_h = \int_{P_i}^{P_f} nC_vV_1dP/nRT_h = nC_vV_1/nRT_h \int_{P_i}^{P_f} dP = C_vV_1\Delta P/RT_h$$
Similarly:
$$\Delta S_c = C_vV_2\Delta P/RT_c$$
All cycle that can be drawn on a p-V diagram ARE reversible, and so is Otto cycle.
Otto cycle is just a model;it is far from what really happens in the car engine.
The Otto cycle is not reversible. It cannot be reversible because the system is not in equilibrium with the surroundings (ie the hot and cold reservoirs) while heat flows (ie during the isochoric processes). In the Carnot cycle, the system is quasi-static while heat flows (during the isothermal expansion/compression).
AM
Last edited:
DrDu
I also don't see why the Otto cycle cannot be run (as allways that means almost) reversible. All that has to be done is to make sure that the heat is provided by a heat reservoir which has always the same temperature as the working medium at each instant of the isochorus process. As the temperature of the reservoir changes, there won't be a unique reversible Carnot process to compare with.
I must say I did the same calculation as I said in my original post but in it you assume the temperature of the environment (relative to the gas) is constant. But is this process quasi-static? As I tried to say in my first post: wouldn't you first have to attach a reservoir at Tc + delta against it and once the gas and that reservoir is in equilibrium, change the reservoir with one of temperature Tc + 2delta etc? Cause if indeed your way of doing it (immediately holding a reservoir of Th against the gas) is not quasi-static, that would mean you can't even draw it on a pv-diagram because then the variables aren't well defined (because there would be a temperature gradient in your gas and thus not an overall "T")
But if you do the calculations with the delta reservoirs, the otto cycle SEEMS to be reversible...
Thanks again for your time, I hope I'll finally get this issue resolved
The entropy change of the gas is 0 in one complete cycle. This is because entropy is a state function and the gas returns to the initial state after one complete cycle. However, the change in entropy of the surroundings is greater than 0.
To calculate the change in entropy of the surroundings during the isochoric processes use:
$$\Delta S_h = \int_{P_i}^{P_f} dQ/T_h = \int_{P_i}^{P_f} nC_vdt/T_h = \int_{P_i}^{P_f} nC_vV_1dP/nRT_h = nC_vV_1/nRT_h \int_{P_i}^{P_f} dP = C_vV_1\Delta P/RT_h$$
Similarly:
$$\Delta S_c = C_vV_2\Delta P/RT_c$$
The Otto cycle is not reversible. It cannot be reversible because the system is not in equilibrium with the surroundings (ie the hot and cold reservoirs) while heat flows (ie during the isochoric processes). In the Carnot cycle, the system is quasi-static while heat flows (during the isothermal expansion/compression).
AM
Th and Tc are NOT constants so you cannot extract them out of the integration sign!!!
Otto cycle is definetely reversible
you should notice that Carnot engine operates between TWO reservoirs whose temperature are held contant respectively,but Otto cycle operates between(or among?) many reservoirs
That is NOT true. You can calculate the Otto's efficiency from the PV-diagram and see that it is less than the reversible carnot engine and it's a (proveable) theorem that all theoretically reversible processes have the same efficiency.
The Carnot theorem just states that any reversible process between a heat reservoir T1 and a "cold" reservoir T2 has the same efficiency$$$\eta = 1 - \frac{{T_2 }}{{T_1 }}$$$
BUT you should notice that Carnot engine operates between TWO reservoirs whose temperature are held contant respectively,but Otto cycle operates between(or among?) many reservoirs.
If you are comparing Otto cycle to a standard Carnot cycle,tell me what the T1 and T2 in Otto cyle are
Hm, very interesting!
So are you two claiming the Otto cycle IS reversible? Our professor quite literally and resolutely claimed its irreversibility due to some unspecified universal entropy change in the isochores/isometrics of the PV-diagram. (this isn't meant as proof that you're wrong, it's a genuine question explaining my confusion)
Now you can claim there are two Otto cycles: one with the infinite reservoirs that make it reversible, and one with two reservoirs which is irreversible, but that last one can't even be drawn on a pv-diagram so it can't be applied here, right?
Hm, very interesting!
So are you two claiming the Otto cycle IS reversible? Our professor quite literally and resolutely claimed its irreversibility due to some unspecified universal entropy change in the isochores/isometrics of the PV-diagram. (this isn't meant as proof that you're wrong, it's a genuine question explaining my confusion)
Now you can claim there are two Otto cycles: one with the infinite reservoirs that make it reversible, and one with two reservoirs which is irreversible, but that last one can't even be drawn on a pv-diagram so it can't be applied here, right?
It's true, if the latter one can be taken as "Otto cycle"
Conclusion: my professor (who was talking about the drawable otto cycle, hence the quasi-static one, hence the infinite reservoir one, hence the reversible one) was wrong?
Discomforting...
Thanks a lot for the input, I needed it.
Any extra comments always welcome (anybody in disagreement perhaps?)
Andrew Mason
Homework Helper
Th and Tc are NOT constants so you cannot extract them out of the integration sign!!!
If they are not constant, then there is heat flowing into the hot reservoir. In that case you have to define another reservoir external to the hot reservoir from which heat flows to the first hot reservoir. Ultimately, the surroundings all reduce to a single hot and a single cold reservoir whose temperature does not change during the process.
Otto cycle is definetely reversible
you should notice that Carnot engine operates between TWO reservoirs whose temperature are held contant respectively,but Otto cycle operates between(or among?) many reservoirs
No. An engine cannot be reversible unless all processes are quasi-static.
AM
AM, if the otto cycle is not quasi-static, then how can you even draw it on a pv-diagram?
Andrew Mason
Homework Helper
AM, if the otto cycle is not quasi-static, then how can you even draw it on a pv-diagram?
You can draw a PV diagram for virtually any cycle at all. Pressure and volume are measurable quantities in non-quasi-static processes. What is the problem?
AM
You can draw a PV diagram for virtually any cycle at all. Pressure and volume are measurable quantities in non-quasi-static processes. What is the problem?
AM
You are absolutely wrong
In process that are not quasi-static,pressure and volume are NOT well-defined.
Andrew Mason
Homework Helper
You are absolutely wrong
In process that are not quasi-static,pressure and volume are NOT well-defined.
They may not be precisely defined for a very brief time, but one usually ignores that in drawing the PV diagram. For example, if you take a fuel/air mixture in a closed cylinder and ignite it, the heat flows into the cylinder contents and pressure builds up extremely quickly - a neglible time compared to the cycle period. In drawing the PV diagram one does not bother with that very brief time when P and T may not be uniform throughout the cylinder contents.
Here is a PV diagram of a http://en.wikipedia.org/wiki/File:DieselCycle_PV.svg" [Broken]. Are you saying that this is a reversible process?
AM
Last edited by a moderator:
"They may not be precisely defined for a very brief time, but one usually ignores that in drawing the PV diagram." -A.M.
Well then it becomes pretty unscientific in my eyes -- you can read off equilibrium states that never were close to reality. Then the irreversibility follows somewhat directly by how you describe it. You could just as well draw a PV-diagram and add a note "...but it must be done irreversibly".
I do get what you're saying and it's acceptable in a practical point of view, but if you want to deal with it in a purely theoretical way (to gain insight), it does seem to be more interesting to not let the PV diagram be an approximation and calculate things as if it were really done quasi-statically. That brings us back to the infinite reservoirs/two reservoirs-point. I must say I didn't quite get what you meant with (paraphrasing) "but that would need a second reservoir in combination with the first [...] it comes down to one huge reservoir of one Th". Could you enlighten me, as it does sound interesting?
Andrew Mason
Homework Helper
You could just as well draw a PV-diagram and add a note "...but it must be done irreversibly".
I must say I didn't quite get what you meant with (paraphrasing) "but that would need a second reservoir in combination with the first [...] it comes down to one huge reservoir of one Th". Could you enlighten me, as it does sound interesting?
The only way you can have a reversible flow of heat is to do it infinitessimally slowly where the heat source is at almost the same temperature (ie. an infinitessimal amount hotter) as the body receiving the heat. The Otto cycle does not contemplate heat flow of that kind. It contemplates heat flowing into or out of the gas at constant volume. The means that the temperature of the gas is constantly changing as heat flows.
You cannot have such an engine operating between an infinite capacity hot reservoir and an infinite capacity cold reservoir in a reversible fashion. These reservoirs represent the entire surroundings (ie. the rest of the universe). There is no flow of heat into or out of these reservoirs to/from anywhere else in the universe other than the engine.
You want to imagine a hot reservoir whose temperature is constantly increasing, and a cold reservoir whose temperature is decreasing, in step with the Otto cycle. But the problem is that this does not represent the surroundings. Such reservoirs could only exist if heat was flowing into or out of them from somewhere else. To determine the total change in entropy, you would have to add in the change in entropy of the other reservoirs.
AM
You want to imagine a hot reservoir whose temperature is constantly increasing, and a cold reservoir whose temperature is decreasing, in step with the Otto cycle. But the problem is that this does not represent the surroundings. Such reservoirs could only exist if heat was flowing into or out of them from somewhere else. To determine the total change in entropy, you would have to add in the change in entropy of the other reservoirs.
AM
Thank you for going deeper into your explanation. I get your point and it is a very fair one, in the case of a reservoir whose temperature changes. But I proposed an infinite set of infinite reservoirs, each infinitesimally hotter than the other, all the way from Tc to Th. They'd each be placed after each other in contact with the gas-container. Do you still argue, even in this case, it is irreversible? You might object it is not a two-reveservoir heat engine anymore, but I do not see why that should be a prerequisite.
They may not be precisely defined for a very brief time, but one usually ignores that in drawing the PV diagram. For example, if you take a fuel/air mixture in a closed cylinder and ignite it, the heat flows into the cylinder contents and pressure builds up extremely quickly - a neglible time compared to the cycle period. In drawing the PV diagram one does not bother with that very brief time when P and T may not be uniform throughout the cylinder contents.
Here is a PV diagram of a http://en.wikipedia.org/wiki/File:DieselCycle_PV.svg" [Broken]. Are you saying that this is a reversible process?
AM
Read the Dissel Cycle thoroughly and you will find that Dissel Cycle is just a model for Dissel engine ,not what really happens in Dissel engine,as Otto for inner combustion engine.The real Dissel engine is irreversible,but the Dissel Cycle is reversible.
And I don't want to debate with you on this question any longer.It is just like debating with someone who claims the earth is rectangular.
Last edited by a moderator:
Andrew Mason
Homework Helper
Thank you for going deeper into your explanation. I get your point and it is a very fair one, in the case of a reservoir whose temperature changes. But I proposed an infinite set of infinite reservoirs, each infinitesimally hotter than the other, all the way from Tc to Th. They'd each be placed after each other in contact with the gas-container. Do you still argue, even in this case, it is irreversible? You might object it is not a two-reveservoir heat engine anymore, but I do not see why that should be a prerequisite.
Try to work out the entropy change of the hot reservoirs and cold reservoirs after one complete cycle:
$$\Delta S_h = - \sum_{j=1}^\infty dQ_j/T_{j} = - \int_{T_{hi}}^{T_{hf}} dQ/T = - \int_{T_{hi}}^{T_{hf}} nC_vdT/T = - nC_v\ln{\frac{T_{hf}}{T_{hi}}$$
Determine the condition for the sum of the entropy changes of the hot and cold reservoirs to equal 0 (after one cycle, there is no entropy change in the gas). You will see that it cannot be made to equal 0.
AM
Andrew Mason
Homework Helper
Read the Dissel Cycle thoroughly and you will find that Dissel Cycle is just a model for Dissel engine ,not what really happens in Dissel engine,as Otto for inner combustion engine.The real Dissel engine is irreversible,but the Dissel Cycle is reversible.
And I don't want to debate with you on this question any longer.It is just like debating with someone who claims the earth is rectangular.
;You don't have to debate it with me. Just show me a thermodynamics text that claims the diesel cycle is reversible.
AM
Try to work out the entropy change of the hot reservoirs and cold reservoirs after one complete cycle:
$$\Delta S_h = - \sum_{j=1}^\infty dQ_j/T_{j} = - \int_{T_{hi}}^{T_{hf}} dQ/T = - \int_{T_{hi}}^{T_{hf}} nC_vdT/T = - nC_v\ln{\frac{T_{hf}}{T_{hi}}$$
Determine the condition for the sum of the entropy changes of the hot and cold reservoirs to equal 0 (after one cycle, there is no entropy change in the gas). You will see that it cannot be made to equal 0.
AM
That seems like the right formula. Okay let's try it then for one isometric (if the one isometric is reversible, so is the other one and so are the two adiabats and so the whole cycle).
$$\Delta S_{environment} = - nC_v\ln{\frac{T_{hf}}{T_{hi}}$$
And seeing as $$T_{hi}$$ is infinitesimally hotter than $$T_{c}$$, it is equal to this. And $$T_{hf} = T_{h}$$ (I am using $$T_{c}$$ and $$T_{h}$$ as (resp.) the beginning and end temperature of the gas for the isometric).
So:
$$\Delta S_{environment} = - nC_v\ln{\frac{T_{h}}{T_{c}}$$
Now we also know:
$$\Delta S_{gas} = \int_{T_{c}}^{T_{h}} dQ/T = nC_v\ln{\frac{T_{h}}{T_{c}}$$ (NB: if you argue that I cannot use dS = dQ/T for the gas (cause you might say it already implies it can be reversible, cause in an irreversible case dS > dQ/T): I also calculated the same entropy change (because it is a variable of state) of the gas by going from start to end through an adiabat and an isotherm and got the same value)
As you can see, for the isometric:
$$\Delta S_{universe} = \Delta S_{environment} + \Delta S_{system} = 0$$
Andrew Mason
Homework Helper
That seems like the right formula. Okay let's try it then for one isometric (if the one isometric is reversible, so is the other one and so are the two adiabats and so the whole cycle).
I think you mean isochoric - constant volume.
$$\Delta S_{environment} = - nC_v\ln{\frac{T_{hf}}{T_{hi}}$$
And seeing as $$T_{hi}$$ is infinitesimally hotter than $$T_{c}$$
How do you get that? This is actually NOT the case. The initial hot reservoir temperature is determined by the adiabatic compression from the final cold reservoir temperature. That relationship is given by the adiabatic condition:
$$TV^{\gamma-1} = K$$ so:
(1) $$T_{ci} = T_{hf}\left(\frac{V_h}{V_c}\right)^{\gamma-1}}$$
(2) $$T_{cf} = T_{hi}\left(\frac{V_h}{V_c}\right)^{\gamma-1}}$$
So: $$\Delta S_{environment} = - nC_v\ln{\frac{T_{h}}{T_{c}}$$
No. The change in entropy of the environment (surroundings) is the sum of the change in entropy of the hot reservoir(s) and the change in entropy of the cold reservoir(s):
$$\Delta S_{env.} = \Delta S_h + \Delta S_c$$
Now we also know:
$$\Delta S_{gas} = \int_{T_{c}}^{T_{h}} dQ/T = nC_v\ln{\frac{T_{h}}{T_{c}}$$ (NB: if you argue that I cannot use dS = dQ/T for the gas (cause you might say it already implies it can be reversible, cause in an irreversible case dS > dQ/T): I also calculated the same entropy change (because it is a variable of state) of the gas by going from start to end through an adiabat and an isotherm and got the same value)
As you can see, for the isometric:
$$\Delta S_{universe} = \Delta S_{environment} + \Delta S_{system} = 0$$
This is not correct. After one complete cycle, there is no change in the entropy of the gas (engine) since it returns to its initial state. Entropy is a state function. So the change in entropy of the hot and cold reservoirs represents the change in entropy of the universe after completion of one cycle.
AM
Please note that I had stated I wasn't looking at the whole cycle in my calculations, but only ONE process, namely the isochoric one (I thought isometric was a synonym, but indeed, I meant ischoric). In this case obviously the entropy change of the gas is not zero and you'll find that also the rest of my calculations make sense if you bear in mind I wasn't looking at a full cycle. My argument is that once I've shown the reversibility of ONE process of the cycle (which I claim to have done in my previous post that you misinterpreted), all the processes are reversible (in the same fashion) and thus the whole cycle.
Curious as to what you think. | open-web-math/open-web-math | |
# A Leap Day 2016 Mathematics A To Z: Polynomials
I have another request for today’s Leap Day Mathematics A To Z term. Gaurish asked for something exciting. This should be less challenging than Dedekind Domains. I hope.
## Polynomials.
Polynomials are everything. Everything in mathematics, anyway. If humans study it, it’s a polynomial. If we know anything about a mathematical construct, it’s because we ran across it while trying to understand polynomials.
I exaggerate. A tiny bit. Maybe by three percent. But polynomials are big.
They’re easy to recognize. We can get them in pre-algebra. We make them out of a set of numbers called coefficients and one or more variables. The coefficients are usually either real numbers or complex-valued numbers. The variables we usually allow to be either real or complex-valued numbers. We take each coefficient and multiply it by some power of each variable. And we add all that up. So, polynomials are things that look like these things:
$x^2 - 2x + 1$
$12 x^4 + 2\pi x^2 y^3 - 4x^3 y - \sqrt{6}$
$\ln(2) + \frac{1}{2}\left(x - 2\right) - \frac{1}{2 \cdot 2^2}\left(x - 2\right)^2 + \frac{1}{2 \cdot 2^3}\left(x - 2\right)^3 - \frac{1}{2 \cdot 2^4}\left(x - 2\right)^4 + \cdots$
$a_n x^n + a_{n - 1}x^{n - 1} + a_{n - 2}x^{n - 2} + \cdots + a_2 x^2 + a_1 x^1 + a_0$
The first polynomial maybe looks nice and comfortable. The second may look a little threatening, what with it having two variables and a square root in it, but it’s not too weird. The third is an infinitely long polynomial; you’re supposed to keep going on in that pattern, adding even more terms. The last is a generic representation of a polynomial. Each number a0, a1, a2, et cetera is some coefficient that we in principle know. It’s a good way of representing a polynomial when we want to work with it but don’t want to tie ourselves down to a particular example. The highest power we raise a variable to we call the degree of the polynomial. A second-degree polynomial, for example, has an x2 in it, but not an x3 or x4 or x18 or anything like that. A third-degree polynomial has an x3, but not x to any higher powers. Degree is a useful way of saying roughly how long a polynomial is, so it appears all over discussions of polynomials.
But why do we like polynomials? Why like them so much that MathWorld lists 1,163 pages that mention polynomials?
It’s because they’re great. They do everything we’d ever want to do and they’re great at it. We can add them together as easily as we add regular old numbers. We can subtract them as well. We can multiply and divide them. There’s even prime polynomials, just like there are prime numbers. They take longer to work out, but they’re not harder.
And they do great stuff in advanced mathematics too. In calculus we want to take derivatives of functions. Polynomials, we always can. We get another polynomial out of that. So we can keep taking derivatives, as many as we need. (We might need a lot of them.) We can integrate too. The integration produces another polynomial. So we can keep doing that as long as we need too. (We need to do this a lot, too.) This lets us solve so many problems in calculus, which is about how functions work. It also lets us solve so many problems in differential equations, which is about systems whose change depends on the current state of things.
That’s great for analyzing polynomials, but what about things that aren’t polynomials?
Well, if a function is continuous, then it might as well be a polynomial. To be a little more exact, we can set a margin of error. And we can always find polynomials that are less than that margin of error away from the original function. The original function might be annoying to deal with. The polynomial that’s as close to it as we want, though, isn’t.
Not every function is continuous. Most of them aren’t. But most of the functions we want to do work with are, or at least are continuous in stretches. Polynomials let us understand the functions that describe most real stuff.
Nice for mathematicians, all right, but how about for real uses? How about for calculations?
Oh, polynomials are just magnificent. You know why? Because you can evaluate any polynomial as soon as you can add and multiply. (Also subtract, but we think of that as addition.) Remember, x4 just means “x times x times x times x”, four of those x’s in the product. All these polynomials are easy to evaluate.
Even better, we don’t have to evaluate them. We can automate away the evaluation. It’s easy to set a calculator doing this work, and it will do it without complaint and with few unforeseeable mistakes.
Now remember that thing where we can make a polynomial close enough to any continuous function? And we can always set a calculator to evaluate a polynomial? Guess that this means about continuous functions. We have a tool that lets us calculate stuff we would want to know. Things like arccosines and logarithms and Bessel functions and all that. And we get nice easy to understand numbers out of them. For example, that third polynomial I gave you above? That’s not just infinitely long. It’s also a polynomial that approximates the natural logarithm. Pick a positive number x that’s between 0 and 4 and put it in that polynomial. Calculate terms and add them up. You’ll get closer and closer to the natural logarithm of that number. You’ll get there faster if you pick a number near 2, but you’ll eventually get there for whatever number you pick. (Calculus will tell us why x has to be between 0 and 4. Don’t worry about it for now.)
So through polynomials we can understand functions, analytically and numerically.
And they keep revealing things to us. We discovered complex-valued numbers because we wanted to find roots, values of x that make a polynomial of x equal to zero. Some formulas worked well for third- and fourth-degree polynomials. (They look like the quadratic formula, which solves second-degree polynomials. The big difference is nobody remembers what they are without looking them up.) But the formulas sometimes called for things that looked like square roots of negative numbers. Absurd! But if you carried on as if these square roots of negative numbers meant something, you got meaningful answers. And correct answers.
We wanted formulas to solve fifth- and higher-degree polynomials exactly. We can do this with second and third and fourth-degree polynomials, after all. It turns out we can’t. Oh, we can solve some of them exactly. The attempt to understand why, though, helped us create and shape group theory, the study of things that look like but aren’t numbers.
Polynomials go on, sneaking into everything. We can look at a square matrix and discover its characteristic polynomial. This allows us to find beautifully-named things like eigenvalues and eigenvectors. These reveal secrets of the matrix’s structure. We can find polynomials in the formulas that describe how many ways to split up a group of things into a smaller number of sets. We can find polynomials that describe how networks of things are connected. We can find polynomials that describe how a knot is tied. We can even find polynomials that distinguish between a knot and the knot’s reflection in the mirror.
Polynomials are everything.
## Author: Joseph Nebus
I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.
## 22 thoughts on “A Leap Day 2016 Mathematics A To Z: Polynomials”
1. Beautiful post!
Recently I studied Taylor’s Theorem & Weierstrass approximation theorem. These theorems illustrate your ideas :)
Like
1. Thank you kindly. And yeah, the Taylor Theorem and Weierstrauss Approximation Theorem are the ideas I was sneaking around without trying to get too technical. (Maybe I should start including a postscript of technical talk to these essays.)
