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# Answer to Riddle #68: Arrangements of Red & Blue Marbles
68. You have 50 red marbles and 50 blue marbles, other wise identical. You have 2 identical jars. A marble will be selected at random, from a jar selected at random. How would you apportion the marbles so as to maximise the chances of a red marble being selected?
Can you prove it?
Interesting puzzle this. For a while I thought there is just no way. But there is, let's take a look.
We're assuming that the selection of the jars is beyond our control. Whomever selects will decide on which jar with a 50 50 distribution. All we can work with is the marbles in the jars.
A formula for working out the probability of picking a red is not so hard to derive. It's going to be .5 times the probability of picking a red from the first jar + .5 times the probability of picking a red from the second jar. And the probability of picking a red marble from any jar is the number of red marbles in the jar over the total number of marbles in the jar. (Note it doesn't say there has to be 50 marbles in each jar!)...
P(R) = .5R1 / (R1 + B1) + .5R2 / (R2 + B2)
If we could get one of the jars up to 100% red without dropping the percentage red in the other jar too far. At this stage I can tell you that the optimal arrangement is to put one red marble in one jar and all the other marbles into the other jar. Plug in the numbers and we get...
P(R) = .5 * 1 + .5 * 49 / 99 = 0.747474
What we have done is ensured that if the first jar is picked there is a 100% chance of picking a red marble, whilst diluting the presence of blue marbles in the second jar as best we can.
## Proof
I'm using the phrase loosely here, rather than a proof it's more of a demonstration. One way of showing that a function is at a maxima is to show that any move away from that point reduces the function's value. Not proof that it is the maximum but a maxima. Since the problem is not really defined for no marbles in one of the jars the only moves we have are to add to the first jar either a red or a blue marble. For those positions the probabilities fall to 0.745 and 0.5 respectively. So we're on track.
Next we have the equation P(R) = .5R1 / (R1 + B1) + .5R2 / (R2 + B2) but actually not all the variables are independent. There are only 50 of each marble so R2 = 50 - R1 and similarly B2 = 50 - B1. This gives us an equation with only two variables...
P(R) = .5R1 / (R1 + B1) + .5(50 - R1) / (100 - R1 - B1)
Which is a help because we can graph it as I've done above. You can interact with the graph or technically a 3D surface plot at plot.ly the R1 axis is on the left the B1 on the right. Note the symmetry as we move from one red marble in jar one to one red marble in jar two. Also that the maxima is right by the minima, this is the reversal of the optimal strategy i.e. putting one blue marble in the jar that had one red marble in it.
## Assumptions
I'm assuming that you have to use all the marbles. If you read the question differently then you could only use one colour, put the rest in your pocket. I'm pretty confident my answer is the one that an interviewer would be looking for.
Google's Bard AI had no problem with this puzzle ChatGPT on the other hand...
If you're curious what Bard made of this puzzle...
If you're curious what ChatGPT made of this puzzle...
© Nigel Coldwell 2004 - – The questions on this site may be reproduced without further permission, I do not claim copyright over them. The answers are mine and may not be reproduced without my expressed prior consent. Please inquire using the link at the top of the page. Secure version of this page.
PayPal
I always think it's arrogant to add a donate button, but it has been requested. If I help you get a job though, you could buy me a pint! - nigel
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binomial distributions
label Statistics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
It has been estimated that only about 40% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies
How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies?
Oct 1st, 2015
W is the number of California resident and has geometric distribution with p =probability of not having supplies = 1-0.4 = 0.6
Hence P(W=x) = (0.4)n-10.6 for n = 1,2,3…….
For example, P(W≥3) = 1-P(W≤2) = 1-W(1)-W(2) = 1- 0.6-0.4(0.6) = 0.7
which is the probability of at least 3 California residents until we find a California resident who does not have adequate earthquake supplies
** P/S: Please remember to give me a review later on once the answer is correct. I would be highly appreciate it. Thank you. **
Oct 2nd, 2015
...
Oct 1st, 2015
...
Oct 1st, 2015
Nov 18th, 2017
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## How do I calculate my body mass index?
Body Mass Index is a simple calculation using a person’s height and weight. The formula is BMI = kg/m2 where kg is a person’s weight in kilograms and m2 is their height in metres squared. A BMI of 25.0 or more is overweight, while the healthy range is 18.5 to 24.9. BMI applies to most adults 18-65 years.
### Is 27 a good body mass index?
A healthy weight is considered to be a BMI of 24 or less. A BMI of 25 to 29.9 is considered overweight. A BMI of 30 and above is considered obese.
Is a body mass index of 23 good?
BMI ranges below 18.5 – you’re in the underweight range. between 18.5 and 24.9 – you’re in the healthy weight range. between 25 and 29.9 – you’re in the overweight range. between 30 and 39.9 – you’re in the obese range.
How is BMI calculated CDC?
How is BMI calculated? With the metric system, the formula for BMI is weight in kilograms divided by height in meters squared. Because height is commonly measured in centimeters, divide height in centimeters by 100 to obtain height in meters.
## What is BMI in community health nursing?
Body mass index – BMI BMI, formerly called the Quetelet index, is a measure for indicating nutritional status in adults. It is defined as a person’s weight in kilograms divided by the square of the person’s height in metres (kg/m2).
### How do you calculate BMI in kg?
With the metric system, the formula for BMI is weight in kilograms divided by height in meters squared. Since height is commonly measured in centimeters, an alternate calculation formula, dividing the weight in kilograms by the height in centimeters squared, and then multiplying the result by 10,000, can be used.
Is BMI calculator accurate?
BMI (body mass index), which is based on the height and weight of a person, is an inaccurate measure of body fat content and does not take into account muscle mass, bone density, overall body composition, and racial and sex differences, say researchers from the Perelman School of Medicine, University of Pennsylvania.
What is BMI PDF?
Body mass index (BMI) is a measure of weight adjusted for height, calculated as weight in kilograms divided by the square of height in meters (kg/m2).
## What is a good BMI for a 55 year old woman?
A BMI of between 18.5 and 24.9 is ideal.
### Does age matter in calculating BMI?
Issues with BMI For example, BMI doesn’t take into consideration your age, sex, or muscle mass, which are all important when it comes to finding your ideal weight. Older adults tend to lose muscle and bone, so more of their body weight is likely to come from fat.
What BMI to look slim?
What exactly counts as being “skinny?” Some healthcare experts believe a BMI in the 15-18 range to be clinically underweight. This seems to fall pretty close to what many everyday people consider to be “skinny” with a BMI of 18 or lower frequently listed as the indicator of someone considered to be slim.
Does age affect BMI?
For adults ages 20 years and older, BMI incorporates weight and height, but it does not take age or sex into account. A woman tends to have more body fat than a man with the same BMI.
## How do I use the Body Mass Index calculator?
The Body Mass Index (BMI) Calculator can be used to calculate BMI value and corresponding weight status while taking age into consideration. Use the “Metric Units” tab for the International System of Units or the “Other Units” tab to convert units into either US or metric units.
### What is the body mass index (BMI) of an adult?
What is the body mass index (BMI)? The body mass index (BMI) is a measure that uses your height and weight to work out if your weight is healthy. The BMI calculation divides an adult’s weight in kilograms by their height in metres squared. For example, A BMI of 25 means 25kg/m2. For most adults, an ideal BMI is in the 18.5 to 24.9 range.
Is there a BMI Calculator for children and teens?
This calculator provides BMI and the corresponding BMI-for-age percentile on a CDC BMI-for-age growth chart. Use this calculator for children and teens, aged 2 through 19 years old. School staff, child care leaders, and other professionals can use this spreadsheet to compute BMI for as many as 2,000 children.
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# Solution to the integral,i.e, expected value of a function of normal variable
1. Jun 26, 2012
### ait.abd
I want to calculate $\int_a^b \frac{1}{\sqrt{2 \pi \sigma^2}} e^{(-(x-\mu)/\sigma^2)} log_2 (1 + e^{-x}) dx$
2. Jun 26, 2012
### D H
Staff Emeritus
You'll need to use numerical techniques. The above integral doesn't have a closed form solution in the elementary functions.
3. Jun 26, 2012
### ait.abd
4. Jun 26, 2012
### chiro
Hey ait.abd and welcome to the forums.
Numerical techniques are ones that give approximate answers in general. You can supply parameters to get a good enough approximation (like for example a number good enough to say 4 decimal places) for the better implementations.
If you are unsure, just use a common package for numerical calculation.
You should probably try searching online for a numeric integrator Java applet, or go to www.wolframalpha.com and enter in your expression to get an approximate answer.
5. Jun 26, 2012
### ait.abd
Thanks chiro. But, wolfram online integrator doesn't work for this expression as it tries to compute the exact expression. I can perform numerical integration but I want answer in terms of $a$ and $b$. Numerical integration will calculate the answer for a particular $a$ and $b$.
6. Jun 26, 2012
### Mute
If you need something analytical, you're going to have to develop some sort of approximation for the integral for certain parameter regimes. For example, for $\sigma \rightarrow 0$, the Gaussian essentially becomes a delta function and you would get
$$\log_2(1+e^{-\mu})\left[\Theta(b-\mu)+\Theta(\mu - a)\right]$$
as the result (the step functions $\Theta$ guarantee that mu is between a and b). So, for $\sigma$ small, a rather crude approximation might be
$$\log_2(1+e^{-\mu}) \int_a^b \frac{dx}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right],$$
where the integral can be evaluated in terms of the error function. There is a more systematic way to generate this approximation called the method of steepest descent. (You'll have to look that up in a book; I'm afraid the wikipedia article isn't very helpful).
You might also be able to write down an infinite series for the integral. However, when I tried this by expanding the logarithm in powers of e^(-x), I got a sum which looks like it doesn't converge, indicating that either I made a mistake in my calculation or that switching the integral and sum isn't valid in this case.
7. Jun 26, 2012
### utkarsh1
Just looking at the form of equation. It seems that complex contour integral MAY work.
8. Jun 26, 2012
### HallsofIvy
Staff Emeritus
The integral in the original post does NOT involve $e^{-x^2}$. It is, rather, of the form $e^{-x}$.
9. Jun 26, 2012
### D H
Staff Emeritus
I suspect that that may have been a typo. Even if it is a typo, it doesn't help. Either way ( exp(-x2) vs exp(-x) ), this function is not integrable in the elementary functions. Since it's not an integral that is widely used, it's dubious that someone has come up with a nifty way to evaluate it.
To ait.abd: You need to learn how to do numerical integration sometime. If this is the right integral, that sometime is now.
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Friday
May 6, 2016
# Homework Help: calculus
Posted by Joseph on Tuesday, May 24, 2011 at 8:13pm.
let m and b be nonzero real numbers
if the line y=mx+b intersects y^2=4px in only one point, show that p=mb
• calculus - Reiny, Tuesday, May 24, 2011 at 8:51pm
sub the straight line into the parabola
(mx+b)^ = 4px
m^2x^2 + 2mbx + b^2 - 4px = 0
m^2x^2 + x(2mb-4p) + b^2 = 0
to have one solution, (one intersection point), the discriminant has to be zero
(2mb-4p)^2 - 4(m^2)(b^2) = 0
4 m^2b^2 - 16mbp + 16 p^2 - 4m^2b^2 = 0
16mbp - 16p^2 = 0
16p(mb - p) = 0
so p = 0 or p = mb , but if p=0 we couldn't have a parabola, so
p = mb
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# Spinner, Math, Probability
## 3D model description
Spinner, Math, Probability
Once before the time of “fidget spinners,” a spinner was frequently used to model random processes for student to feel the idea of uncertainty, data trends, and mathematical structures. Or, perhaps, this mathematical spinner can also serve the purpose of “fidgeting”?
The board by default has a diameter of 80mm, with both numbers (1, 2, 3, 4, 5, 6) and letters (A, B, C, D, E, F). The pointer is designed to be a bit loose for easy spinning and removal.
Here are a few questions for students to explore:
(1). How likely does one get an even number?
(2). What is the probability that one gets a prime number?
(3). If I spin twice, how likely do I get a sum of 8?
(4). If I spin five times, what is the probability do I get ALL prime numbers?
(5) What is the probability does one get a vowel?
Reference:
Bu, L. & Fernandez, M. https://www.maa.org/external_archive/joma/Volume7/Bu/Spinner.html
## 3D printing settings
Once before the time of “fidget spinners,” a spinner was frequently used to model random processes for student to feel the idea of uncertainty, data trends, and mathematical structures, along with dice.
The board by default has a diameter of 80mm, with both numbers (1, 2, 3, 4, 5, 6) and letters (A, B, C, D, E, F).
Lesson Plan and Activity
Pose a few questions about numbers, sum/difference of numbers, or even letters and allow students to spin the math out and discuss their findings.
Here are a few questions for students to explore:
(1). How likely does one get an even number?
(2). What is the probability that one gets a prime number?
(3). If I spin twice, how likely do I get a sum of 8?
(4). If I spin five times, what is the probability do I get ALL prime numbers?
(5) What is the probability does one get a vowel?
(6) If I spin twice, how likely am I going to get a difference of two?
• 3D model format: STL
## Creator
STEAM educator, learning from and working with K-12 STEAM teachers to explore new ideas of teaching and engagement. I firmly believe ART is at the core of STEM learning or all human learning! I owe my ideas and designs to the hundreds of K-12 children and teachers and university professors I have had the pleasure of working with, in multiple disciplines-- math, science,engineering language arts, social studies, early childhood education and more! All mistakes, of course, are mine! There is no warranty or liability whatsoever implied or explicit behind the designs or ideas. They are all posted for their potential educational values.
When working with children, please strictly observe all safety and health procedures! Please refer to the NSTA safety guides: http://www.nsta.org/safety/.
LGBU Contact: LGBU@SIU.EDU
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Functions
# How to use now function in Excel?
## How to use now function in Excel?
1. Enter "NOW" in the cell that is required. Please review the following:
2. At this point, we need to press tab. The image is provided for your consideration down below:
3. Because we have just entered), we have closed the formula, which means that it will now be finished.
## FAQ
### How does now function work in Excel?
Excel's "NOW Function" is a function that may be used to see the time and date as they are right now. It is automatically updated every time the worksheet is accessed, as well as if any changes are made to it. Additionally, you may manually refresh it by hitting the F9 key.
### How do I calculate now in Excel?
To recalculate all of the open worksheets, either click the Calculate Now button or use the F9 key. To recalculate only the currently selected worksheet, click the Calculate Sheet button (or press Shift+F9).
### What does now () function return?
The date and time that is now being kept is what is returned by the NOW() method.
### Why is my now function not working in Excel?
The NOW function's important defining characteristics
In the event that it does not work, check to see if the automatic recalculation option under Formulas > Calculation Options > Automatic is deactivated. Excel will provide a date and time that is accurate when you use the NOW function. Check the formatting of the cell to see why your output is displaying a serial number with a decimal value.
### Can Excel automatically insert current date in a cell?
Insert a date and time for the current day that can be kept up to date automatically. Use one of the following Excel date methods if you want to input today's date in Excel in a way that will always be up to date. =TODAY() will insert the current date into a cell. Using the =NOW() function will insert the current date and time into a cell.
### What is the difference between now and today in Excel?
The Now function provides a date/time value that is equivalent to the current date and time. The Today function provides a date/time value that is equivalent to the current date.
### How many parameters does now () accept?
No parameters are going to be taken in by this procedure. The value of the current date and time is what it hands back to you.
## Conclusion
We hope this article has explained everything you needed to know about "How to use now function in Excel?". If you have any other questions about the Ms Excel software, please take the time to search for additional Excel content in wikiExcel.com. Otherwise, don't hesitate to reach out to us through the contact page.
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# Exam #2 Locations
```Chapter 8 Stock Valuation
Overview
Preferred Stock Characteristics and
Valuation
Common Stock Characteristics
Common Stock as a Financing Tool
Common Stock Valuation
Dividend Discount Model
Preferred Stock Characteristics
Unlike common stock, no ownership interest
Second to debt holders on claim on
company’s assets in the event of bankruptcy.
Annual dividend yield as a percentage of par
value
Preferred dividends must be paid before
common dividends
If cumulative preferred, all missed past
dividends must be paid before common
dividends can be paid.
Preferred Stock Valuation
Promises to pay the same dividend year after
year forever, never matures.
A perpetuity.
Vps = D/kps
Example: GM preferred stock has a \$25 par
value with a 8% dividend yield. What price
would you pay if your required return is 9%?
D = \$25(0.08) = \$2
Vps = \$2/0.09 = \$22.22
Expected Rate of
Return on Preferred
kps =
D
Po
Example
If we know the preferred stock
price is \$40, and the preferred
dividend is \$4.125, the expected
return is:
Example
kps
If we know the preferred stock
price is \$40, and the preferred
dividend is \$4.125, the expected
return is:
D
=
Po
4.125
=
=
40
Example
kps
If we know the preferred stock
price is \$40, and the preferred
dividend is \$4.125, the expected
return is:
D
=
Po
4.125
=
= .1031
40
The Financial Pages:
Preferred Stocks
52 weeks
Yld
Vol
Hi
Lo
Sym
Div % PE 100s
2788 2506 GM pfG 2.28 8.9 … 86
Close
25 53
Dividend: \$2.28 on \$25 par value
= 9.12% dividend rate.
Expected return: 2.28 / 25.53 = 8.9%.
Claim on Income after interest and dividend
payments to creditors and preferred
stockholders.
Represents ownership.
Ownership implies control.
Limited liability.
Stockholders elect directors. = Voting Rights
Directors elect management.
Management’s goal: Maximize stock price.
Stock:
No required fixed payments.
No maturity.
Improves debt ratio, coverage.
with Stock:
Controlling shareholders may lose some
ownership control.
Preemptive Right
Future earnings shared with new
stockholders. Possible EPS Dilution.
Higher flotation costs vs. debt.
Higher component cost of capital.
Too little debt may encourage a takeover bid.
Common Stock Valuation
(Single Holding Period)
You expect XYZ stock to pay a \$5.50
dividend at the end of the year. The
stock price is expected to be \$120 at
that time.
If you require a 15% rate of return,
what would you pay for the stock now?
Common Stock Valuation
(Single Holding Period)
You expect XYZ stock to pay a \$5.50
dividend at the end of the year. The
stock price is expected to be \$120 at
that time.
If you require a 15% rate of return,
what would you pay for the stock now?
?
5.50 + 120
0
1
Common Stock Valuation
(Single Holding Period)
Solution:
Vcs = (5.50/1.15) + (120/1.15)
= 4.783
+ 104.348
= \$109.13
Common Stock Valuation
(Single Holding Period)
Financial Calculator solution:
P/Y =1, I = 15, n=1, FV= 125.50
CPT PV = -109.13
or:
P/Y =1, I = 15, n=1, FV= 120,
PMT = 5.50
CPT PV = -109.13
The Financial Pages:
Common Stocks
52 weeks
Hi Lo Sym
126 87 IBM
Div
.56
63 42 WalMart .28
Yld
Vol
Net
% PE 100s Close Chg
0.6 23 77995 98.12 +0.29
0.5 47 119515 62.01 -0.24
IBMs Dividend Yield = \$0.56/\$98.12 = 0.6%
PE Ratio = Close Price/Earnings Per
Share(EPS)
IBMs Latest EPS = Close/PE = \$98.12/23 =
\$4.27
Stock Valuation
Multiple Holding Periods
Stock Value = PV of Future
Expected Dividends
D3
D1
D2
D
Vcs
...
1
2
3
1 kcs 1 kcs 1 kcs
1 kcs
Stock Valuation: Dividend
Patterns
For Valuation: we will assume stocks fall into
one of the following dividend growth
patterns.
Constant growth rate in dividends
Zero growth rate in dividends, like preferred
stock
“Supernormal” (non-constant) growth rate in
dividends(see Chapter 8 notes in Syllabus
book)
Doh! Doughnuts Stock Valuation
Example: Basic Information
We have found the following information for
Doh! Doughnuts:
current dividend = \$2,
beta of 0.9
T-bill (risk-free) rate = 1.75%
the market risk premium is 9.5%
Using the SML equation to find Doh!’s
required return = krf +(krp)b =
1.75% +(9.5%)0.9 = 10.3% = kcs
Analysts Estimates for Doh!
Doughnuts
NEDFlanders predicts a constant annual growth rate in
dividends and earnings of zero percent (0%)
Barton Kruston Simpson predicts a constant annual
growth rate in dividends and earnings of 8 percent
(8%).
Moe Homer Simpson & Bernard expect a dramatic
growth phase of 20% annually for each of the next 3
years followed by a constant 8% growth rate in year 4
and beyond.
What should be each analyst’s
estimated value of Doh! Doughnuts?
First Analyst: Zero Growth
Stock Valuation
No growth in dividends, so Doh!
Doughnuts will remain at the current
dividend of \$2 forever.
Estimated Value (Vcs)= PV(perpetuity) =
D0/kcs
Doh! Kcs = 10.3%
NEDFlanders Estimate P0 = \$2/.103 =
\$19.42
Constant Growth Stock
Valuation Model
Dividends are expected to grow at an
annual constant rate, g, forever.
D1 = D0(1+g)
Dt = D0(1+g)t
Vcs =
V
cs
=
D0(1+g) =
kcs – g
D0 (1 + g )
k
cs
-
g
D1
kcs – g
=
D
1
k
cs
-
g
Constant Growth Stock
Valuation Model Terms
D0 = today’s (or current) dividend
D1 = expected dividend at the end of
this year(year 1)
kcs = stock’s required rate of return
g = the constant growth rate in
dividends
2nd Analyst: Constant Growth
in Dividends
Current Dividend = \$2
Projected Constant Growth Rate = 8% or
0.08
Kcs = 10.3%
What happens if g > kcs?
D1
Vcs
requires kcs g.
kcs g
If kcs< g, get negative stock price, which is
nonsense.
We can’t use model unless (1) kcs> g and (2)
g is expected to be constant forever.
Assume Doh! Doughnuts current stock price
is \$100.
Required return = 1.75% + 9.5%(0.9) =
10.3%
Let’s assume the 2nd analyst is correct and
Doh! Has a constant growth rate of 8% and
its current dividend is \$2.
What is the stock’s expected return?
Is Doh! Doughnuts’ current stock price in
equilibrium?
Expected Return of Constant
Growth Stocks
Expected Rate of Return = Expected Dividend Yield +
Expected Capital Gains Yield
D1/P0 = D0(1+g)/P0 = Expected Dividend Yield
g = Expected Capital Gains Yield
From our example, D1=\$2(1.08) = 2.16, P0=\$100, g
= 8% or 0.08
D1
\$2.16
k cs
g
8% 2.16% 8% 10.16%
P0
\$100
^
DOH! Doh! Doughnuts
Stock Market Equilibrium
The stock price when the stock’s
expected return = stock’s required
return (CAPM)
D1/P0 + g = krf +(km - krf )b
Expected Return = Required Return
The Effect On the Stock Price
Return
SML
1.75
0.9
Beta
Expected Return needs to rise to the required return of
10.3%. This means the stock price must fall to the the
equilibrium price which yields the required return of
10.3%
New Price = D1/(kcs- g)=\$2.16/(.103 - .08)= \$93.91
At the current price of \$100, Doh! has NPV of
\$93.91 - \$100 = -\$6.09
“Supernormal” Growth Stock
Valuation
Framework: Assume Stock has period
of non-constant growth in dividends
and earnings and then eventually
settles into a normal constant growth
pattern(gc).
0 g1 1 g2 2 g3 3 gc 4 gc 5 gc…
D1
D2
D3
“Supernormal” Growth Period
Constant Growth
Supernormal Growth Valuation
Process
3 Step Process
Estimate Dividends during “supernormal”
growth period.
Estimate Price, which is the PV of the
constant growth dividends, at the end of
“supernormal” growth period which is also
the beginning of the constant growth period.
Find the PV of “supernormal” dividends and
constant growth price. The total of these PVs
= Today’s estimated stock value.
3rd Analyst:“Supernormal”
Growth Stock Valuation for Doh!
“supernormal” growth rate g for years
1-3 = 20% or 0.2
After year 3, Doh! Has constant growth
rate gc = 8% or 0.08
D0 = \$2.00
kcs = 10.3% or .103
0 g = 20% 1 g = 20% 2 g = 20% 3 gc = 8%
\$2.40
\$2.88
PV= P0
10.3%,1
2.17
10.3%,2
2.37
123.51
128.05 = \$128.05 = P0
\$3.46
Fin’l Calculator
\$162.28 = P3 Solution:
\$165.74
CF0=0,C01= 2.40
C02 = 2.88
10.3%,3
C03 = 165.74
I = 10.3
NPV=128.05 = P0
P0 = \$2.40(PVIF10.3%,1)+\$2.88(PVIF10.3%,2)+\$3.46(PVIF10.3%,3)
+ \$162.28(PVIF10.3%,3) = \$128.05
Summary of Doh! Doughnuts
Stock Price Estimates
NEDFlanders: 0% constant growth: P0 =
\$19.42
Barton Kruston Simpson: 8% constant
growth: P0 = \$93.91
Moe Homer Simpson & Bernard: 20%,
3-year supernormal growth followed by
8% constant growth: P0 = \$128.05
Other Valuation Approaches
Our dividend discount models are best for
established dividend paying companies, which
makes it difficult to apply to non-dividend
paying start-up companies.
PE Multiple Approach: Forecast a company’s
earnings per share and multiply this forecast
times the company’s PE ratio.
Value entire firm by finding PV of future
expected Free Cash Flows available to
stockholders, then divide by number of
shares. (Chapter 13)
```
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CC-MAIN-2021-49
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https://masurp.github.io/specr/reference/icc_specs.html
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This function extracts intraclass correlation coefficients (ICC) from a multilevel model. It can be used to decompose the variance in the outcome variable of a specification curve analysis (e.g., the regression coefficients). This approach summarises the relative importance of analytical choices by estimating the share of variance in the outcome (e.g., the regression coefficient) that different analytical choices or combinations therefor account for. To use this approach, one needs to estimate a multilevel model that includes all analytical choices as grouping variables (see examples).
icc_specs(model, percent = TRUE)
Arguments
model
a multilevel (i.e., mixed effects) model that captures the variances of the specification curve.
percent
a logical value indicating whether the ICC should also be printed as percentage. Defaults to TRUE.
Value
a tibble including the grouping variable, the random effect variances, the raw intraclass correlation coefficient (ICC), and the ICC in percent.
References
• Hox, J. J. (2010). Multilevel analysis: techniques and applications. New York: Routledge.
See also
plot_variance() to plot the variance decomposition.
Examples
# Step 1: Run spec curve analysis
results <- run_specs(df = example_data,
y = c("y1", "y2"),
x = c("x1", "x2"),
model = c("lm"))
# Step 2: Estimate a multilevel model without predictors
model <- lme4::lmer(estimate ~ 1 + (1|x) + (1|y), data = results)
# Step 3: Estimate intra-class correlation
icc_specs(model)
#> grp vcov icc percent
#> 1 x 0.669097576 0.0310285827 3.10285827
#> 2 y 20.890686733 0.9687800785 96.87800785
#> 3 Residual 0.004126013 0.0001913388 0.01913388
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CC-MAIN-2022-33
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https://profound-answers.com/what-do-closely-spaced-isobars-on-a-map-indicate-about-the-wind/
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Collection of recommendations and tips
# What do closely spaced isobars on a map indicate about the wind?
## What do closely spaced isobars on a map indicate about the wind?
Closely spaced isobars indicate a steep pressure gra- dient and high winds. Widely spaced isobars indicate a weak pressure gradient and light winds.
## What do isobars on a weather map represent?
Isobars are lines on a weather map that join places of equal pressure. Meteorologists collect information from weather stations, buoys and ships and then draw smooth curves to join the dots.
What would lines very close together indicate on an isobar map?
The spacing between isobars represents a pressure differential between those two isobars. When two isobars are closer together then the pressure changes at a greater rate over distance. It is pressure differences that set the air in motion.
### How does the spacing of isobars on a weather map relate to wind speed quizlet?
How does the spacing of isobars on a weather map generally relate to the speed of wind? Closely spaced isobars indicate a strong pressure gradient and high wind speeds; widely spaced isobars indicate a weak pressure gradient and low wind speeds. The meteorologist comments that the pressure tendency is rising.
### Why are the Isobar lines close together near the area of low pressure?
Isobars are lines/areas of equal pressure represented on a weather map. When isobars become very tightly grouped together it indicates a “tight pressure gradient” (steep slope). The tightly packed isobars are due to the difference in air pressure between between High and Low pressure systems.
What is a closed Isobar?
An isobar is defined as being closed if its numerical value exists on all 576 radial legs.
## What does a an isobar that is a closed circle represent?
What does an isobar that is closed circle represent. They represent high and low pressure area.
How does the spacing of isobars on a weather map generally relate to the speed of wind quizlet?
### What does an isobar represent on a weather map quizlet?
On a weather map, what does an isobar represent? High pressure guarantees nice weather.
### What does it mean when isobars are close together quizlet?
Isobars that are close together indicate a large pressure. A large pressure causes strong winds. Isobars that are spread apart indicate a smaller pressure.
What is the purpose of isobars?
Definition of an Isobar An isobar is a line on a map that shows a meteorologist what the pressure is at the surface of the earth. They are lines that connect equal points of pressure. Isobars can be used to map atmospheric or air pressure in a way that makes it easier to understand.
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https://amp.doubtnut.com/question-answer/find-the-sum-of-all-integers-between-100-and-550-which-are-divisible-by-9-642575654
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# Find the sum of all integers between 100 and 550, which are divisible by 9.
Updated On: 15-3-2021
This browser does not support the video element.
Answer
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
Time Transcript
00:00 - 00:59 find sum of all integers between 100 and 550 which are divisible by 9 in this position the first integer integer integer between 100 and 550 which are divisible by which are divisible by 9 by 9 is equal to 8 and the last integer and last integer which are in which are divisible divisible by 9 and which is between hundred 55549 AP that is equal to 108 109 it goes 2549 and there is D there is equal to b is equal to
01:00 - 01:59 because the numbers which are divisible by 9 the sum of n terms the sum of n terms that is equal to 2 and minus 1 is equal to now but is equal to 108 we have to find out that we have to find out this to find out if we have to find out how to find out how many tens are there we have to find out and terms and also here to find out and terms the formula is a n is equal to a plus and minus D now is equal to 549 a is equal to 108 - 1 MB is equal to 9 now
02:00 - 02:59 this and here and is equal to 50 now have an equal to 50 and put it into the value of Asin Asin what is equal to 2 2A + admin now here we have everything that is required to solve this equation now equal to 50 by 22 into a is equal to 10 8 + equal to 50 - 1 MB is equal to 9 now sold this now solve this 108 into two equal to 216 in in 29 that is equal to 9 into 9 is equal to 14 in in 29 that is equal to
03:00 - 03:59 25 to 1681 1A 936 + 8 is 40241 now let's all this and it becomes the value of 25 into 65757 and the answer is 15025 and deserve final ank you
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https://brilliant.org/practice/function-graphs-level-4-5-challenges/?subtopic=functions&chapter=function-graphs
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Algebra
# Function Graphs: Level 4 Challenges
Given the equation $4x^2+2\sqrt{3}xy+2y^2=1$.
Through what angles $\theta \in [0,\pi]$ should the axes be rotated so that the term $xy$ is removed from the transformed equation?
Give the answer as the sum of all values of $\theta$(in degrees).
The graph above shows the figure created by the function $x^2+(y+x)^2=1$.
Find the area of this tilted ellipse correct to 2 decimal places.
For each integer $n>1$, let $F(n)$ be the number of solutions to the equation $\sin{x} = \sin{(nx)}$ on the interval $[0,\pi]$. What is $\displaystyle\sum_{n=2}^{2007}$$F(n)$?
If $f(x)=\dfrac{x^2}{1-\cos x}$, where $0. Which of the following describes the function of $f(x)$?
If $f(x)$ and $g(x)$ have the same fundamental period, then so does $f(x) + g(x)$.
×
Problem Loading...
Note Loading...
Set Loading...
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# Chemistry
Hi DrBob,
Sorry about that, I wasn't very clear!
A. Standardization of Na2S2O3
-Mass of KIO3 in 100mL = 0.09600g
-Calculate Molarity of KIO3= ?
-4.486 mol/0.1L= 0.004486 M
B. Titration/ Volume of Na2S2O3
1: 38.44 mL
2: 37.23 mL
3: 38.11 mL
Average Volume: 37.93mL
-Calculate molarity of Na2S2O3=?
Not sure what to use as the mass here? Help
C. V of Bleach= 1.020 mL
V of Na2S2O3= 2.010 mL
Calculate Mass of NaOCl/100mL bleach=? Help with this.
-We are told here from the concentrations and volume of the added sodium thiosulfate solution used to titrate, calculate the numbers of moles of the oxidizing agent present, assuming it to be sodium hypochlorite, NaOCl. Use this to calculate the number of grams of sodium hypochlorite that was present in the 0.5mL of bleach.
The equation for I2 is: KIO3 reacts w/ excess KI in the following reaction:
IO3 + 5I + 6H--> 3I2 + 3H20
2S2O3 + I2 --> 2I + S406
Sodium Hypochlorite reacts with sodium thiosulfate in the following reaction:
NaClO + 2H + 2I--> I2 + Cl + H20+ Na
2S203 + I2 --> 2I + S406
Thanks.
1. 👍 0
2. 👎 0
3. 👁 325
1. A. Standardization of Na2S2O3
-Mass of KIO3 in 100mL = 0.09600g
-Calculate Molarity of KIO3= ?
-4.486 mol/0.1L= 0.004486 M
This looks ok to me except for the -4.486. I don't know why the - sign is there unless it's just a typo.
B. Titration/ Volume of Na2S2O3
1: 38.44 mL
2: 37.23 mL
3: 38.11 mL
Average Volume: 37.93mL
-Calculate molarity of Na2S2O3=?
Not sure what to use as the mass here? Help
You have 100 mL of 0.004486M KIO3. Multiply those to find mmoles of KIO3 (mL x M = mmols). Then convert moles KIO3 to moles S2O3^-2. So moles KIO3 x 6 = mmoles thiosulfate. (How do I do that. 1 KIO3 = 3I2 and 3I2 = 6S2O3^-2; therefore, 1/6 mol KIO3 = 0.4486 mmols x (6S2O3/1KIO3) = 0.4486+6 = moles S2O3). M thiosulfate = mmoles/mL
C. V of Bleach= 1.020 mL
V of Na2S2O3= 2.010 mL
Calculate Mass of NaOCl/100mL bleach=? Help with this.
-We are told here from the concentrations and volume of the added sodium thiosulfate solution used to titrate, calculate the numbers of moles of the oxidizing agent present, assuming it to be sodium hypochlorite, NaOCl. Use this to calculate the number of grams of sodium hypochlorite that was present in the 0.5mL of bleach.
mL S2O3 x M S2O3 = mmoles S2O3.
Convert mmols S2O3 to mmoles OCl using the coefficients in the equations used (as I did in the KIO3 and S2O3 above.)
Then mmoles OCl x molar mass NaOCl = g NaOCl and that will be the grams in the volume you took. Convert that to g/100 mL.
The equation for I2 is: KIO3 reacts w/ excess KI in the following reaction:
IO3 + 5I + 6H--> 3I2 + 3H20
2S2O3 + I2 --> 2I + S406
Sodium Hypochlorite reacts with sodium thiosulfate in the following reaction:
NaClO + 2H + 2I--> I2 + Cl + H20+ Na
2S203 + I2 --> 2I + S406
1. 👍 0
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2. Too late at night and I made some typos.
You have 100 mL of 0.004486M KIO3. Multiply those to find mmoles of KIO3 (mL x M = mmols). Then convert moles KIO3 to moles S2O3^-2. So moles KIO3 x 6 = mmoles thiosulfate. I've redone the explanation because I typed a + sign instead o a x sign AND I notice a 1/6 which, although correct, it's better to do it another way.
(How do I do that. 1 KIO3 = 3I2 and 3I2 = 6S2O3^-2; therefore, 1/6 mol KIO3 = 0.4486 mmols x (6S2O3/1KIO3) = 0.4486+6 = moles S2O3). M thiosulfate = mmoles/mL
Here is that section done over.
How did I did that. 1 KIO3 = 3I2 and 3I2 = 6 S2O3; therefore, convert moles KIO3 to moles S2O3 the usual way using the coefficients in the balanced equation. 0.4486 mmoles KIO3 x (6 mols S2O3/1 mole KIO3) = 0.4486 mmoles x 6 = ??mmoles S2O3. THEN, ??mmoles S2O3/mL S2O3 = M S2O3.
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https://python-forum.io/Thread-How-to-switch-table-area-coordinates-in-Python-Camelot-and-Tabula-Py?pid=79801
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How to switch table area coordinates in Python Camelot and Tabula-Py john5 Unladen Swallow Posts: 1 Threads: 1 Joined: May 2019 Reputation: 0 Likes received: 0 #1 May-08-2019, 04:31 PM (This post was last modified: May-08-2019, 04:32 PM by john5. Edited 2 times in total. Edit Reason: spelling mistake in 'All' ) Dear All, I have obtained the coordinates of a table bounding box using Camelot, but I need to use tabula-py to extract the table data, as camelot is only extracting the first line in each table cell, even in lattice mode. I have noticed that when defining the same table region in tabula-py, 2 of the resulting coordinates are largely different form the camelot values (shown in the code sample below). Whilst the 2nd and 4th values in tabula are similar to 1st and third values in camelot, the others are largely different, how can I translate these readings from the camelot values please? I have been trying to use proportion, and to add and subtract values but all were in vain... ```df= tabula.read_pdf(pdf_path, lattice=True, area=(71, 627, 325, 1160), pages=page) #but camelot coordinate values from bounding box are: 631, 518, 1154, 765 ``` « Next Oldest | Next Newest »
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# Statistics data base final project
Attached was the midterm project now i need the final project done
Requirements:
• No more than 2 to 3 pages
• Present the findings using the skillset acquired (topics covered) in class.
• Also include the dataset with the analysis (could be excel or any statistical package). You should provide details of the analysis in an Appendix.
How are the 40 points given: 10 for each Module you choose to apply. (For example, you choose regression to test an association or predict an outcome, you get 10 points for that analysis)
Data Analytics is a subject that can be best appreciated only when applied to a dataset you are familiar with. The aim of this project is to achieve that. Do not view this project as a hurdle in the course, rather a bridge to connect the topics you learnt to your work or subject domain. There are five main modules in this course:
• Module 1 : Normal Distribution (Percentile, distribution of means, and chance of occurrence if we assume normal distribution)
• Module 2 : Confidence Interval Estimation (Including Sample Size determination)
• Module 3 : Inferences from data (Hypothesis testing, i.e., confirming or checking if a claim made about the data. In this module, we dealt with only one sample)
• Module 4 : More Inferences from data (Multiple samples)
• Module 5 : Regression analysis (Both simple and multiple, apart from basic ANOVA)
1. Bring your own data from work (you can remove any private or confidential information, for example: if you are bringing any sales or cost data of an item/product or service – the name can be masked)
2. Use data from your previous work or company you have access to (again you can remove any private/confidential information)
3. Use data from public domain – In today’s world, there is no dearth of structured data. Here are some places where you can get data from:
## Calculate the price of your order
550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
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\$26
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Convert 3.75 pounds into grams? - ThatsAnswer.COM
## Similar Questions
• Answer: the wight in lbs. X 453.6 = the weight in grams
• Answer: 380 grams = 0.837756596 pounds
• Answer: 3000 grams = 6.6138679 pounds1 pound = 453.59237 grams (for cooking, call it 454 grams ;-)
• Answer: No, 10 pounds = 4,535.9237 Grams.
- Hope this helped
• Answer: Lbs (pounds) are a unit of weight.
Ft (feet) are a unit of length.
There is no way to universally convert between the two and make any sense at all.
• Answer: 1 ft. lb. * 1.3558179 = 1 Nm
or, for rough guesstimates, add one third of the ft. lbs.
# Convert 3.75 pounds into grams?
• 3.75 lbs = 1,701 g
Name: *
• Answer: 500 grams = 1 pound 1 ounce
• Answer: 3.75 lbs = 1,701 g
* * * * *
No. Multiply by 100. 1 gram = 100 cg.
• Answer: This is an impossible question. A kilometer is a measure of distance and a gram is a measure of weight . This is like saying how do I convert an apple into an orange.
• Answer: 1 oz = 28.3495 g
→ To convert ounces to grams multiply by 28.3495, thoughmultiplying by 28 will be close enough for small values.
→ To convert grams to ounces divide by 28.3495, though dividing by28 will be close enough for small values.
When recipes are converted for cooking, the value of 1 oz = 25 g isusually used. As the conversion makes 4 oz = 100 g, whereas 100g ismuch closer to 3½ oz, you cannot mix imperial and metric measuresin a recipe.
• Answer: You cannot convert one to the other.Grams is a measure of weight, where millilitres is a measure of volume.
• Answer: We need it for school
• Answer: grams per litre is the same as grams over litres
kilograms per millilitre is the same as kilograms over millilitres
Convert the numerator and denominator separately and then separateout the numbers:
1 kg = 1000 g → 1 g = 0.001 kg
1 l = 1000 ml
→ g/l = (0.001 kg)/(1000 ml) = (0.001/1000) kg/ml = 0.000001 kg/ml= 1/1000000 × kg/ml
Therefore to convert g/l to kg/ml divide by 1,000,000.
• Answer: grams per milliliter is the same as grams over milliliters
kilograms per liter is the same as kilograms over liters
Convert the numerator and denominator separately and then separateout the numbers:
1 kg = 1000 g → 1 g = 0.001 kg
1 l = 1000 ml → 1 ml = 0.001 l
→ g/ml = (0.001 kg)/(0.001 l) = (0.001/0.001) kg/l = 1 kg/l = kg/l
Therefore to convert g/l to kg/ml just change the units as thenumbers are the same.
eg 10 g/ml = 10 kg/l
• Answer: vicodin can consist of 5mg or 10mg of hydrocodone and 325mg of acetaminophen. So to convert mg to gm there are 1000mg to 1 gram
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# Finding the Probability Density Function
1. Jan 22, 2008
### sarujin
1. The problem statement, all variables and given/known data
A dial indicator has a needle that is equally likely to come to rest at an angle between 0 and Pi. Consider the y-coordinate of the needle point (projection on the vertical axis). What is the probability density function (PDF) p(y)?
2. Relevant equations
I know the integral of p(y) over all space has to equal 1. The y-coordinate of the dial is of course radius*sine(theta).
3. The attempt at a solution
The first part of the question asked for the PDF for the angle, which wasn't too difficult. Knowing the integral over all space had to equal 1 and that the probability was a constant I could see that p(theta)=1/Pi . I just can't find any recipe on how to come up with the PDF for the y coordinate, in most cases it is given! I can see that it must be zero at both y=0 and y=r, which suggests to me it is probably sine or sine^2 but I cannot prove it.
Thanks a bunch.
Last edited: Jan 22, 2008
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# Distance between Casper, WY (CPR) and Tucson, AZ (TUS)
Flight distance from Casper to Tucson (Casper–Natrona County International Airport – Tucson International Airport) is 784 miles / 1261 kilometers / 681 nautical miles. Estimated flight time is 1 hour 59 minutes.
Driving distance from Casper (CPR) to Tucson (TUS) is 1176 miles / 1893 kilometers and travel time by car is about 19 hours 24 minutes.
## Map of flight path and driving directions from Casper to Tucson.
Shortest flight path between Casper–Natrona County International Airport (CPR) and Tucson International Airport (TUS).
## How far is Tucson from Casper?
There are several ways to calculate distances between Casper and Tucson. Here are two common methods:
Vincenty's formula (applied above)
• 783.523 miles
• 1260.958 kilometers
• 680.863 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 784.662 miles
• 1262.791 kilometers
• 681.852 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Casper–Natrona County International Airport
City: Casper, WY
Country: United States
IATA Code: CPR
ICAO Code: KCPR
Coordinates: 42°54′28″N, 106°27′50″W
B Tucson International Airport
City: Tucson, AZ
Country: United States
IATA Code: TUS
ICAO Code: KTUS
Coordinates: 32°6′57″N, 110°56′27″W
## Time difference and current local times
The time difference between Casper and Tucson is 1 hour. Tucson is 1 hour behind Casper.
MDT
MST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 133 kg (294 pounds).
## Frequent Flyer Miles Calculator
Casper (CPR) → Tucson (TUS).
Distance:
784
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
784
Round trip?
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## Category: "R:WM Dev Blog"
Now this was straight forward for projecting a planet sphere at a specific distance away so that, no matter what maximum distance the view area is set at, the planet is still visible and scaled appropriately. This all worked and has been working.
The problem I have been running into is the equation to know how much of the planet to show at any one time. The equation for determining the visible surface area angle is the ArcCosine of the Radius of the Planet divided by the Distance the observer is to the center of the planet (which would be somewhere, perhaps high above the surface of the planet, which of course the surface would be the radius). Ok, that works fine. Take the circumference of the planet, divide that by 360, that gives you the distance per degree and you multiply that by what you got from the ArcCos (Yes, typical ACOS gives you radians, not degrees, but the concept still applies). This gives you the distance of the visible arc. Meaning, from the center point that the observer is directly over, how much distance FROM there can that observer see of the sphere. That all works. To get the size of the original triangle from the isohedron is 26 degrees or the ArcTangent of 1/2 (look it up). Multiply that by the same distance per degree as above and you get how big the original isohedron triangles are based on the radius of the planet. Ok, that all works.
Now I wanted to know what subdivision level I should be at to cover that area with a specific set of “rings” of triangles. Keep in mind that all the subdivisions are of that original isohedron triangle. And this kept screwing me up. Here is the solving equation:
Distance of Visible Arc/Number of Rings = Distance of Original Isohedron Triangle/ 2 to the power of X (X being the amount of subdivisions needed based on the number of desired rings).
A little algebra later and you get this:
2 to the power of x = (Distance of Original Isohedron Triangle/(Distance of Visible Arc/Number of Rings))
Now I thought the way to solve this was:
X=Square root of (Distance of Original Isohedron Triangle/(Distance of Visible Arc/Number of Rings))
It isn’t. That doesn’t work. That isn’t how you solve it. It’s not squareroot. It’s LOG to the base of 2. So the answer is finally:
X=log2(Distance of Original Isohedron Triangle/(Distance of Visible Arc/Number of Rings))
Older C didn’t have Log2, but did have Log10. So to emulate it, it is Log10(Distance of Original Isohedron Triangle/(Distance of Visible Arc/Number of Rings))/Log10(2). But it seems newer C has Log2.
## Behold the field which I grow my....Perlin Noise?
First test of the Perlin Noise field. I'm not 100% happy with it, but that is just tweaking later on anyways. I want to get it integrated first. (This is using the internal point dumpster rountine I made so that I can visualize points and objects more easily).
## Sub Div 9 Planets are working...
In this still:
the "land" is being rendered from the outside (the red mesh). You can see the colored triangles that I used for debugging. The mini-triangles represent the points by color: red, green, and blue for points 1, 2 and 3 respectively. So as it subdivides down it is displaying two sets of triangles (Best seen here:)
This is showing the planet sphere in white at the lowest resolution (so you can see what is going on) from the inside out, so it's easier to see what is happening. The "red" main triangle is the point at which the player is to the planet (now behind us). The two sets of triangles represent the subdivisions from the 20 sided sphere (so flat against that first triangle), and the actual triangles ON the sphere as each subdivison is approximated on the real sphere.
This shows how it looks as the resolution of the real sphere goes up.
This matches the resolution of what I'm making te surface out of. Currently, Sub division 6.
This shows the planet at subdivision 9, with still sub division 6 "land". To do sub division 9 on that wide of an area would make the system choke (and doing the whole planet at sub division 9 kills the frame rate, obviously). But the point is, the red "land" represents the visible arc of the sphere at the distance we are from the planet. The monitor says that's 992 miles away. So at 992 miles, that land mass is what you can actually see of the planet (Mars), but because of the curvature of the sphere you can only see so much of that sphere. As one gets closer to the planet, the amount of area goes down drastically. At which point the resolution will go up of the land mass, but the area will go down proportionally. The idea is to keep doing this all the way down to subdivision 19 or 20. Subdivision 9, each triangle is between 2.5 to 5 miles I estimate, so to get it down to around 10 feet, that is subdivision 19. After Sub divison 9, it's going to use either a height map or a perlin noise generated heightmap to handle the points below. I'm working on that part. :)
## Planets are running...Kinda
I am showing a SubDivisioned Planet down to 9 subdivisions. I found the problem since Sub Division 0 looks like origami, but everything else looks fine. If you notice towards the end, you can see the heights embedded in the points. And best of all, no poles!
Note: This is not all of it by any means. I need to go down to subdivision 19, but I needed to get this running completely. Unlike the original "sphere" you saw, this actually has height information in it and can drill down.
## 2 New Fighters
2 new styles of fighter ship
## Issues around designing a planet
While displaying a textured sphere is easy, getting to procede from high orbit to actual land fall and dynamically scale up isn't so easy. By definition, you are defining a whole planet which is a really big place. That means lots of memory or hardrive space to store something on that scale. The inital answer is just use a texture and that works to a point. The PROBLEM with that is, it really won't go down to "walking on the ground" level and it really starts to get distorted at the poles.
Down around the equator, the distortion is so minor you don't consider it, but as you can see at the poles of the planet, the idea of a texture can get messy. That isn't so much a problem (but still is) with a color texture wrapping a sphere, but when you try to do X and Y shifts the concepts no longer work. Even if you simulate it with spherical coordinates (theta and phi, ie. theta being lattitude and phi being longitude), close to the poles, you are litterially running around in circles when you go along the "X" axis.
The current "solve" for this is you don't use X and Y, but a isohedral format.
Now there are no "poles" per se. This is based on a isohedron. (think of a 20 sided die, although the picture is one subdivision down). While this works, if you noticed, some triangles are layed out in a hexigonal pattern and some in a pentagonal pattern. This creates it's own series of problems, so my original idea of just doing hexigons won't really work globally. The isometric triangles still will, however, but now the trick is to get them to play nice. I have an idea via linking triangles together, but again, on planetary scales, that gets very memory intensive. I have ideas for optimisation, but then things get unlinked, so there are still problems to solve.
Of course, none of this solves the issue when you get below the original heightmap threshold, which will have to be handled by something procedural like Perlin noise.
## Reactor Building
Modeling and texturing of the Reactor Building.
## New Ships
Just smoothing out the normals for the new Cap Ship and working on the base model for the freighter.
Full story »
## Terrain Curved!
This took a lot longer than it should have, but I have been working on other things too. What this is showing is the normal hexigonal terrain layout is now properly curved based on distance. You can't see that from the first picture that it's just the original hexigon. The second picture is just showing the heightmap is working. And the third just to show it is the hexigon (just from inside the planet so you can actually see it). The normals aren't correct right this second, but I want to get this posted. So next is alignment and proper approach positioning and scaling.
## Typical Meeting
If you ever wanted to know what a typical meeting looked like from everyone, here is a screen shot. We mostly talk about progress, what should be focused on until the next meeting, and other suggestions as they come up.
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# Equation of astronomical telescope
Angle determines the size of an image of the object formed on the retina. As shown in the figure, the farther the object distance from the eye the smaller the angle and therefore the smaller the image size formed on the retina.
The simple astronomical telescope has two convex lenses; each called the objective lens and the ocular lens. The objective lens has a greater distance from the eye while the ocular lens has a distance closer to the eye. Objective lens functions to bring the image closer to the ocular lens, so the angle becomes larger. Ocular lens function to increase the angle so that the size of the image formed on the retina is greater.
1. The total magnification of the telescope when accommodation of eye is minimum (eye focused at the far point)
1.1 The linear magnifications of the objective lens when accommodation of eye is minimum
The objective lens is a convex lens. Therefore, the equation of the linear magnification of the objective lens is the same as the equation of the linear magnification of the convex lens.
Negative signs only explain the image is inverted so that it can be eliminated from the equation.
When the observer’s eyes focused at the far point (accommodation is minimum) the image produced by the objective lens must be at the second focal point of the objective lens. Thus the image distance of the objective lens (diob) = the focal length of the objective lens (fob). Objects are very far away from the objective lens and are considered infinite therefore the object distance from the objective lens (doob) = infinite.
Based on this equation it is concluded that the linear magnification of the objective lens is close to zero so that it can be ignored. Note that even though the linear magnification is small, the objective lens brings the real image closer to the ocular lens, so the angle between the real image and the ocular lens is larger.
Read : Gravitational field and gravitational field strength
1.2 The angular magnification of the ocular lens when the accommodation of eye is minimum
Ocular lenses function to increase the angle, so the equation of the magnification used is the equation of the angular magnification. The equation of the angular magnification of the ocular lens of the astronomical telescope is different from the equation of the angular magnification of the ocular lens of the microscope because the microscope is used to see objects nearby while the telescope is used to see distant objects.
The equation of the angular magnification:
M = θ’ / θ
Angles is small so tangent θ ≈ θ
θ = hi / fob
θ’ = hi / fok
The angular magnification:
Mok = the angular magnification of the ocular lens, fob = the focal length of the objective lens, fok = the focal length of the ocular lens.
The length of the astronomical telescope (l) = the focal length of the objective lens (fob) + the focal length of the ocular lens (fok). So fob = l – fok or fok = l – fob
1.3 The total angular magnification when accommodation of eye is minimum
There is no linear magnification hence the total angular magnification of the astronomical telescope when the eye focused at the far point, or the accommodation is minimum (M) = the angular magnification of the ocular lens when the eye focused at the far point, or the accommodation is minimum (Mok).
M = the total angular magnification, fob = the focal length of the objective lens, fok = the focal length of the ocular lens, l = distance between the objective lens and the ocular lens = the length of the telescope
2. The total magnification of telescope when accommodation of eye is maximum (eye focused at the near point)
2.1 The linear magnifications of the objective lens when accommodation of eye is maximum
The objective lens is a convex lens. Therefore, the equation of the linear magnification of the objective lens is the same as the equation of the linear magnification of the convex lens.
Negative signs only explain the inverted image so that it can be eliminated from the equation.
When the observer’s eye accommodates maximum, the image produced by the objective lens is between the first focal point of the ocular lens and the lens. Thus the distance of the real image from the objective lens (diob) = length of the telescope (l) – the distance of the real image from the ocular lens (dook). Objects are very far away from the objective lens and are considered infinite therefore the object distance from the objective lens (doob) = infinite.
Based on this equation it is concluded that the linear magnification of the objective lens is close to zero so that it can be ignored.
2.2 The angular magnification of the ocular lens when the accommodation of the eye is maximum
The equation of the angular magnification:
M = θ’ / θ
Angle is small so tangent θ ≈ θ
θ = hi / diob
θ’ = hi / dook
The angular magnification:
Mok = the angular magnification of the ocular lens, sob’ = diob = the image distance from the objective lens, dook = the distance of the real image (image are considered objects) from the ocular lens.
The length of the astronomical telescope (l) = the focal length of the objective lens (diob) + the focal length of the ocular lens (dook). So diob = l – dook or dook = l – diob
2.3 The total angular magnification when accommodation of eye is maximum
There is no linear magnification hence the total angular magnification of the astronomical telescope when the eye focused at the far point, or the accommodation is minimum (M) = The total angular magnification of the ocular lens when the eye focused at the far point, or the accommodation is minimum (Mok).
M = the total angular magnification, diob = the image distance of the objective lens, dook = the distance of the real image (image is considered an object) from the ocular lens, l = the distance between the objective lens and the ocular lens = the length of the telescope.
Read : Image formation by concave mirror
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Gold dram to cubic meters of gold converter
# gold conversion
## Amount: dram (dr) of gold mass Equals: 0.000000092 cubic meters (m3) in gold volume
Calculate cubic meters of gold per dram unit. The gold converter.
TOGGLE : from cubic meters into drams in the other way around.
### Enter a New dram Amount of gold to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
## gold from dram to cubic meter Conversion Results :
Amount : dram (dr) of gold
Equals: 0.000000092 cubic meters (m3) in gold
Fractions: 1/10869565 cubic meters (m3) in gold
CONVERT : between other gold measuring units - complete list.
## Solid Pure 24k Gold Amounts
This calculator tool is based on the pure 24K gold, with Density: 19.282 g/cm3 calculated (24 karat gold grade, finest quality raw and solid gold volume; from native gold, the type we invest -in commodity markets, by trading in forex platform and in commodity future trading. Both the troy and the avoirdupois ounce units are listed under the gold metal main menu. I advice learning from a commodity trading school first. Then buy and sell.) Gold can be found listed either in table among noble metals or with precious metals.
Is it possible to manage numerous calculations for how heavy are other gold volumes all on one page? Yes, all in one Au multiunit calculator makes it possible managing just that.
Convert gold measuring units between dram (dr) and cubic meters (m3) of gold but in the other direction from cubic meters into drams.
conversion result for gold: From Symbol Equals Result To Symbol 1 dram dr = 0.000000092 cubic meters m3
# Precious metals: gold conversion
This online gold from dr into m3 (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this gold calculator are ...
With the above mentioned units calculating service it provides, this gold converter proved to be useful also as a teaching tool:
1. in practicing drams and cubic meters ( dr vs. m3 ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with gold's density values including other physical properties this metal has.
International unit symbols for these two gold measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for dram is: dr
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for cubic meter is: m3
### One dram of gold converted to cubic meter equals to 0.000000092 m3
How many cubic meters of gold are in 1 dram? The answer is: The change of 1 dr ( dram ) unit of a gold amount equals = to 0.000000092 m3 ( cubic meter ) as the equivalent measure for the same gold type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of gold can have a crucial/pivotal role in investments. If there is an exact known measure in dr - drams for gold amount, the rule is that the dram number gets converted into m3 - cubic meters or any other unit of gold absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many cubic meters ( m3 ) of gold are contained in a dram ( 1 dr ). Or, how much in cubic meters of gold is in 1 dram? To link to this gold - dram to cubic meters online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: gold from dram (dr) to cubic meters (m3) metal conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
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http://soultionmanual.blogspot.com/2017/02/chapter-27-exercise-1-introduction-to.html
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## Tuesday, 14 February 2017
### Chapter 27 Exercise 1, Introduction to Java Programming, Tenth Edition Y. Daniel LiangY.
27.1 (Implement MyMap using open addressing with linear probing) Create a new concrete class that implements MyMap using open addressing with linear probing. For simplicity, use f(key) = key % size as the hash function, where size is the hash-table size. Initially, the hash-table size is 4 . The table size is doubled
whenever the load factor exceeds the threshold ( 0.5 ).
public class Exercise01 {
public static void main(String[] args) {
MyMap<String, Integer> map = new MyHashMap<String, Integer>();
map.put("Smith", 30);
map.put("Anderson", 31);
map.put("Lewis", 29);
map.put("Cook", 29);
map.put("Smith", 65);
System.out.println("Entries in map: " + map);
System.out.println("The age for " + "Lewis is " + map.get("Lewis"));
System.out.println("Is Smith in the map? " + map.containsKey("Smith"));
System.out.println("Is age 33 in the map? " + map.containsValue(33));
map.remove("Smith");
System.out.println("Entries in map: " + map);
map.clear();
System.out.println("Entries in map: " + map);
}
static class MyHashMap<K, V> implements MyMap<K, V> {
// Define the default hash table size. Must be a power of 2
private static int DEFAULT_INITIAL_CAPACITY = 4;
// Define the maximum hash table size. 1 << 30 is same as 2^30
private static int MAXIMUM_CAPACITY = 1 << 30;
// Current hash table capacity. Capacity is a power of 2
private int capacity;
private static float DEFAULT_MAX_LOAD_FACTOR = 0.5f;
// Specify a load factor used in the hash table
// The number of entries in the map
private int size = 0;
// Hash table is an array with each cell that is a linked list
MyMap.Entry<K, V>[] table;
/** Construct a map with the default capacity and load factor */
public MyHashMap() {
}
/**
* Construct a map with the specified initial capacity and default load
* factor
*/
public MyHashMap(int initialCapacity) {
}
/**
* Construct a map with the specified initial capacity and load factor
*/
@SuppressWarnings("unchecked")
public MyHashMap(int initialCapacity, float loadFactorThreshold) {
if (initialCapacity > MAXIMUM_CAPACITY)
this.capacity = MAXIMUM_CAPACITY;
else
this.capacity = trimToPowerOf2(initialCapacity);
table = new MyMap.Entry[capacity];
}
@Override
/** Remove all of the entries from this map */
public void clear() {
size = 0;
removeEntries();
}
@Override
/** Return true if the specified key is in the map */
public boolean containsKey(K key) {
int i = hash(key.hashCode());
while (table[i] != null) {
if (table[i].getKey().equals(key))
return true;
i = (i + 1) % capacity;
}
return false;
}
@Override
/** Return true if this map contains the value */
public boolean containsValue(V value) {
for (int i = 0; i < capacity; i++) {
if (table[i] != null) {
if (table[i].getValue().equals(value))
return true;
}
}
return false;
}
@Override
/** Return a set of entries in the map */
public java.util.Set<MyMap.Entry<K, V>> entrySet() {
java.util.Set<MyMap.Entry<K, V>> set = new java.util.HashSet<MyMap.Entry<K, V>>();
for (int i = 0; i < capacity; i++) {
if (table[i] != null) {
}
}
return set;
}
@Override
/** Return the value that matches the specified key */
public V get(K key) {
int i = hash(key.hashCode());
while (table[i] != null) {
if (table[i].getKey().equals(key))
return table[i].getValue();
i = (i + 1) % capacity;
}
return null;
}
@Override
/** Return true if this map contains no entries */
public boolean isEmpty() {
return size == 0;
}
@Override
/** Return a set consisting of the keys in this map */
public java.util.Set<K> keySet() {
java.util.Set<K> set = new java.util.HashSet<K>();
for (int i = 0; i < capacity; i++) {
if (table[i] != null) {
}
}
return set;
}
@Override
/** Add an entry (key, value) into the map */
public V put(K key, V value) {
if (get(key) != null) { // The key is already in the map
int i = hash(key.hashCode());
while (table[i] != null) {
if (table[i].getKey().equals(key)) {
V oldValue = table[i].getValue();
table[i].value = value;
return oldValue;
}
i = (i + 1) % capacity;
}
}
if (size >= capacity * loadFactorThreshold) {
if (capacity == MAXIMUM_CAPACITY)
throw new RuntimeException("Exceeding maximum capacity");
rehash();
}
int i = hash(key.hashCode());
while (table[i] != null) {
i = (i + 1) % capacity;
}
table[i] = new MyMap.Entry<K, V>(key, value);
size++;
return value;
}
@Override
/** Remove the entries for the specified key */
public void remove(K key) {
int i = hash(key.hashCode());
while (table[i] != null) {
if (table[i].getKey().equals(key)) {
size--;
table[i] = null;
}
i = (i + 1) % capacity;
}
}
@Override
/** Return the number of entries in this map */
public int size() {
return size;
}
@Override
/** Return a set consisting of the values in this map */
public java.util.Set<V> values() {
java.util.Set<V> set = new java.util.HashSet<V>();
for (int i = 0; i < capacity; i++) {
if (table[i] != null) {
}
}
return set;
}
/** Hash function */
private int hash(int hashCode) {
return supplementalHash(hashCode) & (capacity - 1);
}
/** Ensure the hashing is evenly distributed */
private static int supplementalHash(int h) {
h ^= (h >>> 20) ^ (h >>> 12);
return h ^ (h >>> 7) ^ (h >>> 4);
}
/** Return a power of 2 for initialCapacity */
private int trimToPowerOf2(int initialCapacity) {
int capacity = 1;
while (capacity < initialCapacity) {
capacity <<= 1;
}
return capacity;
}
/** Remove all entries from each bucket */
private void removeEntries() {
for (int i = 0; i < capacity; i++) {
if (table[i] != null) {
table[i] = null;
}
}
}
/** Rehash the map */
@SuppressWarnings("unchecked")
private void rehash() {
java.util.Set<Entry<K, V>> set = entrySet(); // Get entries
capacity <<= 1; // Double capacity
table = new MyMap.Entry[capacity];
size = 0; // Reset size to 0
for (Entry<K, V> entry : set) {
put(entry.getKey(), entry.getValue()); // Store to new table
}
}
@Override
public String toString() {
StringBuilder builder = new StringBuilder("[");
for (int i = 0; i < capacity; i++) {
if (table[i] != null)
builder.append(table[i]);
}
builder.append("]");
return builder.toString();
}
}
interface MyMap<K, V> {
/** Remove all of the entries from this map */
public void clear();
/** Return true if the specified key is in the map */
public boolean containsKey(K key);
/** Return true if this map contains the specified value */
public boolean containsValue(V value);
/** Return a set of entries in the map */
public java.util.Set<Entry<K, V>> entrySet();
/** Return the first value that matches the specified key */
public V get(K key);
/** Return true if this map contains no entries */
public boolean isEmpty();
/** Return a set consisting of the keys in this map */
public java.util.Set<K> keySet();
/** Add an entry (key, value) into the map */
public V put(K key, V value);
/** Remove the entries for the specified key */
public void remove(K key);
/** Return the number of mappings in this map */
public int size();
/** Return a set consisting of the values in this map */
public java.util.Set<V> values();
/** Define inner class for Entry */
public static class Entry<K, V> {
K key;
V value;
public Entry(K key, V value) {
this.key = key;
this.value = value;
}
public K getKey() {
return key;
}
public V getValue() {
return value;
}
@Override
public String toString() {
return "[" + key + ", " + value + "]";
}
}
}
}
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https://skattefrie-pokersider.com/ex-evaluate-common-logarithms-on-the-calculator/
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# Ex: Evaluate Common Logarithms on the Calculator
– WE WANT TO EVALUATE
EACH LOGARITHM ON A CALCULATOR. THEN WRITE
A EXPONENTIAL EQUATION TO SHOW THE MEANING
OF THE VALUE. NOTICE IN BOTH
OF THESE LOGARITHMS THE BASE IS NOT GIVEN THEREFORE WE KNOW IT’S A COMMON
LOGARITHM OR LOG BASE 10. MOST CALCULATORS
ONLY CONTAIN TWO LOG BUTTONS, LOG FOR COMMON LOG AND LN
FOR NATURAL LOG OR LOG BASE E. SO TO EVALUATE THIS FIRST
LOGARITHM WE JUST PRESS LOG, AND BECAUSE THIS COMMON LOG WE
ALREADY KNOW IT’S LOG BASE 10. SO WE JUST TYPE IN 10,000, CLOSE
PARENTHESIS AND PRESS ENTER. THIS IS EQUAL TO 4. SO THIS IS THE FIRST PART
OF THE QUESTION. THE SECOND PART IS WE WANT
TO WRITE AN EXPONENTIAL EQUATION TO EXPLAIN WHY THIS IS EQUAL
TO 4. WELL THE REASON IT’S EQUAL TO 4 IS BECAUSE OUR BASE 10
RAISED TO THE 4th POWER IS EQUAL TO 10,000. SO THIS EMPHASIS’S THAT
WHEN WE EVALUATE A LOGARITHM, WE’RE ACTUALLY FINDING
AN EXPONENT AND IN THIS CASE,
IT’S TELLING US THAT 10 TO THE 4th
IS EQUAL TO 10,000. NOW WE HAVE THE COMMON LOG
OF 1/10, SO WE’LL GO BACK TO THE
CALCULATOR AND PRESS THE LOG KEY WHICH IS THE COMMON LOG
AND THEN 1/10, SO 1 DIVIDED BY 10, CLOSES
PARENTHESIS, AND PRESS ENTER. THIS IS EQUAL TO -1. AND AGAIN, TO SHOW
WHY THIS IS EQUAL TO -1, WE CAN WRITE
AN EXPONENTIAL EQUATION. SO WE’D HAVE THE BASE 10
RAISED TO THE -1 POWER IS EQUAL TO THE NUMBER OF 1/10. AGAIN, THIS EMPHASIS’S WHEN WE DETERMINE THE VALUE
OF A LOGARITHM WE’RE ACTUALLY FINDING
AN EXPONENT. IN THIS CASE, 10 TO THE -1 POWER
IS EQUAL TO 1/10.
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| 3.96875
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latest
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en
| 0.585384
|
https://www.coursehero.com/file/109155/Chapter-38/
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Chapter 38
# Chapter 38 - 38 Diffraction Patterns and Polarization...
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38 CHAPTER OUTLINE 38.1 Introduction to Diffraction Patterns 38.2 Diffraction Patterns from Narrow Slits 38.3 Resolution of Single-Slit and Circular Apertures 38.4 The Diffraction Grating 38.5 Diffraction of X-Rays by Crystals 38.6 Polarization of Light Waves Diffraction Patterns and Polarization ANSWERS TO QUESTIONS Q38.1 Audible sound has wavelengths on the order of meters or centimeters, while visible light has a wavelength on the order of half a micrometer. In this world of breadbox-sized objects, λ a is large for sound, and sound diffracts around behind walls with doorways. But λ a is a tiny fraction for visible light passing ordinary-size objects or apertures, so light changes its direction by only very small angles when it diffracts. Another way of phrasing the answer: We can see by a small angle around a small obstacle or around the edge of a small opening. The side fringes in Figure 38.1 and the Arago spot in the center of Figure 38.3 show this diffraction. We cannot always hear around corners. Out-of-doors, away from reflecting surfaces, have someone a few meters distant face away from you and whisper. The high-frequency, short- wavelength, information-carrying components of the sound do not diffract around his head enough for you to understand his words. Q38.2 The wavelength of light is extremely small in comparison to the dimensions of your hand, so the diffraction of light around an obstacle the size of your hand is totally negligible. However, sound waves have wavelengths that are comparable to the dimensions of the hand or even larger. Therefore, significant diffraction of sound waves occurs around hand-sized obstacles. Q38.3 If you are using an extended light source, the gray area at the edge of the shadow is the penumbra. A bug looking up from there would see the light source partly but not entirely blocked by the book. If you use a point source of light, hold it and the book motionless, and look at very small angles out from the geometrical edge of the shadow, you may see a series of bright and dark bands produced by diffraction of light at the straight edge, as shown in the diagram. FIG. Q38.3 407
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408 Diffraction Patterns and Polarization Q38.4 An AM radio wave has wavelength on the order of 3 10 1 10 300 8 6 1 × × m s s m ~ . This is large compared to the width of the mouth of a tunnel, so the AM radio waves can reflect from the surrounding ground as if the hole were not there. (In the same way, a metal screen forming the dish of a radio telescope can reflect radio waves as if it were solid, and a hole-riddled screen in the door of a microwave oven keeps the microwaves inside.) The wave does not “see” the hole. Very little of the radio wave energy enters the tunnel, and the AM radio signal fades. An FM radio wave has wavelength a hundred times smaller, on the order of a few meters. This is smaller than the size of the tunnel opening, so the wave can readily enter the opening. (On the other hand, the long wavelength of AM radio waves lets them diffract more around obstacles. Long-wavelength waves can change direction more
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This quiz gives a teacher insight on student learning for the standard 3.MD.5 (concept of area). This quiz includes a short answer question which connects a real-word example problem for using area.
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This is a task from the Illustrative Mathematics website that is one part of a complete illustration of the standard to which it is aligned. Each task has at least one solution and some commentary that addresses important asects of the task and its potential use. Here are the first few lines of the commentary for this task: Find the area of each colored figure. Each grid square is 1 inch long....
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For this task, students view four letters and imagine that each square in the picture measures one centimeter on each side. Students use this information to find the area of each letter.
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This parent guide supports parents in helping their child at home with the 3rd grade Math content.
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Students are introduced to the classification of animals and animal interactions. Students also learn why engineers need to know about animals and how they use that knowledge to design technologies that help other animals and/or humans. This lesson is part of a series of six lessons in which students use their growing understanding of various environments and the engineering design process, to design and create their own model biodome ecosystems.
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Learn to identify one square unit that can be used to measure area in this brief interactive tutorial.
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In this math game, students will explore the concept of covering a region with unit squares and score points for each correct response. Once they complete the round they can advance to the next level where they will multiply the length measurements to find the area of a rectangular region.
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This resource can be used as a center activity to enrich students' understanding of area and perimeter. Links to teacher resources and the solution are located on the left side of the page.
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This lesson is for Grade 3 on math. At Home Learning Lessons are a partnership between the North Carolina Department of Public Instruction, PBS North Carolina, and the William and Ida Friday Institute for Educational Innovation. Each lesson contains a video instructional lesson, a PDF lesson plan with a transcript, and a PDF file of extension activities.
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Day 1:
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Use Bee-bots to trace the perimeter of a rectangle. They will then prove their findings with equations that match the path that they coded.
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Use Bee-bots to “drive” the area of a rectangle. They will then prove their findings with equations that match the path that they coded.
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Students bury various pieces of trash in a plotted area of land outside. After two to three months, they uncover the trash to investigate what types of materials biodegrade in soil.
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This unit contains a curriculum-embedded Common Core aligned task and instructional supports. The task is embedded in a 3-4 week unit on solving real-world problems to reinforce the strategies of tiling, multiplication, addition and area models to calculate the perimeter and area of polygons. In addition, students will need to demonstrate and explain their reasoning of strategies used to solve the task.
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This lesson is designed to challenge students to learn about how geometry and measurement are used in the real world by architects. The students will explore various concepts such as area and perimeter as well as developing benchmarks for measurement. This lesson was developed by NCDPI as part of the Academically and/or Intellectually Gifted Instructional Resources Project. This lesson plan has been vetted at the state level for standards alignment, AIG focus, and content accuracy.
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Students select three different rectangles and determine the number of color tiles needed to cover them. Students sketch and label the rectangles in order from least to greatest area and complete sentence stems about the rectangles.
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Find the area of a shape by counting unit squares
Describe the relationship between area and perimeter
Build shapes with a given area and/or perimeter
Find the area of an irregular shape by decomposing into smaller, regular shapes (e.g., rectangles, triangles, squares)
Determine the scale factor of similar shapes
Generalize how area and perimeter change when scaling shapes
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PhET Interactive Simulations
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What is the derivative of (4x^8-sqrt(x))/(8x^4)?
Aug 1, 2015
${y}^{'} = \frac{1}{16} \cdot \left(32 {x}^{\frac{15}{2}} + 7\right) \cdot {x}^{- \frac{9}{2}}$
Explanation:
Start by rewriting your function like this
$y = \frac{4 {x}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}}{8 \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{4}}}}} - {x}^{\frac{1}{2}} / \left(8 {x}^{4}\right)$
$y = \frac{1}{2} {x}^{4} - \frac{1}{8} \cdot {x}^{- \frac{7}{2}}$
Now you can use the power rule to differentiate $y$
$y = \frac{1}{2} \left[\frac{d}{\mathrm{dx}} \left({x}^{4}\right)\right] - \frac{1}{8} \frac{d}{\mathrm{dx}} \left({x}^{- \frac{7}{2}}\right)$
$y = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} {x}^{3} - \frac{1}{8} \cdot \left(- \frac{7}{2}\right) \cdot {x}^{- \frac{9}{2}}$
$y = 2 {x}^{3} + \frac{7}{16} {x}^{- \frac{9}{2}}$
This can be rewritten as
${y}^{'} = \frac{{x}^{- \frac{9}{2}} \cdot \left(16 \cdot 2 {x}^{\frac{15}{2}} + 7\right)}{16} = \textcolor{g r e e n}{\frac{1}{16} \cdot \left(32 {x}^{\frac{15}{2}} + 7\right) \cdot {x}^{- \frac{9}{2}}}$
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https://www.numbersaplenty.com/321103212301
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Search a number
321103212301 = 17177710629389
BaseRepresentation
bin1001010110000110011…
…11100010111100001101
31010200211020100021202201
410223003033202330031
520230104410243201
6403302432121501
732125144161544
oct4530317427415
91120736307681
10321103212301
111141a7435023
1252294996291
1324383c40391
1411781b6115b
1585451e7801
hex4ac33e2f0d
321103212301 has 8 divisors (see below), whose sum is σ = 340182997560. Its totient is φ = 302044689408.
The previous prime is 321103212299. The next prime is 321103212329. The reversal of 321103212301 is 103212301123.
It can be written as a sum of positive squares in 4 ways, for example, as 71377137225 + 249726075076 = 267165^2 + 499726^2 .
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 321103212301 - 21 = 321103212299 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (321103212361) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 5284486 + ... + 5344903.
It is an arithmetic number, because the mean of its divisors is an integer number (42522874695).
Almost surely, 2321103212301 is an apocalyptic number.
It is an amenable number.
321103212301 is a deficient number, since it is larger than the sum of its proper divisors (19079785259).
321103212301 is a wasteful number, since it uses less digits than its factorization.
321103212301 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 10631183.
The product of its (nonzero) digits is 216, while the sum is 19.
Adding to 321103212301 its reverse (103212301123), we get a palindrome (424315513424).
The spelling of 321103212301 in words is "three hundred twenty-one billion, one hundred three million, two hundred twelve thousand, three hundred one".
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UGC NET Follow
August 12, 2020 2:52 pm 30 pts
101. If y 2x7,find the derivative qf y=2x2 orqT fo (1) x (2 x2a (3) x-2 (4)x 102.If y = x + 3x, find the derivative af y x2+ 3x Bt at yqbTTT 51TT lg: (1 2x+ 12x (2) 3x4x 13) 4x 12x Y2x + 12x3 103.If y = (3x2 +1) (x* + 2x), find the derivative: af y (3x+ 1) (x3 +2x), T srqeTA STT fg (1) 21x+ 15x+2 2 15x+21x +2 (3) 3x + 15x + 2 (4) 6x+21x+ 2 104.If demand function is P 100- 4x and supply function is P = 30 +X, then find the equilibrium price yf HTT P- 100-4x ATT ei 5TT P - 30+x, igTT (1 44 (2) 16 3 14 (4) 30
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101- option C and 102- option- D
• Ashish sharma
103- option B
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# BTU to ccf
Member Posts: 1
I open up the newspaper to find natural gas is at 3.74 mm btu. My utility bill charges me in ccf's at \$ 0.87. My boiler is 100,000 btu's in and 85,000 out and it runs for 15 minutes.
#1 How do you convert from mm btu to ccf ?
#2 Will 85,000 btu's really be transferred into the radiators and piping?
#3 Can I calculate/measure the btu's my boiler produced and confirm it is being transferred using the pipe size ( volume of water ) and temperature rise?
please help, this has caused a number of HEATED arguments. Thank you.
## Comments
• Member Posts: 2,666
This won't help much.
Lat month I used 4 ccf (400 cubic feet) of gas. It averaged 1.04 BTU therms per ccf, so they billed me for 4.17 therms. For this they charged me \$3.35 for the gas and \$2.09 to deliver it, and added a customer charge of \$8.25, so the total cost was \$13.69. Clear? The number of therms per ccf changes every month. The customer charge is constant, so when I start heating the house instead of just my domestic hot water, the customer charge will not be such a large proportion of the total.
• Member Posts: 3,086
Simple
There is 103,00 btu's in a cubic foot of gas. That 3.74mm represents cost of 3.74 million btu's.
100,000 btu boiler just about burns a cubic foot on every fire. Now if it's a 80% AFUE boiler that means your sending 20% of that .87 cents or .17 cents into never ever land.
I doubt your using the entire load. You never need that entire load until you reach the coldest day in your region. You will never know the anwser without doing a heat loss of the home. Sad to say but that .17 cents is probably in reality more like .25 cents or even more.
There was an error rendering this rich post.
• Member Posts: 4,630
The answer
is that there are approximately 950 to 1075 BTU's per cubic foot. The equipment is rated in BTU's the gas meter measures in cubic feet.
On your bill it will state something like this so many CCf at say 1027, the 1027 is the actual BTU content average for the month that the gas company sent out.
The actual transfer of BTU's into the system is called thermal efficiency. If a proper heat loss was done per room and the piping to the rooms is sized correctly along with sufficient radiation for each room then the applied BTU will be pretty close.
Actual measurement of real BTU's would require a BTU flow meter to be accurate.
• Member Posts: 2,398
edited October 2010
Or another way to look at it
Gas is sold typically in Therms. A therm is 100,000 BTUs, no matter what. But the therms are delivered in volumes of gas, measured in cubic feet over varying compositions and heat values.
As Tim pointed out, the therm content per CF does vary. Most natural gas is methane, about 95%, the remainder can be butane, ethane heptane, propane and a host of other minor players, plus inert gasses with no appreciable fuel value. Then they toss in ethyl mercaptan as the odorant we all know and love.
The gas company buys gas on the open market and depending on the market condition, cuts or enhances the natural gas with additives. Butanes are the largest category, but if in short supply and no other additives, or whatever it is, the fuel content in therms is adjusted on your bill. Usually this is fractional, within ten percent.
The gas METER however, measures volume in cubic feet. If the presumed gas heat value is 1,000 BTUs, a therm is 100 cubic feet or CCF. So for discussion purposes, a therm is the same as a CCF.
So in answer to your questions in order:
1. To convert from MMBTU to CCF, divide MMBTU by 10. (10 CCF of 1,000 BTU/CF gas is a million BTUs. Put another way, 1 CF =1,000, 100 CF (or 1 CCF) = 100,000 and 10 CCF = 1,000,000 BTUs. That is input.
2. Will 85% be transferred to radiators and piping? (And your space we hope!). This can vary over time. A boiler that cycles more in warmer weather, (say the 30's), will lose more to jacket losses and start-up losses than the same boiler at zero degrees. The colder weather in this case will be more efficient. Think of the efficiency as a sliding curve, descending as it gets warmer outside from your cold design temperature. But for discussion purposes, 80-85% is doing pretty well for most non-condensing boilers. But it really does vary.
3. Can you calculate this? As Tim said, you really need a BTU meter. This means real-time and continuous measurement and logging of flow rate, temperature of the supply water out and temperature returning, all over time, usually in hours but also over a day divided by 24. This is compared to the fuel rate input.
To be proper about it, the process has to be continuous. If you fire up from a cold start, your delta-T will be wide and your space gains (what you want to do!) will be some minutes, maybe half an hour or more, away.
This is not a good time to pat yourself on the back!
You need to take into account the up-cycles, down-cycles and in-between coasting. Remember too, between firing cycles, your pump still runs and clocks delta-T over time with no fuel input. Another time not to pat yourself on the back!
My \$0.02
Brad
"If you do not know the answer, say, "I do not know the answer", and you will be correct!"
-Ernie White, my Dad
This discussion has been closed.
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# Pecks (US dry) to fluid ounces (UK) (pk to fl oz)
## Convert pecks (US dry) to fluid ounces (UK)
Pecks (US dry) to fluid ounces (UK) converter above calculates how many fluid ounces (UK) are in 'X' pecks (US dry) (where 'X' is the number of pecks (US dry) to convert to fluid ounces (UK)). In order to convert a value from pecks (US dry) to fluid ounces (UK) (from pk to fl oz) just type the number of pk to be converted to fl oz and then click on the 'convert' button.
## Pecks (US dry) to fluid ounces (UK) conversion factor
1 peck (US dry) is equal to 310.0604710147 fluid ounces (UK)
## Pecks (US dry) to fluid ounces (UK) conversion formula
Volume(fl oz) = Volume (pk) × 310.0604710147
Example: Assume there are 466 pecks (US dry). Shown below are the steps to express them in fluid ounces (UK).
Volume(fl oz) = 466 ( pk ) × 310.0604710147 ( fl oz / pk )
Volume(fl oz) = 144488.17949285 fl oz or
466 pk = 144488.17949285 fl oz
466 pecks (US dry) equals 144488.17949285 fluid ounces (UK)
## Pecks (US dry) to fluid ounces (UK) conversion table
pecks (US dry) (pk)fluid ounces (UK) (fl oz)
154650.9070652204
257751.5117753674
3510852.116485514
4513952.721195661
5517053.325905808
6520153.930615955
7523254.535326102
8526355.140036249
9529455.744746396
10532556.349456543
11535656.95416669
12538757.558876837
13541858.163586984
14544958.768297131
15548059.373007278
pecks (US dry) (pk)fluid ounces (UK) (fl oz)
20062012.094202939
30093018.141304409
400124024.18840588
500155030.23550735
600186036.28260882
700217042.32971029
800248048.37681176
900279054.42391323
1000310060.4710147
1100341066.51811617
1200372072.56521764
1300403078.6123191
1400434084.65942057
1500465090.70652204
1600496096.75362351
Versions of the pecks (US dry) to fluid ounces (UK) conversion table. To create a pecks (US dry) to fluid ounces (UK) conversion table for different values, click on the "Create a customized volume conversion table" button.
## Related volume conversions
Back to pecks (US dry) to fluid ounces (UK) conversion
TableFormulaFactorConverterTop
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Simplify the expression: 3 square root of 5/64
3 square root of 5/64 = 3 sqrt (5/64) = 3 * (1/8) sqrt 5 = (3/8) sqrt 5
Question
Updated 1/25/2015 6:32:41 PM
Rating
3
3 square root of 5/64
= 3 sqrt (5/64)
= 3 * (1/8) sqrt 5
= (3/8) sqrt 5
Confirmed by jeifunk [1/25/2015 10:10:36 PM]
Questions asked by the same visitor
Simplify the expression.sqrt of 6 (5 -sqrt of 6 ) + 4 sqrt of 6
Question
Updated 1/25/2015 6:31:26 PM
sqrt 6 (5 -sqrt 6 ) + 4 sqrt 6
= 5 sqrt 6 - 6 + 4 sqrt 6
= (9 sqrt 6) - 6
Confirmed by jeifunk [1/25/2015 10:11:06 PM]
Determine the equation of g(x) that results from translating the function f(x) = x2 + 7 upward 9 units.
Weegy: The equation of g(x) that results from translating the function f(x) = x^2 + 1 upward 2 units is g(x) = x^2 + 3. (More)
Question
Updated 1/25/2015 8:40:14 PM
The equation of g(x) that results from translating the function f(x) = x^2 + 7 upward 9 units is g(x) = x^2 + 16.
Confirmed by jeifunk [1/25/2015 9:57:58 PM]
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#### Polynomial trendline order in Excel 2007
```Any particular reason why the maximum order for a polynomial trendline in
Excel 2007 is 6? This seems rather limiting. Any way of using a higher order
polynomial?
Arnaud
```
0
Arnaud
11/4/2009 2:49:10 PM
excel.charting 18370 articles. 0 followers.
2 Replies
1896 Views
Similar Articles
[PageSpeed] 19
```FWIW, most poly fits of order six that you see in Excel have been
woefully overfitted. Most poly fits should not have an order higher than
two.
- Jon
-------
Jon Peltier
Peltier Technical Services, Inc.
http://peltiertech.com/
Arnaud Miege wrote:
> Any particular reason why the maximum order for a polynomial trendline in
> Excel 2007 is 6? This seems rather limiting. Any way of using a higher order
> polynomial?
>
>
> Arnaud
>
>
```
0
Jon
11/4/2009 3:52:43 PM
```I concur with Jon (who is not know for lightly forgiving Microsoft for any
of its 'trespasses')
In the physical sciences, a fit with more than 4 terms is very rare
One has to ask: what is the meaning of these terms?
Mathematically, any N order pairs of data can be fitted to a N-1 order
polynomial (2 points -> straight line, 3 points -> a parabola, etc) but just
because we can make a fit it does not mean we necessarily know more about
the data.
best wishes
--
Bernard V Liengme
Microsoft Excel MVP
http://people.stfx.ca/bliengme
remove caps from email
"Arnaud Miege" <arnaud.miege@nospam.mathworks.co.uk> wrote in message
news:uzml53VXKHA.4588@TK2MSFTNGP04.phx.gbl...
> Any particular reason why the maximum order for a polynomial trendline in
> Excel 2007 is 6? This seems rather limiting. Any way of using a higher
> order polynomial?
>
>
> Arnaud
>
```
0
Bernard
11/4/2009 10:59:49 PM
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I know that it is possible to rotate content within a cell, but what if I want to rotate the whole worksheet? And yes, I know that I could just set it to landscape mode, but I want to modify it once the rows become columns and the columns become rows. You are limited to 256 columns (and therefore 256 rows). create a new worksheet Select your range on the original worksheet edit|copy go to A1 of the new worksheet edit|paste special|check transpose. MarkRulesTheWorld wrote: > > I know that it is possible to rotate content within a cell, but what if I > want to rotate the whole w...
Is it possible to merge workbooks in excel?
I am currently working on a project with a coworker in which we need to enter quality scores for employees into excel workbooks, with one sheet per group of employees. Is it possible, if we enter data for separate groups, to later merge this data into one master workbook? Thanks! Are you asking if you can move/copy your sheet for Group AAA into another workbook that has a sheet for Group BBB? If so, all you need to do is: 1) open both the source and destination workbooks 2) activate the source workbook 3) select any and all tabs that you want in the destination workbook (Ctrl-click on ...
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# What constitutes a "Sol" cycle in the film "The Martian"
In the film (and book) The Martian, NASA has termed a cycle of time as a "Sol", and it is what Watney and the crew use to refer to what seems like it might be a Martian day.
1. Is a Sol equal to one Martian day?
2. If so, is it explained why do they call it a Sol instead of a Martian Day, or maybe Martian Solar Cycle? (Could it be a cool-sounding abbreviation?)
• VTC as off topic, as this is straight up a question of science fact: en.wikipedia.org/wiki/Timekeeping_on_Mars#Sols Oct 8, 2015 at 4:59
• Copy the question and the answer to the appropriate stack, please. It applies to both the fictional world and the real world, but it still completely applies to the fictional world of The Martian. Oct 8, 2015 at 6:53
• the question is about a term used in a fictional work. it's on-topic on both stacks as OP was apparently not aware that it was a real-life term. Oct 8, 2015 at 10:50
• @phantom42 No more no-topic than questions about how thrust vector calculations can be used to maneuver space vehicles... (off topic here, on topic on physics SE). Oct 9, 2015 at 1:37
While the book uses the term, it does not explain the definition or exact length.
Real-life NASA, however, does.
One sol = one solar day.
Mars Solar Days and 24-hr Clock Convention
Following the long-standing practice originally adopted in 1976 by the Viking Lander missions, the daily variation of Mars solar time is reckoned in terms of a "24-hour" clock, representing a 24-part division of the planet's solar day, along with the traditional sexagesimal subdivisions of 60 minutes and 60 seconds. A Mars solar day has a mean period of 24 hours 39 minutes 35.244 seconds, and is customarily referred to as a "sol" in order to distinguish this from the roughly 3% shorter solar day on Earth. The Mars sidereal day, as measured with respect to the fixed stars, is 24h 37m 22.663s, as compared with 23h 56m 04.0905s for Earth.
• So a Sol is specifically for Mars and not for any other planet? It is not a general term for a solar day on a random planet? Oct 8, 2015 at 14:04
• @VincentAdvocaat You have it, at least in theory. It is the term for a solar day for a planet, and the context should make which planet clear. As a practical matter, it would only apply to Mars. Mercury is tidally-locked to the Sun, and therefore has no solar day. Venus rotates so slowly that it has just less than two sols per Venusian year. Jupiter, Saturn, Uranus, and Neptune are gas giants, and don't have definitely measurable solar days. Pluto isn't a planet anymore, but it does have a solar day. Would that be shortened to "sol" if/when we start operating there? We'll have to see. Oct 8, 2015 at 14:53
• @MontyHarder Mercury is not tidally-locked, that's known since 1965. en.wikipedia.org/wiki/Mercury_%28planet%29 Oct 8, 2015 at 17:10
• @edc65 I stand corrected. Mercury goes in the same category as Venus, only worse: With a 3:2 resonance, a solar day is two (Mercury) years, so it's hardly a subsidiary unit in the way days and sols are to Earth's and Mars' respective years. Oct 8, 2015 at 18:05
• About "sol" being used for other planets: I think it just hasn't come up yet, so "nobody knows if we will use it elsewhere". As a practical matter, Mars is the only body (so far) where we have landed anything, where we have had continuous extended operations on the surface, and where the day/night cycle is highly relevant for operational reasons (illumination, temperature, power, etc.), which leads to the need for a short, unambiguous name for the local solar cycle. I personally favor leaving it a Mars term, but I'm guessing it will be adapted eventually to other bodies, as needed. Oct 10, 2015 at 3:32
A sol is one solar day on Mars, which is a little longer than the 24 hour solar day on Earth. For some interesting perspectives on what it's like to work on the Sol schedule, check out this recent blog post: https://sketchtogether.com/blog/what-the-martian-didnt-show-you-how-scientists-communicate-with-mars.html
Had the term been something longer like "Martian day" it would inevitably get shortened or abbreviated at least sometimes, likely to "day" which could cause confusion.