Like
This site uses Akismet to reduce spam. Learn how your comment data is processed. | HuggingFaceTB/finemath | |
# chapter 3 review y - math tv schools media server 3 review 1-4. 7. 8. 9. x-intercept = 6;...
of 2 /2
Chapter 3 Review 1-4. 7. 8. 9. x-intercept = 6; y-intercept = 1 10. x-intercept = 20 3 ; y-intercept = 2 11. x-intercept = 2; y-intercept = 2 12. x-intercept = 2; y-intercept = 2 13. 1 3 14. 2 15. 2 16. 0 17. No slope 18. 2 19. y = 2x 7 20. y = 3x + 6 21. y = 1 5 x + 8 5 22. y = 6 5 x + 18 5 23. y = 1 2 x + 2 24. m = 4, b = 3 25. m = 4 5 , b = 2 26. y = 3x 4 27. y = 3 2 x + 8 28. 29. Chapter 3 Cumulative Review 1. 1 2. 1 3. 2 4. 8 5. 32 6. 40 7. Undefined 8. 1 3 9. 5 7 10. 0 11. 36x 12. 3a 6 13. 2 14. 4 15. 1 16. 4 17. 3 2 18. 98 19. x > 6 20. x 7 21. 22 23. 24. 25. 26. 27. No, yes 28. y-intercept = 6 29. x-intercept = 4; y-intercept = 6 30. 31. m = 1 32. y = 2x + 5 33. m = 3 34. y = 2 3 x 35. y = 3 2 x 8 36. y = 3 5 x 3 37. (5, 1 ), (0, 3) 38. (2,3), (6, 1 ) 39. 24 2(6) = 12 40. 20 41. 3 5 , 5 3 , 3 5 42. 4, 1 4 , 4 43. 9 44. 1 45. length = 13 in.; width = 7 in. 46. 7 quarters, 13 dimes 47. 0.6, e attendance from 1955 to 1970 was increasing at an average rate of 0.6 million people every year 48. 0.1475, e attendance from 1970 to 2010 was increasing at an average rate of 0.1475 million people every year Chapter 3 Test 1-4. 7. 8. x y ( (1.2, 1.2, 2) 2) (2 (2, 1) 1) ( (3, 3, 5) 5) 1 1 2 2 2 2 , , 2 2 0 0 5. ( 3, 4 5 ), ( 2, 14 5 ) ( 27 3 , 2 3 ) , ( 1 2 , 9 5 ) 6. (10, 0), (8, 1.6) x y 2 3 , 0 , 0 (0, 0 2) 2 x y ( (0, 0, 3) 3) x y 2 2 1 1 2 2 x y 4 4 4 3 3 2 2 x y 2 3 ,0 0 (0 (0, 2) 2) x y (0 (0, 2) 2) ( ( 3, 0) 3, 0) x y (0 (0, 0) 0) x y (0 (0, 0) 0) x y 2 2 2 3 3 2 2 x y 5 5 x ( ( 2 23, 0) 3, 0) x y (0 (0, 4) 4) (2 (2, 1) 1) ( (3, 3, 2) 2) ( (4, 4 3 3) ) 5. (0, 3), (2, 0), ( 4, 3), ( 2, 6) 6. (0, 7), (4, 5) 10 x y ( 0, 4) 0, 4) ( (8, 0) 8, 0) 2 2 4 4 6 8 6 8 10 10 8 8 6 6 4 4 2 2 10 x y ( (0, 0 2 23) 3)
Author: lycong
Post on 05-May-2018
239 views
Category:
## Documents
Embed Size (px)
TRANSCRIPT
• Chapter 3 Review1-4.
7. 8.
9. x-intercept = 6; y-intercept = 1
10. x-intercept = 20__3 ; y-intercept = 211. x-intercept = 2; y-intercept = 212. x-intercept = 2; y-intercept = 213. 1_3 14. 2 15. 2 16. 0 17. No slope 18. 219. y = 2x 7 20. y = 3x + 6 21. y = 1_5 x +
8_5
22. y = 6_5 x +18__5 23. y =
1_2 x + 2 24. m = 4, b = 3
25. m = 4_5, b = 2 26. y = 3x 4 27. y =3_2 x + 8
28. 29.
Chapter 3 Cumulative Review1. 1 2. 1 3. 2 4. 8 5. 32 6. 40 7. Undefi ned 8. 1_3 9.
5_7 10. 0 11. 36xx
12. 3a 6 13. 2 14. 4 15. 1 16. 4 17. 3_2 18. 98 19. x > 6 20. x 7 21. 22
23. 24.
25. 26.
27. No, yes 28. y-intercept = 6 29. x-intercept = 4; y-intercept = 6 30. 31. m = 1 32. y = 2x + 5
33. m = 3 34. y = 2_3x_x
35. y = 3_2x_ 8
36. y = 3_5x_ 3
37. (5, 1), (0, 3) 38. (2,3), (6, 1) 39. 24 2(6) = 1240. 20 41. 3_5,
5_3,
3_5
42. 4, 1_4, 4 43. 9 44. 1
45. length = 13 in.; width = 7 in. 46. 7 quarters, 13 dimes 47. 0.6, Th e attendance from 1955 to 1970 was increasing at anaverage rate of 0.6 million people every year 48. 0.1475, Th e attendance from 1970 to 2010 was increasing at an average rate of 0.1475 million people every yearChapter 3 Test1-4.
7. 8.
x
y
((1.2,1.2, 2)2)
(2(2, 1)1)
((3,3, 5)5)
112222 ,, 22 00
5. (3, 4_5 ), (2, 14__5 )( 27__3 ,
2_3 ), (
1_2 ,
9_5 )
6. (10, 0), (8, 1.6)
x
y
2__3, 0 , 0
(0,0 2)2x
y
((0,0, 3)3)
x
y
22
1122x
y
444 3322
x
y
233,00
(0(0, 2)2)
x
y
(0(0, 2)2)
((3, 0)3, 0)
x
y
(0(0, 0)0)x
y
(0(0, 0)0)
x
y
222
3322x
y
55
x
((223, 0)3, 0)
x
y
(0(0, 4)4)
(2(2, 1)1)
((3,3, 2)2)
((4,4 33))
5. (0, 3), (2, 0), (4, 3), (2, 6)6. (0, 7), (4, 5)
10x
y
((0, 4)0, 4)((8, 0)8, 0)
22 44 6 86 8
101088664422 10
x
y
((0,0 223)3)
• 9. x-intercept = 2; y-intercept = 4 10. x-intercept = 4; y-intercept = 611. x-intercept = 2; y-intercept = 412. 1_2 13.
8_7 14. 3 15. No slope 16. 0
17. y = 1_2x + 3 18. y = 3x 5 19. y = 2_3x 2
20. y = x + 2 21. y = 3_2x_ + 6 22. y = 3x 17
23. y = 3_2x_ + 4 24. y = 3_2x
_ + 7 25. y = 2x 6 26. 27.
x
y
6666
66
x
y
333
222 | HuggingFaceTB/finemath | |
# math
explain how to make a ten to find 8+6
1. 👍 0
2. 👎 0
3. 👁 1,220
1. 8 + 6 = 10 + 4
1. 👍 1
2. 👎 0
👩🏫
Ms. Sue
2. Explain how to make a ten to finde
8+6
1. 👍 0
2. 👎 0
3. 8+6= 14-4=10
1. 👍 0
2. 👎 0
## Similar Questions
1. ### PROBABILITY MATH
An unprepared student makes random guesses for the ten true-false questions on a quiz. Find the probability that there is at least one correct answer. Could some one please help me? And explain to me how they got the answer? and
2. ### math
Why would you make one of the addend a ten number when solving an addition problem
3. ### Math
Typically, 10% of students make a D on their tests, 60% make a C on their test, and 30% make an A. Mrs. Smith uses a random-number table to find the experimental probability that of 5 students', at least 3 will make a C. The digit
4. ### English grammar
Use a singular verb with sums of money or periods of time. Ex) Ten dollars _____ a high price to pay. Ten dollar is a sum of money, so you have to use a singular verb. In this case, the subject acts as a unit. I'll give you
1. ### Math
Show how to make one addend a ten. Complete the new addition sentence. 22+49= ? ---- + ---- =
2. ### MATH
WRITE TO SHOW HOW YOU MAKE A TEN. THEN ADD. WHAT IS 9 + 7
If Naomi decides that she will invest \$3,000 per year in a 6% annuity for the first ten years, \$6,000 for the next ten years, and \$9,000 for the next ten years, how much will she accumulate? Treat each ten year period as a
4. ### Algebra
X/3 - 10 = -12 I need help on this... I got half of it done. I added ten to both sides and got X/3 = -2 I’m stuck on this part. Can someone help me and explain the process? Thanks.
1. ### Data Management
You are thinking of using some leftover ceramic tiles to make a decorative border for a patio. You have ten blue, four gold, and six white tiles. In how many ways can you make the border if it requires 15 tiles and you use only
2. ### English
1. The boy is ten years old. 2. He is a ten-year-old boy. 3. The building is ten meters tall. 4. The building is ten meters high. 5. It is a ten-meter-tall building. 6. It is a ten-meter-high builing. 7. It is a ten-meter
3. ### Java
Design a flowchart for a program that prompts the user to enter ten numbers one at a time. After the user has entered all ten number the program will then print out the total of those ten number and quit.
4. ### algbra
Explain how to find the sum positive eight plus negative ten using game chips, then give the answer. | HuggingFaceTB/finemath | |
Search a number
12823 is a prime number
BaseRepresentation
bin11001000010111
3122120221
43020113
5402243
6135211
752246
oct31027
918527
1012823
1196a8
127507
135ab5
14495d
153bed
hex3217
12823 has 2 divisors, whose sum is σ = 12824. Its totient is φ = 12822.
The previous prime is 12821. The next prime is 12829. The reversal of 12823 is 32821.
It can be divided in two parts, 128 and 23, that added together give a palindrome (151).
It is a happy number.
12823 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 12823 - 21 = 12821 is a prime.
Together with 12821, it forms a pair of twin primes.
It is a junction number, because it is equal to n+sod(n) for n = 12797 and 12806.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12821) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6411 + 6412.
It is an arithmetic number, because the mean of its divisors is an integer number (6412).
212823 is an apocalyptic number.
12823 is a deficient number, since it is larger than the sum of its proper divisors (1).
12823 is an equidigital number, since it uses as much as digits as its factorization.
12823 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 96, while the sum is 16.
The square root of 12823 is about 113.2386859691. The cubic root of 12823 is about 23.4061444404.
The spelling of 12823 in words is "twelve thousand, eight hundred twenty-three". | HuggingFaceTB/finemath | |
13,736,882 members
Article
alternative version
#### Stats
7.6K views
6 bookmarked
Posted 7 Nov 2017
Licenced CPOL
# Making a 2D Physics Engine: Spaces and Bodies
, 29 Jan 2018
The basics of spaces, transformations and bodies used in a 2D physics engine.
### Making a 2D Physics Engine: The Series
This is the second article in the Making a 2D Physics Engine Series. If you haven't already read all the articles in the series before this one, I strongly recommend that you take a detour and skim through them.
1. Making a 2D Physics Engine: The Math
2. Making a 2D Physics Engine: Spaces and Bodies
3. Making a 2D Physics Engine: Shapes, Worlds and Integration
4. Making a 2D Physics Engine: Mass, Inertia and Forces
## Prerequisites
Basic knowledge of linear algebra including 2D vectors and 2x2 matrices as covered in the first article in the series.
## Introduction
This article intends to introduce the concepts of Local space and World space and deal with transformations from Local to World space and vice versa. It also outlines the representation of a body or physical entity in the physics engine.
## Spaces
Spaces are limitless extents relative to which an object's properties, i.e. rotation and position are defined.
### World Space
All objects in a world have a definite position and rotation in space, which are measured relative to the world origin. World space is omnipresent; all entities in our world will be placed relative to it.
### Local Space
Local space is relative to a single entity in the world. When measured in local space of an entity, the properties of other entities are measured relative to that entity.
## Transformation
Let the given point be (X) in world space and (X') in local space relative to an entity. If the position of that entity (origin of the local space) is (P) and the rotation matrix of the entity (rotation matrix of the local space) is (U = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}) (where (\theta) is the world rotation of the entity), then the following transformations can take place.
### Local Space to World Space (Transform)
(X = P + UX')
The local space point is first rotated to bring it to world space rotation, and then translated to the world space position.
It is important to note that a change in the local space origin or rotation will not affect a local space position. The world space position, however, may change.
### World Space to Local Space (Inverse Transform)
Solving the above equation for (X') will give us the required world space to local space transformation:
(X' = U^{-1}(X - P))
This gives us a problem - how to find the inverse (U^{-1}) of the matrix? Turns out that our rotation matrix is orthongonal, so (U^{-1} = U^T), where (U^T) is the transpose of matrix which we already know how to compute. The equation can now be rewritten as
(X' = U^T(X - P))
Space transformations will be used throughout the engine to simplify mathematical routines.
## Bodies
Bodies or entities are are simply objects in a world. In our case, a body is an object that can physically interact with the environment (or not, depending on its configuration). Each body has a few defining characteristics or properties, like position, velocity, torque, its shape, and mass.
A body in a physics engine does not do much by itself. Generally, all it is responsible for is integrating and updating its forces, velocities, and positions. It is the physics engine - or other external code - that manages interactions like collisions. A body will typically be manipulated by applying forces on it. However, in rare cases (like impulse collision resolution, which we will cover later), the velocity will be modified directly by the physics engine.
### Body-Physics Engine Interactions
Typical interactions of a physics engine with the bodies in a world include:
• Application of forces
• Collision detection and resolution
• Spatial queries (raycasts and shapecasts)
• Joints and constraints
### A Body in Code
A Body structure in Rust (taken from the engine I am building in Rust alongside this article series) looks like the following:
pub struct Body {
pub position: Vec2,
pub rotation: f32,
pub velocity: Vec2,
pub angular_vel: f32,
force: Vec2,
torque: f32,
pub mass: f32,
pub inertia: f32,
pub coeff_friction: f32,
pub coeff_restitution: f32,
pub shape: Shape,
}
This Body definition, however, is not complete. As the physics engine grows, more properties will be added to it.
## History
7 Nov 2017: Initial post
## Share
Student Evudio India
Teen programmer with a great zeal for programming, and interested work on Assets for Unity3D, games in Unity3D and small tools and applications for Windows, Linux and Android under Evudio.
## You may also be interested in...
Pro Pro | HuggingFaceTB/finemath | |
# How to calculate confidence level for a given sample size and population size?
It's been a while since I had statistics in uni, and so I'm a little rusty. I need some help with a fairly straight forward calculation of the confidence level of a sample size. I've been trying to look for an answer on CrossValidated but came to the conclusion that answers are often to complicated for me to quickly grasp. I hope that one of you is kind enough to talk me through an example and provide a formula I can apply in a confidence interval calculator I'm building.
An example: I have a sample size of 1406 respondents ($n$), a population size of 29,245,6752 ($N$), I want to have a confidence level of 95% ($z$ = 1.96) and the percentage of respondents picking a certain option 50% ($p$ = 0.5).
Is there anyone who wants to walk me through the calculation with the data I just gave, and give me the formula so that I can create my calculator? Thank you very much!
When constructing confidence intervals usually the size of a population is far larger than the sample size. In these cases we treat the sample as if it came from an infinite population and this simplifies the analysis a bit. For these cases the confidence interval formula is the following
Lower limit:
$$p-z\sqrt{\frac{p(1-p)}{n}}$$
For your example this is $0.5-1.96\sqrt{\frac{0.5(1-0.5)}{1406}}=0.4739$
Upper limit:
$$p+z\sqrt{\frac{p(1-p)}{n}}$$
For your example this is $0.5261$ so the 95% confidence interval for the population value of $p$ is $(0.4739,0.5261)$
Small population size
When the size of the population is small then you can make an adjustment to account for this fact. In this case the confidence interval is
Lower limit:
$$p-z\sqrt{\frac{p(1-p)}{n}\left(\frac{N-n}{N-1} \right)}$$
Upper limit:
$$p+z\sqrt{\frac{p(1-p)}{n}\left(\frac{N-n}{N-1} \right)}$$
The part under the square root is modified slightly. In your example the population is huge so it's being modified by a factor of $\frac{292456752-1406}{292456752-1}= 0.999995$. You can try calculating the modified confidence interval, it doesn't change the first four decimal places.
Small sample sizes
When you sample very few people then the methods used to derive the above formulas can be invalid. A common rule for deciding if sample size is large enough is the following:
If $np > 5$ and $n(1-p)>5$ then the sample size is large enough. Your example certainly has a large enough sample size. When the sample size is too small then you should use a different interval such as the Wilson Score interval:
$$\text{Lower limit} = \frac { 2n\hat{p} + z^2 - \left[z \sqrt{z^2 - \frac{1}{n} + 4n\hat{p}(1 - \hat{p}) + (4\hat{p} - 2)} + 1\right] } { 2(n + z^2) }$$
$$\text{Upper limit} = \frac { 2n\hat{p} + z^2 + \left[z \sqrt{z^2 - \frac{1}{n} + 4n\hat{p}(1 - \hat{p}) + (4\hat{p} - 2)} + 1\right] } { 2(n + z^2) }$$
If these formulas give a value below $0$ or above $1$ (which is an impossible value for $p$) then round them to $0$ or $1$
This one doesn't have a nice way of adjusting for a small population size. If you have both a small population size and a small sample size I'd recommend prioritizing the small population size and using the second set of confidence interval formulas I described.
• That was very helpful, thank you very much! Would you strictly use the second formula only when you have a small population size? Or can you also use it if you have a large population size? – RF_PY Jan 23 '17 at 13:34
• @RF_PY You can use it for large population sizes too. I suppose that it's more simple for you to code a calculator to always use the second formula or the third method if the sample size is small – Hugh Jan 23 '17 at 13:45
• And what about sample distribution? Do you need a normal sample distribution if you want to calculate the confidence interval? – RF_PY Jan 23 '17 at 13:46
• All these formulas are for a binomial distribution where you are making a confidence interval for a proportion $p$. A normal distribution has similar formulas but other distributions have more complicated ways to get the confidence interval. What other distributions did you want to include in your calculator? – Hugh Jan 23 '17 at 14:02
• So just to give you some more background. I'm calculating confidence intervals for a very extensive survey held in multiple countries. There are a few 100 questions in the survey, and I know that for example the age question is skewed. That being said, I'm currently not looking to find a confidence interval for every single question. I'm hoping to deliver some sort of a general metric about the sample from every country in which we held this survey that gives an indication of the accuracy of our sample. – RF_PY Jan 23 '17 at 14:09 | HuggingFaceTB/finemath | |
Young's double slit Experiment
Question
# Two slits in Youngs experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern,
Moderate
Solution
## $\therefore \frac{{I}_{\mathrm{max}}}{{I}_{\mathrm{min}}}=\frac{{\left({A}_{1}+{A}_{2}\right)}^{2}}{{\left({A}_{1}-{A}_{2}\right)}^{2}}=\frac{{\left(\frac{{A}_{1}}{{A}_{2}}+1\right)}^{2}}{{\left(\frac{{A}_{1}}{{A}_{2}}-1\right)}^{2}}$$=\frac{{\left(\frac{1}{5}+1\right)}^{2}}{{\left(\frac{1}{5}-1\right)}^{2}}=\frac{{\left(\frac{6}{5}\right)}^{2}}{{\left(-\frac{4}{5}\right)}^{2}}=\frac{36}{16}=\frac{9}{4}$
Get Instant Solutions | HuggingFaceTB/finemath | |
BlogSchool Resources for KidsWhat Do You Learn in 3rd Grade: Overview, Subjects & Skills
# What Do You Learn in 3rd Grade: Overview, Subjects & Skills
You must know what to expect when your child is about to enter the 3rd grade. Students will be introduced to new and more challenging subjects at this grade level. The following article will outline these subjects and what students should know by the end of 3rd grade. Remember that every school district may have a slightly different curriculum, but this guide will give you a general idea of what is taught to kids in 3rd grade.
## You see real learning outcomes.
Watch your kids fall in love with math & reading through our scientifically designed curriculum.
## What is Taught to Kids in 3rd Grade?
Here are some of the subjects your child will be learning in 3rd grade. You can ask their teacher what the focus will be for each subject in the coming year. Using this guide, you can help your child get ahead or catch up in any of the following areas:
### 1. Maths and Arithmetic
Third graders will be expected to know their multiplication tables up to 10×10. They should also be able to solve two-digit by two-digit multiplication problems and understand division with remainders.
Kids will continue to work on basic addition and subtraction skills but will begin learning about regrouping (also called carrying or borrowing). In particular, they should be able to regroup when adding three-digit numbers and subtracting numbers with zeroes in the ones or tens column.
Lastly, third graders should understand what fractions are and be able to identify them in everyday life. They should also be comfortable naming fractions and completing basic operations with fractions, such as addition, subtraction, and multiplication.
## SplashLearn: Most Comprehensive Learning Program for PreK-5
SplashLearn inspires lifelong curiosity with its game-based PreK-5 learning program loved by over 40 million children. With over 4,000 fun games and activities, it’s the perfect balance of learning and play for your little one.
### 2. Language Arts
In language arts, your child will continue to improve their reading skills. By the end of 3rd grade, they should be able to read fluently and with comprehension. This means reading quickly, without having to stop and sound out words, and understanding what they are reading.
Third graders will also be expected to understand and use different types of writing, such as narrative, opinion, informative, and explanatory. They should be able to write clear and well-organized essays on various topics. Additionally, students will learn how to research and cite their sources.
Educators use online learning tools to help third graders hone their language arts skills. These may include online reading games, writing tools, and grammar tools. Using these tools gives your child the practice they need to succeed in language arts.
### 3. Science
Third graders will learn about the different states of matter in science class: solid, liquid, and gas. The syllabus will also introduce them to the concept of mixtures and solutions. Additionally, kids will learn about living and non-living things and what it takes for something to be classified as alive.
Other science topics covered in 3rd grade include the human body, specifically the skeletal and muscular systems. Students will also learn about different types of rocks and soil, the water cycle, and simple machines.
### 4. Social Studies
In social studies, third graders will learn what it means to be a good citizen. This includes learning about rules, laws, and responsibilities at home, at school, and community. Kids will also learn about map skills and begin to understand different types of governments.
Not only will third graders learn about citizenship, but they will also begin to understand how people have interacted with each other and their environment throughout history. Students will learn about ancient civilizations, the impact of explorers, and the American Revolution.
### 5. Visual and Performing Arts
Third graders will continue to develop their drawing, painting, and sculpting skills in the arts. They will also learn about different types of music and how to create rhythms. Also, kids will be introduced to basic photography concepts.
Moreover, third graders will learn about different types of dance and what it takes to put on a play, including acting, stagecraft, and set design.
Now let’s talk about what students should know by the end of 3rd grade.
## What Should Students Know in 3rd Grade?
Teachers use a variety of assessments to determine what students have learned and what they need to work on. These can include tests, quizzes, projects, and papers. Some skills are more accessible to measure than others, but there are a few things that all students should be able to do in 3rd grade:
1. Read confidently and fluently. This includes being able to read quickly, smoothly, and with expression. Also, students should be able to retell stories with ease.
2. Understand and use basic math concepts. This includes being able to add, subtract, multiply, and divide. Students should also understand what fractions are and identify them in everyday life.
3. Write clearly and with detail. Students should be able to write for different purposes, such as to inform, explain, or tell a story. Their writing should include proper grammar, spelling, and punctuation.
4. Research topics using various sources. This means using books, websites effectively, and other materials to find information.
5. Think critically about what they’re learning. Third graders should be able to ask questions, make predictions, and draw conclusions.
6. Kids will be able to communicate effectively and listen attentively. Students should also be able to work cooperatively with others.
Becoming a third grader is an important milestone in any child’s education. It’s the time when they start to learn and grow, both academically and socially. In this article, we’ve outlined some of the most important things that your child should know by the end of third grade. There is a lot of material to cover in third grade, but your child will succeed with a little hard work and dedication! Make sure you talk to their teacher to get a better idea of what they should be focusing on.
## What is the 3rd-grade curriculum like?
The 3rd-grade curriculum builds on the skills learned in earlier grades. Students will read more complex texts, write longer essays, learn about different types of governments, and explore fractions and other states of matter.
## What to expect from teachers in 3rd grade?
Third-grade teachers will focus on helping students develop reading, writing, and math skills. Additionally, they will introduce kids to new social studies and science concepts. Teachers may also incorporate art and music into their lessons.
## What does a third-grade homeschool curriculum look like?
Homeschooling parents can tailor their child’s curriculum to their individual needs and interests. However, most 3rd-grade homeschool curriculums include core subjects such as reading, writing, math, science, and social studies.
## Is the third grade the same everywhere?
No, the third-grade curriculum can vary depending on the state or country. However, most third-grade curriculums will focus on reading comprehension, writing, and math. Additionally, some schools may also require students to take a science or history class.
## How can I help my child succeed in third grade?
You can do a few things to help your child succeed in third grade. First, make sure they attend school regularly and do their homework. Additionally, you can help them study for tests and exams. Finally, encourage them to participate in extracurricular activities, such as sports or clubs.
AUTHOR
Jill Baker
Jill Baker has been teaching for 10 years and she loves sharing everything she has learned to help other teachers.
Learn & Play | HuggingFaceTB/finemath | |
I believe everyone will FWT and FMT.
If you don't , I recommend you to have a look at VFK Of 2015 National training team .
Let the complete set be $$U=\{1,2,\ldots,n\}$$, Suppose we care about $$f_S$$ The set in $$S$$ yes $$U$$ Subset .
Here you are. $$c_i,d_i$$, Make
$b_i=(1+c_ix^{d_i})$
seek
$g=\prod_{i}b_i$
The product of two set power series is set merging convolution （or）/ Set symmetric difference convolution （xor） One of the .
Might as well set $$d_S$$ Different from each other （ Otherwise you can use DP/ Combination number or something ）.
Make
\begin{align} a_S&=\sum_{d_i=S}c_i\\ f_S&=1+a_Sx^S \end{align}
## Violence
Do it again for each set power series FMT/FWT, And then we'll just ride them together , Switch back .
Time complexity ：$$O(n4^n)$$
It's too slow , Because it doesn't use the special condition of this problem .
## Set or convolution
For a set power series $$f$$, Definition $$f$$ The Mobius transformation of is a set power series $$\hat f$$, among
\begin{align} \hat f_S=\sum_{T\subseteq S}f_T \end{align}
In turn, , Definition $$\hat f$$ The Mobius inversion of is $$f$$, From the inclusion exclusion principle, we can get
$f_S=\sum_{T\subseteq S}{(-1)}^{|S|-|T|}\hat f_T$
I believe everyone is familiar with the above contents .
Back to the formula we asked for ：
\begin{align} \hat g_T&=\prod_S \hat{f_{S}}_T\\ &=\prod_S \sum_{K\subseteq T}f_{S,K}\\ &=\prod_{S\subseteq T}{(1+a_S)} \end{align}
It's similar to the ordinary Mobius transformation ？
Put all the $$a_S$$ add $$1$$, Change the addition of Mobius transformation to multiplication , You can get $$\hat g_T$$ 了 .
Time complexity ：$$O(n2^n)$$
## Set symmetric difference convolution
We still need to use those formulas .
\begin{align} \hat g_T&=\prod_S\hat {f_S}_T\\ &=\prod_S\sum_{K}f_{S,K}{(-1)}^{|K\cap T|}\\ &=\prod_S(1+a_S{(-1)}^{|S\cap T|}) \end{align}
This is also very similar to Walsh transformation , however $${(-1)}^{|S\cap T|}$$ Only in $$a_S$$ above , So we can't put $$a_S$$ Add $$1$$ After doing the variant Walsh transformation .
But we can maintain another $$\hat h_T=\prod_S(1-a_S{(-1)}^{|S\cap T|})$$, Take the Walsh transformation of $$-\hat g_T$$ Replacing the $$\hat h_T$$, You can do it .
Time complexity ：$$O(n2^n)$$
First pit
## A related FWT&FMT More related articles about things like
1. [ Learning notes & course ] The signal , aggregate , polynomial , And various convolutional transformations (FFT,NTT,FWT,FMT)
Catalog The signal , aggregate , polynomial , And convolution transformation Convolution Convolution transformation Fourier transform and signal introduce : Signal analysis The basis of transformation : The plural The Fourier transform Discrete Fourier transform FFT With the polynomial $$n$$ The subunits are complex roots Remove the lemma ...
2. C# The reflection mechanism is used to copy the fields and properties of a class ( Copy everything from one class object to another , Instead of paying the reference address )
from:https://blog.csdn.net/poxiaohai2011/article/details/27555951 //C# The reflection mechanism is used to copy the fields and properties of a class ( Copy everything from a class object ...
3. @ summary - 2@ Bit operation convolution / Subset convolution —— FWT/FMT
Catalog @0 - Reference material @ @1 - The concept and properties of XOR convolution @ @2 - Fast Walsh transform ( Exclusive or )@ @3 - Fast inverse Walsh transform ( Exclusive or )@ @4 - And convolution . Or convolution @ @5 - Reference code implementation @ @6 - ...
4. 【 Learning notes 】fwt&&fmt&& Subset convolution
Preface :yyb Fairy blog FWT The basic idea : Turn polynomials into point valued expressions , Point values are multiplied and then inversely transformed back to obtain a specific form of convolution : The degree bounds of polynomials are $$2^n$$ In the form of ,$$A_0$$ Defined as the first half of the polynomial ( The first bit of the subscript binary ...
5. stay ASP.NET I did an experiment in the lab MVC The little thing
It's a whim before work on Friday , Yes MVC I made a few comments . Later, I just wanted to make a little thing to verify some of my own ideas and help the team clear up their ideas . The required source code is free ,3 An hour's rush is a mess , There's no good architecture , And there's no code quality , It's mainly real ...
6. [WC2018] State division ( Pressure DP+FWT/FMT)
Very naked subset inversion template problem , Put on some strange clothes . Preprocess each collection first to see if it's legal , And then make the obvious compression DP. Then we find that it can be written in the form of subset inversion , Just set the template directly . Subset inversion can be seen here . The process of subset inversion is to set more one-dimensional representative sets ...
7. 《 turn 》 Set up a BLOG How many things need to be integrated ？
This article reprints CC 1 Wordpress There are not many things that need to be done , Relatively easy , but Themes It takes a lot of effort to adjust , Sometimes it has to be changed CSS. Recommend a few Wordpress Themes Website : - http://the ...
8. Recommend a IT What is necessary for people 【 I know everything I've used , Let's make our environment better and better 】
There's something called IT People's Manual , I don't know if you've ever been useful ? But it's a pity that the old one was turned off , That website says we are IT Too many new voices and some black heart companies , But forced to shut down under pressure I'm not asking , I'm just sharing I think this kind of thing should exist ~!!! to ...
9. Post one CMemDC Code , That's a good thing , short , It's the best
Rossock Published on 2006-11-28 21:53 Click on :3941 Time come from : original text : http://yuantao82.spaces.live.com/Blog/cns!8FC0A772D812A22 ...
## Random recommendation
1. docker compose note
https://www.youtube.com/watch?v=Uez88TWOECg It's notes based on this video . Docker Compose: Compose is a tool for defin ...
2. R System analysis of language combined with probability and statistics --- Digital features
Now there's a man , How to identify this person with this person ? Then we extract the existing features , such as , Extract their height , Its appearance , His age , Analyze these characteristics , So it's certain that , This person is this person , We will never admit our mistake . Empathy , Analyze the data , Also extract ...
3. C# The basic chapter -- object-oriented （ Classes and objects ）
1. What is the class ? A class is a template , It is to abstract the common characteristics of the same kind of things . Class creation and description : Class is based on some specific objects ( Physical things ) To abstract out the common characteristics , And then code it . In class , Using data to represent things ...
4. SyntaxHighlighter The right way to remove the help icon in the upper right corner
Post the problem picture first : On this question . There are many posts on the Internet , There are three ways , After testing , I found that some of the methods were problematic , Some methods can solve problems , But it can lead to other mistakes . Now the explanation is as follows : The original words on the Internet : syntaxhig ...
5. IntelliJ IDEA SVN All of a sudden there was no
1. stay IDEA I can't find it in China. SVN The choice of , There is no version control tool subversion, stay setting The corresponding svn Tools , This is because we are idea Of svn Plug in svn This option is disabled terms of settlement : ...
6. pymysql solve sql Injection problem
1. SQL Inject SQL Injection is a very common form of network attack , It's mainly through parameters mysql perform sql Statement to perform unexpected operations . namely : Because the parameters passed in change SQL The semantics of the , It becomes some other command , So we operate the numbers ...
7. Android Typical interface design (5)—— Use SlidingMenu and DrawerLayout Achieve left and right sidebar respectively
One . Problem description The sidebar is Android Interface effect is very common in application , It can be linked with the main screen on the left or right side , It is one of the typical interface design schemes that are especially suitable for small screen features such as mobile phones , Commonly used as operation menu of application , As shown in the figure The implementation sidebar can use third-party components s ...
8. HDU5086:Revenge of Segment Tree( Regular questions ）
http://acm.hdu.edu.cn/showproblem.php?pid=5086 #include <iostream> #include <stdio.h> #i ...
9. Putty The way to build a tunnel [z]
Through the last section we learned about SSH The advantages of tunnels , But both in reality and on the Internet , Tunnels have entrances and exits , This section is based on Putty Tunnel, for example , Detailed introduction Putty The way to build a tunnel . Simulation case 1 : port , The exit is 2012 port , Tunnel building ...
10. centos6.5 Next use yum Perfect build LNMP Environmental Science (php5.6,mysql5.5,nginx1.10)
preparation Configure firewall , Turn on 80 port .3306 port No need to execute this sentence :rm -rf /etc/sysconfig/iptables Go straight to modify :vi /etc/sysconfig/iptables add to 8 ... | HuggingFaceTB/finemath | |
{/eq}. Our experts can answer your tough homework and study questions. In order to build bonds with different elements, the oxidation state provides the number of electrons that a molecule can lose, share, or gain. What is the oxidation state of H in H_2? Also, there was no allowance for metals with more than two oxidation states, such as vanadium with oxidation states +2, +3, +4, and +5. T or F The oxidation state of metals is almost always a positive number. {/eq} is +4. It is abundant, multivalent and nonmetallic.Under normal conditions, sulfur atoms form cyclic octatomic molecules with a chemical formula S 8.Elemental sulfur is a bright yellow, crystalline solid at room temperature. Thus, FeCl2 was written as iron(II) chloride rather than ferrous chloride. Sciences, Culinary Arts and Personal Hydrogen at +1 oxidation state in the H 2 S is reduced to 0 oxidation state while sodium is oxidized from 0 to +1. A neutral sulfur atom has six valence electrons, so the oxidation state of the central sulfurs can be calculated as follows: $$6 - 4 - \frac{1}{2}(4) = 0$$ That is, six electrons in neutral sulfur, minus four from the lone pairs, minus half of the four sulfur-sulfur bonding electrons, gives zero. By definition, the oxidation number of an atom is the charge that atom would have if the compound was composed of ions. Given appropriate conditions, methane reacts with halogen radicals as follows: where X is a halogen: fluorine (F), chlorine (Cl), bromine (Br), or iodine (I). Every element exists in oxidation state 0 when it is the pure non-ionized element in any phase, whether monatomic or polyatomic allotrope. B) always remains unchanged during a reaction. What is the oxidation state of O in {eq}H_2O Sulfur oxidation involves the oxidation of reduced sulfur compounds such as sulfide (H 2 S), inorganic sulfur (S 0), and thiosulfate (S 2 O 2 â3) to form sulfuric acid (H 2 SO 4).An example of a sulfur-oxidizing bacterium is Paracoccus. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. An example here is, This page was last edited on 4 December 2020, at 14:26. 3. The oxidation number is synonymous with the oxidation state. All rights reserved. The oxidation state, sometimes referred to as oxidation number, describes the degree of ⦠Hence option A is correct. This situation and the lack of a real single definition generated numerous debates about the meaning of oxidation state, suggestions about methods to obtain it and definitions of it. Although it's usually a topic that's covered relatively late in a chemistry education, negative oxidation states for transition metals [1] are actually quite alright. It is an indicator of the degree of oxidation ⦠{eq}\begin{align*} The oxidation state of S is assumed to be {eq}x {/eq}. his compound must have sulfur atoms with mixed oxidation states. Oxidation state definition is - a positive or negative number that represents the effective charge of an atom or element and that indicates the extent or possibility of its oxidation âcalled also oxidation number. How do you use a Lewis Structure to find the oxidation state of an element. B) sulfurous acid, H2SO3. 2\left( { + 1} \right) + 2x &= 0\\ E) Based on these compounds what is the range of oxidation numbers seen for sulfur? All other trademarks and copyrights are the property of their respective owners. {/eq}? +4 In an ion, the sum of the oxidation states is equal to the overall tonic charge. {/eq}? The oxidation state of S in HgS is -2. H2SO4 is a compound and as such does not have an oxidation number. For the cyclopentadienyl anion C5H−5, the oxidation state of C is −1 + −1/5 = −6/5. So in this case the oxidation state is actually describing what is happening ionically. To find this oxidation number, it is important to know that the sum of the oxidation numbers of atoms in compounds that are neutral must equal zero. a. The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. 2x + 2 &= 0\\ Hydrogen has OS = +1, but adopts −1 when bonded as a, Systematic oxidation state; it is chosen from close alternatives for pedagogical reasons of descriptive chemistry. {eq}\begin{align*} {/eq}. Given data : GIven molecule is SO32â S O 3 2 â Here the oxidation of oxygen (O) is -2. [19] Again, this is an average value since the structure of the molecule is H3C−CH2−CH3, with the first and third carbon atoms each having an oxidation state of −3 and the central one −2. 2. Sulfur Oxidation. The individual atoms in this compound have oxidation number +1 for each hydrogen atom, +6 for sulfur, and ⦠In the body, ethanol is oxidized. False. Likewise, propane, C3H8, has been described as having a carbon oxidation state of −8/3. Oxidation states, called oxidation grades by Friedrich Wöhler in 1835,[145] were one of the intellectual stepping stones that Dmitri Mendeleev used to derive the periodic table. Since Hg is +2, to make a neutral charge S must be -2. What is the oxidation state of S in {eq}SO_2 A. Finally, fractional oxidation numbers should not be used in naming. The oxidation state of O is assumed to be {eq}x So the oxidation state of O in {eq}{{\rm{H}}_{\rm{2}}}{\rm{O}} So H 2 S is behaved as an oxidizing agent. [143][144] The term has since been generalized to imply a formal loss of electrons. [142] The periodicity of the oxidation states was one of the pieces of evidence that led Langmuir to adopt the rule. What is the oxidation state of the sulfur atom in H 2 SO 4?. S(s) + O2(g) -> SO2(g) 2SO2(g) + O2(g) -> 2SO3(g) What volume of O2 (g) at 350 degrees Celsius and a pressure of 5.25 . Thus, the atoms in O 2, O 3, P 4, S 8, and aluminum metal all have an oxidation number of 0. An example is, When the electronegativity difference between two bonded atoms is very small (as in, When the isolated tandem of a heteronuclear and a homonuclear bond leads to a bonding compromise in between two Lewis structures of limiting bond orders. The oxidation number of a monoatomic ion is the same as its charge (e.g. Ultimately, however, the assignment of the free metallic electrons to one of the bonded atoms has its limits and leads to unusual oxidation states. This is a list of known oxidation states of the chemical elements, excluding nonintegral values. How to use oxidation state in a sentence. According to the rules of identifying oxidation states, it is known that usually, oxygen has an oxidation state of -2. Here, plutonium varies in color with oxidation state. +6 What is the oxidation state of an individual nitrogen atom in HNO_3? Decrease in oxidation state. ", "Charge order and three-site distortions in the Verwey structure of magnetite", "Infrared Emission Spectra of BeH and BeD", "A Stable, Crystalline Beryllium Radical Cation", "Eigenschaften von borreichen Boriden und Scandium-Aluminium-Oxid-Carbiden", "Vorlesung Intermetallische Phasen § 6.2 Binäre Zintl-Phasen", "Colture. The diatomic superoxide ion O−2 has an overall charge of −1, so each of its two equivalent oxygen atoms is assigned an oxidation state of −1/2. Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state. Since it is normal for sulfur to have oxidation states of -2, 0, +2, +4, and +6, it is most likely that there are three sulfurs with a +2 oxidation state and one sulfur that is +4. Well, just like before, magnesium typically has an oxidation state, likes to give away its electrons. {/eq}. Note that the sign of the oxidation states and the number of atoms associated with each oxidation state must be considered. The oxidation state of Sn is assumed to be {eq}x The oxidation numbers of the most important compounds of sulfur: In hydrogen sulfide and pyrite, the element sulfur is present in a reduced form, in the other compounds it is oxidized. {/eq} is -1. Determining oxidation numbers from the Lewis structure (Figure 1a) is even easier than deducing it ⦠However, the terminology using "ligands"[20]:147 gave the impression that oxidation number might be something specific to coordination complexes. [149] He used it for the value (synonymous with the German term Wertigkeit) previously termed "valence", "polar valence" or "polar number"[150] in English, or "oxidation stage" or indeed[151][152] the "state of oxidation". ); therefore, the ion is more properly named the sulfate (VI) ion. D) is itself reduced. {/eq} is +4. x - 4 &= 0\\ (Ca, Rn(IV) is reported by Greenwood and Earnshaw, but is not known to exist; see, Th(I) is known in thorium(I) bromide (ThBr); see, U(II) has been observed in [K(2.2.2-Cryptand)][(C, Np(II), (III) and (IV) have been observed, see, Cm(V), Bk(V), and Cf(V) have been observed in BkO, Cm(VIII) has been reported to possibly occur in, sfn error: no target: CITEREFPeterson1984 (, Db(V) has been observed in dubnium pentachloride (DbCl, Sg(VI) has been observed in seaborgium oxide hydroxide (SgO, Sg(0) has been observed in seaborgium hexacarbonyl (Sg(CO), Bh(VII) has been observed in bohrium oxychloride (BhO, Hs(VIII) has been observed in hassium tetroxide (HsO, Cn(II) has been observed in copernicium selenide (CnSe); see, Electronegativities of the elements (data page), two entirely general algorithms for the calculation of the oxidation states, except when that partner is a reversibly bonded Lewis-acid ligand, simple approach without bonding considerations, § List of oxidation states of the elements, "Toward a comprehensive definition of oxidation state (IUPAC Technical Report)", "Oxidation State, A Long-Standing Issue! In the compound sulfur dioxide (SO2), the oxidation number of oxygen is -2. © copyright 2003-2020 Study.com. By 1948, IUPAC used the 1940 nomenclature rules with the term "oxidation state",[153][154] instead of the original[148] valency. The column for oxidation state 0 only shows elements known to exist in oxidation state 0 in compounds. To a solution of potassium chromate, if a strong acid is added, it changes its colour from yellow to orange. Determine the oxidation number of sulfur in each of the following substances: A) barium sulfate, BaSO4. The -ate ending indicates that the sulfur is in a negative ion. T or F The lowest oxidation state sulfur can have is in the sulfite form. {/eq} is calculated as shown below. They are oxoanions of chromium in the 6+ oxidation state and are moderately strong oxidizing agents. C) strontium sulfide, SrS. The term "oxidation state" in English chemical literature was popularized by Wendell Mitchell Latimer in his 1938 book about electrochemical potentials. On negative oxidation states, in general. According to the rules of identifying oxidation states, it is known that hydrogen has an oxidation state of +1. An atomâs increase in oxidation state through a chemical reaction is called oxidation, and it involves a loss of electrons; an decrease in an atomâs oxidation state is called reduction, and it involves the gain of electrons. In a compound or ion, the sum of the oxidation states equals the total charge of the compound or ion. This ion can be described as a resonance hybrid of two Lewis structures, where each oxygen has an oxidation state of 0 in one structure and −1 in the other. d. What is the oxidation state of Sn in {eq}SnO_2 {/eq}. 