To answer the question "why not Martian Day", the answer is simply for ease of use / to reduce confusion.
"Sol" was coined for the use of real-life teams operating probes and rovers on Mars. As in The Martian, most stuff is solar-powered, so operational schedules depend on Martian sunrise and sunset. They probably would anyway, for reasons of visibility and temperature. If the term was "Martian day" then people would be tempted to abbreviate it to "day", and then there would be perpetual confusion what kind of "day" anyone is talking about, which could lead to costly mistakes. A word that's distinct from "day" and easy to use in conversation solves that problem. And "sol" is as good as any other.
A Mars solar day has a mean period of 24 hours 39 minutes 35.244 seconds, and is customarily referred to as a "sol" in order to distinguish this from the roughly 3% shorter solar day on Earth.
http://www.giss.nasa.gov/tools/mars24/help/notes.html
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# how to simplify expressions
need help on figuring out expressions in math
simplify then exponets:
x/7 - y/z (those are fractions)
answered Dec 4, 2011 by anonymous
How do you simplify the expression: 2p x 3pq?
answered Feb 5, 2012 by anonymous
simplify the expression (2 root 27x(3 root 32).
answered Jul 23, 2012 by anonymous
how to write this expresssion with out getting a syntax error. This ishow I wrote the answer but get syntax error. x+7/18(x-5). (x-7)/(18*9x-5)
answered Feb 15, 2013 by anonymous
X^2-49/3x-21
answered May 9, 2013 by anonymous
x^8/2y*5y^2/x^3
i need to simplify square root inside the square root there is a 45 and a p square it has a 2 i need help please
answered May 27, 2013 by anonymous
This program will not let me type in exponents.
How do you simplify this math expression? -4x5y3 2xy2
answered Jun 25, 2013 by anonymous
-3(3m+6)-2(2m-2)-9m
answered Sep 2, 2013 by Nakisha Baptiste
find the value of the annuity A^1=\$2,000, I=0.07,n=10 round to the nearest cent.
answered Oct 10, 2013 by Jessica
root6(7-root6)+6root6
answered Oct 17, 2013 by anonymous
(c3)2(-3c5)2
answered Oct 28, 2013 by colton
simplify the expression.
-4(x-7)+3(x+9)
answered Oct 31, 2013 by Natalie
8+7(2-6)2
answered Nov 2, 2013 by anonymous
(5rx-4)(5rx-4)
answered Nov 2, 2013 by anonymous
Simplify the expression k − 2(k − (2 − k)) − 2 by writing it without using parentheses.
answered Nov 6, 2013 by Sheryl
11c²-11d²/c²-2cd+d²
answered Nov 20, 2013 by anonymous
Simplify the expression
-2(3-5+2)
answered Dec 14, 2013 by anonymous
7g+g2+6g2 =
implify the expression by combining like terms, if possible. If not possible, select “already simplified.”
A. B. already simplified C. D.
5(x-7)+4x
How do I simplify x2+x(x-5)-x+7
(2x^2-7xy+8y^2)(4x^2+8xy-4y^20
8g/12(g+h)
(3a2+ 9a) + (6a2 + 2a)
what is the answer to a simplify expression
3x6-7
(5a2) (4a5)
5+(x+7)
3/5 to 9
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# SOLVED CS501 Assignment 2 Solution and Discussion
• Assignment No. 02
Semester: Fall 2019
Advance Computer Architecture – CS501
Total Marks: 20
Due Date: 28-11-2019
Please carefully read the following instructions before attempting assignment
OBJECTIVE OF ASSIGNMENT
Objective of this assignment is to increase the learning capabilities of the students about:
• Encoding of Assembly Instructions
• Behavioral RTL description of instruction
• ISA of Modified EAGLE
• Address Bus and Data Bus in Modified EAGLE
RULES FOR ASSIGNMENT
It should be clear that your assignment will get credit only if:
• The assignment is submitted before or on the due date.
• The submitted assignment file is not corrupted or damaged.
• The assignment is not copied (from another student or internet).
• Upload/Submit assignment in your VULMS assignment interface.
• No assignment will be accepted through email.
• Your assignment must be with .doc extension. (Any other format will not be accepted)
NOTE
No assignment will be accepted after the due date via email in any case (whether it is the case of load shedding or internet malfunctioning etc.). Hence, refrain from uploading assignment in the last hour of deadline. It is recommended to upload solution file at least two days before its closing date.
If you find any mistake or confusion in assignment (Question statement), please consult with your instructor before the deadline. After the deadline, no queries will be entertained in this regard. Moreover, keep checking announcements section.
For any query, feel free to email at:
[email protected]
Best of Luck
Question # 01 10 Marks
Consider the below given table consisting of assembly instructions belonging to different processors.
Instruction Processor Hexadecimal Code Behavioral RTL
JPL R5, [26] Falcon - A
STS R7, R2 (100) Falcon - E
STACC R4, 36 Modified EAGLE
DIV R2 EAGLE
SHIFTL R5, R2, 7 FALCON - A
Write machine code (in the hexadecimal representation) and behavioral RTL description for each instruction given in table.
Question # 02 10 Marks
Consider the below given tables belonging to Modified EAGLE architecture.
• Table 1 shows the contents of registers R1, R2 and memory labels A, B, C. The contents of memory labels are memory addresses.
Register Contents Memory Label Address
R1 0015h A AB10h
R2 25CBh B 3320h
C AB0Eh
Table 1: Contents in Registers and Memory labels
• Table 2 represents byte- aligned memory map and shows the value stored at each memory address.
Memory Address Memory Contents Memory Address Memory Contents
AB0Eh 15h 3320h CEh
AB0Fh 20h 3321h 55h
AB10h 56h 3322h 39h
AB11h EFh 3323h 20h
Table 2: Contents at different memory addresses
• Table 3 contains the instructions of an assembly program for Modified EAGLE. You are required to complete Table 3 by writing the values of destination operand, 16-bit address bus and 16-bit data bus after each instruction is executed.
Write the complete steps for calculating the values of Destination Operand, Data Bus & Address Bus after the execution of each instruction.
Instruction Contents stored in Destination Operand Data Bus Address Bus Calculation Steps
LDACC B
SUB R1
LDACC C
STACC A
Table 3: Data Bus and Address Bus Contents for Modified Eagle
• Solution # 1
Instruction Processor Hexadecimal Code Behavioral RTL
JPL R5, [26] Falcon - A 10000 101 00011010, 1000 0101 0001 1010, 851A (R[5] ≥ 0): PC ← PC+ (26-PC);
STS R7, R2 (100) Falcon - E 00101 111 010, 000000000000001100100, 0010 1111 0100 0000, 0000 0000 0110 0100, 2F40 0064 M[R[2]+100] ← R [7]
STACC R4, 36 Modified EAGLE 10111 100 00100100, 1011 1100 0010 0100, BC24 M[R[4] + ([email protected]<7>)©C] ← ACC C represents the constant 36
DIV R2 EAGLE 10000 010, 1000 0010, 82 R[0] ← R[0]/R[2] R[2] ← R[0]%R[2]
SHIFTL R5, R2, 7 FALCON - A 01100 101 010 00111, 0110 0101 0100 0111, 6547 R[5]<15…0> ← R [2]<(15- N)…0>©([email protected]) N represents constant 7
Solution # 2
Instruction Value of Destination Operand Data Bus <15…0> Address Bus (15…0>
LDACC B 55CEh 55CEh 3320h
SUB R1 55B9h N/A N/A
LDACC C 2015h 2015h AB0Eh
ADD R2 45E0h N/A N/A
STACC A 45E0h 45E0h AB10h
Table 3 Data Bus and Address Bus Contents for Modified Eagle
Calculation Steps (Instruction-By-Instruction)
• LDACC B
LDACC stands for load accumulator. In LDACC, the destination operand is accumulator and source operand is the memory location labelled as B. The memory label B points to the memory address 3320h. When this instruction is executed, the value stored at memory address 3320h will be read and loaded in Accumulator register. This address 3320h will be copied into Address Bus which will then read its contents from memory and load the contents at data bus. The operand size in Modified EAGLE is 2-byte. Therefore, the values stored at addresses 3320h and 3321h will be loaded at data bus. These values are CEh and 55h respectively. Since, Modified EAGLE employs Little endian notation hence, the 2-bye value will be read as 55CEh. The value of data bus will also be 55CEh and this will be loaded into destination operand Accumulator.
• SUB R1
SUB R1 means to subtract the value of source operand register R1 which is 0015h, from the destination operand Accumulator (ACC) register which contains 55CEh. The result of subtraction will be stored back into ACC. After subtraction, the value stored in ACC will be 55B9h. Because SUB is not a memory instruction, we are not concerned with the contents of Data Bus or Address Bus because the values we need to execute the instruction are already available in registers. Hence, Data Bus and Address Bus values will be labelled as N/A.
• LDACC C
When this instruction is executed, the value stored at memory address labelled with C is read and loaded in Accumulator register ACC. In this case, the address of the C is AB0Eh which is also the value of address bus. The operands in Modified EAGLE are 2-byte values. The contents at addresses AB0Eh and AB0Fh will be copied into data bus which will then be loaded into ACC register. These contents are 15h and 20h. Due to Little endian notation, the 2-bye value will be 2015h. The value of data bus will be 2015h and same will be loaded in destination register ACC.
When ADD is executed, the value of register R2 is added to ACC register. Hence, after the execution, the ACC register will hold 45E0h. AS usual, ADD is not a memory instruction, so we are not concerned with the values of Data Bus and Address Bus and both are labelled as N/A.
• STACC A
STACC stands for Store ACC. There is one destination operand which is a memory label A. When the instruction is executed, the value of Accumulator register ACC is stored at the memory address labelled by A. The destination memory address will be AB10h. The value of ACC is 45E0h will be stored as address AB10h. However, due to Little-Endian notation, the address will be stored as E0h at memory location AB10h and then 45h at memory location AB11h.
• Solution # 1
Instruction Processor Hexadecimal Code Behavioral RTL
JPL R5, [26] Falcon - A 10000 101 00011010, 1000 0101 0001 1010, 851A (R[5] ≥ 0): PC ← PC+ (26-PC);
STS R7, R2 (100) Falcon - E 00101 111 010, 000000000000001100100, 0010 1111 0100 0000, 0000 0000 0110 0100, 2F40 0064 M[R[2]+100] ← R [7]
STACC R4, 36 Modified EAGLE 10111 100 00100100, 1011 1100 0010 0100, BC24 M[R[4] + ([email protected]<7>)©C] ← ACC C represents the constant 36
DIV R2 EAGLE 10000 010, 1000 0010, 82 R[0] ← R[0]/R[2] R[2] ← R[0]%R[2]
SHIFTL R5, R2, 7 FALCON - A 01100 101 010 00111, 0110 0101 0100 0111, 6547 R[5]<15…0> ← R [2]<(15- N)…0>©([email protected]) N represents constant 7
Solution # 2
Instruction Value of Destination Operand Data Bus <15…0> Address Bus (15…0>
LDACC B 55CEh 55CEh 3320h
SUB R1 55B9h N/A N/A
LDACC C 2015h 2015h AB0Eh
ADD R2 45E0h N/A N/A
STACC A 45E0h 45E0h AB10h
Table 3 Data Bus and Address Bus Contents for Modified Eagle
Calculation Steps (Instruction-By-Instruction)
• LDACC B
LDACC stands for load accumulator. In LDACC, the destination operand is accumulator and source operand is the memory location labelled as B. The memory label B points to the memory address 3320h. When this instruction is executed, the value stored at memory address 3320h will be read and loaded in Accumulator register. This address 3320h will be copied into Address Bus which will then read its contents from memory and load the contents at data bus. The operand size in Modified EAGLE is 2-byte. Therefore, the values stored at addresses 3320h and 3321h will be loaded at data bus. These values are CEh and 55h respectively. Since, Modified EAGLE employs Little endian notation hence, the 2-bye value will be read as 55CEh. The value of data bus will also be 55CEh and this will be loaded into destination operand Accumulator.
• SUB R1
SUB R1 means to subtract the value of source operand register R1 which is 0015h, from the destination operand Accumulator (ACC) register which contains 55CEh. The result of subtraction will be stored back into ACC. After subtraction, the value stored in ACC will be 55B9h. Because SUB is not a memory instruction, we are not concerned with the contents of Data Bus or Address Bus because the values we need to execute the instruction are already available in registers. Hence, Data Bus and Address Bus values will be labelled as N/A.
• LDACC C
When this instruction is executed, the value stored at memory address labelled with C is read and loaded in Accumulator register ACC. In this case, the address of the C is AB0Eh which is also the value of address bus. The operands in Modified EAGLE are 2-byte values. The contents at addresses AB0Eh and AB0Fh will be copied into data bus which will then be loaded into ACC register. These contents are 15h and 20h. Due to Little endian notation, the 2-bye value will be 2015h. The value of data bus will be 2015h and same will be loaded in destination register ACC.
When ADD is executed, the value of register R2 is added to ACC register. Hence, after the execution, the ACC register will hold 45E0h. AS usual, ADD is not a memory instruction, so we are not concerned with the values of Data Bus and Address Bus and both are labelled as N/A.
• STACC A
STACC stands for Store ACC. There is one destination operand which is a memory label A. When the instruction is executed, the value of Accumulator register ACC is stored at the memory address labelled by A. The destination memory address will be AB10h. The value of ACC is 45E0h will be stored as address AB10h. However, due to Little-Endian notation, the address will be stored as E0h at memory location AB10h and then 45h at memory location AB11h.
• @zareen This is not a right assignment solution
this is not a solution just idea how to solve the question.
cs501 assignment 2 idea solution…docx
• @zareen This is not a right assignment solution
this is not a solution just idea how to solve the question.
• @zareen This is not a right assignment solution
• jump instruction
• Control Instruction
• We have looked at the various jump instructions in our study of the FALCON-A. Using that knowledge, this problem can be solved easily.
• Solution
Addressing modes relate to the way architectures specify the address of the objects they access. These objects may be constants and registers, in addition to memory locations.
• Fields in the FALCON-A InstructionsWe now use the RTL naming operator to name the various fields of the RTL instructions.Naming the fields appropriately helps us make the study of the behavior of a processormore readable.op<4…0>:= IR<15…11>:operation code fieldra<2…0> := IR<10…8>:target register fieldrb<2…0> := IR<7…5>:operand or address indexrc<2…0> := IR<4…2>:second operandc1<4…0> := IR<4…0>:short displacement fieldc2<7…0> := IR<7…0>:long displacement or the immediate field
• A better method is to use the addi instruction with the constant set to 0.
• This instruction is to load a register from an input/output device. The effective address of the I/O device has to be calculated before it is accessed to read the word into the destination register ra, as shown in the example:in R5, R4(100)In RTL:R[5]←IO[R[4]+100]
• We can also find the machine code for these instructions. The machine code (in the hexadecimal representation) is given for these instructions in the given table.
2
1
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3
3
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LeetCode in Swift: Binary Tree Level Order Traversal
Problem Statement
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
return its level order traversal as:
My Solution in Swift
This is how TreeNode class looks like:
I used tail recursion technique to make the recursive approach faster. The following is my recursive version of Breadth-first traversal (aka level order traversal) of a binary search tree:
The iterative version of Breadth-first traversal (aka level order traversal) of a binary search tree:
Even using -O compilation flag, the recursive version is still slower than the iterative counterpart with 33 ms vs 6 ms. This doesn’t follow behaviours as depicted in Depth-first traversals (NLR, LNR, LRN). The main reason is that the recursive version has to pass intermediate results deep down along with those recursive function calls. Besides, the overhead of function calls is not negligible.
Try It Yourself
1: each links to a blog post of mine that is dedicated to the problem
2: total execution time of a solution on my MacBook Pro (Late 2013, 2.6 GHz Intel Core i7, 16 GB 1600 MHz DDR3). Each solution is compiled with following command:
`\$ swiftc -O -sdk `xcrun --show-sdk-path --sdk macosx` json.swift main.swift -o mySolution`
The total execution time is the average of 10 runs.
3: these test cases are semi-automatically :P retrieved from LeetCode Online Judge system and are kept in JSON format
4: each Xcode project includes everything (my Swift solution to a problem, its JSON test cases and a driver code to test the solution on those test cases)
Problem1 Time2 Test Cases3 My Xcode Project4 Binary Tree Level Order Traversal 5.631ms
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http://reference.wolfram.com/legacy/v7/ref/GeodesyData.html
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This is documentation for Mathematica 7, which was
based on an earlier version of the Wolfram Language.
# GeodesyData
GeodesyData["name", "property"] gives the value of the specified property for a named geodetic datum or reference ellipsoid. GeodesyData[{a, b}, "property"]gives the value of the property for the ellipsoid with semimajor axis a and semiminor axis b. GeodesyData[obj, {"property", coords}]gives the value of the property at the specified coordinates.
• gives a list of all available named geodetic datums and reference ellipsoids.
• Geodetic datums are specified by standard names such as "NAD27" and "ITRF00".
• Reference ellipsoids are given standard names such as "Clarke1866" and "GRS80".
• GeodesyData["Datum"] gives the names of all available named geodetic datums; GeodesyData["ReferenceEllipsoid"] all available named reference ellipsoids.
• GeodesyData["datum", "ReferenceEllipsoid"] gives the name of the reference ellipsoid associated with the specified datum.
• Basic geometrical properties include:
"EllipsoidParameters" ellipsoid parameters "InverseFlattening" inverse flattening of the ellipsoid "SemimajorAxis" length of the semimajor axis (equatorial radius) "SemiminorAxis" length of the semiminor axis (or polar radius)
"AuthalicRadius" radius of a sphere of the same surface "Eccentricity" first eccentricity of the ellipsoid "Flattening" flattening of the ellipsoid "MeanMassRadius" mean mass radius "MeanRadius" geometric mean radius "MeridianQuadrant" length of the meridian from the equator to the pole "SecondEccentricity" second eccentricity of the ellipsoid "VolumetricRadius" radius of a sphere of equal volume
• Coordinate-dependent properties include:
{"MeridionalArc",lat1,lat2} length of the meridian between latitudes lat1 and lat2 {"MeridionalCurvatureRadius",lat} radius of curvature in the meridian at latitude lat {"PrimeVerticalCurvatureRadius",lat} radius of curvature in the prime vertical at latitude lat {"NormalSectionCurvatureRadius",lat,a} radius of curvature at latitude lat in the direction azimuth a
• Properties converting from geodetic latitude to alternative forms of latitude include:
{"ReducedLatitude",lat} parametric latitude of an equivalent point on a sphere {"GeocentricLatitude",lat} angle between the equatorial plane and a line from the center {"AuthalicLatitude",lat} latitude of an equivalent point on the authalic sphere {"ConformalLatitude",lat} conformal projection of a geodesic latitude lat {"IsometricLatitude",lat} isometric latitude referred to lat {"RectifyingLatitude",lat} projection preserving distance between meridians
• Properties converting from alternative forms of latitude to geodetic latitude include:
{"FromReducedLatitude",lat} parametric latitude of an equivalent point on a sphere {"FromGeocentricLatitude",lat} angle between the equatorial plane and a line from the center {"FromAuthalicLatitude",lat} latitude of an equivalent point on the authalic sphere {"FromConformalLatitude",lat} conformal projection of a geodesic latitude lat {"FromIsometricLatitude",lat} isometric latitude referred to lat {"FromRectifyingLatitude",lat} projection preserving distance between meridians
• Other properties include:
"AlternateNames" alternate English names "StandardName" Mathematica standard name "Name" English name "Properties" available properties
• Reference ellipsoids can be specified by semiaxes {a, b} or by semimajor axis and inverse flattening {a, {invf}}.
• GeodesyData gives symbolic results if the parameters of a reference ellipsoid are given symbolically.
• GeodesyData[{datum1, datum2}] gives rules for the parameters used to transform datum1 to datum2 .
• GeodesyData[{datum1, datum2}, "param"] gives the specified parameter for transforming from datum1 to datum2.
• Transformation parameters for datums include:
"ParameterDefinitionYear" decimal year when parameters values were defined "Rotation" rotation angles in milliarc seconds "RotationDerivative" rate of change of rotation in milliarc seconds per year "Scale" transformation scale factor "ScaleDerivative" annual change of transformation scale factor "Translation" translation vector given in meters "TranslationDerivative" rate of change of translation vector in meters per year
Semiaxes of Clarke1866:
Semimajor axis and inverse flattening of GRS80:
Eccentricity of the GRS80 reference ellipsoid:
Full name of the ITRF00 datum:
Alternate names for ITRF00:
Names of all properties of the ITRF00 datum:
This gives the 14 parameters required to transform from ITRF00 to NAD83CORS96:
Semiaxes of Clarke1866:
Out[1]=
Semimajor axis and inverse flattening of GRS80:
Out[1]=
Eccentricity of the GRS80 reference ellipsoid:
Out[1]=
Full name of the ITRF00 datum:
Out[1]=
Alternate names for ITRF00:
Out[1]=
Names of all properties of the ITRF00 datum:
Out[1]=
This gives the 14 parameters required to transform from ITRF00 to NAD83CORS96:
Out[1]//TableForm=
New in 7
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'
# Search results
Found 903 matches
Area-based particle size
Particle size is a notion introduced for comparing dimensions of solid particles (flecks), liquid particles (droplets), or gaseous particles (bubbles).
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Hyperbolic law of cosines - 1st law
In hyperbolic geometry, the law of cosines is a pair of theorems relating the sides and angles of triangles on a hyperbolic plane, analogous to the planar ... more
Settling velocity (Stokes law)
Stokes’ law can be used to calculate the viscosity of a fluid. Stokes’ law is also important in the study for Viscous Drag , Terminal Velocity ... more
Ball Screw - Preload Drag Torque
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
Ball Screw - Driving Torque
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
Ball Screw - Frictional Resistance
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
Ball Screw - Tensile Compressive Load
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
In mathematics, a 3-sphere is a higher-dimensional analogue of a sphere. It consists of the set of points equidistant from a fixed central point in ... more
Law of sines at the hyperbolic triangle
A hyperbolic triangle is a triangle in the hyperbolic plane. It consists of three line segments called sides or edges and three points called angles or ... more
Orbital Period - as a function of central body's density
The orbital period is the time taken for a given object to make one complete orbit around another object.
When mentioned without further ... more
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https://www.analystforum.com/t/how-long-would-it-take-you-to-do-all-the-eoc-cfai-questions/108450
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# How long would it take you to do all the EOC CFAI questions?
I am trying to pace myself. I haven’t counted how many questions are there but I was thinking of doing perhaps several sets of questions per day starting from Book 1 to Book 6.
Put differently, how long did it take you guys to do it?
It takes a lot of time.
I remember Level 1 questions took me around a month to complete.
This was with both Wiley and CFA EOC and In text examle questions.
I am guessing it will take around the same time.
You could expect atleast 15 days to go with it. Especially if you are doing it for the first time.
I decided since I already did them in my first pass of the curriculum , and understood them, my time would be better spent doing mocks and topic Tests. If you started now you wouldn’t be done until May. Not worth it.
I think i’ll be done by May and use entire month of May for mocks, topic tests, and Schweser exams.
omfg a lonnnnng time
i counted for each case question, like 6, it takes me about 30 min to go through a set during the first run
I recently completed them all for the first time. This acted as a second pass at the material, some of which hadn’t been looked at for 3+ months. It was very useful.
There are only 40 chapters you should actually do EOCs for - exclude the ones without 3 answer questions e.g. QM and exclude any non item-set style questions.
On average 1 hour per EOC, some much quicker and some slower. So 40 hours.
I’ve got an Excel for tracking progress. I didn’t make it to count EOC PP, but it has a count of each.
The sum of that column is 1083 (mean = 23.0; standard dev = 14.7, mode = 6, ). If question has a part A, Part B, I could each of the parts as separate questions (so, 1A, 1B, 1C, 2, 3 = 5 questions) – but I was inconsistent at that at the start so my number may not be perfect.
1083 * 3 min per = … anyone. Anyone have a calculator? I suck at math. Say … about a year.
I got an Excel for portal topic tests.
Mean 64,82. Median 58,5. Mode 50. Average solving minutes per vignette 14,79. Didn’t calculate stdevpt but seems there are few positive outliers.
CFAI EOCs vignettes are much more easier to solve. I am currently doing review of entire material and solving them again. Average 75 % sucess with about avr 8 minutes per vignette. There are EOCs with only 1 vignette with 6-8 questions but there are some chapters with even 4 or more 6Q vignettes. The recorder is first chapter EOCs in Corp. finance. 50 questions overall.
I figure it would take me at least 20 minutes.
Maybe even 30.
I’ll found out in 2 weeks, hopefully.
Still haven’t gone through the practice bootcamp.
wait, easier to solve compare to what
^compared to portal topic tests. Is it not clearly and logical stated above?
wait whats the difference, maybe I am just being dumb, whats a portal topic test
down right side click on your candidate resources
then click on topic tests and mock exams
you will find tests for each of 10 CFA L 2 topics and mock exams
Actually regarding the CFAI Top test, do you guys know if we can print them out? Looks like we can only do them online and they’ll also time you.
Regarding the question on EOC questions, for item-set questions I can still do them around 20 min / set, but I really hate those questions that are NOT multiple choice question. I’m thinking of skipping them and just get more item-set questions from sources like FinQuiz etc.
oh wow, i have not started on those yet, sh**
Don’t worry. You’ve got a plenty of time…
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https://www.convert-measurement-units.com/convert+Piconewton+meter+to+Inch-ounces-force.php
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Convert pNm to inoz (Piconewton meter to Inch-ounces-force)
## Piconewton meter into Inch-ounces-force
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Piconewton+meter+to+Inch-ounces-force.php
## How many Inch-ounces-force make 1 Piconewton meter?
1 Piconewton meter [pNm] = 0.000 000 000 141 643 059 490 08 Inch-ounces-force [inoz] - Measurement calculator that can be used to convert Piconewton meter to Inch-ounces-force, among others.
# Convert Piconewton meter to Inch-ounces-force (pNm to inoz):
1. Choose the right category from the selection list, in this case 'Torque'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Piconewton meter [pNm]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Inch-ounces-force [inoz]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '410 Piconewton meter'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Piconewton meter' or 'pNm'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Torque'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '91 pNm to inoz' or '54 pNm into inoz' or '1 Piconewton meter -> Inch-ounces-force' or '15 pNm = inoz' or '63 Piconewton meter to inoz' or '34 pNm to Inch-ounces-force' or '31 Piconewton meter into Inch-ounces-force'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(70 * 63) pNm'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '410 Piconewton meter + 1230 Inch-ounces-force' or '83mm x 41cm x 98dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 2.856 099 974 009 5×1030. For this form of presentation, the number will be segmented into an exponent, here 30, and the actual number, here 2.856 099 974 009 5. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 2.856 099 974 009 5E+30. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 2 856 099 974 009 500 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
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https://forum.appacademy.io/t/explain-invalid-or-dont-care-inputs-in-boolean-algebra/2550
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# Explain Invalid or "don't care" inputs in Boolean Algebra
As far as practical usage in truth tables/kmaps, I understand perfectly that they can be either 0 or 1 at the engineer’s whim in order to create the most simplified expression. What I am confused about is when or what exact set of criteria would cause an input to be considered an invalid input.
in the lesson, it says the following
“There are certain conditions where a function may not be completely specified, meaning there may be some inputs that are undefined for the function”
what exactly are these certain conditions? they seem to vary depending on the function. Is this something we will cover later on?
These would be inputs which, for the purposes of your function, would never occur; because the point of the kmap is to build out the simplest function (for the sake of the simplest circuitry), eliminating impossible inputs simplifies your problem.
Imagine you are building a circuit whose output is to turn an air conditioning on and off in your home. You want to conserve power, so you want the A/C to only turn on only if the temperature is above a certain threshold, there is at least one person present in your house, and all the doors are closed (that way the cold air doesn’t escape outside). Now, let’s imagine that you are a person who is rather protective of your home, so you are never going to leave your door open if there isn’t somebody inside. Thus, for your use case, all inputs in which “someone is home” is false and “a door is open” is true are irrelevant for your circuit as they will never occur. Therefore, you should flip those impossible outputs to true or false depending on what makes the simplest circuit.
From my understanding, in all simple boolean expressions (which is what we are working with here), inputs will only be invalid based on the context in which you are using the circuit: what is your use case and, in that use case, what things can we rule out?
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https://webdevelopersfieldguide.com/204sxq1m/rV1260mF/
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1st Grade Math Worksheets: Free Math Worksheets First Grade Addition Number Bonds Sum For 1st Free Math Worksheets First Grade Addition Number Bonds Sum Of Printable Games Children Problems Without Regrouping Reading. Math Sheets Grade 2 A To Z Worksheets For Kindergarten Changing Subtraction To Addition Worksheets | Webdevelopersfieldguide
# Free Math Worksheets First Grade Addition Number Bonds Sum For 1st Free Math Worksheets First Grade Addition Number Bonds Sum Of Printable Games Children Problems Without Regrouping Reading
Published at Monday, 17 August 2020. 1st Grade Math Worksheets. By .
Most volumes begin with an explanation of basic arithmetic operations namely: addition, subtraction, multiplication, and division. Reference tables are supplied to provide clues for quick mental arithmetic and mastery of math facts. When ready to be tested, the student can select a drill, which has 10 questions and are selected from a database of number pairs for calculation. The Basic Level volumes use simple single digit numbers and the interactive math software at the Advanced Level uses mostly double digit numbers for math practice problems. Each drill is then scored and timed with the results saved. With the test records, students can follow their own progress and adults who may be supervising can monitor progress and assess if there are any learning issues that require intervention.
There are many opportunities to teach your child how to count. You probably already have books with numbers and pictures, and you can count things with your child all the time. There are counting games and blocks with numbers on them, wall charts and a wide variety of tools to help you teach your child the basic principles of math. Mathematics worksheets can help you take that initial learning further to introduce the basic principles of math to your child, at a stage in their lives where they are eager to learn and able to absorb new information quickly and easily.
If the materials do not specifically indicate "brain-based," determine if they are at least brain-friendly. This would mean that you are looking for lots of color, material interesting to the child, many varied activities-especially involving movement, and using several of the senses. I saw one company whose worksheets included the instruction to "say the number out loud as you..." This is very good! Speaking out loud is very important for learning to occur. Ideally, all worksheets should include this instruction. If you can not find any that do, then you need to add that instruction yourself.
File name:
### Free Math Worksheets First Grade Addition Number Bonds Sum For 1st Free Math Worksheets First Grade Addition Number Bonds Sum Of Printable Games Children Problems Without Regrouping Reading
Image Size: 2048 x 1024 Pixels
File Type: Image/jpg
Total Gallery: 52 Pictures
File Size: 160 kb
## Math Printable First Graders
### Comments of Math Worksheets For 1st Grade
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https://studylib.net/doc/7369029/handout
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# handout
```Description and Pseudocode for the Dynamic Programing Algorithm
to solve the string alignment problem
The String alignment problem asks us to compute the “best” alignment of two strings
with given penalties for character mismatch and or character skipping.
For example if S1 = “VINTNER” and S2 = "WRITERS” and the cost of a mismatch is 1
and the cost of a skip is 1 then one possible optimal alignment is given by:
W R I T – E R S
V I N T N E R The penalty for this alignment is 5 since there are 3 mismatched characters and 2 skipped
characters.
We present a dynamic programming algorithm which solves this problem using a table
containing the solutions of subproblems. T(i, j) will solve the problem using the first i
characters of S1 and the first j characters of S2.
The algorithm:
Input Section
1) Input S1 and S2, two strings. (eg. S1 = “WRITERS”, S2 = “VINTNER”)
2) Compute n to be the length of S1 and m of S2 originally. (eg. n = 7, m = 7
3) Pad the beginning of each with a blank to simplify things.
(e.g. S1 = “_WRITERS” and S2 = “_VINTNER”)
4) Get skipPenalty and mismatchPenalty. (from the file or user depending on the
program)
Initialization
5) Create an initially empty n+1 by m+1 array, T.
6) fill the first column with multiples of skip penalty
i.e., T(i, 0) = i*skipPenalty for i from 0 to n inclusive
7) fill the first row with
i.e. T(0, j) = j*skipPenalty for j from 0 to m inclusive
Main Loop(s)
8) for i going from 1 to n inclusive
9)
10)
for j going from 1 to m inclusive
T(i, j) = the minimum of the following four things:
T(i-1, j) + skipPenalty,
// skipping a char in S2
T(i, j-1) + skipPenalty,
// skipping a char in S1
T(i-1, j-1)
// only if S1(i)=S2(j) ie match
T(i-1, j-1) + mismatchPenalty // only if S1(i) ≠ S2(j) ie mismatch
Output
11) The answer is sitting in T(n,m)
--------------------------------------------------------------------------------------------------------Bonus!! With another table we can figure out the alignment that produced the best result.
We use a second table to keep track of which of the four choices we made in step ten.
5b) create a second n+1 by m+1 array P of String
6b) initialize the 0th row of P to all “left“ this indicates T(0,j) came from T(0, j-1) +1
7b) initialize the first column to all “up”, this indicates the T(i,0) came from T(i-1,0) + 1
10)
If the first choice is the minimum put “up” in P(i, j)
If the second choice is the minimum put “left” in P(i, j)
If one of the last two are minimum put “diag” in P(i, j)
Recovering the path
12) set ans1 and ans2 to “” ( empty strings) and set i and j to n and m
13) while i>0 or j >0 do
14)
if P(i,j) is “”up”:
ans1 = S1(i) + ans1; ans2 = “-“ + ans2; i = i -1;
else if P(i,j) is “left”: ans1 = “-“+ans1; ans2 = S2(j) + ans2; j = j-1
else:
ans1 = S1(i) + ans 1; ans2 = S2(j)+ans2; j = j-1; i = i-1
ans1 and ans2 printed one above the other show the proper alignment
While this is a well-known algorithm, the immediate source of this handout and example
is: Gusfield, Dan. Algorithms on Strings, Trees and Sequences: Computer Science and
Computational Biology. Cambridge University Press, 1997.
```
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Employment costs rose 2.8 percent in the 12 months that : GMAT Sentence Correction (SC)
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# Employment costs rose 2.8 percent in the 12 months that
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Employment costs rose 2.8 percent in the 12 months that [#permalink]
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Employment costs rose 2.8 percent in the 12 months that ended in September, slightly less than they did in the year that ended in the previous quarter.
A. less than they did
B. less than it did
C. less than they were
D. lower than
E. lower than they were
If you have any questions
you can ask an expert
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06 Apr 2006, 23:31
One more for A. But can someone please explain the proper usage of lower and lesser to me? Always end up gettting confused....
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07 Apr 2006, 06:01
Employment costs rose 2.8 percent in the 12 months that ended in September, slightly less than they did in the year that ended in the previous quarter.
A. less than they did
B. less than it did
C. less than they were
D. lower than
E. lower than they were -> CORRECT! because we are talking about rise and fall, and not less or more!
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07 Apr 2006, 06:11
I think we use lower when we talk in qualitative terms, in size persay.
and less in qualitative terms say difference.
Please notice usage of less and lower in below sentence.
eg.
"The human, no less than the lower forms of life, is a product of the evolutionary process"
Hope this helps!
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07 Apr 2006, 09:43
"Less" will be appropriate here. "Employment costs" is plural.
So, D, B, E are out.
Between C and A, I choose A because of parallelism.
What's the OA?
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08 Apr 2006, 10:46
^ A ^
Employment costs rose 2.8 percent in the 12 months that ended in September, slightly less than they did in the year that ended in the previous quarter.
"they" and "did" are necessary here as I made the parts bold above..
>>"less" is a quantifier so it is an adv. it is correct to say "..costs less than..".
>>"lower" is and adj. it would be correct to say:
Employment costs were lower than... or
A lower cost of..
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08 Apr 2006, 11:03
One more for 'A'.
Selene has already explained.
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08 Apr 2006, 11:11
OA is A.
Great explanation Selene.
selene wrote:
^ A ^
Employment costs rose 2.8 percent in the 12 months that ended in September, slightly less than they did in the year that ended in the previous quarter.
"they" and "did" are necessary here as I made the parts bold above..
>>"less" is a quantifier so it is an adv. it is correct to say "..costs less than..".
>>"lower" is and adj. it would be correct to say:
Employment costs were lower than... or
A lower cost of..
08 Apr 2006, 11:11
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# Employment costs rose 2.8 percent in the 12 months that
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Cody
# Problem 283. Give the Shortest Path Through The Maze
Solution 35322
Submitted on 9 Feb 2012 by Martijn
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% assert(isequal(solve_maze([3 2;1 2]),[1 0;2 3]))
2 Pass
%% assert(isequal(solve_maze([3 2 2;3 3 1;1 1 1]),[1 2 3;0 0 4;0 0 5]))
3 Pass
%% assert(isequal(solve_maze([3 3 2 2 2;1 0 1 2 1;3 2 3 3 0;1 0 2 2 0;1 1 1 0 1]),[1 4 5 6 7;2 3 0 0 8;0 0 0 0 9;0 0 0 0 10;0 0 0 0 11]))
4 Pass
%% assert(isequal(solve_maze([3 2 2 3 3 3 2 3;3 1 2 1 1 3 1 3;1 0 3 0 0 1 3 3;1 1 0 2 0 0 1 1;3 1 3 3 2 1 2 1;3 2 2 1 3 0 1 0;3 3 2 3 1 2 2 0;1 3 3 0 0 2 0 2;1 2 3 1 0 2 1 3;1 1 3 2 3 1 1 0]),[1 2 0 0 0 0 0 0;0 3 0 0 0 0 0 0;0 4 0 0 0 0 0 0;0 5 6 7 8 9 0 0;0 0 0 0 0 10 11 0;0 0 0 0 12 11 12 13;0 0 0 0 13 14 15 14;0 0 0 0 14 15 16 0;0 0 0 0 0 0 17 0;0 0 0 0 0 0 18 19]))
5 Pass
%% assert(isequal(solve_maze([3 2 2 3 2 3 3 2 3 3 3 2 3 2 3 2 3;3 0 3 3 1 2 2 3 3 0 2 1 0 1 2 2 0;1 0 2 3 2 1 1 0 0 3 1 0 2 1 1 0 2;3 1 1 2 0 0 2 0 2 3 3 1 0 3 1 0 3;1 0 1 0 3 0 1 0 0 2 2 1 0 2 1 2 1;1 2 0 0 3 1 2 1 2 3 1 1 0 1 3 1 1;3 1 1 3 2 2 2 2 2 1 2 3 2 0 3 1 2;1 3 0 1 0 3 0 2 1 3 2 2 2 1 1 1 2;1 3 3 1 1 1 2 3 2 2 2 2 3 3 2 0 1;3 1 3 0 2 0 2 0 1 0 3 1 2 3 2 0 0;1 1 0 2 3 3 1 2 2 1 3 0 2 3 2 2 0;3 0 0 2 3 3 0 0 2 2 0 0 1 3 1 3 1]),[1 2 0 0 0 0 0 0 0 0 0 0 21 22 0 0 0;0 3 0 0 0 0 0 0 14 15 16 19 20 23 24 0 0;0 4 5 0 0 0 0 12 13 0 17 18 0 0 25 0 0;0 0 6 7 8 9 10 11 0 0 0 0 0 0 26 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 27 28 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 29 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 30 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 31 32;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 32 33;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 34;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 35;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 36]))
6 Pass
%% assert(isequal(solve_maze([3 2 3 3 2 2 2 3 2 3 3 2 2 2 2 2 2 2 2 3 3 2 3 2 3;1 2 2 2 1 1 1 3 0 3 1 1 0 2 3 0 2 0 0 3 3 0 3 1 3;3 1 3 2 3 1 2 2 2 3 3 0 3 3 3 3 3 3 0 0 2 0 3 0 3;3 1 1 1 3 3 3 1 0 1 1 3 3 2 2 2 3 1 3 1 3 1 3 1 0;3 1 3 0 1 3 2 1 1 2 1 0 3 1 3 2 1 0 1 0 2 0 3 2 0;3 0 2 2 2 3 2 1 2 2 0 3 0 0 1 1 0 1 3 1 3 0 2 0 0;3 3 0 0 2 0 3 0 1 1 2 3 1 0 3 1 3 3 0 1 2 3 1 3 2;1 0 2 0 3 1 1 0 3 2 0 0 0 0 0 0 1 3 3 2 2 0 0 2 2;1 2 1 3 2 1 3 1 0 3 2 3 3 3 1 0 1 2 0 1 3 3 0 2 1;1 0 3 2 0 0 3 3 2 2 0 0 3 3 0 1 2 0 2 1 2 0 2 3 3;3 0 2 3 3 2 3 2 1 3 3 1 2 0 3 2 1 0 2 0 1 3 1 2 0;1 0 1 1 2 2 1 2 2 1 3 1 0 3 3 2 3 2 1 2 3 3 3 1 2;1 2 1 0 0 3 3 3 0 1 2 1 1 2 2 0 2 3 3 3 1 0 3 1 3;3 3 2 3 1 2 2 2 1 0 2 0 3 1 3 0 0 2 1 3 0 1 3 0 1;1 2 2 0 0 3 1 1 1 3 1 0 3 3 1 0 2 0 3 3 1 0 0 0 3;1 2 1 2 1 1 0 0 2 0 3 0 0 1 0 0 1 3 0 0 1 3 2 0 1;3 3 1 2 0 1 1 1 1 2 2 0 1 2 1 1 3 0 0 0 1 1 3 0 0;3 2 2 0 0 0 0 3 2 2 1 1 0 0 2 3 1 3 3 3 0 3 1 3 0;1 2 2 0 3 1 2 0 3 1 3 2 3 3 2 1 2 3 3 2 2 1 2 0 0;1 3 1 3 2 1 3 1 2 1 0 0 0 2 1 3 0 3 0 2 3 0 0 0 0]),[1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 7 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 10 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;12 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;13 14 0 0 0 0 0 0 0 0 0 0 0 0 43 44 0 0 0 0 0 0 0 0 0;14 15 0 0 0 0 0 0 0 0 0 0 0 41 42 45 46 47 0 51 52 53 54 0 0;0 16 17 0 0 0 0 0 0 0 0 38 39 40 0 0 47 48 49 50 0 0 55 56 0;0 0 18 0 0 0 0 0 0 0 0 37 38 0 0 0 0 0 0 0 0 0 0 57 0;0 0 19 20 21 0 0 0 0 0 0 36 0 0 0 0 0 0 0 0 0 0 0 58 0;0 0 0 0 22 23 24 25 0 0 0 35 0 0 0 0 0 0 0 0 0 0 0 59 0;0 0 0 0 0 0 25 26 0 0 0 34 0 0 0 0 0 0 0 0 0 0 0 60 0;0 0 0 0 0 0 26 27 28 0 0 33 0 0 0 0 0 0 0 0 0 0 0 61 0;0 0 0 0 0 0 0 0 29 30 31 32 0 0 0 0 0 0 0 0 0 0 0 62 63;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 64;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 65;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 66]))
7 Pass
%% assert(isequal(solve_maze([3 2 2 3 3 3 2 3 2 3 2 3 2 3 2 2 2 2 3 3 3 2 2 2 2 2 2 2 3 3 2 2 2 2 2 2 2 3 3 3 2 2 3 2 3;3 2 0 1 3 2 1 0 1 2 2 1 2 1 1 0 2 1 0 2 1 3 3 2 3 3 0 0 3 2 3 1 3 1 3 3 0 3 1 0 0 2 1 1 0;1 1 3 0 3 3 0 1 2 2 3 0 0 2 2 3 3 0 0 1 3 3 3 1 3 2 3 1 3 3 1 0 0 0 0 2 2 2 0 3 3 0 1 1 2;1 0 2 2 0 1 0 2 2 3 2 1 0 3 1 3 1 1 1 3 3 2 1 1 2 0 1 1 2 1 3 0 3 3 0 1 1 2 2 3 1 1 1 1 3;1 1 0 0 0 3 0 2 2 0 0 2 1 3 2 3 2 1 3 2 0 2 3 0 0 0 3 0 0 1 0 1 0 2 0 1 1 1 2 3 0 2 1 3 1;3 3 1 1 2 0 3 0 0 3 2 1 1 1 3 1 3 0 1 0 1 0 3 1 3 1 2 3 1 2 1 1 3 3 2 1 2 1 3 0 2 3 2 0 2;3 1 2 0 0 3 3 1 1 2 1 0 3 3 3 0 2 1 3 0 3 0 0 2 2 3 2 1 0 1 3 2 0 2 3 3 3 0 1 2 1 1 1 0 2;3 1 2 0 0 3 1 0 3 3 2 3 2 0 2 3 2 2 0 0 2 0 1 1 3 1 1 3 2 0 1 0 0 2 2 1 2 1 0 0 3 0 2 2 1;3 0 0 2 0 0 1 0 3 2 2 0 0 3 1 1 2 0 0 0 3 2 0 2 1 1 2 0 3 2 3 2 0 3 3 3 1 3 2 1 0 2 0 1 2;1 0 1 0 1 2 3 2 0 1 0 0 0 1 1 2 1 3 0 2 0 0 0 3 2 0 0 1 2 0 2 1 2 3 2 1 0 3 2 0 2 1 1 2 2;3 0 0 1 0 2 0 2 0 2 3 2 3 0 2 3 3 1 0 2 3 2 0 3 0 0 2 0 2 3 0 3 1 0 3 1 0 2 2 1 0 1 2 2 3;1 3 3 1 2 3 1 2 2 3 3 3 3 0 2 1 2 1 1 3 0 2 2 2 0 3 3 2 0 3 0 0 0 0 3 2 2 0 0 1 3 3 0 0 0;1 0 3 1 2 1 0 3 0 0 0 0 3 2 3 0 2 2 0 0 1 1 1 2 3 3 1 0 2 1 2 1 3 1 2 0 2 3 0 0 0 3 1 0 3;3 2 0 0 1 1 2 3 1 1 2 1 2 2 0 3 2 1 2 0 0 1 3 1 3 2 2 3 3 3 3 0 3 1 2 0 0 1 1 3 3 2 2 1 2;1 3 3 1 3 0 0 2 0 3 0 3 1 2 2 3 0 3 0 2 0 3 1 1 0 2 3 2 1 2 0 0 2 0 3 1 0 3 2 1 0 1 1 1 2;3 1 3 3 2 1 2 1 1 3 1 3 1 0 3 3 2 3 1 3 0 0 1 2 1 3 0 2 2 2 2 1 2 3 2 0 3 2 0 2 1 2 1 1 2;1 2 2 1 3 2 0 3 1 1 1 0 2 3 1 2 0 0 2 0 0 2 3 2 3 1 1 3 2 1 1 3 3 3 0 1 3 3 1 0 2 0 1 1 0;1 2 0 0 2 1 1 0 2 2 0 2 0 3 1 2 2 0 2 0 3 2 0 0 2 1 1 1 0 1 2 0 0 2 3 1 0 0 1 2 0 3 2 1 2;1 2 2 1 0 2 1 1 3 1 2 3 1 2 2 3 1 2 2 2 1 1 0 0 1 2 0 2 0 3 2 1 0 2 3 1 1 0 2 1 1 0 0 0 2;3 0 2 0 1 1 2 0 3 3 3 3 0 1 0 1 3 1 1 2 1 1 1 3 0 0 3 0 0 3 1 0 2 3 2 0 2 2 0 3 0 1 1 3 0;3 1 3 0 2 1 3 1 0 3 0 3 1 2 1 3 2 3 0 1 0 3 3 3 0 1 2 0 0 1 2 2 3 2 0 2 1 2 2 3 1 0 0 3 1;3 1 2 0 1 2 3 0 1 0 0 1 2 0 1 1 1 2 1 1 3 1 1 2 0 1 3 0 3 0 0 0 0 1 1 2 1 3 0 2 0 3 2 2 1;3 0 2 1 0 3 2 2 1 0 2 3 1 1 2 1 1 1 1 3 3 2 2 3 2 1 2 2 0 1 3 3 0 3 3 0 1 2 0 0 3 2 1 0 1;3 3 2 2 3 3 2 0 2 1 1 3 3 3 1 2 0 3 0 1 1 3 3 1 0 3 0 1 0 2 1 2 2 0 0 0 2 1 2 1 0 3 0 3 2;1 0 3 2 1 0 2 3 1 1 3 1 3 2 2 2 3 0 2 1 3 2 1 1 1 3 1 0 0 0 0 0 3 2 3 3 3 1 2 2 1 2 0 3 0;1 0 3 3 3 2 2 3 0 0 0 3 3 3 0 0 3 3 0 0 2 3 1 3 3 3 1 1 3 2 3 0 0 2 2 1 0 0 2 3 3 3 3 3 0;3 0 1 1 0 1 2 3 1 3 1 3 0 3 3 1 2 3 3 2 3 3 2 3 1 3 2 1 0 1 0 1 3 1 0 2 2 0 0 3 1 0 3 0 0;1 3 3 2 2 2 1 0 1 0 3 3 2 0 2 3 0 1 0 0 3 1 0 1 2 1 0 2 1 3 2 3 2 0 0 2 0 3 2 1 3 1 0 0 3;3 2 3 2 2 3 1 3 0 1 2 0 3 3 2 3 3 1 0 1 1 1 0 2 2 2 3 3 2 0 2 0 1 1 0 2 0 1 0 2 1 0 3 2 0;3 0 3 1 0 3 3 3 0 2 1 2 2 2 0 2 0 3 1 1 0 2 2 0 0 2 2 1 0 1 0 0 2 3 3 0 0 0 0 2 2 0 3 2 0;3 1 3 0 1 2 3 3 1 1 3 1 2 0 1 1 0 3 1 1 0 2 3 1 3 2 1 1 3 0 3 3 3 0 1 3 2 0 0 3 2 3 2 0 1;3 2 2 2 2 3 3 0 1 2 1 2 3 3 2 2 2 2 3 0 1 2 0 3 2 2 3 0 2 2 3 1 1 0 0 3 2 3 0 0 1 2 2 0 0;3 1 3 2 1 3 1 2 3 2 2 2 2 0 1 3 0 2 3 0 0 3 3 2 3 3 1 2 1 1 0 0 3 2 0 3 0 2 1 0 2 3 2 1 3;3 1 0 0 1 2 3 2 2 3 1 0 1 2 2 0 3 2 1 0 1 1 0 2 1 3 0 1 1 0 1 2 1 3 1 3 3 0 2 0 0 1 0 0 2;3 0 0 3 0 2 2 1 1 0 0 2 0 3 0 2 1 0 3 2 0 3 3 3 3 0 3 0 0 1 2 0 1 3 2 3 0 3 0 2 3 1 2 0 0;3 3 0 2 0 0 2 0 0 0 1 2 2 0 0 1 0 2 2 1 0 2 2 2 3 3 2 3 0 2 0 2 1 2 0 1 1 3 3 0 3 2 1 2 3;3 2 0 2 1 3 3 0 3 0 1 1 0 3 0 2 0 3 1 2 0 0 3 2 2 1 1 2 1 0 2 0 0 1 3 2 2 2 2 1 2 1 2 3 1;1 3 1 2 2 0 2 0 3 3 3 2 1 0 2 3 1 2 3 3 2 3 2 1 2 0 3 3 1 1 1 0 3 2 0 0 1 2 2 2 0 2 1 3 3;1 3 3 3 0 0 3 2 2 1 2 0 2 3 1 2 0 2 1 3 3 2 0 1 3 0 0 2 3 1 3 3 1 0 0 0 2 2 3 3 2 0 1 2 1;3 3 2 0 0 0 1 2 0 0 3 1 2 1 0 1 0 2 1 0 2 1 2 2 2 1 0 1 1 3 1 3 2 3 3 1 3 2 0 3 3 0 2 2 2;1 1 1 0 3 1 0 0 3 2 3 2 2 0 2 2 3 0 1 2 0 1 3 3 2 1 3 1 1 0 2 1 0 0 0 0 2 3 1 3 0 1 1 2 0;3 1 0 1 0 1 3 3 2 2 3 2 1 3 2 0 1 1 2 0 0 2 2 2 2 1 1 3 3 1 0 3 0 2 0 1 2 2 2 2 2 0 1 3 1;1 0 3 1 1 2 3 1 3 1 0 1 1 1 3 0 2 3 1 3 1 0 3 3 3 2 0 1 3 3 1 3 0 0 2 3 0 2 3 1 3 2 1 3 2;3 0 3 2 3 3 2 3 3 1 0 0 3 2 0 1 0 3 1 1 2 0 0 3 2 1 1 2 2 1 1 1 2 1 3 1 3 3 3 2 0 1 0 2 2;1 3 3 1 3 2 2 3 2 0 2 3 3 1 3 1 0 3 0 1 2 2 1 1 1 2 2 2 0 3 0 2 1 0 3 1 2 0 1 0 0 3 0 3 0;3 2 3 1 0 0 0 0 2 3 1 3 3 2 2 1 2 1 3 2 2 1 0 2 2 1 0 3 1 3 1 3 2 0 1 3 2 2 3 0 2 1 3 1 3;1 3 3 1 3 0 3 1 0 3 1 1 2 0 3 2 1 2 1 2 3 0 1 2 3 0 1 2 1 0 0 3 1 3 0 2 3 2 2 2 0 1 0 3 3;3 0 0 0 0 0 3 0 0 1 3 0 1 3 3 3 1 0 1 2 1 0 1 0 3 1 0 1 2 1 3 1 0 2 0 3 0 3 0 2 2 0 2 0 3;3 2 0 3 2 3 3 1 1 3 0 0 0 3 3 2 2 3 1 3 3 2 1 3 3 2 0 2 0 3 2 0 1 1 2 2 1 1 2 0 0 2 0 1 1;1 1 1 2 0 1 0 0 2 0 2 2 3 3 3 3 0 3 0 3 1 3 3 1 3 0 3 3 0 1 0 1 1 2 2 0 3 3 0 3 0 3 3 0 0]),[1 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 5 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 7 8 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 8 9 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 11 12 0 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# Difference between revisions of "Talk:Average"
Notice postied on EdPoor's page:
I noticed that you edited out my (correct) version of average, and replaced it with the previously (incorrect) one, labelling my version as obscure. This confirms for me what I have been thinking for some time, that the entire CP project is doomed, unless you and other Sysops get a bit more open-minded, and read the entries that others have written and are prepared to learn, rather than be dogmatic. I defined Average, exactly as I would define it to my students; the point is, that in common parlance, people use average as the middle or most likely of set of data, without actually understanding that the idea is problematic. As it turns out, the most sensible 'middle' is actually the median, and the most common is the mode; however, neither of these are what people normally refer to as "the average"; notmally they calculate it using the arithmetic mean. However that is neither the most common, nor the middle. Can you provide an explanation what the arithmetic mean ACTUALLY is - what is it attempting to measure in everyday language??? It's extremely difficult to explain.