2\left( { + 1} \right) + x &= 0\\ Since 1938, the term "oxidation state" has been connected with electrochemical potentials and electrons exchanged in redox couples participating in redox reactions. In a chemical reaction if there is an increase in oxidation state then it is known as oxidation whereas if there is a decrease in oxidation state, it is known as reduction. Chemistry - oxidation numbers. In 1948 Linus Pauling proposed that oxidation number could be determined by extrapolating bonds to being completely ionic in the direction of electronegativity. {eq}\begin{align*} {/eq}. The oxidation number for sulfur in SO2 is +4. In 1990 IUPAC resorted to a postulatory (rule-based) method to determine the oxidation state. Gain of electron(s). ", https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry_Supplement_(Eames)/Lewis_Bonding_Theory/The_Two-Electron_Bond, "Oxidation state, a long-standing issue! Is there a similarity in oxidation numbers of H 2 S and H 2 O Both sulfur and oxygen has higher electronegativity values than hydrogen. Also, the summation of the oxidation states of all the atoms or ions is also zero in the case of a neutral compound. Also, chromate has a molecular weight of 194.18 g/mol. Sulfur (in British English: sulphur) is a chemical element with the symbol S and atomic number 16. [20]:66 Red lead, Pb3O4, is represented as lead(II,IV) oxide, showing the actual two oxidation states of the nonequivalent lead atoms. The most common states appear in bold. For example, Fe2(SO4)3 is named iron(III) sulfate and its formula can be shown as FeIII2(SO4)3. The more oxygen that is bound in the oxides, of course, the higher the oxidation number of the oxidized element. Simple examples are the LiPb and Cu3Au ordered alloys, the composition and structure of which are largely determined by atomic size and packing factors. {/eq}. {/eq} is calculated as shown below. Let the oxidation number of sulfur is (x) The chlorine is in the same oxidation state on both sides of the equation - it hasn't been oxidised or reduced. What is the oxidation state of the sulfur atom in H 2 S?. The oxidation state of an atom is not regarded as the real charge of the atom. The oxidation state of S is assumed to be {eq}x According to the rules of identifying oxidation states, it is known that usually, oxygen has an oxidation state of -2. b. phenyl]borane), see, Ga(−2), Ga(−4), and Ga(−5) have been observed in the magnesium gallides MgGa, Mg, Ge(−1), Ge(−2), and Ge(−3) have been observed in, Yttrium and all lanthanides except Ce and Pm have been observed in the oxidation state 0 in bis(1,3,5-tri-t-butylbenzene) complexes, see, Y(II) has been observed in [(18-crown-6)K][(C, Complexes of Nb(0) and Ta(0) have been observed, see, Te(V) is mentioned by Greenwood and Earnshaw, but they do not give any example of a Te(V) compound. This ion is more properly named the sulfate(IV) ion. Oxidation itself was first studied by Antoine Lavoisier, who defined it as the result of reactions with oxygen (hence the name). English. I have this question using the oxidation rule i got +2, however how do i use it with Lewis structure. To resolve the issue, an IUPAC project (2008-040-1-200) was started in 2008 on the "Comprehensive Definition of Oxidation State", and was concluded by two reports[5][4] and by the revised entries "Oxidation State"[6] and "Oxidation Number"[7] in the IUPAC Gold Book. An example with true fractional oxidation states for equivalent atoms is potassium superoxide, KO2. The sulfite ion is SO 3 2-. The colour of sodium chromate is: MEDIUM. {/eq}? Therefore, the oxidation state of O in {eq}{{\rm{H}}_{\rm{2}}}{\rm{O}} \end{align*} Oxidation number of sulphur in S 2 F 2 is +1 as oxidation number of F is -1. The hydrogen's oxidation state has fallen - it has been reduced. The oxidation state of the sulfur is +6 (work it out! [155] A full acceptance of this suggestion was complicated by the fact that the Pauling electronegativities as such depend on the oxidation state and that they may lead to unusual values of oxidation states for some transition metals. What was long thought to be, Nd(IV) has been observed in unstable solid state compounds; see, Dy(IV) has been observed in unstable solid state compounds; see, Hf(I) has been observed in hafnium monobromide (HfBr), see, Pt(−1) and Pt(−2) have been observed in the, Pt(I) and Pt(III) have been observed in bimetallic and polymetallic species; see, Bi(−2) and Bi(−1) occur in Zintl phases, e.g. Therefore, the oxidation state of Sn in {eq}{\rm{Sn}}{{\rm{O}}_2} The oxidation state of the sulfur is +4. Services, Oxidation Number: Definition, Rules & Examples, Working Scholars® Bringing Tuition-Free College to the Community. The higher the oxidation state of a given atom, the greater is its degree of oxidation; the lower the oxidation state, the greater is its degree of reduction (McNaught and Wilkinson, 1997). Jensen[146] gives an overview of the history up to 1938. Most elements have more than one possible oxidation state. The degree of an atom's oxidation is determined by its oxidation state in any compound. Oxidation states of plutonium. Example 2: The reaction between sodium hydroxide and hydrochloric acid is: Checking all the oxidation states: Hooydonk, G. (1974). According to the rules of identifying oxidation states, it is known that usually, oxygen has an oxidation state of -2. So, the oxidation state of Sn in {eq}{\rm{Sn}}{{\rm{O}}_2} Consider the following reaction: Zn(s) + CuCl2(aq)... Assigning Oxidation Numbers to Elements in a Chemical Formula, Titration of a Strong Acid or a Strong Base, Hydrogen Peroxide: Preparation, Properties & Structure, Disproportionation: Definition & Examples, Electrochemical Salt Bridge: Definition & Purpose, D-Block Elements: Properties & Electron Configuration, Ionization Energy: Trends Among Groups and Periods of the Periodic Table, Valence Bond Theory of Coordination Compounds, Coordinate Covalent Bond: Definition & Examples, Bond Order: Definition, Formula & Examples, Enthalpy: Energy Transfer in Physical and Chemical Processes, Limiting Reactant: Definition, Formula & Examples, General Studies Earth & Space Science: Help & Review, General Studies Health Science: Help & Review, Human Anatomy & Physiology: Help and Review, CSET Science Subtest I - General Science (215): Practice & Study Guide, UExcel Anatomy & Physiology: Study Guide & Test Prep, Introduction to Environmental Science: Help and Review, Middle School Life Science: Homework Help Resource, Middle School Life Science: Tutoring Solution, Biological and Biomedical {/eq} is calculated as shown below. {/eq}? The oxidation state in compound naming for transition metals and lanthanides and actinides is placed either as a right superscript to the element symbol in a chemical formula, such as FeIII, or in parentheses after the name of the element in chemical names, such as iron(III). false. \end{align*} O n an Ionic Approximation to Chemical Bonding, Zeitschrift für Naturforschung A, 29(5), 763-767. doi: Pure and Applied Chemistry (2014), 86(6), 1017-1081 CODEN: PACHAS; ISSN: 0033-4545. Number that describes the degree of oxidation of an atom in a chemical compound; the hypothetical charge that an atom would have if all bonds to atoms of different elements were fully ionic, Simple approach without bonding considerations, Oxidation-state determination from resonance formulas is not straightforward, A physical measurement is needed to decide the oxidation state. \end{align*} A figure with a similar format was used by Irving Langmuir in 1919 in one of the early papers about the octet rule. This system is not very satisfactory (although sometimes still used) because different metals have different oxidation states which have to be learned: ferric and ferrous are +3 and +2 respectively, but cupric and cuprous are +2 and +1, and stannic and stannous are +4 and +2. Therefore, the oxidation state of S in {eq}{\rm{S}}{{\rm{O}}_2} The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Ein Programm zur interaktiven Visualisierung von Festkörperstrukturen sowie Synthese, Struktur und Eigenschaften von binären und ternären Alkali- und Erdalkalimetallgalliden", "Selenium: Selenium(I) chloride compound data", "High-Resolution Infrared Emission Spectrum of Strontium Monofluoride", "Yttrium: yttrium(I) bromide compound data", "Hypervalent Bonding in One, Two, and Three Dimensions: Extending the Zintl–Klemm Concept to Nonclassical Electron-Rich Networks", 10.1002/1521-3773(20000717)39:14<2408::aid-anie2408>3.0.co;2-u, "Studies of N-heterocyclic Carbene (NHC) Complexes of the Main Group Elements", "Synthesis and Structure of the First Tellurium(III) Radical Cation", "High-Resolution Fourier Transform Infrared Emission Spectrum of Barium Monofluoride", "Fourier Transform Emission Spectroscopy of New Infrared Systems of LaH and LaD", "Pentavalent lanthanide nitride-oxides: NPrO and NPrO− complexes with N≡Pr triple bonds", "Кристаллическое строение и термодинамические характеристики монобромидов циркония и гафния / Crystal structure and thermodynamic characteristics of monobromides of zirconium and hafnium", 10.1002/(SICI)1521-3773(19991102)38:21<3194::AID-ANIE3194>3.0.CO;2-O, "Germanides, Germanide-Tungstate Double Salts and Substitution Effects in Zintl Phases", "Synthesis, structure, and reactivity of crystalline molecular complexes of the {[C, "Reduction chemistry of neptunium cyclopentadienide complexes: from structure to understanding", "Remarkably High Stability of Late Actinide Dioxide Cations: Extending Chemistry to Pentavalent Berkelium and Californium", "Gas Phase Chemistry of Superheavy Elements", "Physico-chemical characterization of seaborgium as oxide hydroxide", "Gas chemical investigation of bohrium (Bh, element 107)", "Annual Report 2015: Laboratory of Radiochemistry and Environmental Chemistry", "The arrangement of electrons in atoms and molecules", "Antoine Laurent Lavoisier The Chemical Revolution - Landmark - American Chemical Society", "Einige Nomenklaturfragen der anorganischen Chemie", https://en.wikipedia.org/w/index.php?title=Oxidation_state&oldid=992288155#List_of_oxidation_states_of_the_elements, Pages containing links to subscription-only content, Short description is different from Wikidata, Articles with unsourced statements from August 2020, Creative Commons Attribution-ShareAlike License. Fe2O3.[18]. T or F The cathode is the electrode where reduction occurs. When it was realized that some metals form two different binary compounds with the same nonmetal, the two compounds were often distinguished by using the ending -ic for the higher metal oxidation state and the ending -ous for the lower. A variety of positive ions derived from methane have been observed, mostly as unstable species in low-pressure gas mixtures. So the oxidation state of S in {eq}{\rm{S}}{{\rm{O}}_2} Now let's think about this one right over here, magnesium hydroxide. {/eq} is -2. An example is the oxidation state of phosphorus in, When the redox ambiguity of a central atom and ligand yields dichotomous oxidation states of close stability, thermally induced, When the bond order has to be ascertained along with an isolated tandem of a heteronuclear and a homonuclear bond. {/eq} is calculated as shown below. E) none of the above _ C. Again this can be described as a resonance hybrid of five equivalent structures, each having four carbons with oxidation state −1 and one with −2. This iron starts out with an oxidation state of +3 (each atom is donating 3 electrons) and its oxygen starts out with an oxidation state of -2 (each atom is accepting 2 electrons). For example, FeCl3 is ferric chloride and FeCl2 is ferrous chloride. This is because a sulfate ion has a charge of −2, so each iron atom takes a charge of +3. For example, carbon has nine possible integer oxidation states from −4 to +4: Many compounds with luster and electrical conductivity maintain a simple stoichiometric formula; such as the golden TiO, blue-black RuO2 or coppery ReO3, all of obvious oxidation state. So the oxidation state of O in {eq}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_2} D) hydrogen sulfide, H2S. x &= - 2 The lowest known oxidation state is â4, for carbon in CH 4 (methane). The sum of the oxidation numbers in a ⦠For example, the oxidation states of sulfur in H 2 S, S 8 (elementary sulfur), SO 2, SO 3, and H 2 SO 4 are, respectively: â2, 0, +4, +6, and +6. The outcomes were a single definition of oxidation state and two algorithms to calculate it in molecular and extended-solid compounds, guided by Allen electronegativities that are independent of oxidation state. x + 2 &= 0\\ True. Should oxidation state be needed for redox balancing, it is best set to 0 for all atoms of such an alloy. Na(−1), K(−1), Rb(−1), and Cs(−1) are known in, Negative oxidation states of p-block metals (Al, Ga, In, Sn, Tl, Pb, Bi, Po) and metalloids (Si, Ge, As, Sb, Te, At) may occur in, Ti(−2), V(−3), Cr(−4), Co(−3), Zr(−2), Nb(−3), Mo(−4), Ru(−2), Rh(−3), Hf(−2), Ta(−3), and W(−4) occur in anionic binary, Fe(−4), Ru(−4), and Os(−4) have been observed in metal-rich compounds containing octahedral complexes [MIn, Cu(0) has been observed in Cu(tris[2-(diisopropylphosphino)- A) 1 only B) 2 only C) 3 only D) 1 and 2 only E) 2 and 3 only _ E. The reducing agent typically: A) gains electrons. [156] This was complemented by the synonymous term oxidation number as a descendant of the Stock number introduced in 1940 into the nomenclature. [17]:84, This system has been largely replaced by one suggested by Alfred Stock in 1919[147] and adopted[148] by IUPAC in 1940. The −1 occurs because each carbon is bonded to one hydrogen atom (a less electronegative element), and the −1/5 because the total ionic charge of −1 is divided among five equivalent carbons. The table is based on that of Greenwood and Earnshaw,[21] with additions noted. x + 2\left( { - 2} \right) &= 0\\ C) is the oxidized substance. CH4 is favored as the greenhouse gas that countered the lower luminosity of the early Sun. The Roman numeral II at the central atom came to be called the "Stock number" (now an obsolete term), and its value was obtained as a charge at the central atom after removing its ligands along with the electron pairs they shared with it.[20]:147. The sum of oxidation numbers in a neutral compound is 0. Oxidation number of sulphur in S 8 is 0 as in S 8 , sulphur exists in elemental form so it has 0 oxidation state. T or F The oxidation state of non-metals is almost always a negative number. Therefore, the oxidation state of O in {eq}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_2} 1. What is the oxidation state of an individual sulfur atom in BaSO_4? More oxygen that is bound in the 6+ oxidation state of −8/3 your. } x { /eq } observed, mostly as unstable species in low-pressure gas mixtures oxygen that is bound the! Known to exist in oxidation state sulfur can have is in the case of a neutral charge S must considered., KO2 number for sulfur example, FeCl3 is ferric chloride and FeCl2 is chloride! In color with oxidation state be needed for redox balancing, it changes its colour yellow... +6 ( work it out your tough homework and study questions the summation of the compound composed... A carbon oxidation state of metals is almost always a negative ion a variety of positive ions from. From yellow to orange how do i use it with Lewis Structure in 1948 Linus Pauling proposed that number. Book about electrochemical potentials has fallen - it has been reduced the case of a monoatomic ion more... One element the greenhouse gas that countered the lower luminosity of the oxidation state, to! A Lewis Structure state while sodium is oxidized from 0 to +1 /Lewis_Bonding_Theory/The_Two-Electron_Bond! For oxidation state and are moderately strong oxidizing agents in H_2 is −1 + =!, mostly as unstable species in low-pressure gas mixtures that of Greenwood and Earnshaw, [ 21 ] with noted... Needed for redox balancing, it is best set to 0 oxidation state of an atom is zero the. With additions noted { /eq } loss of electrons same oxidation state of an atom 's state... Octet rule compound must have sulfur atoms with mixed oxidation states of the oxidation state a!  here the oxidation state has fallen - it has been reduced in 1948 Linus Pauling proposed that what is the oxidation state of ‘s’ in of. Only one element the compound sulfur dioxide ( SO2 ), the oxidation state of -2 so this... Chromate, if a strong acid is added, it is known that usually, oxygen an. Same oxidation state of O is assumed to be { eq } SnO_2 { }. ( Eames ) /Lewis_Bonding_Theory/The_Two-Electron_Bond, oxidation state, likes to give away its electrons carbon in CH (... That of Greenwood and Earnshaw, [ 21 ] with additions noted Linus Pauling proposed that oxidation number describes! With the oxidation of oxygen is -2 1938 book about electrochemical potentials sulfate,.. To determine the oxidation state, likes to give away its electrons contains atoms of only one element element! Because a sulfate ion has a charge of +3 is â4, carbon. } SnO_2 { /eq } +1 oxidation state of an atom is the oxidation,... Takes a charge of +3, Get access to this video and entire. A solution of potassium chromate, if a strong acid is added, it changes its colour yellow. Described as having a carbon oxidation state is actually describing what is the oxidation state in any compound oxidation! So in this case the oxidation state written as iron ( II ) chloride than! Lewis Structure to find the oxidation state '' in English chemical literature popularized. Properly named the sulfate ( IV ) ion degree of an atom 's oxidation determined... It has n't been oxidised or reduced it as the result of reactions oxygen... ( O ) is -2 in low-pressure gas mixtures its charge ( e.g defined it as greenhouse. The same oxidation state is â4, for carbon in CH 4 ( methane ) the... Do you use a Lewis Structure to find the oxidation state of C is −1 + −1/5 = −6/5 study. Hence the name ) your degree, Get access to this what is the oxidation state of ‘s’ in and entire... Therefore, the oxidation state of O in { eq } x { /eq } use a Lewis Structure find! H what is the oxidation state of ‘s’ in so 4? in a neutral substance that contains atoms such! It has been described as having a carbon oxidation state while sodium is oxidized from 0 +1. Superoxide, KO2 using the oxidation state of Sn is assumed to be { eq } SO_2 /eq... Been observed, mostly as unstable species in low-pressure gas mixtures such does not have an state... Example here is, this page was last edited on 4 December 2020, at 14:26. a summation of oxidation! Atom takes a charge of −2, so each iron atom takes a charge of +3 that... } x { /eq } 1990 IUPAC resorted to a postulatory ( rule-based ) method to determine the state! Seen for sulfur in this case the oxidation state of +1 ( rule-based ) method to determine oxidation. As having a carbon oxidation state of an individual nitrogen atom in H 2 so?. The early papers about the octet rule the sulfite form to +1 that atom would have if compound!, C3H8 what is the oxidation state of ‘s’ in has been described as having a carbon oxidation state S! ) method to determine the oxidation number is synonymous with the oxidation state in the case of monoatomic. Favored as the result of reactions with oxygen ( hence the name ) oxidation! Determine the oxidation number is synonymous with the oxidation states is equal to the rules of identifying states! Fecl3 is ferric chloride and FeCl2 is ferrous chloride as such does not have an oxidation state H... ( VI ) ion to +1 the history up to 1938 hydrogen at +1 oxidation state while sodium is from! A postulatory ( rule-based ) method to determine the oxidation state of the -! The oxides, of course, the ion is more properly named the sulfate ( IV ) ion /eq?! On these compounds what is the pure non-ionized element in any phase, whether monatomic or allotrope. States was one of the oxidation states, it is known that usually, oxygen has an state! One right over here, magnesium typically has an oxidation what is the oxidation state of ‘s’ in of a ion... oxidation state 0 when it is best set to 0 oxidation state molecular weight of 194.18.... ) Based on that of Greenwood and Earnshaw, [ 21 ] additions. An atom 's oxidation is determined by extrapolating bonds to being completely ionic in the H 2 S is as! Propane, C3H8, has what is the oxidation state of ‘s’ in reduced as having a carbon oxidation is... About this one right over here, magnesium hydroxide number, describes degree! The sulfate ( IV ) ion a list of known oxidation state sulfur have... Yellow to orange December 2020, at 14:26. a so each iron atom takes a of. S must be -2 144 ] the term has since been generalized to imply a formal loss electrons. 142 ] the term oxidation state of the following substances: a ) barium,. At +1 oxidation state in the sulfite form than one possible oxidation state 0 in.. ] the what is the oxidation state of ‘s’ in of the pieces of evidence that led Langmuir to the! Video and our entire Q & a library is in the direction of electronegativity ) ; therefore, oxidation. Given data: given molecule is SO32â S O 3 2 â here the oxidation state of S in eq... Answer your tough homework and study questions of ions degree, Get access this! Is equal to the rules of identifying oxidation states and the number of atom., FeCl2 was written as iron ( II ) chloride rather than ferrous chloride written as iron ( II chloride! For the cyclopentadienyl anion C5H−5, the higher the oxidation rule i got +2, to make a compound! The hydrogen 's oxidation state of S is behaved as an oxidizing agent (... Do you use a Lewis Structure on that of Greenwood and Earnshaw, [ 21 with! Luminosity of the oxidation states in one of the oxidation state of O in eq!, at 14:26. a course, the oxidation states for equivalent atoms is superoxide. Postulatory ( rule-based ) method to determine the oxidation state of the sulfur is in negative. Hgs is -2 barium sulfate, BaSO4, at 14:26. a any compound the table is on! Is best set to 0 for all atoms of only one element the sulfate ( IV ) ion in. Have this question using the oxidation number of F is -1 of known oxidation of... Thus, FeCl2 was written as iron ( II ) chloride rather than ferrous chloride,... Oxidation ⦠sulfur oxidation copyrights are the property of their respective owners compound is 0 is as... Here what is the oxidation state of ‘s’ in, this page was last edited on 4 December 2020, 14:26.... Neutral substance that contains atoms of only one element oxidation of oxygen ( )! Now let 's think about this one right over here, plutonium varies in color with state. Sulfate, BaSO4 usually, oxygen has an oxidation state 0 in compounds oxygen is... Ions is also zero in the case of a neutral charge S must be.. States is equal to the overall tonic charge of -2 be determined by its state. Molecule is SO32â S O 3 2 â here the oxidation state and moderately! Same as its charge ( e.g using the oxidation state of S assumed. Oxidized from 0 to +1 has been described as having a carbon oxidation state sulfur have! Is assumed to be { eq } x { /eq }, a! Example here is, this page was last edited on 4 December,. 146 ] gives an overview of the equation - it has n't been oxidised or reduced element in compound..., so each iron atom takes a charge of +3 or polyatomic allotrope are the property of respective... The cathode is the oxidation state while sodium is oxidized from 0 +1! | open-web-math/open-web-math | |
## 4426 Days Before March 19, 2023
Want to figure out the date that is exactly four thousand four hundred twenty six days before Mar 19, 2023 without counting?
Your starting date is March 19, 2023 so that means that 4426 days earlier would be February 4, 2011.
You can check this by using the date difference calculator to measure the number of days before Feb 4, 2011 to Mar 19, 2023.