In fact, what the arithmetic mean is, goes something like: "If all the data were the same, but you had the same total as you had before, then the data value you get would be called the average". The aritmetic mean of 10, 20 30 is 20.
The geometric mean you get when you ask the same question for rates of interest (10%, 20% and 30% have a mean of 18.17%), and the harmonic mean you get when you ask the same question about speeds (10kmph, 20kmph, 30kmph have an mean of 16.3kmph) .
If anyone who was editing the average pages knew anything about descriptive statistics, this page should say something a lot different.
You certainly should not be reverting such pages, because you clearly don't have sufficient mathematical understanding to appreciate the nuances.
SeanTheSheep 03:05, 14 May 2007 (EDT)
The point about all of this is that calculations made on data sets are statistical estimates of something or other. If they are not, then there is no point to them. In this particular case, the average is an estimate of the central location or central tendency of the data, i.e. the answer to the question whereabouts is the data located?
A definition of average, which tells someone how to calculate one of the measures for this without explaining what we are trying to do, or discussing what the problems are in trying to capture such a thing is wrong.
The main reason that we almost uniformly use the arithmetic mean as the average is quite difficult to grasp: There is an important result in "Expectation Algebra" which says if you estimate the overall 'expected' mean of a probability distribution by using the arithmetic mean of a data set derived from that distribution, then the expected value of the distribution of arithmetic means is the mean that you started with; i.e. that the arithmetic mean is an unbiased estimator of the distribution mean (it is also a consistent estimator, and, as it turns out the most efficient estimator). Those are the reasons it is used, and that result underpins what is probably the most important theorem in the whole of statistics: The Central Limit Theorem.
However, I am not suggesting that we clutter up the definition with all of that. What I am saying though is that there are lots of different ways of trying to find the middle of a set of data, and that saying that average is (add them up and divide by the number of numbers) is inapproprate on three counts:
• firstly that it ignores what the idea of average is trying to do,
• secondly that it does not discuss whether "average" defined in this manner actually captures the required notion
• thirdly it does not discuss in what situations an arithmetic average is, or is not appropriate.
--SeanTheSheep 04:31, 14 May 2007 (EDT)
## some random math stuff in case anyone wants to use it
Human 17:13, 14 May 2007 (EDT)
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# Notes I: Sets and Functions: Solutions
Problem 1:
If $\{ A_{i}\}$ and $\{ B_{j}\}$ are two classes of sets such that $\{ A_{i}\} \subseteq \{ B_{j}\}$, prove that $\bigcup_{i}A_{i} \subseteq \bigcup_{i}B_{i}$ and $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$.
The first part is clear as the hypothesis says the set of sets $\{ A_{i}\}$ is a subset of the set of sets $\{ B_{j}\}$. So, quite clearly $\bigcup_{i}A_{i} \subseteq \bigcup_{j}A_{j}$. For the second part, we can use the generalized De Morgan’s laws. (Note: if you try to derive it straight, you will prove exactly what is opposite of the required result :-)). For the second part note that $X \subseteq Y \Longleftrightarrow Y^{'} \bigcap X^{'}$. Using the result in the first part, that is, consider $\bigcup_{i}A_{i} \subseteq \bigcup_{j}A_{j}$ and now take the complement of both sides so we get: $\bigcup_{i}A_{i} \subseteq \bigcup_{j}B_{j} \Longleftrightarrow (\bigcup_{j}B_{j})^{'} \subseteq (\bigcup_{i}A_{i})^{'}$, that is, $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$. QED.
Problem 2:
The difference between two sets A and B, denoted by $A-B$, is the set of all elements in A and not in B, thus $A-B = A \bigcap B^{'}$. Show the following:
(a)$A-B = A - (A \bigcap B) = (A \bigcup B) -B$.
Answer (a): $A-B$ is the set of elements $x \in A$, but not in B. Hence, $A-B = A-(A \bigcap B)$, simply by definition. Again, if we consider elements in A or B or both, and then take away elements of B, we are just left with elements of A only, but not of B. That is, $A-B = A-(A \bigcap B) = (A \bigcup B) - B$. QED. ps: A venn diagram may help visualize v well.
(b) TPT: $(A-B) - C = A - (B \bigcup C)$ (PS: a Venn diagram can help visualize and perhaps, guide the writing of the proof also).
Let $x \in A-B$, but $x \notin C$. Then, $x \in A$, but $x \notin B$, or $x \notin C$, or not in both B and C. That is, $x \in A-(B \bigcup C)$. By reversing the argument, we get the other subset relationship. Hence, $(A-B)-C = A- (B \bigcap C)$. QED.
(c) TPT: $A - (B-C) = (A-B) \bigcup (A \bigcap C)$.
Answer (c): Let $x \in A-(B-C)$. That is, $x \in A$ and $x \notin (B-C)$. That is, $x \in A, x \notin B, x \in C$. That is, $x \in A, x \notin B$, or x could be both in A and C. That is, $x \in (A-B) \bigcup (A \bigcap C)$. Simply reversing the arguments, we get the reverse subset relation. Hence, the two sets are equal. Hence, $A-(B-C) = (A-B) \bigcup (A \bigcap C)$. QED.
(d) TPT: $(A \bigcup B) - C = (A-C) \bigcup (B-C)$.
answer (d): let $x \in A\bigcup B$, but $x \notin C$. Then, $x \in A$, or $x \in B$, or x is an element of both A and B, but x is not an element of C. Which clearly means, x is an element of A but not C, OR x is an element of B but not C. That is, $x \in (A-C) \bigcup (B-C)$ (upon reversing the arguments, we get the other subset relations. Hence, the two sets are equal). QED.
problem (e): TPT: $A-(B \bigcup C) = (A-B) \bigcap (A-C)$ (once again, note that for set theoretic relations with up to three sets, venn diagrams are helpful to visualize and guide the proofs..)
answer (c): let $x \in A$, but not in $B \bigcup C$. That is, x is an element of A, but not of B, not of C, or not both of B and C. That is, x is an element of A not of B, and x is an element of A not of C. That is, $x \in (A-B) \bigcap (A-C)$. Reversing the arguments, we get the other subset relationship. Hence, the two sets are equal. Hence, $A - (B \bigcup C) = (A-B) \bigcap (B-C)$. QED.
Problem 3:
The symmetric difference of two sets A and B, denoted by $A \triangle B$, is defined by $A \triangle B = (A-B) \bigcup (B-A)$; it is the union of the difference of two sets in opposite orders. Prove the following:
3(a) Symmetric difference is associative: $A \triangle (B \triangle C) = (A \triangle B) \triangle C$.
3(b): TPT: $A \triangle \phi = A$; and $A \triangle A = \phi$. (this is some sort of ‘existence of additive inverse of ‘in a symmetric relation’).
Answer 3b: both results are obvious from definition.
3(c): TPT: Symmetric difference is commutative operation: $A \triangle B = B\triangle A$.
answer 3c: By definition, LHS is $(A-B)\bigcup (B-A) = (B-A) \bigcup (A-B)$, which is RHS. Hence, done.
3(d): TPT: Some sort of distributive law: $A \bigcap (B \triangle C) = (A \bigcap B) \triangle (A \bigcap C)$.
answer 3d: Let $x \in A$, and also simultaneously $x \in (B \triangle C)$. That is, $x \in A$, and $x \notin (B \bigcap C)$. That is, $x \in A, x \notin (A \bigcap B)\bigcap (B \bigcap C)$. That is, $x \in RHS$. Reversing the arguments, we get the other subset relation. Hence, the two sets are equal. Hence, we can say “intersection distributes over symmetric difference.” QED.
Problem 4:
A ring of sets is a non-empty class A of sets such that if A and B are in A then $A \triangle B$ is in A and $A \bigcap B$ is also in A. Show that A must contain the empty set, $A \bigcup B$ and $A-B$. Also, show that if a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets. Also, prove that a Boolean algebra of sets is a ring of sets.
Answer 4(i): TPT: A must also contain the empty set, the set $A \bigcup B$ and the set $A-B$. As $A \triangle B$ is in A, so also $A \triangle A = \phi \in A$; now, it is already given that the ring of sets is non-empty, so it contains a non empty universal set also, call it U. We know from problem 2 in this blog that $A - B = A \bigcap B^{'}$. As U exists, so does $B^{'}$ by definition and so $A-B$ is in A. Also, as $A-B$ exists, $A-B^{'}$ exists in A, but $A-B^{'}=A\bigcap B$ also exists in A. QED.
Answer 4(ii): TPT: If a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets.
Let A be any non-empty class of sets. Clearly, it therefore has a non empty universal set. As $\phi \in U$, so also $\phi$ is contained in A. Hence, the complement of a set A is also in A. Now, $A \bigcup B \in A$, so $U - (A \bigcup B)$ is in A, that is, by De Morgan law, $B^{'} \bigcap A^{'}$ is in A. That is, again, $A \bigcap B$ is in A. Now, since $A-B$ is in A, hence, $B-A$ is in A, and since $A \bigcup B$ is in A, then $(A-B) \bigcup (B-A)$ is in A also. Hence, A is a ring of sets.
4(iii) TPT: Show that a boolean algebra of sets is a ring of sets.
Answer 4(iii); Firstly, as a boolean algebra of sets is non-empty, and it also contains the empty set and the Universal set. As A is in A implies that $A^{'}$ is in A, so $A \bigcup B^{'}$ is also in A (by definition of Boolean algebra), but this is precisely that $A-B$ is in A. Now, as $A \triangle B = (A-B) \bigcup (B-A)$ clearly is also in A. Hence, a Boolean algebra of sets is a ring of sets.
Problem 5:
Show that the class of all finite subsets (including the empty set) of an infinite set is a ring of sets but is not a Boolean algebra.
In problem 4 above, we showed that if a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets. The given class is a subclass of all an infinite set such that it contains finite subsets only; that is, again, we know that finite unions of finite subsets and finite intersections of finite subsets also lie in that subclass, but the complements might be infinite sets so that it need not always be a Boolean algebra of sets.
Problem 6:
Show that the class of all finite unions of closed-open intervals on the real line is a ring of sets but is not a Boolean algebra of sets.
Same reasoning as above tells us that it is indeed a ring of sets; but consider for example the closed-open interval $[a,b)$, the complement of this is not in that class as it is an infinite interval so that it is not a Boolean algebra of sets.
Problem 7:
Assuming that the Universal set U is non empty, show that Boolean algebras of sets can be described as rings of sets which contain U.
As $U \neq \phi$, by definition of Boolean algebras, $U^{'}$ is in this class of sets, so that the empty set is also a part of this class of sets. Again, by definition of Boolean algebras, the union of two sets of this class is also in this set so also the intersection of any two sets of this class is also in this class. So, for any two sets A and B, $A-B$ is also in this set (as complement of B exists), and so also, $A \triangle B = (A-B) \bigcup (B-A)$ is also in this set. In other words, this is also a ring of sets.
Regards,
Nalin Pithwa
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# coordinate geomoetry
Find the midpoint of PQ if P(-7, 11) and Q(1, -4).
Find the area of the rectangle with vertices B(8, 0), T(2, -9), R(-1, -7) and C(5, 2).
Find the area of a triangle with vertices (0, 0), E(3, 1) and F(-2, 6).
Given a rhombus with points E(-4, 1), F(2, 3), G(4, 9) and H(-2, 7), find the slopes of the diagonals.
A line passes through points (-2, -1) and (4, 3). Where does the line intersect the x-axis?
I will be happy to critique your thinking and work on this. I wont do if for you.
Send you the answerx?? Who exactly is getting the education here? :)
Give us some idea where the difficulty is and what you've tried. We'll give advice then. K?
1. 👍 0
2. 👎 0
3. 👁 71
1. Under root (X2-X1) the whole square +(Y2-Y1) the whole square.
Under root (1-(-7))whole square + (-4-11)whole square.
Under root (1+7)whole square + (-15)whole square.
Under root 8 square + 225
Under root 64 + 225
Under root 289
square root of17 is 289
SO ROOT OF 289 IS 17
1. 👍 0
2. 👎 0
2. 1/2Ix1(y2-y3)+x2(y3-y1)+x3(y1-y2)I
1/2I0(1-6)+3(6-0)+(-2)(0-1)I
1/2I0+18-2I
1/2I16I
1/2X16
1. 👍 0
2. 👎 0
3. I dont know so why wont you tell us the answer
1. 👍 0
2. 👎 0
posted by Mulan
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calculus high school
posted by .
7y^4+x^3y+x=4
using implicit differentation, how do i solve this problem. it is a textbook example but the partwhere i "use the chain rule on the first term" doesn't make sense to me because they get
d/dx 7y^4+(x^3dy/dx +y d/dx x^3)+d/dx x
• calculus high school -
I get
28y^3(dy/dx) + x^3(dy/dx) + y(3x^2) + 1 = 0
dy/dx(28y^3 + x^3) = - 1 - 3(x^2)(y)
dy/dx = (- 1 - 3(x^2)(y))/(28y^3 + x^3)
I used he chain rule on the first term, the product rule on the second term, the others were routine
• calculus high school -
thanks. you got the answer right. but i don't understand how the chain rule is used iin this problem.
• calculus high school -
in 7y^4, according to "chain rule"
exponent times front coefficient ---> 4(7) = 28
keep the base, reduce the exponent by 1 ----y^4 becomes y^3
times the derivative of the base, the base is y, so derivative of y is called dy/dx
result ---> 28y^3(dy/dx)
• calculus high school -
isn't that the power rule?
anyway, ow do you derive the second term?
• calculus high school -
never mind, the second term is derived using the product rule
• calculus high school -
Chain Rule with single variable:
d/dx [3(2x-4)^2]
=6(x-4)(2)=12x-48
Chain Rule with multiple variables:
d/dx [3y^2]
=6y(dy/dx)
The y and the (x-4) terms serve a similar function. You technically use the chain rule all the time. The derivative of a variable, like x, is simply 1. So taking the derivative of 3x^2 is really easy.
d/dx [3x^2]
=6x(1)
But (2x-4) and y aren't the same as x, and neither are the derivatives. So the chain rule will produce something appreciable. The derivative of 2x-4 is 2, so it affects the expression (whereas the derivative of x is simply one, and multiplying by one changes nothing). Y is also not equal to x. Because we are differentiating with respect to x, d/dx of y cant be 1. And we don't know anything else about y, so all we can do is say that dy/dx equals... well dy/dx.
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# NumPy Kronecker Delta | What is NumPy.kron()?
Hello geeks and welcome in this article, we will cover the NumPy Kronecker delta function. Along with that, for an overall better understanding, we will also look at its syntax and parameter. Then we will see the application of all the theory part through a couple of examples. While writing our program, we represent our function as NumPy.kron(). But at first, let us try to understand what does the Kronecker means in general. We can understand the Kronecker function as the operation on the 2 matrices of arbitrary sizes resulting in a block matrix in mathematical terms. To denote this operation, the symbol “⊗” is used. So we can conclude that NumPy Kronecker delta or NumPy.kron() helps us by finding the Kronecker product of 2 Input arrays. Next, we will look at the syntax associated with the function.
Contents
## SYNTAX OF NUMPY KRONECKER DELTA
``numpy.kron`(a, b)`
Here we can see that the function has a straightforward syntax and has only 2 parameters. In the next section, we will cover its parameter as well as the return type.
## PARAMETER OF NUMPY KRONECKER DELTA
a,b: array_like
This parameter represents the 2 input arrays of which the Kronecker product needs to be calculated.
## RETURN
out: ndarray
The function returns a ndarray representing the Kronecker Product of our input.
## EXAMPLES
We have covered all the necessary theory associated with our function. Also, we have understood what Kronecker product means in general. In this section, we will look at various examples and understand how the function works. We will start with an elementary level example and gradually move our way to more complicated examples.
```#input
import numpy as ppool
a=[1,23,4]
b=[2,45,5]
print(ppool.kron(a,b))
```
Output:
``[ 2 45 5 46 1035 115 8 180 20]``
Before moving ahead with the explanation of the above example, I would like to make a few things clear. Kronecker multiplication is totally different from when compared to matrix multiplication in general. Now, let’s get back to the example. Here we have at first imported the NumPy module. Then we have defined 2 arrays of which we wish to get the Kronecker product. In the next step, we have used the print statement along with our syntax. Here our output justifies the input properly. We can understand it in this way, the general Kronecker product representation is [a11 b11, a11 b12,……… so on] and we compare it to our input the answer is justified.
From the above example, it is quite evident how it is different from the original matrix product. As for matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. But as in the above example, we can see that we performed operation for two 1*3 matrix. One difference that we can spot is the general matrix multiplication representation is [a11b11+a12b21 + a13b31 + ……… so on].
Now as we are done with this let us look at one more example.
```#input
import numpy as ppool
a=[[1,23,4],
[20,45,9],
[34,8,6]]
b=[[2,45,5],
[5,8,7]]
print(ppool.kron(a,b))
```
Output:
``````[[ 2 45 5 46 1035 115 8 180 20]
[ 5 8 7 115 184 161 20 32 28]
[ 40 900 100 90 2025 225 18 405 45]
[ 100 160 140 225 360 315 45 72 63]
[ 68 1530 170 16 360 40 12 270 30]
[ 170 272 238 40 64 56 30 48 42]]``````
Here we can look at another example. In this particular case, we have followed all the steps similar to that of the first example. But instead of having a 1-d array as that of the 1st example, we went for two different 2-d arrays. The output here justifies our input. Just Imagine doing such humungous calculations by hand.
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Q:
# What are common wide-flange beam sizes?
A:
Some common American wide-flange beam sizes are W 27 x 178, W 24 x 162, W 21 x 147 and W 18 X 119. Standardized methods for sizes specify the dimensions through two measurements: the depth of the beam and the amount of weight in pounds per feet. Some smaller common wide-flange beam sizes are W 6 x 25, W 5 x 19 and W 4 x 13.
## Keep Learning
The depth size in such beams could also be termed as height, and standard depth sizes could also have varying width sizes and pound-per-feet ratios. Web thicknesses, which is the thickness of the stem size in the typical I-shaped beam, may also vary for standard depth sizes. For example, the W 27 X 178 size has a width of 14.09 inches, a web thickness of 0.725 inches and a pound-to-foot ratio of 178, but the W 27 x 84 has a width of only 9.93 inches, a web thickness of 0.460 inches and a pound-to-foot ratio of 84.
Some depth measurements on the standard beam size chart have more variations than others when it comes to pound-to-foot ratios. For example, the standard beam size with a depth size of 12 inches has 21 variations, and the standard beam size with a depth size of 5 inches has only two variations.
Sources:
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# FormulaF - Math 421:01 and 02 Spring 2008 Final Exam Review...
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Math 421:01 and 02 — Spring 2008 Final Exam Review and Formula Sheet Trigonometry The formula e it D cos t C i sin t (with i 2 D 1 ) connects exponential and trigonometric functions. Consequences include the basic identities cos C ˇ/ D cos ˛ cos ˇ sin ˛ sin ˇ sin C ˇ/ D sin ˛ cos ˇ C cos ˛ sin ˇ that are used to prove the orthogonality of the functions f 1; . cos nx; sin nx W n D 1; 2; : : :/ g on the interval < x < . Laplace transforms The Laplace transform of a function of t is a function of a new variable s defined by ˇ f f .t/ g D Z 1 0 f .t/e st dt This definition shows that the Laplace transform is a linear operator . In describing properties of the Laplace transform, it is conventional to write F.s/ for ˇ f f .t/ g . Some useful properties (in addition to linearity) are f .t/ F.s/ 1 1 s t n 1 s n C 1 cos t s s 2 C 1 sin t 1 s 2 C 1 f 0 .t/ sF.s/ f .0/ f .bt/ 1 b F .s=b/ e at f .t/ F.s a/ tf .t/ F 0 .s/ f .t a/ U .t a/ e as F.s/ ı.t a/ e as .f g/.t/ F.s/G.s/ In the last formula, .f g/.t/ denotes the convolution of f .t/ and g.t/ which is defined by .f g/.t/ D Z t 0 f . /g.t / d
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Math 421:01 and 02, May 13-14, 2008, p. 2 Calculation of Laplace transforms or inverse transforms typically requires several steps taken from this table. To apply these rules, functions must be found so that the given expression has the form shown in a line of this table. Ordinary Differential Equations There are two families of differential equations that are easy to solve exactly . ) Linear equations with constant coefficients . Usually, the solutions have the form y D e cx . . ) Euler equations , whose terms are combinations with constant coefficients of x k d k y=dx k for several integer k . Usually, the solutions have the form y D x c . To solve these equations, try the correct form of solution. The equation reduces to an algebraic equation in c . If a root c is complex, the exponential is usually converted to a form containing trigonometric functions. For Euler equations, this also requires the identity x c D e c ln x . If the algebraic equation has repeated roots , this method does not give enough solutions. If c is a repeated root, a second solution is xe cx in the constant coefficient case, or x c ln x in the Euler equation case. A full characterization of solutions will not be given here, since it will be easy to check any tentative solution constructed from this principle. Fourier series The Fourier series of a function f .x/ on the interval p < x < p is given by a 0 2 C 1 X n D 1 a n cos n x p C b n sin n x p where a 0 D 1 p Z p p f .x/ dx a n D 1 p Z p p f .x/ cos n x p dx b n D 1 p Z p p f .x/ sin n x p dx . / The constant term a 0 =2 is the average value of f .x/ , and Z p p cos 2 n x p dx D p Z p p sin 2 n x p dx D p so that the Fourier series of f .x/ D 1 , f .x/ D cos n x=p or f .x/ D sin n x=p is a single term identical to f .x/ .
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# Distance between Tianshui (THQ) and Tunxi (TXN)
Flight distance from Tianshui to Tunxi (Tianshui Maijishan Airport – Huangshan Tunxi International Airport) is 798 miles / 1285 kilometers / 694 nautical miles. Estimated flight time is 2 hours 0 minutes.
Driving distance from Tianshui (THQ) to Tunxi (TXN) is 936 miles / 1506 kilometers and travel time by car is about 16 hours 28 minutes.
## Map of flight path and driving directions from Tianshui to Tunxi.
Shortest flight path between Tianshui Maijishan Airport (THQ) and Huangshan Tunxi International Airport (TXN).
## How far is Tunxi from Tianshui?
There are several ways to calculate distances between Tianshui and Tunxi. Here are two common methods:
Vincenty's formula (applied above)
• 798.355 miles
• 1284.828 kilometers
• 693.752 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 797.376 miles
• 1283.252 kilometers
• 692.900 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Tianshui Maijishan Airport
City: Tianshui
Country: China
IATA Code: THQ
ICAO Code: ZLTS
Coordinates: 34°33′33″N, 105°51′36″E
B Huangshan Tunxi International Airport
City: Tunxi
Country: China
IATA Code: TXN
ICAO Code: ZSTX
Coordinates: 29°43′59″N, 118°15′21″E
## Time difference and current local times
There is no time difference between Tianshui and Tunxi.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 135 kg (297 pounds).
## Frequent Flyer Miles Calculator
Tianshui (THQ) → Tunxi (TXN).
Distance:
798
Elite level bonus:
0
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0
### In total
Total frequent flyer miles:
798
Round trip?
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# How many integer points lie between points A and B on the
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How many integer points lie between points A and B on the [#permalink] 03 Nov 2007, 17:48
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How many integer points lie between points A and B on the line segment AB, if A is (5, 7) and B is (10, -3)?
(A) 4
(B) 5
(C) 6
(D) 10
(E) 15
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Ravshonbek wrote:
How many integer points lie between points A and B on the line segment AB, if A is (5, 7) and B is (10, -3)?
(A) 4
(B) 5
(C) 6
(D) 10
(E) 15
I would say 15.
Along y axis 7+3=10 distinct integer points example 7,6,5,4,...0,-1,-2,-3
Along x axis 10-5=5 distinct points exmaple 5,6,7,8,9,10
Total 15 distinct x,y points
Anser E
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Hi,
The equation of the line is y=-2*x+17 substituting for x= 6,7,8 and 9 we get values for y 5,3,1 and -2 then the ans to the question is 4
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Get the equation of the line which pass through the points A and B....
we get 2x+y=17 now start substituting x and y values where the x lies between +5 and +10.........5<x<10 and find all the possible integer solutions of y....similarly find all possible integral solutions of x where y ranges from +7 and -3...i.e 7<y<-3.......
so we have 4 such possible integral solutions....
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# How many integer points lie between points A and B on the
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# How many consecutive positive integers can you make using exactly four instances of the digit '4'?
Starting at 1 (which is 4 - $\sqrt4$ - 4/4), how many consecutive integers can you make using exactly four instances of the digit '4'?
Basic rules:
• Any operator symbol is "free".
• Any printed '4' counts toward your four '4's.
• Other digits are NOT allowed.
• Other characters like letters or miscellaneous punctuation are NOT allowed, unless you can provide some citation of an accepted mathematical definition.
Allowed operators (non-exhaustive):
• - Subtraction or Negation
• * Multiplication
• / Division
• $\sqrt4$ Square root (ignore the implicit '2' there)
• $\sqrt[4]4$ Radical (in this case you've used two '4's)
• ^ Exponentiation
• ! Factorial
• ? Terminal function (4? = 4 + 3 + 2 + 1)
• 44 Concatenation (which in this case consumes two '4's)
• |4| Absolute value
• . Decimal point
• If you can find a way to use calc, trig, matrices, whatever, by all means please do
-
Last time I tried this I allowed the ̇ operator e.g. 4/.4̇ is 36. – Neil Feb 26 at 19:56
@question_asker Yes, I typoed - it should say 4 * 4 / .4̇. – Neil Feb 26 at 22:16
If you allow any other symbol with an accepted mathematical definition, then doesn't that mean we have free and unlimited use of $-e^{i \pi}$, $\ln e$, $-\cos \pi$, and other such digitless ways of constructing 1? It seems like we don't even need the abstruse math of some of the answers below. – ruakh Feb 29 at 4:53
@ruakh I assume that we have access to any mathematical definition but not to any mathematical constant. So, no $\pi$, no $e$, no $i$. – MariusSiuram Feb 29 at 8:41
All of them!
How?
For every positive integer $n$,$$\underbrace{\sec\arctan}_{n^2-1\text{ times}}\,\frac{44}{44}=n$$otherwise written as$$\sec\arctan(\sec\arctan(...\sec\arctan(\frac{44}{44})))$$so all positive integers can be made with four fours. (Idea from this answer.)
-
There goes a whole category of puzzles :). +1 – Lawrence Feb 26 at 4:34
@question_asker That's shorthand for writing out $\sec\arctan$ that many times. For example, when $n=2$ the expression is $\sec\arctan\sec\arctan\sec\arctan\frac{44}{44}$. – f'' Feb 26 at 13:27
@StephanBijzitter There wouldn't be a fifth 4 used. When n = 4, the expression has 15 occurrences of sec arctan: sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan sec arctan$\frac{44}{44}$ – Brian J Feb 26 at 14:12
@StephanBijzitter - You never write 4^2-1. f"s answer only gives you the "pseudo-code" for how to write it. You would actually write sec arctan 15 times, followed by 44/44, i.e. sec arctan sec arctan sec arctan (...11 more times...) sec arctan 44/44 – Sphinxxx Feb 26 at 15:28
@Stephan: His answer is instructions for what to write down, not what you should actually write down. Sure, you use $n$ to figure out how many times you should write $\sec\arctan$, but that doesn't mean you ever write down $n$. – Deusovi Feb 26 at 18:04
Well, for someone who's not a mathematics afficionado, the last bullet point (and its applications) seems like alien talk. So here's a list I've compiled which went until 40.
1 = 4-sqrt(4)-4/4
2 = 4-4+4-sqrt(4)
3 = (4+4+4)/4
4 = 4-sqrt(4)-sqrt(4)+4
5 = 4+sqrt(4)-4/4
6 = 4+sqrt(4)-4+4
7 = 4+sqrt(4)+4/4
8 = 4+4+4-4
9 = 4+4+4/4
10 = 4+4+4-sqrt(4)
11 = 4/.4+4/4
12 = 4+4+sqrt(4)+sqrt(4)
13 = 4?+4-4/4
14 = 4?+4+4-4
15 = 4?+4+4/4
16 = 4*4-4+4
17 = 4*4-4/4
18 = 4*4-4/sqrt(4)
19 = 4?+4?-4/4
20 = 4?+4?-4+4
21 = 4?+4?+4/4
22 = 4?+4?+4/sqrt(4)
23 = 4!-sqrt(4)+4/4
24 = 4*4+4+4
25 = 4!+sqrt(4)-4/4
26 = 4!+sqrt(4)-4+4
27 = 4!+sqrt(4)+4/4
28 = 4!+4-4+4
29 = 4!+4-4/4
30 = 4!+4+4/sqrt(4)
31 = 4??-4!+4-4
32 = 4!+4+sqrt(4)+sqrt(4)
33 = 4??-4!+4-sqrt(4)
34 = 4*4*sqrt(4)+sqrt(4)
35 = 4??-4!+sqrt(4)-sqrt(4)
36 = 4!+4+4+4
37 = 4??-4!+4+sqrt(4)
38 = 4!+4*4-sqrt(4)
39 = 4??-4?-4-sqrt(4)
40 = 4!+4*sqrt(sqrt(4^4))
I'm not taking away from the brilliance of the top voted answer, just thought someone would enjoy going about it in this way.
I'm not certain if 41 can (or cannot) be done excluding the functions that fall under the last bullet point. I will turn this into community wiki if people can contribute to extend this list.
Check out MariusSiuram's answer (and its edit history) for a longer answer and an approach to extend this list. I've decided to take an early retirement at 40. ;)
-
I don't know what the ? operator is, but it seems that 4? is equal to 10. If this is true, then (4?) * 4 + 4/4 would make 41. – Svalorzen Feb 26 at 14:29
@Svalorzen if that is accepted, then (4?) * 4.4 - sqrt(4) would make 42. – Piotr Pytlik Feb 26 at 14:40
I compiled almost the same list without seeing your answer. I think that I forgot to refresh the page, silly me. Well, it was a nice ride :) – MariusSiuram Feb 26 at 15:01
## Prefix edit
Why not allow the double factorial? well, let's use it. For the record:
\begin{align} 0!! &= 1 \\ 5!! &= 3 \cdot 5 = 15 \\ 6!! &= 2 \cdot 4 \cdot 6 = 48 \\ \end{align}
Also the choose operator (also known as binomial coefficient).
# Modular approach
I had some fun doing it the long way, but then I decided to jump into a more modular / exploitable strategy. I will build it for a little bit, so bear with me for now :)
First, let's make the following list with one four:
\begin{align} 2 &= \sqrt 4 \\ 3 &= \left(\sqrt 4\right)? \\ 4 &= 4 \\ 6 &= \left(\sqrt 4\right)?! \\ 8 &= 4!! \\ 10 &= 4? \\ 21 &= \left(\sqrt 4\right)?!? \\ 24 &= 4! \\ 36 &= \left(4!!\right)? \\ 48 &= \left(\left(\sqrt 4\right)?!\right)!! = 6!!\\ 55 &= 4?? \end{align}
I can also consider $4 = \sqrt 4 + \sqrt 4$, so there is no need to consider the "extra fours".
From now on, I will use the fancy "one-four" substitution, and maybe some results will use less than four fours. But the translation from a compact equation to a "four-fours" equation is immediate.
Let's consider the following list of two-four numbers (I purposely omit the ones that can be obtained with a signel four):
\begin{align} 0 &= 4 - 4 \\ 1 &= \frac{4}{4} \\ 5 &= 2 + 3 \\ 7 &= 4 + 3 \\ 9 &= 3 + 6 \\ 11 &= 21 - 10 \\ 12 &= 10 + 2 \\ 13 &= 10 + 3 \\ 14 &= 10 + 4 \\ 15 &= 21 - 6 \\ 16 &= 4 \cdot 4 \\ 17 &= 21 - 4 \\ 18 &= 21 - 3 \\ 19 &= 21 - 2 \\ 20 &= 24 - 4 \\ 22 &= 24 - 2 \\ 23 &= 21 + 2 \\ 25 &= 21 + 4 \\ 26 &= 24 + 2 \\ 27 &= 24 + 3 \\ 28 &= 24 + 4 \end{align}
Up to this point we can get any integer below 28 with only two fours. My strategy will be obtaining numbers by combining the "high part" and "low part". So, given any two-four "high number" $n$ we can generate all integers between $n - 28$ and $n + 28$. And the resulting formula uses up to four fours.
We can trivially consider the following property:
$$\forall n < 55: \quad (n+1)? - n? < 56$$
So we can have a "dense-enough" set of high numbers by simply using the $?$ operator to the list of "two-four numbers". The maximum number at the moment is:
$$434 = 28? + 28 = (4! + 4)? + 4! + 4$$
To continue, we should pick integers with a maximum distance of 56 between them. The next integer should be at most 463, because $462 - 28 = 434$.
Next hand-picked "high numbers" (credit to @f'' for most of them!): \begin{align} 441 &= 21^2 \\ 465 &= (24 + 6)? \\ 504 &= 21*24 \\ 550 &= 55*10 \\ 600 &= 24?+24? \\ 630 &= \binom{36}{2} \\ 665 &= 6! - 55 \\ 720 &= 6! \\ 775 &= 6! + 55 \end{align}
My original post contained a exhaustive list up to 132 and some odd holes up to 148, holes that user @f fixed in the comments, so credit for him for that. But now that I present the alternative strategy, the original post seems overweighted and slow to load :(
-
I haven't found any 2-four numbers between 600 and 665, but the gap (629 to 636) can be filled in other ways: $(\frac{10}{2})^4$ uses three fours to make 625, and the fourth one can add 4 (629), 6 (631), or 10 (635). Similarly, $21*3*10$ is three fours to make 630, and the fourth one can add 2 (632), 3 (633), 4 (634), or 6 (636). – f'' Feb 28 at 17:01
I'm not going to do this, because last time I tried it (about 40 years ago) I was still growing the list after three weeks. But I do remember making great use of the fact that dividing by the square root of point four recurring is equivalent to multiplying by 1.5. – Michael Kay Feb 28 at 18:45
Solution for any odd number of fours different from one:
$$\underbrace{\sec\arctan}_{(n+4)^2-1\text{ times}}\,\frac{44 \cdots}{44 \cdots} - 4=n$$
-
This formula will make any positive integer $n$ from four fours:
$-\sqrt4\frac{\ln\left[\left(\ln\underbrace{\sqrt{\sqrt{\cdots\sqrt4}}}_{n}\right) / \ln4\right]}{\ln{4}}$.
If we allow the number $44$ to count as two fours, then we can also have:
$\underbrace{\sec\arctan\cdots\sec\arctan}_{n^2-1}\,\frac{44}{44}$.
-
For $n=4$ your square root version is 4 for 4: 4 4s, 4 logs, 4 parentheses, 4 nested roots – humn Feb 26 at 19:32
@f'' ‘s solution can be generalized for any positive even number of 4s.
But what about odd numbers of 4s?
One 4:
$$\begin{matrix} \underbrace{ \sec\arctan }_{ n^2-1 ~ \text{times} } \, \biggl( \cdots \root\of{\root\of{ \surd 4 }\,} \, \biggr) \end{matrix} = n$$
Three 4s to get 4 (or whatever number of nested √ s):
$${ \ln \ln 4 - \ln \ln \root\of{\root\of{\root\of{ \surd 4 }\,}\,} \over \ln \root\of 4 } = 4$$
-
4/(sqrt(4)sqrt(4)) - Three 4's to get 1, leaving you with an even number of fours left. Multiply the answer from f''. – Taemyr Feb 26 at 9:07
"Run-time error 14: out of string space" is not an answer. – immibis Feb 26 at 12:35
Okay, deleted the nonsolution with infinitely many 4s. The point was that, even though infinitely many such solutions exist, no such answer can actually be written down. – humn Feb 26 at 18:35
All of them. Using S(n) the successor function used in the Peano axioms to define all natural numbers. And is equivalent to S(n) = n+1
1 = 44/44
2 = S(44/44)
3 = S( S( 44/44 ) )
n+1 = S( n )
-
I applaud your minimalism! – JamesFaix Mar 1 at 12:42
In C, 1 = (44/44), 2 = (44/44)++, 3 = ((44/44)++)++... – JamesFaix Mar 1 at 12:46
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# Question
Experiment: Box 0 contains 2 colored balls and one white ball
Box 1 contains 1 color ball and 2 white balls
Given the color of the ball pick which box it came from.
X = color or white ball
Pr[B0] = 1/3 Pr[B1] = 2/3
Pr[white|B0] = 1/3 Pr[white|B1] = 2/3 Pr[color|B0] = 2/3 Pr[color|B2] = 1/3
MAP Rule: Pick the Box i (i = 0,1) with largest Pr[x|Bi]*Pr[Bi]
if x = white Pr[white|B0]*P[B0] = (1/3)*(1/3) = 1/9
Pr[white|B1]*P[B1] = (2/3)*(2/3) = 4/9
if x = color Pr[color|B0]*P[B0] = (2/3)*(1/3) = 2/9
Pr[color|B1]*P[B1] = (1/3)*(2/3) = 2/9
If x = white then pick Box 1 If x = color then pick either box
## Alumni Liaison
Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.
Dr. Paul Garrett
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# Complete the following sentence: Density of water is______. At 40C. - Physics
Fill in the Blanks
Complete the following sentence:
The density of water is______. At 4°C.
#### Solution
The density of water is Maximum At 4°C.
Concept: Reactivity of Metals with Cold Water, Hot Water and Steam (With Products Formed).
Is there an error in this question or solution?
#### APPEARS IN
Frank Class 9 Physics ICSE
Chapter 5 Heat
Thermal Expansion | Q 12.1 | Page 203
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# 7 Helpful Posts for Teaching Multiplication
Teaching multiplication can take many forms. Teachers are always finding new and engaging ways of teaching multiplication. Whether it be using manipulatives, arrays, songs, tips or strategies, there is no one way of teaching multiplication.