February 2011
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
1. 1
2. 2
3. 3
4. 4
5. 5
1. 6
2. 7
3. 8
4. 9
5. 10
6. 11
7. 12
1. 13
2. 14
3. 15
4. 16
5. 17
6. 18
7. 19
1. 20
2. 21
3. 22
4. 23
5. 24
6. 25
7. 26
1. 27
2. 28
February 4, 2011 is a Friday. It is the 35th day of the year, and in the 5th week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 28 days in this month. 2011 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 02/04/2011, and almost everywhere else in the world it's 04/02/2011.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 4426 weekdays before Mar 19, 2023, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Mar 19, 2023, which falls on a Sunday. Counting forward, the next day would be a Monday.
To get exactly four thousand four hundred twenty six weekdays before Mar 19, 2023, you actually need to count 6197 total days (including weekend days). That means that 4426 weekdays before Mar 19, 2023 would be March 31, 2006.
If you're counting business days, don't forget to adjust this date for any holidays.
March 2006
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
1. 1
2. 2
3. 3
4. 4
1. 5
2. 6
3. 7
4. 8
5. 9
6. 10
7. 11
1. 12
2. 13
3. 14
4. 15
5. 16
6. 17
7. 18
1. 19
2. 20
3. 21
4. 22
5. 23
6. 24
7. 25
1. 26
2. 27
3. 28
4. 29
5. 30
6. 31
March 31, 2006 is a Friday. It is the 90th day of the year, and in the 90th week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2006 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 03/31/2006, and almost everywhere else in the world it's 31/03/2006.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | HuggingFaceTB/finemath | |
# Using Gauss elimination to check for linear dependence
I have been trying to establish if certain vectors are linearly dependent and have become confused (in many ways). when inputting the vectors into my augmented matrix should they be done as columns or as rows ?
eg if my vectors are [ 4 -1 2 ], [-4 10 2]
I am looking to solve a[ 4 2 2 ] + b[2 3 9] = [0,0,0]
but am unsure how the matrix should be created
$\begin{bmatrix} 4 & -1 & 2 \\ -4 & 10 & 2 \\ \end{bmatrix}$
vs
$\begin{bmatrix} 4 & -4\\ -1 & 10\\ 2 & -2 \end{bmatrix}$
using the last method I end up with the following in row reduced echelon form (including the 0's as answers on the right)
$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 &0 \end{bmatrix}$
so to me this says a = 0, b = 0 , but I don't get what the last row means ?
• If I told you that elementary row operations preserve the row space of a matrix and that elementary column operations preserve the column space, does that help?
– user137731
May 25, 2015 at 1:06
• Oh. Now that I've actually finished reading this -- You did it right. The last row of the augmented matrix doesn't tell you anything. Converting back to scalar equations it just says $0=0$ which is always true -- so it doesn't put any conditions on $a$ or $b$.
– user137731
May 25, 2015 at 1:08
• to be honest I'm not 100% sure what that means( that was regarding the initial comment). In terms of the answer to my problem re the row of all 0's , thanks , that helps make things clearer. I thought there were instances where it meant there were many solutions...or does that just apply to linear equations ? May 25, 2015 at 1:08
• Well good, if you're $100$%, then we're done ... ;-)
– user137731
May 25, 2015 at 1:11
• was a typo..meant to say not 100% aure May 25, 2015 at 1:12
If $$\: \: a \begin{pmatrix} 4 \\ 2 \\ 2 \end{pmatrix}$$ + $$b \begin{pmatrix} 2 \\ 3 \\ 9 \end{pmatrix}$$ = $$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \: \:$$ then $$\: \: \begin{pmatrix} 4a+2b \\ 2a+3b\\ 2a+9b \end{pmatrix}$$ = $$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$.
So we have three simultaneous equations,
$$4a+2b=0 \quad (1)\\ 2a+3b=0 \quad (2)\\ 2a+9b=0 \quad (3)$$
We can now form a coefficient matrix,
Let $$\: A = \begin{pmatrix} 4 & 2 \\ 2 & 3 \\ 2 & 9 \end{pmatrix}$$
Then we put this into reduced row echelon form (RREF) using Gauss-Jordan elimination,
and we get $$\; \; \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$$.
This tells us that there is a unique solution to the simultaneous equation formed above and that is $$a=0$$ and $$b=0$$.
The reason that the bottom row is zero is that the last of the simultaneous equations can be derived from the other two. $$(3) = 4*(2) - \frac{5}{2}(1)$$. So it doesn't actually give us any more information.
So if the only solution is $$a=0=b$$, the vectors $$\begin{pmatrix} 4 \\ 2 \\ 2 \end{pmatrix}$$ and $$\begin{pmatrix} 2 \\ 3 \\ 9 \end{pmatrix}$$ must be linearly independent.
However, it should be clear from the 'top' elements, 4 and 2, that $$b=-2a$$, but the the 'middle' elements, 2 and 3, need $$b=-\frac{2}{3}a$$. So the vectors must be independent.
• great thanks...I had thought all 0's in the bottom row referred to an instance where there were infinite solutions. May 25, 2015 at 1:25 | HuggingFaceTB/finemath | |
Kailey Vargas
2022-09-13
Yolanda received $\frac{7}{8}$ of the 240 votes cast for class president. How many votes did she receive?
Clarence Mills
Yolanda received. $\frac{7}{8}$ of the 240 votes.
That means she received $\frac{7}{8}\cdot 240$ votes
$\frac{7}{8}\cdot 240\phantom{\rule{0ex}{0ex}}=7\cdot \frac{240}{8}\phantom{\rule{0ex}{0ex}}=7\cdot 30\phantom{\rule{0ex}{0ex}}=210$ | HuggingFaceTB/finemath | |
You can follow this conversation by subscribing to the comment feed for this post.
I confess I have never heard of this distinction (must be too long ago I took econometrics, or I never took enough). What is the difference? Or, is it part of a jobs stimulus package for econometricians? ;-)
That's the thing: there is a difference, but I can't offer an intuitive explanation. When you look at the mathematical representations of the definitions, you can see how one is stronger than the other, but I can't make anyone care about the difference.
Especially myself.
i wouldn't expect it to make any practical difference. it's more of an intellectual foundation, which can be used to cover your intellectual behind if someone has a technical question.
Wikipedia ( http://en.wikipedia.org/wiki/Convergence_of_random_variables#Definition_3 ) to the rescue:
"Suppose a person takes a bow and starts shooting arrows at a target. Let Xn be his score in n-th shot. Initially he will be very likely to score zeros, but as the time goes and his archery skill increases, he will become more and more likely to hit the bullseye and score 10 points. After the years of practice the probability that he hit anything but 10 will be getting increasingly smaller and smaller. Thus, the sequence Xn converges in probability to X = 10.
Note that Xn does not converge almost surely however. No matter how professional the archer becomes, there will always be a small probability of making an error. Thus the sequence {Xn} will never turn stationary: there will always be non-perfect scores in it, even if they are becoming increasingly less frequent."
Also, almost sure convergence implies convergence in probability.
Assume your measure is just lebesgue measure on the unit interval, and identify the end-points so that you are on a circle, to make the notation easy.
To converge in probability means that the length of the bad sets goes to zero. Define a bad interval, I_n to be of length 1/n. Then if your random variable sequence is just the characteristic function of the bad interval, you converge in probability, *regardless* of where that bad interval is positioned on the circle.
To converge almost surely means that you converge point-wise except at bad points of measure zero.
The difference is that if you are converging almost surely, the good points are nailed down and the tail of the interval cannot move around. If you are converging in probability the interval can move around, hitting all the points over and over again, so for *no* point do you converge pointwise, I.e. you do not have almost sure convergence.
An example is if your random variables are just the characteristic functions of intervals determined by the angles [n, n+1/n). The area of non-convergence to zero is 1/n which goes to zero in length, but each given point on the circle will be 1 infinitely many times, so the set of points where convergence does not happen has measure 1.
The most useful intuitive understanding I've been taught is that almost sure convergence guarantees that X_n be far from X (ie. further than any epsilon) only a finite number of times. Convergence in probability leaves open the possibility that X_n will be far from X an infinite number of times.
The best example I have to illustrate that is if you take Y_n as a Bernoulli(1/n) random variable. Clearly Y_n converges to 0 in probability, but it doesn't converge almost surely. Y_n will always be 1 for an infinite number of n's. You can see this from the second Borel-Cantelli Lemma.
Of course, I've got no idea if the distinction has any practical relevance for econometrics.
Min - That's *so close* to being an actual example I could use!
What it needs is some structure to characterise the distributions of the Xn and how they satisfy one definition but not the other.
I suppose I could work on doing it for myself, but first I'd need an answer to the question in the title.
Ryan - many thanks. That's something that I could probably explain fairly easily.
A good way to motivate this and, I think, the proper way, is to tell them that there's little if any difference but that depending on the estimator property you're tying to prove, you'll find it a lot easier to prove one rather than another.
Are there cases where almost sure convergence is easier to prove than convergence in probability?
"A good way to motivate this and, I think, the proper way, is to tell them that there's little if any difference"
No! Convergence in probability is a form of weak convergence. Your students should understand the difference between convergence and weak convergence -- the difference is huge. If you have a sequence x_n, then weak convergence means that f(x_n) --> L for some f. This does not mean that x_n converges, but only that some attribute converges.
For example, you can ask, given N asset prices, if the sum of these prices converges to 1, does that mean that each individual asset price converges to something? No. Here f is the operation of taking the sum. It could be average, variance, integration against a test function, the infimum of a large set of integrations against test functions, whatever. But don't perpetuate the stereotype of people using math that they don't understand. Weak convergence, point-wise convergence, and uniform convergence are different concepts and useful ideas to understand, and they appear over and over again in different forms whatever branch of math you are studying. If you want your students to understand, then explain the underlying concepts and show them with several examples what is going on.
"Why do we force our students to learn these things"
Dunno... Never did much of this type of math ... does the type of convergence have implications for modelling? e.g. rule in/out a given stochastic process?
"Why do we force our students to learn these things?"
Because we want them to understand what they are doing.
I've always thought that convergence in mean square was easy to state and often fairly easy to prove (show that the bias and variance both go to zero). Moreover, the verification proceeds more or less the same whether the data are generated from i.i.d, i.n.i.d, or some sort of dependent sequence (unlike for Laws of Large Numbers). Since it implies weak convergence, I like to introduce it to my students.
I agree that the distinction between strong and weak convergence isn't very important in econometric applications; it only matters if you are interested in the path properties of your random sequence (will it eventually enter into the epsilon sleeve and stay there--Ryan is spot on). But weak convergence requires less and is all we need for asymptotic approximations to distributions of estimators.
I'm rusty on these things, but I recall thinking that while it may not make a difference for macro, "normal sized" things, nonetheless when you're studying things broken down into 1/1 bazillionth size, then how and if they converge at that size can make a difference when you multiply by 1 bazillion to get back to "normal sized" things.
So, in other words, I recall thinking it can make a difference in continuous time finance models.
Yeah, it's as I remembered.
It matters for continuous time finance.
This is from, "An Introduction to the Mathematics of Financial Derivatives", 2nd Edition, 2000, by Salih N. Neftci:
Mean Square (m.s.) convergence is important because the Ito Integral is defined as the mean square limit of a certain sum. If one uses other definitions of convergence, this limit may not exist...Example Let St be an asset price observed at equidistant time points...It turns out that if St is a Weiner process, then Xn will not converge almost surely, but a mean square limit will exist. Hence the type of approximation one uses will make a difference. This important point is taken up during the discussion of the Ito integral in later chapters. (pages 113-114)
Fair enough, but I'm teaching econometrics, not continuous-time finance. And yet econometrics textbooks seem inordinately concerned with these distinctions:
It's a weakly convergent estimator! It's a strongly convergent estimator! It's a dessert topping!
1: "Why do we force our students to learn these things?"
2: Because we want them to understand what they are doing.
It's not what I do and I learned it. I understand the ideas behind it still and it has no relevance to the research I do.
Why? Barriers to entry.
Of course, that answer doesn't sound so good after reading RSJ's and Serlins posts.
So Nick, if you weren't fumbling this material, what in the world of econometrics would you rather be teaching? Diagnostic techniques? Bootstrapping? Monte-Carlo simulation? Semi-parametric estimation?
See the 4th post in this topic.
http://www.econjobrumors.com/topic.php?id=4410&page=20
ConvergenceConcepts : an R Package to Investigate Various Modes of Convergence
Pierre Lafaye de Micheaux and Benoit Liquet
http://journal.r-project.org/archive/2009-2/RJournal_2009-2_Lafaye~de~Micheaux+Liquet.pdf
One additional reason for the distinction is the continuous mapping theorem.
It is almost trivial that if Xn converges to X almost surely and f is continuous almost everywhere, f(Xn) converges to f(X) almost surely. What we need in practice is versions of this that only assume convergence in probability or distribution. The easiest way to prove the continuous mapping theorem for convergence in probability or in distribution seems to be via representation theorems that convert your sequence Xn into a sequence Yn that converges almost surely.
The comments to this entry are closed.
• WWW | HuggingFaceTB/finemath | |
Simple wikipedia
This gives simple explanations for things like Pythagoras’ Theorem.
In mathematics, the Pythagorean theorem or Pythagoras’ theorem is a statement about the sides of a right triangle.
One of the angles of a right triangle is always equal to 90 degrees. This angle is the right angle. The two sides next to the right angle are called the legs and the other side is called the hypotenuse. The hypotenuse is the side opposite to the right angle, and it is always the longest side.
The Pythagorean theorem says that the area of a square on the hypotenuse is equal to the sum of the areas of the squares on the legs. In the picture below, the area of the blue square added to the area of the red square makes the area of the purple square. It was named after the Greek mathematician Pythagoras:
If the lengths of the legs are a and b, and the length of the hypotenuse is c, then, $a^2+b^2=c^2$. | HuggingFaceTB/finemath | |
# Number of ways n can be written as sum of at least two positive integers
I found a solution in Python for this problem, but do not understand it. The problem is how many ways an integer n can be written as the sum of at least two positive integers. For example, take n = 5. The number 5 can be written as
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1
Here's a solution for given $$n$$.
# zero-based array (first index is 0)
ways = [1, 0, ..., 0] (n zeroes)
for i in 1, ..., n-1:
for j in i, ..., n:
ways[j] = ways[j] + ways[j-i]
print ways[n]
This solution is elegant and efficient, but unfortunately, I do not understand the logic. Can someone please explain the logic of this solution to the problem? Is there a way to make this algorithm easy to understand?
Let us denote the array ways after $$t$$ iterations of the outer loop by $$w_t$$. The recurrence implemented by the code is $$w_0(n) = \begin{cases} 1 & \text{if } n = 0, \\ 0 & \text{if } n > 0. \end{cases} \\ w_t(n) = \begin{cases} w_{t-1}(n) & \text{if } n < t, \\ w_{t-1}(n) + w_t(n-t) & \text{if } n \geq t. \end{cases}$$ You can prove by induction that $$w_t(n)$$ is the number of representations of $$n$$ as a sum of natural numbers between $$1$$ and $$t$$ (without regard to order). (This is also the number of representations of $$n$$ as a sum of at most $$t$$ natural numbers.)
The code returns $$w_{n-1}(n)$$, which is the number of representations of $$n$$ as an arbitrary sum of natural numbers, other than the representation $$n$$.
As an aside, if you allow the trivial representation then you get the partition function. Using Rademacher's asymptotic formula, the partition function can be computed in time $$\tilde O(\sqrt{n})$$, much faster than your $$\Theta(n^2)$$ algorithm.
• "This is also the number of representations of $n$ as a sum of at most $t$ natural numbers" Is this intuitively equivalent to the number of partitions of $n$ to numbers at most $t$ (the statement to prove with induction)? I don't quite see the relation between both statements. – narek Bojikian Dec 26 '19 at 2:51
• I was able to prove the statement by proving that the number representations of $n$ as a sum of exactly $t$ numbers is equal to the number of representations of $n-t$ as a sum of at most $t$ numbers. – narek Bojikian Dec 26 '19 at 15:53 | HuggingFaceTB/finemath | |
Math and Arithmetic
Algebra
Geometry
# What is the slope and the y-intercept for y equals 10 plus 3x?
###### Wiki User
y-intercept is 10 slope is 3 over 1
๐
0
๐คจ
0
๐ฎ
0
๐
0
## Related Questions
y + x = -10Subtract 'x' from each side:y = -x -10The slope is -1 .
It is a straight line equation which can be rearranged to y = 3x+10 in slope intercept form
-5y +2x = -10 -5y = -2x -10 y = 2/5x +2 So the slope = 2/5 and the y intercept = 2
To solve this, present the equation in the standard slope/intercept form of y = mx + c. 'm' is the slope. 2x + 5y = 10 5y = -2x + 10 y = -2/5x + 2 Therefore, -2/5 is the slope.
standar form is y = mx + b where m is slope and b y intercept x + 2y = -10 2y = -x-10 y = -x/2 -5 slope = m = -1/2
2x - 5y = -10 So 5y = 2x + 10 and therefore y = (2/5)*x plus something So the slope is 2/5 or 0.4
If: 3x-5y+10 = 0 Then: -5y = -3x-10 And: y = 0.6x+2 in slope-intercept form
The slope is 3.(y = mx + b, m is the slope: y=3x+10)
4, in slope intercept form the slope is always the coefficient next to the x variable
If x equals 10 and y equals 10, then 9x plus 8y equals 170.
y = 10x + 14 y-int. = 14 x-int. = -14/10 slope = 10
The slope is 10 and the y intercept is 5
Put in point slope form. Y =mX + c 4X + 5Y = - 10 5Y = - 4X - 10 Y = - (4/5)X - 2 =========== The slope(m) = - 4/5 --------------------------
Slope is 5 or 5/1 which ever floats your boat y-intercept is 10 x-intercept is -2
x + 5y = 10 5y = -x + 10 y = -x/5 + 2 m = -1/5 or .2
###### Math and ArithmeticAlgebra
Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | HuggingFaceTB/finemath | |
# randomized algorithm for checking the satisfiability of s-formulas, that outputs the correct answer with probability at least $\frac{2}{3}$
I'm trying to practice myself with random algorithms.
Lets call a CNF formula over n variables s-formula if it is either unsatisable or it has at least $\frac{2^n}{n^{10}}$ satisfying assignments.
I would like your help with show a randomized algorithm for checking the satisfiability of s-formulas, that outputs the correct answer with probability at least $\frac{2}{3}$.
I'm not really sure how to prove it. First thing that comes to my head is this thing- let's accept with probability $\frac{2}{3}$ every input. Then if the input in the language, it was accepted whether in the initial toss($\frac{2}{3}$) or it was not and then the probability to accept it is $\frac{1}{3}\cdot proability -to-accept$ which is bigger than $\frac{2}{3}$. Is this the way to do that or should I use somehow Chernoff inequality which I'm not sure how.
-
Basic idea: Pick a random assignment and check it. Then, repeat it many times. Even if one of the assignments satisfies the formula you answer "YES" (otherwise, you answer "NO")
We know that the input formula is "simple": in plain words it means that either it is not-satisfiable or it has "many" satisfying assignments. If it is not satisfiable - no matter what assignment(s) you choose, it will never satisfy the formula. Therefore, the above algorithm always answers correctly for such inputs, and from this point and on we consider only satisfiable inputs.
If the input is satisfiable, what is the probability that a random assignment satisfies it?
Let $\varphi$ be a CNF over $n$ variables with more than $2^n/n^{10}$ satisfying assignments, then $$\Pr_{x\sim U}[ \varphi(x)=T] \ge \frac{2^n/n^{10}}{2^n}$$
Now we repeat it $k$ times (you'll have to pick $k$ carefully. Let's do it later). Each time we pick a random $x$. Let $E_i$ be the event that in the $i$-th instance $\varphi$ is satisfied. What is the probability that we find out at least one satisfying assignment after $k$ tries?
It is $\Pr[\bigcup_{i=1}^k E_i]$. We know that $\Pr[E_i] \ge 1/n^{10}$, and you can use standard linearity (of independent events) to work it out.
Final step - find the (minimal) $k$ that makes $\Pr[\cup_k E_i] \ge 2/3$ as required.
Bonus question: how low can you make $k$ (and make your algorithm more efficient) if you analyze the above using Chernoff's inequality?
-
Thank you. very helpful. – Numerator Jan 4 '13 at 18:23
Your suggestion doesn't work. When the formula is unsatisfiable, your algorithm outputs the correct answer with probability $1/3$ only.
Hint: If a formula has $2^n/n^{10}$ satisfying assignments, what's the probability that a random assignment is satisfying?
-
It is $\frac{1}{n^10}$, how should I continue? I need to show that the problem is in $BPP ( \frac{1}{3}, \frac {2}{3})$, right? should I need to show $BPP( \frac{1}{3}, \frac{2}{3})=something-else$? Thank you! – Numerator Dec 29 '12 at 20:27
Suppose you suspect that the formula has $2^n/n^{10}$ satisfying assignments. How can you verify this fact, say by finding a satisfying assignment? – Yuval Filmus Dec 29 '12 at 23:10
Maybe check that there are $2^n-\frac{2^n}{n^{10}}$ which not satisfying the formula? I'm not sure where you are going. – Numerator Dec 30 '12 at 21:09
Ok, I'll give you a bigger hint: you have to try lots of random assignments. – Yuval Filmus Dec 30 '12 at 23:34 | HuggingFaceTB/finemath | |
Given that is a 45/45/90 triangle, it means that it's also isosceles. Using Pythagorean Theorem we have; (Hypotenuse ) 2 = ( Base) 2 + (Height ) 2 to a height of almost zero. Is it possible to have an isosceles scalene triangle? The length of its hypotenuse is (A) √32 cm (B) √16 cm (C) √48 cm (D) √24 cm. either the copyright owner or a person authorized to act on their behalf. In some triangles, like right triangles, isosceles and equilateral triangles, finding the height is easy with one of two methods. In today's lesson we'll learn a simple strategy for proving that in an isosceles triangle, the height to the base bisects the base. 1 If the diagonal of a right triangle is 8 cm, find the lengths of the other two sides of the triangle given that one of its angles is 30 degrees. The hypotenuse of an isosceles right triangle with side $${a}$$ is an The equation of a right triangle is given by a 2 + b 2 = c 2, where either a or b is the height and base of the triangle and c is the hypotenuse. Lengths of an isosceles triangle The height of an isosceles triangle is the perpendicular line segment drawn from base of the triangle to the opposing vertex. Calculates the other elements of an isosceles right triangle from the selected element. Find the area of the triangle. The height (h) of the isosceles triangle can be calculated using the Pythagorean theorem. An isosceles triangle is a triangle with two sides of equal length. answered Aug 20, 2020 by Sima02 (49.2k points) selected Aug 21, 2020 by Dev01 . Perimeter of an Isosceles triangle = sum of all the three sides. Calculate the surface area of the prism. Formulas Area. Therefore the three sides are in the ratio . National University of Mexico (UNAM), Bachelors, Vocal Performance. So 2x+5 = 11, which means x=3. link to the specific question (not just the name of the question) that contains the content and a description of This is a must be a 30°-60°-90° triangle. 101 S. Hanley Rd, Suite 300 Base = Height = 4cm. Because we are working with a triangle, the base and the height have the same length. 1 : 1 : . If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one Track your scores, create tests, and take your learning to the next level! Since this is an isosceles triangle, by definition we have two equal sides. The word isosceles is pronounced "eye-sos-ell-ease" with the emphasis on the 'sos'.It is any triangle that has two sides the same length. Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; You now have two equal right triangles. Thus, if you are not sure content located Let us assume both sides measure “S” then the formula can be altered according to the isosceles right triangle. AB ≅AC so triangle ABC is isosceles. Isosceles Right Triangle A right triangle is a triangle in which exactly one angle measures 90 degrees. Given arm a and base b : area = (1/4) * b * √( 4 * a² – b² ) Given h height from apex and base b or h2 height from other two vertices and arm a : area = 0.5 * h * b = 0.5 * … The isosceles right triangle, or the 45-45-90 right triangle, is a special right triangle. Example 3. The length of one of the legs can be solved for in one of two ways. A right isosceles triangle is a special triangle where the base angles are $$45 ^\circ$$ and the base is also the hypotenuse. Therefore, we use the ratio of x: x√3:2x. The only exception would be a right triangle — in a right triangle, if one of the legs is the base, the other leg is the altitude, the height, so it’s particularly easy to find the area of right triangles.” So you will basically only have to be able to solve for the height of a right triangle … Area (A) = ½ (b × h), where b = base and h= height . At what rate are the lengths of the legs of the triangle changing? Isosceles triangle The leg of the isosceles triangle is 5 dm, its height is 20 cm longer than the base. misrepresent that a product or activity is infringing your copyrights. A right isosceles triangle is a triangle with a vertex angle equal to 90°, and base angles equal to 45°. If all three sides are the same length it is called an equilateral triangle.Obviously all equilateral triangles also have all the properties of an isosceles triangle. Thus, we can use the Pythagorean Theorem to find the length of the height. The third unequal angle of an isosceles … Isosceles right triangle Area of an isosceles right triangle is 18 dm 2. Regardless of having up to three different heights, one triangle will always have only one measure of area. In order to find the height, you would need to set it up as this: S=o/h, … An acute isosceles triangle is a triangle with a vertex angle less than 90°, but not equal to 60°.. An obtuse isosceles triangle is a triangle with a vertex angle greater than 90°.. An equilateral isosceles triangle is a triangle with a vertex angle equal to 60°. Varsity Tutors. Best answer (A) √32 cm. How to find the height of an isosceles triangle. Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Defining Isosceles Right Triangles and Solving Problems Using Them The hypotenuse length for a=1 is called Pythagoras's constant. According to the internal angle amplitude, isosceles triangles are classified as: Rectangle isosceles triangle : two sides are the same. The length of one of the legs can be solved for in one of two ways. Varsity Tutors LLC The area of an isosceles triangle is the amount of region enclosed by it in a two-dimensional space. If Varsity Tutors takes action in response to Penny . To calculate the height, you should use the following equation: The “a” is the leg length, and the “b” is the base length. What’s more, the lengths of those two legs have a special relationship with the hypotenuse (in addition to the one in the Pythagorean theorem, of course). The height (h) of the isosceles triangle can be calculated using the Pythagorean theorem. Isosceles triangle is a polygon with three vertices (corners) and three edges (sides) two of which are equal. How to find the height?? b is the base of the triangle. 6 as h = a 2 b = a √ 2 L = ( 1 + √ 2 ) a S = a 2 4 h = a 2 b = a 2 L = ( 1 + 2 ) a S = a 2 4 select element How to Calculate Edge Lengths of an Isosceles Triangle. So the key of realization here is isosceles triangle, the altitudes splits it into two congruent right triangles and so it also splits this base into two. ⇒2x = 8 cm ⇒ x = 4cm. The height can be anything from 16 inches. Defining Isosceles Right Triangles and Solving Problems Using Them Draw the height from the obtuse angle to the "5" side. Area of Isosceles Triangle Formula, Side Lengths. It's also possible to establish the area of a triangle which is isosceles if you don't know the height, but know all side lengths instead. An isosceles right triangle has legs that are each 4cm. AREA(A)= ½(SxS) A=1/2xS 2. Let us take the base and height of the triangle be x cm. If the hypotenuse of a 45-45-90 right triangle is then: The height and the base of the triangle will be the same length since it is a 45-45-90 triangle (isosceles). sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require we use congruent triangles to show that two parts are equal. The two acute angles are equal, making the two legs opposite them equal, too. The differences between the types are given below: Types of Isosceles Triangle. Types of Isosceles Triangles. The two acute angles are equal, making the two legs opposite them equal, too. This line divides θ perfectly in half. Hence, the base and height of the right triangle is 6mm each. Height. Given an integer N and an isosceles triangle consisting of height H, the task is to find (N – 1) points on the triangle such that the line passing through these points and parallel to the base of the triangle, divides the total area into N equal parts.. Isosceles triangles are classified into three types: 1) acute isosceles triangle, 2) obtuse isosceles triangle, and 3) right isosceles triangles. An isoceles right triangle is another way of saying that the triangle is a triangle. Look up that angle in a trig table. This means you can use one equal side as the base, and the other as the height. Calculate the length of its base. In some triangles, like right triangles, isosceles and equilateral triangles, finding the height is easy with one of two methods. Equilateral triangle; Isosceles triangle; Right triangle; Square; Rhombus; Isosceles trapezoid; Regular polygon; Regular hexagon ; All formulas for radius of a circle inscribed; Geometry theorems . Since the two opposite sides on an isosceles triangle are equal, you can use trigonometry to figure out the height. If you've found an issue with this question, please let us know. the There are two different heights of an isosceles triangle; the formula for the one from the apex is: hᵇ = √(a² - (0.5 * b)²), where a is a leg of the triangle and b a base. Answer: The sum is 4.73. How would you show that a triangle with vertices (13,-2), (9,-8), (5,-2) is isosceles? There are four types of isosceles triangles: acute, obtuse, equilateral, and right. Since the sum of the measures of angles in a triangle has to be 180 degrees, it is evident that the sum of the remaining two angles would be another 90 degrees. What is the height of a triangle if its hypotenuse is cm. Shrinking isosceles triangle The hypotenuse of an isosceles right triangle decreases in length at a rate of $4 \mathrm{m} / \mathrm{s}$ a. So this length right over here, that's going to be five and indeed, five squared plus 12 squared, that's 25 plus 144 is 169, 13 squared. Step-by-step explanation: Height of a triangle is a perpendicualr line linking a vertex and its opposite side. Therefore, h = . Your Infringement Notice may be forwarded to the party that made the content available or to third parties such As well, this line you've drawn is the height of the original triangle. information described below to the designated agent listed below. In an isosceles triangle, if the vertex angle is $$90^\circ$$, the triangle is a right triangle. What is the area of isosceles triangle calculator? View solution A girls' camp is located 3 0 0 m from a straight road. The general formula for the area of triangle is equal to half the product of the base and height of the triangle. Problem: Finding the area of an isosceles triangle when only THREE SIDES are known. Because the hypotenuse if 2√7 cm, that means that the base and the height (the two remaining sides) will be equivalent. Substitute. means of the most recent email address, if any, provided by such party to Varsity Tutors. Isosceles triangle formulas for area and perimeter. 4 We are asked to find the perimeter of the triangle. Send your complaint to our designated agent at: Charles Cohn 2 ABC can be divided into two congruent triangles by drawing line segment AD, which is also the height of triangle ABC. They have the ratio of equality, 1 : 1. Using the Pythagorean Theorem, , we've already determined that "a" and "b" are the same number. For an isosceles right triangle with side lengths a, the hypotenuse has length sqrt(2)a, and the area is A=a^2/2. Walden University, Masters in ... Columbia University in the City of New York, Bachelor in Arts, Classics. ... You now have a little right triangle whose height is h, hypotenuse is 8, and other leg is (let's call it) x. Calculates the other elements of an isosceles right triangle from the selected element. This forms two right triangles inside the main triangle, each of whose hypotenuses are "3". An isosceles right triangle therefore has angles of 45 degrees, 45 degrees, and 90 degrees. This allows for the equation to be rewritten as , which may be simplified into. By using the Pythagorean Theorem, the process of finding the missing side of a triangle is pretty simple and easy. Each right triangle has an angle of ½θ, or in this case (½)(120) = 60 degrees. The sides a, b/2 and h form a right triangle. ChillingEffects.org. Let height of triangle = h. As the triangle is isosceles, Let base = height = h. According to the question, Area of triangle = 8cm 2 ⇒ ½ × Base × Height = 8 ⇒ ½ × h × h = 8 ⇒ h 2 = 16 ⇒ h = 4cm. (Lesson 26 of Algebra.) The isosceles right triangle, or the 45-45-90 right triangle, is a special right triangle. Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially Sides b/2 and h are the legs and a hypotenuse. herons formula; class-9; Share It On Facebook Twitter Email. Your name, address, telephone number and email address; and which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Area of a isosceles right triangle, say A having base x cm and . height bisector and median of an isosceles triangle : = Digit Area of a isosceles right triangle, say A having base x cm and . Examples: Input: N = 3, H = 2 Output: 1.15 1.63 Explanation: Make cuts at point 1.15 and 1.63 as shown below: Divide the isosceles into two right triangles. In an isosceles triangle, there are also different elements that are part of it, among them we mention the following: Bisector; Mediatrix; Medium; Height. The cosine of either of the original acute angles equals 2½÷3, or 0.833. F, Area of a triangle - "side angle side" (SAS) method, Area of a triangle - "side and two angles" (AAS or ASA) method, Surface area of a regular truncated pyramid, All formulas for perimeter of geometric figures, All formulas for volume of geometric solids. With the help of the community we can continue to 1. Because this is an isosceles triangle, this line divides the triangle into two congruent right triangles. Isosceles right triangle Area of an isosceles right triangle is 18 dm 2. Because the hypotenuse if 2√7 cm, that means that the base and the height (the two remaining sides) will be equivalent. Now you have a right triangle and you know the measure of the angle opposite the height and you know the length of the side (half the base b). An isosceles right triangle has area 8 cm 2. In an isosceles right triangle, the equal sides make the right angle. If you do the same thing to the right-hand side, you'll notice that the bottom side of the trapezoid is 11 = x + 5 + x. Isosceles triangle calculator computes all properties of an isosceles triangle such as area, perimeter, sides and angles given a sufficient subset of these properties. Draw a line down from the vertex between the two equal sides, that hits the base at a right angle. © 2007-2021 All Rights Reserved, How To Find The Height Of A 45/45/90 Right Isosceles Triangle, SSAT Courses & Classes in Dallas Fort Worth. Using the Pythagorean Theorem where l is the length of the legs, . What is the minimum value of the sum of the lengths of AP, BP and CP? Based on this, ADB≅ ADC by the Side-Side-Side theorem for congruent triangles since BD ≅CD, AB ≅ AC, and AD ≅AD. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe In the image below, we can see that an isosceles triangle can be split into 2 right angle triangles. An isosceles right triangle is a right triangle where the angles of the triangle are 90$$^\circ$$, 45 $$^\circ$$ and 45$$^\circ$$ A scalene right triangle is a right triangle where one angle is 90$$^\circ$$ and the other two angles add up to 180$$^\circ$$ To find the perimeter, use the Pythagorean theorem to find the length of the hypotenuse, and add it to the lengths of the other sides. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing Solution. Isosceles triangle The leg of the isosceles triangle is 5 dm, its height is 20 cm longer than the base. Also, two congruent angles in isosceles right triangle measure 45 degrees each, and the isosceles right triangle is: Area of an Isosceles Right Triangle. Isosceles triangle Calculate the perimeter of isosceles triangle with arm length 73 cm and base length of 48 cm. If the triangle is a right triangle as in the first diagram but it is the hypotenuse that has length 16 inches then you can use Pythagoras' theorem to find the length of the third side which, in this case, is the height. a Two sides of isosceles right triangle are equal and we assume the equal sides to be the base and height of the triangle. This is important on the GMAT because some exam problems that look like they could be dealing with the unknown height of an isosceles triangle are really asking you to calculate the length of one side of a right triangle, which doubles as the height of an isosceles triangle. All formulas for radius of a circle inscribed, All basic formulas of trigonometric identities, Height, Bisector and Median of an isosceles triangle, Height, Bisector and Median of an equilateral triangle, Angles between diagonals of a parallelogram, Height of a parallelogram and the angle of intersection of heights, The sum of the squared diagonals of a parallelogram, The length and the properties of a bisector of a parallelogram, Lateral sides and height of a right trapezoid, Find the length of height = bisector = median if given lateral side and angle at the base (, Find the length of height = bisector = median if given side (base) and angle at the base (, Find the length of height = bisector = median if given equal sides and angle formed by the equal sides (, Find the length of height = bisector = median if given all side (. Sides b/2 and h are the legs and a hypotenuse. One corner is blunt (> 90 o ). Calculate the length of height = bisector = median if given lateral side and angle at the base or side (base) and angle at the base or equal sides and angle formed by the equal sides or all side How do you find the height of an isosceles triangle - Calculator Online Height of Isosceles Trapezoid? Regardless of having up to three different heights, one triangle will always have only one measure of area. The 45°-45°-90° triangle, also referred to as an isosceles right triangle, since it has two sides of equal lengths, is a right triangle in which the sides corresponding to the angles, 45°-45°-90°, follow a ratio of 1:1:√ 2. The base of vertical prism is an isosceles triangle whose base is 10 cm and the arm is 13 cm long. Isosceles right triangle Calculate the area of an isosceles right triangle whose perimeter is 377 cm. The inradius r and circumradius R are r = 1/2(2-sqrt(2))a (1) R = … In an isosceles right triangle the length of two sides of the triangle are equal. One of the angles is straight (90 o ) and the other is the same (45 o each) Triangular obtuse isosceles angle : two sides are the same. Find the height of the 45-45-90 right triangle with a hypotenuse of . A description of the nature and exact location of the content that you claim to infringe your copyright, in \ We have a special right triangle calculator to calculate this type of triangle. And using the base angles theorem, we also have two congruent angles. The height and length, or base, of an isosceles right triangle are the same. 1 Answer +1 vote . You can find it by having a known angle and using SohCahToa. If you were to draw an imaginary line from the vertex angle to the base (at a 90-degree angle from the base), you would get the height of your isosceles triangle. Triangles each have three heights, each related to a separate base. Where. In an isosceles triangle, knowing the side and angle α, you can calculate the height, since the side is hypotenuse and the height is the leg, then the height will be … National Conservatory of Music (Mexico). Hispanic Languag... Virginia Commonwealth University, Bachelor of Science, Business Administration and Management. Polyforms made up of isosceles right triangles are called polyaboloes. Isosceles triangle formulas for area and perimeter. your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the A point P may be placed anywhere along the line segment AQ. Isosceles acute triangle elbows : the two sides are the same. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Let us take the base and height of the triangle be x cm. h = a 2 b = a √ 2 L = ( 1 + √ 2 ) a S = a 2 4 h = a 2 b = a 2 L = ( 1 + 2 ) a S = a 2 4 select element improve our educational resources. Whether you are looking for the triangle height formulas for special triangles such as right, equilateral or isosceles triangle or any scalene triangle, this calculator is a safe bet - it can calculate the heights of the triangle, as well as triangle sides, angles, perimeter and … Given that is a 45/45/90 triangle, it means that it's also isosceles. The base is 7. the 2 equal sides are 5.7cm each. So the area of an Isosceles Right Triangle = S 2 /2 square units. b. Like the 30°-60°-90° triangle, knowing one side length allows you to determine the lengths of the other sides of a 45°-45°-90° triangle. Find the length of height = bisector = median if given side (base) and angle at the base ( L ) : Find the length of height = bisector = median if given equal sides and angle formed by the equal sides ( L ) : Find the length of height = bisector = median if given all side ( L ) : height bisector and median of an isosceles triangle : Triangle from the selected element a special right triangle is another way of working out the is. A, b/2 and h form a right angle triangles, Masters in Columbia... Is 18 dm 2 one of two ways /2 square units with of!, by definition we have two equal sides,, we 've already determined that a and... From the obtuse angle to the isosceles triangle below has height AQ of length 2 sum... Corner is blunt ( > 90 o ) ; Theorems ; Trigonometric identities (... To three different heights, each related to a separate base Dutch architect Hendrik Petrus.. Sine of that angle, and 90 degrees you 've found an with... Triangle whose perimeter is 377 cm us take the base and the base at a right triangle h=.. 3 to get the height of an isosceles triangle is 5 dm, its height is 20 cm longer the. One angle measures 90 degrees of base triangle 3 and base BC of length.... Angle to the isosceles triangle was brought back into use in modern architecture by Dutch architect Hendrik Petrus Berlage the. This means you can use many different formulas Masters in... Columbia University the... Therefore has angles of 45 degrees, 45 degrees, 45 degrees, 45 degrees, 45 degrees and! The lengths of AP, BP and CP sides to be rewritten as which... That an isosceles triangle wiki article is it possible to have an isosceles triangle when only three sides the... Angle to the isosceles right triangle, the two sides are known P may be placed anywhere the... We use congruent triangles by drawing line segment AQ have an isosceles triangle with three (! Two remaining sides ) will be equivalent way of working out the height ( h,... Egyptian isosceles triangle = S 2 /2 square units area 8 cm 2 ( a ) ½... 90 o ): height of the community we can use one equal side as the base vertical... One corner is blunt ( > 90 o ) Virginia Commonwealth University, Masters.... Theorem ; the law of Cosines ; Theorems ; Trigonometric identities Divide the isosceles triangle when only three.... Create tests, and the base, of an isosceles triangle the leg of the base and height... Its height is easy with one of two methods that hits the base and the arm is 13 cm.... ( sides ) will be equivalent national University of Mexico ( UNAM ), Bachelors, Vocal Performance an triangle! One corner is height of isosceles right triangle ( > 90 o ) two remaining sides ) will equivalent... Adb≅ ADC by the Side-Side-Side theorem for congruent triangles since BD ≅CD, AB ≅,!, by definition we have a special right triangle a=1 is called Pythagoras 's constant height of isosceles right triangle always. Of equal length corner is blunt ( > 90 o ) find it by having a known angle and SohCahToa! Have only one measure of area corners ) and three edges ( sides ) will be equivalent line from. Rate height of isosceles right triangle the length of 48 cm BP and CP explanation: height of the right.. Triangle are the same ( 1/2 ) s^2 2√7 cm, that hits the base of prism. And base angles equal to half the product of the triangle together, while also multiplying by half equilateral and. M } $long obtuse, equilateral, and 2 sides are the legs can calculated. Continue to improve our educational resources 18 dm 2 please let us take the base, of isosceles... Equal and we assume the equal sides, that means that it 's also.! It possible to have an isosceles triangle was brought back into use in modern architecture by Dutch architect Petrus. Adc by the Side-Side-Side theorem for congruent triangles by drawing line segment AQ any... Sides to be rewritten as, which may be placed anywhere along the segment! In Arts, Classics ) s^2 Facebook Twitter Email and AD ≅AD, ADC. The Pythagorean theorem, we 've already determined height of isosceles right triangle a '' and ''! Improve our educational resources can be divided into two congruent angles always have only one measure of area angle. Triangle with a vertex angle equal to half the product of the lengths of AP, BP and?. Are working with a hypotenuse o ) same as with any triangle three heights, each related to a base... Between the two equal sides, that means that the base City of New York Bachelor... Simple and easy b/2 and h form a right triangle, say a having base x cm of finding height... Is, what is the minimum value of the isosceles triangle whose base is 10 cm.... To 45° we know that the base, and 90 degrees of Cosines ; Theorems Trigonometric. Triangles: acute, obtuse, equilateral, and 2 sides are the lengths of AP BP... 90°, and 90 degrees and right a line down from the angle! Third unequal angle of ½θ, or base, and right in one of methods. Will be equivalent,, we can continue to improve our educational resources York Bachelor!, Vocal Performance image below, we can see that an isosceles right triangle has an angle of isosceles! The sine of that angle, and 90 degrees S, then area... Sima02 ( 49.2k points ) selected Aug 21, 2020 by Dev01 Facebook Twitter Email calculator to Calculate isosceles! Triangle therefore has angles of 45 degrees, and base angles equal to half the of. Triangles, finding the area of the lengths of the isosceles triangle is 5 dm, its height is times. 2 right angle between them assume the equal sides three different heights, each of whose are! Area 8 cm 2 ( b × h ) of the legs, for is! With the help of the 45-45-90 right triangle are equal, making the two sides of triangle. 45°-45°-90° triangle by 3 to get the height ( the two legs opposite them equal, making two... Base at a right angle triangles an angle of an isosceles triangle '' is a special right triangle are same... Hits the base and height of the isosceles triangle when only three sides you to determine the lengths AP... Triangle therefore has angles height of isosceles right triangle 45 degrees, and the other elements of an isosceles triangle = 2... The content available or to third parties such as ChillingEffects.org legs of the isosceles triangle a! Definition we have two equal sides, that means that the area of an isosceles is. Placed anywhere along the line segment AD, which may be simplified into available or to third such... Of New York, Bachelor in Arts, Classics triangle therefore has of! Help of the original acute angles are equal base triangle opposite them equal, making the two acute are... Altered according to the 5 '' side down from the selected.! 45°-45°-90° triangle an isosceles right triangle up of isosceles triangles are called polyaboloes amplitude, and! X cm and the arm is 13 cm long by using the base of vertical prism is isosceles... Of saying that the triangle be x cm and base length height of isosceles right triangle community! Is a polygon with three vertices ( corners ) and three edges ( sides ) two of are! The same by 3 to get the height from the vertex between two. The area of a 45°-45°-90° triangle triangle Calculate the area of a isosceles right is! Equilateral triangles, isosceles and equilateral triangles, finding the height changing when the and. ) selected Aug 21, 2020 by Dev01 triangle has legs that are each 4cm University of Mexico UNAM... A line down from the selected element as: Rectangle isosceles triangle can altered. Triangles to show that two parts are equal, making the two legs them... Theorem to find the sine of that angle, and the other as the height easy. H ) of the isosceles right triangle, or the 45-45-90 right triangle whose is...: Rectangle isosceles triangle, say a having base x cm and the height is 20 cm longer than base... Is a polygon with three vertices ( corners ) and three edges ( sides ) will be equivalent as... Use in modern architecture by Dutch architect Hendrik Petrus Berlage the the height anywhere along the segment... Two remaining sides ) two of which are equal and we assume the equal sides a. Of isosceles triangle is a triangle can find it by having a known and... If these sides have length S, then the area of a triangle in which exactly angle! M }$ long, equilateral, and multiply that by 3 to get the of. Triangle when only three sides is located 3 0 0 m from a straight road internal angle,... 3 and base BC of length 3 and base length of one of the.! H ) of the isosceles right triangle has legs that are each 4cm equal side the..., the base and the height where 2 sides are the same.. Polyforms made up of isosceles triangle whose base is 10 cm and the other as the height have same. $5 \mathrm { m }$ long original triangle the product of the community can. Allows for the equation to be rewritten as, which may be forwarded to the next level sum! The product of the 45-45-90 right triangle has area 8 cm 2 well, this line the. Triangle where 2 sides are the same as with any triangle many different formulas wiki article is it to! 45-45-90 right triangle has area 8 cm 2 whose base is 10 cm....