In this roundup post, I’ve highlighted 7 helpful posts to inform your teaching of multiplication. Each article also highlights helpful resources (some paid, some free).
UPDATE: November 2020
STOP! I wrote another blog post with ideas for teaching multiplication virtually when we switched to distance learning. You can see it HERE!
## Teaching Multiplication with 3 Proven Strategies
In this post, I’ve written how I use 3 proven techniques for teaching multiplication as a concept: using equal groups, arrays, and a number line. I use a hands-on learning approach that focuses on using manipulatives to understand multiplication before any symbols or numbers are even introduced.
I don’t move on until I know each student has an understanding of the concept of multiplication.
## Multiplication Practice Made Fun and Easy
After students understand multiplication and you’ve been at it for a while, do you ever feel as a teacher that some students just have a very difficult time recalling the multiplication facts no matter what? Or they learn them and forget them?
In this post, I explain my frustration with this and what I did to fix it. Shameless promotion plug. Yes, I created a new resource that’s now on TpT. But it worked so well I had to share!
## Don’t Forget the Distributive Property of Multiplication!
If you want to get your students ready for higher levels of math (especially middle school algebra!), they need to learn how to use the Distributive Property of Multiplication. When California adopted the Common Core State Standards for Mathematics, I was genuinely baffled by how I was going to teach this standard to eight-year-olds.
Eventually, I did some research and some experimenting. Once my oldest son hit middle school, I saw the important role this particular property of multiplication played in algebra.
Don’t overlook this property. Teach it so that students understand that it is used to break down more complex numbers into easier units to solve an equation.
You can check out this post I wrote on how I break down the Distributive Property of Multiplication to my students.
## Teaching Multiplication – The Year Long Plan
In this post, I give a detailed explanation of how I approach multiplication all year long. You can just teach it and forget it. You’ve got to support students all year long.
The post details how I teach the concept of multiplication, the Distributive Property of Multiplication, the strategies and tips I give to learning the tables and finally, fluency strategies for math fact fluency.
It’s a long post but illustrates the need to support students all year long because understanding and using multiplication is more than just memorization!
## More Ideas from Other Teacher-Bloggers
We all know what happens to third graders when they come back from summer vacation and start fourth grade. Multiplication? What was that?
My colleague over at Enjoy Teaching with Brenda Kovich has a very informative post on keeping multiplication going in fourth grade. She has some great hints for learning the multiplication facts for fourth graders!
`|`
The Caffeine Queen Teacher has a developed a unique way to teach two digits by two digits multiplication using shapes. Read her blog post to see how she uses this visual process of multiplication with great success!
|
Over at Elementary Matters, read this post on using memorization techniques to learn math facts. She offers 10 brain tricks for learning math facts. Did you know that music helps the brain organize information during practice? Great tip!
|
## Looking For More Ideas for Multiplication?
Follow my Pinterest Board for Multiplication Strategies!
Subscribe to the newsletter to receive this FREE Guide to Achieving Multiplication Fluency. Get it now by signing up for my newsletter below!
I hope you enjoyed this roundup post.
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# What's wrong with Python code ?? Bhaskara [closed]
0
``````import math
a = int(input('Digite um valor para A '))
b = int(input('Digite um valor para B '))
c = int(input('Digute um valor para C '))
#(-b +- {-b² - 4 * a * c} ) / 2 * c
operacao = -b
raiz = (operacao * operacao) - 4 * a * c
raiz = math.sqrt(raiz)
divisao = 2 * c
# (operacao +- raiz) / divisao
x1 = (operacao + raiz) / divisao
x2 = (operacao - raiz) / divisao
print('X1 = {}'.format(x1))
print('X2 = {}'.format(x2))
if(__name__ in '__main__'):
``````
Consider the indented code correctly
asked by anonymous 07.07.2018 / 19:37
2
Your denominator is incorrect. The denominator in the Bháskara formula is `2*a` , not `2*c` :
``````# Errado: divisao = 2 * c
divisao = 2 * a # Certo
``````
In addition, you will have problems when the roots are imaginary, because `math.sqrt` only accepts 0 and positive numbers. I suggest you implement a check before attempting to apply `math.sqrt` to a possibly negative number, or treat the exception.
07.07.2018 / 19:47
2
You are changing the quadratic coefficient `a` by the constant coefficient `c` in the denominator of the formula:
Youarealsonotverifyingthattherootispartofthesetofrealnumbersbytestingwhetherthediscriminatorislessthan`0`.
``importmathdefBhaskara(a,b,c):delta=(b**2)-(4*a*c)if(delta<0):return(None,None)#RaizNegativax=math.sqrt(delta)x1=(-b+x)/(2*a)x2=(-b-x)/(2*a)return(x1,x2)if(__name__in'__main__'):a=int(input('DigiteumvalorparaA:'))b=int(input('DigiteumvalorparaB:'))c=int(input('DiguteumvalorparaC:'))x=Bhaskara(a,b,c)print('X1={}'.format(x[0]))print('X2={}'.format(x[1]))``
Test#1:
``DigiteumvalorparaA:2DigiteumvalorparaB:7DiguteumvalorparaC:3X1=-0.5X2=-3.0``
Test#2:
``````Digite um valor para A: -3
Digite um valor para B: 2
Digute um valor para C: -1
X1 = None
X2 = None
``````
08.07.2018 / 06:47
0
It is not necessary to use the math module because it is equivalent to `delta**(1/2)` , the code below solves the equation where `x1` and `x2` are complex containing the real and imaginary part ( if there is ) of the roots.
``````def Bhaskara():
a = int(input('Digite um valor para A '))
b = int(input('Digite um valor para B '))
c = int(input('Digute um valor para C '))
delta = (b ** 2) - (4 * a * c)
x1 = (-b + delta ** (1 / 2)) / (2 * a)
x2 = (-b - delta ** (1 / 2)) / (2 * a)
print('X1 = {}'.format(x1))
print('X2 = {}'.format(x2))
print(type(x1)) # Verificação do tipo da variável
if __name__ in '__main__':
``````
07.07.2018 / 21:20
Encoding problems ___ ___ erkimt The third parameter filter_input is mandatory in PHP? ______ qstntxt ___
%pre%
I took a look at the PHP manual but did not say anything, I saw a guy not using the third parameter and then I was in doubt if it is mandatory or not
______ azszpr313414 ___
%pre%
These %code% indicate optional arguments, in PHP there are predefined arguments, when you omit in your use php passes the default value, which in the specific case of this function would be %code% , the code you posted shows that the person has changed %code% by %code% , that is, each one is for one thing, just see the values supported in link
As it says in the doc:
If omitted, %code% will be used, which is equivalent to %code% . This will result in no filtering taking place by default.
If omitted, it will use %code% , which is equivalent to %code% . This will result in an unfiltered value.
Just to note, %code% is part of the link , this filter in the case escapes / converts characters such as: %code% , %code% , %code% , among others that the ASCII value is less than 32, ie something like:
%pre%
Will print this:
%pre%
What are HTML entities, which when rendered on the page actually display %code% , but without affecting HTML.
______ azszpr313406 ___
The documentation of the %code% function says the following about the third parameter:
"If omitted, FILTER_DEFAULT will be used, which is equivalent to FILTER_UNSAFE_RAW. This will result in no filtering taking place by default."
That is, the third parameter is not required and can be omitted. Although this does not filter the value in fact.
___
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# 13.10: Binary Numbers
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
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$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
The countdown example has three parts: (1) it checks the base case, (2) displays something, and (3) makes a recursive call. What do you think happens if you reverse steps 2 and 3, making the recursive call before displaying?
public static void countup(int n) {
if (n == 0) {
System.out.println("Blastoff!");
} else {
countup(n - 1);
System.out.println(n);
}
}
The stack diagram is the same as before, and the method is still called n times. But now the System.out.println happens just before each recursive call returns. As a result, it counts up instead of down:
Blastoff!
1
2
3
This behavior comes in handy when it is easier to compute results in reverse order. For example, to convert a decimal integer into its binary representation, you repeatedly divide the number by two:
23 / 2 is 11 remainder 1
11 / 2 is 5 remainder 1
5 / 2 is 2 remainder 1
2 / 2 is 1 remainder 0
1 / 2 is 0 remainder 1
Reading these remainders from bottom to top, 23 in binary is 10111. For more background about binary numbers, see http://www.mathsisfun.com/binary-number-system.html.
Here is a recursive method that displays the binary representation of any positive integer:
public static void displayBinary(int value) {
if (value > 0) {
displayBinary(value / 2);
System.out.print(value % 2);
}
}
If value is zero, displayBinary does nothing (that’s the base case). If the argument is positive, the method divides it by two and calls displayBinary recursively. When the recursive call returns, the method displays one digit of the result and returns (again).
The leftmost digit is at the bottom of the stack, so it gets displayed first. The rightmost digit, at the top of the stack, gets displayed last. After invoking displayBinary, we use println to complete the output.
displayBinary(23);
System.out.println();
// output is 10111
Learning to think recursively is an important aspect of learning to think like a computer scientist. Many algorithms can be written concisely with recursive methods that perform computations on the way down, on the way up, or both.
This page titled 13.10: Binary Numbers is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Allen B. Downey (Green Tea Press) .
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# Proposal: merge Data.Functor.Coproduct into transformers
Andreas Abel andreas.abel at ifi.lmu.de
Sun Dec 16 15:36:32 CET 2012
```Standing up against the dictator...
I like neither 'Product' nor 'Sum'. For one, they are ambiguous already
in type theory. They mean something different in the context of simple
types and the context of dependent types:
Sum Product
simple types set of tagged elements set of pairs
dependent types set of pairs set of functions
Further they have this number-connotation. Sum and Product are
mathematician's terms inspired by the cardinality of the sets they
construct (in the simply typed setting), or the likeness to summation
and product terms (in the dependently typed setting).
IMHO, programming is much more based on logics and information theory
than on mathematics and numbers. The logical content of 'Sum' is making
a decision (left or right) and the one of 'Product' is adjoining two
things, putting them next to each other. Disjunction and conjunction
would be the logically correct terms, but we can say it simpler. I
think the current Haskell approach to speak of 'Either' and pairing is
the good one, free of mathematical burden, and it can be extended to
higher-order kinds:
Either :: * -> * -> *
Either1 :: (* -> *) -> (* -> *) -> (* -> *)
Either2 :: (* -> * -> *) -> (* -> * -> *) -> (* -> * -> *)
(,) :: * -> * -> *
Pair1 :: (* -> *) -> (* -> *) -> (* -> *)
Pair2 :: (* -> * -> *) -> (* -> * -> *) -> (* -> * -> *)
Cheers,
Andreas
On 16.12.12 1:57 AM, Ross Paterson wrote:
> On Sat, Dec 15, 2012 at 09:24:48PM +0000, Henning Thielemann wrote:
>> Since I proposed that and became aware of the newtype solution in the
>> meantime, I change my mind to:
>>
>>> data Sum f g a = Sum { getSum :: Either (f a) (g a) }
>
> OK, let's do that. It matches the treatment of Product.
> (So now they'll both clash with Data.Monoid.)
>
> _______________________________________________
> Libraries mailing list
>
--
Andreas Abel <>< Du bist der geliebte Mensch.
Theoretical Computer Science, University of Munich
Oettingenstr. 67, D-80538 Munich, GERMANY
andreas.abel at ifi.lmu.de
http://www2.tcs.ifi.lmu.de/~abel/
```
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# What Is Math?
## By Wendy Petti
What is math? It might seem obvious: We hope we know what we're teaching -- and that our students know what they're learning! But responses to that question can be surprisingly diverse.
Before you read on, please take a few minutes to reflect and record your own thoughts on "What is math?"
Illustration courtesy of Wendy Petti.
I'm a math educator -- I teach grade 4 math, I've created a math Web site, and I write about topics in math education. Yet I cannot easily express what math is. I'm in good company. Bertrand Russell has quipped, "Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true." Math teacher Sanderson M. Smith observes that "students can gain a tremendous appreciation for mathematics if they understand that the question 'What is mathematics?' has been analyzed and debated since the time of the Pythagoreans, around 550 B.C. "
It is worth pursuing a clear understanding of the meaning and scope of mathematics so that we might provide our students with a richer learning experience and help them more fully appreciate the beauty and power of mathematics.
## STUDENTS AND TEACHERS REFLECT ON "WHAT IS MATH?"
In my math classes, no matter how much attention we pay to the five content standards and five process standards, many of my students seem to focus on number and operation when they share their thoughts and feelings about math. One of my students had an "aha!" moment the morning after a student-led math night: "I look at math a different way now. I've usually thought of math as something we need to do, to know how much money we need to pay, things like that. But this morning I woke up and said, 'Wow! Math is everywhere!'"
A 2005 Math Cats writing contest on "What is math?" produced these reflections:
• From an elementary school volunteer: "Math is more than a subject we learn in school. Math is every breath we take and every second of the day. From the moment we wake up in the morning, math is the core of everything we do..."
• From a first-grade teacher:
"Math is you.
Math is me.
Math is everything we see!
Infinity and beyond our wildest dreams
Math encompasses all extremes!"
• From a 13-year-old: "Math is the entire world simplified on a piece of paper... Math is ingeniousness morphed into a tiny simple formula so we can harness its fantastic powers."
• From a 12-year-old: "Math is the universal language of the world."
• From an 11-year-old: "No one can live without math; it means different things to different people. But to me it means love, liberty, learning. I could keep going but to sum it up, math is my life."
## WHAT DO THE EXPERTS SAY?
The Merriam-Webster dictionary defines mathematics as "the science of numbers and their operations, interrelations, combinations, generalizations, and abstractions and of space configurations and their structure, measurement, transformations, and generalizations."
Professor Robert H. Lewis writes: "Mathematics is not about answers, it's about processes." He observes that, far too often, the way in which students learn math is equivalent to building a scaffold without ever constructing the building that the scaffolding is intended to support. "The real 'building' in the mathematics sense is the true mathematical understanding, the true ability to think, perceive, and analyze mathematically."
Professor Dave Moursund suggests that mathematics can be divided into three overlapping categories:
• "mathematics as a human endeavor" -- the human development and use of mathematics in measurement, dance, music, and so on.
• "mathematics as an academic discipline" to be studied
• "mathematics as an interdisciplinary language and a tool"
He notes that most school mathematics focuses on the third category, but that we need to embrace all three. "Relatively few K-12 teachers study enough mathematics so that they understand and appreciate the breadth, depth, complexity, and beauty of the discipline."
Ask Dr. Math at the Math Forum has answered the question What is math? in various thought-provoking ways over the years. Here are a few snippets.
• "Stripped to its barest essence, mathematics is the derivation of theorems from axioms. ... Mathematics is a collection of extended, collaborative games of 'what if', played by mathematicians who make up sets of rules (axioms) and then explore the consequences (theorems) of following those rules."
• "To me, mathematics is a discipline that seeks understanding of the patterns and structures of constructs of the human mind. Understanding has no end to its depth, and mathematics seeks the highest standards of understanding by demanding rigor in its foundations and in its development..."
• "If you asked a dozen mathematicians... what math is, you would probably get a dozen answers...My own favorite definition is that math is the study of abstractions. That is, we isolate one or a few features of some kind of object for study, and see what we can learn about the behavior of those features while ignoring everything else about them: features like number, shape, or direction."
• "I would say math is both an art (a system of skills used to do calculations, devise proofs, and so on) and a science (a system of knowledge about numbers, shapes, and other abstract entities, and a way of gaining that knowledge). That is, it's both a way of doing (an art) and a way of learning (a science)."
In Mathematics: The Science of Pattern, Keith Devlin describes Egyptian and Babylonian mathematics up to 500 B.C. as "the study of number," and the era of Greek mathematics from 500 B.C. to 300 A.D. as "the study of number and shape." In the mid-1600's, with the invention of calculus by Newton and Leibniz, mathematics became "the study of number, shape, motion, change, and space." As interest grew in mathematics itself, "by the end of the nineteenth century, mathematics had become the study of number, shape, motion, change, and space, and of the mathematical tools that are used in this study." In the past century, mathematical activity and knowledge has increased more than a thousand fold. Devlin describes mathematics as the "science of patterns... patterns of counting, patterns of reasoning and communicating, patterns of motion and change, patterns of shape, patterns of symmetry and regularity, and patterns of position."
What does mathematics mean to you and your students? How do they respond to the experts' characterizations of mathematics? The discussion is sure to give you new insights into your students' thinking and to give them new insights into the meaning of mathematics.
## ADDITIONAL RESOURCES
Mathematics: The Most Misunderstood Subject, by Robert H. Lewis, July 2000.
What Is Mathematics?, by Dave Moursand
What Is Mathematics?, by Sanderson M. Smith
What Is Math? from The Math Forum - Ask Dr. Math:
What is math? Math Cats' writing contest winners
Mathematics: The Science of Patterns: The Search for Order in Life, Mind, and the Universe, by Keith Devlin, Scientific American Library, 1997.
The Math Process Standards Series, edited by Susan O'Connell. Heinemann, 2007-2008.
In each grade band, five books promote mathematical thinking through a focus on one of the five process standards (problem-solving, communication, reasoning and proof, representation, and connections); each book comes with a CD of editable resource materials.
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# Making RigidBody face direction for movement
so i have a gameobject with a rigidbody attached and it currently moves fine with the code i have, the problem is that when it moves from input currently it doesnt face the direction its moving, i tried to fix this by adding some rotation code but this messes up the movement as it turns does stuff like the player moving forward whilst facing the opposite way, so my question is how do i make the gameobject face the way its going so that what ever direction is pressed the player rotates to that direction then goes forward in that direction, if that makes sense.
heres my code below with the non working rotation code included for abit of context
``````void FixedUpdate()
{
myMove = new Vector3(ControllerInputs.controllerInputs.LeftLR, 0,
ControllerInputs.controllerInputs.LeftUD);
if (grounded)
{
Quaternion rotation = Quaternion.LookRotation(myMove, Vector3.up);
transform.rotation = rotation;
targetVelocity = myMove;
targetVelocity = transform.TransformDirection (targetVelocity);
targetVelocity *= speed;
Vector3 velocity = rigidbody.velocity;
Vector3 velocityChange = (targetVelocity - velocity);
velocityChange.x = Mathf.Clamp (velocityChange.x, -maxVelocityChange,
maxVelocityChange);
velocityChange.z = Mathf.Clamp (velocityChange.z, -maxVelocityChange,
maxVelocityChange);
velocityChange.y = 0;
if (canJump)
{
if(Input.GetButtonDown ("Jump"))
{
rigidbody.velocity = new Vector3 (velocity.x, CalculateJumpVerticalSpeed (), velocity.z);
}
rigidbody.AddForce (new Vector3 (0, -gravity * rigidbody.mass, 0));
grounded = false;
}
void OnCollisionStay(Collision col)
{
if(col.transform.name == "Plane")
grounded = true;
}
float CalculateJumpVerticalSpeed()
{
return Mathf.Sqrt (2 * jumpHeight * gravity);
}
``````
after some trial and error i have the code doing what i wanted, i deleted the line changing target velocity into local coordinates and this seemed to fix the problem.
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# Sufficient statistic, specifics/intuition problems
I'm teaching myself some statistics for fun and I have some confusion regarding sufficient statistics. I'll write out my confusions in list format:
1. If a distribution has $$n$$ parameters then will it have $$n$$ sufficient statistics?
2. Is there any sort of direct correspondence between the sufficient statistics and the parameters? Or do the sufficient statistics just serve as a pool of "information" so that we can recreate the setting so we can calculate the same estimates for the parameters of the underlying distribution.
3. Do all distributions have sufficient statistics? ie. can the factorization theorem ever fail?
4. Using our sample of data, we assume a distribution that the data is most likely to be from and then can calculate estimates (e.g. the MLE) for the parameters for the distribution. Sufficient statistics are a way to be able to calculate the same estimates for the parameters without having to rely on the data itself, right?
5. Will all sets of sufficient statistics have a minimal sufficient statistic?
This is the material which I am using to try to understand the topic matter here.
From what I understand we have a factorization theorem which separates the joint distribution into two functions, but I do not understand how we are able to extract the sufficient statistic after factorizing the distribution into our functions.
1. The Poisson question given in this example had a clear factorization, but then it was stated that the sufficient statistics were the sample mean and the sample sum. How did we know that those were the sufficient statistics just by looking at the form of the first equation?
2. How is it possible to conduct the same MLE estimates using sufficient statistics if the second equation of the factorization result will sometimes depend on the data values $$X_i$$ themselves? For instance in the Poisson case the second function depended on the inverse of the product of the factorials of the data, and we would no longer have the data!
3. Why would the sample size $$n$$ not be a sufficient statistic, in relation to the Poisson example on the webpage? We would require $$n$$ to reconstruct certain parts of the first function so why is it not a sufficient statistic as well?
• Just a quick clarifying question - which "angle" are you coming at sufficiency from? Maximum likelihood? Bayesian? Maximum entropy? Sampling theory? Something else? Commented Jan 12, 2014 at 13:29
• I was coming from a standpoint of MLE, sorry if my posting wasn't the greatest, it's my first post on this forum! Commented Jan 13, 2014 at 6:56
You'd probably benefit from reading about sufficiency in any textbook on theoretical statistics, where most of these questions will be covered in detail. Briefly ...
1. Not necessarily. Those are special cases: of distributions where the support (the range of values the data can take) doesn't depend on the unknown parameter(s), only those in the exponential family have a sufficent statistic of the same dimensionality as the number of parameters. So for estimating the shape & scale of a Weibull distribution or the location & scale of a logistic distribution from independent observations, the order statistic (the whole set of observations disregarding their sequence) is minimal sufficient—you can't reduce it further without losing information about the parameters. Where the support does depend on the unknown parameter(s) it varies: for a uniform distribution on $(0,\theta)$, the sample maximum is sufficient for $\theta$; for a uniform distribution on $(\theta-1,\theta+1)$ the sample minimum and maximum are together sufficient.
2. I don't know what you mean by "direct correspondence"; the alternative you give seems a fair way to describe sufficient statistics.
3. Yes: trivially the data as a whole are sufficient. (If you hear someone say there's no sufficient statistic they mean there's no low-dimensional one.)
4. Yes, that's the idea. (What's left—the distribution of the data conditional on the sufficient statistic—can be used for checking the distributional assumption independently of the unknown parameter(s).)
5. Apparently not, though I gather the counter-examples are not distributions you're likely to want to use in practice. [It'd be nice if anyone could explain this without getting too heavily into measure theory.]
In response to the further questions ...
1. The first factor, $\mathrm{e}^{-n\lambda}\cdot\lambda^{\sum{x_i}}$, depends on $\lambda$ only through $\sum x_i$. So any one-to-one function of $\sum x_i$ is sufficient: $\sum x_i$, $\sum x_i/n$, $(\sum x_i)^2$, & so on.
2. The second factor, $\tfrac{1}{x_1! x_2! \ldots x_n!}$, doesn't depend on $\lambda$ & so won't affect the value of $\lambda$ at which $f(x;\lambda)$ is a maximum. Derive the MLE & see for yourself.
3. The sample size $n$ is a known constant rather than a realized value of a random variable, so isn't considered part of the sufficient statistic; the same goes for known parameters other than the ones you want to infer things about.
† In this case squaring is one-to-one because $\sum x_i$ is always positive.
‡ When $n$ is a realized value of the random variable $N$, then it will be part of the sufficient statistic, $(\sum x_i,n)$. Say you choose a sample size of 10 or 100 by tossing a coin: $n$ tells you nothing about the value of $\theta$ but does affect how precisely you can estimate it; in this case it's called an ancillary complement to $\sum x_i$ & inference can proceed by conditioning on its realized value—in effect ignoring that it might have come out different.
• I'd love to see the counterexamples to 5. I tried to prove the opposite for a while with Zorn's Lemma but it breaks down at one point. But from what I've gathered the counterexample should be really whacky. Do you have any reference point where I could find it? I don't mind it being heavy on measure theory. Commented Jan 13, 2014 at 0:55
• @sjm.majewski: Lehmann gives Pitcher (1957), "Sets of measures not admitting necessary and sufficient statistics or subfields", Ann. Math. Statist., 28, p267-268; and Landers & Rogge (1973). "On sufficiency and invariance", Ann. Statist., 1, p543-544. Commented Jan 14, 2014 at 22:08
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# PHYSICS
posted by on .
Two objects (m1 = 4.90 kg and m2 = 2.75 kg) are connected by a light string passing over a light, frictionless pulley as in the figure below. The 4.90-kg object is released from rest at a point h = 4.00 m above the table.
(a) Determine the speed of each object when the two pass each other.
m/s
(b) Determine the speed of each object at the moment the 4.90-kg object hits the table.
m/s
(c) How much higher does the 2.75-kg object travel after the 4.90-kg object hits the table?
m
#### Answer This Question
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The dollars price of an ounce of gold
pdf
Since the ounce of gold has passed the \$ 300 I wrote 26 subjects on its price. I had 100% success, so I hope that if will be the 27th. I keep here only the technical analysis of the price of one ounce of gold; I will explain the fundamentals of gold in my 2009 report about world production of gold.
The record price for one ounce of gold to 1005 dollars.
The ounce of gold rose to \$ 1005 in March 2008 to establish its record, and then a decline has occurred to 692 dollars in October 2008.
The gold price has again risen to 989 dollars in February 2009, to a level close to its previous record.
At what level the price of the ounce is there correct?
Thanks to Fibonacci retracements (indicators), we isolated four possible downside:
38.2% retracement i.e drop to 876 dollars.
50% retracement i.e drop to 841 dollars.
62.8% retracement i.e a drop to 803 dollars.
100% retracement i.e a drop to 692 dollars.
You have noted that ounce of gold is already at its 38.2% retracement to 876 dollars one
ounce.
- The retracement of 50% is the 1st Summit of 1980 to 843 dollars one ounce
and that of 100% in the second summit of 1980 to 697 dollars one ounce, it forms a sort of channel of purchase.
- If you are looking for an entry point to the gold, you have it between 876 and 692 dollars the ounce. Ideally you can find it at the bottom of the horizontal channel of the 1980s, between 843 and 692 dollars.
The next record gold prices.
After the correction of an ounce gold price is complete, it will rise beyond its previous record to reach area to 1300 dollars per ounce simply by a pendulum effect.
The road map is as follows: we buy between 876 and 692 dollars in the area of the old peaks of 1980 during the current decline then to cover 1300 dollars.
The summit of 1980 to 843 dollars, adjusted for very optimistic "official" inflation, gives us
2 158 dollars to be truly at the same level as 1980. "Inflation is like toothpaste: once out of the tube, it is impossible to return them." Otto Pöhl
In a few weeks, I will publish an overview on the state of gold production in the world in 2009 as I had already done in 2008 and 2007.
Dr Thomas Chaize
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Answers to math homework is a mathematical instrument that assists to solve math equations. We can solve math word problems.
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Forum
Apr 12th, 2024, 9:03am
Pages: 1
Why do ring oscillators have poor Quality factor? (Read 2343 times)
Venky_analog Junior Member Offline Analog Design is my trade. Math is my love. Posts: 10 TX, USA Why do ring oscillators have poor Quality factor? Nov 14th, 2015, 2:49pm I am trying to reason out why ring oscillators have lower quality factors than LC oscillators, but I am not able to find an explanation for this. Your answer would be of great help to me. Thanks in advance. Back to top Either die a white noise or live long enough to see yourself become the 1/f noise. IP Logged
raja.cedt Senior Fellow Offline Posts: 1516 Germany Re: Why do ring oscillators have poor Quality factor? Reply #1 - Nov 16th, 2015, 2:45am Hello---Please derive an expression for open loop Quality factor for an N-stage ring oscillator, when number of stages (N) tends to infinity, Quality factor equals to Pi/2, means around 1.5. where as in case of LC vco if you are close to the resonance frequency you will always get much higher than this(above 10, if you are below 10Ghz and around 5 in the 20-30ghz carrier frequency range and assume you are not limited by varactor Quality factor). Intuitively in case of LC tank inductor energy will be transferred into capacitor energy almost, rest of the energy from supply where as in case of ring oscillator, every node will be discharges in every cycle and energy being pulled from supply to charge it again, hence more power loss so less Q. Hope this helps----Thanks,Raj. Back to top IP Logged
loose-electron Senior Fellow Offline Best Design Tool = Capable Designers Posts: 1638 San Diego California Re: Why do ring oscillators have poor Quality factor? Reply #2 - Dec 8th, 2015, 11:04pm google:ali hajimiri ring oscillator Qshould be useful Back to top Jerry Twomeywww.effectiveelectrons.comRead My Electronic Design Column Here Contract IC-PCB-System Design - Analog, Mixed Signal, RF & Medical IP Logged
subtr Community Member Offline Analog Enthusiast Posts: 72 India Re: Why do ring oscillators have poor Quality factor? Reply #3 - Jan 9th, 2017, 10:25am I'll give an answer which is not really a mathematical proof. But this is some kind of rough intuition. There is something called Van der Pol Oscillator. It's not a real oscillator, but it's a model of oscillator. It's a differential equation which models both relaxation as well as sinusoidal oscillators. The equation has a parameter mu : https://en.wikipedia.org/wiki/Van_der_Pol_oscillatorYou can look at the first equation in the wikipedia link. It looks like a complicated equation. Solving it requires non linear dynamics, phase plane etc. But let's linearize it at some point and see. The equation is non linear because there is a dependence for the first order term with amplitude. A linear system would not have it. By assuming amplitude is small, we will find that poles lie in the RHS and when amplitude goes high above 1, the poles will go to LHS making it damp. Thus the poles move between RHS and LHS every cycle in a crude sense.If this movement is large, then it's almost a relaxation oscillator. The value of mu thus at zero makes the movement very small, but at higher values makes it jump like a ball in the tennis court. Now what has this to do with the Q? 1. Q is in a sense how good the pole sits on the jw axis. 2.Secondly Q in a non linear fashion is : How much say does the amplitude have in deciding the oscillator frequency. Clearly if mu is non zero, frequency depends on amplitude which is the non linearity. If amplitude decides the frequency you probably can make it oscillate at a different frequency by forcing it. You can call it a sloppy oscillator which agrees to oscillate at a forced frequency. This concept is called injection locking. This sloppy oscillator is more prone to phase noise because amplitude has say on its dynamics. If the value of mu was less, external forcing becomes less powerful in injecting signal. A linear oscillator cannot be injection locked.Now ring oscillator is a relaxation oscillator and has a very bad linearity. In LCVCO, the frequency is decided clearly by the LC which are linear components. The current in the LC is still a square wave. But the LC does a good job of filtering the harmonics and selecting the fundamental. The cross coupled pair(XCP) is just providing a -gm to compensate for the resistance in the LC. These are my intuitions. I'm not sure if you would find these in any materials. I have not validated this, but after a lot of thinking this is what I felt it really could be. Back to top « Last Edit: Jan 10th, 2017, 1:14am by subtr » RegardsSubtr IP Logged
kumar.g Community Member Offline Posts: 51 Frankfurt, Germany Re: Why do ring oscillators have poor Quality factor? Reply #4 - Dec 4th, 2017, 1:30am An another way of interpretation is that Ring oscillator contains several harmonics and the power is less concentrated on the fundamental harmonic. Whereas in the LC VCO there is just the fundamental harmonic and to a lesser extent 2nd and 3rd harmonic. So almost all the power is used up in the fundamental harmonic content. Back to top IP Logged
iVenky Senior Member Offline Posts: 101 Silicon Valley Re: Why do ring oscillators have poor Quality factor? Reply #5 - Feb 3rd, 2019, 12:44pm Hi Raja,Thanks for the reply. I tried deriving but I am making something wrong.Q=2pi * maximum energy stored/ energy dissipated per cyclesince the max energy stored in the cap is CV2/2 and energy lost in the resistors as heat every cycle is CV2, I get Q < pi ~ 3.14. What's wrong here? (is it something to do with the harmonics?) And, also I don't see an obvious dependance on 'N' since for each cycle, all inverters store and dissipate the amount described before.Regards,iVenky Back to top IP Logged
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# Why do nearsighted people see better with their glasses *rotated*?
If you are nearsighted (like me), you may have noticed that if you tilt your glasses, you can see distant objects more clear than with normally-positioned glasses. If you already see completely clear, you can distance your glasses a little more from your eyes and then do it. To do so, rotate the temples while keeping the nosepads fixed on your nose, as is shown in the figures.
As I said, starting with your glasses farther than normal from your eyes, you can observe the effect for near objects too. (By distant, I mean more than 10 meters and by near I mean where you can't see clear without glasses)
Note that if you rotate more than enough, it will distort the light completely. Start from a small $\theta$ and increase it until you see blurry, distant objects more clear. (You should be able to observe this at $\theta\approx20^\circ$ or maybe a little more)
When looking at distant objects, light rays that encounter lenses are parallel, and it seems the effect happens because of oblique incidence of light with lenses:
The optical effect of oblique incidence for convex lenses is called coma, and is shown here (from Wikipedia):
I am looking for an explanation of how this effect for concave lenses (that are used for nearsightedness) causes to see better.
One last point: It seems they use plano-concave or convexo-concave lenses (yellowed lenses below) for glasses instead of biconcave ones.
• I'd suggest that you update your question to indicate that it may be specific to particular eyes and that some people (e.g. me!) benefit from rotating about a different axis (e.g. the vertical axis). Plus it only works in one direction. Also, note that increasing the distance between the lens and the eye (as per your first paragraph) changes the focal length of the lens-eye system, which may cause a different effect entirely. – Tom Shaw Aug 2 '13 at 13:10
• Part of the missing puzzle is that the real model is a concave lens (glasses) which are tilted into a biconvex lens (eye). I think you'll also find the focal distance changes slightly in this case. In addition there can be an astigmatism (regular or irregular) in your eye that might be improved using this tilted distortion. – user6972 Oct 2 '13 at 7:37
• @ChrisMueller It's me that should thank you for the answer. It was almost 5 years since the first time I wanted to calculate the effective focal length of a tilted lens, but never done it, until I saw your answer! – Mo_ Apr 22 '14 at 13:33
• As the question is protected, I can't answer, but I've discovered that my problem with this was due to an incorrect Pantoscopic tilt. See The Real Details of Vertex, Tilt and Wrap for advice to opticians and Pantoscopic tilt induced higher order aberrations... for a paper on the subject. – Mark Booth Apr 4 at 9:32
This is a real effect, but it doesn't have anything to do with coma or any of the optical aberrations. It is caused by the fact that the effective focal length shortens as you tilt a lens. When your eyesight gets worse, you need a stronger focal length lens in your eyeglasses, and tilting the lenses has this effect. The problem with doing this all of the time is that it introduces distortions such as the coma you've pointed out in your question.
This presentation and this journal paper show the the effective focal length of a tilted lens from ray tracing simulations and from theory respectively. From the paper, the focal length for the tangential focal point (up and down as you look through your glasses) and the sagittal focal point (left and right as you look forward) are given from the paper by \begin{align} f_{tan}&=f_0\frac{n-1}{n}\frac{\cos\theta\sqrt{n^2-\sin^2\theta}} {\sqrt{n^2-\sin^2\theta}-\cos\theta}\\ f_{sag}&=f_0(n-1)\frac{1}{\sqrt{n^2-\sin^2\theta}-\cos\theta} \end{align} where $n$ is the index of refraction of the material and $f_0$ is the original focal length. I've plotted $f/f_0$ for both of these in the figure below for an index of refraction of 1.5. I believe that this is special to a lens which has equal radii of curvature on both sides (bi-convex or bi-concave), but the results for other types of lenses will have similar outcomes.
• As I mentioned in the comments above...Visual Acuity (VA) can be improved in some cases where coma is added to an astigmatism. "Adding coma (0.23 μm for 6-mm pupil) to astigmatism resulted in a clear increase of VA in 6 subjects, consistently with theoretical optical predictions, while VA decreased when coma was added to astigmatism in 7 subjects." So in Mostafa's case his titled glass could in fact be more clear due to introduced coma. journalofvision.org/content/11/2/5.full – user6972 Apr 16 '14 at 4:25
• @user6972 I don't disagree with you that it is possible to compensate aberrations in the eye with the correct aberrations in the lens of one's glasses. This is done all of the time with astigmatism. I don't think that this is a dominant effect when compared with the focal length shortening. Specifically in the case mentioned by Mostafa of starting by putting the glasses farther from the eyes than normal and then tilting them to correct for the extra distance. – Chris Mueller Apr 16 '14 at 21:16
• I read "If you already see completely clear, you can distance your glasses a little more from your eyes and then do it." As the coma corrected his weak astigmatism. – user6972 Apr 17 '14 at 2:05
Your basic premise that one can see better with the glasses tilted is false. If the lenses have the right correction for your eyes, then tilting them will make things worse.
The reason this does work often is that the lenses are not at the right correction. In young nearsighted (myopic) people, the myopia usually gets worse with age. The glasses may have had the right correction 2 years ago, but meanwhile the eyes have gotten a little more myopic and a stronger correction is required.
Tilting the lenses makes light pass thru them in a way that effectively makes the lens seem stronger at that angle. If distant objects are a little blurry due to more myopia than the lenses are correcting, then tilting the lenses will make horizontal edges sharper. It does nothing for vertical edges. But, overall the image will appear sharper.
• This is exactly what I'm asking: How "light pass thru them in a way that effectively makes the lens seem stronger at that angle"? – Mo_ Jul 14 '13 at 12:35
• @Mostafa This answer isn't quite right. It is possible you have an astigmatism that is not well corrected for because it requires inducing some coma. You might see better in the distance with your tilted glass, but most of the time the added coma distortion would be worse than your astigmatism. So your prescribed lenses may not be optimized for you at far distances, but rather for most optical conditions. – user6972 Apr 16 '14 at 4:40
• I am short-sighted, and have verified for myself that tilting the glasses sharpens the image. So this can't be right... – QuantumDot Apr 16 '14 at 20:33
• @Quant: Short-sighted people would go for the immediate fix of tilting the glasses instead of updating the perscription. Those near-sighted people that take the longer view would get updated glasses, after which tilting them would make things worse. If your presecription is correct, then any modification to the correction, like changing the effective power of the lens by tilting it, would obviously make things worse. – Olin Lathrop Apr 16 '14 at 22:07
• I have just gotten lenses that are a bit blurry for some reason the optician cannot determine. So I went to my eye doctor and he determined that my eyes have not changed - they are still at the same prescription as before, and he checked the glasses - they are at the correct prescription. But when I tilt the glasses, my vision does appear sharper. – pacoverflow Jul 24 '17 at 19:12
When you place two lenses one behind the other, the effective focal length is a function of their distance. This is the principle behind a zoom lens - with the same pieces of glass, a zoom lens achieves a range of distances.
What you are describing is the simplest case of a zoom lens - just two elements. The first element is the cornea + lens in the eye - for a myopic person, this has a focal length that is slightly too short for the distance to the retina, which is why close-up things are in focus but distant objects are fuzzy. To correct this, you add a second lens with a negative focal length - a diverging lens.
The task then is to compute the apparent focal length of the combination, and show that it depends on the distance between the lenses.
From http://en.wikipedia.org/wiki/Lens_(optics)#Compound_lenses you can see
1) when two lenses touch, their compound focal length is found by
$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$$
2) when there is a distance $d$ between the lenses, this complicates to
$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}-\frac{d}{f_1 f_2}$$
Rearranging things and taking the focal distance from the second element (the eye - this is the focal length of interest) you get the "Back focal length" or BFL:
$$\mbox{BFL} = \frac{f_2 (d - f_1) } { d - (f_1 +f_2) }$$
For a typical eye, $f_2$ is about 1.7 cm. If you have mild myopia your prescription might be -2 diopters, or $f_2=-0.5m$ since $diopter = 1/f$. When you move the glasses away from your eye, the focal length of the compounds system actually gets shorter - this means for a myopic person that they will work "less well":
Now when you tilt your glasses, you "compress" the vertical dimension a little bit - in effect, you increase the radius of curvature in the vertical direction (although not in the horizontal). The net result is that you make the lens stronger in one direction - it becomes a slightly astigmatic lens (cylindrical), with greater strength vertically.
On average then, this makes a stronger lens. And if your eyes have a slight cylindrical component of aberration that is not fully corrected by your glasses, this will really help.
In my case, I have strong astigmatism (although no myopia). I find that slightly rotating my glasses (which changes the cylindrical axis direction) can help with focus - as will pressing against the side of my eye (which distorts the eye ball and has the same effect).
In short - what you are doing when you move one lens relative to the other is change the optical properties. Tilting adds a cylindrical component, while moving them apart changes the compound focal length.
Whether this helps you depends on the gap between your prescription and your aberration.
My theory is that this effect is not seen by most myopes. Rather, it indicates that you and I have an uncorrected higher-order aberration in the eye, i.e. coma, which is corrected by inducing the opposite amount of coma in the spectacle lens by rotation.
I have noticed that rotating my left spectacle lens about the vertical (not horizontal) axis by $10 ^\circ$ in a particular direction gives me much crisper vision. I've also noticed that when I look at a distant circular light source (like a traffic light) with the lens unrotated, it looks somewhat like this image of uncorrected coma. The image is much better with the lens rotated.
Note that the sphere and cylinder prescriptions in the (unrotated) lens were confirmed yesterday by an optometrist as being the best I could get -- but simply by yawing the lens I can see much better.
I'd appreciate it if someone familiar with optics could confirm whether or not coma can actually be corrected in this way. Everything I can find online about higher-order aberrations in the eye suggests that they can only be corrected by refractive surgery or contact lenses. In addition, could a lens be constructed to perform the same correction without rotation?
• Yes it can. Many prescriptions are a balancing act of corrective actions. It is rare to be able to compensate for all the optical conditions especially with astigmatism. – user6972 Apr 16 '14 at 4:48
In the image you have posted, the "rotated lens" results to be a "worse lens" (very poor converging ray concentration) but a stronger one, with a nearer focal plane (if you only consider upper 1/3 of rays, you can try to see a blurred convergence point 30% nearer), wich may feel more suitable if your graduation has got worse.
## protected by Qmechanic♦Oct 2 '13 at 6:49
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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# Top What Is Notation in Math Reviews!
Posted on Posted in Uncategorized
# What You Don’t Know About What Is Notation in Math
You may select unique formats for the multiplication troubles and the selection of numbers to use. At precisely the same time, function notation is an easy concept that isn’t difficult to use. It’s now the normal symbol for integration.
Specifically, though other philosophies of mathematics permit objects that may be proved to exist even though they can’t be constructed, intuitionism allows only mathematical objects that you can high school book reports actually construct. You may select various variables to customize these place value worksheets for your requirements. Determine the approximate measures of angles in degrees to the closest whole number by means of a protractor.
These papers won’t only provide you with a rough idea of what things to expect on your own test, but they permit you to see which areas you perform well in and where you have to improve. On this page you can discover a number of fun math equation games that middle school and higher school students may play online. No matter how long you’ve got for studying, it’s possible to produce a balanced SAT study program.
These completely free worksheets and activities cover virtually all topics https://royalessays.co.uk in a standard Pre-Algebra course. These on-line flashcards make it simple to review concepts that your son or daughter is learning, in short bursts, if needed, or in longer study sessions. There are lots of math games readily available on the internet that require children to solve math problems to move ahead.
Fourth grade students also fix a number of measurement issues. New kindergarten worksheets will be used weekly. Utilize our service to locate an elementary math tutor or maybe to ask a math question.
## New Ideas Into What Is Notation in Math Never Before Revealed
You can select the sort of issues you want on the test and after that just use a random number generator to alter the numbers somewhat. An extensive tutorial is supplied, together with a set of worked examples illustrating the way the library is utilized to conduct statistical tests. Expanded form is a means to compose a number such that all the place value elements of the number are separated.
The workbooks include tons of additional practice complications, so they’re an excellent way to solidify what you just learned https://biology.barnard.edu/ in that section. It’s possible to learn mathematics by studying the info on this internet page and in the many amazing books and multimedia materials. Worksheets are an extremely versatile tool and there are a lot of different forms of worksheets, based on the topic in question. Click the usual core topic title to look at all available worksheets.
Instead, decide on a goal like I will take my Math GED test June. This assessment day isn’t included in the range of days allotted to lessons. Learn how to count pounds and pence, coins utilised in the uk.
For those students searching for a challenge, there’s an entire line of open college courses, too. It is going to be a challenging year since they are presented with a number of new and complicated concepts. Instead, below are some premade activities that you could share with your students so they return in the fall prepared to learn.
It is possible to make worksheets to accompany virtually every lesson conceivable. There are hundreds and hundreds of math worksheets to be found on the net and you are able to find totally free worksheets anywhere. Emphasizes using mathematics manipulatives for modeling the fundamental concepts.
The history of mathematics can be viewed as an ever-increasing collection of abstractions. The distinction is very important. Utilize our service to discover a calculus tutor.
If you are getting ready the test the traditional way and with just one group of questions, it is going to take you anything from half an hour to a few hours to do it, based on the intricacy of the subject issue. Think about it as a trade off you drop a specific amount of work, but you acquire life. The great thing is they are videos which means they may be watched over and over again.
Fear of math might affect your opportunity to succeed in life. As you answer each question, make an effort not to examine the appropriate answers immediately. Multiple-select questions ask the student to decide on each one of the appropriate answer choices among numerous choices.
## Key Pieces of What Is Notation in Math
The purpose of the game is to compare the fractions and click the one with the maximum value. And on account of the associative property, you could also change how you group the numbers. Keep going until each of the numbers are reduced to 1.
In fact, the majority of the moment, students simultaneously employ a number of the practices since they take part in mathematical experiences. It has plenty of guidance on just what to say and do. This is another link that you may want to post to your teacher website or email to parents at the start of the year.
Within this tutorial, you will see the fundamentals of mathematical notation you can come across when reading descriptions of techniques in machine learning. Taking note of the time that it took your learner to finish the test is a good means to get started planning your next study session. Students need to understand and understand the advantages together with the limitations of various tools.
Each plays a significant role in mathematical abilities. For children to be successful in mathematics, several brain functions will need to work together. The simple fact that math skills aren’t necessarily learned sequentially suggests that natural development is extremely tough to chart and, thus, problems are equally tricky to pin down.