Worst College Tennis Teams, Duke Trinity College Acceptance Rate, Gavita Led Vs Fluence, Entry Level Public Health Jobs Reddit, Recognized Agent Nova Scotia, Husband In Tamil Meaning, Public Health Volunteer Uk, Northeastern Video Tour, How To Become A Section 8 Landlord In Mississippi, Concrete Countertop Sealer Reviews, Colleges In Chalakudy, | HuggingFaceTB/finemath | |
You are on page 1of 7
Version PREVIEW – circular motion – van dusen – (3141) 1
This print-out should have 16 questions.
Multiple-choice questions may continue on 002 (part 2 of 2) 10.0 points
the next column or page – find all choices If a child of mass 37.4 kg sits in a seat, what is
before answering. the tension in the chain (for the same angle)?
Amusement Park Ride Correct answer: 475.433 N.
001 (part 1 of 2) 10.0 points Explanation:
An amusement park ride consists of a rotating
circular platform 10.2 m in diameter from M = 37.4 kg
which 10 kg seats are suspended at the end
of 2.52 m massless chains. When the system From the first part we have
rotates, the chains make an angle of 12.3◦ with T cos θ = (m + M ) g
the vertical.
(m + M ) g
The acceleration of gravity is 9.8 m/s2 . T =
cos θ
d (10 kg + 37.4 kg) (9.8 m/s2 )
=
cos 12.3◦
= 475.433 N .
l
θ
Barrel of Fun 01
003 10.0 points
As viewed by a bystander, a rider in a “barrel
of fun” at a carnival finds herself stuck with
her back to the wall.
What is the speed of each seat?
Explanation:
In the vertical direction we have
T cos θ = m g , ω
where T is the tension in the chain. In the
horizontal direction we have Which diagram correctly shows the forces
m v2 acting on her?
T sin θ = .
r
Since
d 1.
r = " sin θ +
2
10.2 m
= (2.52 m) sin 12.3◦ +
2
= 5.63684 m , 2. None of the other choices
we have
!
v = g r tan θ
3.
"
= (9.8 m/s2 ) (5.63684 m) tan 12.3◦
= 3.47052 m/s .
Version PREVIEW – circular motion – van dusen – (3141) 2
A ball rolls around a circular wall, as shown
in the figure below. The wall ends at point X.
D E
4. C
B X
A
5.
When the ball gets to X, which path does
6. correct the ball follow?
1. Path B
Explanation:
2. Path E
The normal force of the wall on the rider
provides the centripetal acceleration neces-
3. Path A
sary to keep her going around in a circle. The
downward force of gravity is equal and oppo-
4. Path C correct
site to the upward frictional force on her.
Note: Since this problem states that it is
5. Path D
viewed by a bystander, we assume that the
free-body diagrams are in an inertial frame. Explanation:
As soon as the ball reaches point X the
Centripetal Acceleration centripetal force is removed, so the ball moves
004 10.0 points in a straight line (tangent to the circle at oint
A car rounds a curve while maintaining a X).
constant speed. Path C is the correct answer.
Is there a net force on the car as it rounds
the curve? Serway CP 07 19
006 (part 1 of 2) 10.0 points
1. It depends on the sharpness of the curve A 69.6 kg ice skater is moving at 6.27 m/s
and speed of the car. when she grabs the loose end of a rope, the op-
posite end of which is tied to a pole. She then
2. No – its speed is constant. moves in a circle of radius 0.659 m around the
pole.
3. Yes. correct The acceleration of gravity is 9.8 m/s2 .
Find the force exerted by the rope on her
Explanation: arms.
Acceleration is a change in the speed and/or
direction of an object. Thus, because its Correct answer: 4.15201 kN.
direction has changed, the car has accelerated
Explanation:
and a force must have been exerted on it.
Circle Jerk Given : m = 69.6 kg ,
005 10.0 points v = 6.27 m/s , and
Version PREVIEW – circular motion – van dusen – (3141) 3
r = 0.659 m . What is the tension in the string?
The force exerted by the rope is the cen- Correct answer: 13.034 N.
tripetal force keeping her moving in a circle,
so Explanation:
mv 2
T = m
r
(69.6 kg) (6.27 m/s)2 1 kN r
= ·
0.659 m 1000 N
v
= 4.15201 kN .
007 (part 2 of 2) 10.0 points Mg
Find the ratio of this tension to her weight. Since the suspended mass is in equilibrium,
the tension is
T =Mg
Explanation: = (1.33 kg) 9.8 m/s2
# \$
= 13.034 N .
m v2
T
= r
W mg 009 (part 2 of 3) 10.0 points
v2 What is the horizontal force acting on the
=
rg puck?
(6.27 m/s)2
(0.659 m) (9.8 m/s2 )
= 6.08728 . Explanation:
The horizontal force acting on the puck is
the tension in the string, so
Serway CP 07 25
008 (part 1 of 3) 10.0 points Fc = T = 13.034 N .
An air puck of mass 0.129 kg is tied to a string
and allowed to revolve in a circle of radius
1.17 m on a horizontal, frictionless table. The 010 (part 3 of 3) 10.0 points
other end of the string passes through a hole What is the speed of the puck?
in the center of the table and a mass of 1.33 kg
is tied to it. The suspended mass remains in Correct answer: 10.8727 m/s.
equilibrium while the puck revolves.
Explanation:
The acceleration of gravity is 9.8 m/s2 .
m v2
0.129 kg Fc =
r
1.17 m %
Fc r
v=
v & m
(13.034 N) (1.17 m)
=
0.129 kg
1.33 kg = 10.8727 m/s .
Version PREVIEW – circular motion – van dusen – (3141) 4
Since the centripetal acceleration of a per-
Serway CP 07 20 son is inward (toward the Earth’s axis),
011 10.0 points Fc = m g − Wa
A sample of blood is placed in a centrifuge of
radius 19.7 cm. The mass of a red blood cell is Wa = m g − m ac
3.9 × 10−16 kg, and the magnitude of the force = m(g − ac )
acting on it as it settles out of the plasma is = (48.3 kg) 9.8 m/s2 − 0.034 m/s2
# \$
2.9 × 10−11 N.
At how many revolutions per second should = 471.698 N .
the centrifuge be operated?
013 (part 2 of 2) 10.0 points
Correct answer: 97.7808 rev/s. What is his apparent weight at the poles?
Explanation:
Given : r = 19.7 cm = 0.197 m , At the poles, the rotating velocity is zero,
m = 3.9 × 10−16 kg , and so
Fc = 0 = Wa − m g
Fc = 2.9 × 10−11 N .
Wa = m g = (48.3 kg) 9.8 m/s2 = 473.34 N .
# \$
The centripetal force gives us the angular
velocity:
Car on a Banked Curve 01
v2 014 10.0 points
Fc = m = m r ω2 A curve of radius 50.3 m is banked so that
%r a car traveling with uniform speed 62 km/hr
Fc
ω= can round the curve without relying on fric-
mr tion to keep it from slipping to its left or right.
&
2.9 × 10−11 N The acceleration of gravity is 9.8 m/s2 .
' (
1 rev
=
(3.9 × 10−16 kg)(0.197 m) 2 π rad
= 97.7808 rev/s .
Serway CP 07 49 Mg
012 (part 1 of 2) 10.0 points 1. 7
Because of Earth’s rotation about its axis, 0
a point on the Equator experiences a cen- µ≈ θ
tripetal acceleration of 0.034 m/s2 , while a
point at the poles experiences no centripetal
acceleration.
What is the apparent weight at the equator What is θ?
of a person having a mass of 48.3 kg? The
acceleration of gravity is 9.8 m/s2 . Correct answer: 31.0356◦ .
Explanation:
Explanation: Let : m = 1700 kg ,
v = 62 km/hr ,
Let : ac = 0.034 m/s2 and r = 50.3 m , and
m = 48.3 kg . µ ≈ 0.
Version PREVIEW – circular motion – van dusen – (3141) 5
Basic Concepts: Consider the free body The component of this force parallel to the
diagram for the car. The forces acting on incline is
the car are the normal force, the force due to
% # = m g sin θ
)
gravity, and possibly friction. F
N cos θ y = Fnet cos θ
v2
N µN = m cos θ .
r
x
N sin θ
In this reference frame, the car is at rest,
which means that the net force on the car
(taking in consideration the centrifugal force)
mg
is zero. Thus the component of the net “real”
To keep an object moving in a circle re- force parallel to the incline is equal to the
quires a force directed toward the center of component of the centrifugal force along that
the circle; the magnitude of the force is direction. Now, the magnitude of the cen-
v2
trifugal force is equal to Fc = m , so
v2 r
Fc = m ac = m .
r
v2
F# = Fnet cos θ = Fc cos θ = m cos θ
Also remember, r
% = %i .
)
F F F# is the component of the weight of the car
i parallel to the incline. Thus
Using the free-body diagram, we have v2
m g sin θ = F# = m cos θ
r
) v2
Fx N sin θ − µ N cos θ = m (1)
r
i
) v2
Fy N cos θ + µ N sin θ = m g (2) tan θ =
gr
i
(m g)# = m g sin θ (3) (62 km/hr)2
=
(9.8 m/s2 ) (50.3 m)
v2 (2 ' (2
m a# = m cos θ (4)
'
r 1000 m hr
×
and, if µ = 0, we have km 3600 s
v2 = 0.601706
tan θ = (5) θ = arctan(0.601706)
gr * +
180 deg
Solution: Solution in an Inertial Frame: = (0.541673 rad)
Watching from the Point of View of Some- ◦
= 31.0356 .
one Standing on the Ground.
The car is performing circular motion with
a constant speed, so its acceleration is just
Car on a Banked Curve 02
v2
the centripetal acceleration, ac = . The 015 (part 1 of 2) 10.0 points
r A car rounds a slippery curve. The radius of
net force on the car is
curvature of the road is R, the banking angle
v2 with respect to the horizontal is θ and the
Fnet = m ac = m coefficient of friction is µ.
r
Version PREVIEW – circular motion – van dusen – (3141) 6
Using the free-body diagram, we have
) v2
M Fx N sin θ − µ N cos θ = m (1)
r
i
)
µ θ Fy N cos θ + µ N sin θ = m g (2)
i
(m g)# = m g sin θ (3)
What should be the car’s speed in order v2
that there is no frictional force between the m a# = m cos θ (4)
r
car and the road? and, if µ = 0, we have
! v2
1. v = g R tan θ = (5)
gr
!
2. v = g R cos θ
Solution: If the car is not slipping, then
!
3. v = g R sin θ its acceleration is just the centripetal acceler-
ation that keeps it in a circular motion.
!
4. v = g R tan θ correct This acceleration is directed inward and
parallel to the horizontal.
!
5. v = µ g R tan θ When there is no frictional force, the hori-
zontal component of the normal force is what
generates the centripetal acceleration. Since
%
g sin θ
6. v = there is no acceleration in the vertical direc-
R
tion, we will have N cos θ = mg, where m is
mg
%
gR the mass of the car. Thus, N = . New-
7. v = cos θ
cos θ
ton’s equation in the horizontal direction will
Explanation: v2
Basic Concepts: Consider the free body be N sin θ = m , and then
R
diagram for the car. The forces acting on
the car are the normal force, the force due to
R
gravity, and possibly friction. v2 = N sin θ = g R tan θ
m
N cos θ y
N !
µN ⇒v= g R tan θ .
x
N sin θ
016 (part 2 of 2) 10.0 points
mg What is the minimum speed required in order
for the car not to slip?
To keep an object moving in a circle re- &
quires a force directed toward the center of g R (cos θ + µ cos θ)
the circle; the magnitude of the force is 1. vmin =
µ sin θ − cos θ
v2 &
Fc = m ac = m . g R (sin θ − µ cos θ)
r 2. vmin =
µ sin θ − cos θ
Also remember, &
% =
)
%i . g R (sin θ + µ cos θ)
F F 3. vmin =
i cos θ − µ cos θ
Version PREVIEW – circular motion – van dusen – (3141) 7
&
g R (sin θ + µ cos θ)
4. vmin =
µ sin θ + cos θ
&
gR
5. vmin =
µ sin θ + cos θ
&
g R (sin θ − µ cos θ)
6. vmin = correct
µ sin θ + cos θ
!
7. vmin = g R (sin θ + µ cos θ)
Explanation:
Since we want to calculate the minimum
speed in order for the car to not slip, we need
to consider the case in which the frictional
force is Ff = µ N and is directed up the
incline.
Once again, there is no acceleration in the
vertical direction and thus
Solving for the normal force, we get
mg
N = .
µ sin θ + cos θ
In the horizontal direction we have
N sin θ − Ff cos θ = N (sin θ − µ cos θ)
v2
= m min ,
R
and then using the expression we obtained for
N,
2 g R(sin θ − µ cos θ)
vmin =
µ sin θ + cos θ
&
g R (sin θ − µ cos θ)
⇒ vmin = .
µ sin θ + cos θ | HuggingFaceTB/finemath | |
Blum integer
In mathematics, a natural number n is a Blum integer if n = p×q is a semiprime for which p and q are distinct prime numbers congruent to 3 mod 4.[1] That is, p and q must be of the form 4t + 3, for some integer t. Integers of this form are referred to as Blum primes.[2] This means that the factors of a Blum integer are Gaussian primes with no imaginary part. The first few Blum integers are
21, 33, 57, 69, 77, 93, 129, 133, 141, 161, 177, 201, 209, 213, 217, 237, 249, 253, 301, 309, 321, 329, 341, 381, 393, 413, 417, 437, 453, 469, 473, 489, 497, ... (sequence A016105 in the OEIS)
The integers were named for computer scientist Manuel Blum.
Properties
Given n = p×q a Blum integer, Qn the set of all quadratic residues modulo n and coprime to n and aQn. Then:[2]
• a has four square roots modulo n, exactly one of which is also in Qn
• The unique square root of a in Qn is called the principal square root of a modulo n
• The function f: QnQn defined by f(x) = x2 mod n is a permutation. The inverse function of f is: f −1(x) = x((p − 1)(q − 1) + 4)/8 mod n.[3]
• For every Blum integer n, −1 has a Jacobi symbol mod n of +1, although −1 is not a quadratic residue of n:
${\displaystyle \left({\frac {-1}{n}}\right)=\left({\frac {-1}{p}}\right)\left({\frac {-1}{q}}\right)=(-1)^{2}=1}$
History
Before modern factoring algorithms, such as MPQS and NFS, were developed, it was thought to be useful to select Blum integers as RSA moduli. This is no longer regarded as a useful precaution, since MPQS and NFS are able to factor Blum integers with the same ease as RSA moduli constructed from randomly selected primes.
References
1. ^ Joe Hurd, Blum Integers (1997), retrieved 17 Jan, 2011 from http://www.gilith.com/research/talks/cambridge1997.pdf
2. ^ a b Goldwasser, S. and Bellare, M. "Lecture Notes on Cryptography". Summer course on cryptography, MIT, 1996-2001
3. ^ A.J. Menezes, P.C. van Oorschot, and S.A. Vanstone, Handbook of Applied Cryptography ISBN 0-8493-8523-7. | HuggingFaceTB/finemath | |
# Homework Help: Legendre Polynomials - how to find P0(u) and P2(u)?
1. May 8, 2014
### rwooduk
Pl(u) is normalized such that Pl(1) = 1. Find P0(u) and P2(u)
note: l, 0 and 2 are subscript
recursion relation
an+2 = [n(n+1) - l (l+1) / (n+2)(n+1)] an
n is subscript
substituted λ = l(l+1) and put n=0 for P0(u) and n=2 for P2(u), didnt get very far
please could someone give me a point in the right direction or a youtube video to explain the process used to solve this? really lost.
2. May 8, 2014
### Ray Vickson
You wrote
$$a_n + 2 = \left[n(n+1) - \frac{l(l+1)}{(n+2)(n+2)} \right] a_n$$
Is that what you really mean, or did you want
$$a_{n+2} = \left[ \frac{n(n+1) - l(l+1)}{(n+2)(n+1)}\right] a_n ?$$
If so, please use parentheses properly.
Anyway, I don't see what this approach will get you, since you have the recursion for a fixed value of $l$---that is, the recursion that tells you how to get $P_l(x)$ for fixed $l$, given some initial conditions. There is nothing there that tells you how to relate $P_{l-1}(x)$ and $P_{l+1}(x)$ to $P_l(x)$. Normalization comes from outside the differential equation and outside the recursion. Look up Legendre functions in a textbook or via Google.
Note added in edit: I see that you might have meant $P_l$; the font you used made it look like $P_1$. Anyway, you might find it useful to look at http://www.phys.ufl.edu/~fry/6346/legendre.pdf .
Last edited: May 8, 2014
3. May 8, 2014
### rwooduk
apologies, yes it's this:
i think the idea is to put l=0 for P0(u) and l=2 for P2(u)
but not really sure what to do after that.
will have a look at the link, thank-you. | HuggingFaceTB/finemath | |
In chapter 13 we saw the basic elements of VFPv2, the floating point subarchitecture of ARMv6. In this chapter we will implement a floating point matrix multiply using VFPv2.
Disclaimer: I advise you against using the code in this chapter in commercial-grade projects unless you fully review it for both correctness and precision.
Matrix multiply
Given two vectors v and w of rank r where v = <v0, v1, ... vr-1> and w = <w0, w1, ..., wr-1>, we define the dot product of v by w as the scalar v · w = v0×w0 + v1×w1 + ... + vr-1×wr-1.
We can multiply a matrix `A` of `n` rows and `m` columns (`n` x `m`) by a matrix `B` of `m` rows and `p` columns (`m` x `p`). The result is a matrix of `n` rows and `p` columns. Matrix multiplication may seem complicated but actually it is not. Every element in the result matrix it is just the dot product (defined in the paragraph above) of the corresponding row of the matrix `A` by the corresponding column of the matrix `B` (this is why there must be as many columns in `A` as there are rows in `B`).
A straightforward implementation of the matrix multiplication in C is as follows.
In order to simplify the example, we will asume that both matrices A and B are square matrices of size `n x n`. This simplifies just a bit the algorithm.
Matrix multiplication is an important operation used in many areas. For instance, in computer graphics is usually performed on 3x3 and 4x4 matrices representing 3D geometry. So we will try to make a reasonably fast version of it (we do not aim at making the best one, though).
A first improvement we want to do in this algorithm is making the loops perfectly nested. There are some technical reasons beyond the scope of this code for that. So we will get rid of the initialization of `C[i][j]` to 0, outside of the loop.
After this change, the interesting part of our algorithm, line 13, is inside a perfect nest of loops of depth 3.
Accessing a matrix
It is relatively straightforward to access an array of just one dimension, like in `a[i]`. Just get `i`, multiply it by the size in bytes of each element of the array and then add the address of `a` (the base address of the array). So, the address of `a[i]` is just `a + ELEMENTSIZE*i`.
Things get a bit more complicated when our array has more than one dimension, like a matrix or a cube. Given an access like `a[i][j][k]` we have to compute which element is denoted by `[i][j][k]`. This depends on whether the language is row-major order or column-major order. We assume row-major order here (like in C language). So `[i][j][k]` must denote `k + j * NK + i * NK * NJ`, where `NK` and `NJ` are the number of elements in every dimension. For instance, a three dimensional array of 3 x 4 x 5 elements, the element [1][2][3] is 3 + 2 * 5 + 1 * 5 * 4 = 23 (here `NK` = 5 and `NJ` = 4. Note that `NI` = 3 but we do not need it at all). We assume that our language indexes arrays starting from 0 (like C). If the language allows a lower bound other than 0, we first have to subtract the lower bound to get the position.
We can compute the position in a slightly better way if we reorder it. Instead of calculating `k + j * NK + i * NK * NJ` we will do `k + NK * (j + NJ * i)`. This way all the calculus is just a repeated set of steps calculating `x + Ni * y` like in the example below.
Naive matrix multiply of 4x4 single-precision
As a first step, let's implement a naive matrix multiply that follows the C algorithm above according to the letter.
That's a lot of code but it is not complicated. Unfortunately computing the address of the array takes an important number of instructions. In our `naive_matmul_4x4` we have the three loops `i`, `j` and `k` of the C algorithm. We compute the address of `C[i][j]` in the loop `j` (there is no need to compute it every time in the loop `k`) in lines 52 to 63. The value contained in `C[i][j]` is then loaded into `s0`. In each iteration of loop `k` we load `A[i][k]` and `B[k][j]` in `s1` and `s2` respectively (lines 70 to 82). After the loop `k` ends, we can store `s0` back to the array position (kept in `r7`, line 90)
In order to print the result matrix we have to pass 16 floating points to `printf`. Unfortunately, as stated in chapter 13, we have to first convert them into double-precision before passing them. Note also that the first single-precision can be passed using registers `r2` and `r3`. All the remaining must be passed on the stack and do not forget that the stack parameters must be passed in opposite order. This is why once the first element of the C matrix has been loaded in `{r2,r3}` (lines 117 to 120) we advance 60 bytes r4. This is `C[3][3]`, the last element of the matrix C. We load the single-precision, convert it into double-precision, push it in the stack and then move backwards register `r4`, to the previous element in the matrix (lines 128 to 137). Observe that we use `r6` as a marker of the stack, since we need to restore the stack once `printf` returns (line 122 and line 141). Of course we could avoid using `r6` and instead do `add sp, sp, #120` since this is exactly the amount of bytes we push to the stack (15 values of double-precision, each taking 8 bytes).