## New Questions About What Is Notation in Math
Power and exponent may be used interchangeably. Employing standard form is likely to make the values simpler to see, and in this event, it’ll just create numbers with a couple of zeroes (the exponents have a tiny enough absolute value). Also utilize factor tree method to discover the GCF.
There are two methods to take a look at fractions. In different words any number will remain divisible by all its factors. How to round off an entire number.
Convert the remainders to base 16 (which you might need to think of with respect to decimal numbers, or you may use your fingers and a few toes) and compose the digits in reverse order. Many models may also work as clickers. How to take part of a fraction.
## Whatever They Told You About What Is Notation in Math Is Dead Wrong…And Here’s Why
Saxon packages include all you need to teach 1 child. The colors that ought to be used are coded! 1 easy means to do this is by pressing the appropriate arrow key.
I really like the video segments and I’ve loved having the ability to print off the added problems too. Any position in the 2D plane can be recognized by means of a pair of numbers within this manner. It is essential that each time you introduce a child to a brand-new idea, you explain it well and in detail.
Algebrator is simply awesome. You get to determine when to study. Set to be a classic.
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https://www.thepoorcoder.com/leetcode-remove-duplicates-from-sorted-list-solution/
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# Leetcode - Remove Duplicates from Sorted List Solution
Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.
Example 1:
Output: [1,2]
Example 2:
Output: [1,2,3]
Constraints:
• The number of nodes in the list is in the range [0, 300].
• -100 <= Node.val <= 100
• The list is guaranteed to be sorted in ascending order.
## Solution in python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
while curr_node and curr_node.next:
next_node = curr_node.next
while next_node and curr_node.val == next_node.val:
next_node = next_node.next
curr_node.next = next_node
curr_node = next_node
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https://socratic.org/questions/what-are-all-the-zeroes-of-the-function-f-x-x-2-169#574129
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# What are all the zeroes of the function f(x) = x^2-169?
Mar 15, 2018
The zeroes of f(x) are $\pm$ 13
#### Explanation:
let f(x) = 0
${x}^{2}$ - 169 = 0
${x}^{2}$ = 169
take square root of both sides
sqrt${x}^{2}$ =$\pm$sqrt169
x = $\pm$13
$\therefore$The zeroes of f(x) are $\pm$13
Mar 15, 2018
$x = \pm 13$
#### Explanation:
$\text{to find the zeros set } f \left(x\right) = 0$
$\Rightarrow f \left(x\right) = {x}^{2} - 169 = 0$
$\Rightarrow {x}^{2} = 169$
$\textcolor{b l u e}{\text{take the square root of both sides}}$
$\Rightarrow x = \pm \sqrt{169} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$
$\Rightarrow x = \pm 13 \leftarrow \textcolor{b l u e}{\text{are the zeros}}$
Mar 15, 2018
$f \left(x\right)$ has exactly two zeroes: $+ 13$ and $- 13$.
#### Explanation:
We call zero of a function to those values of $x$ such that $f \left(x\right) = 0$. We call also roots in polynomial functions.
In our case, we have to resolve ${x}^{2} - 169 = 0$
Transposing terms, we have ${x}^{2} = 169$. the square root of both sides give us
$\sqrt{{x}^{2}} = x = \pm \sqrt{169} = \pm 13$ because
(+13)·(+13)=13^2=169 and
(-13)·(-13)=(-13)^2=169
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http://blog.targettestprep.com/gmat-official-guide-ps-33-the-dots-on-the-graph-above-indicate-the-weights/
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# The dots on the graph above indicate the weights…
## Solution:
The easiest way to solve this problem is to actually count the dots on the graph that fit the given criteria. We must find cars that weigh more than 2,500 pounds and get more than 22 miles per gallon. Since the weight is in “hundreds of pounds,” a reading of 20 is actually 2,000 pounds, a reading of 25 is actually 2,500 pounds, and a reading of 30 is actually 3,000 pounds.
To find the cars that weigh more than 2,500 pounds and get more than 22 miles per gallon, we need to locate the dots to the right of 25 on the horizontal axis and above 22 on the vertical axis. When we do this we count 5 dots.
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https://people.maths.bris.ac.uk/~matyd/GroupNames/193/S3xD17.html
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Copied to
clipboard
## G = S3×D17order 204 = 22·3·17
### Direct product of S3 and D17
Aliases: S3×D17, D51⋊C2, C31D34, C171D6, C51⋊C22, (S3×C17)⋊C2, (C3×D17)⋊C2, SmallGroup(204,7)
Series: Derived Chief Lower central Upper central
Derived series C1 — C51 — S3×D17
Chief series C1 — C17 — C51 — C3×D17 — S3×D17
Lower central C51 — S3×D17
Upper central C1
Generators and relations for S3×D17
G = < a,b,c,d | a3=b2=c17=d2=1, bab=a-1, ac=ca, ad=da, bc=cb, bd=db, dcd=c-1 >
3C2
17C2
51C2
51C22
17C6
17S3
3C34
3D17
17D6
3D34
Character table of S3×D17
class 1 2A 2B 2C 3 6 17A 17B 17C 17D 17E 17F 17G 17H 34A 34B 34C 34D 34E 34F 34G 34H 51A 51B 51C 51D 51E 51F 51G 51H size 1 3 17 51 2 34 2 2 2 2 2 2 2 2 6 6 6 6 6 6 6 6 4 4 4 4 4 4 4 4 ρ1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 trivial ρ2 1 -1 1 -1 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1 1 linear of order 2 ρ3 1 1 -1 -1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 linear of order 2 ρ4 1 -1 -1 1 1 -1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1 1 linear of order 2 ρ5 2 0 2 0 -1 -1 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1 -1 -1 orthogonal lifted from S3 ρ6 2 0 -2 0 -1 1 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1 -1 -1 orthogonal lifted from D6 ρ7 2 -2 0 0 2 0 ζ1714+ζ173 ζ1716+ζ17 ζ1711+ζ176 ζ179+ζ178 ζ1712+ζ175 ζ1710+ζ177 ζ1713+ζ174 ζ1715+ζ172 -ζ1711-ζ176 -ζ1712-ζ175 -ζ1710-ζ177 -ζ1715-ζ172 -ζ1714-ζ173 -ζ179-ζ178 -ζ1713-ζ174 -ζ1716-ζ17 ζ179+ζ178 ζ1713+ζ174 ζ1716+ζ17 ζ1711+ζ176 ζ1712+ζ175 ζ1710+ζ177 ζ1715+ζ172 ζ1714+ζ173 orthogonal lifted from D34 ρ8 2 -2 0 0 2 0 ζ1716+ζ17 ζ1711+ζ176 ζ1715+ζ172 ζ1714+ζ173 ζ1713+ζ174 ζ179+ζ178 ζ1710+ζ177 ζ1712+ζ175 -ζ1715-ζ172 -ζ1713-ζ174 -ζ179-ζ178 -ζ1712-ζ175 -ζ1716-ζ17 -ζ1714-ζ173 -ζ1710-ζ177 -ζ1711-ζ176 ζ1714+ζ173 ζ1710+ζ177 ζ1711+ζ176 ζ1715+ζ172 ζ1713+ζ174 ζ179+ζ178 ζ1712+ζ175 ζ1716+ζ17 orthogonal lifted from D34 ρ9 2 2 0 0 2 0 ζ1712+ζ175 ζ1713+ζ174 ζ1710+ζ177 ζ1715+ζ172 ζ1714+ζ173 ζ1711+ζ176 ζ1716+ζ17 ζ179+ζ178 ζ1710+ζ177 ζ1714+ζ173 ζ1711+ζ176 ζ179+ζ178 ζ1712+ζ175 ζ1715+ζ172 ζ1716+ζ17 ζ1713+ζ174 ζ1715+ζ172 ζ1716+ζ17 ζ1713+ζ174 ζ1710+ζ177 ζ1714+ζ173 ζ1711+ζ176 ζ179+ζ178 ζ1712+ζ175 orthogonal lifted from D17 ρ10 2 2 0 0 2 0 ζ1714+ζ173 ζ1716+ζ17 ζ1711+ζ176 ζ179+ζ178 ζ1712+ζ175 ζ1710+ζ177 ζ1713+ζ174 ζ1715+ζ172 ζ1711+ζ176 ζ1712+ζ175 ζ1710+ζ177 ζ1715+ζ172 ζ1714+ζ173 ζ179+ζ178 ζ1713+ζ174 ζ1716+ζ17 ζ179+ζ178 ζ1713+ζ174 ζ1716+ζ17 ζ1711+ζ176 ζ1712+ζ175 ζ1710+ζ177 ζ1715+ζ172 ζ1714+ζ173 orthogonal lifted from D17 ρ11 2 -2 0 0 2 0 ζ1711+ζ176 ζ1715+ζ172 ζ1712+ζ175 ζ1716+ζ17 ζ1710+ζ177 ζ1714+ζ173 ζ179+ζ178 ζ1713+ζ174 -ζ1712-ζ175 -ζ1710-ζ177 -ζ1714-ζ173 -ζ1713-ζ174 -ζ1711-ζ176 -ζ1716-ζ17 -ζ179-ζ178 -ζ1715-ζ172 ζ1716+ζ17 ζ179+ζ178 ζ1715+ζ172 ζ1712+ζ175 ζ1710+ζ177 ζ1714+ζ173 ζ1713+ζ174 ζ1711+ζ176 orthogonal lifted from D34 ρ12 2 2 0 0 2 0 ζ1710+ζ177 ζ179+ζ178 ζ1714+ζ173 ζ1713+ζ174 ζ1711+ζ176 ζ1712+ζ175 ζ1715+ζ172 ζ1716+ζ17 ζ1714+ζ173 ζ1711+ζ176 ζ1712+ζ175 ζ1716+ζ17 ζ1710+ζ177 ζ1713+ζ174 ζ1715+ζ172 ζ179+ζ178 ζ1713+ζ174 ζ1715+ζ172 ζ179+ζ178 ζ1714+ζ173 ζ1711+ζ176 ζ1712+ζ175 ζ1716+ζ17 ζ1710+ζ177 orthogonal lifted from D17 ρ13 2 -2 0 0 2 0 ζ1710+ζ177 ζ179+ζ178 ζ1714+ζ173 ζ1713+ζ174 ζ1711+ζ176 ζ1712+ζ175 ζ1715+ζ172 ζ1716+ζ17 -ζ1714-ζ173 -ζ1711-ζ176 -ζ1712-ζ175 -ζ1716-ζ17 -ζ1710-ζ177 -ζ1713-ζ174 -ζ1715-ζ172 -ζ179-ζ178 ζ1713+ζ174 ζ1715+ζ172 ζ179+ζ178 ζ1714+ζ173 ζ1711+ζ176 ζ1712+ζ175 ζ1716+ζ17 ζ1710+ζ177 orthogonal lifted from D34 ρ14 2 2 0 0 2 0 ζ1716+ζ17 ζ1711+ζ176 ζ1715+ζ172 ζ1714+ζ173 ζ1713+ζ174 ζ179+ζ178 ζ1710+ζ177 ζ1712+ζ175 ζ1715+ζ172 ζ1713+ζ174 ζ179+ζ178 ζ1712+ζ175 ζ1716+ζ17 ζ1714+ζ173 ζ1710+ζ177 ζ1711+ζ176 ζ1714+ζ173 ζ1710+ζ177 ζ1711+ζ176 ζ1715+ζ172 ζ1713+ζ174 ζ179+ζ178 ζ1712+ζ175 ζ1716+ζ17 orthogonal lifted from D17 ρ15 2 -2 0 0 2 0 ζ1712+ζ175 ζ1713+ζ174 ζ1710+ζ177 ζ1715+ζ172 ζ1714+ζ173 ζ1711+ζ176 ζ1716+ζ17 ζ179+ζ178 -ζ1710-ζ177 -ζ1714-ζ173 -ζ1711-ζ176 -ζ179-ζ178 -ζ1712-ζ175 -ζ1715-ζ172 -ζ1716-ζ17 -ζ1713-ζ174 ζ1715+ζ172 ζ1716+ζ17 ζ1713+ζ174 ζ1710+ζ177 ζ1714+ζ173 ζ1711+ζ176 ζ179+ζ178 ζ1712+ζ175 orthogonal lifted from D34 ρ16 2 2 0 0 2 0 ζ1713+ζ174 ζ1710+ζ177 ζ179+ζ178 ζ1712+ζ175 ζ1716+ζ17 ζ1715+ζ172 ζ1711+ζ176 ζ1714+ζ173 ζ179+ζ178 ζ1716+ζ17 ζ1715+ζ172 ζ1714+ζ173 ζ1713+ζ174 ζ1712+ζ175 ζ1711+ζ176 ζ1710+ζ177 ζ1712+ζ175 ζ1711+ζ176 ζ1710+ζ177 ζ179+ζ178 ζ1716+ζ17 ζ1715+ζ172 ζ1714+ζ173 ζ1713+ζ174 orthogonal lifted from D17 ρ17 2 -2 0 0 2 0 ζ1715+ζ172 ζ1712+ζ175 ζ1713+ζ174 ζ1711+ζ176 ζ179+ζ178 ζ1716+ζ17 ζ1714+ζ173 ζ1710+ζ177 -ζ1713-ζ174 -ζ179-ζ178 -ζ1716-ζ17 -ζ1710-ζ177 -ζ1715-ζ172 -ζ1711-ζ176 -ζ1714-ζ173 -ζ1712-ζ175 ζ1711+ζ176 ζ1714+ζ173 ζ1712+ζ175 ζ1713+ζ174 ζ179+ζ178 ζ1716+ζ17 ζ1710+ζ177 ζ1715+ζ172 orthogonal lifted from D34 ρ18 2 2 0 0 2 0 ζ179+ζ178 ζ1714+ζ173 ζ1716+ζ17 ζ1710+ζ177 ζ1715+ζ172 ζ1713+ζ174 ζ1712+ζ175 ζ1711+ζ176 ζ1716+ζ17 ζ1715+ζ172 ζ1713+ζ174 ζ1711+ζ176 ζ179+ζ178 ζ1710+ζ177 ζ1712+ζ175 ζ1714+ζ173 ζ1710+ζ177 ζ1712+ζ175 ζ1714+ζ173 ζ1716+ζ17 ζ1715+ζ172 ζ1713+ζ174 ζ1711+ζ176 ζ179+ζ178 orthogonal lifted from D17 ρ19 2 2 0 0 2 0 ζ1711+ζ176 ζ1715+ζ172 ζ1712+ζ175 ζ1716+ζ17 ζ1710+ζ177 ζ1714+ζ173 ζ179+ζ178 ζ1713+ζ174 ζ1712+ζ175 ζ1710+ζ177 ζ1714+ζ173 ζ1713+ζ174 ζ1711+ζ176 ζ1716+ζ17 ζ179+ζ178 ζ1715+ζ172 ζ1716+ζ17 ζ179+ζ178 ζ1715+ζ172 ζ1712+ζ175 ζ1710+ζ177 ζ1714+ζ173 ζ1713+ζ174 ζ1711+ζ176 orthogonal lifted from D17 ρ20 2 2 0 0 2 0 ζ1715+ζ172 ζ1712+ζ175 ζ1713+ζ174 ζ1711+ζ176 ζ179+ζ178 ζ1716+ζ17 ζ1714+ζ173 ζ1710+ζ177 ζ1713+ζ174 ζ179+ζ178 ζ1716+ζ17 ζ1710+ζ177 ζ1715+ζ172 ζ1711+ζ176 ζ1714+ζ173 ζ1712+ζ175 ζ1711+ζ176 ζ1714+ζ173 ζ1712+ζ175 ζ1713+ζ174 ζ179+ζ178 ζ1716+ζ17 ζ1710+ζ177 ζ1715+ζ172 orthogonal lifted from D17 ρ21 2 -2 0 0 2 0 ζ179+ζ178 ζ1714+ζ173 ζ1716+ζ17 ζ1710+ζ177 ζ1715+ζ172 ζ1713+ζ174 ζ1712+ζ175 ζ1711+ζ176 -ζ1716-ζ17 -ζ1715-ζ172 -ζ1713-ζ174 -ζ1711-ζ176 -ζ179-ζ178 -ζ1710-ζ177 -ζ1712-ζ175 -ζ1714-ζ173 ζ1710+ζ177 ζ1712+ζ175 ζ1714+ζ173 ζ1716+ζ17 ζ1715+ζ172 ζ1713+ζ174 ζ1711+ζ176 ζ179+ζ178 orthogonal lifted from D34 ρ22 2 -2 0 0 2 0 ζ1713+ζ174 ζ1710+ζ177 ζ179+ζ178 ζ1712+ζ175 ζ1716+ζ17 ζ1715+ζ172 ζ1711+ζ176 ζ1714+ζ173 -ζ179-ζ178 -ζ1716-ζ17 -ζ1715-ζ172 -ζ1714-ζ173 -ζ1713-ζ174 -ζ1712-ζ175 -ζ1711-ζ176 -ζ1710-ζ177 ζ1712+ζ175 ζ1711+ζ176 ζ1710+ζ177 ζ179+ζ178 ζ1716+ζ17 ζ1715+ζ172 ζ1714+ζ173 ζ1713+ζ174 orthogonal lifted from D34 ρ23 4 0 0 0 -2 0 2ζ1715+2ζ172 2ζ1712+2ζ175 2ζ1713+2ζ174 2ζ1711+2ζ176 2ζ179+2ζ178 2ζ1716+2ζ17 2ζ1714+2ζ173 2ζ1710+2ζ177 0 0 0 0 0 0 0 0 -ζ1711-ζ176 -ζ1714-ζ173 -ζ1712-ζ175 -ζ1713-ζ174 -ζ179-ζ178 -ζ1716-ζ17 -ζ1710-ζ177 -ζ1715-ζ172 orthogonal faithful ρ24 4 0 0 0 -2 0 2ζ1714+2ζ173 2ζ1716+2ζ17 2ζ1711+2ζ176 2ζ179+2ζ178 2ζ1712+2ζ175 2ζ1710+2ζ177 2ζ1713+2ζ174 2ζ1715+2ζ172 0 0 0 0 0 0 0 0 -ζ179-ζ178 -ζ1713-ζ174 -ζ1716-ζ17 -ζ1711-ζ176 -ζ1712-ζ175 -ζ1710-ζ177 -ζ1715-ζ172 -ζ1714-ζ173 orthogonal faithful ρ25 4 0 0 0 -2 0 2ζ1713+2ζ174 2ζ1710+2ζ177 2ζ179+2ζ178 2ζ1712+2ζ175 2ζ1716+2ζ17 2ζ1715+2ζ172 2ζ1711+2ζ176 2ζ1714+2ζ173 0 0 0 0 0 0 0 0 -ζ1712-ζ175 -ζ1711-ζ176 -ζ1710-ζ177 -ζ179-ζ178 -ζ1716-ζ17 -ζ1715-ζ172 -ζ1714-ζ173 -ζ1713-ζ174 orthogonal faithful ρ26 4 0 0 0 -2 0 2ζ1711+2ζ176 2ζ1715+2ζ172 2ζ1712+2ζ175 2ζ1716+2ζ17 2ζ1710+2ζ177 2ζ1714+2ζ173 2ζ179+2ζ178 2ζ1713+2ζ174 0 0 0 0 0 0 0 0 -ζ1716-ζ17 -ζ179-ζ178 -ζ1715-ζ172 -ζ1712-ζ175 -ζ1710-ζ177 -ζ1714-ζ173 -ζ1713-ζ174 -ζ1711-ζ176 orthogonal faithful ρ27 4 0 0 0 -2 0 2ζ1712+2ζ175 2ζ1713+2ζ174 2ζ1710+2ζ177 2ζ1715+2ζ172 2ζ1714+2ζ173 2ζ1711+2ζ176 2ζ1716+2ζ17 2ζ179+2ζ178 0 0 0 0 0 0 0 0 -ζ1715-ζ172 -ζ1716-ζ17 -ζ1713-ζ174 -ζ1710-ζ177 -ζ1714-ζ173 -ζ1711-ζ176 -ζ179-ζ178 -ζ1712-ζ175 orthogonal faithful ρ28 4 0 0 0 -2 0 2ζ1716+2ζ17 2ζ1711+2ζ176 2ζ1715+2ζ172 2ζ1714+2ζ173 2ζ1713+2ζ174 2ζ179+2ζ178 2ζ1710+2ζ177 2ζ1712+2ζ175 0 0 0 0 0 0 0 0 -ζ1714-ζ173 -ζ1710-ζ177 -ζ1711-ζ176 -ζ1715-ζ172 -ζ1713-ζ174 -ζ179-ζ178 -ζ1712-ζ175 -ζ1716-ζ17 orthogonal faithful ρ29 4 0 0 0 -2 0 2ζ179+2ζ178 2ζ1714+2ζ173 2ζ1716+2ζ17 2ζ1710+2ζ177 2ζ1715+2ζ172 2ζ1713+2ζ174 2ζ1712+2ζ175 2ζ1711+2ζ176 0 0 0 0 0 0 0 0 -ζ1710-ζ177 -ζ1712-ζ175 -ζ1714-ζ173 -ζ1716-ζ17 -ζ1715-ζ172 -ζ1713-ζ174 -ζ1711-ζ176 -ζ179-ζ178 orthogonal faithful ρ30 4 0 0 0 -2 0 2ζ1710+2ζ177 2ζ179+2ζ178 2ζ1714+2ζ173 2ζ1713+2ζ174 2ζ1711+2ζ176 2ζ1712+2ζ175 2ζ1715+2ζ172 2ζ1716+2ζ17 0 0 0 0 0 0 0 0 -ζ1713-ζ174 -ζ1715-ζ172 -ζ179-ζ178 -ζ1714-ζ173 -ζ1711-ζ176 -ζ1712-ζ175 -ζ1716-ζ17 -ζ1710-ζ177 orthogonal faithful
Smallest permutation representation of S3×D17
On 51 points
Generators in S51
(1 20 43)(2 21 44)(3 22 45)(4 23 46)(5 24 47)(6 25 48)(7 26 49)(8 27 50)(9 28 51)(10 29 35)(11 30 36)(12 31 37)(13 32 38)(14 33 39)(15 34 40)(16 18 41)(17 19 42)
(18 41)(19 42)(20 43)(21 44)(22 45)(23 46)(24 47)(25 48)(26 49)(27 50)(28 51)(29 35)(30 36)(31 37)(32 38)(33 39)(34 40)
(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17)(18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34)(35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51)
(1 17)(2 16)(3 15)(4 14)(5 13)(6 12)(7 11)(8 10)(18 21)(19 20)(22 34)(23 33)(24 32)(25 31)(26 30)(27 29)(35 50)(36 49)(37 48)(38 47)(39 46)(40 45)(41 44)(42 43)
G:=sub<Sym(51)| (1,20,43)(2,21,44)(3,22,45)(4,23,46)(5,24,47)(6,25,48)(7,26,49)(8,27,50)(9,28,51)(10,29,35)(11,30,36)(12,31,37)(13,32,38)(14,33,39)(15,34,40)(16,18,41)(17,19,42), (18,41)(19,42)(20,43)(21,44)(22,45)(23,46)(24,47)(25,48)(26,49)(27,50)(28,51)(29,35)(30,36)(31,37)(32,38)(33,39)(34,40), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17)(18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34)(35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51), (1,17)(2,16)(3,15)(4,14)(5,13)(6,12)(7,11)(8,10)(18,21)(19,20)(22,34)(23,33)(24,32)(25,31)(26,30)(27,29)(35,50)(36,49)(37,48)(38,47)(39,46)(40,45)(41,44)(42,43)>;
G:=Group( (1,20,43)(2,21,44)(3,22,45)(4,23,46)(5,24,47)(6,25,48)(7,26,49)(8,27,50)(9,28,51)(10,29,35)(11,30,36)(12,31,37)(13,32,38)(14,33,39)(15,34,40)(16,18,41)(17,19,42), (18,41)(19,42)(20,43)(21,44)(22,45)(23,46)(24,47)(25,48)(26,49)(27,50)(28,51)(29,35)(30,36)(31,37)(32,38)(33,39)(34,40), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17)(18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34)(35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51), (1,17)(2,16)(3,15)(4,14)(5,13)(6,12)(7,11)(8,10)(18,21)(19,20)(22,34)(23,33)(24,32)(25,31)(26,30)(27,29)(35,50)(36,49)(37,48)(38,47)(39,46)(40,45)(41,44)(42,43) );
G=PermutationGroup([[(1,20,43),(2,21,44),(3,22,45),(4,23,46),(5,24,47),(6,25,48),(7,26,49),(8,27,50),(9,28,51),(10,29,35),(11,30,36),(12,31,37),(13,32,38),(14,33,39),(15,34,40),(16,18,41),(17,19,42)], [(18,41),(19,42),(20,43),(21,44),(22,45),(23,46),(24,47),(25,48),(26,49),(27,50),(28,51),(29,35),(30,36),(31,37),(32,38),(33,39),(34,40)], [(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),(18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34),(35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51)], [(1,17),(2,16),(3,15),(4,14),(5,13),(6,12),(7,11),(8,10),(18,21),(19,20),(22,34),(23,33),(24,32),(25,31),(26,30),(27,29),(35,50),(36,49),(37,48),(38,47),(39,46),(40,45),(41,44),(42,43)]])
S3×D17 is a maximal quotient of D512C4 C51⋊D4 C3⋊D68 C17⋊D12 C51⋊Q8
Matrix representation of S3×D17 in GL4(𝔽103) generated by
1 0 0 0 0 1 0 0 0 0 0 102 0 0 1 102
,
1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0
,
0 1 0 0 102 95 0 0 0 0 1 0 0 0 0 1
,
0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1
G:=sub<GL(4,GF(103))| [1,0,0,0,0,1,0,0,0,0,0,1,0,0,102,102],[1,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0],[0,102,0,0,1,95,0,0,0,0,1,0,0,0,0,1],[0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,1] >;
S3×D17 in GAP, Magma, Sage, TeX
S_3\times D_{17}
% in TeX
G:=Group("S3xD17");
// GroupNames label
G:=SmallGroup(204,7);
// by ID
G=gap.SmallGroup(204,7);
# by ID
G:=PCGroup([4,-2,-2,-3,-17,54,3075]);
// Polycyclic
G:=Group<a,b,c,d|a^3=b^2=c^17=d^2=1,b*a*b=a^-1,a*c=c*a,a*d=d*a,b*c=c*b,b*d=d*b,d*c*d=c^-1>;
// generators/relations
Export
×
𝔽
| 6,777
| 11,527
|
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## Sequential Games V
Final look at sequential games
Today we will take one last look at solving sequential games. Before we do that though let’s solve the example at the end of the previous blog.
Let p represent the probability that your boss writes you up and 1-p be the probability that your boss does not write you up. The probability distribution would be -10p + 40(1-p) = 20.
-10p + 40 – 40p = 20 =>
-50p = -20 =>
p = -20/-50 = 2/5
1-p = 3/5
So you will choose to shirk if the probability of your boss writing you up is less than 2/5. Or you will choose to shirt if the probability of your boss not writing you up is greater than 3/5. Of course you would never shirk to begin with and always work hard right?
In most of our examples we have looked at sequential games that involved two players. In the following example there will be four participants. In particular, Yogurtland is looking to establish a shop in a new market. However, three potential competitors (Mickey’s Yogurt, Yum Yum Yogurt, and Big Kahuna) are also debating whether to enter the market or not.
The payoffs are organized from top to bottom representing Yogurtland, Yum Yum, Mickey’s, and Big Kahuna respectively. We can solve this game using backwards induction. Let’s start with the options for Big Kahuna.
Scenarios for Big Kahuna (a lot):
Enter, Enter, Enter, Enter = 0
Enter, Enter, Enter, No = 1
Enter, Enter, No, Enter = 3
Enter, Enter, No, No = 1
Enter, No, Enter, Enter = 3
Enter, No, Enter, No = 2
Enter, No, No, Enter = 3
Enter, No, No, No = 4
No, No, Enter, Enter = 4
No, No, Enter, No = 4
No, No, No, Enter = 3
No, No, No, No = 0
No, Enter, Enter, Enter = 2
No, Enter, Enter, No = 4
No, Enter, No, Enter = 3
No, Enter, No, No = 1
The preferences of Big Kahuna have been highlighted in red based on the payouts from each outcome. I also placed in bold the preferred actions of Big Kahuna above. Now let’s look at Mickey’s. Since Mickey’s assumes Big Kahuna will make optimal decisions, Mickey’s options are the following:
Enter, Enter, Enter, No = 3
Enter, Enter, No, Enter = 1
Enter, No, Enter, Enter = 2
Enter, No, No, No = 1
No, No, Enter, Enter = 1
No, No, Enter, No = 1
No, No, No, Enter = 2
No, Enter, Enter, No = 1
No, Enter, No, Enter = 0
Using backwards induction, Mickey’s optimal choices will be in blue. Let’s continue the process by identifying Yum Yum’s options.
Scenarios for Yum Yum:
Enter, Enter, Enter, No = 2
Enter, No, Enter, Enter = 3
No, No, No, Enter = 1
No, Enter, Enter, No = 3
Yum Yum’s optimal choices are now highlighted in green. As you can see, the options for yogurt shops becomes less as we use backwards induction. This leaves Yogurtland with a decision to make, enter or do not enter the market.
Their scenarios are:
Enter, No, Enter, Enter = 2
No, Enter, Enter, No = 2
Based on Yogurtland’s options, the company is indifferent between entering the market or not entering the market since their payoffs will be the same. So the solution to this game is (2,3,2,3) and (2,3,1,4) or (Enter, No, Enter, Enter) and (No, Enter, Enter, No).
## Sequential Games IV
To be or not to be?
Last time we looked at the dilemma of Charlie Brown attempting to kick a football and Lucy pulling the football away from him (to this day he still tries). In this scenario I modified the game tree to where only Charlie knows his payoffs but not Lucy’s. I introduced probability distribution into the game to determine the probability of Charlie choosing to kick the football based on the potential payoffs. Let’s apply probability distribution to other instances of sequential games.
A great example comes from the humanities. One of William Shakespeare’s most famous plays was Hamlet. In Hamlet, the title character seeks revenge against his uncle King Claudius who, according to the ghost of Hamlet’s father, murdered Hamlet’s father and took the throne for himself. We can represent a sequential game as follows:
Hamlet has the choice of either living or killing himself (to be or not to be). If he chooses to kill himself (not to be) then he will receive a payoff of -50 since he will be unable to avenge his father’s killer. Now if he chooses “to be” then Claudius can either let Hamlet live or kill him.
If Hamlet is allowed to live he will receive a payoff of 200 since he will avenge his father’s death by killing Claudius. However, if Claudius kills Hamlet then Hamlet’s payoff is -100 since he was unable to avenge his father and he was killed by his father’s killer (double negative).
Since we do not know the payoffs of Claudius, we can only determine the potential outcome through probability distribution. First let p represent the probability that Claudius kills Hamlet and 1-p be the probability that Claudius does not kill Hamlet. The probability distribution of this game would be:
-100p + 200(1-p) = -50 =>
-100p + 200 – 200p = -50 =>
-300p = -250 =>
p = 5/6
1-p = 1/6
So Hamlet will choose “to be” if the probability of Claudius killing Hamlet is less than 5/6 or if the probability of Claudius letting Hamlet live is greater than 1/6. Of course, things don’t look good for Hamlet in this situation. More than likely Claudius would want to kill Hamlet. But we cannot make that assumption since we do not know Claudius’s payouts.
Try to solve the following game below. In this scenario, you are currently at work and have a tendency to shirk (neglect your job) instead of doing your job (untrue of course). Now if you decide to shirk your boss would either write you up or not (more than likely write you up). Solve for p, 1-p, and explain the probability of you choosing to shirk.
## Solving Sequential Games III
Why does Charlie always kick?
Today we are going to analyze one of the greatest and puzzling dilemmas in cartoon animation history. Why does Charlie Brown always attempt to kick the football that Lucy is holding for him? In every instance she always pulls the ball away from him and he ends up going airborne from missing the ball. Let’s see if we can understand Charlie’s reasoning in a sequential game.
First we will represent the game in the following game tree:
In this game Charlie has the option of either kicking the football or not. If he chooses not to kick the football then his payout will be 5 utils (happiness) while Lucy’s will be 0 (no bumps and bruises for Charlie). Now if Charlie chooses to kick the ball then he is at the mercy of Lucy. If Lucy allows Charlie to kick the ball then the payouts will be 15 for Charlie and 5 for Lucy. If Lucy pulls the football then Charlie’s payout is -5 since he will miss the kick and go flying into the air. Lucy will achieve a payout of 10 utils since she enjoys watching Charlie suffer (never liked Lucy).
Let’s solve this game under the assumption of complete rationality:
The rational choice for Lucy is to not allow Charlie to kick since 10 > 5. Charlie should assume that Lucy is a rational decision maker and will opt to not allow him to kick. As such, Charlie should just walk away knowing that he did not give Lucy the satisfaction of missing the kick. But we know this is not the case as time and time again Charlie attempts to kick the football.
It’s possible to assume that Charlie always makes an irrational decision while Lucy always makes the rational choice. Lucy may know that Charlie is irrational hence her reason for always offering Charlie the opportunity to kick the ball. But what if we didn’t factor in Lucy payouts? Let’s assume Charlie does not know Lucy’s payouts and only knows his own as shown below.
How would Charlie choose now? Let’s solve this using a probability distribution similar to mixed strategies. Let p be the probability that Lucy doesn’t allow Charlie to kick and 1-p be the probability that Lucy allows Charlie to kick. The equation would appear as follows:
-5p + 15(1-p) = 5
Now let’s solve for p and 1-p:
-5p + 15(1-p) = 5 =>
-5p + 15 – 15p = 5 =>
-20p = -10 =>
p = -10/-20 or 1/2
1-p = 1 – 1/2 = 1/2
The way to interpret this is Charlie will choose to kick the ball as long as the probability of Lucy not allowing him to kick is less than 1/2. Or we can say Charlie will choose to kick the ball as long as the probability of Lucy letting him kick is greater than 1/2. So Charlie is still looking at a 50/50 split.
Based on Charlie’s history though we can only assume that he makes irrational decisions while Lucy always makes the rational choice. For some reason the desire for Charlie to kick that ball beats out all of the future body aches and pains. Poor Charlie.
## Solving Sequential Games II
Where to set up shop?
In my previous blog I introduced the extensive form of games that are used in illustrating sequential games (order matters). These games are solved using backwards induction. Today we are going to analyze a scenario of where a restaurant should open up shop taking into consideration a competitor looking to open shop as well.
Pieology and Blaze Pizza are competing for world domination. It’s hard to miss these locations in town as they are popping up everywhere. Due to their long term strategy for aggressive expansion, Pieology is looking to open a location at a new shopping center. However, Blaze Pizza wants to capture market share in the new shopping center as well. The owners of the shopping center are willing to give Pieology first choice for the exact location (first-mover advantage). Which location should Pieology choose?
Observe the game tree above. Pielogy has the option of either choosing the West, South, or East side of the shopping center. Once Pieology makes there choice of location, Blaze Pizza will come in and establish a new location based on Pieology’s choice. Using backwards induction we can solve for the optimal choice for Pieology to maximize profit. If Pieology chooses West, Blaze Pizza can select either West, South, or East. Blaze Pizza will prefer to set up a location at the East side of the shopping center since 3 (East) > 2 (South) > 1(West).
Now let’s assume Pieology chooses to build the new location on the South side of the shopping center. Blaze Pizza will evaluate their potential profits based on the possible locations. Blaze Pizza’s preference is to build a new location on the East side of the shopping center due to 3 (East) > 2 (West) > 1 (South).
Similar analysis can be used to determine Blaze Pizza’s optimal choice if Pieology chooses the East side of the shopping center. Based on the possible profits, Blaze Pizza will select the West side of the shopping center since 3 (West) > 2 (South) > 1 (East).
Based on backwards induction what would be the optimal choice for Pieology?
If Pielogy chooses West: (West, East) = (4,3)
If Pieology chooses South: (South, East) = (3,3)
If Pieology chooses East: (East, West) = (5,3)
As you can see Pieology will always have higher profits than Blaze Pizza due to their first-mover advantage. However, their is an optimal location for Pieology. Based on the payouts (profits), Pieology will choose to establish their new location on the East side of the shopping center since 5 (East) > 4 (West) > 3 (South).
To optimize profits for both businesses, it is best that they establish locations furthest from each other at the new shopping center. We will continue to look at sequential games in the next blog which will include more examples as we continue to solve these types of games. I look forward to your feedback in the comments below.
## Introduction to Sequential Games
Overview of solving sequential games in extensive form
We are going to switch gears and analyze sequential games. The format will be different but the principles of game theory that we reviewed will be the same. I am going to introduce the concept of backward induction to solve these type of games. Similar to simultaneous games, the sequential games will be complete and with perfect information.
Let’s look at the following example in extensive form:
Going forward I will illustrate extensive form games in a game tree as shown above. In this example we have two siblings, one age 4 and the other age 2. On the table there is a plate with only two cookies remaining. Player 1 represents the older sibling and chooses an action a¹ from his or her set of options A¹. In this game the older sibling has the actions of “share” or “take”. Player 2 represents the younger sibling and observes the action of the older brother and then chooses an action a² (“share” or “take”) from his or her set of options A².
So if the older sibling chooses “take” then the payoff will be two cookies for the older sibling and none for the younger sibling (how cruel). If the older sibling chooses the option of “share”, then the younger sibling can choose either “take” or “share.” Now we can introduce the concept of backwards induction. We basically look at the options of the younger sibling first and work our way backwards.
Before we start, it is important to remember that players will look for the optimal choice based on their sets of options (similar to the simultaneous games we analyzed). So the younger sibling has the option of choosing the action “share” or “take”. If the younger sibling chooses “share” then both siblings will end up with one cookie. If the younger sibling chooses “take” then the younger sibling gets two cookies and the older sibling gets nothing. Since we are assuming rational players, the younger sibling will choose “take” since 2 > 1.
Knowing that the younger sibling will choose “take” the optimal choice for the older sibling is to choose “take” since 2 > 0. So this game ends where the older sibling will approach the table and grab both cookies since it is assumed the younger sibling would do the same thing given the opportunity. It appears that both siblings do not understand the concept of sharing and equality. Although a simple example, this game represents the basics of backward induction and solving sequential games with complete information.
Going forward we will look at larger games with more players, actions, and options. I look forward to your feedback in the comments below and feel free to follow me on Twitter and connect via LinkedIn or Facebook or both.
## Mixed Strategy Nash Equilibrium IV
We will focus on solving more games that involve pure strategy Nash Equilibrium and mixed strategy Nash Equilibrium. It is important that we understand how to solve these games as it lays the foundation for further topics in game theory. This is especially the case if you are interested in advancing your studies. In these problems the goal is to identify the Nash Equilibrium and not necessarily the payoffs.
q 1-q 1\2 Red Blue p Red 1,4 2,3 1-p Blue 3,2 1,4
Player 1 chooses Blue when Player 2 chooses Red.
Player 1 chooses Red when Player 2 chooses Blue.
Player 2 chooses Red when Player 1 chooses Red.
Player 2 chooses Blue when Player 1 chooses Blue.
There is no pure strategy Nash Equilibrium in this game. However, let’s see if a mixed strategy Nash Equilibrium exists. Similar to previous examples, we need to solve for p, 1-p, q, and 1-q to determine the mixed strategies.
For Player 1:
4p + 2(1-p) = 3p + 4(1-p)
=> 4p + 2 – 2p = 3p + 4 – 4p
=> 2p + 2 = 4 – p
=> 3p = 2
=> p = 2/3, 1-p = (1 – 2/3) = 1/3
For Player 2:
1q + 2(1-q) = 3q + 1(1-q)
=> q + 2 -2q = 3q +1 – q
=> 2 – q = 2q + 1
=> 1 = 3q
=> q = 1/3, 1-q = (1 – 1/3) = 2/3
Summary: Pure strategy Nash Equilibrium does not exist. Mixed strategy for Player 1 (Red = 2/3, Blue = 1/3) and Player 2 (Red = 1/3, Blue = 2/3)
Problem #2
q 1-q 1\2 Left Right p Up 5,6 4,1 1-p Down 3,2 5,3
Player 1 chooses Up when Player 2 chooses Left.
Player 1 chooses Down when Player 2 chooses Right.
Player 2 chooses Left when Player 1 chooses Up.
Player 2 chooses Right when Player 1 chooses Down.
The pure strategy Nash Equilibrium for this game is (Up, Left) and (Down, Right). Now let’s determine if a mixed strategy Nash Equilibrium exists.
For Player 1:
6p + 2(1-p) = 1p + 3(1-p)
=> 6p + 2 – 2p = p + 3 – 3p
=> 4p + 2 = 3 – 2p
=> 6p = 1
=> p = 1/6, 1-p = (1 – 1/6) = 5/6
For Player 2:
5q + 4(1-q) = 3q + 5(1-q)
=> 5q + 4 – 4q = 3q + 5 – 5q
=> q + 4 = 5 – 2q
=> 3q = 1
=> q = 1/3, 1-q = (1-1/3) = 2/3
Summary: In this game we have both pure strategy Nash Equilibrium and a mixed strategy Nash equilibrium. Pure strategies are (Up, Left) and (Down, Right). Mixed strategies are Player 1 (Up = 1/6, Down = 5/6) and Player 2 (Left = 1/3, Right = 2/3).
Problem #3:
q 1-q 1\2 Yellow Green p Yellow 5,4 2,4 1-p Green 1,4 3,3
Player 1 chooses Yellow when Player 2 chooses Yellow.
Player 1 chooses Green when Player 2 chooses Green.
Player 2 is indifferent between Yellow and Green when Player 1 chooses Yellow.
Player 2 chooses Yellow when Player 1 chooses Green.
There is only one pure strategy Nash Equilibrium in this game (Yellow, Yellow). Now for the mixed strategy Nash equilibrium:
For Player 1:
4p + 4(1-p) = 4p + 3(1-p)
=> 4p + 4 – 4p = 4p + 3 – 3p
=> 4 = 3 +p
=> p = 1, 1-p = (1 – 1) = 0
What does it mean for p = 1? Remember in a probability distribution the sum of the probabilities must equal 100%. This means there is no mixed strategy Nash Equilibrium for Player 1. So Player 1’s optimal choice is to always choose the pure strategy of Red.
For Player 2:
5q + 2(1-q) = 1q + 3(1-q)
=> 5q + 2 – 2q = q + 3 – 3q
=> 3q + 2 = 3 – 2q
=> 5q = 1
=> q = 1/5, 1-q = (1 – 1/5) = 4/5
Summary: The pure strategy Nash Equilibrium is (Yellow,Yellow). Even though Player 1 does not have a mixed strategy Nash Equilibrium, Player 2’s mixed strategy is (Yellow = 1/5, Green = 4/5)
For additional practice I recommend you look for available problems online. Feel free to leave comments or feedback below.
## Mixed Strategy Nash Equilibrium III
Is a mixed strategy optimal?
So far we have solved and analyzed games with only a mixed strategy Nash Equilibrium. Today we will analyze a scenario where both mixed and pure strategies exist. On occasion my wife and I are too lazy to cook and choose to go out to dinner. However, this is where the fun really begins. Like other couples, sometimes we don’t always agree on where to eat and have our own preferences.
This can be represented in the following game:
Me\Wife Mexican BBQ Mexican 2,3 0,0 BBQ 0,0 3,2
On this night, I prefer to eat BBQ (love tri tip) over my wife’s preference of Mexican food. As you can see in this game, I would rather eat Mexican food over eating alone and my wife would rather have BBQ over eating alone. Eating alone represents a payoff of 0. In this game we have two pure strategy Nash Equilibrium represented as (Mexican, Mexican) and (BBQ, BBQ).
As we have seen in the past few blogs a mixed strategy Nash Equilibrium exists as well. Let’s solve for this equilibrium.
q 1-q Me\Wife Mexican BBQ p Mexican 2,3 0,0 1-p BBQ 0,0 3,2
Using the information above we will first solve for p and 1-p:
3(p) + 0(1-p) = 0(p) + 2(1-p)
=> 3p = 2 – 2p
=> 5p = 2
=> p = 2/5 and 1-p = 3/5
Now let’s solve for q and 1-q:
2(q) + 0(1-q) = 0(q) + 3(1-q)
=> 2q = 3 – 3q
=> 5q = 3
=> q = 3/5 and 1-q = 2/5
Here is an update of the game matrix with the mixed strategies.
q = 3/5 1-q = 2/5 Me\Wife Mexican BBQ p = 2/5 Mexican 2,3 0,0 1-p = 3/5 BBQ 0,0 3,2
With the probabilities of each strategy being chosen solved, let’s determine the payouts for both my wife and I choosing a mixed strategy. We first need to determine the probability of us choosing:
(Mexican, Mexican) = (p)(q) = (2/5)(3/5) = 6/25
(Mexican, BBQ) = (p)(1-q) = (2/5)(2/5) = 4/25
(BBQ, Mexican) = (1-p)(q) = (3/5)(3/5) = 9/25
(BBQ, BBQ) = (1-p)(1-q) = (3/5)(2/5) = 6/25
Next we need to take summation of each individual player’s pure strategy payoff multiplied by the corresponding probability in that quadrant.
My payoff: (2)(6/25) + (0)(9/25) + (0)(4/25) + (3)(6/25) = 1.2
Wife’s payoff: (3)(6/25) + (0)(9/25) + (0)(4/25) + (2)(6/25) = 1.2
For this game my payoffs are (3, 2, and 1.2) for (BBQ, Mexican, and Mixed) and my wife’s are (3, 2, and 1.2) for (Mexican, BBQ, and Mixed). Looking at these payoffs, is a mixed strategy Nash Equilibrium optimal? We clearly don’t want to eat dinner alone (payoff of 0) and we run the risk of this occurring if we choose a mixed strategy. In addition, even though I favor BBQ over Mexican food, Mexican food with my wife is a better payoff than choosing a mix.
I can summarize my preferences as BBQ > Mexican > Mixed (3 > 2 > 1.2). Since we assume rational decisions, my wife will approach this game similarly. Her preferences are Mexican > BBQ > Mixed (3 > 2 > 1.2). It is not optimal for us to choose a mixed strategy. Plus, why would you want to choose a mixed strategy on your wife or husband in the first place? Are you trying to trick/confuse him or her? Sometimes it is best to concede to your spouse’s preference for the night and he or she will return the favor another night.