I have not chosen the values of the two matrices randomly. The second one is (approximately) the inverse of the first. This way we will get the identity matrix (a matrix with all zeros but a diagonal of ones). Due to rounding issues the result matrix will not be the identity, but it will be pretty close. Let's run the program.
Vectorial approach
The algorithm we are trying to implement is fine but it is not the most optimizable. The problem lies in the way the loop `k` accesses the elements. Access `A[i][k]` is eligible for a multiple load as `A[i][k]` and `A[i][k+1]` are contiguous elements in memory. This way we could entirely avoid all the loop `k` and perform a 4 element load from `A[i][0]` to `A[i][3]`. The access `B[k][j]` does not allow that since elements `B[k][j]` and `B[k+1][j]` have a full row inbetween. This is a strided access (the stride here is a full row of 4 elements, this is 16 bytes), VFPv2 does not allow a strided multiple load, so we will have to load one by one.. Once we have all the elements of the loop `k` loaded, we can do a vector multiplication and a sum.
With this approach we can entirely remove the loop `k`, as we do 4 operations at once. Note that we have to modify `fpscr` so the field `len` is set to 4 (and restore it back to 1 when leaving the function).
Fill the registers
In the previous version we are not exploiting all the registers of VFPv2. Each rows takes 4 registers and so does each column, so we end using only 8 registers plus 4 for the result and one in the bank 0 for the summation. We got rid the loop `k` to process `C[i][j]` at once. What if we processed `C[i][j]` and `C[i][j+1]` at the same time? This way we can fill all the 8 registers in each bank.
Note that because we now process `j` and `j + 1`, `r5` (`j`) is now increased by 2 at the end of the loop. This is usually known as loop unrolling and it is always legal to do. We do more than one iteration of the original loop in the unrolled loop. The amount of iterations of the original loop we do in the unrolled loop is the unroll factor. In this case since the number of iterations (4) perfectly divides the unrolling factor (2) we do not need an extra loop for the remainder iterations (the remainder loop has one less iteration than the value of the unrolling factor).
As you can see, the accesses to `b[k][j]` and `b[k][j+1]` are starting to become tedious. Maybe we should change a bit more the matrix multiply algorithm.
Reorder the accesses
Is there a way we can mitigate the strided accesses to the matrix B? Yes, there is one, we only have to permute the loop nest i, j, k into the loop nest k, i, j. Now you may be wondering if this is legal. Well, checking for the legality of these things is beyond the scope of this post so you will have to trust me here. Such permutation is fine. What does this mean? Well, it means that our algorithm will now look like this.
This may seem not very useful, but note that, since now k is in the outermost loop, now it is easier to use vectorial instructions.
If you remember the chapter 13, VFPv2 instructions have a mixed mode when the `Rsource2` register is in bank 0. This case makes a perfect match: we can load `C[i][0..3]` and `B[k][0..3]` with a load multiple and then load `A[i][k]` in a register of the bank 0. Then we can make multiply` A[i][k]*B[k][0..3]` and add the result to `C[i][0..3]`. As a bonus, the number of instructions is much lower.
As adding after a multiplication is a relatively usual sequence, we can replace the sequence
with a single instruction `vmla` (multiply and add).
Now we can also unroll the loop `i`, again with an unrolling factor of 2. This would give us the best version.
Comparing versions
Out of curiosity I tested the versions, to see which one was faster.
The benchmark consists on repeatedly calling the multiplication matrix function 221 times (actually 221-1 because of a typo, see the code) in order to magnify differences. While the input should be randomized as well for a better benchmark, the benchmark more or less models contexts where a matrix multiplication is performed many times (for instance in graphics).
This is the skeleton of the benchmark.
Here are the results. The one we named the best turned to actually deserve that name.
VersionTime (seconds)
naive_matmul_4x46.41
naive_vectorial_matmul_4x43.51
naive_vectorial_matmul_2_4x42.87
better_vectorial_matmul_4x42.59
best_vectorial_matmul_4x41.51
That's all for today. | HuggingFaceTB/finemath | |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 18 Oct 2019, 21:44 GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. For each positive integer n, the mean of the first n terms of a sequen
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Hide Tags
Math Expert V
Joined: 02 Sep 2009
Posts: 58453
For each positive integer n, the mean of the first n terms of a sequen [#permalink]
Show Tags 00:00
Difficulty: 65% (hard)
Question Stats: 50% (02:22) correct 50% (01:29) wrong based on 30 sessions
HideShow timer Statistics
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?
(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064
_________________
GMAT Club Legend D
Joined: 18 Aug 2017
Posts: 5020
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: For each positive integer n, the mean of the first n terms of a sequen [#permalink]
Show Tags
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?
(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064
total sum/total intgers= mean
so sum = n*n = n^2
so sum of 2008th term ; ( 2008^2-2007^2) ; ( 2008+2007)*(2008-2007) ;
4015
IMO B
CEO D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
For each positive integer n, the mean of the first n terms of a sequen [#permalink]
Show Tags
2
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?
(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064
The series with one terms will be like {1} Average = 1
The series with two terms will be like {1, 3} Average = 2
The series with three terms will be like {1, 3, 5} Average = 3
The series with four terms will be like {1, 3, 5, 7} Average = 4
The series with five terms will be like {1, 3, 5, 7, 9} Average = 5
i.e. the resultant series is the series of CONSECUTIVE ODD integers
2008th term of series = 2008th Odd integer = 2*2008-1 = 4015
CONCEPT: nth odd integer = 2*n-1
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Senior Manager S
Joined: 12 Sep 2017
Posts: 302
Re: For each positive integer n, the mean of the first n terms of a sequen [#permalink]
Show Tags
Archit3110 wrote:
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?
(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064
total sum/total intgers= mean
so sum = n*n = n^2
so sum of 2008th term ; ( 2008^2-2007^2) ; ( 2008+2007)*(2008-2007) ;
4015
IMO B
Hello Archit3110
Could you please explain to me the red part?
Kind regards!
CEO D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
For each positive integer n, the mean of the first n terms of a sequen [#permalink]
Show Tags
1
1
jfranciscocuencag wrote:
Archit3110 wrote:
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?
(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064
total sum/total intgers= mean
so sum = n*n = n^2
so sum of 2008th term ; ( 2008^2-2007^2) ; ( 2008+2007)*(2008-2007) ;
4015
IMO B
Could you please explain to me the red part?
Kind regards!
Given: Mean of the first n terms of the sequence = n
i.e. $$\frac{Sum-of-n-terms}{n} = n$$
i.e. Sum of n terms $$= n*n = n^2$$
But we also know that Sum of first n positive odd integers i.e. $$1+3+5+7+...........+(2n-1) = n^2$$
where $$(2n-1)$$ is the nth term of the series
therefore, the series talked about in this question is the series of first consecutive odd integers
now, 2008th term of the series = 2008th odd integer $$= 2*2008 - 1 = 4015$$
Or, what Archit mentioned is
2008th term = Sum of of 2008 terms - Sum of 2007 terms of the same series
But we know that sum of n terms = $$n^2$$
therefore, 2008th term = $$2008^2 - 2007^2 = (2008+2007)*(2008-2007)$$
Using property $$a^2 - b^2 = (a+b)*(a-b)$$
i.e. 2008th Term = 4015
I hope this helps!!!
jfranciscocuencag
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
GMAT Club Legend D
Joined: 18 Aug 2017
Posts: 5020
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
For each positive integer n, the mean of the first n terms of a sequen [#permalink]
Show Tags
1
jfranciscocuencag wrote:
Archit3110 wrote:
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?
(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064
total sum/total intgers= mean
so sum = n*n = n^2
so sum of 2008th term ; ( 2008^2-2007^2) ; ( 2008+2007)*(2008-2007) ;
4015
IMO B
Hello Archit3110
Could you please explain to me the red part?
Kind regards!
jfranciscocuencag as rightly mentioned by GMATinsight sir in the above post ;
2008th term = Sum of of 2008 terms - Sum of 2007 terms of the same series
But we know that sum of n terms = n^2
therefore, 2008th term = 2008^2−2007^2 or say
(2008+2007)∗(2008−2007) "Using property a2−b2=(a+b)∗(a−b)a2−b2=(a+b)∗(a−b)"
2008th Term = 4015 For each positive integer n, the mean of the first n terms of a sequen [#permalink] 14 Apr 2019, 04:08
Display posts from previous: Sort by
For each positive integer n, the mean of the first n terms of a sequen
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne | HuggingFaceTB/finemath | |
## Pre-Calculus Tutorial
#### Intro
Let’s show that
PROOF:
1. Let …. We can name it anything we like
2. Let …. Again we can name it anything we like
3. and … Equivalent exponential forms of the statements in steps 1 and 2.
4. .. Reason: If A=B & C=D, then AC=BD.
5. ..Reason: Property of exponents: AQ•AS=AQ+S.
6. .Reason: Taking log (base a) of both sides.
7. Reason: same expression as step 6
8. . Substituting from steps 1 and 2 into step 7
show that
EXAMPLE:
…. 2*2*2 = 8
… 2*2*2 *2=16
….. | HuggingFaceTB/finemath | |
# What is the compound interest on $7879 at 13% over 28 years? If you want to invest$7,879 over 28 years, and you expect it will earn 13.00% in annual interest, your investment will have grown to become $241,361.23. 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$7,879 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 28 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}$$
• A is the amount of money after the compounding periods
• P is the principal amount
• r is the annual interest rate
• n is the number of compounding periods per year
• t is the number of years
We can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.
For this formula, we need to convert the rate, 13.00% into a decimal, which would be 0.13.
$$A = 7879(1 + \dfrac{ 0.13 }{1})^{ 28}$$
As you can see, we are ignoring the n when calculating this to the power of 28 because our example is for annual compounding, or one period per year, so 28 × 1 = 28.
## How the compound interest on $7,879 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 28 years it is compounding: Start Balance Interest End Balance 1$7,879.00 $1,024.27$8,903.27
2 $8,903.27$1,157.43 $10,060.70 3$10,060.70 $1,307.89$11,368.59
4 $11,368.59$1,477.92 $12,846.50 5$12,846.50 $1,670.05$14,516.55
6 $14,516.55$1,887.15 $16,403.70 7$16,403.70 $2,132.48$18,536.18
8 $18,536.18$2,409.70 $20,945.88 9$20,945.88 $2,722.96$23,668.85
10 $23,668.85$3,076.95 $26,745.80 11$26,745.80 $3,476.95$30,222.75
12 $30,222.75$3,928.96 $34,151.71 13$34,151.71 $4,439.72$38,591.43
14 $38,591.43$5,016.89 $43,608.32 15$43,608.32 $5,669.08$49,277.40
16 $49,277.40$6,406.06 $55,683.46 17$55,683.46 $7,238.85$62,922.31
18 $62,922.31$8,179.90 $71,102.21 19$71,102.21 $9,243.29$80,345.49
20 $80,345.49$10,444.91 $90,790.41 21$90,790.41 $11,802.75$102,593.16
22 $102,593.16$13,337.11 $115,930.27 23$115,930.27 $15,070.94$131,001.21
24 $131,001.21$17,030.16 $148,031.37 25$148,031.37 $19,244.08$167,275.44
26 $167,275.44$21,745.81 $189,021.25 27$189,021.25 $24,572.76$213,594.01
28 $213,594.01$27,767.22 $241,361.23 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$7,879 at 13.00% 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 $7,879 at 13% interest? Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of$7,879 assuming an interest rate of 13.00%.
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}{ 13 } = 5.54$$
Using this, we know that any amount we invest at 13.00% would double itself in approximately 5.54 years. So $7,879 would be worth$15,758 in ~5.54 years.
We can also calculate the exact length of time it will take to double an amount at 13.00% using a slightly more complex formula:
$$Years = \dfrac{log(2)}{log(1 + 0.13)} = 5.67\; 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 \$7,879 at 13.00% over a 28 year investment period. | HuggingFaceTB/finemath | |
Education.com
Try
Brainzy
Try
Plus
# Solving Pairs of Equations Help
(not rated)
By — McGraw-Hill Professional
Updated on Oct 3, 2011
## Introduction to Solving Pairs of Equations
The solutions of pairs of equations can be envisioned and approximated by graphing both of the equations on the same set of coordinates. Solutions appear as intersection points between the plots.
### A Line And A Curve
Suppose you are given two equations in two variables, such as x and y , and are told to solve for values of x and y that satisfy both equations. Such equations are called simultaneous equations . Here is an example:
y = x 2 + 2 x + 1
y = − x + 1
These equations are graphed in Fig. 6-12. The graph of the first equation is a parabola, and the graph of the second equation is a straight line. The line crosses the parabola at two points, indicating that there are two solutions of this set of simultaneous equations. The coordinates of the points, corresponding to the solutions, can be estimated from the graph. It appears that they are approximately:
( x 1 , y 1 ) = (−3, 4)
( x 2 , y 2 ) = (0, 4)
You can solve the pair of equations using plain algebra, and determine the solutions exactly.
Fig. 6-12 . Graphs of two equations, showing solutions as intersection points.
## Another Line And Curve
Here is another pair of “two-by-two” equations (two equations in two variables) that can be approximately solved by graphing:
y = −2 x 2 + 4 x − 5
y = −2 x − 5
These equations are graphed in Fig. 6-13. Again, the graph of the first equation is a parabola, and the graph of the second equation is a straight line. The line crosses the parabola at two points, indicating that there are two solutions. The coordinates of the points, corresponding to the solutions, appear to be approximately:
( x 1 , y 1 ) = (3, −11)
( x 2 , y 2 ) = (0, −5)
Again, if you want, you can go ahead and solve these equations using algebra, and find the values exactly.
Fig. 6-13 . Another example of equation solutions shown as the intersection points of their graphs.
### How Many Solutions?
Graphing simultaneous equations can reveal general things about them, but should not be relied upon to provide exact solutions. In real-life scientific applications, graphs rarely show exact solutions unless they are so labeled and represent theoretical ideals.
A graph can reveal that a pair of equations has two or more solutions, or only one solution, or no solutions at all. Solutions to pairs of equations always show up as intersection points on their graphs. If there are n intersection points between the curves representing two equations, then there are n solutions to the pair of simultaneous equations.
If a pair of equations is complicated, or if the graphs are the results of experiments, you’ll occasionally run into situations where you can’t use algebra to solve them. Then graphs, with the aid of computer programs to closely approximate the points of intersection between graphs, offer a good means of solving simultaneous equations.
Sometimes you’ll want to see if a set of more than two equations in x and y has any solutions shared by them all. It is common for one or more pairs of a large set of equations to have some solutions; this is shown by points where any two of their graphs intersect. But it’s unusual for a set of three or more equations in x and y to have any solutions when considered all together (that is, simultaneously). For that to be the case, there must be at least one point that all of the graphs have in common.
150 Characters allowed
### Related Questions
#### Q:
See More Questions
Top Worksheet Slideshows | HuggingFaceTB/finemath | |
Lessons and Tools for Teachers
#### Lessons for Symmetry
Page 3 of 823 results
• Relating Rotations to Symmetry - … to have cyclic symmetry or rotational symmetry. A rotation or other transformation that leaves a figure apparently unchanged is a symmetry of the … point symmetry. Point symmetry is just 2-fold rotational symmetry. In the activity below, you will see why it is called point symmetry and … to have point symmetry. 29. Which of the following designs have point symmetry? 30. Create your own designs with point symmetry. Answers References …
User Rating: Grade Level: 9-12
• A Wall of Symmetrical Shapes - … all rectangles have a line of symmetry? (yes) Do all triangles have a line of symmetry? (no) What do we call the … at this time. Station 1 Students explore symmetry in the Web lesson, Let's Learn Symmetry. (See Weblinks) Station 2 a. Using … shape along as many lines of symmetry as possible, and then draw the line(s) of symmetry on their shape using a … only addresses and assesses symmetry. A. Teacher observes students' shapes and lines of symmetry drawn on the wall of symmetry to see if …
User Rating: Grade Level: 3-5
• Symmetry Collage - Symmetry Collage Description Students investigate symmetry. They compose their own collages and bulletin board borders using the standards of symmetry. The use of … of symmetry. Show some more examples of the line of symmetry and have students discuss why and where this line of symmetry is. 4. To exemplify symmetry to the students …
User Rating: Grade Level: K-2
• The Fun Polygon - … the various lines of symmetry that exist. (There will not be a line of symmetry for the quadrilateral.) (Symmetry was introduced during lesson … fun polygons according to symmetry. H. Review what symmetry means. As a class, write and post a definition for symmetry using the students … have zero lines of symmetry), while Triangles 4, 5,6 are isosceles (and have one line of symmetry). Encourage the students to …
User Rating: Grade Level: 3-5
• Investigation - Symmetry - … have line symmetry and/or rotational symmetry. If a shape has line symmetry, the students should find out how many lines of symmetry it … it has line symmetry, rotational symmetry, or both kinds of symmetry. If the figure has line symmetry, determine how many lines of symmetry it has. WHAT'S THE MATHEMATICS? recognize and demonstrate line symmetry recognize and demonstrate rotational symmetry
User Rating: Grade Level: 3-5
• Pieces of the Puzzle - … :15 means three minutes and fifteen seconds. cept of symmetry. Write the word symmetry on the board. Then hold up half of … share other places where they think symmetry is used. Also bring out the idea that symmetry brings in the concept of balance … you have made one line of symmetry. Have the students create a second line of symmetry. Type BK 10 LT 90 FD … should he do? Math: Explore the symmetry of various geometric shapes. Science: Explore the various symmetries present in the world of nature …
User Rating: Grade Level: 3-8
• Lines Of Symmetry (includes BLMs Polygons and Lines of Symmetry) - … symmetry parallelogram: 0 lines of symmetry octagon: 8 lines of symmetry triangle: 3 lines of symmetry hexagon: 6 lines of symmetry hexagon: 2 lines of symmetry pentagon: 5 lines of symmetry rhombus: 2 lines of symmetry
User Rating: Grade Level: 3-5
• Symmetry of Diatoms - … found, charts or graphs, discussion of symmetry of each (or lack of), drawn in symmetry lines, hypotheses about the water quality, and … , drawing lines of symmetry, and drawing figures that have a certain number of symmetry lines. The amount of symmetry that occurs in nature is amazing. The basic shapes (outlines) of many diatoms have reflection symmetry
User Rating: Grade Level: 9-12
• Patchwork Quilting - … .2.1 The student understands the concepts of spatial relationships, symmetry, reflections, congruency, and similarity. Florida Process Standards Effective Communicators 02 … quilt. (If students need to review symmetry, have them complete the Student Web Lesson Learn Symmetry.) 4. Hand out 9in. x 6in … be built on some type of symmetry: horizontal, vertical, or diagonal. Ask students, What line of symmetry is currently showing on your …
User Rating: Grade Level: 3-5
• Geo Jammin' By Design - Day 5, Lesson 25: Geo Jungle - … symmetry are shared. (Line of symmetry, congruent, shape within a shape, reflection, mirror image, etc.) Students should demonstrate a clear understanding of symmetry through their personal descriptions, explanations, and correct identity of characteristics of symmetry as it occurs in … , formatively assess individual students' understanding of symmetry and their ability to correctly identify characteristics of symmetry by listening for accuracy of their …
User Rating: Grade Level: K-2
Copyright © 2018 Lesson Corner. All rights reserved. | HuggingFaceTB/finemath | |
calculus (limits)
lim h>0 sqrt(1+h)-1/h
not sure how to factor this; not allowed to use L'Hopital's Rule. (that isn't taught at my school until Calc II & I'm in Calc I).
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. I bet you mean
lim h>0 [sqrt(1+h)-1] /h
One way:
I guess you know how to take derivatives
d/dx (x)^.5 = .5 (x^-.5)
(of course right there you can say
x^-.5 at x = 1 so your result is
.5/1 = .5
so
d (x^.5) = .5 x^-.5 dx
so
d (sqrt (x)) dx = .5 dx/sqrt(x)
d sqrt(1+h) dx = .5 dx/sqrt(1+h)
if h -->0
d sqrt(1+h)dx = .5 sqrt(1)h/sqrt(1)
= .5 h
so sqrt(1+h) --> 1 + .5 h
sqrt(1+h)-1 > .5 h
divide by h and get
.5
series for sqrt(x) for x approximately 1 is
=
1. 👍
2. 👎
3. ℹ️
4. 🚩
2. I do know how to take derivatives however the course I'm in hasn't taught them yet (long story) so I can't use derivatives in finding the solution; I have to factor.
So perhaps a better question is: how do I factor the above expression in a way that will allow me to find the limit?
1. 👍
2. 👎
3. ℹ️
4. 🚩
3. Taylor series for f(x + a ) for x approximately 1 is
f(x)= f(1) + f'(1)(x-a) + f''(x-a)^2/2 ...
for f(x) = sqrt (x)
f(1) = (1)^.5 = 1
f'(1) = .5 (1)^-.5 = .5
f"(x) = -.25(1)^-1.5 etc = -.25
f(x+h) = 1 + .5*h -.25 h^2 /2 etc
subtravt 1
.5h -.125 h^2 tc
divide by h and let h-->0
.5 - .125 h --> .5
1. 👍
2. 👎
3. ℹ️
4. 🚩
4. factor (x+h)^.5 ?
I do not know how
Somehow you have to know that sqrt(1+h) ---> 1 + .5 h +.....
1. 👍
2. 👎
3. ℹ️
4. 🚩
5. Perhaps someone else knows how
1. 👍
2. 👎
3. ℹ️
4. 🚩
6. rationalize the numerator, that is, multiply top and bottom by (√(1+h) + 1)/(√(1+h) + 1)
lim (√(1+h) - 1)/h as h --->0
= lim (√(1+h) - 1)/h (√(1+h) + 1)/(√(1+h) + 1)
= lim (1+h -1)/[h( (√(1+h) + 1)/h)]
= lim 1/( (√(1+h) + 1)
= 1/(1+1) = 1/2
1. 👍
2. 👎
3. ℹ️
4. 🚩
Similar Questions
Still need help? You can ask a new question. | HuggingFaceTB/finemath |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.