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https://support.office.com/en-us/article/ASIN-function-81fb95e5-6d6f-48c4-bc45-58f955c6d347?CTT=5&origin=HA010342655&CorrelationId=72794e28-ec6b-4292-b647-1eeb1e5e18ab&ui=en-US&rs=en-US&ad=US
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ASIN function - Office Support
# ASIN function
This article describes the formula syntax and usage of the ASIN function in Microsoft Excel.
## Description
Returns the arcsine, or inverse sine, of a number. The arcsine is the angle whose sine is number. The returned angle is given in radians in the range -pi/2 to pi/2.
## Syntax
ASIN(number)
The ASIN function syntax has the following arguments:
• Number Required. The sine of the angle you want and must be from -1 to 1.
## Remark
To express the arcsine in degrees, multiply the result by 180/PI( ) or use the DEGREES function.
## Example
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Formula Description Result =ASIN(-0.5) Arcsine of -0.5 in radians, -pi/6 -0.523598776 =ASIN(-0.5)*180/PI() Arcsine of -0.5 in degrees -30 =DEGREES(ASIN(-0.5)) Arcsine of -0.5 in degrees -30
Applies To: Excel 2016, Excel 2010, Excel Starter, Excel 2013, Excel Online, Excel 2016 for Mac, Excel for Mac 2011, Excel 2007
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# Solve 1-(x-3)/3⩽1/2
#### skeeter
##### Well-known member
MHB Math Helper
$1 - \dfrac{x-3}{3} \le \dfrac{1}{2}$
multiply every term by the common denominator, $6$ ...
$6 - 2(x-3) \le 3$
can you finish it?
Yes, thank you
-2x - 6 ≤ -3
-2x ≤ -9
x ≤ 9/2
#### skeeter
##### Well-known member
MHB Math Helper
Yes, thank you
-2x - 6 ≤ -3
-2x ≤ -9
x ≤ 9/2
... what happens when you divide both sides of an inequality by a negative number ???
#### HallsofIvy
##### Well-known member
MHB Math Helper
Yes, thank you
-2x - 6 ≤ -3
-2x ≤ -9
x ≤ 9/2
If x= -2 then x is certainly less than 9/2= 4 but -2x- 6= 4- 6= -2 is not less than -3 so that can't be right.
#### frctl
##### New member
Correction
6 - 2(x - 3) ≤ 3
-2x - 6 ≤ -3
-2x ≤ -3
flip inequality sign
x ≥ 3/2
#### skeeter
##### Well-known member
MHB Math Helper
Correction
6 - 2(x - 3) ≤ 3
-2x - 6 ≤ -3 should be -2x + 6 < -3
-2x ≤ -3
flip inequality sign
x ≥ 3/2
correction again
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http://tushar-mehta.com/misc_tutorials/project_euler/euler014.html
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Project Euler Problem 14
You are on the Home/Other Tutorials/Project Euler/Problem 14 page
Web This Site
# Project Euler - Problem 14
More about Project Euler.
## Problem description
The following iterative sequence is defined for the set of positive integers:
n n/2 (n is even)
n 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 40 20 10 16 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
## Solution
I delayed tackling this problem because the brute force approach did not appeal to me. I expected it to take well over the expectation of the duration for a Project Euler problem, which happens to be 1 minute.
The solution I finally came up with was a recursive routine that populates a table with results as they become available. The benefit of this approach is that as soon as we find an existing entry in the table, we know the length of the chain from that number on out and can use it. Since each even number will be halved in the first step, if we build the data base from 1 to 999,999, we will have the answer for each even number in just 1 step! For odd numbers, we will populate the table for numbers greater than the starting number itself. Here's an example. Consider the chain in the statement of the problem. In computing the length of the chain starting with 13, we also discover that the chain starting with 40 has a length of 9 and the chain starting with 20 has a length of 8. Further, we would not even have to compute the entire chain for 20 since the next number, 10, has already been analyzed (remember, we started from 1). So, the length of the chain for 13 requires building only 3 values (40, 20, and 10). In addition, we can update our table for the chain length for starting values 20 and 40.
As it turns out the best way to implement this method is to use a recursive approach. This way we let the OS / compiler do the heavy lifting of keeping track of what counter goes with what value.
The code below ran in about 50 seconds.
```Option Explicit
Function CheckOneNbr(ByVal aNbr As Variant, SeqCount As Collection) As Integer
Debug.Assert TypeName(aNbr) = "Decimal"
On Error GoTo AddData
CheckOneNbr = SeqCount(CStr(aNbr))
Exit Function
On Error GoTo 0
If CDec(Right(aNbr, 1)) Mod 2 = 0 Then CheckOneNbr = CheckOneNbr(aNbr / 2, SeqCount) + 1 _
Else CheckOneNbr = CheckOneNbr(3 * aNbr + 1, SeqCount) + 1
End Function
Sub Euler014()
Dim I As Long
Dim ProcTime As Single, MaxRslt As Integer, MaxStartNbr As Long
Dim SeqCount As Collection
ProcTime = Timer
Set SeqCount = New Collection
MaxRslt = 1: MaxStartNbr = 1
For I = 2 To 1000000 - 1
CheckOneNbr CDec(I), SeqCount
If MaxRslt < SeqCount(CStr(I)) Then
MaxRslt = SeqCount(CStr(I))
MaxStartNbr = I
End If
Next I
Debug.Print MaxRslt, MaxStartNbr, Timer - ProcTime
End Sub```
Given how long it took, I entertained the possibility that the recursive solution was not the way to go. So, I wrote and checked the performance of the "flat" routine. It ran in about 120 seconds on my laptop.
```Sub Euler014s()
Dim I As Long
Dim ProcTime As Single, MaxRslt As Integer, MaxStartNbr As Long
ProcTime = Timer
MaxRslt = 1: MaxStartNbr = 1
For I = 2 To 1000000 - 1
Dim Rslt As Variant, ThisSeqLen As Integer
Rslt = CDec(I): ThisSeqLen = 1
Do While Rslt <> 1
If Right(Rslt, 1) Mod 2 = 0 Then Rslt = Rslt / 2 _
Else Rslt = 3 * Rslt + 1
ThisSeqLen = ThisSeqLen + 1
Loop
If MaxRslt < ThisSeqLen Then MaxRslt = ThisSeqLen: MaxStartNbr = I
Next I
Debug.Print MaxRslt, MaxStartNbr, Timer - ProcTime
End Sub```
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https://www.vedantu.com/question-answer/goal-is-a-quadrilateral-in-which-goparallel-al-class-10-maths-cbse-60a6449f7f6dcb5d8fbaf038
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# GOAL is a quadrilateral in which $GO\parallel AL$. If $\angle G=\angle O={{40}^{\circ }}$ what are the measures of $\angle A$ and $\angle L$.
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Hint: In this problem, we are given that GOAL is a quadrilateral in which $GO\parallel AL$. If $\angle G=\angle O={{40}^{\circ }}$, we have to find the measures of $\angle A$ and $\angle L$. Here we can first draw the diagram. Here we have to use the interior angle property, where the sum of the internal and external angle in the same vertex is ${{180}^{\circ }}$. By using this property we can find the required angles.
Here, we are given that GOAL is a quadrilateral in which $GO\parallel AL$. If $\angle G=\angle O={{40}^{\circ }}$, we have to find the measures of $\angle A$ and $\angle L$.
We can now draw the diagram.
Here we can use the property of interior angle.
We know that the interior angle property is where the sum of the internal and external angle in the same vertex is ${{180}^{\circ }}$.
We can now find the measure of $\angle A$,
$\Rightarrow \angle A+\angle O={{180}^{\circ }}$
We can now substitute the given measure and simplify it, we get
\begin{align} & \Rightarrow \angle A+{{40}^{\circ }}={{180}^{\circ }} \\ & \Rightarrow \angle A={{180}^{\circ }}-{{40}^{\circ }}={{140}^{\circ }} \\ \end{align}
The measure of $\angle A={{140}^{\circ }}$
We can now find the measure of $\angle L$,
$\Rightarrow \angle G+\angle L={{180}^{\circ }}$
We can now substitute the given measure and simplify it, we get
\begin{align} & \Rightarrow \angle L+{{40}^{\circ }}={{180}^{\circ }} \\ & \Rightarrow \angle L={{180}^{\circ }}-{{40}^{\circ }}={{140}^{\circ }} \\ \end{align}
The measure of $\angle L={{140}^{\circ }}$
Therefore, the measure of$\angle A=\angle L={{140}^{\circ }}$
Note: We should always remember the interior angle property that the interior angle property is where the sum of internal and external angle in the same vertex is ${{180}^{\circ }}$. So if we have one of the ongles, then we can substitute in the sum to get the value of another angle.
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## lbartman.com - the pro math teacher
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# Multiplying Fractions And Mixed Numbers Worksheets
Public on 02 Oct, 2016 by Cyun Lee
### multiplying fractions and mixed numbers education
Name : __________________
Seat Num. : __________________
Date : __________________
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show printable version !!!hide the show
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### Teachers That Give
I'm so thankful you have chosen to stop by today. I'm teaming up with some fabulous bloggers for this Teachers That Give blog hop! I'm so thankful for the teaching community and am thrilled to have an opportunity to give back. Today I have a new reading craft to share with you and a chance to win a \$25 TpT gift card!
As your students are reading and responding to literature this month, you can use this gift craft to keep things fresh and exciting.
When opened, these little gifts reveals five spaces where students can respond to their text. I have included lined templates and blank templates which students can mix-and-match.
Here's a look at how I fold each flap in to create the final product.
And the final product:
Because this is a gift craft, I have also included printable gift tags. These can be printed on colored paper to keep things festive.
Each of these gifts can be made using 9x12 construction paper and the printable templates. You can also grab 12x12 scrapbook paper and cut them down to size. I used the excess strips of scrapbook paper to create a decorative wrapping to the outside of the gifts.
You can find the templates, directions, and reading response ideas for this FREE reading craft in my TpT shop.
Now that you've grabbed your freebie, you can also enter to win a \$25 TpT gift card! This one will end on Friday, December 2nd.
a Rafflecopter giveaway
Don't forget to check out these blogs for seasonal ideas, freebies, and even more chances to win gift cards!
Happy Teaching!
### Turkey Trios (Freebie for 3-Addends)
Are your students working on finding the sum of 3-addends this month? Here's a great way to get your students up and moving during math time!
If you are a Common Core classroom, 3-addends is a first grade standard:
CCSS 1.0A.A.2: Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
However, my 2nd grade students still saw 3-addends in their daily work when using our curriculum materials. We focused on strategies for solving (look for tens partners, doubles, number neighbors, etc.). Most of my students were able to tackle these problems, but some still needed more practice so I put together these Turkey Trios.
There are three ways I have used this product in my own classroom in order to meet the needs of my students: whole group, small group, and independent work.
When the whole class needed practice, I printed the cards onto colored cardstock, cut them, and placed them around the classroom. Students carried their recording sheets on a clipboard and completed the task cards at their own pace. When students finished, I could quickly check their work using the answer key and send them back to make corrections, as needed.
When I only had a handful of students that needed practice, I could work with them at the back table. Instead of having the students walk around the room to work, I had the cards on a ring. We could tackle each card together and focus on which strategies to use. This was a perfect time to offer tools like number lines and 20-frames.
I have also used these cards for independent practice. When printing, I select to print 4 per page. This way the students have all of the cards and the recording sheet in one place while they work. I had students color in the turkey feathers used and then record the numbers on the recording sheet form.
You can grab this freebie here:
Happy Teaching!
### 5 Tips for Successful Conferences
Conferences are a great time to connect with families, but they can be stressful! Here are some tips I keep in mind every time I'm getting ready for conference week.
It can be very frustrating to put time and effort into preparing for a conference where the family doesn't even show up! One of the best ways to ensure your families make it to their conference is to send out more than one reminder.
Whether you are discussing student behaviors, work habits, or academic progress, make sure you have student work samples and assessment data to back it up! You might even grab a copy of the previous report card just in case.
One of the best pieces of advice I was ever given was to start and end with a positive. Families need to know that you care about their child and see them in a positive light.
Everyone's time is valuable! Keep a checklist of all of the items you need to discuss during each conference and use this checklist to keep yourself on track. Save time at the end for questions or concerns. If a family shows up late or there are more concerns than you originally thought, offer to schedule another conference on a different day. If you let one conference run over, it will have a domino effect on the rest of your day!
Take notes on any questions or concerns brought up during the conference. When families offer tips or suggestions that have worked in previous years, write them down! You will thank yourself later when some of the conversations begin to fade and your notes will remind you of all of the items you promised you would follow-up on.
One last thing to keep in mind as you approach any conference:
Have a great day!
### The Very Stuffed Turkey (Turkey Craft & Book Companion)
Are you still looking for a great read aloud before Thanksgiving? You should grab a copy of The Very Stuffed Turkey by Katharine Kenah!
Picture books are an amazing resource to introduce vocabulary words and to check for comprehension. The amount of text and content is manageable and the pictures support your struggling readers. You will find vocabulary word cards as well as a match-up page in this pack. You'll also find comprehension questions that can be used for group discussion or for students to respond to in writing. I also like to review story elements of picture books. Often I will give each table group a spinner and have the students work together to identify the elements.
After reading I like to deepen our connections with the text using writing prompts that get my students thinking about and beyond the text. I have the students glue their prompts into their notebooks. By the end of the year, these notebooks become a wonderful way for students to see how much they have grown as readers during the school year.
This book is jam packed with verbs! I love using this sorting activity to make sure my students can identify and distinguish the verbs from familiar nouns used in the text.
If you're looking for that something extra, you'll have to check out this turkey craft that doubles as a reading or writing response booklet! There are three writing templates included: full sheet of lines, half a sheet of lines, and a blank template. You can pick and choose which template will work best for your readers!
You can find all of these resources here:
### The Littlest Pilgrim (Book Companion Freebie)
The Littlest Pilgrim by Brandi Dougherty is a sweet little story about friendship that I love to read before Thanksgiving. Here are some resources you can use in your classroom, too!
If you aren't familiar with the story, Mini is the littlest Pilgrim in her village. She is too little to sew, too little to bake, and too little to fish. But she's not too little to make a friend!
Because this story is short, it is the perfect opportunity for my students to practice summarizing. We use the frame Somebody-Wanted-But-So-Then to guide us:
There is a student printable for this included in the pack.
I often use comprehension questions to check for understanding and writing prompts to have my students respond to the text. These cards are perfect because I can pick and choose the questions or prompts I want and students glue them into their black and white notebooks:
You can find all of these resources in this mini book companion
]
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# Recursion Theory/Incompleteness Theorems: Computability of sets of formulas in first order logic
I am struggling with the following two problems:
1. Suppose that $$M$$ is a structure with finite universe and finite alphabet. Show that the set of formulas $$\{\varphi$$ $$\mid$$ for every $$M$$-assignment $$\nu$$ of the variables, $$(M,\nu) \models \varphi\}$$ is computable.
2. Give an example of a finite language such that the set of formulas $$\{\varphi$$ $$\mid$$ for every finite structure $$M$$ which interprets that language and every $$M$$-assignment $$\nu$$ of the variables, $$(M,\nu) \models \varphi\}$$ is not computable.
Note that in the above problems, "$$(M,\nu) \models \varphi$$" is to be read as "$$\varphi$$ holds in $$M$$ when the variables of $$\varphi$$ are evaluated in $$M$$ according to $$\nu$$."
Now let $$V:= \{\varphi \mid \varphi$$ is a validity $$\}$$, where "A formula $$\varphi$$ is a validity" means "For all $$(M,\nu)$$, if $$(M,\nu)$$ interprets all the nonlogical symbols of $$\varphi$$, then $$(M,\nu) \models \varphi$$." I know that $$V \geq_m H$$ and therefore $$V$$ is not computable. The example sought by the second problem would seem to involve this theorem.
Unfortunately I can't see much further beyond this. Would anyone be so kind to share any hints or remarks that would help me begin to understand how to work towards a solution of these?
Thank you so much!
• How do you define computable? Surely you can come up with an algorithm that solves 1. – Andrés E. Caicedo Nov 17 '18 at 21:23
• Hi, thanks for your response. The definition of a computable set is this: en.wikipedia.org/wiki/Recursive_set – Rebecca Bonham Nov 17 '18 at 21:34
• Intuitively, I understand the idea of algorithmically deciding whether each $\varphi$ is belongs to the set defined in the first problem, but am having trouble translating that idea into a computable function, i.e. exhibiting an algorithm, if that makes sense. – Rebecca Bonham Nov 17 '18 at 21:37
• For part 1, what level of detail do you need to go into to describe the algorithm? If you are given a finite model, a formula $\phi$, it is a finite task to check whether $\phi$ holds under every interpretation of its free variables in $M$. (because it is a finite task to enumerate all the interpretations and then, for each interpretation it is a finite task to check whether $\phi$ holds). – Rob Arthan Nov 17 '18 at 23:41
• If pressed to give more details of something like this in my own work, I would write some mathematical style pseudo-code. I can't really comment on how much explicit detail is required in your case. – Rob Arthan Nov 17 '18 at 23:53
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Expand menu menu_open Minimize Go to startpage Home History history History expand_more
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An equation can be rewritten by changing all signs. This is equivalent to multiply or divide both sides by $\text{-}1.$ \begin{aligned} \text{-} x=3 \quad &\Leftrightarrow \quad x=\text{-} 3 \\ \text{-} x^2 -7x+4=0 \quad &\Leftrightarrow \quad x^2+7x-4=0 \end{aligned}
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Question:-
time estimates of an activity in a PERT network are :
Optimistic time t0 = 9 days; pessimistic time tp = 21 days and most likely time tm = 15 days. The approximates probability of completion of the activity in 13 days is :
Option (A)
16%
Option (B)
34%
Option(C)
50%
Option(D)
84%
Correct Option:
(A)
Question Solution:
texpected = ((t0 + 4tm + tp) / 6)
= {(9+(4 x 15)+21) / 6} = 15
S.D. = ((b-a) / 6) = ((21-9) / 6) = 2
Z = ((x-`x)/ s) = ((13-15)/2) = -1
P(-1) = 0.1586
~ 0.16
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# BigIntIsIntegral
### Related Docs: trait BigIntIsIntegral | package Numeric
#### implicit object BigIntIsIntegral extends BigIntIsIntegral with BigIntOrdering
Source
Numeric.scala
Linear Supertypes
Ordering
1. Alphabetic
2. By inheritance
Inherited
1. BigIntIsIntegral
2. BigIntOrdering
3. BigIntIsIntegral
4. Integral
5. Numeric
6. Ordering
7. PartialOrdering
8. Equiv
9. Serializable
10. Serializable
11. Comparator
12. AnyRef
13. Any
1. Hide All
2. Show all
Visibility
1. Public
2. All
### Type Members
1. #### class IntegralOps extends Ops
Definition Classes
Integral
2. #### class Ops extends AnyRef
Definition Classes
Numeric
### Value Members
1. #### final def !=(arg0: Any): Boolean
Test two objects for inequality.
Test two objects for inequality.
returns
`true` if !(this == that), false otherwise.
Definition Classes
AnyRef → Any
2. #### final def ##(): Int
Equivalent to `x.hashCode` except for boxed numeric types and `null`.
Equivalent to `x.hashCode` except for boxed numeric types and `null`. For numerics, it returns a hash value which is consistent with value equality: if two value type instances compare as true, then ## will produce the same hash value for each of them. For `null` returns a hashcode where `null.hashCode` throws a `NullPointerException`.
returns
a hash value consistent with ==
Definition Classes
AnyRef → Any
3. #### final def ==(arg0: Any): Boolean
The expression `x == that` is equivalent to `if (x eq null) that eq null else x.equals(that)`.
The expression `x == that` is equivalent to `if (x eq null) that eq null else x.equals(that)`.
arg0
the object to compare against this object for equality.
returns
`true` if the receiver object is equivalent to the argument; `false` otherwise.
Definition Classes
AnyRef → Any
4. #### def abs(x: BigInt): BigInt
Definition Classes
Numeric
5. #### final def asInstanceOf[T0]: T0
Cast the receiver object to be of type `T0`.
Cast the receiver object to be of type `T0`.
Note that the success of a cast at runtime is modulo Scala's erasure semantics. Therefore the expression `1.asInstanceOf[String]` will throw a `ClassCastException` at runtime, while the expression `List(1).asInstanceOf[List[String]]` will not. In the latter example, because the type argument is erased as part of compilation it is not possible to check whether the contents of the list are of the requested type.
returns
Definition Classes
Any
Exceptions thrown
`ClassCastException` if the receiver object is not an instance of the erasure of type `T0`.
6. #### def clone(): AnyRef
Create a copy of the receiver object.
Create a copy of the receiver object.
The default implementation of the `clone` method is platform dependent.
returns
a copy of the receiver object.
Attributes
protected[java.lang]
Definition Classes
AnyRef
Annotations
( ... )
Note
not specified by SLS as a member of AnyRef
7. #### def compare(x: BigInt, y: BigInt): Int
Returns an integer whose sign communicates how x compares to y.
Returns an integer whose sign communicates how x compares to y.
The result sign has the following meaning:
• negative if x < y
• positive if x > y
• zero otherwise (if x == y)
Definition Classes
BigIntOrderingOrdering → Comparator
8. #### final def eq(arg0: AnyRef): Boolean
Tests whether the argument (`arg0`) is a reference to the receiver object (`this`).
Tests whether the argument (`arg0`) is a reference to the receiver object (`this`).
The `eq` method implements an equivalence relation on non-null instances of `AnyRef`, and has three additional properties:
• It is consistent: for any non-null instances `x` and `y` of type `AnyRef`, multiple invocations of `x.eq(y)` consistently returns `true` or consistently returns `false`.
• For any non-null instance `x` of type `AnyRef`, `x.eq(null)` and `null.eq(x)` returns `false`.
• `null.eq(null)` returns `true`.
When overriding the `equals` or `hashCode` methods, it is important to ensure that their behavior is consistent with reference equality. Therefore, if two objects are references to each other (`o1 eq o2`), they should be equal to each other (`o1 == o2`) and they should hash to the same value (`o1.hashCode == o2.hashCode`).
returns
`true` if the argument is a reference to the receiver object; `false` otherwise.
Definition Classes
AnyRef
9. #### def equals(arg0: Any): Boolean
The equality method for reference types.
The equality method for reference types. Default implementation delegates to `eq`.
See also `equals` in scala.Any.
returns
`true` if the receiver object is equivalent to the argument; `false` otherwise.
Definition Classes
AnyRef → Any
10. #### def equiv(x: BigInt, y: BigInt): Boolean
Return true if `x` == `y` in the ordering.
Return true if `x` == `y` in the ordering.
Definition Classes
OrderingPartialOrderingEquiv
11. #### def finalize(): Unit
Called by the garbage collector on the receiver object when there are no more references to the object.
Called by the garbage collector on the receiver object when there are no more references to the object.
The details of when and if the `finalize` method is invoked, as well as the interaction between `finalize` and non-local returns and exceptions, are all platform dependent.
Attributes
protected[java.lang]
Definition Classes
AnyRef
Annotations
( classOf[java.lang.Throwable] )
Note
not specified by SLS as a member of AnyRef
12. #### def fromInt(x: Int): BigInt
Definition Classes
BigIntIsIntegralNumeric
13. #### final def getClass(): Class[_]
A representation that corresponds to the dynamic class of the receiver object.
A representation that corresponds to the dynamic class of the receiver object.
The nature of the representation is platform dependent.
returns
a representation that corresponds to the dynamic class of the receiver object.
Definition Classes
AnyRef → Any
Note
not specified by SLS as a member of AnyRef
14. #### def gt(x: BigInt, y: BigInt): Boolean
Return true if `x` > `y` in the ordering.
Return true if `x` > `y` in the ordering.
Definition Classes
OrderingPartialOrdering
15. #### def gteq(x: BigInt, y: BigInt): Boolean
Return true if `x` >= `y` in the ordering.
Return true if `x` >= `y` in the ordering.
Definition Classes
OrderingPartialOrdering
16. #### def hashCode(): Int
The hashCode method for reference types.
The hashCode method for reference types. See hashCode in scala.Any.
returns
the hash code value for this object.
Definition Classes
AnyRef → Any
17. #### final def isInstanceOf[T0]: Boolean
Test whether the dynamic type of the receiver object is `T0`.
Test whether the dynamic type of the receiver object is `T0`.
Note that the result of the test is modulo Scala's erasure semantics. Therefore the expression `1.isInstanceOf[String]` will return `false`, while the expression `List(1).isInstanceOf[List[String]]` will return `true`. In the latter example, because the type argument is erased as part of compilation it is not possible to check whether the contents of the list are of the specified type.
returns
`true` if the receiver object is an instance of erasure of type `T0`; `false` otherwise.
Definition Classes
Any
18. #### def lt(x: BigInt, y: BigInt): Boolean
Return true if `x` < `y` in the ordering.
Return true if `x` < `y` in the ordering.
Definition Classes
OrderingPartialOrdering
19. #### def lteq(x: BigInt, y: BigInt): Boolean
Return true if `x` <= `y` in the ordering.
Return true if `x` <= `y` in the ordering.
Definition Classes
OrderingPartialOrdering
20. #### def max(x: BigInt, y: BigInt): BigInt
Return `x` if `x` >= `y`, otherwise `y`.
Return `x` if `x` >= `y`, otherwise `y`.
Definition Classes
Ordering
21. #### def min(x: BigInt, y: BigInt): BigInt
Return `x` if `x` <= `y`, otherwise `y`.
Return `x` if `x` <= `y`, otherwise `y`.
Definition Classes
Ordering
22. #### def minus(x: BigInt, y: BigInt): BigInt
Definition Classes
BigIntIsIntegralNumeric
23. #### implicit def mkNumericOps(lhs: BigInt): IntegralOps
Definition Classes
IntegralNumeric
24. #### implicit def mkOrderingOps(lhs: BigInt): BigIntIsIntegral.Ops
This implicit method augments `T` with the comparison operators defined in `scala.math.Ordering.Ops`.
This implicit method augments `T` with the comparison operators defined in `scala.math.Ordering.Ops`.
Definition Classes
Ordering
25. #### final def ne(arg0: AnyRef): Boolean
Equivalent to `!(this eq that)`.
Equivalent to `!(this eq that)`.
returns
`true` if the argument is not a reference to the receiver object; `false` otherwise.
Definition Classes
AnyRef
26. #### def negate(x: BigInt): BigInt
Definition Classes
BigIntIsIntegralNumeric
27. #### final def notify(): Unit
Wakes up a single thread that is waiting on the receiver object's monitor.
Wakes up a single thread that is waiting on the receiver object's monitor.
Definition Classes
AnyRef
Note
not specified by SLS as a member of AnyRef
28. #### final def notifyAll(): Unit
Wakes up all threads that are waiting on the receiver object's monitor.
Wakes up all threads that are waiting on the receiver object's monitor.
Definition Classes
AnyRef
Note
not specified by SLS as a member of AnyRef
29. #### def on[U](f: (U) ⇒ BigInt): Ordering[U]
Given f, a function from U into T, creates an Ordering[U] whose compare function is equivalent to:
Given f, a function from U into T, creates an Ordering[U] whose compare function is equivalent to:
`def compare(x:U, y:U) = Ordering[T].compare(f(x), f(y))`
Definition Classes
Ordering
30. #### def one: BigInt
Definition Classes
Numeric
31. #### def plus(x: BigInt, y: BigInt): BigInt
Definition Classes
BigIntIsIntegralNumeric
32. #### def quot(x: BigInt, y: BigInt): BigInt
Definition Classes
BigIntIsIntegralIntegral
33. #### def rem(x: BigInt, y: BigInt): BigInt
Definition Classes
BigIntIsIntegralIntegral
34. #### def reverse: Ordering[BigInt]
Return the opposite ordering of this one.
Return the opposite ordering of this one.
Definition Classes
OrderingPartialOrdering
35. #### def signum(x: BigInt): Int
Definition Classes
Numeric
36. #### final def synchronized[T0](arg0: ⇒ T0): T0
Definition Classes
AnyRef
37. #### def times(x: BigInt, y: BigInt): BigInt
Definition Classes
BigIntIsIntegralNumeric
38. #### def toDouble(x: BigInt): Double
Definition Classes
BigIntIsIntegralNumeric
39. #### def toFloat(x: BigInt): Float
Definition Classes
BigIntIsIntegralNumeric
40. #### def toInt(x: BigInt): Int
Definition Classes
BigIntIsIntegralNumeric
41. #### def toLong(x: BigInt): Long
Definition Classes
BigIntIsIntegralNumeric
42. #### def toString(): String
Creates a String representation of this object.
Creates a String representation of this object. The default representation is platform dependent. On the java platform it is the concatenation of the class name, "@", and the object's hashcode in hexadecimal.
returns
a String representation of the object.
Definition Classes
AnyRef → Any
43. #### def tryCompare(x: BigInt, y: BigInt): Some[Int]
Returns whether a comparison between `x` and `y` is defined, and if so the result of `compare(x, y)`.
Returns whether a comparison between `x` and `y` is defined, and if so the result of `compare(x, y)`.
Definition Classes
OrderingPartialOrdering
44. #### final def wait(): Unit
Definition Classes
AnyRef
Annotations
( ... )
45. #### final def wait(arg0: Long, arg1: Int): Unit
Definition Classes
AnyRef
Annotations
( ... )
46. #### final def wait(arg0: Long): Unit
Definition Classes
AnyRef
Annotations
( ... )
47. #### def zero: BigInt
Definition Classes
Numeric
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Bookkeeping
# Annuity Table: Overview, Examples, and Formulas
Say you own a fixed annuity that pays a set amount of \$10,000 every year. The terms of your contract state that you will hold the annuity for seven years at a guaranteed effective interest rate of 3.25%. You’ve owned the annuity for five years and now have two annual payments left. You can then look up the present value interest factor in the table and use this value as a factor in calculating the present value of an annuity, series of payments.
A 10-year annuity paying \$3,500 per year at a 5% discount rate gives a present value of approximately \$27,026. This was calculated by finding the cell in the Year 10 row and 5% column (7.7217) and multiplying it by \$3,500. Using an annuity calculator or a financial spreadsheet set up for calculating the present value of an annuity is often more precise than using the preset annuity table.
## Rate Table For the Present Value of an Ordinary Annuity of 1
You want to sell five years’ worth of payments (\$5,000) and the secondary market buying company applies a 10% discount rate. Therefore, the present value of five \$1,000 structured settlement payments is worth roughly \$3,790.75 when a 10% discount rate is applied. It lets you compare the amount you would receive from an annuity’s series of payments over time to the value of what you would receive for a lump sum payment for the annuity right now. The formulas described above make it possible—and relatively easy, if you don’t mind the math—to determine the present or future value of either an ordinary annuity or an annuity due.
The “section 7520 rate,” which is 120 percent of the applicable Federal mid-term rate in effect for the month when the valuation date falls, rounded to the nearest 2 tenths of a percent. Get instant access to lessons taught by experienced Specialized Tax Services STS accounting method: PwC private equity pros and bulge bracket investment bankers including financial statement modeling, DCF, M&A, LBO, Comps and Excel Modeling. John Egan is a freelance writer, editor and content marketing strategist in Austin, Texas.
## Understanding the Value of an Annuity
Bonds are often ordinary annuities because they are paid at the end of a period. Payments are made at the end of every period into an account until the bond matures. The interest rate and period of time before maturity are also fixed. An annuity due is the type of annuity that requires a payment at the beginning of a period. A car payment or house payment would be good examples of an annuity due. You make a payment at the first of each month, and each month thereafter on the same date, until the end of the defined term.
Therefore, while the decision is not clear-cut, the process still aids in decision-making since calculating the present value of these annuities takes the time value of money into account. This gives a baseline and provides clear, money-based outcomes depending on the choices. For example, if the person doesn’t need any money for https://turbo-tax.org/best-law-firm-accounting-software-in-2023/ the foreseeable future, then investing that \$50,000 for 6 years might be the best choice. However, if they’d rather receive \$10,000 yearly instead of waiting for the entire 6 years, they might choose that. The effect of the discount rate on the future value of an annuity is the opposite of how it works with the present value.
## Get 5 FREE Video Lessons With Uncommon Insights To Accelerate Your Financial Growth
Annuities are either lump-sum payments or multiple payments made at regular intervals. The deposits made to savings accounts, monthly rent payments, and retirement pensions are considered annuities. The payments received from an annuity are reported as income, and the amount of tax to be paid depends on the product. For example, if we wanted to determine the present value of receiving \$2,000 per year for 7 years at an 8% discount rate, we simply multiply \$2,000 by 5.2064, giving us approximately \$10,413. (5.2064 was obtained from the cell in the Year 7 row and 8% column).
• Use this calculator to find the present value of annuities due, ordinary regular annuities, growing annuities and perpetuities.
• Simply select the correct interest rate and number of periods to find your factor in the intersecting cell.
• Any variations you find among present value tables for ordinary annuities are due to rounding.
• The present value of an annuity is the amount of money needed today to cover future annuity payments.
• Our expert reviewers hold advanced degrees and certifications and have years of experience with personal finances, retirement planning and investments.
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https://documen.tv/question/a-rock-is-thrown-straight-upward-with-an-initial-velocity-of-20-m-s-where-the-downward-accelerat-17830434-49/
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## A rock is thrown straight upward with an initial velocity of 20 m/s where the downward acceleration due to gravity is -10 m/s^2. How long is
Question
A rock is thrown straight upward with an initial velocity of 20 m/s where the downward acceleration due to gravity is -10 m/s^2. How long is the ball in the air before it returns to the thrower’s hand?
in progress 0
4 months 2021-07-14T03:00:47+00:00 1 Answers 2 views 0
4 seconds
Explanation:
Given:
Δy = 0 m
v₀ = 20 m/s
a = -10 m/s²
Find: t
Δy = v₀ t + ½ at²
0 m = (20 m/s) t + ½ (-10 m/s²) t²
0 = 20t − 5t²
0 = 5t (4 − t)
t = 4
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http://mathhelpforum.com/calculus/52729-hausdroff.html
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1. Hausdroff
If $x,y$ are two distinct points in a metric space $X$ then there exists $r > 0$ such that $B_{r}(x)$ and $B_{r}(y)$ are disjoint.
So let $p = d(x,y) > 0$. Choose $r = p/2$. Then $B_{r}(x)$ and $B_{r}(y)$ are disjoint for all $l \leq r$?
Is this right? e.g. $r = \sup A$ where $A = \{l: B_{r}(l) \ \text{and} \ B_{l}(y) \ \text{are disjoint} \}$ ?
2. Originally Posted by particlejohn
If $x,y$ are two distinct points in a metric space $X$ then there exists $r > 0$ such that $B_{r}(x)$ and $B_{r}(y)$ are disjoint.
So let $p = d(x,y) > 0$. Choose $r = p/2$. Then $B_{\color{red}l}(x)$ and $B_{\color{red}l}(y)$ are disjoint for all $l \leq r$?
If they're open balls, yep.
Is this right? e.g. $r = \sup A$ where $A = \{l: B_{r}(l) \ \text{and} \ B_{l}(y) \ \text{are disjoint} \}$ ?
Isn't there a problem with the indices?
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http://h3maths.edublogs.org/tag/1-equals-3/
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## Welcome to H3 Maths
Blog Support for Growing Mathematicians
## Posts tagged with 1 equals 3
### Can 1 = 3?
October26
by posted under Uncategorized | tagged under , , | Comments Off on Can 1 = 3?
### 1 = 3 What? Call the Math Doctor!
September24
Yes, you read about it here on H3! In Year 10 you learn the basic laws when working with logarithms. Do you remember what a logarithm is? If not, read more here from an earlier post. Then, take one of these laws (the log of a power) and you can prove, in about as many […]
by posted under Uncategorized | tagged under , , , , , , , | Comments Off on 1 = 3 What? Call the Math Doctor!
#### Post Support
NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]
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https://www.teacherspayteachers.com/Product/Unknown-Angles-Bingo-Game-PPT-with-Blank-Bingo-Card-7GB5-886942
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Total:
\$0.00
# Unknown Angles Bingo Game PPT with Blank Bingo Card 7.G.B.5
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
Compressed Zip File
How to unzip files.
2.88 MB | 40 pages
### PRODUCT DESCRIPTION
**These problems are all different than the other Bingo game in my store for the same standards. This version has updated graphics as well.
Included in this purchase is a 40 slide PowerPoint presentation for "calling" bingo with a different angle problem on each slide. There are 12 problems where the student must find the missing complementary angle, 12 problems with a missing supplementary angle, 5 problems with a missing vertical angle, and 7 problems with a missing adjacent angle. This is great for practicing 7.G.B.5: Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.
See picture previews for example.
Also included is a blank Bingo card (prints 2 per page). Simply print enough blank Bingo cards for everyone in your class. At the beginning of the presentation is a slide listing the 36 different angles used in the PPT. Students randomly choose 24 of the 36 angles and fill in their own Bingo card. (Make sure to tell them to put them in random order!)
Another option is to copy and paste the angles from the slide into one of the many free Bingo Card Generators online and print the cards for your class.
Once the Bingo cards are filled in/created, you are ready to play. Just click through the PowerPoint in order or rearrange the slides randomly making sure to keep track of the amounts you have shown so you can check the winner! I started using this method of playing Bingo years ago with my classes. The kids LOVE to fill in their own cards.
Can be played over and over again! Of course, this can just be used as a review without playing it as a Bingo game.
Total Pages
40
N/A
Teaching Duration
N/A
### Average Ratings
4.0
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https://www.physicsforums.com/threads/electronics-box-in-the-heat-situation.239964/
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# Electronics box in the heat situation
i'm designing a little project but i'm running into some environmental heat issues that i'm trying to work out. i just need a little clarification and insight from anyone willing.
I have this box made of sheet steel that is going to be mounted to the side of a building. the box will theoretically be air tight so that none of the environmental elements will harm the components inside. the components inside include electronics that will do up to 70W of power and are rated to work in an environmental condition (inside the box) up to 60 C. The thing i'm trying to figure out is what temperature it would have to be outside of the box (i'm first trying to do this without considering direct sunlight because that's really difficult to do) where it would put the box temperature over 60 C and destroy the components.
this is what i'm thinking. after some reasonable amount of time of letting the electronics run, some sort of equilibrium will occur between the temperatures inside and outside the box. at that point the rate of heat transfer due to conduction through the sheet metal will be equal to the rate of heat production by the electronics (70W).
so using fourier's law (the rate of heat conduction):
dQ/dt = k*A*(T_in - T_out) / x
where:
dQ/dt = rate of conduction
k = steel's conductivity constant
A = surface area exposed to environment
T_in = temp inside box
T_out = temp outside box
x = thickness of sheet metal
we can conclude that the temperature inside the box will only be slightly hotter than the temp outside.
So if what i did was true then the temperature would have to reach a really high 58 C or so to compromise the electronics functioning capacity?
berkeman
Mentor
You are going to need electronics rated higher than 60C to sit in a sealed metal box outside. Standard Industrial temperature range is -40C to +85C, and even +85C rated components may be problematic in a metal box outside in the sun. Why are you trying to use Commercial-grade components in an application that is at least Industrial-grade, and maybe even Automotive-grade?
the components are always a matter of price and availability to what is required for the project.
but can you comment on my analysis taking into account that the box will always be in the shade?
berkeman
Mentor
If it will always be in the shade, that will help some. Thermal conduction and heat rise for electronics is generally characterized in terms of a thermal conductivity theta, in units of degrees C per watt. A good heat sink will have a fairly low Theta (a few degrees C per watt or less), and straight convection cooling with stagnant air (like in your box) will have a much higher theta (many degrees C per watt).
Are you directly heat sinking this 70W to the metal box, or is only through the air inside? It's probably easiest to measure the theta numbers, as opposed to calculating them. They depend on how components are mounted to metal structures and such. BTW, your power components should have theta(J,A) (junction to ambient) numbers in their datasheets, for their various package and mounting options. All of the thetas add up to give you the overall temperature rise. You can look at a heat sink manufacturer's website for more info on the theta numbers and calculations.
i understand what your saying but i'm not sure if we're on the right page or if i'm just completely wrong. the components that i'm using are rated at 60C ambient temperature, meaning the temp inside the box. if they give a rating like that i'm assuming that they take into account the worst scenario of cooling (stagnant like you said) so that the components will function in an environment of 60C, not an internal operating temperature of 60C.
am i making any sense?
berkeman
Mentor
i understand what your saying but i'm not sure if we're on the right page or if i'm just completely wrong. the components that i'm using are rated at 60C ambient temperature, meaning the temp inside the box. if they give a rating like that i'm assuming that they take into account the worst scenario of cooling (stagnant like you said) so that the components will function in an environment of 60C, not an internal operating temperature of 60C.
am i making any sense?
I'm speaking from a designer's standpoint -- I design circuits and modules with components (transistors, resistors, ICs, etc.), and those components typically have either Industrial temperature range parts (-40C to +85C) or Commercial temperature range parts (0C to +70C). That temperature rating is for the body temperature of most parts, for example. There will generally be a different max junction temperature for semiconductor parts, higher than that max body temperature, and it's my job as the circuit/product designer to be sure that I'm using adequate cooling to ensure that the max junction temperature is not exceeded when the ambient environment is at max (70C or 85C).
It sounds like maybe you are dealing with an already-designed product that you want to put into an additional metal box for operation outside? 60C is a strange temperature rating as I mentioned, since it does not match standard component temperature operating ranges. If it is a product already, it assumes some ambient conditions associated with that 60C rating. Does it assume no heat sinking and only stagnant air up to 60C? If so, then you can just put 70W of light bulbs in the box and test the temperature of the air in the box after it stabilizes to see if it gets over 60C.
So when I design a circuit module or product, it is my job to use the theta(J,A) numbers for the semiconductor parts, using whatever cooling method I have (dead air, forced air, heat sinking, or whatever), and calculate
yes i am using prefabricated parts that i'll assemble to make my system. i'm working off the spec sheets of each part which says max ambient of 60C.
unfortunately i can't test the temps in the box because, 1) i don't have the box yet, and 2) i'm not in the environment that i'm going to be assembling it. that's why i'm going through all this trouble working out the math.
berkeman
Mentor
I think I'll move this thread to the ME/Aero forum for a bit to get some different views. There should be some folks there who do thermal conductivity work for metal enclosures in outside environments. If that doesn't work, I may move it again to EE.
I design equipment for the military that is fairly similar to what you're describing (electronic components in metal enclosures).
My immediate suggestions would be to A) vent the bottom of the box to allow cooler air in while still disallowing moisture (rain or snow) an entrance way and B) possibly provide a fan and/or heat sink over the known warmest components to help alleviate internal temperature rise. Also, if possible, consider making your enclosure out of aluminum, not steel. The coefficient of conduction is much higher (~180 W/m-K, depending on your source) than steel (12-27 W/m-K, depending on your source and material).
Your analysis in the first post was simply relying on the ambient air and the thermal conduction of the material to produce an inside temperature for the box. You are not taking into account heat produced from the components that becomes 'trapped' in the box.
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http://sagemath.org/doc/reference/combinat/sage/combinat/sf/homogeneous.html
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# Homogeneous symmetric functions¶
By this we mean the basis formed of the complete homogeneous symmetric functions $$h_\lambda$$, not an arbitrary graded basis.
class sage.combinat.sf.homogeneous.SymmetricFunctionAlgebra_homogeneous(Sym)
A class of methods specific to the homogeneous basis of symmetric functions.
INPUT:
• self – a homogeneous basis of symmetric functions
• Sym – an instance of the ring of symmetric functions
TESTS:
sage: h = SymmetricFunctions(QQ).e()
True
sage: TestSuite(h).run(skip=['_test_associativity', '_test_distributivity', '_test_prod'])
sage: TestSuite(h).run(elements = [h[1,1]+h[2], h[1]+2*h[1,1]])
class Element(M, x)
Create a combinatorial module element. This should never be called directly, but only through the parent combinatorial free module’s __call__() method.
TESTS:
sage: F = CombinatorialFreeModule(QQ, ['a','b','c'])
sage: B = F.basis()
sage: f = B['a'] + 3*B['c']; f
B['a'] + 3*B['c']
True
expand(n, alphabet='x')
Expand the symmetric function self as a symmetric polynomial in n variables.
INPUT:
• n – a nonnegative integer
• alphabet – (default: 'x') a variable for the expansion
OUTPUT:
A monomial expansion of self in the $$n$$ variables labelled by alphabet.
EXAMPLES:
sage: h = SymmetricFunctions(QQ).h()
sage: h([3]).expand(2)
x0^3 + x0^2*x1 + x0*x1^2 + x1^3
sage: h([1,1,1]).expand(2)
x0^3 + 3*x0^2*x1 + 3*x0*x1^2 + x1^3
sage: h([2,1]).expand(3)
x0^3 + 2*x0^2*x1 + 2*x0*x1^2 + x1^3 + 2*x0^2*x2 + 3*x0*x1*x2 + 2*x1^2*x2 + 2*x0*x2^2 + 2*x1*x2^2 + x2^3
sage: h([3]).expand(2,alphabet='y')
y0^3 + y0^2*y1 + y0*y1^2 + y1^3
sage: h([3]).expand(2,alphabet='x,y')
x^3 + x^2*y + x*y^2 + y^3
sage: h([3]).expand(3,alphabet='x,y,z')
x^3 + x^2*y + x*y^2 + y^3 + x^2*z + x*y*z + y^2*z + x*z^2 + y*z^2 + z^3
sage: (h([]) + 2*h([1])).expand(3)
2*x0 + 2*x1 + 2*x2 + 1
sage: h([1]).expand(0)
0
sage: (3*h([])).expand(0)
3
omega()
Return the image of self under the omega automorphism.
The omega automorphism is defined to be the unique algebra endomorphism $$\omega$$ of the ring of symmetric functions that satisfies $$\omega(e_k) = h_k$$ for all positive integers $$k$$ (where $$e_k$$ stands for the $$k$$-th elementary symmetric function, and $$h_k$$ stands for the $$k$$-th complete homogeneous symmetric function). It furthermore is a Hopf algebra endomorphism and an involution, and it is also known as the omega involution. It sends the power-sum symmetric function $$p_k$$ to $$(-1)^{k-1} p_k$$ for every positive integer $$k$$.
The images of some bases under the omega automorphism are given by
$\omega(e_{\lambda}) = h_{\lambda}, \qquad \omega(h_{\lambda}) = e_{\lambda}, \qquad \omega(p_{\lambda}) = (-1)^{|\lambda| - \ell(\lambda)} p_{\lambda}, \qquad \omega(s_{\lambda}) = s_{\lambda^{\prime}},$
where $$\lambda$$ is any partition, where $$\ell(\lambda)$$ denotes the length (length()) of the partition $$\lambda$$, where $$\lambda^{\prime}$$ denotes the conjugate partition (conjugate()) of $$\lambda$$, and where the usual notations for bases are used ($$e$$ = elementary, $$h$$ = complete homogeneous, $$p$$ = powersum, $$s$$ = Schur).
omega_involution() is a synonym for the omega() method.
OUTPUT:
• the image of self under the omega automorphism
EXAMPLES:
sage: h = SymmetricFunctions(QQ).h()
sage: a = h([2,1]); a
h[2, 1]
sage: a.omega()
h[1, 1, 1] - h[2, 1]
sage: e = SymmetricFunctions(QQ).e()
sage: e(h([2,1]).omega())
e[2, 1]
omega_involution()
Return the image of self under the omega automorphism.
The omega automorphism is defined to be the unique algebra endomorphism $$\omega$$ of the ring of symmetric functions that satisfies $$\omega(e_k) = h_k$$ for all positive integers $$k$$ (where $$e_k$$ stands for the $$k$$-th elementary symmetric function, and $$h_k$$ stands for the $$k$$-th complete homogeneous symmetric function). It furthermore is a Hopf algebra endomorphism and an involution, and it is also known as the omega involution. It sends the power-sum symmetric function $$p_k$$ to $$(-1)^{k-1} p_k$$ for every positive integer $$k$$.
The images of some bases under the omega automorphism are given by
$\omega(e_{\lambda}) = h_{\lambda}, \qquad \omega(h_{\lambda}) = e_{\lambda}, \qquad \omega(p_{\lambda}) = (-1)^{|\lambda| - \ell(\lambda)} p_{\lambda}, \qquad \omega(s_{\lambda}) = s_{\lambda^{\prime}},$
where $$\lambda$$ is any partition, where $$\ell(\lambda)$$ denotes the length (length()) of the partition $$\lambda$$, where $$\lambda^{\prime}$$ denotes the conjugate partition (conjugate()) of $$\lambda$$, and where the usual notations for bases are used ($$e$$ = elementary, $$h$$ = complete homogeneous, $$p$$ = powersum, $$s$$ = Schur).
omega_involution() is a synonym for the omega() method.
OUTPUT:
• the image of self under the omega automorphism
EXAMPLES:
sage: h = SymmetricFunctions(QQ).h()
sage: a = h([2,1]); a
h[2, 1]
sage: a.omega()
h[1, 1, 1] - h[2, 1]
sage: e = SymmetricFunctions(QQ).e()
sage: e(h([2,1]).omega())
e[2, 1]
SymmetricFunctionAlgebra_homogeneous.coproduct_on_generators(i)
Returns the coproduct on $$h_i$$.
INPUT:
• self – a homogeneous basis of symmetric functions
• i – a nonnegative integer
OUTPUT:
• the sum $$\sum_{r=0}^i h_r \otimes h_{i-r}$$
EXAMPLES:
sage: Sym = SymmetricFunctions(QQ)
sage: h = Sym.homogeneous()
sage: h.coproduct_on_generators(2)
h[] # h[2] + h[1] # h[1] + h[2] # h[]
sage: h.coproduct_on_generators(0)
h[] # h[]
#### Previous topic
Hall-Littlewood Polynomials
#### Next topic
Jack Symmetric Functions
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# Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed in the expansion ?
This question was previously asked in
NCL Assistant Foreman Mechanical (Trainee) 27 Dec 2020 Official Paper
View all Northern Coalfields Limited Papers >
1. 10 Joule
2. 80 Joule
3. Zero
4. -80 Joule
Option 3 : Zero
Free
NCL Assistant Foreman EE Full Test 1
3.7 K Users
100 Questions 100 Marks 90 Mins
## Detailed Solution
Explanation:
In case of expansion, work is done by the system. Now, as we know that,
W = -Pext, ΔV
As the gas is expanding is vaccum which has no pressure, i.e.,
Pext. = 0
∴ W = 0
Hence no work is done.
From first law of thermodynamics,
ΔU = q + W
As the system is working at constant temperature, i.e., isothermally.
∴ ΔU = 0
Hence q = -W = 0
Hence no heat is absorbed in the expansion.
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https://pypi.org/project/EMD-signal/
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Implementation of the Empirical Mode Decomposition (EMD) and its variations
PyEMD
Introduction
Python implementation of the Empirical Mode Decomposition (EMD). The package contains multiple EMD variations and intends to deliver more in time.
EMD variations
• Ensemble EMD (EEMD),
• "Complete Ensemble EMD" (CEEMDAN)
• different settings and configurations of vanilla EMD.
• Image decomposition (EMD2D & BEMD) (experimental, no support)
• Just-in-time compiled EMD (JitEMD)
PyEMD allows you to use different splines for envelopes, stopping criteria and extrema interpolations.
Available splines
• Natural cubic (default)
• Pointwise cubic
• Hermite cubic
• Akima
• PChip
• Linear
Available stopping criteria
• Cauchy convergence (default)
• Fixed number of iterations
• Number of consecutive proto-imfs
Extrema detection
• Discrete extrema (default)
• Parabolic interpolation
Installation
PyPi (recommended)
The quickest way to install package is through pip.
$pip install EMD-signal From source In case, if you only want to use EMD and its variations, the best way to install PyEMD is through pip. However, if you want to modify the code, anyhow you might want to download the code and build package yourself. The source is publicaly available and hosted on GitHub. To download the code you can either go to the source code page and click Code -> Download ZIP, or use git command line$ git clone https://github.com/laszukdawid/PyEMD
Installing package from source is done using command line:
EMD2D/BEMD
Unfortunately, this is Experimental and we can't guarantee that the output is meaningful. The simplest use is to pass image as monochromatic numpy 2D array. Sample as with the other modules one can use the default setting of an instance or, more explicitly, use the emd2d() method.
from PyEMD.EMD2d import EMD2D #, BEMD
import numpy as np
x, y = np.arange(128), np.arange(128).reshape((-1,1))
img = np.sin(0.1*x)*np.cos(0.2*y)
emd2d = EMD2D() # BEMD() also works
IMFs_2D = emd2d(img)
F.A.Q
Why is EEMD/CEEMDAN so slow?
Unfortunately, that's their nature. They execute EMD multiple times every time with slightly modified version. Added noise can cause a creation of many extrema which will decrease performance of the natural cubic spline. For some tweaks on how to deal with that please see Speedup tricks in the documentation.
Contact
Feel free to contact me with any questions, requests or simply to say hi. It's always nice to know that I've helped someone or made their work easier. Contributing to the project is also acceptable and warmly welcomed.
Citation
If you found this package useful and would like to cite it in your work please use the following structure:
@misc{pyemd,
author = {Laszuk, Dawid},
title = {Python implementation of Empirical Mode Decomposition algorithm},
year = {2017},
publisher = {GitHub},
journal = {GitHub Repository},
howpublished = {\url{https://github.com/laszukdawid/PyEMD}},
doi = {10.5281/zenodo.5459184}
}
Release history Release notifications | RSS feed
Uploaded source
Uploaded py3
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+0
# points of discontinuity question
0
37
5
What are the points of discontinuity? Are they all removable? TIA!
y = (x+5)(x+3) / x2+8x+15
Guest Apr 17, 2018
Sort:
#3
+2
The defined function is equivalent to
\(y=(x+5)(x+3)/[(x+5)(x+3)],\)
so the denominator is zero, and therefore \(y\) is undefined, at \(x = -5\) and \(x = -3\). For all other values of \(x \) we have \(y=1\), so we see that \(y\) is continuous everywhere it is defined. The questioner therefore has made a mistake, since continuity is only meaningful where a function is defined (see https://en.wikipedia.org/wiki/Classification_of_discontinuities#Removable_discontinuity). The points \(x = -5\) and \(x = -3\) are actually removable singularities which can simply be removed by defining \(y(-5)=y(-3)=1\) so that \(y(x)=1\) for all \(x\), and \(y(x)\) is then continuous everywhere.
Guest Apr 17, 2018
#4
0
thank you so much for replying! so the points are x = -5 and x = -3 and they are removable? just to clarify.
Guest Apr 17, 2018
#5
+21
0
I now wonder, did you understand how to solve these equations?
Let me try to explain to you, first you do the multiplication action between the parentheses. after that you have a square equation that you decide with the help of the formula the discriminant of the quadratic polynomial ax+ bx + c is b2- 4ac
This and any other information you can find on the homework help online and there to ask any questions you are interested in, I'm sure that they will help you and explain.
AKM17 Apr 17, 2018
### 26 Online Users
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# : Quadrilaterals and Their Properties
## Presentation on theme: ": Quadrilaterals and Their Properties"— Presentation transcript:
6.4 6.5 6.6: Quadrilaterals and Their Properties
Objectives: Be able to use properties of sides and angles of rhombuses, rectangles, squares, trapezoids and kites. Be able to use properties of diagonals of rhombuses, rectangles and squares. Be able to identify quadrilaterals based on limited information
A parallelogram with four congruent sides. Rhombus A parallelogram with four right angles. Rectangle A parallelogram with four congruent sides, and four right angles. Square
Corollaries Rhombus Corollary: A quadrilateral is a rhombus if and only if it has four congruent sides. Rectangle Corollary: A quadrilateral is a rectangle if and only if it has four right angles. Square Corollary: A quadrilateral is a square if and only if it is a rhombus and a rectangle. You can use these to prove that a quadrilateral is a rhombus, rectangle or square without proving first that the quadrilateral is a parallelogram.
Example: 1) Decide whether the statement is always, sometimes, or never. A. A rectangle is a square. B. A square is a rhombus.
Theorems Theorem 6.11 Theorem 6.12 Theorem 6.13
A parallelogram is a rhombus if and only if its diagonals are perpendicular. A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles. Theorem 6.12 Theorem 6.13 A parallelogram is a rectangle if and only if its diagonals are congruent.
Examples: 2) Which of the following quadrilaterals have the given property? All sides are congruent. All angles are congruent. The diagonals are congruent. Opposite angles are congruent. Parallelogram Rectangle Rhombus Square
Example: 3) In the diagram at the right, PQRS is a rhombus. What
is the value of y?
Trapezoids A trapezoid is a quadrilateral with exactly one pair of parallel sides. Bases: The parallel sides of a trapezoid. Legs: The nonparallel sides of the trapezoid. Isosceles Trapezoid: A trapezoid whose legs are congruent. Midsegment: A segment that connects the midpoints of the legs and that is parallel to each base. Its length is one half the sum of the lengths of the bases. Base Midsegment Leg Leg Base Angles Base
Isosceles Trapezoids A trapezoid that has congruent legs.
Theorem 6.14 Theorem 6.15 Theorem 6.16
A B If a trapezoid is isosceles, then each pair of base angles is congruent. D C A B If a trapezoid has a pair of congruent base angles, then it is an isosceles trapezoid. D C A B A trapezoid is isosceles if and only if its diagonals are congruent. C D
Example C D E F
Theorem 6.17: Midsegment of a trapezoid
The midsegment of a trapezoid is the segment that connects the midpoints of its legs. Theorem 6.17: Midsegment of a trapezoid The midsegment of a trapezoid is parallel to each base and its length is one half the sums of the lengths of the bases. MN║AD, MN║BC MN = ½ (AD + BC)
Example: 5) Find the length of the midsegment RT.
Definition A kite is a quadrilateral that has two pairs of consecutive congruent sides, but opposite sides are not congruent.
Kite Theorems Theorem 6.18 If a quadrilateral is a kite, then its diagonals are perpendicular. AC BD Theorem 6.19 If a quadrilateral is a kite, then exactly one pair of opposite angles is congruent. A ≅ C, B ≅ D
Example 6) Find the lengths of all four sides of the kite.
Example 7) Find mG and mJ in the diagram at the right. 132° 60°
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Next Article in Journal
One-Sign Order Parameter in Iron Based Superconductor
Next Article in Special Issue
Following Knots down Their Energy Gradients
Previous Article in Journal
Hidden Symmetries in Simple Graphs
Previous Article in Special Issue
Intrinsic Symmetry Groups of Links with 8 and Fewer Crossings
Symmetry 2012, 4(1), 225-250; https://doi.org/10.3390/sym4010225
Article
Classical Knot Theory
University of South Alabama, Department of Mathematics and Statistics, ILB 325, Mobile, AL 36608, USA
Received: 3 February 2012; in revised form: 1 March 2012 / Accepted: 1 March 2012 / Published: 7 March 2012
## Abstract
:
This paper is a very brief introduction to knot theory. It describes knot coloring by quandles, the fundamental group of a knot complement, and handle-decompositions of knot complements.
Keywords:
knots; quandles; fundamental groups; handles; knot colorings; symmetry; surfaces; Klein bottle; projective plane
## 1. Introduction
Over the past several months, Slavik Jablan asked me to contribute to this volume. I hesitated since I have very little to say that is new, that which is new does not fit into this venue, and I have been procrastinating rather than writing the newer results. The current article contains one of those results: the handle decomposition that yields the Alexander–Briggs [1] presentation of the knot group. Masahico Saito, Dan Silver, Susan Williams and I will be writing more about this presentation as it relates to virtual knots and knot colorings shortly after this article is complete. The article that I present to you now could be written by any other author in the current volume. Many could do a better job than I, for I am focusing upon some well-known properties of classical knots and the space that surrounds them.
Specifically, the article focuses upon the so-called handle decomposition of the knot complement. These handle-theoretic techniques are central to geometric topology. My hope is that the details presented here will aid an uninitiated reader in mastering them.
This paper was developed as an introductory series of lectures that I gave to the beginning graduate students at Kyongpook National University, Daegu, Korea, during a sabbatical semester from the University of South Alabama. My visit in Korea is being funded by a grant from the Brain-Pool Trust. I would like to thank my host, Professor Yongju Bae for his hospitality. I also would like to thank my long-time collaborator Masahico Saito for our discussions about these matters and related things. Finally, many thanks to the referees for their helpful and kind comments.
Here is an outline. The second section follows this introduction. In it, I indicate that the three knots that are illustrated in Figure 1 are distinct. The discussion there focuses upon Reidemeister moves, quandle colorings, and the quandle structure of a group under the conjugation operation. The third section gives an outline of the definition of the fundamental group of a knot complement. I have included some of the diagrams that are used to demonstrate the homotopy equivalences necessary for the well-definedness of the invariant. Many texts (e.g., [2,3,4]) present this material in detail. The fourth section sketches handle theory. It describes handle decompositions of a few surfaces (sphere, torus, projective plane, Klein bottle), illustrates handle sliding in two dimensions, and discusses turning the handle decomposition of the torus upside-down. In the fifth section, the discussion turns to decompositions of the trefoil and the figure-eight knot complements. The knot diagrams are annotated to manipulate these decompositions. In the last section, I describe an alternative handle decomposition that can be used to present the fundamental group of a knot. I call this presentation the Alexander–Briggs presentation since a careful reading of the last section of their seminal paper [1] indicates that these authors understood this handle decomposition even if handles had not yet been defined during that era.
This paper does not cover any post-Jones (e.g., [5,6]) invariants and it does not cover the Alexander polynomial even though the latter can easily be gleaned from some of the discussions herein. For a wonderful historical survey see [7]. I am writing with a deadline in mind. My goal is a self-contained opus that contains the ideas and visual imagery that occupies my current state of consciousness.
Let us begin.
## 2. The Unknot, the Trefoil, and the Figure-8 Knot Are Distinct
The goal of this section will be to demonstrate that the three knots that are depicted in Figure 1 are distinct. The embedded curves represented are called the unknot, the trefoil knot, and the Figure-8 knot as indicated in the figure. First, here are some basic notions.
A classical knot is a (smooth or piecewise-linear locally-flat) embedding of a circle $S 1 = { z ∈ C : | z | = 1 }$ into 3-dimensional space. Two such knots are said to be equivalent if one can be continuously deformed into the other without breaking or cutting. More precisely, $f 1 : S 1 ↪ R 3$ and $f 2 : S 1 ↪ R 3$ are equivalent if and only if there is an orientation preserving homeomorphism of pairs $( R 3 , f 1 ( S 1 ) ) → ( R 3 , f 2 ( S 1 ) )$. We often replace $R 3$ with $S 3 = { ( x , y , z , w ) : x 2 + y 2 + z 2 + w 2 = 1 }$ which is the one-point compactification of 3-space.
A classical knot is depicted via a diagram as indicated in Figure 2. In such a diagram a generic projection of the knot to plane is chosen; when two arcs project to transversely intersecting arcs in the the plane, crossing information is depicted by breaking the arc that is closer to the plane of projection than an observer is. Some additional standard terminology is indicated in that figure. Figure 3 indicates crossing conventions. From a diagram, a local picture of a crossing can be reproduced via choosing a coordinate system at the intersection points and lifting the under-arc (which lies along the y-axis) to the segment ${ ( 0 , y , 1 ) : | y | ≤ 1 }$, and lifting the over-arc to the segment ${ ( x , 0 , 2 ) : | x | ≤ 1 }$ as in Figure 4. Observe that if the orientations of these segments coincide with the orientations of the axes, then the crossing so depicted is positive.
Knot diagrams are studied via an equivalence relation on diagrams. The relation is generated by the local moves that are indicated in Figure 5 and which are called Reidemeister moves. These moves reflect the motion of a knot in space in the sense that if a knot is generically moved through space, then the projection of the motion is encapsulated by a sequence of applications of moves to the projection. In order to experience the moves directly, I suggest making a knot out of a rigid material such as thick wire and watch the knot as you move your point of view. It is not difficult to experience each of the Reidemeister moves in one’s visual sphere.
With these preliminary matters established, we demonstrate that the trefoil is not the unknot. To do so, we color the arcs of the trefoil using three distinct colors as indicated in Figure 6. In general a knot is said to be 3-colorable if each of the arcs in a representative diagram can be assigned a color such that either all three colors are coincident at a crossing or only one color is incident. This relationship is indicated in Figure 7.
Since 3-colorability is defined in terms of a representative diagram, we must show that it is invariant under the Reidemeister moves. Observe that for a type I move (top of Figure 5) all three colors must coincide. For a type II move (middle of the figure), either the two disentangled arcs have different colors or they have the same color. In the latter case, the short arc that is introduced has the same color as these two arcs do. In the former case, the short arc has the third color.
For the type III move there are five cases to consider:
1.
Case $( a , a , a )$: In this case, the three arcs from left to right on the left-hand-side of the move all have the same color at the top of the diagram. All the arcs on either side of the move are the same color.
2.
Case $( a , b , c )$: the three arcs from left to right on the left-hand-side of the move all have different colors at the top of the diagram. The top of the arcs on the right hand side are similarly colored. At the bottom of the diagram on either side the colors are $( c , a , c )$ from left to right.
3.
Case $( a , a , b )$: the first two arcs at the top of either the left or right side of the move have the same colors and the last arc is colored differently. At the bottom of either side the colors are $( b , c , c )$ from left to right.
4.
Case $( a , b , b )$: The two arcs on the right have the same color, and the first arc is colored differently.
5.
Case $( c , a , c )$: This case is obtained from case $( a , b , c )$ by turning it upside-down.
Figure 8 illustrates.
From these illustrations, we can see that if a diagram is 3-colorable, then any diagram of the knot is 3-colorable in a way that uses three distinct colors. Moreover, any coloring of the unknot is monochromatic. Thus the unknot is distinct from the trefoil. Any attempt to 3-color the figure-8 knot with three distinct colors will fail. We leave that as an exercise for the reader.
To demonstrate that the figure-8 knot is distinct from the other examples, we formalize the idea of colorability be demonstrating that crossings in a knot diagram can be used to axiomatize an algebraic system: a set with a binary operation that is assumed to satisfy three axioms which correspond to the Reidemeister moves. Specifically, if an over-crossing arc is oriented, then a homunculus standing on the over-arc holding out its (all homunculi in this paper are gender neutral) left hand points towards an under-arc as in Figure 9. The under-arc on the right is labeled a, the over-arc is labeled b, and the target arc is labeled with a product $a ◃ b$. We read $a ◃ b$ as “a acted upon by b”.
To continue, we define a set X that has a binary operation $◃ : X × X → X$ (written in in-fix notation) a quandle if the following three axioms hold:
I
for any $a ∈ X$, we have $a ◃ a = a$;
II
for each $a , b ∈ X$, there is a unique $c ∈ X$ such that $c ◃ b = a$;
III
for each $a , b , c ∈ X$, we have $( a ◃ b ) ◃ c = ( a ◃ c ) ◃ ( b ◃ c )$;
As indicated in Figure 10, the axioms correspond to the Reidemeister moves in the sense that if a knot diagram is colored by elements of a quandle X (in such a way that the under-arc towards which the humunculus’ left hand points receives a product), then any diagram related by a sequence of Reidemeister moves will also be colored.
The first example of a quandle that we have in mind is the set $X = { 0 , 1 , 2 }$ with $a ◃ b = 2 b − a ( mod 3 )$. This product yields the 3-colorability conditions of the previous example. The quandle here is called $R 3$—the dihedral quandle of order 3.
If X is a subset of a group that is invariant under conjugation by its elements, then X is a quandle under the operation $x ◃ y = y − 1 x y$. For example, let $G = S 4 = S 4 { 0 , 1 , 2 , 3 }$ denote the group of permutations of the set ${ 0 , 1 , 2 , 3 }$. Let $X = { ( 1 , 2 , 3 ) , ( 0 , 3 , 2 ) , ( 0 , 1 , 3 ) , ( 0 , 2 , 1 ) }$ denote the subset of “orientedof ” 3-cycles. Label each such 3-cycle by the element that it fixes: $0 ↔ ( 1 , 2 , 3 )$, $1 ↔ ( 0 , 3 , 2 )$, $2 ↔ ( 0 , 1 , 3 )$, and $3 ↔ ( 0 , 2 , 1 )$. Then the quandle table for X is given in Table 1.
This quandle is called $QS 4$the tetrahedral quandle. Figure 11 indicates that the figure-8 knot can be colored by $QS 4$. Thus the figure-8 knot is distinct from the unknot. Even though the trefoil knot can be colored by $QS 4$, it is distinct from the figure-8 knot since the latter cannot be colored by $R 3$.
We state the classical result:
Theorem 2.1
The unknot, the trefoil knot, and the figure 8 re distinct.
In fact, finite quandles are very good at distinguishing knots. Each classical knot has an associated fundamental quandle that is somewhat stronger than its fundamental group. The development of the quandle as a knot invariant occurred independently in papers by Matveev [8] and Joyce [9] about twenty years before the turn of the century.
The dihedral quandle and the tetrahedral quandle are both examples of quandles that are defined geometrically. The dihedral quandle $R n$ consists of the reflections of an n-gon composed under conjugation. The tetrahedral quandle is defined similarly as rotations of the tetrahedron through a vertex. In general, a (connected) quandle exhibits aspects of symmetry since it can be defined in terms of a binary operation on a set of cosets of a group of automorphisms of the quandle. This description appears elsewhere [8,9,10] and would take us far away from the current purposes of the paper.
## 3. The Fundamental Group
Let $K : S 1 ↪ S 3$ denote a smooth or PL-locally-flat embedding of a circle into the 3-dimensional sphere $S 3 = { ( x , y , z , w ) ∈ R 4 : x 2 + y 2 + z 2 + w 2 = 1 }$. For brevity, we write the image of the embedding as K, and we speak of the knot K. The smooth or PL-locally-flat condition suffices to provide a tubular neighborhood of the knot. This is a smooth embedding of a solid torus $N : S 1 × D 2 ↪ S 3$ that is a tubular neighborhood of the knot. That is, letting $D 2 = { ( x , y ) ∈ R 2 : x 2 + y 2 ≤ 1 }$ denote the 2-dimensional disk, then the knot is embedded as the core ${ 0 } × S 1$ of this solid torus. The knot exterior is the space $E = E ( K ) = S 3 ∖ int ( N )$. It has a torus ($S 1 × S 1$) as its boundary. We fix a base point of the exterior, and call the point *.
The fundamental group of the knot exterior E based at the point * is defined as the set of homotopy classes of maps of pairs $γ : ( [ 0 , 1 ] , { 0 , 1 } ) → ( E , ∗ )$ where the homotopies are required to fix the boundary points at the base point, and the group structure is induced from path multiplication. This definition requires much explication.
First, a homotopy between paths $γ 0$ and $γ 1$ is a map $H : [ 0 , 1 ] × [ 0 , 1 ] → E$ such that $H ( s , i ) = γ i ( s )$ for $i = 0 , 1$. That the homotopy fixes the boundary means that $H ( 0 , t ) = H ( 1 , t ) = ∗$ for all $t ∈ [ 0 , 1 ]$. We say that two paths are homotopic if there is a homotopy between them. This induces an equivalence relation on the set of paths; we call an equivalence class a loop in space. If $α , β : ( [ 0 , 1 ] , { 0 , 1 } ) → ( E , ∗ )$ are a pair of paths, then they can be multiplied by the rule
$α · β ( s ) = α ( 2 s ) if s ∈ [ 0 , 1 / 2 ] β ( 2 s − 1 ) if s ∈ [ 1 / 2 , 1 ]$
This means that a particle traveling along the composition first travels along $α$ at double speed and then along $β$ at double speed.
We define a group structure on the set of homotopy classes of loops by declaring the composition to be induced by path multiplication, the identity element is the equivalence class of the constant path, and the inverse of $α$ is to traverse $α$ backwards (specifically $α − 1 ( s ) = α ( 1 − s )$). The illustrations in Figure 12 indicate the geometric notions, and outline the homotopies that are needed to demonstrate that $α · α − 1$ is homotopic to the constant map, and that · induces an associative product on equivalence classes. In this figure, the exterior of the trefoil knot is shown as the space that is interior to the torus. It is not easy to make the intuitive leap from the complement of the knot as depicted in a non-compact 3-dimensional space to the concise picture given.
Following the discussion about handles in the next section, we will demonstrate that the fundamental group can be generated by the set of loops that encircle each of the arcs that appear in the diagram. We will also demonstrate that the set of relations in such a knot group presentationi.e., a presentation of the fundamental group of the exterior of the knot given in terms of generators and relations—are all of the form $c = b − 1 a b$. Figure 13 indicates the relationship. The triangular loop labeled b on the left of the figure is homotopic to the loop labeled b on the right. The square at the bottom of the figure indicates that there is a homotopy from c to $b − 1 a b$. The main portion of the homotopy is that square.
The connection between the quandle colorings given in the second section and the fundamental group is that the colorings induce representations of the fundamental group into a permutation group. For the trefoil, the representation is into the permutations of $0 , 1 , 2$; for the figure-8 knot the representation is into the group of alternating permutations of $0 , 1 , 2 , 3$.
Figure 13 indicates that the relations of the form $c = b − 1 a b$ exist at any crossing. However, it is necessary to demonstrate that such relations suffice. To this end, we decompose the knot complement into a cell complex called a handle decomposition that realizes the exterior as a thickening of a 2-dimensional complex. Here are some preliminary aspects on handle theory.
## 4. Handles
For any non-negative integer k, let $D k = { ( x 1 , … , x k ) ∈ R k : ∑ j = 1 k x j 2 ≤ 1 }$ denote the k-dimensional disk. Let $S k − 1 = { ( x 1 , … , x k ) ∈ R k : ∑ j = 1 k x j 2 = 1 }$ denote the $( k − 1 )$-dimensional sphere which is the boundary of the disk. In case $k = 0$, the k-disk is a point and its boundary is empty. Here we will concentrate on the cases in which $k ∈ { 0 , … , n }$ and $n = 1 , 2 , 3$.
A k-handle is a subset of a topological space that is homeomorphic to $D k × D n − k$. An explicit identification between the subset and this product of disks is assumed throughout. There are several important subsets of the k-handle.
• The core disk is the subset $D k × { 0 }$.
• The attaching sphere is the subset $S k − 1 × { 0 }$.
• The co-core disk is the disk ${ 0 } × D n − k$.
• The belt sphere is the sphere ${ 0 } × S n − k − 1$.
• The attaching region is $S k − 1 × D n − k$.
• The belt region is $D k × S n − k − 1$.
Of course, a k-handle is also an $( n − k )$-handle. When we consider it as such, we are turning the handle decomposition upside-down. Every handle is homeomorphic to an n-disk. But we are interested in the incidence relations among the handles in a given topological space (such as a surface or the complement of a knot). The incidence relations can be determined by orienting the belt sphere of a k-handle and computing a signed intersection between this sphere and the oriented attaching sphere of a $( k + 1 )$-handle. Our illustrations will attempt to make these ideas more clear. The attaching sphere and the belt sphere are often called the A-sphere or B-sphere, respectively. The basic terminology was introduced in the book [11].
Figure 14 indicates handles in various dimensions and the important spheres and disks. Recall that a 0-disk is a point with empty boundary.
In dimension 2, compact surfaces are classified according to their genus and the number of boundary components. The orientable surfaces that do not have boundary are in the ordered list that begins: sphere, torus, genus two surface, and that continues without end. It is instructive to give handle decompositions of some of these. Standard decompositions for the sphere and torus are given in Figure 15. The sphere is decomposed as a 0-handle and a 2-handle. The torus is decomposed as a 0-handle, two 1-handles, and one 2-handle. On the right of the standard handle decomposition of the torus, the same decomposition with the roles of the handles reversed is indicated. We will use this decomposition as part of the decomposition that is associated to the Alexander–Briggs presentation. In the dual decomposition the large rectangular region that wraps around most of the torus is to be considered a 0-handle. The 1-handles are the same, but their cores and co-cores have been switched, and the 2-handle is the button like disk that holds the configuration together.
Figure 16 indicates another standard picture of the handle decomposition of the torus. This time the 2-handle has been removed, and the meridional and longitudinal 1-handles appear as “basket" handles attached to the disk. In the lower figure, the Klein bottle is indicated with a 2-handle missing, and it has been constructed as a schematic of the connected sum of two projective planes. The projective planes having been obtained as a single 0-handle, a 1-handle attached with a twist, and the 2-handle to be attached on the outside. In the next two illustrations a topological deformation occurs in which the Möbius band on the left “slides over” the Möbius band on the right. The sliding is achieved via the method of excavating a thin disk from the surface. At the bottom left of the illustration, the resulting 1-handle is bent upwards and rounded out. The resulting decomposition is an untwisted band nestled with a twisted band as illustrated on the bottom right.
Boy’s surface (Figure 17) is an immersion of the projective plane that is immersed in 3-dimensional space. It was discovered by David Hilbert’s student Werner Boy much to the surprise of Hilbert. It enjoys many remarkable properties including a 3-fold symmetry. The symmetry here is indicated by means of primary colors: red, blue, and yellow. These were chosen to emulate the colors and symmetries expressed in some traditional Korean fans (Figure 18) in homage to my hosts.
## 5. Handles in Dimension $n = 3$
Handle techniques are often used in the study of 3 and 4 dimensional manifolds. In 3-dimensions, the handle decomposition coincides with what is also called the Heegaard decomposition of the 3-dimensional manifold. As an example, the complement of a knot is a 3-dimensional manifold with boundary. We will use the knot diagram to construct a handle decomposition of the knot complement. In this decomposition, the arcs correspond to 1-handles and the crossings (under arcs) correspond to 2-handles.
In order to help develop this description in general, we begin with a decomposition of the unknot (Figure 19). A 0-handle appears at the top of the diagram. A 1-handle is attached in such a way that its core disk corresponds to the disk that is bounded by the circle in space. The 1-handle then corresponds to a maximal point of the curve as it appears in space. A 2-handle is attached at the minimal point of the curve. The attaching sphere for the 2-handle does not intersect the core disk of the 1-handle. A 3-handle can be attached on the outside to complete the construction of the complement.
To better understand this decomposition (and those which follow) the picture will be rearranged. Consider first the complement of a neighborhood of the maximal point of the circle. It is homeomorphic to a cube with a hole drilled through it as in the left of Figure 20. Observe that the cube from which a hole has been drilled is homeomorphic to a 3-ball to which a 1-handle has been attached. Figure 20 indicates the correspondences. Next observe that the attaching sphere for the 2-handle (which corresponds to the minimum point in the diagram) does not intersect the belt sphere of the 1-handle. The core disk of the 2-handle envelops the two ends of the excavated hole. The 3-handle consists of the space in which the observer sits.
Alternatively, the attaching sphere for the 2-handle can be slid to the top of the cube in the figure, and the 3-handle is nestled inside the resulting purse.
## 6. Handles in 3-Dimensions
A knot diagram easily lends itself to the construction of a handle decomposition of the complement. The arcs of the diagram correspond to holes that have been drilled from the space above the diagram as indicated in Figure 20. Such “holes removed" can be reinterpreted as 1-handles that are attached to a 0-handle that is envisioned as lying above the plane in which the diagram is drawn. The crossings—representing arcs that go under the plane of the diagram—define 2-handles.
The most simple case that can be depicted and that includes a crossing is that of the unknot with one crossing. The handle decomposition is represented in Figure 21. This figure indicates that the attaching sphere for the 2 handle (which is an S-shape in the figure) intersects the belt sphere of the 1-handle in a sequence $1 , − 1 , − 1 , 1$. Here there is one 1-handle (thus labeled “1”) and is oriented from the lower right towards the upper left of the planar picture, and the S-curve is oriented clockwise. Clearly, the intersections between the 1 and 2 handles can be removed since there are a pair of disks co-bounded by segments of each of these handles.
The intersections between 1-handles and 2-handles can be indicated completely within a planar diagram built from the knot diagram. Figure 22 demonstrates one such diagram associated to the trefoil. Over arcs correspond to the segments of the co-cores of the 1-handles that intersect the plane of the diagram. The attaching spheres for the 2-handles encircle the pair of segments of the under crossing.
In the diagram of Figure 22, we can read the intersection of the attaching sphere of the 2-handles with the co-cores of the 1-handles. In this diagram, the attaching spheres for the 2-handles are oriented counterclockwise. So the intersection sequences are (up to cyclic permutation) as follows:
$A : 1 , 3 − 1 , 2 − 1 , 3$
$B : 3 , 2 − 1 , 1 − 1 , 2$
$C : 2 , 1 − 1 , 3 − 1 , 1$
Any path in space is homotopic to a path that misses the 3-handle. Thus any loop in space can be moved into the union of the 0, 1, and 2-handles. We realize, therefore, that the intersection sequences A, B and C define a presentation for the fundamental group of the trefoil. Specifically, the fundamental group is given:
$〈 x , y , z : x z − 1 y − 1 z , z y − 1 x − 1 y , y x − 1 z − 1 x 〉$
Just as in the case of surfaces, handles in the decompositions of knot exteriors (or any 3-dimensional manifold) can be slid over each other. Moreover, it is possible to cancel i-handles against $( i + 1 )$-handles provided they intersect geometrically once. A sequence of handle slides and cancelations is depicted in Figure 23. Let the diagrams in the figure be labeled by row/column indices. The transition from $( 1 , 1 )$ to $( 1 , 2 )$ is simply a topological isotopy. From $( 1 , 2 )$ to $( 1 , 3 )$ handle A slides over handle C. From $( 1 , 3 )$ to $( 2 , 1 )$ handle A slides over handle B. From $( 2 , 1 )$ to $( 2 , 2 )$ handle A is homotoped to have no intersections with the 1-handles. At this point handle A can cancel with the 3-handle on the outside. The move from $( 2 , 2 )$ to $( 2 , 3 )$ is this cancelation together with handle C sliding over B. From $( 2 , 3 )$ to $( 3 , 1 )$ handle 3 cancels with B. The move from $( 3 , 1 )$ to $( 3 , 2 )$ is a homotopy.
The presentation for the fundamental group reduces to
$〈 x , y : y x − 1 y − 1 x − 1 y x 〉 = 〈 x , y : x y x = y x y 〉$
Figure 24 indicates a similar sequence of handle slides that moves the handle decomposition of the figure-8 knot to a decomposition with one 2-handle and two 1-handles. The last illustration in this sequence is related to the so-called 2-bridge presentation of the knot. From this representation one can determine that the 2-fold branched cover is the 3-dimensional manifold which is known as the lens space $L ( 5 , 2 )$: this manifold has a handle decomposition with one 0-handle, one 1-handle and one 2-handle that is wrapped around the boundary torus twice longitudinally and five times meridionally.
Some general properties can be proven following a close examination of these examples. First of all, it is always the case that one of the 2-handles can be cancelled against the 3-handle on the outside. This can be seen by mimicking the handle slides in these examples. Any particular 2-handle’s attaching sphere can be slid to envelope the remaining attaching spheres. Consequently, the knot complement is, in essence, a 2-dimensional complex that can be formed by attaching as many 1-handles as over-arcs, and one fewer 2-handle.
Often the handle decomposition can be simplified. In particular, whenever the attaching sphere for a 2-handle intersects the belt sphere for a 1-handle exactly once and this belt sphere has no other intersections, then this pair of handles can be removed. The removal corresponds to removing a generator of the fundamental group and replacing all occurrences of that generator with a sequence that is obtained from intersection sequence of the attaching sphere of the 2-handle.
All of the aspects of group presentations for classical knots and the relationship between this presentation (the Wirtinger presentation) and the handle decomposition induced from the under-arcs can be found in various spots in the literature.
There are many other methods to obtain a presentation for the fundamental group of a knot complement, and even though many important knot invariants can be obtained from the fundamental group (and the closely related fundamental quandle), other invariants such as the Jones polynomial [6] and the HOMPLYPT (or FLY THOMP) [5] polynomial are obtained diagrammatically, but their relationship to other traditional topological invariants is difficult to discern. There is not enough time or space within this article to spend time on these. However, there is an alternative method for obtaining a presentation of the fundamental group that will close the article.
## 7. The Alexander–Briggs Presentation
The Alexander-Briggs presentation of the fundamental group is obtained from a handle decomposition of the knot complement in which most (all but two) of the 1-handles are associated to the crossings, and most of the 2-handles (all but one) are associated to the regions. In future work with Masahico Saito, Dan Silver, and Susan Williams, I will develop this presentation in the context of virtual knots and their Alexander invariants.
In the initial stage of this decomposition, the complement is built out from the boundary torus. The “upside-down” decomposition of the torus that is depicted on the lower right side of Figure 15 is thickened to a decomposition of the torus times an interval. The 0-handle envelopes the torus. The 1-handles are spatial regular neighborhoods of the letter C, with the co-core disks being planar neighborhoods. These 1-handles correspond to the longitude and meridian of the knot, but be aware that since the cores are short arcs, that which is apparently a longitude is a segment of the co-core disk of the meridian and vice versa. Figure 25 contains the details.
At each crossing a 1-handle is attached. It is a pillar erected between neighborhoods of arcs at the crossing. The core disk runs from the lower arc to the upper arc, and the belt sphere is oriented in a counterclockwise fashion. The attaching sphere lies on the 0-handle that envelopes the boundary torus. Figure 26 indicates the details.
There is a 2-handle that lies in a neighborhood of the boundary. In Figure 25 this is represented as a small cube. Its attaching sphere intersects the belt sphere of the longitude and meridian in the commutator $M L M − 1 L − 1$ up to orientation and cyclic permutation.
The remaining 2-handles are represented by the regions in the diagram. See Figure 2 for the definition of regions. The attaching sphere for the regions intersects the pillars at the crossings. Figure 27 indicates the intersection of one such attaching sphere in a neighborhood of the crossing. A schematic diagram on the right side of the illustration indicates how this attaching sphere and those of the adjacent regions can be drawn at a knot diagram. In the schematic, the co-core disk of the meridian is illustrated as thick segments. Each region including the unbounded region of a planar diagram serves as the core disk of a 2-handle.
To complete the construction of the complement a pair of 3-handles are attached to the complex—one 3-handle above the plane of the diagram and one 3-handle below the plane of the diagram.
Figure 28 indicate the handle decomposition of the trefoil knot. From this decomposition we can deduce a presentation for the fundamental group with generators m, L, A, B, and C. There are six relations that are read from the attaching spheres of the 2-handles. These are:
0.
$L − 1 M − 1 L M = 1$.
1.
$L A M B M C M = 1$.
2.
$C B A = 1$.
3.
$A B M = 1$.
4.
$B C M = 1$.
5.
$L A M C = 1$.
From the relations (5) and (1), we obtain $C − 1 B M C M = 1$. From (3) and (4) we obtain $A M − 1 C − 1 M = 1$. Thus $C = M A M − 1$, and $B = A − 1 M − 1$. Substituting,
$C − 1 B M C M = [ M A − 1 M − 1 ] [ A − 1 M − 1 ] M [ M A M − 1 ] M = 1 M A − 1 M − 1 A − 1 M A = 1 A − 1 M − 1 A − 1 = M − 1 A − 1 M − 1$
Consequently, the group is presented as
$〈 A , M : A M A = M A M 〉$
Since the longitude can be written as $L = C − 1 M − 1 A − 1$, further substitutions give that the relation $L M L − 1 M − 1 = 1$ is a consequence of the defining relation $A M A = M A M$.
We close this discussion with the remark that the dot notation in [1] can be chosen to coincide with the incidence of the meridional co-core and the regions with some crossing conventions. Thus there is a calculus to go from this presentation of the fundamental group to a presentation of the Alexander module. This topic and more will be the subject of a subsequent paper.
## 8. Conclusions
There is a close connection between the fundamental quandle of a classical knot and its fundamental group. In fact, the former is given as coset space of the latter where the subgroup is coincident with a peripheral subgroup of the fundamental group. Different presentations of the fundamental group can be used to derive differing features of the knot complement. In any case, the knot complement can be thought of in terms of a 2-dimensional complex on which it deformation retracts. In general, knot theory appeals to our visual sensibilities. In this paper, I have attempted to make some of the visual aspects of the theory more apparent. I hope that I have succeeded.
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Figure 1. The unknot, the trefoil knot, and the figure-8 knot.
Figure 1. The unknot, the trefoil knot, and the figure-8 knot.
Figure 2. A diagram of the Figure 8 knot.
Figure 2. A diagram of the Figure 8 knot.
Figure 3. The projection of a crossing and point of view.
Figure 3. The projection of a crossing and point of view.
Figure 4. Lifting a crossing to standard position in space.
Figure 4. Lifting a crossing to standard position in space.
Figure 5. The Reidemeister moves.
Figure 5. The Reidemeister moves.
Figure 6. The trefoil knot is 3-colorable.
Figure 6. The trefoil knot is 3-colorable.
Figure 7. The 3-colorability condition.
Figure 7. The 3-colorability condition.
Figure 8. The type III move and colorability.
Figure 8. The type III move and colorability.
Figure 9. The under-arc towards which the left hand points received the product.
Figure 9. The under-arc towards which the left hand points received the product.
Figure 10. The axioms of a quandle correspond to the Reidemeister moves.
Figure 10. The axioms of a quandle correspond to the Reidemeister moves.
Figure 11. The figure-8 knot can be colored by $Q S 4$.
Figure 11. The figure-8 knot can be colored by $Q S 4$.
Figure 12. Ideas that are used to demonstrate the fundamental group and necessary homotopies.
Figure 12. Ideas that are used to demonstrate the fundamental group and necessary homotopies.
Figure 13. The Wirtinger relations at a crossing.
Figure 13. The Wirtinger relations at a crossing.
Figure 14. Handles in dimensions 0 through 3.
Figure 14. Handles in dimensions 0 through 3.
Figure 15. Handle decompositions for the sphere and the torus.
Figure 15. Handle decompositions for the sphere and the torus.
Figure 16. The torus and two decompositions of the Klein bottle.
Figure 16. The torus and two decompositions of the Klein bottle.
Figure 17. Boy’s surface, front, back, and as a handle decomposition.
Figure 17. Boy’s surface, front, back, and as a handle decomposition.
Figure 19. A handle decomposition of the unknot.
Figure 19. A handle decomposition of the unknot.
Figure 20. Re-interpreting this handle decomposition of the unknot.
Figure 20. Re-interpreting this handle decomposition of the unknot.
Figure 21. Another handle decomposition of the unknot.
Figure 21. Another handle decomposition of the unknot.
Figure 22. A handle decomposition of the trefoil knot.
Figure 22. A handle decomposition of the trefoil knot.
Figure 23. Handle sliding and cancelation of the trefoil complement.
Figure 23. Handle sliding and cancelation of the trefoil complement.
Figure 24. Handle sliding on the figure-8 knot complement.
Figure 24. Handle sliding on the figure-8 knot complement.
Figure 25. The decomposition of the knot complement in the neighborhood of the boundary.
Figure 25. The decomposition of the knot complement in the neighborhood of the boundary.
Figure 26. At each crossing a 1-handle is attached to the 0-handle that envelopes the boundary.
Figure 26. At each crossing a 1-handle is attached to the 0-handle that envelopes the boundary.
Figure 27. The attaching spheres of the four 2-handles that are incident to the pillar are shown in the associated schematic.
Figure 27. The attaching spheres of the four 2-handles that are incident to the pillar are shown in the associated schematic.
Figure 28. The decomposition of the trefoil complement.
Figure 28. The decomposition of the trefoil complement.
Table 1. The tetrahedral quandle.
Table 1. The tetrahedral quandle.
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# Activity 8.5 Class 9 Science Chapter 8 Motion
Activity 8.5 Class 9 Science Chapter 8 Motion
Activity 8.5 asks us to look at the chart and see which vehicle (A or B) has uniform speed.
Vehicle A is traveling with uniform motion.
#### Explanation:
Here we see that every 15 minutes A travels 10 km always. While in the case of B, it is irregular. Sometimes B traves 7 km in the 15 minutes, while in the next 15 minutes it travels 4km.
#### Inference:
This activity wants us to know about uniform and non-uniform speed. It wants us to show that if a vehicle travels the same distance in the same unit of time, the vehicle is in uniform motion.
#### Implication
While studying motion we come into the concept of acceleration. An object with uniform motion is with zero acceleration. While an object with non-uniform motion is either accelerating or decelerating.
| 204
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CC-MAIN-2024-22
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latest
|
en
| 0.919715
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